====== 4. Analysis of direct current networks ======

<callout>
<WRAP right>
<imgcaption imageNo1 | examples for networks>
</imgcaption>
{{drawio>Examples Networks}}
</WRAP>

Network analysis plays a central role in electrical engineering. It is so important because it can be used to simplify what at first sight appear to be complicated circuits and systems to such an extent that they can be understood and results derived from them.

In addition, networks also occur in other areas, for example the force flow through a truss or the heat flow through individual hardware elements (<imgref FigNo1>). The concepts shown below can also be applied to these networks.

On the {{wpde>network_analysis_(electrical_engineering)|wiki page on network analysis}} the different methods are described very well in a compact way
</callout>

<callout>

=== Goals ===

After this lesson, you should:

  - <del>be able to determine the number of nodes, number of (tree and connecting) branches, and number of meshes.</del>
  - <del>be able to construct a complete tree from an electrical network.</del>
  - <del>be able to understand the branch current method, mesh current method, and node potential method.</del>
  - Understand and be able to apply the superposition procedure.
</callout>

<callout type="danger" icon="true">

Due to the shortened semester, only the subchapter [[analysis_of_dc_grids#overlay_procedure_superposition_principle|4.5]] is relevant for WiSe2020.

</callout>


===== 4.1 Preliminary work on network analysis =====

==== Preparation of the circuit ====

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<imgcaption imageNo10 | Preparing the circuit>
</imgcaption>
{{drawio>PreparingTheCircuit}}
</WRAP>

Before the network analysis can be tackled, the circuit must be suitably prepared (cf. <imgref FigNo10>):
  - Clarify what is given and what is sought
  - Draw a circuit
  - Add counting arrows. If not already given, then:
    - First draw current and voltage arrows at all sources according to the generator arrow system.
    - Afterwards define the current arrows at the remaining branches as you like.
    - Finally, draw the voltage arrows at the loads according to the load arrow system.
  - Select suitable current and voltage designations. If not already given, then:
    - Count indices continuously, i.e. one number per element (source or load).
    - Do not insert any signs in front of the designators in the circuit.

In real applications it is useful to specify the number of variables ("what is wanted?"), parameters ("what can be adjusted?", e.g. potentiometer) and known quantities ("what is given?"). \\ This makes it clear how many equations are needed. This seems to become difficult for larger networks - but a trick for this is presented below.

It often helps to draw the drawing several times (at least in your head) to have enough space for the identifiers (cf. <imgref imageNo10> below).

~PAGEBREAK~ ~CLEARFIX~~
==== Graph and Trees ====

<WRAP right>
<imgcaption imageNo11 | Graph of a network>
</imgcaption>
{{drawio>GraphOfANetwork}}
</WRAP>

In the chapter [[simple_dc_circuits#nodes_branches_and_meshes|2. simple dc_circuits]] the terms nodes, branches and meshes have already been explained. These will now be expanded here to better explain the various network analysis methods in the following.
In <imgref FigNo11> the **graph** of the example network is drawn. We had already seen this one too, but without knowing that this is called a graph! \\
But the important thing is: In this graph only the (real) nodes are drawn. Nodes are by definition the connection of __more than two__ branches. Accordingly, the connection between $R_{10}$ and $R_7$ is __not a node__ ((sometimes such connections are called "fake nodes"))! For this reason also the blue circle as sign for nodes is omitted here.

A concept that has not yet appeared is that of the complete tree. For this, some (mathematical) graph theory is needed. There, too, the terms nodes and meshes are used as before. A **tree** here is a special kind of graph. The graph in <imgref imageNo11> shows several meshes. \\
Now a tree is characterized precisely by the fact that it contains __no__ meshes. Three different trees are drawn in the picture. From a given network, many different trees can be created (depending on the number of nodes). \\
Among the different trees, there are now some in which each node connects two or fewer meshes.((Here we now depart from the previous electrotechnical notion of node (= connecting more than 2 branches). The mathematical notion of node does not have this restriction)) These are called **complete trees** (occasionally also {{wpde>Hamilton circle problem|Hamilton way}}). Complete trees can also be understood as this shows a path through the network where all nodes are visited only exactly once.

Tree 3 in <imgref imageNo11> is now just one of the possible complete trees of this network.

The branches in complete trees are now distinguished according to their membership:
  * **tree branches** belong to the complete tree (solid lines in <imgref imageNo11>).
  * **Connecting branches** do not belong to the complete tree (dotted lines in <imgref imageNo11>).

Why does the swing to graph theory make sense now? The trick is that by defining the complete tree, all meshes have just been removed. Conversely, a new (independent) mesh can be created by each connecting branch. So if the number of independent mesh equations $m$ is sought, this is just equal to the number of connecting branches.

To do this, proceed as follows:
  - Determine the number of (real) nodes $k$.
  - Determine the number of branches $z$
  - The number of tree branches $b$ is now $k-1$. (each node is traversed only once; at the last node there is no further branch).
  - The number of connecting branches $v$ is given by "All branches minus tree branches": $v = z - b = z - k + 1$

Thus, the number of independent mesh equations $m$ is findable by counting the nodes $k$ and branches $z$ over $m = v = z - k + 1$.

This explanation can also be heard again in [[https://www.youtube.com/watch?v=c7z1pRCzEuw|this video]] and is explained again clearly via [[https://studyflix.de/informatik/euler-und-hamiltonkreis-1287|StudyFlix]].

~~PAGEBREAK~~ ~CLEARFIX~~
===== 4.2 Branch Current Procedure =====

<WRAP right>
<imgcaption imageNo12 | example circuit>
</imgcaption>
{{drawio>example circuit}}
</WRAP>

In the branch current method now "simply times" (almost) all equations of the circuit. Specifically, for each node and each __independent__ mesh, the node and mesh equations are written down:
  * for all nodes k respectively the equation: $\sum_{k=0}^{N_k}{I_k}=0$
  * for all independent meshes m respectively the equation: $\sum_{m=0}^{N_m}{U_m}=0$ \\ Here the number $m$ (as mentioned in the previous subsection) can be determined by the number of nodes and branches.

This forms a linear system of equations. This can then be considered as a matrix equation and solved with the rules of (mathematical) art.


<WRAP onlyprint>
For the example (<imgref imageNo12>), these would be the equations:

~~PAGEBREAK~~~CLEARFIX~~~

The matrices still need to be corrected for the voltage and current sources!!!

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=== Example of nodal equations ===
\begin{align*}
\sum\limits_{k=0}^{N_k}{I_k}=0 \.
 \end{align*}

Setting up the individual equations:
\begin{align*}
\scriptsize\text{node 'a'} & \scriptsize : -I_0 - I_9 - I_7 = 0 \.
\scriptsize\text{node 'b'} & \scriptsize : +I_0 - I_1 - I_3 = 0 \.
\scriptsize\text{node 'c'} & \scriptsize : + I_1 - I_2 - I_4 = 0 \.
\scriptsize\text{node 'd'} & \scriptsize : - I_5 + I_4 - I_{11} = 0 \.
\scriptsize\text{node 'e'} & \scriptsize : + I_5 + I_6 - I_7 = 0 \.
\scriptsize\text{node 'f'} & \scriptsize : - I_2 + I_3 - I_6 + I_9 + I_{11} = 0
\end{align*}

Sorting streams into columns:
\begin{align*}
\begin{smallmatrix}
\text{node 'a'}: & -I_0 & & & & & & & - I_7 & - I_9 & & = 0 \.
\text{node 'b'}: & +I_0 & - I_1 & & - I_3 & & & & & & = 0 \.
\text{node 'c'}: & & + I_1 &- I_2 & & - I_4 & & & & & = 0 \.
\text{node 'd'}: & & & & + I_4 & - I_5 & & & - I_{11} & = 0 \\
\text{node 'e'}: & & & & & + I_5 & + I_6 & - I_7 & & = 0 \.
\text{node 'f'}: & & & - I_2 & + I_3 & & - I_6 & & + I_9 & + I_{11} & = 0 \\
\end{smallmatrix}
\end{align*}

Setting up the matrix:
\begin{align*}
 
\left( \begin{smallmatrix}
-1 & 0 & 0 & 0 & 0 & 0 & 0 & -1 & -1 & 0 \\
+1 & -1 & 0 & -1 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & +1 & -1 & 0 & -1 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & +1 & -1 & 0 & 0 & 0 & -1 \\
0 & 0 & 0 & 0 & 0 & +1 & +1& -1 & 0 & 0 \\
0 & 0 & -1 & +1 & 0 & 0 & -1& 0 & +1 & +1 \\
\end{smallmatrix} \right)  \cdot
\left( \begin{smallmatrix}
I_0 \\ I_1 \\ I_2 \ I_3 \ I_4 \ I_5 \ I_6 \ I_7 \ I_9 \ I_{11}
\end{smallmatrix} \right) = \vec{0}
\end{align*}

</WRAP><WRAP column half>
=== Example of mesh equations ===
\begin{align*}
\sum\limits_{m=0}^{N_m}{U_m}=0 \.
 \end{align*}

Setting up the individual equations:
\begin{align}
\scriptsize\text{mesh 'abf'} & \scriptsize : -U_0 + U_3 - U_9 = 0 \.
\scriptsize\text{mesh 'bcf'} & \scriptsize : +U_1 - U_2 - U_3 = 0 \.
\scriptsize\text{mesh 'cdf'} & \scriptsize : + U_2 + U_4 - U_{11} = 0 \.
\scriptsize\text{mesh 'def'} & \scriptsize : + U_5 - U_6 + U_{11} = 0 \.
\scriptsize\text{mesh 'eaf'} & \scriptsize : + U_6 - U_7 - U_{10} + U_9 = 0 \.
\quad \
\end{align*}

Sorting stresses into columns:
\begin{align*}
\begin{mallmatrix}
\text{mesh 'abf'}: &-U_0 & & + U_3 & & & & & - U_9 & & = 0 \.
\text{mesh 'bcf'}: & & + U_1 & - U_2 & - U_3 & & & & & & & & = 0 \.
\text{mesh 'cdf'}: & & + U_2 & & + U_4 & & & & & & & - U_{11}& = 0 \.
\text{mesh 'def'}: & & & & & + U_5 & - U_6 & & & & + U_{11} & = 0 \\
\text{mesh 'eaf'}: & & & & & & + U_6 & - U_7 - U_{10} & & - U_9 & & & = 0 \\
\quad \\ \quad \
\end{smallmatrix}
\end{align*}

Set up the matrix, but note $U_m = R_x \cdot I_m$:
\begin{align*}
 
\left( \begin{smallmatrix}
-R_0 & 0 & 0 & +R_3 & 0 & 0 & -R_9 & 0 & 0 \\.
0 & +R_1 & -R_2 & -R_3 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & +R_2 & 0 & +R_4 & 0 & 0 & 0 & -R_{11}  \\
0 & 0 & 0 & 0 & +R_5 &-R_6 & 0 & 0 & +R_{11}   \\
0 & 0 & 0 & 0 & 0 &+R_6 &-R_7-U_{10}& -R_9 & 0 \
\end{smallmatrix} \right) \cdot
\left( \begin{smallmatrix}
I_0 \\ I_1 \\ I_2 \ I_3 \ I_4 \ I_5 \ I_6 \ I_7 \ I_9 \ I_{11}
\end{smallmatrix} \right) = \vec{0}
\quad \\
\end{align*}
</WRAP></WRAP>
</WRAP>

These matrices can be solved using, for example, the {{wpde>Gau%C3%9Fsches_Eliminationsverfahren#Example|Gaußes Eliminationsverfahren}}.
~~PAGEBREAK~~ ~CLEARFIX~~

=== another example in videos ===

<WRAP group> <WRAP half column>

In video 1 the following steps are described:\\ \\
1. writing down the given circuit and sizes
2. drawing in and designating the knots
3. draw in and label the meshes\\

</WRAP> <WRAP half column>
Branch current analysis 1/4
{{youtube>CE1tEhpPJd0}}
</WRAP> </WRAP>
<WRAP group> <WRAP half column>

Video 2 describes the following steps: \\ \\
4. draw and label the branch currents \\
5. drawing in and designating the branch voltages \\

</WRAP> <WRAP half column>
Branch current analysis 2/4
{{youtube>wZENsTIte_Q}}
</WRAP> </WRAP>
<WRAP group> <WRAP half column>

Video 3 describes the following steps: \\ \\
6. set up node equations and mesh equations \
7. convert to matrix notation \

</WRAP> <WRAP half column>
Branch current analysis 3/4
{{youtube>dLqJ0vfKLLQ}}
</WRAP> </WRAP>
<WRAP group> <WRAP half column>

In the [[https://www.youtube.com/watch?v=YlSlyby_4PY|Video 4]] (not embedded here), the following steps are described: \\ \\
8. inserting the numerical values \\
9. calculating the matrix with a calculator \

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<panel type="info" title="Task 4.2.1 Exercise Task">
<WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>

{{youtube>gkJfKFuuyr8}}

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<panel type="info" title="Task 4.2.2 Exercise Task">
<WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>

{{youtube>ueKmNw2dtlI}}

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===== 4.3 Mesh flow method =====

In the mesh flow method, only for all meshes m each equation: $\sum\limits_{j=0}^{N_j}{U_j}=0$ are considered. However, these are represented in the form $R\cdot I = U $.

The advantage here is that the number of equations to be solved is reduced to the number of independent mesh currents.

These can also be considered as matrix equations and can be solved with the rules of (mathematical) art.


<WRAP group> <WRAP half column>

In video 1, the mesh flow method is applied by means of an example.

<WRAP important>Important: Although the video explains the application super, but contains a small error at minute 6:50. The sign of the voltages on the right side must be inverted in each case. This was also explained correctly a few seconds before.</WRAP>

</WRAP> <WRAP half column>
Mesh stream analysis
{{youtube>CebyoWnsarI}}
</WRAP> </WRAP>
<WRAP group> <WRAP half column>

Also in video 2, the mesh flow method is applied by means of an example.

</WRAP> <WRAP half column>
Mesh stream analysis
{{youtube>rRls0ySxbMA}}
</WRAP> </WRAP>
<WRAP group> <WRAP half column>

In the [[https://www.youtube.com/watch?v=__gUrBuJBes|Video 3]] (not embedded here) shows in detail how the mesh current method can be derived.

</WRAP> <WRAP half column>
</WRAP> </WRAP>

===== 4.4 Nodal potential method =====

In the nodal potential method, only the equation: $\sum\limits_{i=0}^{N_i}{I_i}=0$ are considered for all nodes k respectively. However, these are expressed in the form ${1\over R} \cdot U = I $ and $G \cdot U = I $ respectively.

The advantage here is that the number of equations to be solved is reduced to the number of existing nodes (minus 1).

These can also be considered as matrix equations and can be solved with the rules of (mathematical) art.

<WRAP group> <WRAP half column>

In Video 1, the idea behind node potential analysis is simply explained.

</WRAP> <WRAP half column>
Simple example of node potential analysis
{{youtube>fSg7OVRiN4c}}
</WRAP> </WRAP>
<WRAP group> <WRAP half column>

Video 2 also uses the nodal potential method with an example.

</WRAP> <WRAP half column>
more complex example of a node potential analysis
{{youtube>hnlPFAvIhkY}}
</WRAP> </WRAP>
<WRAP group> <WRAP half column>

In the [[https://www.youtube.com/watch?v=tpeu84Zq63g|Video 3]] (not embedded here) shows in detail how the nodal potential method can be derived.

</WRAP> <WRAP half column>
</WRAP> </WRAP>

===== 4.5 Superposition method / Superposition principle =====

The superposition principle shall first be illustrated by some examples

<callout title="Example 1 - from consulting industry interviews">.

**Task**: Three students are to fill a pool. If Alice were to fill it alone, she would need 2 days. Bob would need 3 days and Carol would need 4 days. How long would it take all three to fill a pool if they helped together? \\ \\

The question sounds far off topic at first, but is directly related. The point is that to solve it, filling the pool is assumed to be linear. So Alice will fill $1 \over 2$, Bob $1 \over 3$, and Carol $1 \over 4$ of the pool per day. So on the first day, ${1 \over 2}+{1 \over 3}+{1 \over 4} = {{6 + 4 + 3} \over 12} = {13 \over 12}$ of the pool filled. \\ So the three of them need ${12 \over 13}$ of a day. \\ \\
However, this solution path is only possible because in linear systems the partial results can be added.
</callout>

<callout title="Example 2 - Spring Force and Travel">

<WRAP right>
<imgcaption imageNo02 | mechanical spring>
</imgcaption>
{{drawio>mechanicalspring}}
</WRAP>

**Task**:A mechanical, linear spring is deflected with masses $m_1$ and $m_2$ in the Earth's gravitational field (see <imgref FigNo02>). What is the magnitude of the deflection if both masses are attached simultaneously? \\ \\

Again, a linear law is used here:
\begin{align*}
\vec{s}= f(\vec{F}) = - D \cdot \vec{F}
\end{align*}

The (seemingly trivial) approach applies here:
\begin{align*}
\vec{s}_{1+2} = f(\vec{F_1} + \vec{F_2}) &= - D \cdot (\vec{F_1} + \vec{F_2}) \\
&= - D \cdot \vec{F_1} - D \cdot \vec{F_2} \\
&= f(\vec{F_1}) + f(\vec{F_2}) \
&= \vec{s_1} + \vec{s_2}
\end{align*}
</callout>

<callout icon="fa fa-exclamation" color="red" title="Notice:">
In a physical system in which effect and cause are linearly related, the effect of each cause can first be determined separately. The total effect is then the sum of the individual effects.
</callout>

For electrical engineering this principle was described by {{wpde>Hermann_von_Helmholtz}}:

> The currents in the branches in a linear network are equal to the sum of the partial currents in the branches concerned caused by the individual sources.

<WRAP group> <WRAP half column>
Thus, in the superposition method, the current (or voltage) sought in a circuit with multiple sources can be viewed as a superposition of the resulting currents (or voltages) of the individual sources.

The "recipe" for the overlay is as follows:
  - Choose next source ''x''
  - Replace all ideal sources with their respective equivalent resistors:
    - ideal voltage sources by short circuits
    - ideal current sources by an open line
  - Calculate the partial currents sought in the branches considered.
  - Go to the next source ''x=x+1''(( ''x=x+1'' is not meant mathematically, but procedurally as in the programming language C)) and to point 2, as long as the partial currents of all sources have not been calculated.
  - Add up the partial currents in the branches under consideration, observing the correct sign.

This procedure is explained again in more detail using examples in the two videos on the right.

</WRAP> <WRAP half column>
Simple view of the superposition principle
{{youtube>tP6kinOsxp4?start=25}}

more complex example of the superposition method
{{youtube>vMuy1xr-ECM}}
</WRAP> </WRAP>

==PAGEBREAK~~CLEARFIX~~
=== Example ===

<WRAP right>
<imgcaption imageNo03 | example circuit with superposition>
</imgcaption>
{{drawio>ExampleCircuitSuperposition}}
<WRAP>



~~PAGEBREAK~~~CLEARFIX~~~

<panel type="info" title="Task 4.5.1 Converting a bipolar signal to a unipolar signal">
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<WRAP right>{{url>https://www.falstad.com/circuit/circuitjs.html?running=false&cct=$+1+3e-7+63.8+50+5+43%0Ar+-48+224+64+224+0+10000%0Ar+64+224+64+304+0+50000%0Ar+64+224+64+160+0+2000%0Ag+64+304+64+320+0%0AR+64+160+64+128+0+0+40+5+0+0+0.5%0Av+-192+304+-192+240+0+1+40+20+0+0+0.5%0Ag+-192+304+-192+320+0%0Aw+-192+240+-192+224+0%0Ar+-192+224+-96+224+0+1000%0A368+64+224+224+224+0+0%0Ab+-256+144+-112+341+0%0Ax+-252+367+-84+397+4+24+bipolare%5CsQuelle%5Cs%5C%5Cn(z.B.%5CsSensor)%0Ax+92+193+109+196+4+24+R%0Ax+110+207+123+210+4+24+1%0Ax+112+273+125+276+4+24+2%0Ax+94+259+111+262+4+24+R%0Ax+-18+207+-5+210+4+24+3%0Ax+-36+193+-19+196+4+24+R%0Ax+182+361+355+391+4+24+unipolare%5CsSenke%5C%5Cn(z.B.%5CsuC))%0A370+-96+224+-48+224+1+0%0Ax+-188+184+-171+187+4+24+R%0Ax+-169+193+-156+196+4+24+q%0Ao+9+1024+0+4098+20+6.4+0+2+5+0%0A 730,400 noborder}}
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Imagine you want to develop a circuit that conditions a sensor signal so that it can be processed by a microcontroller. The sensor signal is in the range $U_{sens} \in [-15...15V]$, the microcontroller input can read values in the range $U_{uC} \in [0...3.3V]$. The sensor can supply a maximum current of $I_{sens,max}=1mA$. For the internal resistance of the microcontroller input applies: $R_{uC} \rightarrow \infty$

For conditioning, the input signal is to be fed via the series resistor $R_3$ to the center potential of a voltage divider $R_1 - R_2$ with $R_1$ against $U_{uC,max}$ (similar circuit see in simulation on the right).

  - Find the relationship between $R_1$, $R_2$ and $R_3$ using superposition.
  - Find the relationship between $R_1$, $R_2$ and $R_3$ using star-delta transformation.
  - What is the input resistance $R_{in}(R_1, R_2,R_3)$ of the circuit (viewed from the sensor)?
  - What is the maximum allowed input resistance $R_{in}(R_1, R_2,R_3)$ for the sensor to still deliver current?
  - Determine suitable values for $R_1$, $R_2$ and $R_3$.
  - What values for $R^0_1$, $R^0_2$, and $R^0_3$ from the [[https://de.wikipedia.org/wiki/E-Reihe|E24 series]] can be used to do this?

</WRAP></WRAP></panel>


{{page>task_4.5.2_with_calculation&nofooter}}
{{page>aufgabe_4.5.3&nofooter}}
{{page>aufgabe_4.5.4&nofooter}}

<panel type="info" title="Task 1.7.6: temperature-dependent resistance of a winding (exam task, approx 6% of a 60-minute exam, WS2020)"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>

On the rotor of an asynchronous motor, the windings are designed in copper.
The length of the winding wire is 40 m.
The diameter is 0.4 mm.

When the motor is started, it is uniformly cooled down to the ambient temperature of 20°C.
During operation, the windings on the rotor have a temperature of 90°C. \\
$\alpha_{Cu,20°C}=0,0039 \frac{1}{K}$ \
$\beta_{Cu,20°C}=0,6 \cdot 10^{-6}  \frac{1}{K^2}$ \
$\rho_{Cu,20°C}=0.0178 \frac{\Omega mm^2}{m}$

Use both the linear and quadratic temperature coefficients!
1. determine the resistance of the line for $T = 20°C$.

<button size="xs" type="link" collapse="Solution_1_7_6_1_Solution path">{{icon>eye}} Solution path</button><collapse id="Solution_1_7_6_1_Solution path" collapsed="true">

\begin{align*}
R_{20°C} &= \rho_{Cu,20°C} \cdot \frac{l}{A} && | \text{with } A = r^2 \cdot \pi = \frac{1}{4} d^2 \cdot \pi \.
R_{20°C} &= \rho_{Cu,20°C} \cdot \frac{4 \cdot l}{d^2 \cdot \pi} && \\
R_{20°C} &= 0.0178 \frac{\Omega mm^2}{m} \cdot \frac{4 \cdot 40m}{(0.4mm)^2 \cdot \pi} && \\
\end{align*}

</collapse>

<button size="xs" type="link" collapse="Solution_1_7_6_1_Final result">{{icon>eye}} Final result</button><collapse id="Solution_1_7_6_1_Final result" collapsed="true">
\begin{align*}
R_{20°C} &= 5.666 \Omega -> 5.7 \Omega \.
\end{align*}
 \\
</collapse>


2. what is the increase in resistance $\Delta R$ between $20°C$ and $90°C$ for a winding?

<button size="xs" type="link" collapse="Solution_1_7_6_2_Solution Path">{{icon>eye}} Solution path</button><collapse id="Solution_1_7_6_2_Solution path" collapsed="true">

\begin{align*}
R_{90°C} &= R_{20°C} \cdot ( 1 + \alpha_{Cu,20°C} \cdot \Delta T + \beta_{Cu,20°C} \cdot \Delta T^2 ) && | \text{with } \Delta T = T_2 - T_1 = 90°C - 20°C = 70 °C = 70 K\\
\Delta R &= R_{20°C} \cdot ( \alpha_{Cu,20°C} \cdot \Delta T + \beta_{Cu,20°C} \cdot \Delta T^2 ) \.
\Delta R &= 5.666 \Omega \cdot ( 0.0039 \frac{1}{K} \cdot 70K + 0.6 \cdot 10^{-6}  \frac{1}{K^2} \cdot (70K)^2 ) \\
\end{align*}

</collapse>

<button size="xs" type="link" collapse="Solution_1_7_6_2_Final result">{{icon>eye}} Final result</button><collapse id="Solution_1_7_6_2_Final result" collapsed="true">
\begin{align*}
\Delta R &= 1.56 \Omega -> 1.6 \Omega \.
\end{align*}
 \\
</collapse>

</WRAP></WRAP></panel>
<panel type="info" title="Task 2.7.10 - Variation: Simplifying Circuits III (written exam task, approx 8% of a 60 minute written exam, WS2020)"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>

<WRAP right>
{{elektrotechnik_1:schaltung_klws2020_2_2_3.jpg?400}}
</WRAP>

Given is the adjoining circuit with \\
$R_1=5 \Omega$\\
$R_2=20 \Omega$\\
$R_3=10 \Omega$\\
and the switch $S$.

1. determine the total resistance $R_{ges}$ between A and B by summing the resistances with the switch $S$ open.

2. now let the voltage from A to B be: $U_{AB}=U_0= 30 V$. What is the current $I$?

</WRAP></WRAP></panel><panel type="info" title="Task 2.7.7 Simplifying Circuits (exam task, about 8% of a 60-minute exam, WS2020)"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>

<WRAP right>
{{elektrotechnik_1:schaltung_klws2020_2_1_1.jpg?400}}
</WRAP>

Given is the adjoining circuit with \\
$R_1=10 \Omega$\\
$R_2=20 \Omega$\\
$R_3=5 \Omega$\\
and the switch $S$.

1. determine the total resistance $R_{ges}$ between A and B by summing the resistances with the switch $S$ open.

<button size="xs" type="link" collapse="Solution_2_7_7_1_Tips">{icon>eye}} Tips for solving</button><collapse id="Solution_2_7_7_1_Tips" collapsed="true">
  * How can the circuit be better represented or pulled apart?
  * The switch should be replaced by an open wire in this case.
</collapse>

<button size="xs" type="link" collapse="Solution_2_7_7_1_Solution Path">{{icon>eye}} Solution path</button><collapse id="Solution_2_7_7_1_Solution path" collapsed="true">

First of all, it is a good idea to reshape the circuit so that the actual structure becomes visible. \\
For this purpose, the individual branches can be highlighted in color and interpreted as a "conductive rubber band". \\
This results in:


{{elektrotechnik_1:schaltung_klws2020_2_1_1_loesung1.jpg?300}}
{{elektrotechnik_1:schaltung_klws2020_2_1_1_loesung2.jpg?250}}

Thus $R_3$ and $R_3$ can be combined to $R_{33} = 2 \cdot R_3 = R_1$, yielding a left and a right voltage divider. \\
Now it is visible that in the left and right voltage divider the same potential is at the respective branch, or at the node K1 (green) and K2 (pink).

Thus, the total resistance can be calculated as $R_{ges} = (2 \cdot R_1)||(2 \cdot R_1)$. \\
However, by symmetry, nodes K1 and K2 can also be short-circuited. Thus, $R_{ges} = 2 \cdot \left( R_1||R_1 \right)$ also holds.

</collapse>

<button size="xs" type="link" collapse="Solution_2_7_7_1_Final result">{{icon>eye}} Final result</button><collapse id="Solution_2_7_7_1_Final result" collapsed="true">
\begin{align*}
R_{ges} &= 2 \cdot \left( 10 \Omega || 10 \Omega \right) = 10 \Omega
\end{align*}
 \\
</collapse>

2. what is the total resistance when switch $S$ is closed?

<button size="xs" type="link" collapse="Solution_5_1_3_2_FinalResult">{{icon>eye}} Final result</button><collapse id="Solution_5_1_3_2_Final result" collapsed="true">
Due to symmetry, the potentials at K1 and K2 are equal. Thus, no current flows across resistor $R_2$ even when the switch is closed. \\
So the resistance remains the same. \\ \\
</collapse>

</WRAP></WRAP></panel>
<panel type="info" title="Task 2.7.8: Simplifying Circuits II (written exam task, approx 8% of a 60-minute written exam, WS2020)"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>

<WRAP right>
{{elektrotechnik_1:schaltung_klws2020_2_2_1.jpg?400}}
</WRAP>

Given is the adjoining circuit with \\
$R_1=5 \Omega$\\
$R_2=10 \Omega$\\
$R_3=20 \Omega$\\
and the switch $S$.

1. determine the total resistance $R_{ges}$ between A and B by summing the resistances with the switch $S$ open.

<button size="xs" type="link" collapse="Solution_2_7_8_1_Tips">{icon>eye}} Tips for solving</button><collapse id="Solution_2_7_8_1_Tips" collapsed="true">
  * How can the circuit be better represented or pulled apart?
  * The switch should be replaced by an open line.
  * Does this result in equal potentials at different nodes that can be cleverly used?
</collapse>

<button size="xs" type="link" collapse="Solution_2_7_8_1_Solution path">{{icon>eye}} Solution path</button><collapse id="Solution_2_7_8_1_Solution path" collapsed="true">

First of all, it is a good idea to reshape the circuit so that the actual structure becomes visible. \\
For this purpose, the individual branches can be highlighted in color and interpreted as a "conductive rubber band". \\
It can be seen that the two resistors $R_3$ at the top left and bottom right are each shorted.
The result is thus:

{{elektrotechnik_1:schaltung_klws2020_2_2_1_loesung1.jpg?350}}
{{elektrotechnik_1:schaltung_klws2020_2_2_1_loesung2.jpg?300}}

Here it helps to consider the potential of the nodes K1, K2 and K3. For K2, the resistances $R_2 || R_3 || R_2$ must be combined at the top and bottom. Thus, the same resistance values at the top and bottom result. Also at the nodes K1 and K2 the same resistance values at the top and at the bottom result. With the same ratios of the resistances at K1, K2 and K3 respectively, it can be concluded that no current flows across the resistors $R_3$ between K1 and K2 or K2 and K3. Thus, these do not contribute to the total resistance. In such a case, a short circuit or an open line can be freely chosen between the relevant nodes for the calculation. In the following an open line is chosen. Additionally the parallel strings can be reordered. \\
This results in:

{{elektrotechnik_1:schaltung_klws2020_2_2_1_loesung3.jpg?300}}

\begin{align*}
R_{ges} &= \left( \left( 2 \cdot R_2 \right) || \left( 2 \cdot R_2 \right) \right) \quad && || \quad \left( \left( 2 \cdot R_3 \right) || \left( 2 \cdot R_3 \right) || \left( 2 \cdot R_3 \right) \right)  \\
R_{ges} &= R_2 \quad && || \quad \left( R_3 || \left( 2 \cdot R_3 \right) \right)  \\
R_{ges} &= R_2 \quad && || \quad \frac{R_3 \cdot 2 R_3}{R_3 + 2 R_3}   \\
R_{ges} &= R_2 \quad && || \quad \frac{2}{3}\cdot R_3 \
R_{ges} &= \frac{R_2 \cdot \frac{2}{3}\cdot R_3}{R_2 + \frac{2}{3}\cdot R_3} = \frac{R_2 \cdot R_3}{\frac{3}{2}\cdot R_2 + R_3}   \\ \\
\end{align*}

</collapse>

<button size="xs" type="link" collapse="Solution_2_7_8_1_Final result">{{icon>eye}} Final result</button><collapse id="Solution_2_7_8_1_Final result" collapsed="true">
\begin{align*}
R_{ges} &= \frac{10 \Omega \cdot 20 \Omega}{\frac{3}{2}\cdot 10 \Omega + 20 \Omega} = 5.7143 \Omega -> 5.7 \Omega \.
\end{align*}
 \\
</collapse>

2. now let the voltage from A to B be: $U_{AB}=U_0= 20 V$. What is the current $I$?

<button size="xs" type="link" collapse="Solution_2_7_8_2_Solution Path">{{icon>eye}} Solution path</button><collapse id="Solution_2_7_8_2_Solution path" collapsed="true">
The partial current $I$ is obtained directly from the voltage $U_0$:
\begin{align*}
I &=\frac{U_0}{2 \cdot R_3}
\end{align*}
 \\
</collapse>

<button size="xs" type="link" collapse="Solution_2_7_8_2_Final result">{{icon>eye}} Final result</button><collapse id="Solution_2_7_8_2_Final result" collapsed="true">
\begin{align*}
I =\frac{20V}{2 \cdot 20 \Omega} = 0.5 A
\end{align*}
 \\

</collapse>

</WRAP></WRAP></panel>
<panel type="info" title="Task 2.7.9 - Variation: Simplifying Circuits II (written exam task, approx 8% of a 60 minute written exam, WS2020)"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>

<WRAP right>
{{elektrotechnik_1:schaltung_klws2020_2_2_2.jpg?400}}
</WRAP>

Given is the adjoining circuit with \\
$R_1=10 \Omega$\\
$R_2=20 \Omega$\\
$R_3=5 \Omega$\\
and the switch $S$.

1. determine the total resistance $R_{ges}$ between A and B by summing the resistances with the switch $S$ open.

<button size="xs" type="link" collapse="Solution_2_7_9_1_Final result">{{icon>eye}} Final result</button><collapse id="Solution_2_7_9_1_Final result" collapsed="true">
\begin{align*}
R_{ges} &= \frac{20 \Omega \cdot 5 \Omega}{\frac{3}{2}\cdot 20 \Omega + 5 \Omega} = 2.858 \Omega -> 2.9 \Omega \.
\end{align*}
 \\
</collapse>

2. now let the voltage from A to B be: $U_{AB}=U_0= 10 V$. What is the current $I$?

<button size="xs" type="link" collapse="Solution_2_7_9_2_Final result">{{icon>eye}} Final result</button><collapse id="Solution_2_7_9_2_Final result" collapsed="true">
\begin{align*}
I =\frac{10V}{2 \cdot 5 \Omega} = 1 A
\end{align*}
 \\

</collapse>

</WRAP></WRAP></panel>
<panel type="info" title="Task 3.1.3: Load on linear source (exam task, about 8% of a 60-minute exam, WS2020)"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>


Given is a linear current source which supplies a resistive load $R_L=10\Omega$.
The current at the load is $I_L=2A$. The short-circuit current is $5 A$.

1. draw the equivalent circuit diagram of the structure.

<button size="xs" type="link" collapse="Solution_3_1_3_1_Tips">{icon>eye}} Tips for solving</button><collapse id="Solution_3_1_3_1_Tips" collapsed="true">
  * What does a linear current source look like?
  * How is the load connected?
</collapse>

<button size="xs" type="link" collapse="Solution_3_1_3_1_FinalResult">{{icon>eye}} Final result</button><collapse id="Solution_3_1_3_1_Final result" collapsed="true">
{{elektrotechnik_1:schaltung_klws2020_2_4_1.jpg?200}}
 \\
</collapse>

2. what is the internal conductance of the source?

<button size="xs" type="link" collapse="Solution_3_1_3_2_Tips">{icon>eye}} Tips for solving</button><collapse id="Solution_3_1_3_2_Tips" collapsed="true">
  * In the short-circuit case, the short-circuit current flows through $R_L$ only.
  * To simplify the calculation, it is convenient to convert the linear current source into a linear voltage source.
</collapse>

<button size="xs" type="link" collapse="solution_3_1_3_2_solution_path">{{icon>eye}} Solution path</button><collapse id="Solution_3_1_3_2_Solution path" collapsed="true">

Converting the linear current source to a linear voltage source results in an open circuit voltage $U_{LL}=R_i \cdot I_{KS}$. \\
The conversion also converts the circuit from a parallel circuit to a series circuit.  \\
Thus, the voltage $U_{LL}$ is across the voltage divider of $R_i$ and $R_L$: $U_{LL} = U_i + U_L$ \\
With the load resistor $R_L$, there is a current of $I_L$ through the series circuit. \\
With the given $R_L$ and $I_L$, the voltage $U_L$ across the load can be calculated. \\
The remaining voltage $U_i$ is applied to the internal resistor $R_i$, through which the current $I_L$ also flows.
This results in the conductance $G_i$:

\begin{align*}
U_{LL} &= U_i + U_L \\
R_i \cdot I_{KS} &= R_i \cdot I_L + R_L \cdot I_L \
R_i \cdot I_{KS} - R_i \cdot I_L &= R_L \cdot I_L \
R_i \cdot (I_{KS} - I_L) &= R_L \cdot I_L \\
R_i &= R_L \cdot \frac{I_L}{I_{KS} - I_L}   \\
G_i &= \frac{I_{KS} - I_L}{R_L \cdot I_L}   \\

\end{align*}

</collapse>

<button size="xs" type="link" collapse="Solution_3_1_3_2_FinalResult">{{icon>eye}} Final result</button><collapse id="Solution_3_1_3_2_Final result" collapsed="true">
\begin{align*}
G_i &= \frac{5A - 2A}{10 \Omega \cdot 2A} = 0.15 S \.
\end{align*}
 \\
</collapse>

3. what power does the load consume?

<button size="xs" type="link" collapse="Solution_3_1_3_final_result">{{icon>eye}} Final result</button><collapse id="Solution_3_1_3_3_Final result" collapsed="true">
\begin{align*}
P = R_L \cdot I_L^2 = 10 \Omega \cdot (2A)^2 = 40 W \.
\end{align*}

 \\
</collapse>

</WRAP></WRAP></panel>
<panel type="info" title="Task 4.5.2: Open-circuit voltage across superposition (written exam task, about 12% of a 60-minute written exam, WS2020)"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>

<WRAP right>
{{elektrotechnik_1:schaltung_klws2020_2_3_1.jpg?400}}
</WRAP>

Given is the adjoining circuit with \\
$R_1=5 \Omega$\\
U_1=2 V
$I_2=1 A$\\
$R_3=20 \Omega$\\
$U_3=8 V$\\
$R_4=10 \Omega$

Determine the open circuit voltage between A and B using the superposition principle.

<button size="xs" type="link" collapse="Solution_4_5_2_1_Tips">{icon>eye}} Tips for solving</button><collapse id="Solution_4_5_2_1_Tips" collapsed="true">
  * What are the individual circuits by which the effects (voltage between A and B) of each source can be calculated? \\ By which equivalent resistance must a current or a voltage source be replaced in the calculation of the individual effects?
  * Where are the open-circuit voltages located when considering the individual effects?
</collapse>

<button size="xs" type="link" collapse="solution_4_5_2_1_solution_path">{{icon>eye}} Solution path</button><collapse id="Solution_4_5_2_1_Solution path" collapsed="true">

First, the individual circuits must be created, from which the effect of the individual sources between points A and B can be determined.
\\ \\
**(Voltage)Source $U_1$**

  * Replace current source $I_2$ with short circuit
  * Replace voltage source $U_3$ with open circuit

{{elektrotechnik_1:schaltung_klws2020_2_3_1_q1.jpg?400}}

If the components are moved, the circuit can be better understood:

{{elektrotechnik_1:schaltung_klws2020_2_3_1_q1_1.jpg?300}}

It can be seen that in the no-load case there is no current flowing through any resistor. The following therefore applies to the effect: $U_{AB,1} = U_1$

**(Current)Source $I_2$**

  * Replace voltage source $U_1$ with open line
  * Replace voltage source $U_3$ with open line

{{elektrotechnik_1:schaltung_klws2020_2_3_1_q2.jpg?400}}

Again, components can be moved to better understand the circuit:

{{elektrotechnik_1:schaltung_klws2020_2_3_1_q2_1.jpg?300}}

Here the current source $I_2$ generates the voltage $U_{AB_2}$ at the resistor $R_2$: $U_{AB,2} = - R_1 \cdot I_2$


**(Voltage) source $U_3$**

  * Replace voltage source $U_1$ with open circuit
  * Replace current source $I_2$ with short circuit

{{elektrotechnik_1:schaltung_klws2020_2_3_1_q3.jpg?400}}

Also here the circuit becomes more understandable by moving the components:

{{elektrotechnik_1:schaltung_klws2020_2_3_1_q3_1.jpg?300}}

In this circuit, an unloaded voltage divider results across $R_3$ and $R_4$ in the no-load case. No current flows across resistor $R_1$ during no-load operation. \\
It results in:

\begin{align*}
U_{AB,3} = \frac{R_4}{R_3 + R_4} \cdot U_3
\end{align*}

\\ \\
**Resultant stress**

\begin{align*}
U_{AB} &= U_1 - R_1 \cdot I_2 + \frac{R_4}{R_3 + R_4} \cdot U_3 \\
\end{align*}

</collapse>

<button size="xs" type="link" collapse="Solution_4_5_2_1_FinalResult">{{icon>eye}} Final result</button><collapse id="Solution_4_5_2_1_Final result" collapsed="true">
\begin{align*}
U_{AB} &= 2 V - 5 \Omega \cdot 1 A + \frac{10 \Omega}{20 \Omega + 10 \Omega} \cdot 8 V = 0.333... V -> 0.3 V \
\end{align*}
 \\
</collapse>

</WRAP></WRAP></panel>
<panel type="info" title="Task 4.5.3 - Variation: open-circuit voltage across superposition (written exam task, about 12% of a 60-minute written exam, WS2020)"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>

<WRAP right>
{{elektrotechnik_1:schaltung_klws2020_2_3_2.jpg?400}}
</WRAP>

Given is the adjoining circuit with \\
$R_1=5 \Omega$\\
U_1=2 V
$I_2=1 A$\\
$R_3=20 \Omega$\\
$U_3=8 V$\\
$R_4=10 \Omega$

Determine the open circuit voltage between A and B using the superposition principle.

<button size="xs" type="link" collapse="Solution_4_5_3_1_FinalResult">{{icon>eye}} Final result</button><collapse id="Solution_4_5_3_1_Final result" collapsed="true">
\begin{align*}
U_{AB} = 29.333... V -> 29.3 V \
\end{align*}
 \\
</collapse>

</WRAP></WRAP></panel>
<panel type="info" title="Task 4.5.4 - Variation: open-circuit voltage across superposition (written exam task, about 12% of a 60-minute written exam, WS2020)"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>

<WRAP right>
{{elektrotechnik_1:schaltung_klws2020_2_3_3.jpg?400}}
</WRAP>

Given is the adjoining circuit with \\
$I_1=2 A$\\
$R_2=5 \Omega$\\
$R_3=20 \Omega$\\
$U_3=1 V$\\
$R_4=10 \Omega$ \\
$U_4=3 V$\\

Determine the open circuit voltage between A and B using the superposition principle.

</WRAP></WRAP></panel>
<panel type="info" title="Task 5.1.3 Forces on charges (written exam task, approx 8% of a 60 minute written exam, WS2020)"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>

<WRAP right>
{{electrical engineering_1:coulomb force geometryi.jpg?400}}
</WRAP>

Given is an arrangement of electric charges in a vacuum (see picture on the right). \\
The charges have the following values: \\
$Q_1=7 μC$ (point charge) \
$Q_2=5 μC$ (point charge) \
$Q_3=0 C$ (infinitely extended surface charge)

$\varepsilon_0=8,854\cdot 10^{-12}  F/m$ , $\varepsilon_r=1$

1. calculate the magnitude of the force of $Q_2$ on $Q_1$, without the force effect of $Q_3$.

<button size="xs" type="link" collapse="Solution_5_1_3_1_Tips">{icon>eye}} Tips for solving</button><collapse id="Loesung_5_1_3_1_Tips" collapsed="true">
  * What is the equation to be used for the force effect of charges?
  * How can the distance between the two charges be determined?
</collapse>

<button size="xs" type="link" collapse="solution_5_1_3_1_solution_path">{{icon>eye}} Solution path</button><collapse id="Solution_5_1_3_1_Solution path" collapsed="true">
\begin{align*}
F_C &= {{1} \over {4\pi\cdot\varepsilon}} \cdot {{Q_1 \cdot Q_2} \over {r^2}} \quad && | \text{with } r=\sqrt{\delta x^2 + \delta y^2}  \\
F_C &= {{1} \over {4\pi\cdot\varepsilon}} \cdot {{Q_1 \cdot Q_2} \over {\delta x^2 + \delta y^2}} \quad && | \text{insert numeric values, read distances: } \Delta x = 5dm, \Delta y = 3dm \.
F_C &= {{1} \over {4\pi\cdot 8,854\cdot 10^{-12}  F/m}} \cdot {{7 \cdot 10^{-6} C \cdot 5 \cdot 10^{-6} C} \over { (0.5m)^2 + (0.2m)^2}}
\end{align*}
</collapse>

<button size="xs" type="link" collapse="Solution_5_1_3_1_FinalResult">{{icon>eye}} Final result</button><collapse id="Solution_5_1_3_1_Final result" collapsed="true">
\begin{align*}
 |F_C| = 1.084 N -> 1.1 N
\end{align*}
 \\
</collapse>

2. is this force attractive or repulsive?

<button size="xs" type="link" collapse="Solution_5_1_3_2_Tips">{icon>eye}} Tips for solving</button><collapse id="Solution_5_1_3_2_Tips" collapsed="true">
  * What force effect do equally or oppositely charged bodies exhibit on each other?
</collapse>

<button size="xs" type="link" collapse="Solution_5_1_3_2_FinalResult">{{icon>eye}} Final result</button><collapse id="Solution_5_1_3_2_Final result" collapsed="true">
The force is repulsive because both charges have the same sign. \\ \\ \\
</collapse>

Now let $Q_2=0$ and the surface charge $Q_3$ be shaped in such a way that a homogeneous electric field with $E_3=100 kV/m$ results. \\ What force (magnitude) now results on $Q_1$?

<button size="xs" type="link" collapse="Solution_5_1_3_Tips">{{icon>eye}} Tips for solving</button><collapse id="Loesung_5_1_3_Tips" collapsed="true">
  * Which equation should be used for the force action in a homogeneous field?
</collapse>

<button size="xs" type="link" collapse="solution_5_1_3_3_solution_path">{{icon>eye}} Solution path</button><collapse id="Solution_5_1_3_3_Solution path" collapsed="true">
\begin{align*}
F_C &= E \cdot Q_1 \quad && | \text{insert numeric values} \\
F_C &= 100 \cdot 10^3 V/m \cdot 7 \cdot 10^{-6} C
\end{align*}
</collapse>

<button size="xs" type="link" collapse="solution_5_1_3_3_final_result">{{icon>eye}} Final result</button><collapse id="Solution_5_1_3_3_Final result" collapsed="true">
\begin{align*}
 |F_C| = 0.7 N
\end{align*} \\
</collapse>


</WRAP></WRAP></panel>
<panel type="info" title="Task 5.1.4 Variation: forces on charges (exam task, approx 8% of a 60-minute exam, WS2020)"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>

<WRAP right>
{{electrical engineering_1:coulomb force geometryii.jpg?400}}
</WRAP>

Given is an arrangement of electric charges in a vacuum (see picture on the right). \\
The charges have the following values: \\
$Q_1=5 μC$ (point charge) \
$Q_2=-10 μC$ (point charge) \
$Q_3=0 C$ (infinitely extended surface charge)

$\varepsilon_0=8,854\cdot 10^{-12}  F/m$ , $\varepsilon_r=1$

1. calculate the magnitude of the force of $Q_2$ on $Q_1$, without the force effect of $Q_3$.

<button size="xs" type="link" collapse="Solution_5_1_4_1_FinalResult">{{icon>eye}} Final result</button><collapse id="solution_5_1_4_1_final-result" collapsed="true">
\begin{align*}
 |F_C| = 1.321 N -> 1.3 N
\end{align*}
 \\
</collapse>

2. is this force attractive or repulsive?

<button size="xs" type="link" collapse="Solution_5_1_4_2_FinalResult">{{icon>eye}} Final result</button><collapse id="Solution_5_1_4_2_Final result" collapsed="true">
The force is repulsive because the charges have the same sign. \\ \\ \\
</collapse>

Now let $Q_2=0$ and the surface charge $Q_3$ be shaped in such a way that a homogeneous electric field with $E_3=500 kV/m$ results. \\ What force (magnitude) now results on $Q_1$?


<button size="xs" type="link" collapse="solution_5_1_4_3_final_result">{{icon>eye}} Final result</button><collapse id="Solution_5_1_4_3_Final result" collapsed="true">
\begin{align*}
 |F_C| = 2.5 N
\end{align*} \\
</collapse>


</WRAP></WRAP></panel>
<panel type="info" title="Task 5.1.5 Variation: forces on charges (exam task, approx 8% of a 60-minute exam, WS2020)"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>

<WRAP right>
{{electrical engineering_1:coulomb force geometryiii.jpg?400}}
</WRAP>

Given is an arrangement of electric charges in a vacuum (see picture on the right). \\
The charges have the following values: \\
$Q_1=2 μC$ (point charge) \
$Q_2=-4 μC$ (point charge) \
$Q_3=0 C$ (infinitely extended surface charge)

$\varepsilon_0=8,854\cdot 10^{-12}  F/m$ , $\varepsilon_r=1$

1. calculate the magnitude of the force of $Q_2$ on $Q_1$, without the force effect of $Q_3$.

<button size="xs" type="link" collapse="Solution_5_1_5_1_FinalResult">{{icon>eye}} Final result</button><collapse id="solution_5_1_5_1_final-result" collapsed="true">
\begin{align*}
 |F_C| = 0.3595 N -> 0.36 N
\end{align*}
 \\
</collapse>

2. is this force attractive or repulsive?

<button size="xs" type="link" collapse="Solution_5_1_5_2_FinalResult">{{icon>eye}} Final result</button><collapse id="Solution_5_1_5_2_Final result" collapsed="true">
The force is attractive because the charges have different signs. \\ \\ \\
</collapse>

Now let $Q_2=0$ and the surface charge $Q_3$ be shaped in such a way that a homogeneous electric field with $E_3=200 kV/m$ results. \\ What force (magnitude) now results on $Q_1$?


<button size="xs" type="link" collapse="Solution_5_1_5_3_Final_Result">{{icon>eye}} Final result</button><collapse id="Solution_5_1_5_3_Final result" collapsed="true">
\begin{align*}
 |F_C| = 0.4 N
\end{align*} \\
</collapse>


</WRAP></WRAP></panel>
<panel type="info" title="Task 5.2.1 Multiple Forces on a Charge I (written exam task, about 8% of a 60-minute written exam, WS2020)"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>

<WRAP right>
{{electrical engineering_1:kraefteadditiongeometriei.jpg?400}}
</WRAP>

Given is the arrangement of electric charges in the picture on the right. \\
The following force effects result: \\
$F_{01}=-5 N$ \
$F_{02}=-6 N$ \
$F_{03}=+3 N$

Calculate the magnitude of the resulting force.

<button size="xs" type="link" collapse="Solution_5_2_1_1_Tips">{{icon>eye}} Tips for solving</button><collapse id="Solution_5_2_1_1_Tips" collapsed="true">.
  * How must the forces be prepared so that they can actually be added?
</collapse>

<button size="xs" type="link" collapse="solution_5_2_1_1_solution_path">{{icon>eye}} Solution path</button><collapse id="Solution_5_2_1_1_Solution path" collapsed="true">
\begin{align*}
F_0 &= |\vec{F_0}| \quad \text{ with }  \vec{F_0} = \left( \begin{matrix}{F_{x,0}}\ {F_{y,0}} \end{matrix} \right) = \left( \begin{matrix}  \sum\limits_{n} F_{x,0n} \sum\limits_{n} F_{y,0n}    \end{matrix} \right} \\
F_0 &= \sqrt{ \left(\sum\limits_{n} F_{x,0n} \right)^2 + \left(\sum\limits_{n} F_{y,0n} \right)^2 } \\
\end{align*}

The existing forces must be decomposed into coordinates. Here the orthogonal coordinates ($x$ and $y$) are recommended. \\
Let the coordinate system be such that the origin is in $Q_0$ with the x-axis in the direction of Q_3 and the y-axis correspondingly perpendicular to it. \\
For coordinate decomposition, the angles $alpha_{0n}$ of the forces to the x-axis are necessary. \\
These result in the chosen coordinate system from the coordinates of the charges: $\alpha_{0n} = atan(\frac{\delta y}{\delta x})$ \
$\alpha_{01} = atan(\frac{3}{1})= 1.249 = 71.6°$ \\
$\alpha_{02} = atan(\frac{4}{3})= 0.927 = 53.1°$\
$\alpha_{03} = atan(\frac{0}{3})= 0= 0°$ \

Then the decomposed forces add up to: \\ \\

\begin{align*}
F_{x,0} &= F_{x,01} + F_{x,02} + F_{x,03} && | \quad \text{with } F_{x,0n} = F_{0n} \cdot sin(\alpha_{0n}) \
F_{x,0} &= (-5N) \cdot sin(71,6°) + (-6N) \cdot sin(53,1°) + (+3N) \cdot sin(0°) \
F_{x,0} &= -2.18 N \\ \

F_{y,0} &= F_{x,01} + F_{x,02} + F_{x,03} && | \quad \text{with } F_{y,0n} = F_{0n} \cdot cos(\alpha_{0n}) \.
F_{y,0} &= (-5N) \cdot cos(71,6°) + (-6N) \cdot cos(53,1°) + (+3N) \cdot cos(0°) \
F_{y,0} &= -9.54 N \\ \

\end{align*}


</collapse>

<button size="xs" type="link" collapse="Solution_5_2_1_1_FinalResult">{{icon>eye}} Final result</button><collapse id="Solution_5_2_1_1_Final result" collapsed="true">
\begin{align*}
F_0 &= \sqrt{ (-2.18 N)^2 + (-9.54 N)^2 } = 9.79 N -> 9.8 N \.
\end{align*}
 \\
</collapse>

</WRAP></WRAP></panel>
<panel type="info" title="Task 5.2.2 Variation: multiple forces on a charge II (written exam task, approx 8% of a 60 minute written exam, WS2020)"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>

<WRAP right>
{{electrical engineering_1:kraefteadditiongeometrieii.jpg?400}}
</WRAP>

Given is the arrangement of electric charges in the picture on the right. \\
The following force effects result: \\
$F_{01}=-5 N$ \
$F_{02}=-6 N$ \
$F_{03}=+3 N$

Calculate the magnitude of the resulting force.

<button size="xs" type="link" collapse="Solution_5_2_2_1_FinalResult">{{icon>eye}} Final result</button><collapse id="solution_5_2_2_1_final-result" collapsed="true">
\begin{align*}
|F_0| &= \sqrt{ (-0.418 N)^2 + (-9.264 N)^2 } = 9.274 N -> 9.3 N \.
\end{align*}
 \\
</collapse>

</WRAP></WRAP></panel>
<panel type="info" title="Task 5.2.3 Variation: multiple forces on a charge III (exam task, approx 8% of a 60 minute exam, WS2020)"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>

<WRAP right>
{{electrotechnology_1:kraefteadditiongeometryiii.jpg?400}}
</WRAP>

Given is the arrangement of electric charges in the picture on the right. \\
The following force effects result: \\
$F_{01}=+2 N$ \
$F_{02}=-3 N$ \
$F_{03}=+4 N$

Calculate the magnitude of the resulting force.

<button size="xs" type="link" collapse="Solution_5_2_3_1_FinalResult">{{icon>eye}} Final result</button><collapse id="Solution_5_2_3_1_Final result" collapsed="true">
\begin{align*}
|F_0| &= \sqrt{ (2.12 N)^2 + (0.38 N)^2 } = 2.16 N -> 2.2 N\\
\end{align*}
 \\
</collapse>

</WRAP></WRAP></panel>
<panel type="info" title="Task 5.4.2 Field strength in different geometry I (exam task, approx 6% of a 60 minute exam)"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>

<WRAP right>{{electrical engineering_1:field strength geometryi.jpg?400}}</WRAP>

At picture right side is shown an arrangement of ideal metallic conductors (grey) with given charge. At white is drawn a dielectricum (e.g. vacuum).
Several, marked areas are marked by green dotted frames, which partly are inside of objects.

Clearly arrange the designated areas according to ascending field strength (magnitude)!
Also indicate if the designated areas have the same field strength in terms of magnitude.

<button size="xs" type="link" collapse="Solution_5_4_2_Tips">{{icon>eye}} Tips for solving</button><collapse id="Solution_5_4_2_Tips" collapsed="true">.
  * What is the field in a room completely surrounded by a conducting conductor?
  * How does the field behave inside a conductor?
  * Does the field strength increase or decrease as one charge moves away from another charge?
  * Is the field at at a peak higher or lower?
</collapse>

<button size="xs" type="link" collapse="solution_5_4_2_solution_path">{{icon>eye}} Solution path</button><collapse id="Solution_5_4_2_Solution path" collapsed="true">
  - At $b$ and $d$, no field is measurable because the surrounded conductor is at a constant field. It results in no potential difference and therefore no field.
  - At $c$ a field (magnitude >0) is measurable, which points from the charge ($+1C$) to the elongated conductor ($-2C$). Due to the tip there is an excess charge and thus a higher field.
  - At $a$ a field (magnitude >0) is measurable, which points from the charge ($+1C$) to the elongated conductor ($-2C$).

</collapse>

<button size="xs" type="link" collapse="Solution_5_4_2_FinalResult">{{icon>eye}} Final result</button><collapse id="solution_5_4_2_final_result" collapsed="true">
$ b = d < a < c $
</collapse>

</WRAP></WRAP></panel><panel type="info" title="Task 5.4.3 Field strength in different geometry II (exam task, about 6% of a 60-minute exam)"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>

<WRAP right>{{electrical engineering_1:field strength geometryii.jpg?400}}</WRAP>

At picture right side is shown an arrangement of ideal metallic conductors (grey) with given charge. At white is drawn a dielectricum (e.g. vacuum).
Several, marked areas are marked by green dotted frames, which partly are inside of objects.

Clearly arrange the designated areas according to ascending field strength (magnitude)!
Also indicate if the designated areas have the same field strength in terms of magnitude.

<button size="xs" type="link" collapse="Solution_5_4_3_FinalResult">{{icon>eye}} Final result</button><collapse id="solution_5_4_3_final_result" collapsed="true">
$ a = c < d < b $
</collapse>

</WRAP></WRAP></panel><panel type="info" title="Task 5.4.4 Field strength in different geometry III (exam task, about 6% of a 60-minute exam)"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>

<WRAP right>{{electrical engineering_1:field strength geometryiii.jpg?400}}</WRAP>

At picture right side is shown an arrangement of ideal metallic conductors (grey) with given charge. At white is drawn a dielectricum (e.g. vacuum).
Several, marked areas are marked by green dotted frames, which partly are inside of objects.

Clearly arrange the designated areas according to ascending field strength (magnitude)!
Also indicate if the designated areas have the same field strength in terms of magnitude.

<button size="xs" type="link" collapse="solution_5_4_4_final_result">{{icon>eye}} Final result</button><collapse id="solution_5_4_4_final_result" collapsed="true">
$ a = c < b < d $
</collapse>

</WRAP></WRAP></panel><panel type="info" title="Task 5.9.3: temperature-dependent resistance of a winding (exam task, about 6% of a 60-minute exam, WS2020)"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>

<WRAP right>
{{elektrotechnik_1:schaltung_klws2020_3_1_1.jpg?200}}
</WRAP>

Determine the capacitance $C$ for the plate capacitor drawn on the right with the following data:
  * rectangular electrodes with edge lengths of $6 cm$ and $8 cm$.
  * distance between the plates: $2 mm$
  * dielectric A:
    * $\varepsilon_{r,A} = 1 (air)$
    * thickness $d_A = 1.5 mm$
  * Dielectric B:
    * $\varepsilon_{r,B} = 100 (ice)$
    * thickness $d_B = 0.5 mm$

$\varepsilon_{0} = 8.854 \cdot 10^{-12}  F/m$

<button size="xs" type="link" collapse="Solution_5_9_3_1_Tips">{{icon>eye}} Tips</button><collapse id="Solution_5_9_3_1_Tips" collapsed="true">
  * What circuit can be used to replace a layered structure with different dielectrics?
</collapse>

<button size="xs" type="link" collapse="solution_5_9_3_1_solution_path">{{icon>eye}} Solution path</button><collapse id="Solution_5_9_3_1_Solution path" collapsed="true">

The total capacitance $C$ can be divided into a partial capacitance $C_A$ and a $C_B$. These are connected in series. \\
This results in: $C = \frac{C_A \cdot C_B}{C_A + C_B}$ \\ \\

The partial capacity $C_A$ can be calculated by
\begin{align*}
C_A &= \varepsilon_{0} \varepsilon_{r,A} \cdot \frac{A}{d_A} && | \text{with } A = 3 cm \cdot 5cm = 6 \cdot 10^{-2} \cdot 8 \cdot 10^{-2} m^2 = 48 \cdot 10^{-4} m^2\\.
C_A &= 8.854 \cdot 10^{-12}  F/m \cdot \frac{48 \cdot 10^{-4} m^2}{1.5 \cdot 10^{-3} m} \\
C_A &= 28.33 \cdot 10^{-12} F \\
\end{align*}

The partial capacitance $C_B$ can be calculated by
\begin{align*}
C_B &= \varepsilon_{0} \varepsilon_{r,B} \cdot \frac{B}{d_B} \\
C_B &= 100 \cdot 8,854 \cdot 10^{-12}  F/m \cdot \frac{48 \cdot 10^{-4} m^2}{0.5 \cdot 10^{-3} m} \\
C_B &= 8,500 \cdot 10^{-9} F \\
\end{align*}

</collapse>

<button size="xs" type="link" collapse="Solution_5_9_3_1_FinalResult">{{icon>eye}} Final result</button><collapse id="Solution_5_9_3_1_Final result" collapsed="true">
\begin{align*}
C = 28.24 \cdot 10^{-12} F -> 28pF
\end{align*}
 \\
</collapse>


</WRAP></WRAP></panel>
<panel type="info" title="Task 7.2.6: temperature-dependent resistance of a winding (written exam task, approx. 11% of a 60-minute exam, WS2020)"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>

<WRAP right>
{{elektrotechnik_1:schaltung_klws2020_3_2_1.jpg?400}}
</WRAP>

Given is the adjoining circuit with \\
  * $U = 10 V$
  * $I = 4 mA$
  * $R_1 = 100 \Omega, R_2 = 80 \Omega, R_3 = 50 \Omega, R_4 = 10 \Omega$
  * $C = 40 nF$
At the beginning the capacitor is discharged, all switches are open.
Switch S1 is closed at time t = 0 s.

1. determine the time constant $\tau$ for this charging process.

<button size="xs" type="link" collapse="Solution_7_2_6_1_Tips">{icon>eye}} Tips</button><collapse id="Solution_7_2_6_1_Tips" collapsed="true">
  * What equivalent circuit results from the switch position?
  * By what quantities can $\tau$ be determined?
  * Through what does the charging current flow?
</collapse>

<button size="xs" type="link" collapse="solution_7_2_6_1_solution_path">{{icon>eye}} Solution path</button><collapse id="Solution_7_2_6_1_Solution path" collapsed="true">

This results in a series connection of $R_1$, $R_2$ and $C$, which is fed by $U$.
Thus the time constant $\tau$ becomes:
\begin{align*}
\tau &= (R_1 + R_2) \cdot C \.
\tau &= 180 \Omega \cdot 40 nF
\end{align*}

</collapse>

<button size="xs" type="link" collapse="Solution_7_2_6_1_FinalResult">{{icon>eye}} Final result</button><collapse id="Solution_7_2_6_1_Final result" collapsed="true">
\begin{align*}
\tau = 7.2 µs
\end{align*}
 \\
</collapse>

2. what is the voltage across capacitor $C$ at time $t=10 µs$?

<button size="xs" type="link" collapse="solution_7_2_6_2_solution_path">{{icon>eye}} Solution path</button><collapse id="Solution_7_2_6_2_Solution path" collapsed="true">

It holds:
\begin{align*}
U_C(t) = U \cdot (1 - e^{-t/\tau}) \.
U_C(t) = 10 V \cdot (1 - e^{-10 µs/7,2 µs})
\end{align*}

</collapse>

<button size="xs" type="link" collapse="Solution_7_2_6_2_FinalResult">{{icon>eye}} Final result</button><collapse id="Solution_7_2_6_2_Final result" collapsed="true">
\begin{align*}
U_C(t) = 7.506 V -> 7.5 V
\end{align*}
 \\
</collapse>

3. what is the energy in the capacitor when it is fully charged?

<button size="xs" type="link" collapse="solution_7_2_6_3_solution_path">{{icon>eye}} Solution path</button><collapse id="Solution_7_2_6_3_Solution path" collapsed="true">

\begin{align*}
W_C &= \frac{1}{2}CU^2 \
   &= \frac{1}{2} \cdot 40nF \cdot (10V)^2
\end{align*}

</collapse>

<button size="xs" type="link" collapse="Solution_7_2_6_3_FinalResult">{{icon>eye}} Final result</button><collapse id="Solution_7_2_6_3_Final result" collapsed="true">
\begin{align*}
W_C = 2 µJ
\end{align*}
 \\
</collapse>

Determine the new time constant that is effective when switch S1 is opened and S2 is closed at the same time after complete charging.

<button size="xs" type="link" collapse="solution_7_2_6_4_solution_path">{{icon>eye}} Solution path</button><collapse id="Solution_7_2_6_4_Solution path" collapsed="true">

Here, the capacitor $C$ discharges through the resistors $R_2$ and $R_3$ connected in series.
\begin{align*}
\tau &= (R_2 + R_3) \cdot C \.
\tau &= 130 \Omega \cdot 40 nF
\end{align*}

</collapse>

<button size="xs" type="link" collapse="Solution_7_2_6_4_FinalResult">{{icon>eye}} Final result</button><collapse id="Solution_7_2_6_4_Final result" collapsed="true">
\begin{align*}
\tau = 5.2 µs
\end{align*}
 \\
</collapse>

After the capacitor has been completely discharged, all switches are opened again.
Switch S4 is closed for $t = 1μs$. \\
What is the voltage across C?

<button size="xs" type="link" collapse="Solution_7_2_6_5_Tips">{{icon>eye}} Tips</button><collapse id="Solution_7_2_6_5_Tips" collapsed="true">
  * The current source results in a continuous flow of charges into the capacitor.
  * The resistances along the way are irrelevant to the current into the capacitor. They only increase the voltage necessary to drive the current in an ideal current source.
</collapse>

<button size="xs" type="link" collapse="solution_7_2_6_5_solution_path">{{icon>eye}} Solution path</button><collapse id="Solution_7_2_6_5_Solution path" collapsed="true">

The voltage $U_C$ is generally given by: $U_C = \frac{Q}{C}$. In this case, the constant current I generates the charge $Q = \int I dt = I \cdot t$
\begin{align*}
U_C(t) &= \frac{Q}{C} \\
U_C(t) &= \frac{I \cdot t}{C} \\
U_C(1μs) &= \frac{4mA \cdot 1μs}{40nF} = \frac{4 \cdot 10^{-3}A \cdot 1\cdot 10^{-6}s}{40\cdot 10^{-9}F} \\

\end{align*}

</collapse>

<button size="xs" type="link" collapse="Solution_7_2_6_5_Final_Result">{{icon>eye}} Final result</button><collapse id="Solution_7_2_6_5_Final result" collapsed="true">
\begin{align*}
U_C(1μs) &= 1V \.
\end{align*}
 \\
</collapse>


</WRAP></WRAP></panel>====== 5. The Electrostatic Field ======

<WRAP right>
<imgcaption imageNo01 | John Tra-Voltage >
</imgcaption> <WRAP>.
{{url>https://phet.colorado.edu/sims/html/john-travoltage/latest/john-travoltage_de.html 500,400 noborder}}
</WRAP>

From everyday life it is known that there are different charges and effects of charge. In <imgref imageNo01>, the charge of a person is seen by charge separation between the sole of the foot and the floor. The movement of the foot creates a negative excess charge in the person, which is gradually distributed throughout the body. If a pointed part of the body (e.g. finger) is brought into the vicinity of a charge reservoir without excess charges, a current can flow even through the air.

In the first chapter we had already considered the charge as the central quantity of electricity and understood it as a multiple of the elementary charge. The mutual force action ([[Fundamentals_and_basic_concepts#coulomb-force|the Coulomb-force]]) was already derived there. This is to be explained now more near.


First, however, a differentiation of various terms:
  - **{wpde>electricity}}** describes as an umbrella term all phenomena of moving and resting charges. \\ \\
  - **{wpde>Electrostaticity}}** describes the phenomena of charges at rest and thus of electric fields which do not change in time. Thus, there is no time dependence of the electrical quantities. \ Mathematically, ${{df}\over{dt}}=0$ holds for any function of the electric quantities. \\ \\
  - **{wpde>Electrodynamics}}** describes the phenomena of moving charges. Thus electrodynamics includes both electric fields that change with time and magnetic fields. As a justification for the latter, it shall suffice here at first that magnetic fields are based on a current or on a charge movement. \\ Mathematically here is no longer valid for every function of the electric quantities necessarily that the derivative is equal to zero. \\ Electrodynamics is not considered in this chapter and is introduced step by step in the following chapters and in [[:Electrical engineering 2:start]].

At this chapter, only electrostatics are considered. The magnetic fields are therefore excluded here for the time being.

~~PAGEBREAK~~~CLEARFIX~~~

===== 5.1 Electric field strength and field lines =====

<callout>

=== Goals ===

After this lesson, you should:

  - know that an electric field is formed around a charge.
  - be able to sketch the field lines of the electric field.
  - be able to represent the field strength vectors in a sketch when given several charges.
  - be able to determine the resulting field strength vector by superimposing several field strength vectors using vector calculus.
  - Be able to determine the force on a charge in an electrostatic field by applying Coulomb's law. Specifically:
      - the force vector in coordinate representation
      - the magnitude of the force vector
      - the angle of the force vector

</callout>

<WRAP right>
<imgcaption imageNo02 | setup for own experiments >
</imgcaption> <WRAP>.
{{url>https://phet.colorado.edu/sims/html/charges-and-fields/latest/charges-and-fields_de.html 500,400 noborder}} \\
Take a charge ($+1nC$) and position it. \ Measure the field across a sample charge (a sensor).
</WRAP>

The simulation in <imgref BildNr02> was already briefly considered in the first chapter. Here, however, another point is to be dealt with.

In the simulation, please position a negative charge $Q$ in the middle and deactivate electric field. The latter is done via the hook on the right. Now the situation is close to reality, because a charge shows no effect at first sight.

For impact analysis, a sample charge $q$ is placed in the vicinity of the existing charge $Q$ (in the simulation, the sample charge is called "sensors"). It is observed that the charge $Q$ causes a force on the sample charge. This force can be determined at any point in space with magnitude and direction. It acts in space in a similar way to gravity. The description of the state in space changed by the charge $Q$ is described with the help of a field.

The concept of the field shall now be briefly considered in a little more detail.
  - The introduction of the field separates the cause from the effect.
    - The charge $Q$ causes the field in space.
    - The charge $q$ in space feels a force as an effect of the field.
    - This distinction becomes important again in this chapter. \\ Also in electrodynamics this distinction becomes clear: the field there corresponds to photons, i.e. to a transmission of effects with the finite (light)speed $c$.
  - As with physical quantities, there are different-dimensional fields:
    - In a **scalar field**, a single number is assigned to each point in space. \\ E.g.
      - temperature field $T(\vec{x})$ on the weather map or in an object
      - pressure field $p(\vec{x})$
    - In a **vector field**, each point in space is assigned several numbers in the form of a vector. This reflects the action along the spatial coordinates. \\ For example.
      - gravitational field $\vec{g}(\vec{x})$ pointing to the center of mass of the object.
      - electric field $\vec{E}(\vec{x})$
      - magnetic field $\vec{H}(\vec{x})$
  - If each point in space is associated with a two- or more-dimensional physical quantity - that is, a tensor - then this field is called a tensor field. Tensor fields are relevant in mechanics (e.g., stress tensor) but are not necessary for electrical engineering.
Vector fields can be stated as:
  - Effects along spatial axes $x$,$y$ and $z$ (Cartesian coordinate system).
  - Effect in magnitude and direction vector (polar coordinate system)

~~PAGEBREAK~~~CLEARFIX~~~
<callout icon="fa fa-exclamation" color="red" title="Remember:">
  - Fields describe a physical state of space.
  - Here, a physical quantity is assigned to each point in space.
  - The electrostatic field is described by a vector field.
</callout>

==== The electric field ====

Thus, to determine the electric field, a measure of the strength of the field is now needed. From the first chapter the Coulomb-force between two charges $Q_1$ and $Q_2$ is known:

\begin{align*}
F_C = {{1} \over {4\pi\cdot\varepsilon}} \cdot {{Q_1 \cdot Q_2}} \over {r^2}}
\end{align*}

In order to obtain a measure of the strength of the electric field, the force on a (fictitious) sample charge $q$ is now considered.

\begin{align*}
F_C &= {{1} \over {4\pi\cdot\varepsilon}} \cdot {{Q_1 \cdot q} \over {r^2}} \\
    &= \anderbrace{{1} \over {4\pi\cdot\varepsilon}} \cdot {{Q_1} \over {r^2}}_\text{=independent of q} \cdot q \\
\end{align*}

The left part is therefore a measure of the strength of the field, i.e. independent of the size of the sample charge $q$. The strength of the electric field is thus given by

<WRAP centeralign>
$E = {{1} \over {4\pi\cdot\varepsilon}} \cdot {{Q_1} \over {r^2}} \quad$ with $[E]={{[F]}\over{[q]}}=1 {{N}\over{As}}=1 {{N\cdot m}\over{As \cdot m}} = 1 {{V \cdot A \cdot s}\over{As \cdot m}} = 1 {{V}\over{m}}$
</WRAP>

The result is therefore
\begin{align*}
\boxed{F_C = E \cdot q}
\end{align*}


<callout icon="fa fa-exclamation" color="red" title="Notice:">

  - The **p**rocharge $q$ is always considered to be **p**positive. It is used only as a thought experiment and has no repercussion on the sampled charge $Q$.
  - The sampled charge is a point charge.
</callout>

<callout icon="fa fa-exclamation" color="red" title="Notice:">

A charge $Q$ generates at a measuring point $P$ an electric field strength $\vec{E(Q)}$, which is given by
  - the magnitude $|E|=\Bigl| {{1} \over {4\pi\cdot\varepsilon}} \cdot {{Q_1} \over {r^2}} \Bigl| $ and
  - the direction of the force $\vec{F_C}$ on a sample charge at rest at the measurement point $P$. </callout> This is given by the unit vector $\vec{e_r}={{\vec{F_C}}\over{|F_C|}$ in that direction.
</callout>

The direction of the electric field is switchable in <imgref ImageNo02> via the "Electric Field" option on the right. \\
The electric field can also be viewed again in [[https://www.youtube.com/watch?v=7k77nro5mAo|this video]].

==== Electric field lines ====

<WRAP right>
<imgcaption imageNo03 | examples of field lines>
</imgcaption> <WRAP right>
{{drawio>ExamplesForFieldLines}} \\
\\
Superposition of fields \\ ([[https://creativecommons.org/licenses/by-sa/4.0/deed.de|CC-BY-SA 4.0]]: [[https://lx3.mint-kolleg.kit.edu/onlinekursphysik/html/sectionx1.1.0.html|MINT bridge course]])

{{youtube>VGsZIbDwS8s}}

<panel>
<imgcaption imageNo00 | superposition example>
</imgcaption> <panel>
<collapse id="geogebra" collapsed="true"><well>
{{url>https://www.geogebra.org/material/iframe/id/qIXZJKqj/width/500/height/500/border/888888/rc/false/ai/false/sdz/false/smb/false/stb/false/stbh/true/ld/false/sri/true/at/preferhtml5 500,500 noborder}}
</well></collapse>

<collapse id="geogebra" collapsed="false">
<button type="warning" collapse="geogebra">To view the animation: click here!</button>
</collapse>
 
</panel>

</WRAP>

Electric field lines result as the (fictitious) path of a sample charge. Thus also electric field lines of several charges can be determined. However, these also result from a superposition of the individual effects - i.e. field strengths - at a measuring point $P$.

The superposition is sketched in <imgref imageNo03> and can be viewed again in the simulation in <imgref imageNo00>. In addition, this is described again in more detail in the video on the right.

<callout icon="fa fa-exclamation" color="red" title="Merke:">
  - The electrostatic field is a source field. This means there are sources and sinks. \\ This is equal to: The electric field lines have a beginning (at a positive charge) and an end (at a negative charge).
  - From the field line diagrams, the following can be obtained:
    - Direction of the field ($\hat{=}$ tangent to the field line).
    - Magnitude of the field ($\hat{=}$ number of field lines per unit area).
  - The magnitude of the field strength along a field line is usually __not__ constant.
</callout>

~~PAGEBREAK~~~CLEARFIX~~~

==== Tasks ====
<panel type="info" title="Task 5.1.1 Electric field example tasks"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>

{{youtube>YarLgwyV7e4}}
([[https://creativecommons.org/licenses/by-sa/4.0/deed.de|CC-BY-SA 4.0]]: [[https://lx3.mint-kolleg.kit.edu/onlinekursphysik/html/sectionx1.1.0.html|MINT bridge course]])

</WRAP></WRAP></panel>

<panel type="info" title="Task 5.1.2 Field lines"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>

<WRAP right>
<imgcaption imageNo04 | Task on field lines>
</imgcaption> <WRAP>.
{{drawio>TaskOnFieldLines}} \\
</WRAP>

Sketch the field line plot for the charge configurations given in <imgref FigNo04>. \\
Notice:
  * It is the __overlaid__ image that is sought.
  * Make sure that it is a source field.

</WRAP></WRAP></panel>

{{page>task_5.1.3_with_invoice&nofooter}}
{{page>aufgabe_5.1.4&nofooter}}
{{page>aufgabe_5.1.5&nofooter}}
===== 5.2 Electric charge and Coulomb force (reloaded) =====

<callout>

=== Goals ===

After this lesson, you should:

  - Be able to determine the direction of the forces using given charges.
  - be able to represent the acting force vectors in a sketch.
  - be able to determine a force vector by superimposing several force vectors using vector calculus.
  - be able to state the following quantities for a force vector:
      - Force vector in coordinate representation
      - magnitude of the force vector
      - Angle of the force vector

</callout>

The [[Fundamentals_and_basic_concepts#coulomb-force|electric charge and Coulomb force]] has already been described in chapter 1. However, some points are to be caught up here to it.

==== Direction of the Coulomb force and superposition ====

<WRAP right>
<imgcaption imageNo06 | direction of coulomb force>
</imgcaption> <WRAP>.
{{drawio>DirectionOfCoulombforce}} \\
</WRAP>

In the case of the force, the direction has been considered so far, e.g. direction towards the sample charge, but for future explanations it is important to include the cause-effect in the naming. In <imgref FigNo06> (a) and (b) the convention is shown again: A force $\vec{_{21}}$ acts on charge $Q_2$ and is caused by charge $Q_1$. As a mnemonic you can remember "tip-foot" (first the effect, then the cause).

Furthermore, several forces on a charge can be superimposed to a resulting force. Strictly speaking, it must hold that $\varepsilon$ is constant in the structure. For example, the resultant force in <imgref Fig.#06> Fig. (c) on $Q_3$ becomes equal to: $\vec{F_3}= \vec{F_{31}}+\vec{F_{32}}$.

~~PAGEBREAK~~ ~CLEARFIX~~
==== Geometric Distribution of Charges ====


In previous chapters only single charges (e.g. $Q_1$, $Q_2$) were considered.  
  * The charge $Q$ was previously reduced to a **point charge**.  <WRAP right> <imgcaption imageNo05 | Field lines of various extended charged objects></imgcaption> \ {{youtube>9fVEpwtrKhQ}} \</WRAP> This can be used, for example, for the elementary charge or for extended charged objects from a large distance. The distance is sufficiently large if the ratio between the largest object extent and the distance to the measurement point $P$ is small.
  * If the charges are lined up along a line, this is called a **line charge**. \ Examples of this are a straight trace on a circuit board or a piece of wire. Furthermore, this also applies to an extended charged object, which has exactly an extension that is no longer small in relation to the distance. For this purpose, the charge $Q$ is considered to be distributed over the line. Thus, a (line) charge density $\rho_l$ can be determined: <WRAP centeralign>$\rho_l = {{Q}\over{l}}$</WRAP> or, in the case of different charge densities on subsections: <WRAP centeralign>$\rho_l = {{\Delta Q}\over{\Delta l}} \rightarrow \rho_l(l)={{d}\over{dl}} Q(l)$</WRAP>
  * A **area charge** is spoken of when the charge is to be regarded as distributed over an area. \ Examples of this are the floor or a plate of a capacitor. Again, an extended charged object can be considered if there are two extensions which are no longer small in relation to the distance (e.g. surface of the earth). Again, a (surface) charge density $\rho_A$ can be determined: <WRAP centeralign>$\rho_A = {{Q}\over{A}}$</WRAP> or if there are different charge densities on partial surfaces: <WRAP centeralign>$\rho_A = {{\Delta Q}\over{\Delta A}} \rightarrow \rho_A(A) ={{d}\over{dA}} Q(A)={{d}\over{dx}}{{d}\over{dy}} Q(A)$</WRAP>
  * Finally, a **space charge** is the term for charges that span a volume. \here examples are plasmas or charges in extended objects (e.g. in semiconductor). as with the other charge distributions, a (space) charge density $\rho_V$ can be calculated here: <WRAP centeralign>$\rho_V = {{Q}\over{V}}$</WRAP> or for different charge density in partial volumes: <WRAP centeralign>$\rho_V = {{\Delta Q}\over{\Delta V}} \rightarrow \rho_V(V) ={{d}\over{dV}} Q(V)={{d}\over{dx}}{{d}\over{dy}}{{d}\over{dz}} Q(V)$</WRAP>

==== Types of fields depending on the charge distribution ====

There are two different types of fields:

<WRAP group><WRAP column half>
In **homogeneous fields**, magnitude and direction are constant throughout the field range. This field form is idealized to exist within plate capacitors. e.g., in the plate capacitor (<imgref Fig.#07>), or in the vicinity of widely extended bodies.
</WRAP><WRAP column half>For **inhomogeneous fields**, the magnitude and/or direction of the field strength changes from place to place. This is the rule in real systems, even the field of a point charge is inhomogeneous (<imgref FigNo08>).  
</WRAP> </WRAP>
<WRAP group><WRAP column half>

<WRAP>
<imgcaption imageNo07 | Field lines of a homogeneous field>
</imgcaption> <WRAP>.
{{drawio>FieldLinesOfAHomogeneousField}} \\
</WRAP>

</WRAP><WRAP column half>

<WRAP>
<imgcaption imageNo08 | Field lines of an inhomogeneous field>
</imgcaption> <WRAP>.
{{drawio>FieldLinesOfAnInhomogeneousField}} \\
</WRAP>

</WRAP> </WRAP>

==== Tasks ====

{{page>task_5.2.1_with_invoice&nofooter}}
{{page>aufgabe_5.2.2&nofooter}}
{{page>aufgabe_5.2.3&nofooter}}
=====5.3 Work and Potential =====

<callout>

=== Goals ===

After this lesson, you should:

  - know how work is defined in the electrostatic field.
  - be able to describe when work occurs and when it does not when a charge is moving.
  - know the definition of electric voltage and be able to calculate it in an electric field.
  - understand why the calculation of voltage is independent of displacement.
  - know what a potential difference is and recognise or be able to state equipotential surfaces (lines).
  - be able to determine a potential curve for a given arrangement.

</callout>

<callout type="danger" icon="true">

In the following, only a few brief illustrations of the concepts are given. \\
A detailed explanation can be found in the KIT bridge course. It is recommended to work through this independently. \\ \\
In particular, this applies to:
  * Chapter "[[https://lx3.mint-kolleg.kit.edu/onlinekursphysik/html/1.4.1/xcontent1.html|4.1.2 electric field]]" **from video 221** to the end of the tasks.
  * Chapter "[[https://lx3.mint-kolleg.kit.edu/onlinekursphysik/html/1.4.1/xcontent2.html|4.1.3 work, potential, voltage]]" to the end of the tasks and the Additional Content
</callout>

==== Energy required to displace a charge in the field ====

<WRAP right>
<imgcaption imageNo09 | Observation of work in a homogeneous electric field>
</imgcaption>
{{drawio>WorkInHomField}}
<WRAP>

First, the situation of a charge in a homogeneous electric field shall be considered. As we have seen so far, the magnitude of E is constant and the field lines are parallel. Now a positive charge $q$ is to be brought into this field.

If this charge would be free movable (e.g. electron in vacuum or in extended conductor) it would be accelerated along field lines. Thus its kinetic energy increases. Because the whole system of plates (for field generation) and charge however does not change its energetic state - thermodynamically the system is closed. From this follows: if the kinetic energy increases, the potential energy must decrease.

From mechanics is known, here done work (thus energy done by forces) is defined by force along a way. \\
In a homogeneous field, the following holds for a force action producing motion along a field line from $A$ to $B$ (see <imgref FigNo09>):
\begin{align*}
W_{AB} = F_C \cdot s
\end{align*}


For a motion parallel to a field line (i.e. from $A$ to $C$) $W_{AB}=0$ results. This situation is similar to the movement of a weight in the gravitational field at the same height. There, too, no energy is released or absorbed.
For any direction through the field the part of the path has to be considered, which is parallel to the field lines. This results from the angle $\alpha$ between $\vec{F}$ and $\vec{s}$:
\begin{align*}
W_{AB} = F_C \cdot s \cdot cos(\alpha) = \vec{F_C}\cdot \vec{s}
\end{align*}

The work $W_{AB}$ here describes the energy difference experienced by the charge $q$. \\
Similar to the electric field, we now look for a quantity that is independent of the (sample) charge $q$ in order to describe the energy component. This is done by the **potential** $\varphi$ (also potential). The potential in a homogeneous field is defined as:

\begin{align}}
\varphi_{AB} = {{W_{AB}}\over{q}} = {{F_C \cdot s}\over{q}} = {{E \cdot q \cdot s}\over{q}} = E \cdot s_{AB}
\end{align}}

<callout icon="fa fa-exclamation" color="red" title="Notice:">
The potential $\varphi$ in an $E$-field is the ability to do work $W$.
</callout>

To obtain a general approach to __in__homogeneous fields and arbitrary paths $s_{AB}$, it helps (as is so often the case) to decompose the problem into small parts. In the concrete case, these are small path segments on which the field can be assumed to be homogeneous. These are to be assumed to be infinitesimally small in the extreme case (i.e., from $s$ to $\delta s$ to $ds$):

\begin{align}
W_{AB} = \vec{F_C}\cdot \vec{s} \quad \rightarrow \quad \Delta W = \vec{F_C}\cdot \Delta \vec{s}\quad \rightarrow \quad dW = \vec{F_C}\cdot d \vec{s}
\end{align}

The total energy now results from the sum or integration of these path sections:

\begin{align*}
W_{AB} &= \int_{W_A}^{W_B} dW \
      &= \int_{A}^{B} \vec{F_C}\cdot d \vec{s} \\
      &= \int_{A}^{B} q \cdot \vec{E} \cdot d \vec{s}  \\
      &= q \cdot \int_{A}^{B} \vec{E} \cdot d \vec{s}
\end{align*}

The potential is therewith:

\begin{align}
\varphi_{AB} &= {{W_{AB}}\over{q}}
      &= \int_{A}^{B} \vec{E} \cdot d \vec{s}
\end{align*}

The potential difference $\varphi_{AB}$ is also called voltage $U_{AB}$. The voltage is measured in volts.

It is interesting that it does not matter which way the integration takes place. So it doesn't matter how the charge gets from $A$ to $B$, the energy or potential difference is always the same. This follows from the fact that a charge $q$ at a point $A$ in the field has a unique potential energy. No matter how this charge is moved to a point $B$ and back again: as soon as it gets back to the point $A$, it has the same energy again. So the potential difference of the way there and back must be equal in magnitude.

This concept has already been applied as a mesh theorem in circuits (see [[simple_dc-circuits#the_mesh_theorem_2_kirchhoff_equation|chapter 2]]). However, it is also valid in other structures and arbitrary electrostatic fields.

<callout>

The equation $\varphi_{AB} = \int_{A}^{B} \vec{E} \cdot d \vec{s}$ can be used and applied depending on the geometry present.
As an example, consider the situation of a charge moving from one electrode to another inside a capacitor:

\begin{align}
\varphi_{AB} &= \int_{A}^{B} \vec{E} \cdot d \vec{s} \quad && | \vec{E} \text{ and } d\vec{s} \text{ run parallel } \\
\varphi_{AB} &= \int_{A}^{B} E \cdot ds \quad && | \text{E=const.} \\
\varphi &= E \cdot \int_{0}^{d} ds \quad && | s \text{ starts counting at the negative disk. } d \text{ denotes the distance between the two plates }\
\varphi &= E \cdot d \quad && | \varphi_{AB} \text{ corresponds to the voltage applied to the capacitor } U \
U &= E \cdot d
\end{align*}

</callout>


<callout icon="fa fa-exclamation" color="red" title="Merke:">
Returning to the starting point from any point $A$ after a closed circuit, the __circuit stress__ along the closed path is 0.
A closed path is mathematically expressed as a ring integral:

\begin{align}
\varphi = \oint \vec{E} \cdot d \vec{s} = 0
\end{align}

Or spoken differently: In the electrostatic field there are no self-contained field lines.

</callout>

A field $\vec{X}$ which satisfies the condition $\oint \vec{X} \cdot d \vec{s}=0$ is called __vortex-free__ or __potential field__.

From the potential difference, or the voltage, the work in the electrostatic field results with:

\begin{align}}
\boxed{W_{AB}= q \cdot U_{AB}}
\end{align*}

==== Equipotential lines ====

<WRAP right>
<imgcaption imageNo99 | Indicating elevation contours on a map>
{{electrotechnics_1:elevationlinesonthelandmap.jpg?400}}
</imgcaption>
</WRAP>

If a charge $q$ becomes perpendicular to the field lines, it experiences neither energy gain nor loss. The voltage along this path is $0V$. All points between which the voltage of $0V$ is applied are at the same potential level. The connection of these points is called:
  * equipotential lines for a 2-dimensional representation of the field.
  * equipotential surfaces for a 3-dimensional field

This corresponds in the gravity field to a movement on the same contour line. The contour lines are often drawn in (hiking) maps, cf. <imgref BildNr99>. If one moves along the contour lines, no work is done.

The equipotential surfaces are usually drawn with a fixed step size, e.g. $1V$, $2V$, $3V$, ... . Since the electric field is higher near charges, equipotential surfaces are also closer together there.
In <imgref FigNo98>, the equipotential surfaces of a point charge are shown.
<WRAP right>
<imgcaption imageNo98 | Field lines and equipotential surfaces of a point charge>
</imgcaption> <WRAP>.
{{drawio>fieldlinesequipotentialsurfaces}} \\
</WRAP>

~PAGEBREAK~ ~CLEARFIX~~
==== Reference potential ====

=====5.4 Conductors in the electrostatic field =====

<callout>

=== Goals ===

After this lesson, you should:

  - know that no current flows in a conductor in an electrostatic field.
  - know how charges in a conductor are distributed in the electrostatic field.
  - Be able to sketch the field lines at the surface of the conductor.
  - Understand the effect of the influence of an external electric field.

</callout>

Up to now, charges were considered which were either rigid or freely movable. At the following, charges at an electric conductor are to look at. These are only free movable within the conductor. At first an ideal conductor without resistance is considered.

==== Stationary state of charges without external field ====

<WRAP right>
<imgcaption imageNo10 | Viewing a charged metal ball>
</imgcaption>
{{drawio>LoadedMetalBall}}
<WRAP>

In the first thought experiment, a conductor (e.g. a metal plate) is charged, see <imgref FigNo10>. The additional charges create an electric field. Thus, a resultant force acts on each charge. The cause of this force is the fields of the surrounding electric charges. So the charges repel and move apart. \\
The movement of the charge continues until a force equilibrium is reached. In this steady state, there is no longer a resultant force acting on the single charge. In <imgref FigNo10> this can be seen on the right: the repulsive forces of the charges are counteracted by the attractive forces of the atomic shells. \\
Results:
  * The charge carriers are distributed on the surface.
  * Due to the dispersion of the charges, the interior of the conductor is free of fields.
  * All field lines are perpendicular to the surface. Because: if they were not, there would be a tangential component of the field, i.e. along the surface. Thus a force would act on charge carriers and they would move accordingly.


==== Influenza ====

<WRAP right>
<imgcaption imageNo11 | Viewing the influential charge separation>
</imgcaption>
{{drawio>InfluenceAnMetalObject}}
<WRAP>

In the second thought experiment, an uncharged conductor (e.g. a metal plate) is brought into an electrostatic field (<imgref FigNo11>). The external field or the resulting Coulomb force causes the moving charge carriers to be displaced. \\
Results:
  * The charge carriers are still distributed on the surface.
  * Now an equilibrium is reached, when just so many charges have moved, that the field strength inside the conductor disappears (again).
  * The field lines leave the surface again at right angles. Again, a tangential component would cause a charge shift in the metal.

This effect of charge displacement in conductive objects by an electrostatic field is called **influence**. Influence charges can be separated (<imgref FigNo11> right). If we look at the separated influence charges without the external field, their field is again just as strong in magnitude as the external field only opposite.

<callout icon="fa fa-exclamation" color="red" title="Merke:">
  - The seat of an influenced charge is always the conductor surface. This results in a surface charge density $\varrho_A = {{\Delta Q}\over{\Delta A}}$
  - The conductor surface in the electrostatic field is always an equipotential surface. Thus, the field lines always originate and terminate perpendicularly on conductor surfaces.
  - The interior of the conductor is always field-free (Faraday effect: metallic enclosures shield electric fields).
</callout>

How can the conductor surface be an equipotential surface despite different charge on both sides? Equipotential surfaces are defined only by the fact that the movement of a charge along such a surface does not require/produce a change in energy. Since the interior of the conductor is field free, movement there can occur without a change in energy. As the potential between two points is independent of the path between them, a path along the surface is also possible without energy expenditure.

~~PAGEBREAK~~~CLEARFIX~~~

==== Tasks ====

Application of Influence: Protective bag against electrostatic charge / discharge (cf. [[https://www.youtube.com/watch?v=imdtXcnywb8&t=600s|Video]])

<panel type="info" title="Task 5.4.1 Simulation"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>

In the simulation program of [[https://www.falstad.com/vector2de/|Falstad]] the courses of equipotential surfaces and electric field strength at different objects can be represented.

  - Open the simulation program via the link
  - Select: "Setup: cylinder in field", "Floor: equipotentials" and "Display: Field Vectors".
  - The field of an infinitely long cylinder in a homogeneous electric field is now displayed in section. The solid lines show the equipotential surfaces. The small arrows show the electric field strength.
  - What can be said about the potential distribution on the cylinder?
  - On the left half the field lines enter the body, on the right half they leave the body. What can be said about the charge carrier distribution at the surface? Check also the representation "Color: charge"!
  - Is there an electric field inside the body? Check also the diagram "Floor: Field lines"!
  - Is this cylinder metallic, semiconducting or insulating?

</WRAP></WRAP></panel>

{{page>task_5.4.2_with_invoice&nofooter}}
{{page>aufgabe_5.4.3&nofooter}}
{{page>aufgabe_5.4.4&nofooter}}


=====5.5 The electric displacement flux and Gaussian theorem of electrostatics =====

<WRAP right>
The electric displacement (flux) density
{{youtube>UqzXWU6TsQY}}
</WRAP>

<callout>

=== Goals ===

After this lesson, you should:

  - know how to get the electric displacement flux from single charges
  - be able to state for a given area the displacement flux density of an arrangement
  - know the general meaning of Gauss' theorem of electrostatics
  - be able to choose a closed enveloping surface appropriately and apply Gauss' theorem

</callout>

<WRAP right>
<imgcaption imageNo12 | Influenced charge separation and displacement flow>
</imgcaption>
{{drawio>displacement flow}}
<WRAP>

Now we want to consider the situation at the two conductive plates with the area $\Delta A$ in the electrostatic field $\vec{E}$ in vacuum a little more exactly. For this purpose, the plates shall first be brought into the field separately. As written in <imgref FigNo12> on the left, the influence in a single plate is not considered. Rather, we are now interested in what happens when the plates are brought together. In this case, graphically speaking, just for each field line ending on the pair of plates, a single charge must move from one plate to the other. This ability to separate charges (i.e. to generate influence) is another ability of space.

In the previous arrangement (homogeneous field, all surfaces parallel to each other), the surface charge density $\varrho_A = {{\Delta Q}\over{\Delta A}}$ thus influenced is proportional to the external field $E$. It holds:

\begin{align*}
\varrho_A = {{\Delta Q}\over{\Delta A}} \sim E \\
\varrho_A = {{\Delta Q}\over{\Delta A}} = \varepsilon \cdot E
\end{align*}

The **displacement (flux) density** is now defined as.

\begin{align}
\boxed{\vec{D} = \varepsilon \cdot \vec{E}}
\end{align}

The displacement (flux) density has the unit "charge per area", i.e. $As/m^2$. The displacement (flux) density is also a field. It points in the same direction as the electrostatic field $\vec{E}$.

Why is now a second field introduced? This shall become clearer in the following, but first it shall be considered again how the electrostatic field $\vec{E}$ was defined. This resulted from the Coulomb force, i.e. the __action on a sample charge__. The displacement density, on the other hand, is not described by an action, but __caused by charges__.
The two are related by the above equation.
It will be shown in later sub-chapters that the different influences from the same cause of the field can produce different effects on other charges.

The **permittivity** (or dielectric conductivity) $\varepsilon$ thus results as a constant of proportionality between $D$-field and $E$-field. The inverse ${{1}\over{\varepsilon}}$ is a measure of how much effect ($E$-field) is available from the cause ($D$-field) at a point. In vacuum, $\varepsilon= \varepsilon_0$, the electric field constant.

~~PAGEBREAK~~ ~CLEARFIX~~
==== General relationship between charge Q and displacement density D ====

Up to now, only a homogeneous field and an observation surface perpendicular to the field lines were considered. Thus only equipotential surfaces (e.g. a metal foil) were considered.
In that case it was found that the charge is equal to the displacement density on the surface: $\Delta Q = D\cdot \Delta A$.

This formula is now to be extended to arbitrary surfaces and inhomogeneous fields.
As with the potential and other physical problems, the problem is to be broken down into smaller subproblems, solved and then summed up. For this purpose a small area element $\Delta A = \Delta x \cdot \Delta y$ is needed. In addition, the position of the area in space should be taken into account. This is possible if the cross product is chosen: $\Delta \vec{A} = \Delta \vec{x} \cdot \Delta \vec{y}$, since so is the surface normal. In what follows, the cross product will be relevant to the calculation, but the consequences of the cross product will be:
  * The magnitude of $\Delta \vec{A}$ is equal to the area $\Delta A$.
  * The direction of $\Delta \vec{A}$ is perpendicular to the area.
In addition, let $\Delta A$ now become infinitesimally small, that is, $dA = dx \cdot dy$.

=== 1. problem: inhomogeneity → solution: shrink area ===
First, we shall still assume an observation surface perpendicular to the field lines, but an inhomogeneous field. In the inhomogeneous field, the magnitude of $D$ is no longer constant. In order to correct this, $dA$ is chosen so small that just "only one field line" passes through the surface. In this case D is homogeneous again. Thus holds:

$Q = D\cdot A$

\begin{align*}
Q = D\cdot A \quad \rightarrow \quad dQ = D\cdot dA
\end{align*}

=== 2nd problem: arbitrary surface → solution: vectors ===
Now assume an arbitrary surface. Thus the $\vec{D}$-field no longer penetrates through the surface at right angles. But for the influence only the rectangular part was relevant. So only this part has to be considered. This results from consideration of the cosine of the angle between (right-angled) surface normal and $\vec{D}$-field:

\begin{align*}
dQ = D\cdot dA \quad \rightarrow \quad dQ = D\cdot dA \cdot cos(\alpha) = \vec{D} \cdot d \vec{A}
\end{align*}

=== 3. summing up ===
Since so far only infinitesimally small surface pieces were considered must now be integrated again to a total surface. If a closed enveloping surface around a body is chosen, the result is:

\begin{align}
\boxed{\int dQ = \iint_{\text{hull}} \vec{D} \cdot d \vec{A} = Q}
\end{align}}

The "sum" of the $D$-field emanating over the surface is thus just as large as the sum of the charges contained therein, since the charges are just the sources of this field. This can be compared vividly with a bordered swamp area with water sources and sinks:
  * The sources in the marsh correspond to the positive charges, the sinks to the negative charges. The formed water corresponds to the $D$-field.
  * The sum of all sources and sinks equals in this case just the water stepping over the edge.


==== Applications ====

Are calculated in the course.

=== Spherical capacitor ===

Spherical capacitors are now rarely found in practical applications. In the {{wpde>Van-de-Graaff generator}}, spherical capacitors are used to store the high DC voltages. The earth also represents a spherical capacitor. In this context, the electric field strength of $100...300 V/m$ in the atmosphere is remarkable, since several hundred volts would have to be present between head and foot (for resolution, see the article [[https://www.wissenschaft.de/umwelt-natur/spannung-liegt-in-der-luft/|Spannung lieg in der Luft]] in Bild der Wissenschaft).

=== Plate capacitor ===

The relation between the $E$-field and the voltage $U$ on the ideal plate capacitor is to be derived from the integral of the displacement density $\vec{D}$:
\begin{align*}
Q = \iint_{\text{shell}} \vec{D} \cdot d \vec{A}
\end{align}}

<callout icon="fa fa-info" color="grey" title="Outlook">

The consideration of the displacement-flux-density also solved a problem, which arose quite for at electric circuits: From considerations about magnetic fields the following quite obvious sounding fact can be led: In a series-connected, switched circuit, the current at each point is the same. But if this series circuit contains a capacitor, no electric current can flow inside! The solution is to understand a temporal change of the displacement flux also as a current, which can be generated a magnetic field (thus vortex). Mathematically, vortices are described via the {{wpde>rotation of a vector field}} - a multidimensional differential operator. A deeper {{wpde>displacement_stream#derivation_of_a_contradiction|derivation and solution}} is not considered in the first semester. However, the application will show that the above equation plays a central role in electrical engineering. It is part of the so-called {{wpde>Maxwell equations}}.

</callout>




~~PAGEBREAK~~~CLEARFIX~~~

=====5.6 Non-conductor in electrostatic field =====

<callout>

=== Goals ===

After this lesson, you should:

  - know the two field-describing quantities of the electrostatic field
  - be able to describe and apply the relationship between these two quantities via the material law
  - understand the effect of an electrostatic field on an insulator
  - know what the effect of dielectric polarisation does
  - be able to relate the term dielectric strength to a property of insulators and know what it means

</callout>

<WRAP right>
<imgcaption imageNo13 | Thought experiment: changing the material between the plates>
</imgcaption>
{{drawio>ExperimentDielectric1}}
</WRAP>
==== material-law-of-electrostatics ====

First of all, a thought experiment is to be carried out again (see <imgref Fig.No.13>):
  - First a charged plate capacitor in vacuum is assumed, which is separated from the voltage source after charging.
  - Next, the intermediate region is to be filled with a material.

Think about how $E$ and $D$ would change before you unfold the subsection.\\
Why might which of the two quantities change?

~~PAGEBREAK~~~CLEARFIX~~~

--> The material law of electrostatics #

<WRAP right>
<imgcaption imageNo14 | Thought experiment: changing the material between the plates>
</imgcaption>
{{drawio>ExperimentDielectric2}}
</WRAP>

You may have considered what happens to the charge $Q$ on the plates. This charge cannot leave the plates. So $Q = \iint_{\text{envelope}} \vec{D} \cdot d \vec{A}$ cannot change. \\
Since the sheath as a fictitious surface around an electrode does not change either, $\vec{D}$ cannot change either.

On the other hand, polarizable materials in the capacitor can align themselves. This dampens the effective field. Maybe you remember what the "acting field" was: the $E$-field. So the $E$-field becomes smaller (see <imgref pictureNo14>).

Previously:

\begin{align*}
D = \varepsilon_0 \cdot E
\end{align*}

The determined change is packed into the material constant $\varepsilon_r$. This gives the **material law of electrostatics**:

\begin{align*}
\boxed{D = \varepsilon_r \cdot \varepsilon_0 \cdot E}
\end{align*}

Since the charge $Q$ cannot vanish from the capacitor in this experimental setup and thus $D$ remains constant, the $E$ field must become smaller for $\varepsilon_r>1$.

<imgref FigNo14> is drawn here in a simplified way: the alignable molecules are evenly distributed over the material and are thus also evenly aligned. Accordingly, the E-field is uniformly attenuated.

~~PAGEBREAK~~~CLEARFIX~~~
<callout icon="fa fa-exclamation" color="red" title="Remember:">
  - The material constant $\varepsilon_r$ is called relative permittivity, relative permittivity, or dielectric constant.
  - Relative permittivity is unitless and indicates how much the electric field strength decreases with the presence of material for the same charge.
  - The relative permittivity $\varepsilon_r$ is always greater than or equal to 1 for dielectrics (i.e., nonconductors).
  - The relative permittivity depends on the polarizability of the material, i.e. the possibility to align the molecules in the field. Correspondingly, relative permittivity depends on frequency and often direction and temperature.
</callout>

<callout icon="fa fa-info" color="grey" title="Outlook">

If now the relative permittivity $\varepsilon_r$ depends on the possibility to align the molecules in the field, the following interesting relation arises: if frequencies are "caught", at which the oscillation of the molecule can build up, the energy of the external field is absorbed by the molecule. This build-up is similar to the shattering of a wine glass at a suitable irradiated frequency and is called resonance. Materials can be analysed on the basis of the resonance frequencies. These resonance frequencies are enormously high (1 GHz to 1'000'000 GHz) and in these frequencies the $E$-field detaches from the conductor. This may sound strange, but it becomes a bit more illustrative in the 2nd semester with the resonant circuit. For the 1st semester it is more than sufficient that in the range of 1'000'000 GHz is the visual light, which is obviously not bound to a conductor. But this also makes clear that the relative permittivity $\varepsilon_r$ for high frequencies also has to do with the absorption (and reflection) of electromagnetic waves.

</callout>

<WRAP right 30em>

<tabcaption tab01| relative permittivity>
^ material^ relative permittivity \ $\varepsilon_r$ for low frequencies ^
| air | 1.0006 |
| paper | 2|
| hard paper | 5 |
| glass | 6...8 |
| PE, PP | 2,3 |
| PS | 2.5 |
| water (20°C) | 80 |
</tabcaption>
</WRAP>

Some values of the relative permittivity $\varepsilon_r$ for dielectrics are given in <tabref tab01>.

~~PAGEBREAK~~~CLEARFIX~~~

<--

==== Dielectric strength of dielectrics ====

<WRAP right 30em>

<tabcaption tab02| Dielectric strength>
^ Material^ Dielectric strength $E_0$ in kV/mm ^
| air | 0.1...0.3 |
| SF6 gas| 8 |
| vacuum| 20...30|
| insulating oils| 5...30|
| quartz| 30...40 |
| PP,PE | 50 |
| PS | 100 |
| distilled water | 70 |
</tabcaption>
</WRAP>

  * The dielectrics act as insulators. The flow of current is therefore prevented
  * The ability to insulate is dependent on the material.
  * If a maximum field strength $E_0$ is exceeded, the insulating ability is eliminated
    * One says: The insulator breaks down. This means that above this field strength a current can flow through the insulator.
    * Examples are: Lightning in a thunderstorm, ignition spark, glow lamp in a {{wpde>https://de.wikipedia.org/wiki/Spannungspr%C3%BCfer#Verl%C3%A4sslichkeit_und_Zul%C3%A4ssigkeit|phase tester}}
    * The maximum field strength $E_0$ is called ** dielectric strength**.
    * $E_0$ depends on the material (see <tabref tab02>), but also on other factors (temperature, humidity, ...).

~~PAGEBREAK~~ ~~CLEARFIX~~~
==== tasks====

<panel type="info" title="Task 5.6.1 Thought Experiment"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>

Consider what would have happened if the plates had not been detached from the voltage source in the above thought experiment (<imgref FigNo13>).

</WRAP></WRAP></panel>

=====5.7 Capacitors =====

<callout>

=== Goals ===

After this lesson, you should:

  - know what a capacitor is and how capacitance is defined
  - know the basic equations for calculating a capacitor and be able to apply them
  - be able to imagine a plate capacitor and know examples of its use You also have an idea of what a cylindrical or spherical capacitor looks like and what examples of its use there are
  - know the characteristics of the E-field, D-field and electric potential in the three types of capacitors presented here

</callout>


==== Capacity ====

  * A capacitor is defined by the fact that there are two electrodes (= conductive areas), which are separated by a dielectric (= non-conductor).
  * This makes it possible to build up an electric field in the capacitor without charge carriers moving through the dielectric.
  * The characteristic of the capacitor is the capacity $C$.
  * In addition to the capacitance, every capacitor also has a resistance and an inductance. However, both of these are usually very small.
  * Examples are
    * the electrical component "capacitor",
    * an open switch,
    * a wire to ground,
    * a human being
$\rightarrow$ Thus, for any arrangement of two conductors separated by an insulating material, a capacitance can be specified.

The capacity $C$ can be derived as follows:
  - It is known that $U = \int \vec{E} d \vec{s} = E \cdot l$ and hence $E= {{U}\over{l}}$ or $D= \varepsilon_0 \cdot \varepsilon_r \cdot {{U}\over{l}}$.
  - Furthermore, $\iint_{\text{envelope}} \vec{D} \cdot d \vec{A} = Q$ by the idealized form of the plate capacitor: $Q=D \cdot A$.
  - Thus, the charge $Q$ is given by: \begin{align*} Q = \varepsilon_0 \cdot \varepsilon_r \cdot {{U}\over{l}} \cdot A \end{align*}
  - This means that $Q \sim U$, given the geometry (i.e., $A$ and $d$) and the dielectric ($\varepsilon_r $).
  - So it is reasonable to determine a proportionality factor ${{Q}\over{U}}$.

The capacitance $C$ of an idealized plate capacitor is defined as

\begin{align}
\boxed{C = \varepsilon_0 \cdot \varepsilon_r \cdot {{A}\over{l}} = {{Q}\over{U}}
\end{align*}

This relationship can be examined in more detail in the following simulation:

-->capacitor lab#

If the simulation is not displayed optimally, [[https://phet.colorado.edu/sims/cheerpj/capacitor-lab/latest/capacitor-lab.html?simulation=capacitor-lab&locale=de|this link]] can be used.

{{url>https://phet.colorado.edu/sims/cheerpj/capacitor-lab/latest/capacitor-lab.html?simulation=capacitor-lab&locale=de 900,800 noborder}}

<--


==== Designs and types of capacitors ====

<WRAP right>
<imgcaption imageNo17 | geometry of capacitors>
</imgcaption>
{{drawio>GeometryCapacitors}}
</WRAP>

To calculate the capacitances of different designs, the definition equations of $\vec{D}$ and $\vec{E}$ are used. This can be viewed in detail e.g. in [[https://www.youtube.com/watch?v=6iQAgtikt5o|this video]]. \\ Based on the geometry, different equations result (see also <imgref FigNo17>).

^^Shape of the capacitor^Parameter^Equation for the capacitance^
|plate capacitor| area $A$ of plate \distance $l$ between plates | \begin{align*}C = \varepsilon_0 \cdot \varepsilon_r \cdot {{A}\over{l}} \end{align*}|
|cylinder capacitor |radius of outer conductor $R_a$ \radius of inner conductor $R_i$ \length $l$| \begin{align*}C = \varepsilon_0 \cdot \varepsilon_r \cdot 2\pi {{l}\over{ln \left({{R_a}\over{R_i}}\right)}} \end{align*}|
|spherical capacitor |radius of outer spherical conductor $R_a$ \radius of inner spherical conductor $R_i$| \begin{align*}C = \varepsilon_0 \cdot \varepsilon_r \cdot 4 \pi {{R_i \cdot R_a}\over{R_a - R_i}} \end{align*}|

~~PAGEBREAK~~~CLEARFIX~~~

<WRAP right>
<imgcaption imageNo16 | Structural shapes of capacitors>
</imgcaption>
{{drawio>DesignsCapacitors}}
</WRAP>

In <imgref imageNo16> different designs of capacitors can be seen:
  - **{wpde>variable_capacitor|rotary_capacitor}}** (also variable capacitor or trim capacitor).
    - A variable capacitor consists of two sets of plates: a fixed set and a movable set (stator and rotor). These represent the two electrodes.
    - The movable set can be rotated radially into the fixed set. This covers a certain area $A$.
    - The size of the area is increased by the number of plates. Nevertheless, only small capacities are possible because of the necessary distance.
    - Air is usually used as the dielectric, occasionally small plastic or ceramic plates are used to increase the dielectric constant.
  - **{wpde>Ceramic capacitor#Multilayer ceramic chip capacitors_(MLCC)|multilayer capacitor}}**
    - In the multilayer capacitor, there are again two electrodes. Here, too, the area $A$ (and thus the capacitance $C$) is multiplied by the finger-shaped interlocking.
    - Ceramic is used here as the dielectric.
    - The multilayer ceramic capacitor is also called KerKo or MLCC.
    - The variant shown in (2) is an SMD variant (surface mound device).
  - Disk capacitor
    - A ceramic is also used as dielectric for the disk capacitor. This is positioned as a round disc between two electrodes.
    - Disc capacitors are designed for higher voltages, but have a low capacitance (in the microfarad range).
  - **{wpde>Electrolytic capacitor}}** (also called electrolytic capacitor)
    - In electrolytic capacitors, the dielectric is an oxide layer formed on the metallic electrode. the second electrode is the liquid or solid electrolyte.
    - Different metals can be used as the oxidized electrode, e.g. aluminium, tantalum or niobium.
    - Because the oxide layer is very thin, a very high capacity results (depending on the size: up to a few millifarads).
    - Important for the application is that it is a polarized capacitor. I.e. it may only be operated in one direction with DC voltage. Otherwise, a current can flow through the capacitor, which destroys it and is usually accompanied by an explosive expansion of the electrolyte. To avoid reverse polarity, the negative pole is marked with a dash.
    - The electrolytic capacitor is built up wound and often has a cross-shaped predetermined breaking point at the top for gas leakage.
  - **{wpde>Plastic film capacitor}}** (also called Folko).
    - A material similar to a "chip bag" is used as an insulator: a plastic film with a thin, metallized layer.
    - The construction shows a high pulse load capacity and low internal ohmic losses.
    - In the event of electrical breakdown, the foil enables "self-healing": the metal coating evaporates locally around the breakdown. Thus the short-circuit is cancelled again
    - With some manufacturers this type is called MKS (__M__metallized foil__c__capacitor, Polye__s__ter).
  - **{wpde>Supercapacitor}}** (engl. Super-Caps)
    - As a dielectric is - similar to the electrolytic capacitor - very thin. In the actual sense, there is no dielectric at all.
    - The charges are not only stored in the electrode, but - similar to a battery - the charges are transferred into the electrolyte. Due to the polarization of the charges, they surround themselves with a thin (atomic) electrolyte layer. The charges then accumulate at the other electrode.
    - Supercapacitors can achieve very large capacitance values (up to the kilofarad range), but only have a low maximum voltage

~~PAGEBREAK~~~CLEARFIX~~~

<WRAP right 30em>
<imgcaption imageNo17 | types of capacitors>
{{electrotechnology_1:capacitorsmd.jpg}}{{electrotechnology_1:capacitorsht.jpg}}
</imgcaption>
</WRAP>

In <imgref pictureNo17> are shown different capacitors:
  - Above two SMD capacitors
    - On the left a $100\mu F$ electrolytic capacitor
    - On the right a $100nF$ MLCC in the commonly used {{wpde>chip style}} 0603 (1.6mm x 0.8mm)
  - below different THT capacitors (__T__hrough __H__ole __T__echnology)
    - a big electrolytic capacitor with $10mF$ in blue, the positive terminal is marked with $+$
    - in the second row is a Kerko with $33pF$ and two Folkos with $1,5\mu F$ each
    - in the bottom row you can see a trim capacitor with about $30pF$ and a tantalum electrolytic capacitor and another electrolytic capacitor
Various conventions]] have been established for designating the capacitance value of a capacitor [[https://www.elektronik-kompendium.de/sites/bau/1109061.htm|various conventions]].

\\ \\
<WRAP centeralign>
Electrolytic capacitors can explode!

{{youtube>IMpj1R3we_U}}
</WRAP>

<callout icon="fa fa-exclamation" color="red" title="Notice:">
  - There are polarized capacitors. With these, the installation direction and current flow must be observed, as otherwise an explosion can occur.
  - Depending on the application - and the required size, dielectric strength and capacitance - different types of capacitors are used.
  - The calculation of the capacitance is usually __not__ via $C = \varepsilon_0 \cdot \varepsilon_r \cdot {{A}\over{l}} $ . The capacitance value is given.
  - The capacitance value often varies by more than $\pm 10\%$. I.e. a calculation accurate to several decimal places is rarely necessary/possible.

  - The charge current seems to be able to flow through the capacitor because the charges added to one side induce correspondingly opposite charges on the other side.

</callout>

~~PAGEBREAK~~~CLEARFIX~~~


=====5.8 Interconnection of capacities =====

<callout>

=== Goals ===

After this lesson, you should:

  - be able to recognise a series connection of capacitors and distinguish it from a parallel connection
  - be able to calculate the resulting total capacitance of a series or parallel circuit
  - know how the total charge is distributed among the individual capacitors in a parallel circuit
  - be able to determine the voltage across a single capacitor in a series circuit

</callout>

====Capacitor series connection====

<WRAP right>{{url>https://www.falstad.com/circuit/circuitjs.html?running=false&ctz=CQAgjCAMB0lwrFWAmZBmMbJgGzYCwAc+yAnIciGmiIvmgFADGIyA7JWGzqx+KZUhJs4aMniRk+UjNky2DMIQhEhYfPhCqqaHkKww4R48Z5ZwQwyKtgGAZV6cNjkKT1QQAMwCGAGwDOAKYeyAwA7lTErjzablDhWoRCcdrqmpAJ2tRmOIQ6egniPHFcxQUAbuDc+VVlHkL4QoiWSPAMAE61NeyU2fXQbQDmXX2lVLnxEWhRcciEeXEZU-C9ui59Sy5pW9W2ET2skHkHUmrMLqcX8O6GEGBiEuImxoz7fCd8WJq2LAdf68gWlZEMhoNJSBJnnA5oU+Ns0CtwM5NicjlREYC8hkAA7opyaBGUTHgeoMXGEw55CkHCCWMl49aI0ak5arHJUtabanvHmCWFE64XPgouECC5okW9QHrTkMAAerDAiDG6E4bFIWlYmgAwvLWGhOMotYbNN88rYFehNJcpNb6JqlKw9cguPqQWhrdNNVIQLrLWANWQ2PqzHBvTrnWssET8KG6OAqXr1MHAZx4CnIPHHQBHJPmAPB5OHBpakAAVT1ujTwbQ6vAeHD5crddwvUIGtwWfNzcD4kivbyZqdCqrrGltd7NDtTZHOCJhBrdbIJZ9FZHS7F00DEIdif93zm+rUGEbdmdrt0PCt4w104ASgwgA 600,600 noborder}}
</WRAP>

If capacitors are connected in series, the charging current $I$ into the individual capacitors $C_1 ... C_n$ is equal.
Thus, the charges absorbed $\delta Q$ are also equal:
\begin{align*}
\Delta Q = \Delta Q_1 = \Delta Q_2 = ... = \Delta Q_n
\end{align*}

Furthermore, after charging, a voltage is formed across the series circuit which corresponds to the source voltage $U_q$. This results from the addition of the partial voltages across the individual capacitors.
\begin{align*}
U_q = U_1 + U_2 + ... + U_n = \sum_{k=1}^n U_k
\end{align*}

It holds for the voltage $U_k = \Large{{Q_k}\over{C_k}}$. \\
If all capacitors are initially discharged, then $U_k = \Large{{\Delta Q}\over{C_k}}$ holds.
Thus
\begin{align*}
U_q &= &U_1 &+ &U_2 &+ &... &+ &U_n &= \sum_{k=1}^n U_k \.
U_q &= &{\Delta Q}\over{C_1}} &+ &{{{Delta Q}\over{C_2}} &+ &... &+ &{\Delta Q}\over{C_3}} &= \sum_{k=1}^n {{1}\over{C_k}}\cdot \Delta Q \\
{{1}\over{C_{ges}}\cdot \Delta Q &= &&&&&&&&\sum_{k=1}^n {{1}\over{C_k}}\cdot \Delta Q
\end{align*}

Thus, for the series connection of capacitors $C_1 ... C_n$ :

\begin{align*}
\boxed{ {{1}\over{C_{ges}} = \sum_{k=1}^n {{1}\over{C_k}} }
\end{align*}
\begin{align}}
\boxed{ \Delta Q_k = const.}
\end{align*}

For initially uncharged capacitors, (voltage divider for capacitors) holds:
\begin{align*}
\boxed{Q = Q_k}
\end{align*}
\begin{align*}
\boxed{U_{ges} \cdot C_{ges} = U_{k} \cdot C_{k} }
\end{align}

In the simulation on the right, besides the parallel connected capacitors $C_1$, $C_2$,$C_3$, an ideal voltage source $U_q$, a resistor $R$, a switch $S$ and a lamp are installed.
  * The switch $S$ allows the voltage source to charge the capacitors.
  * The resistor $R$ is necessary because the simulation cannot represent instantaneous charging. The resistor limits the charging current to a maximum value. \\ Further details about the resistor are described in the chapter [[Switching operations on RC combinations]].
  * The capacitors can be discharged again via the lamp.

This derivation is also well explained, for example, in [[https://www.youtube.com/watch?v=9-Bp9Cvr7Jg|this video]].

~~PAGEBREAK~~ ~CLEARFIX~~
====parallel-circuit-capacitors====

<WRAP right>{{url>https://www.falstad.com/circuit/circuitjs.html?running=false&ctz=CQAgjCAMB0l3EaQMxjAFhQJi8gnBgKx4DsJ45hI6yAUAMYhaQAcFAbE6+HllFLEh900PGPETJtMCwjJ2bDOhDy2WQp0jhcI5OnUtkhdGCwsWy5HHAD4kSQ8m34j13loBlLovTLmbPE1+ADMAQwAbAGcAU34sWgB3FWYQQJUFVM1E9IDOVXBfKGz89U4zNQ0ipPRckABaGsyigDd6xtK22q0tTBAqboFCWgBzTqZKhrVzKuStNJxlNMhs-3HOUwqspNWlbw5wbI2C5SOwEnWGbUVzq54+bsEIETcHbMbd95uwS8-Od95+Eg4FRni8JCtuLtyscZqsOtCOssAA5MaZQ6YdRAzdqVI6Iw5mO63VbLbaQgFw7ik6iKAHtKlvWpHJYEpkAllknwnQm7amnQrvQrLABOe0pm0Bj1oorxkMJiOc8VF9LYOKCQPQ2UmaxpOuWAA96ixyGBCOQ6iQtGB2D0mMoAKq0Q11dh4cDsCB1Yju5QnNgARydqIoyBU1z+4EUQbAVtdTBNJGEdpAAGEg3hQ9a3WgrYnqMm04a0BAzpmsCWbn6mNGSJm42A8Lmk1hlIWeBXQ8wKxGZCpaEA 700,500 noborder}}
</WRAP>

If capacitors are connected in parallel, the voltage $U$ across the individual capacitors $C_1 ... C_n$ is equal.
It is therefore valid:

\begin{align*}
U_q = U_1 = U_2 = ... = U_n
\end{align*}

Furthermore, during charging, the total charge $\Delta Q$ from the source is distributed to the individual capacitors.
This gives the following for the individual charges absorbed:
\begin{align*}
\Delta Q = \Delta Q_1 + \Delta Q_2 + ... + \Delta Q_n = \sum_{k=1}^n \Delta Q_k
\end{align*}

If all capacitors are initially discharged, then $Q_k = \delta Q_k = C_k \cdot U$ \.
Thus
\begin{align*}
\Delta Q &= & Q_1 &+ & Q_2 &+ &... &+ & Q_n &= \sum_{k=1}^n Q_k \.
\Delta Q &= &C_1 \cdot U &+ &C_2 \cdot U &+ &... &+ &C_n \cdot U &= \sum_{k=1}^n C_k \cdot U \
C_{ges} \cdot U &= &&&&&&&& \sum_{k=1}^n C_k \cdot U \
\end{align*}

Thus, for the parallel connection of capacitors $C_1 ... C_n$ :
<WRAP>
\begin{align*}
\boxed{ C_{ges} = \sum_{k=1}^n C_k }
\end{align*}
\begin{align*}
\boxed{ U_k = const}
\end{align*}
</WRAP>

For initially uncharged capacitors, (charge divider for capacitors) holds:
\begin{align*}
\boxed{\delta Q = \sum_{k=1}^n Q_k}
\end{align*}
\begin{align}
\boxed{ {{Q_k}\over{C_k}} = {{\Delta Q}\over{C_{ges}} }
\end{align}}

In the simulation on the right, again besides the parallel connected capacitors $C_1$, $C_2$,$C_3$, an ideal voltage source $U_q$, a resistor $R$, a switch $S$ and a lamp are installed.

This derivation is also well explained, for example, in [[https://www.youtube.com/watch?v=fH-9pUeEpZU|this video]].

~~PAGEBREAK~~~CLEARFIX~~~

====Tasks====

<panel type="info" title="Task 5.8.1 Calculating a circuit of different capacitors"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>

See https://www.youtube.com/watch?v=vSeSHAmpd4Y
</WRAP></WRAP></panel>

=====5.9 Interfaces of dielectrics =====

<callout>

=== Goals ===

After this lesson, you should:

  - be able to recognise a stratification of dielectrics and distinguish between a transverse stratification and a longitudinal stratification
  - know which quantity remains constant in the case of transverse layering
  - know the constant quantity for a longitudinal layering as well
  - be familiar with the equivalent circuits for transverse and longitudinal layering
  - be able to calculate the total capacitance of a capacitor with stratification
  - know the law of refraction at interfaces for the field lines in the electrostatic field.

</callout>


<WRAP right>
<imgcaption imageNo18 | Types of stratification in capacitors>
</imgcaption>
{{drawio>LayeringCapacitors}}
</WRAP>

Up to now was assumed only one dielectricum resp. only vacuum within capacitor. Now is looked at more detailed, how multi-layered construction between sheets affects capacity. Thereby several dielectrics build boundary layers between each other. Different variants can be distinguished (<imgref FigNo18>):
  - **transverse layering**: There are different dielectrics perpendicular to the field lines. \\ Thus, the boundary layers are parallel to the capacitor plates.
  - **Longitudinal layering**: There are different dielectrics parallel to the field lines. \ So the boundary layers are perpendicular to the capacitor plates.
  - **Elongitudinal Stratification**: The boundary layers are neither parallel nor perpendicular to the capacitor plates.

~~PAGEBREAK~~~CLEARFIX~~~

==== Cross-layering ====

<WRAP right 40em>
<imgcaption imageNo19 | cross-layered capacitor>
</imgcaption>
{{drawio>cross-layeredcapacitor}}
</WRAP>

First, the situation is considered that the boundary layers are parallel to the electrode surfaces. A voltage $U$ is applied to the structure from the outside. \\
The stratification is now parallel to equipotential surfaces. In particular, the boundary layers are then also equipotential surfaces. \\
The boundary layers can be replaced by an infinitesimally thin conductor layer (metal foil). The voltage $U$ can then be divided into several partial areas:

\begin{align*}
U = \int \limits_{total inside \range} \! \! \vec{E} \cdot d \vec{s} = E_1 \cdot d_1 + E_2 \cdot d_2 + E_3 \cdot d_3
\tag{5.9.1}
\end{align*}

Since there are only polarized charges in the dielectrics and no free charges, the $\vec{D}$ field is constant between the electrodes.

\begin{align*}
Q = \iint_{A} \vec{D} \cdot d \vec{A} = const.
\end{align}

Now, in the setup, the area $A$ of the boundary layers is also constant. Thus:

\begin{align}
\vec{D_1} \cdot \vec{A} & = & \vec{D_2} \cdot \vec{A} & = & \vec{D_3} \cdot \vec{A} & \quad \quad & | \vec{D_k} & \parallel \vec{A} \\
     D_1 \cdot A & = & D_2 \cdot A & = & D_3 \cdot A & \quad \quad \quad & | \:\: A & = const. \\
     D_1 & = & D_2 & = & D_3 & \quad \quad \quad & | D_k & = \varepsilon_{rk} \varepsilon_0 \cdot E_k \
     \varepsilon_{r1} \varepsilon_0 \cdot E_1 &= &\varepsilon_{r2} \varepsilon_0 \cdot E_2 &= &\varepsilon_{r3} \varepsilon_0 \cdot E_3 \\
\end{align}
\begin{align}
\boxed{ \varepsilon_{r1}  \cdot E_1 = \varepsilon_{r2}  \cdot E_2 = \varepsilon_{r3}  \cdot E_3 }
\tag{5.9.2}
\end{align*}

Using $(5.9.1)$ and $(5.9.2)$, we can also derive the following relationship:
\begin{align*}
E_2 = & {{\varepsilon_{r1}}\over{\varepsilon_{r2}}\cdot E_1 , \quad E_3 = {{\varepsilon_{r1}}\over{\varepsilon_{r3}}\cdot E_1 \.
\end{align*}
\begin{align*}
U = & E_1 \cdot d_1 + & E_2 & \cdot d_2 + & E_3 & \cdot d_3 \.
U = & E_1 \cdot d_1 + & {{\varepsilon_{r1}}\over{\varepsilon_{r2}}\cdot E_1 & \cdot d_2 + & {{\varepsilon_{r1}}\over{\varepsilon_{r3}}\cdot E_1 & \cdot d_3 \
\end{align*}
\begin{align*}
U = & E_1 \cdot (d_1 + {{\varepsilon_{r1}}\over{\varepsilon_{r2}} \cdot d_2 + {{\varepsilon_{r1}}\over{\varepsilon_{r3}}\cdot d_3 ) \\
E_1 = & {{U}\over{ d_1 + \large{{\varepsilon_{r1}}\over{\varepsilon_{r2}} \cdot d_2 + \large{{\varepsilon_{r1}}\over{\varepsilon_{r3}}\cdot d_3 }}
\end{align*}
\begin{align}}
\boxed{ E_1 = {{U}\over{ \sum_{k=1}^n \large{{\varepsilon_{r1}}\over{\varepsilon_{rk}} \cdot d_k}} } \quad \text{and} \; E_k = {{\varepsilon_{r1}}\over{\varepsilon_{rk}}\cdot E_1
\end{align}}


The situation can also be transferred to a coaxial structure of a cylindrical capacitor or concentric structure of spherical capacitors.

<callout icon="fa fa-exclamation" color="red" title="Merke:">
Cross-stratification results in:
  - A cross-stratification can be considered as a series connection of partial capacitors with respective thicknesses $d_k$ and dielectric constant $\varepsilon_{rk}$.
  - The flux density is constant in the capacitor
  - Considering the fields __along the field line__ - that is, perpendicular to the interface, or the normal components $E_n$ and $D_n$ of the fields - the following holds:
    - The normal component of the electric field strength $E_n$ changes abruptly at the interface.
    - The normal component of the flux density $D_n$ is continuous at the interface: $D_{n1} = D_{n2}$
</callout>

~~PAGEBREAK~~~CLEARFIX~~~

==== Longitudinal layering ====

<WRAP right 40em>
<imgcaption imageNo20 | longitudinal layered capacitor>
</imgcaption>
{{drawio>longitudinal stratified capacitor}}
</WRAP>

Now the boundary layers should be perpendicular to the electrode surfaces. Again a voltage $U$ is applied to the structure from the outside. \\
The stratification is now perpendicular to equipotential surfaces. However, the same voltage is applied to each dielectric. Thus it is valid:

\begin{align*}
U = \int \limits_{total inside \range} \! \! \vec{E} \cdot d \vec{s} = E_1 \cdot d = E_2 \cdot d = E_3 \cdot d
\end{align}

Since $d$ is the same for all dielectrics, $\large{ E_1 = E_2 = E_3 = {{U}\over{d}} }$

with the displacement flux density $D_k = \varepsilon_{rk} \varepsilon_{0} \cdot E_k$ results:

\begin{align*}
{ { D_1 } \over { \varepsilon_{r1} } } = { { D_2 } \over { \varepsilon_{r2} } } = { { D_3 } \over { \varepsilon_{r3} } } = { { D_k } \over { \varepsilon_{rk} } }
\end{align*}

Since the displacement flux density is just equal to the local surface charge density, the charge will no longer be uniformly distributed over the electrodes. \\
Where a stronger polarization is possible, the $E$-field is damped in the dielectric. For a constant $E$-field, more charges must accumulate there. \\
Concretely, more charges accumulate just around the dielectric constant $\varepsilon_{rk}$.

This situation can also be transferred to a coaxial structure of a cylindrical capacitor or concentric structure of spherical capacitors.

<callout icon="fa fa-exclamation" color="red" title="Remember:">
In the case of longitudinal stratification, the result is:
  - A longitudinal stratification can be viewed as a parallel connection of partial capacitors with respective areas $A_k$ and dielectric constant $\varepsilon_{rk}$.
  - The electric field strength in the capacitor is constant.
  - Considering the fields __transverse to the field lines__ - that is, perpendicular to the interface, or the tangential components $E_t$ and $D_t$ of the fields - the following holds:
    - The tangential components of the flux density $D_t$ changes abruptly at the interface.
    - The tangential components of the electric field strength $E_t$ is continuous at the interface: $E_{t1} = E_{t2}$
</callout>
~~PAGEBREAK~~ ~CLEARFIX~~

==== any stratification====

<WRAP right 30em>
<imgcaption imageNo21 | arbitrary layered capacitor>
</imgcaption>
{{drawio>anylayeredcapacitor}}
</WRAP>

With arbitrary stratification, simple observation is no longer possible. \\
However, some hints can be derived from the previous types of stratification:
  * Electric field strength $\vec{E}$:
    * The normal component $E_{n}$ is discontinuous at the interface: $\varepsilon_{r1} \cdot E_{n1} = \varepsilon_{r2} \cdot E_{n2}$
    * The tangential component $E_{t}$ is continuous at the interface: $ E_{t1} = E_{t2}$
  * Electric displacement flux density $\vec{D}$:
    * The normal component $D_{n}$ is continuous at the interface: $ D_{n1} = D_{n2}$
    * The tangent component $D_{t}$ is discontinuous at the interface: $ {{1}\over \Large{\varepsilon_{r1}}\cdot D_{t1} = {{1}\over \Large{\varepsilon_{r2}} \cdot D_{t1} $

Since $\vec{D} = \varepsilon_{0} \varepsilon_{r} \cdot \vec{E}$ the direction of the fields must be the same. \\
Using the fields, we can now derive the change in the angle:

\begin{align}
\boxed { { tan \alpha_1 } \over { tan \alpha_2 } } = { { \varepsilon_{r1} } \over { \varepsilon_{r2}  } } }
\end{align*}

The formula obtained represents the law of refraction of the field line at interfaces. There is also a hint that for electromagnetic waves (like visible light) the refractive index might depend on the dielectric constant. In fact, this is the case. However, in the calculation presented here, electrostatic fields were assumed. In the case of electromagnetic waves, the distribution of energy between the two fields must be taken into account. This is not considered in detail in this course.

~~PAGEBREAK~~~CLEARFIX~~~


Different dielectrics in the capacitor

{{youtube>Uyddvk23iHY}}

{{youtube>0ZxbPGKA2Po}}

~~PAGEBREAK~~~CLEARFIX~~~

====Tasks====

<panel type="info" title="Task 5.9.1 Layered Capacitor Task"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>

{{youtube>wFSzB68OHzI}}

</WRAP></WRAP></panel>


<panel type="info" title="Task 5.9.2 Capacitor with glass plate"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>

<WRAP right>
<imgcaption imageNo15 | Structure of a capacitor with glass plate>
</imgcaption>
{{drawio>CapacitorWithGlassPlate}}
</WRAP>

Two parallel capacitor plates face each other with a distance $d_K = 10mm$. A voltage of $U = 3'000V$ is applied to the capacitor. Parallel to the capacitor plates there is a glass plate ($\varepsilon_{r,G}=8$) with a thickness $d_G = 3mm$ in the capacitor.

  - Calculate the partial voltages $U_G$ in the glass and $U_L$ in the air gap.
  - What is the maximum thickness of the glass pane if the field strength $E_{0,G} =12 kV/cm$ must not exceed.

~~PAGEBREAK~~ ~CLEARFIX~~
</WRAP></WRAP></panel>

{{page>task_5.9.3_with_calculation&nofooter}}


=====5.10 Summary =====

<WRAP right>
<imgcaption imageNo16 | summary of electrostatics>
</imgcaption>
{{drawio>SummaryElectrostatic}}
</WRAP>

~~PAGEBREAK~~~CLEARFIX~~~

====== Further links ======

  * [[https://lx3.mint-kolleg.kit.edu/onlinekursphysik/html/1.4.1/modstart.html|Online Bridge Course Physics KIT]]: This semi-interactive course contains some of the information from my course. Furthermore, videos, exercises and more can be found there

====== 6. The steady-state electric flow field ======

<callout>

At electrostatic field in principle no charges in motion were considered. Now the motion of charges shall be considered explicitly.

The electric flow field here describes how charge carriers move together (collectively). The stationary flow field describes the charge carrier movement if a **direct voltage** is the cause of the movement. A constant direct current then flows in the stationary electric flow field. Thus there is no time dependence of the current:

$\large{{dI}{dt}}=0$

Also important is: Up to now was considered, charges did move through a field or could be moved in future. Now just the moment of the movement is considered.

</callout>

===== 6.1 Current strength and flow field =====

<callout>

=== Goals ===

After this lesson, you should:

  - be able to sketch the flow field in a constricted and rectilinear conductor.
  - Be able to determine the flow velocity of electrons.
  - know the integral notation of the electric current.

</callout>

<WRAP right>
<imgcaption imageNo01 | part of a ladder >
</imgcaption>
{{drawio>charges_in_ladder}}
</WRAP>

==== Current intensity and current density in the simple case ====

The current strength was previously understood as "charge per time" ($I={{dQ}\over{dt}}$). Microscopically, electric current is the directed motion of electric charge carriers. In the chapter [[basics_and_basic_concepts#charge_and_current|basics and basic_concepts]] we have already discussed the picture of the charge carrier current penetrating through a cross-sectional area $A$ (see <imgref FigNo01>).
Furthermore, we had quite practically applied Ohm's law with $R = {{U}\over{I}}$ in DC. Now we know that from the electrostatic field the voltage can be derived from the electric field strength. But what about the current?

For this purpose, the packet $dQ$ of charges is considered, which will pass the area $A$ in the future in the period $dt$. These charges are located in a partial volume element $dV$, which is given by the area $A$ to be traversed and a partial section $dx$: $dV = A \cdot dx$. The amount of charges per volume is given by the charge carrier density, specifically for metals by the electron density $n_e$. The electron density $n_e$ gives the number of free electrons per unit volume a. For copper, for example, this is approximately $n_e(Cu)=8.47 \cdot 10^{19} {{1}\over{mm^3}}$.

~~PAGEBREAK~~~CLEARFIX~~~

<WRAP right>
<imgcaption imageNo02 | Charges in a partial volume in the conductor>
</imgcaption> <WRAP>.
{{drawio>ChargesInLadderVolume}} \\
</WRAP>

The flowing charges contained in the partial volume element $dV$ are then (with elementary charge $e_0$):

\begin{align*}
dQ = n_e \cdot e_0 \cdot A \cdot dx
\end{align*}

The current is then given by $I={{dQ}\over{dt}}$:

\begin{align*}
dQ = n_e \cdot e_0 \cdot A \cdot {{dx}\cdot{dt}} = n_e \cdot e_0 \cdot A \cdot v_e
\end{align*}

This gives an electron velocity $v_e$ of:

\begin{align*}
v_e = {{dx}\cdot{dt}} = {{I}\over{n_e \cdot e_0 \cdot A }}
\end{align*}

In contrast to the considerations in electrostatics, the charge inertia now travels with finite velocities.
With regard to the electron velocity $v_e \sim {{I}\over{A}}$ it is obvious to determine a current density $S$ (related to the area):

\begin{align}}
\boxed{S = {{I}\over{A}}
\end{align*}

In some books, the letter $J$ is also used for current density.

 
~~PAGEBREAK~~~CLEARFIX~~~

==== Field lines and equipotential surfaces of the electric flow field ====

<WRAP right>
<imgcaption imageNo03 | Field lines and equipotential surfaces of the electric flow field>.
</imgcaption> <WRAP>.
{{drawio>fieldlineelectricflowfield}} \\
</WRAP>

As with the electrostatic field, a homogeneous field form and the inhomogeneous field form are to be contrasted:

  - Homogeneous flow field \\ e.g. conductor with constant cross-section
    - Field lines of the current run parallel
    - equipotential surfaces
      - are always perpendicular to each other, because the potential energy of a charge depends only on its position along the path
      - are equidistant due to the constant electric field causing the current and the homogeneous geometry
    - Current $I = S \cdot A$ is constant \$rightarrow$ charge carriers have the same velocity $v$
  - Inhomogeneous current field \\fuse or taper in wire
    - Field lines of the current are not parallel
    - Current $I = S \cdot A$ must also be constant, because the charge does not disappear / is generated, but the area $A$ becomes smaller \$ $\rightarrow$ thus the current density $S$ and the velocity $v$ at the constriction must become larger
    - Equipotential surfaces
      - are again perpendicular to this.
      - now show a compression at the constriction

But why is there a compression of the equipotential surfaces at the bottleneck? This means that there is a large potential difference, i.e. a large voltage. So this already sounds a little bit plausible. This will be looked at in more detail in a moment.

The current density was determined only for a constant cross-sectional area $A$, through which a homogeneous current, thus also a homogeneous flow field, passes at right angles. Now, however, a general approach for the electric current strength is to be found.

For this purpose, instead of a constant current density $S$ over a vertical, straight cross-sectional area $A$, a varying current density $S(A)$ over many small partial areas $dA$ is considered. Thus, if the subareas are sufficiently small, a constant current density over the subarea can again be obtained. It then becomes from

\begin{align*}
I = S \cdot A \rightarrow dI = S \cdot dA
\end{align*}

The total current over a larger area $A$ is thus given as:

\begin{align*}
I = \int dI = \iint_A S \cdot dA
\end{align*}

But what was not considered here: The chosen area $A$ does not necessarily have to be perpendicular to the current density $S$. To take this into account, the (partial) surface normal vector $d\vec{A}$ can be used. If only the part of the current density $\vec{S}$ is to be considered which acts in the direction of $d\vec{A}$, this can be determined via the scalar product:

\begin{align}
I = \int dI = \iint_A \vec{S} \cdot d\vec{A}
\end{align}

This represents the integral notation of the electric current strength. This can be used to determine the current strength in any field.

==== General Material Law ====

For a "pragmatic" derivation of the general material law for the current density, the compression of the equipotential surfaces at the constriction shall be discussed again. Between two equipotential surfaces there is a voltage difference $\Delta U$. If one chooses this sufficiently small, the transition of $\Delta U \rightarrow dU$ results again. However, the same current $I$ must always flow through the potential surfaces in the conductor. Ohm's law then gives the partial resistance $dR$ between the two equipotential surfaces:

\begin{align*}
dU = I \cdot dR \tag{6.1.1}
\end{align*}

The individual quantities are now to be considered for infinitesimally small parts.
For $I$ an equation about a density - the current density - was already found:

\begin{align}
I = S \cdot A \tag{6.1.2}
\end{align}

But also $R$ has already been expressed by a "density" - the resistivity $\varrho$: $ R = \varrho \cdot {{l}\over{A}}$

If a conductor made of the same material is considered, the resistivity $\varrho$ is the same everywhere. But if there is now a partial element $ds$ along the conductor, where the cross-section $A$ is smaller, the resistance $dR$ of this partial element also changes. The partial resistance is then:

\begin{align*}
dR = \varrho \cdot {{ds}\over{A}} \tag{6.1.3}
\end{align}

**In concrete terms, this means for the bottleneck: The resistance increases at the bottleneck. Thus the voltage drop also increases there. Thus there are also more equipotential surfaces there. **

The enrichment of the equipotential surfaces would be solved with it. Interestingly, however, with the thought model now also for a __homogeneous body__ the general material law can be explained.
To do this, one inserts equation $(6.1.2)$ and $(6.1.3)$ into $(6.1.)$.
Then it follows:

\begin{align*}
dU = I \cdot dR = S \cdot A \cdot \varrho \cdot {{ds}\over{{A}} = \varrho \cdot S \cdot ds \
\end{align*}

If now the electric field strength is inserted as $E={{dU}\over{ds}}$, one obtains:

\begin{align}}
E = {{dU}\over{ds}} = \varrho \cdot S
\end{align*}

With a more detailed (and mathematically correct) derivation you get:

\begin{align}
\boxed{\vec{E} = \varrho \cdot \vec{S} }
\end{align}

This equation expresses how the electric field $\vec{E}$ and the (steady-state) electric flow field $\vec{S}$ are related: both point in the same direction. For a given electric field $\vec{E}$ in a homogeneous conductor, the smaller the resistivity $\varrho$, the larger the flow field $\vec{S}$.

~~PAGEBREAK~~~CLEARFIX~~~

==== Task ====

<panel type="info" title="Task 6.1.1 calculated exercises in video"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>

Examples of electrical current density
{{youtube>9vLvzM9eGxY}}

</WRAP></WRAP></panel>


<panel type="info" title="Task 6.1.2 Electron velocity in copper"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>

In a conductor made of copper with cross-sectional area $A$, the current $I = 20A$ flows. \\
Let further be given the electron density $n_e(Cu)=8.47 \cdot 10^{19} {{1}\over{mm^3}}$ and the magnitude of the elementary charge $e_0 = 1.602 \cdot 10^{-19} As$

  - What is the mean flow velocity $v_{e,1}$ of electrons when the cross-sectional area of the conductor is $A = 1.5mm^2$?
  - What is the mean flow velocity $v_{e,1}$ of the electrons when the cross-sectional area of the conductor is $A = 1.0mm^2$?

</WRAP></WRAP></panel>


=====6.2 Gaussian flow field theorem =====

<callout>

==== Targets ====

After this lesson, you should:

  - know which quantities are comparable for the electrostatic field and the flow field.
  - be able to explain the displacement current on the basis of enveloping surfaces.
  - understand how current can flow "through" a capacitor.

</callout>




==== Video ====

Why does an electron flow through a capacitor?
{{youtube>8tXXlI0gHVk}}





====== Tasks ======
<panel type="info" title="Task 6.2.1 Simulation"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>

In the simulation program of [[http://www.falstad.com/emstatic/|Falstad]] can be represented by equipotential surfaces, electric field strength and current density in different objects.

  - Open the simulation program via the link
  - Select: "Setup: Wire w/ Current" and "Show Current (j)".
  - You will now see a finite conductor with charge carriers starting at the top end and arriving at the bottom end.
  - We now want to observe what happens when the conductor is tapered.
    - To do this, select "Mouse = Clear Square". You can now use the left mouse button to remove parts from the conducting material. The aim should be, that in the middle of the ladder there is only a one box wide line, on a length of at least 10 boxes. If you want to add conductive material again, this is possible with "Mouse = Add - Conductor".
    - Consider why more equipotential lines are now accumulating as the conductor is tapered.
    - If you additionally draw in the E-field with "Show E/j", you will see that it is stronger along the taper. This can be checked with the slider "Brightness". Why is this?
  - Select "Setup: Current in 2D 1", "Show E/rho/j". Why doesn't the cavity behave like a Faraday cage here?

</WRAP></WRAP></panel>

====== 2. Simple DC circuits ======

<WRAP right>
<imgcaption imageNo91 | Circuit example>
</imgcaption>
{{drawio>ExampleCircuit}}
</WRAP>

So far, only simple circuits consisting of a source and a load connected by wires have been considered. \\ In the following, more complicated circuit arrangements will be analysed. These initially contain only one source, but several lines and many ohmic loads (cf. <imgref Fig.No.91>).

~~PAGEBREAK~~~CLEARFIX~~~

===== 2.1 ideal components =====

<callout>

=== Goals ===

After this lesson, you should:

  - know the representation of ideal current and voltage sources in the U-I diagram.
  - know the internal resistance of ideal current and voltage sources.
  - know the symbol of ideal current and voltage sources.
  - know the properties of the ideal resistance and the ideal connection.

</callout>

Every electrical circuit consists of three elements:

  - **Consumers**: Consumers convert electrical energy into energy that is not purely electrical. \\ e.g.
    - into electrostatic energy (capacitor)
    - into magnetostatic energy (magnet)
    - into electromagnetic energy (LED, light bulb)
    - into mechanical energy (loudspeaker, motor)
    - into chemical energy (charging an accumulator)
  - **Generators (sources)**: Generators convert energy from another form of energy into electrical energy. (e.g. generator, battery, photovoltaic)
  - **Connections** (or power grid): the interconnections link consumers to sources.

These elements will be considered in more detail below.

==== Consumer ====

<WRAP right>
<imgcaption imageNo4 | current-voltage characteristic examples>
</imgcaption>
{{drawio>ExamplesCurrentVoltageCurve}}
</WRAP>

  * The colloquial term consumer in electrical engineering stands for an electrical consumer - i.e. a component that converts electrical energy into another form of energy.
  * A resistor is often also referred to as a consumer. In addition to pure ohmic consumers, however, there are also ohmic-inductive consumers (e.g. coils in a motor) or ohmic-capacitive consumers (e.g. various power supplies using capacitors at the output). Accordingly, the equation of resistor and load is incorrect.
  * **Current-voltage characteristics** (cf. <imgref FigNo4>).
    * Current-voltage characteristics of a load always pass through the origin, since without current there is no voltage, and vice versa.
    * Ohmic loads have a linear current-voltage characteristic which can be described by a single numerical value. \\ The slope in the $U$-$I$-characteristic is the conductivity: $I = G \cdot U = {{U}\over{R}}$

~~PAGEBREAK~~~CLEARFIX~~~

==== Sources ====

<WRAP right>

Ideal sources
{{youtube>IZDh_EUuhRs}}

\\
<WRAP group><WRAP column 45%>
<imgcaption imageNo6 | ideal voltage source>
</imgcaption>
{{drawio>ideal voltage source}}
<WRAP>
<WRAP column 45%>

<imgcaption imageNo7 | ideal power source>
</imgcaption>
{{drawio>idealpower-source}}

</WRAP></WRAP></WRAP>

  * Sources act as generators of electrical energy
  * Ideal and real sources are distinguished. \\ The real sources are described in the following chapter ([[Linear sources and dipoles]]).

The **ideal voltage source** generates a defined output voltage $U_q$.
In order to maintain this voltage, it can supply an arbitrary current.
The current-voltage characteristic curve also represents this (see <imgref FigNo6>). \\
The schematic shows a circuit with two terminals. In the circuit, the two terminals are shorted. \\
Another circuit symbol shows the negative terminal of the voltage source as a "thick minus", the positive terminal is drawn wider.

The **ideal current source** generates a defined output current $I_q$.
For this current to flow, any voltage is possible at its terminals.
The current-voltage characteristic also represents this (see <imgref FigNo7>). \\
The schematic shows a circuit with two terminals. In the circle, the two terminals are left open and a line is drawn perpendicular to them.

~~PAGEBREAK~~~CLEARFIX~~~

==== Connections ====

  * The ideal connection line is resistance-free and transmits current and voltage instantaneously.
  * Real influences (e.g. voltage drop) of connections are taken into account via separately drawn components (e.g. ohmic resistance).


===== 2.2 Reference arrows and first consideration of a DC circuit =====


<WRAP right>
<imgcaption imageNo5 | example of a circuit>
</imgcaption> <WRAP right>
{{url>https://www.falstad.com/circuit/circuitjs.html?running=false&ctz=CQAgjCAMB0l3BWcMBMcUHYMGZIA4UA2ATmIxEJRCQBZsAoAJxDqsJvA0JGO8hBRpITTt14hsacMSr9Mw5pP5gZEwnnA0OyhHBGsKHGuJQI+LDHubHup7kumyQBKxKkqqNMPfVRwQhRYTMxZvNQ1+dUCUGmVVbzlYvzAAkQcPASTsbHMzVxj+bO4vHwiLEQKJHP8NTCcweHoANxqBDCowLh5zfiT+JH4YMARiYeM6SBl6AGUBPFr8TLlF5RAAMwBDABsAZwBTPxR6AHc5hY101WikwQ4EtqdsYmFT+8qU2vaoE-BCOS+Pg9vgAPAQjaQQTBIFRGTIgab0UHYGjkFIQbAIDQpWFSI5Ip7gFISFGE2gSfgASXo2Es4U0HGweDuKCxfheEiZrQM7BBbTuxFqeCQaAgHHc9CAA 600,500 noborder}}
</WRAP>

<callout>
=== Targets ===

After this lesson, you should:

  - Be able to apply and distinguish between the producer and consumer reference arrow systems.
</callout>

In the chapter Basics the conventional direction sense of [[basics_and_basic_concepts#definition_of_the_current_direction|currents]] and [[basics_and_basic_concepts#voltage_between_two_points|voltages]] was already discussed. Unfortunately, in the case of meshed networks, it is often not possible to see in which direction the conventional directional sense of all currents and voltages runs before the calculation.

in <imgref FigNo5> such a meshed network is shown. In this circuit, a switch $S_1$ and a current $I_2$ are marked. \\
If the resistor is changed via the switch $S_1$, the direction of the current $I_2$ changes.

~~PAGEBREAK~~~CLEARFIX~~~
<callout icon="fa fa-exclamation" color="red" title="Remember:">
<WRAP right>
<imgcaption imageNo1 | reference arrows>
</imgcaption>
{{drawio>reference arrows1}}
</WRAP>

  * **Before calculation** means that the __reference arrows__ for currents and voltages are set as desired.
  * **After calculation** means.
    * $I>0$: The reference arrow reflects the conventional directional sense of the current.
    * $I<0$: The reference arrow points in the opposite direction to the conventional directional sense of the current.
  * Reference arrows of the current are drawn **__in__** the line trace whenever possible.
</callout>

==== Producer and consumer (reference) arrow systems ====

<WRAP group><WRAP column 45%>
<callout color="grey">
<WRAP right>
<imgcaption imageNo2 | creator arrow system>
</imgcaption>
{{drawio>generator arrow system}}
<WRAP>
=== Producer arrow system ===

With **sources** (or generators), energy is taken from the environment and made available to the circuit. \\
For generators, the arrow__foot__ of current is attached to arrow__head__ of voltage. Voltage and current arrows are antiparallel ($\uparrow \downarrow$).

For producers:

$P_{1} = U_{12} \cdot I_1 \stackrel{!}{>} 0$

The power transfer from the environment to the power grid __via the generator or the generator arrow system__ is calculated positively.

</callout>
</WRAP><WRAP column 45%>

<callout color="grey">
<WRAP left>
<imgcaption imageNo3 | consumer arrow system>
</imgcaption>
{{drawio>Consumer Arrow System}}
<WRAP>
=== Consumer arrow system===

In the case of **consumers**, energy is taken from the circuit and made available to the environment. \\
For loads, the arrow__feet__ or arrow__tips__ of current and voltage are related. Voltage and current arrows are parallel ($\uparrow \uparrow$).

For consumers applies:

$P_{3} = U_{34} \cdot I_3 \stackrel{!}{>} 0$

The power transfer from the power grid to the environment __via the consumer or the consumer arrow system__ is also calculated positively.

</callout>
</WRAP></WRAP>

<WRAP right>
The counting arrow system
{{youtube>ceWNJQpgy8E}}
</WRAP>

~~PAGEBREAK~~~CLEARFIX~~~


===== 2.3 Knots, branches and meshes =====

<WRAP right>
Explanation of the different network structures \\ (graphs and trees are not needed until later chapters).
{{youtube>c7z1pRCzEuw}}
</WRAP>

<callout>
=== Targets ===

After this lesson, you should:

  - be able to identify the nodes, branches and meshes in a circuit.
  - Be able to represent a circuit in a clearer way.

</callout>

~~PAGEBREAK~~~CLEARFIX~~~

<WRAP right>
<imgcaption imageNo0 | Circuits and power grids>
</imgcaption> <WRAP>.
{{drawio>circuits_power_grids}}

<imgcaption imageNo8 | Knots, twigs and meshes>
</imgcaption> \
{{drawio>NodesBranchesMeshes}}
</WRAP>


Electrical circuits typically have the structure of networks. Networks consist of two elementary structural elements:

  - <fc #cd5c5c>**branches/edges**</fc>: connections between two nodes.
  - <fc #6495ed>**nodes**</fc>: connection points of multiple branches

In the case of electrical circuits, note:

  - <fc #cd5c5c>**Branches**</fc> contain at least one component.
  - <fc #6495ed>**Nodes**</fc> connect more than two branches and can also be spatially extended.

Branches in electrical networks are called dipoles.
Their behaviour is described by current-voltage characteristics and explained in more detail in the chapter [[Linear Sources and Two-Poles]].

In addition, another term is to be explained: \\
A **<fc #ffa500>mesh</fc>** is a closed path in the net. That means a mesh starts and ends at the same node and runs over at least one more node.

Since a voltmeter can also be present as a component between two nodes, it is also possible to close a mesh by specifying a voltage (cf. $U_1$ in <imgref FigNo8>).

~~PAGEBREAK~~ ~~CLEARFIX~~
In contrast to the other cause-effect relationships, in networked circuits the entire behavior almost always changes when a change occurs in a branch / at a node. \\ This is comparable to other changes in other networks, e.g. a traffic jam in the road network, due to which other roads experience a higher load. For electrical engineering, this means that in the case of changing circuits, the focus is often on determining the interrelationships (formulas, current-voltage characteristics) and not on a single numerical value.

~~PAGEBREAK~~~CLEARFIX~~~

==== Simplifications ====

<WRAP right>
<imgcaption imageNo9 | example of converting a circuit>
{{electrotechnology_1:conversion-circuit.gif}}
</imgcaption>
<WRAP>

With the knowledge of nodes, branches and meshes, circuits can be simplified.
Circuits can be reshaped arbitrarily, as long as all branches remain at the same nodes after the reshaping.
The <imgref FigureNo9> shows how such a reshaping is possible.

For practical tasks, repeated trial and error can be useful.
It is important to check afterwards that the same components are connected to each node as before the conversion.

More examples can be found in the following video.
{{youtube>bB19bhd6iSI}}

~~PAGEBREAK~~~CLEARFIX~~~

<panel type="info" title="Task 2.3.1 Branches and Nodes"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>
<WRAP right>
<imgcaption imageNo70 | twigs and nodes>
</imgcaption>
{{drawio>BranchesandNodes}}
</WRAP>

For the markers in the circuits in <imgref imageNo70>, specify whether it is a branch, a node, or neither.

</WRAP></WRAP></panel>

<panel type="info" title="Task 2.3.2 Simplifications of Circuits"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>
<WRAP right>
<imgcaption imageNo71 | branches and nodes>
</imgcaption>
{{drawio>CircuitsSimplify}}
</WRAP>

Simplify the circuits in <imgref imageNo71>.
</WRAP></WRAP></panel>

===== 2.4 Kirchhoff equations =====


<WRAP right>
Illustration and application of Kirchhoff's laws
{{youtube>qkKaNsHdxgc?start=6}}
</WRAP>

<callout>
=== Targets ===

After this lesson you should:
  - Know and be able to apply Kirchhoff's equations or knot and mesh theorems.
</callout>

~~PAGEBREAK~~ ~CLEARFIX~~
==== The Knot Theorem (1st Kirchhoff Equation) ====

The node theorem formulates in the language of mathematics the experience that no charge "accumulations" occur in electrical conductors.
This is of particular relevance at a network node (<imgref Fig.No.10>).
To formulate the equation, the reference arrows of the currents at this network node are all set in the same way.
That is: all point away from or towards the node.

<callout icon="fa fa-exclamation" color="red" title="Notice:">
<WRAP right>
<imgcaption imageNo10 | node set>
</imgcaption>
{{drawio>node set}}
</WRAP>

The sum of all currents coming from the nodes must be zero.

$\boxed{I_1 + I_2 + I_3 + ... + I_n = \sum_{x=1}^{n} I_x=0}$

From now on, the following definition applies:
  * Currents whose current arrows point towards the node are added in the calculation.
  * Currents whose current arrows point away from the node are subtracted in the calculation.
</callout>

<WRAP right>
<imgcaption imageNo11 | parallel circuit>
</imgcaption>
{{drawio>parallel circuit}}
</WRAP>

=== Parallel connection of resistors ===

The total resistance for resistors connected in parallel can be derived from the node set (<imgref FigNo11>):

Since the same voltage $U_{ab}$ is dropped across all resistors, the node theorem holds:

$\large{{U_{ab}}\over{R_1}}+ {{U_{ab}}\over{R_2}}+ ... + {{U_{ab}}\over{R_n}}= {{U_{ab}}\over{R_{substitute}}$

$\rightarrow \large{{1}\over{R_1}}+ {{1}\over{R_2}}+ ... + {{1}\over{R_n}}= {{1}\over{R_{substitute}} = \sum_{x=1}^{n} {{1}\over{R_x}}$

For resistors connected in parallel, the conductance $G_{substitute}$ is the sum of the individual conductance values: $G_{substitute} = \sum_{x=1}^{n} {G_x}$

The following applies in general: The equivalent resistance of a parallel circuit is always smaller than the smallest resistance.

Specifically, for two parallel resistors $R_1$ and $R_2$, $R_{substitute}= \large{{R_1 \cdot R_2}\over{R_1 + R_2}}$

=== Current divider ===

<WRAP right>
Derivation of the current divider with further considerations.
{{youtube>JgHwK1jMauM}}
<WRAP>

The current divider rule can also be derived from the node theorem. \\
This states that for resistors $R_1, ... R_n$ their currents $I_1, ... I_n$ behave just like the conductances $G_1, ... G_n$ through which they flow.

$\large{{I_1}\over{I_g}} = {{G_1}\over{G_g}}$

$\large{{I_1}\over{I_2}} = {{G_1}\over{G_2}}$

~~PAGEBREAK~~~CLEARFIX~~~


<panel type="info" title="Task 2.4.1 Current divider"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>

<WRAP right>
<imgcaption imageNo85| Current divider>
</imgcaption> <WRAP>.
{{url>https://www.falstad.com/circuit/circuitjs.html?running=false&ctz=CQAgjCAMB0l3BWKsBMA2AzAgnAdjBgBxhgK7q64gKTXWQBQATiCgmuLh2xxpACxRwcBmHKt2nbpLApCQqkiS0V0JADUA9gBsALgEMA5gFMGhibwzTeCFFAYB3CyD6CeLq+GYvbUn3dchMHhHfxcBMIxPMDNIzywA30YnQLEbOzT7J3dM1K4s50ywNFpMxhZi2hQUQUrnUpCnOtSSjw4Y82bo1oSCupyeiMYAZ3BW6trWsvAQADN9bWHTIA 600,400 noborder}}
</WRAP>

In the simulation in <imgref FigNo85> a current divider can be seen. The currents behave there just inversely proportional to the flowed through resistances.

  - What currents would you expect in each branch if the input voltage were lowered from $5V$ to $3.3V$? __After__ you had considered your result, you can adjust accordingly by moving the slider ''Voltage'' (bottom right of the simulation).
  - Think about what would happen if you __were__ to flip the switch __before__ you flip the switch. \\ After you flip the switch, how can you explain the current in the branch?

</WRAP></WRAP></panel>

<panel type="info" title="Task 2.4.2 two resistors"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>

Two resistors of $18\Omega$ and $2 \Omega$ are connected in parallel. The total current the resistors is $3A$. \\ Calculate the total resistance and the current division.

</WRAP></WRAP></panel>
\\
~~PAGEBREAK~~ ~CLEARFIX~~
==== The Mesh Theorem (2nd Kirchhoff Equation) ====

Also the mesh theorem describes in mathematical language a practical experience:
Between two points $a$ and $b$ of a network there is only one potential difference.
The potential difference is thus in particular independent of the path along which a network is traversed between the two points $1$ and $2$.
This can be described by considering meshes.

<callout icon="fa fa-exclamation" color="red" title="Notice:">
<WRAP right>
<imgcaption imageNo12 | stitch set>
</imgcaption>
{{drawio>mesh set}}
</WRAP>

In any mesh of an electrical network, the sum of all voltages is zero (<imgref FigNo12>):

$\boxed{U_{1} + U_{2} + ... + U_{n} = \sum_{x=1}^{n} U_x = 0}$

For the calculation, a sense of circulation must be defined. This can be chosen arbitrarily at first.
However, the following definition then applies:
  * Stresses whose stress arrows point in the direction of rotation are added in the calculation.
  * Stresses whose stress arrows point against the direction of circulation are subtracted in the calculation.
</callout>

=== Proof of the Mesh Theorem ===

Expressing the voltages in <imgref FigNo12> by the potentials in the nodes, we get:

$U_{12}= \varphi_1 - \varphi_2 $ \\
$U_{23}= \varphi_2 - \varphi_3 $ \
♪U_{34}= \varphi_3 - \varphi_4 ♪ \
$U_{41}= \varphi_4 - \varphi_1 $

If these stresses are inserted into the mesh equation, then

$U_{12}+U_{23}+U_{34}+U_{41} = 0$ \.

=== Series connection of resistors ===

<WRAP right>
<imgcaption imageNo13 | series connection>
</imgcaption>
{{drawio>series circuit}}
</WRAP>

Using the mesh theorem, the total resistance of a series circuit (<imref FigNo13>) can be easily determined:

$U_1 + U_2 + ... + U_n = U_g$

$R_1 \cdot I_1 + R_2 \cdot I_2 + ... + R_n \cdot I_n = R_{substitute} \cdot I $

Since in series connection the current through all resistors must be the same - i.e. $I_1 = I_2 = ... = I$ - the result is:

$R_1 + R_2 + ... + R_n = R_{substitute} = \sum_{x=1}^{n} R_x $

The following applies in general: The equivalent resistance of a series circuit is always greater than the greatest resistance.

~~PAGEBREAK~~~CLEARFIX~~~

<panel type="info" title="Task 2.4.3 three resistors"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>

Given three equal resistors of $20k\Omega$ each. \\
Which values are realizable by arbitrary interconnection of one to three resistors?

</WRAP></WRAP></panel>

===== 2.5 unloaded and loaded voltage divider =====

==== The Unloaded Voltage Divider ====
<WRAP right>
Derivation of the unloaded voltage divider
{{youtube>AmjaKLkPovg}}
</WRAP>

<callout>

=== Goals ===

After this lesson, you should:

  - be able to distinguish between the loaded and unloaded voltage divider.
  - Be able to describe the differences between loaded and unloaded voltage dividers.
  -

</callout>

<WRAP right>
<imgcaption imageNo14 | unloaded voltage divider>
</imgcaption>
{{drawio>unloadedvoltagedivider}}
<WRAP>

Especially the series connection of two resistors $R_1$ and $R_2$ shall be considered now.
This situation occurs in many practical applications (e.g. potentiometer).
In <imgref Fig.No.14> this circuit is shown.

Via the mesh equation we get

$\boxed{ {{U_1}\over{U}} = {{R_1}\over{R_1 + R_2}} }$

The ratio $k={{R_1}\over{R_1 + R_2}}$ also corresponds to the position at a potentiometer.

~~PAGEBREAK~~~CLEARFIX~~
<panel type="info" title="Task 2.5.1 unloaded voltage divider"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>

<WRAP right>
<imgcaption imageNo81| unloaded voltage divider>
</imgcaption> <WRAP>.
{{url>https://www.falstad.com/circuit/circuitjs.html?running=false&ctz=CQAgjCAMB0l3BWKsAsB2ATAZgGw4QhmimJGOiApJSClgFABu4O1RGL1YaOUfK1JNWHQE9bihBZIk7rywY2CXhFJxkSAEoBTAM4BLXQBcAhgDsAxtvoB3EIoAc9zFJnOOkW68nspDny6edgpsGJJY-vZhUPQADn4+0REBHBDCcfaQTnKZTr5pMXZgrOA8uaW8QZzu5b6eAObeNdI+Dk6eimgJUZICob0gADq6w6O6AKoA+gD2AK5GXo4VPctVLcv5ZZ66KzmbleAgAGYmADa61kA 500,400 noborder}}
</WRAP>

In the simulation in <imgref FigNo81>, an unloaded voltage divider in the form of a potentiometer can be seen. The ideal voltage source provides $5V$. The potentiometer has a total resistance of $1K\Omega$. In the configuration shown, this is divided out to $500 \Omega$ and $500 \Omega$.

  - What voltage ''U_out'' would you expect if the switch were closed? After you have considered your result, you can check it by closing the switch.
  - First consider what would happen if you __changed__ the distribution of the resistors by moving the slider ("intermediate tap")? \\ You can check your assumption by using the slider at the bottom right of the simulation.
  - At which position do you get a ''U_out'' of $3.5V$?

</WRAP></WRAP></panel>


~~PAGEBREAK~~ ~CLEARFIX~~
==== The Loaded Voltage Divider ====


<WRAP right>
<imgcaption imageNo15 | loaded voltage divider>
</imgcaption>
{{drawio>loadedvoltagedivider}}
</WRAP>

If - in contrast to the above unloaded voltage divider - a load $R_L$ is connected to the output terminals (<imgref FigNo15>), it influences the output voltage.

A circuit analysis reveals:

$ U_1 = \LARGE{{U} \over{1 + {{R_2}\over{R_L}} + {{R_2}\over{R_1}} }}  }$

or on a potentiometer with $k$ and $R_s = R_1 + R_2$:

$ U_1 = \LARGE{{k \cdot U} \over { 1 + k \cdot (1-k) \cdot{{R_s}\over{R_L}}  }}  }$


<WRAP right>
<imgcaption imageNo65 | voltage waveform of loaded voltage divider>
</imgcaption>
{{drawio>VoltageCurveLoadedVoltageDivider}}
</WRAP>

<imgref FigNo65> shows the ratio of the output voltage $U_1$ to the input voltage $U$ (y-axis), with respect to the ratio $k={{R_1}\over{R_1 + R_2}}$. In principle, this is similar to <imgref FigNo14>, but here it has another dimension: several graphs are plotted. These differ by the ratio ${{R_s}\over{R_L}}$.

What does this diagram say now? This will be shown by an example. First, assume an __unloaded voltage divider__ with $R_2 = 4 k\Omega$ and $R_1 = 6 k\Omega$, and an input voltage of $10V$. Thus $k = 0.6$, $R_s = 10k\Omega$ and $U_1 = 6V$. \\ Now, this voltage divider is loaded with a load resistor. If this is at $R_L = R_1 = 10 k\Omega$, $k$ reduces to about $0.48$ and $U_1$ to $4.8V$ - so the output voltage collapses. For $R_L = 4k\Omega$, $k$ becomes even smaller to $k=0.375$ and $U_1 = 3.75V$. If the load $R_L$ is only one tenth of the resistor $R_s=R_1 + R_2$, $k=0.18$ and $U_1=1.8V$. The output voltage of the unloaded voltage divider ($6V$) thus became less than one third.

~~PAGEBREAK~~~CLEARFIX~~~

<panel type="info" title="Task 2.5.2 loaded voltage divider"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>

From the circuit in <imgref FigNo15>, determine the above equation $ U_1 = {{k \cdot U} \over { 1 + k \cdot (1-k) \cdot{{R_s}\over{R_L}}}}$ where $k={{R_1}\over{R_1 + R_2}}$ and $R_s = R_1 + R_2$.

</WRAP></WRAP></panel>

<panel type="info" title="Task 2.5.3 loaded voltage divider"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>

<WRAP right>
<imgcaption imageNo82| loaded voltage divider>
</imgcaption> <WRAP>.
{{url>https://www.falstad.com/circuit/circuitjs.html?running=false&ctz=CQAgjCAMB0l3BWcMBMcUHYMGZIA4UA2ATmIxAUgpABZsAoAN3BppExRbbA0Kn5pUkVEdAT0ebXN14hsadgj4Qw8KGJAAlAKYBnAJa6ALgEMAdgGNt9AO7t87DJ2mPOkW6-CyXPPu7suHHIInEH+cgooKFIh7NFQ9AAOcnhsUVKpnhAi9ABOwaHx2LFhyHAexaFOKWnV4ZVxGWnx4WCsnmh4nq3tvvZdfe7JnV58I0HZCXYjfemjCbqN8xOyVBAAZiYANrrWaOTYmXOCVMcgADq6l9e6AKoA+gD2AK5G9ADmcpC1zt9fbO4gA 500,400 noborder}}
</WRAP>


In the simulation in <imgref FigNo82>, a loaded voltage divider in the form of a potentiometer can be seen. The ideal voltage source provides $5V$. The potentiometer has a total resistance of $1K\Omega$. In the configuration shown, this is divided out to $500 \Omega$ and $500 \Omega$. The load resistance has a magnitude of $R_L = 1 k\Omega$.

  - What voltage ''U_OUT'' do you expect if the switch would be closed? Here you have to calculate something! __After__ you had calculated your result, you can check it by closing the switch.
  - At which division you get $3.5V$. First determine the result for a calculation.__ Then check it by moving the slider at the bottom right of the simulation.

</WRAP></WRAP></panel>

<panel type="info" title="Task 2.5.4 Application of the loaded voltage divider - motor"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>

<WRAP right>
<imgcaption imageNo16| Sketch of the structure>
</imgcaption> <WRAP>.
{{drawio>MotorAmVoltageDivider}}
<WRAP>

You wanted to test a micromotor for a small robot. Using the maximum current and internal resistance ($R_M = 5\Omega$), you calculate that it can be operated at a maximum of $U_{M,max}=4V$. A colleague suggested that you can get $4V$ from a $9V$ block battery via the setup in <imgref FigNo16>.

  - First calculate the maximum current $I_{M,max}$ of the motor.
  - Draw the corresponding electrical circuit with the motor connected as an ohmic resistor.
  - At the maximum current, the motor should be able to deliver a torque of $M= 100mNm$. What torque would the motor deliver if you implement the setup like this? (Assumption: The torque of the motor increases proportionally to the motor current).
  - What might a setup with a potentiometer look like that would actually allow you to set a voltage between $0.5V$ to $4V$ on the motor? What resistance value should the potentiometer have?
  - Build and test your circuit in the simulation below. For an introduction to online simulation, see: [[electronic_circuit_technology:0 tools#Online Circuit Simulator]]. \\ You will essentially need the following tips for this setup:
    - Routing connections can be activated via the menu ''Draw >> Insert Connection (wire)''. Afterwards you have to click on the start point and then drag it to the end point.
    - Note that connections can only ever be connected to connection points. The red marked node at the $5 \Omega$ resistor indicates that it is not connected. This could be moved one grid step to the left, as there is a connection point there.
    - Pressing the ''<ESC>'' key can be used to disable component insertion.
    - With a right click on a component you can copy it or change values like the resistance via ''Edit...''.

<WRAP right>
<imgcaption imageNo83| Simulation for engine buildup>
</imgcaption> <WRAP>.
{{url>https://www.falstad.com/circuit/circuitjs.html?running=false&ctz=CQAgjCAMB0l3BWKsDsKAc64YMyRWACyRIKQhKE4BQAbiIeiHoQ0wEzuvnnEgCcUITATUATuH7sQnVu0wyuQ0QHM2zSK0bMcANijUA7upYgsG7kfWyz5G+2pgUrJ-tOvmCaRDDxkSACUAUwBnAEsQgBcAQwA7AGMgq3ZIDhRpFI4lSCswKUUXFH0bHKA 800,400 noborder}}
</WRAP>

</WRAP></WRAP></panel>

<panel type="info" title="Task 2.5.5 Examples of the calculation of loaded stress dividers"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>

<WRAP group><WRAP column half>
Voltage divider, series resistor (series resistance) and shunt resistor.
{{youtube>XRBiYdMhiF0}}

</WRAP><WRAP column half>
Voltage divider exercise
{{youtube>6NG4uOMDU7Y}}
</WRAP></WRAP>

</WRAP></WRAP></panel>

<panel type="info" title="Task 2.5.6 Example of a loaded voltage divider: Explanation without calculation"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>

{{youtube>hRVOZIb8-6g}}

</WRAP></WRAP></panel>

~~PAGEBREAK~~~CLEARFIX~~~


===== 2.6 Star-delta connection =====

<WRAP right>

<imgcaption ImageNo98 | Circuit Example>
</imgcaption>
{{drawio>ExampleCircuit}}

<imgcaption ImageNo17| Converting parallel circuit to series circuit>.
</imgcaption> \\
{{url>https://www.falstad.com/circuit/circuitjs.html?running=false&ctz=CQAgjCAMB0l3BWKsDsKAc64YMyRWACyRIIRKE4BQATiALQ4BM4T6Dz4KAbFON92gDhVAOYcWTJoQkhpkKFQBKssGBaMWATl4KFxEEj3IEVAO6r1qtuHMMwWlmo0OnhGWGX3H4K-VcgOnz6CkbB0KYW-j5EMv7cCrGKUVIe7t5uMpBiGXKQ7NGSBtlRAUz5DKl57CX2CdW55TW0dYnp8QpNfGDwLYXg7WUVib109FVJ49INI3B2jAhOfjiLAx5eC0saK9q6fAZhehE5mw2n8opjO2scq10KOI-Qj9RXTjyyVYkCQsLcVEA 600,300 noborder}}

</WRAP>

<callout>

=== Goals ===

After this lesson, you should:

  - be able to convert triangular meshes into a star shape (and vice versa)
</callout>

At the beginning of the chapter an example of a network was shown (<imgref BildNr91>). Here, however, one does not come directly to the solution with the set of nodes and meshes. However it is visible after that many triangular meshes or star-shaped nodes are present there (<imgref BildNr98>). These shall now be discussed in more detail.

First of all, a summary of the previous findings. Using the node and mesh theorem, it became clear that an equivalent resistance can be determined from both a series and a parallel circuit. If one considers the equivalent resistance as a black box - i.e. the inner expansion is unknown - it could be interpreted by both types of circuit (<imgref Fig. No. 17>).

So how does this help us in the case of a triangular mesh?

Also in this case, one can provide a blackbox. However, this should always behave in the same way as the triangular mesh, i.e. any applied voltages should produce equal currents. \\
In other words, the resistances that can be measured between two terminals must be identical for both circuits.

For this purpose, the different resistances between the individual nodes $a$, $b$ and $c$ are now to be considered, see <imgref Fig.No.18>. It is to be found out how a delta circuit can be developed from a star circuit (and vice versa).



~~PAGEBREAK~~~CLEARFIX~~~

<WRAP right>
<imgcaption imageNo18| Star-triangle transformation>
</imgcaption> <WRAP>.
{{url>https://www.falstad.com/circuit/circuitjs.html?running=false&ctz=CQAgjA7CAMB00IQVhvCEAcGGYMzQjABZoUkA2EFI3AKACUQBaYo8CSlsAJncuhggSQwQLhJaAd2as+M8gMiUwtAE7NuEXko3c2eseERqZHOU03azi4+pZmdFvSAOCwx6U3xtH3LHJVGL2gfMy9cZWtBYRRDcSldfS1ElwpwBhTLTP9DYV442Akg2VwMNi5eUrZcgWrRQoSWIjYq+QFWlTtuZyIwTm62InI63FHYUYnRkydBvo1k1vaxyYnpnVaZoTml3HGVuk9vEA2-DGOy9KCjjdHKReiBWPqJTwGhbBShnwzNog-fgCc-AeLlEqBebXO5XWF2gjVOUPmlVh8P8SABvAsyXR+UafUeGNMlBxMB+bwgIRSWjiQgE+We8OcFPKb3I-kCzCOJPCxMJNSoYPih0p3IRJM6RKohM25GcOz2K2mrOcWN4JPl+zoXWcspZOuSGv2eOaUsx+NNlxkJu5YB4Fv5dTEDQAHvNcFQ7SxyFAkGBqOAkAJAAGktFdFgB7u6AI0bJcuF4+jY9AAnaH5j5ZTGM-6wGcAIaos6+ZI6OGec2OHSltMWDDu3PRiwKcCEwYCQAzpDWG+AMI2+spW7SQJ2w9wIqkwBowJGkIMhyOZBgUH13fZl+RKD4zgAjGvcdypTHcaCT7hIHNnADGjP0aU2rjho+SY-6SEjuGjW5Al4LYfNL6tU8PxEXMQG3a8-wwU9uD7NIxwTFwk1TUcaHAP4pyLX0RGcFM926AMUAsaBtCQM4vzzXdR1wH0MCPO1SP9HDkKnYlexkIELS-bdHw0AE2FlfpyGjWVP0QkBcKfd1yFGDQ3xAaTNx7EA8x4oioA3aFaPk71sKQrsMEob1Jy9dSIDnUDL1Uvx1MpJt+IIERhAXLEziQdAY1cj5yP0903L7CBfK0rj9Jswi+nssj5z3dC2X6YC2VEiy93HISWWneSASgNthxveTkhlHIaygqghOOaM3IQ0CAGUABcAFNVAAOwAZ0vAALPMABsaoAV0agBzGtMHAchJ2bcKQLOAARVQ6oASzqy8AGtWo67q+sGsMzGowj3WoyLQIAIU6vNlu3AB7Z1aCAA 600,800 noborder}}
\\ \\
Calculation of the transformation formulas: star connection in delta connection
{{youtube>AFSWn5xR8tE}}

</WRAP>

==== Delta connection ====

In the delta connection, the 3 resistors $R_{ab}^1$, $R_{bc}^1$ and $R_{ca}^1$ are connected in a mesh.

For the resistors between the two connections (e.g. $a$ and $b$), the third one ($c$) is considered as not connected. This results in a parallel connection of the direct star resistor $R_{ab}^1$ with the series connection of the other two star resistors $R_{ca}^1 + R_{bc}^1$:

$R_{ab} = R_{ab}^1 || (R_{ca}^1 + R_{bc}^1) $ \\
$R_{ab} = {{R_{ab}^1 \cdot (R_{ca}^1 + R_{bc}^1)}\over{R_{ab}^1 + (R_{ca}^1 + R_{bc}^1)}} = {{R_{ab}^1 \cdot (R_{ca}^1 + R_{bc}^1)}\over{R_{ab}^1 + R_{ca}^1 + R_{bc}^1} $ \\

The same applies to the other connections. This results in:

\begin{align*}
R_{ab} = {{R_{ab}^1 \cdot (R_{ca}^1 + R_{bc}^1)}\over{R_{ab}^1 + R_{ca}^1 + R_{bc}^1}  \\
R_{bc} = {{R_{bc}^1 \cdot (R_{ab}^1 + R_{ca}^1)}\over{R_{ab}^1 + R_{ca}^1 + R_{bc}^1}  \\
R_{ca} = {{R_{ca}^1 \cdot (R_{bc}^1 + R_{ab}^1)}\over{R_{ab}^1 + R_{ca}^1 + R_{bc}^1} \tag{2.6.1}  \end{align*}

==== Star connection ====

The resistors between the connections must now be similar to those in the star circuit. Also in the star connection 3 resistors are connected, but in a star shape. So the star resistors are all connected with another node $0$ in the middle: $R_{a0}^1$, $R_{b0}^1$ and $R_{c0}^1$.

Again, the same procedure is used as for the delta connection: the resistance between two connections (e.g. $a$ and $b$) is determined, the further connection ($c$) is considered to be open. The resistance of the further terminal ($R_{c0}^1$) is only connected at one side. Thus no current flows through it - it is thus not to be considered. It results:

\begin{align*}
R_{ab} = R_{a0}^1 + R_{b0}^1 \\
R_{bc} = R_{b0}^1 + R_{c0}^1 \
R_{ca} = R_{c0}^1 + R_{a0}^1 \tag{2.6.2}
\end{align*}

From equations $(2.6.1)$ and $(2.6.2)$ one obtains:

\begin{align} R_{ab} = {{R_{ab}^1 \cdot (R_{ca}^1 + R_{bc}^1)}\over{R_{ab}^1 + R_{ca}^1 + R_{bc}^1}} = R_{a0}^1 + R_{b0}^1 \tag{2.6.3} \end{align}
\begin{align} R_{bc} = {{R_{bc}^1 \cdot (R_{ab}^1 + R_{ca}^1)}\over{R_{ab}^1 + R_{ca}^1 + R_{bc}^1} = R_{b0}^1 + R_{c0}^1 \tag{2.6.4} \end{align}
\begin{align} R_{ca} = {{R_{ca}^1 \cdot (R_{bc}^1 + R_{ab}^1)}\over{R_{ab}^1 + R_{ca}^1 + R_{bc}^1} = R_{c0}^1 + R_{a0}^1 \tag{2.6.5} \end{align}
\begin{align} R_{ca} = {{R_{ca}^1 \cdot (R_{bc}^1 + R_{ab}^1)}\over{R_{ab}^1 + R_{ca}^1 + R_{bc}^1}} = R_{c0}^1 + R_{a0}^1 \tag{2.6.5} \end{align}


Equations $(2.6.3)$ to $(2.6.5)$ can now be cleverly combined so that there is only one resistor on one side. \\
A variation is to write the formulas as ${{1}\over{2}} \cdot \left( (2.6.3) + (2.6.4) - (2.6.5) \right)$ or ${{1}\over{2}} \cdot \left(R_{ab} + R_{bc} - R_{ca}\right)$ to combine. This gives $R_{b0}^1$ \.

\{{align}}
{{1}\over{2}} \cdot \left( {{R_{ab}^1 \cdot (R_{ca}^1 + R_{bc}^1)}\over{R_{ab}^1 + R_{ca}^1 + R_{bc}^1} + {{R_{bc}^1 \cdot (R_{ab}^1 + R_{ca}^1)}\over{R_{ab}^1 + R_{ca}^1 + R_{bc}^1} - {{R_{ca}^1 \cdot (R_{bc}^1 + R_{ab}^1)}\over{R_{ab}^1 + R_{ca}^1 + R_{bc}^1}} \right) &= {{1}\over{2}} \cdot \left( R_{a0}^1 + R_{b0}^1 + R_{b0}^1 + R_{c0}^1 - R_{c0}^1 - R_{a0}^1 \right) \.

{{1}\over{2}} \cdot \left( {{R_{ab}^1 \cdot (R_{ca}^1 + R_{bc}^1)} + {R_{bc}^1 \cdot (R_{ab}^1 + R_{ca}^1)} - {R_{ca}^1 \cdot (R_{bc}^1 + R_{ab}^1)}\over{R_{ab}^1 + R_{ca}^1 + R_{bc}^1} \right) &= {{1}\over{2}} \cdot \left( 2 \cdot R_{b0}^1 \right) \\.

{{1}\over{2}} \cdot \left( {{R_{ab}^1 R_{ca}^1 + R_{ab}^1 R_{bc}^1 + R_{bc}^1 R_{ab}^1 + R_{bc}^1 R_{ca}^1 - R_{ca}^1 R_{bc}^1 - R_{ca}^1 R_{ab}^1}\over{R_{ab}^1 + R_{ca}^1 + R_{bc}^1}} \right) &= R_{b0}^1 \

{{1}\over{2}} \cdot \left( {{ 2 \cdot R_{ab}^1 R_{bc}^1 }\over{R_{ab}^1 + R_{ca}^1 + R_{bc}^1}} \right) &= R_{b0}^1 \

{{ R_{ab}^1 R_{bc}^1 }\over{R_{ab}^1 + R_{ca}^1 + R_{bc}^1} &= R_{b0}^1 }\over{R_{ca}^1 + R_{bc}^1}

\end{align*}

Similarly, one can resolve to $R_{a0}^1$ and $R_{c0}^1$, and with a slightly modified approach to $R_{ab}^1$, $R_{bc}^1$ and $R_{ca}^1$.

==== star-triangle-transformation ====
<callout icon="fa fa-exclamation" color="red" title="Merke:">

<WRAP group><WRAP column half>
If a **delta connection is to be converted into a star connection**, the star resistors can be determined via:

\begin{align}}
    \color{lightgray}{\boxed{ \color{black}{\begin{array}{} \text{stare} \text{at port x} \end{array} }} &=
 {{ \color{lightgray}{\boxed{ \color{black}{\begin{array}{} \text{product of} \text{at port x lying} \text{delta resistors} \end{array} }} } \over
  { \color{lightgray}{\boxed{ \color{black}{\begin{array}{} \text{sum of all} \text{triangle resistors} \end{array} }}}}} \\
\\
\text{alias}

R_{a0}^1 &= {{ R_{ca}^1 \cdot R_{ab}^1 }\over{R_{ab}^1 + R_{ca}^1 + R_{bc}^1}} \\
R_{b0}^1 &= {{ R_{ab}^1 \cdot R_{bc}^1 }\over{R_{ab}^1 + R_{ca}^1 + R_{bc}^1} \\
R_{c0}^1 &= {{ R_{bc}^1 \cdot R_{ca}^1 }\over{R_{ab}^1 + R_{ca}^1 + R_{bc}^1}}
\end{align*}

</WRAP><WRAP column half>
If a **star connection is to be converted to a delta connection**, the delta resistors can be determined via:

\begin{align}}
    \color{lightgray}{\boxed{ \color{black}{\begin{array}{} \text{triangle resistance} \text{between} \text{connections x and y } \end{array} }} &=
 {{ \color{lightgray}{\boxed{ \color{black}{\begin{array}{} \text{total of all products} \text{between two} \text{different star resistors} \end{array} }} } \over
  { \color{lightgray}{\boxed{ \color{black}{\begin{array}{} \text{stare} \text{opposite{x and y} \end{array} }}}}} \\
\\
\text{alias:} \quad{alias}

R_{ab}^1 &= {{ R_{a0}^1 \cdot R_{b0}^1 +R_{b0}^1 \cdot R_{c0}^1 +R_{c0}^1 \cdot R_{a0}^1 }\over{ R_{c0}^1}} \\
R_{bc}^1 &= {{ R_{a0}^1 \cdot R_{b0}^1 +R_{b0}^1 \cdot R_{c0}^1 +R_{c0}^1 \cdot R_{a0}^1 }\over{ R_{a0}^1}} \\
R_{ca}^1 &= {{ R_{a0}^1 \cdot R_{b0}^1 +R_{b0}^1 \cdot R_{c0}^1 +R_{c0}^1 \cdot R_{a0}^1 }\over{ R_{b0}^1}}
\end{align*}

</WRAP></WRAP>
</callout>

~~PAGEBREAK~~~CLEARFIX~~~

<panel type="info" title="Task 2.6.1 Application of triangle-star conversion"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>

{{youtube>YDpNFEWkN9U}}{{youtube>s7NqWI_ZSt4}}

</WRAP></WRAP></panel>

<panel type="info" title="Task 2.6.2 more difficult task with star-triangle transformation"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>

{{youtube>poP4a0y0oLU}}

</WRAP></WRAP></panel>

===== 2.7 Group circuit of resistors =====

<callout>

=== Goals ===

After this lesson, you should:

  - be able to simplify circuits consisting only of resistors.
  - Be able to calculate the voltages and currents in circuits with a voltage source and several resistors.
  - Be able to simplify symmetrical circuits.

</callout>

In this subchapter a methodology is discussed, which should help to reshape circuits. In subchapter [[#2.6 Star-Delta-Circuit]] towards the end a network was already transformed in a way, that it does not contain triangular meshes any more. Now this procedure shall be systematized.
Starting point are tasks, where for a resistor network the total resistance, total current or total voltage has to be calculated.

==== simple example ====

<WRAP right>
<imgcaption imageNo89 | example of a circuit>
</imgcaption>
{{drawio>ExampleCircuit2}}
</WRAP>


An example of such a circuit is given in <imgref FigNo89>. Here $I_0$ is wanted. This current can be found by the (given) voltage $U_0$ and the total resistance between the terminals $a$ and $b$. So we are looking for $R_{ab}$.

As already described in the previous subchapters, partial circuits can also be converted into equivalent resistors step by step. It is important to note that these partial circuits for conversion into equivalent resistors may only ever have two connections (= two nodes to the "outside world").


<WRAP right>
<imgcaption imageNo88 | Step by step solution of the example >
</imgcaption>
{{drawio>ExampleCircuit2Solution}}
</WRAP>
~~PAGEBREAK~~ ~CLEARFIX~~

<imgref FigNo88> shows the step-by-step conversion of the equivalent resistors in this example. \\ As a result of the equivalent resistance one gets:

\begin{align*}
R_g = R_{12345} &= R_{12}||R_{345} = R_{12}||(R_3+R_{45}) = (R_1||R_2)||(R_3+R_4||R_5) \\
&= {{ {R_1 \cdot R_2}\over{R_1 + R_2}} \cdot (R_3 + {{R_4 \cdot R_5}\over{R_4 + R_5}}) }\over{ {{R_1 \cdot R_2}\over{R_1 + R_2}} +R_3 + {{R_4 \cdot R_5}\over{R_4 + R_5}} }} \quad \quad \quad \quad \bigg\rvert \cdot{{(R_1 + R_2) \cdot (R_4 + R_5)}\over{(R_1 + R_2) \cdot (R_4 + R_5)}} \\

&= {{ R_1 \cdot R_2 \cdot (R_3 + {{R_4 \cdot R_5}\over{R_4 + R_5}}) \cdot (R_4 + R_5) } \over { R_1 \cdot R_2\cdot(R_4 + R_5) +R_3 + R_4 \cdot R_5 \cdot (R_1 + R_2)}} \\
&= {{ R_1 \cdot R_2 \cdot (R_3 \cdot (R_4 + R_5) + R_4 \cdot R_5) } \over { R_1 \cdot R_2\cdot(R_4 + R_5) +R_3 + R_4 \cdot R_5 \cdot (R_1 + R_2)}} \\
  \end{align*}

==== Example with triangle-star transformation ====

With the triangle-star-transformation now also the initial example can be transformed. For more complicated circuits, the repeated triangle-star-transformation with subsequent combining of the resistors is useful, until the resulting circuit is easily calculable with node and mesh theorem (<imgref FigNo92>). Here a calculation is omitted - it is recommended to calculate here with intermediate results for the transformed resistors.

<WRAP right>
<imgcaption imageNo92 | example circuit conversion>
</imgcaption>
{{drawio>ExampleCircuitConverted}}
<WRAP>

==== Example with symmetries in the circuit ====

A certain special case concerns possible symmetries in circuits. If these3 are present, a further simplification can be made.

<WRAP right>
<imgcaption imageNo40| Example with symmetries in the circuit>
</imgcaption> <WRAP>.
{{url>https://www.falstad.com/circuit/circuitjs.html?running=false&ctz=CQAgjA7CAMB00IQVhvCEAcGGYMzQjABZoUkA2EFI3AKACcQBaSS15gJgg-AkunCIGzduxZ8QXHgLBCA7s3xFelJhywrwtAEqLoysblxsJAgSSowrcJLQVqNYqZujC13Te57OZQxuMpnL0kPXwQ3dmMMTg8oq1lw-.2cg5ziw1392IMjyaPTaAHNmbBA4phKiMH5aAGdOCw4LNWhojhb4kAAzAEMAGxqAUztOdrbogJdhtQam5NC3EqCStMFE4oEynLzVjPWQngmfHYWZCQn2fP8SsUXQ4-tGjeMYnjSpsdL9EQkfWvqn1QfJQdHr9Ib2YjKR7fQLzCFgbztcqnap1FhEKFNMAIyTtGRdPqDKaQ0rPCZveGvL7I-YwP7o5RGVTYqnKfGgokAD2Y5BQHFwyiYSFwklwUAM0W0tG5TF5pQ4KCFItw6hAEpAUpl+A2HAO6miKuo4El0r0An5YE4Gn5ynVms4HAAnKVHXrIC6jWATTKwAKXSKWLqXZQ7aaWNjJOQA47nRxyCHjRqw76UFVFQiRVVoqGZfyZORBRwveByOLE-bxJa2oKwEhYwQ1eXk-HPumkIz2jmRlCoyJXZGE8WpZsJGJRDizLQytczirwKrJ9PzXcjq4ZeQvnGA3XvC2u0KW-hFUh259s02ZUhCLjBSf62Wh2Go-nBeRi1UH97mMLU3KhfyS3PR9LwjLdmCIGMB0bYDvz9XBHW3IN4MHL9ZSdF0DjfKB4M9VCIAQUVLVla0-X3fCdUwjRDWg1CiDlaimCIYV5SA2i5RtcDmP5T8k38IgZ1Ufjl2kY4+IE79R1MUTwNuA4r1hESEl2RjllyCTKBWJS3CE2kVI2NT8iKPTSjU4zKmqfx5N0qzNL8HkPCCUtXgM6ShQkMonJM7YtMsn4PFlBy7h8+yeDETyji0+xcmiMKPAuHRmHwgwzggZ4LisCwUDMVBbHsJLPGizQwDcKyxCsiK7LckxVE89KtLiJlSRDKpJAoawpzUuYeHM+dtlcOIPiBL4WXa7k00kR0oEgbCCK7fBlXdXBa3lcwLyg3UoUwSQuBopMxowVMkBkSbwGYrsuGdSBWkdS0vUtOaWq4ZVVS4FC9tKOi1WgZ1cCQbMcXOjFSggPk2rFVj3t9ULIJJNpurWtoRXUc1lvUEU5sg37xQjYxVpg4xYzrUoMCrch4cfRhfu8RpSiQbwNEuFbwH7BU2H7Rmqcka1ue8uzcAwRkBaZ-m2WkkXhaF8X8gUMB+1l0KN22tlCmJwW1bVOddn1LnWioun4js+XWrZ142gNtZtc5jbacUuzWeN5nvDaxmqnzN2lfN1wZaB12S3d1xdARGLFcdNhFeyzLrBy4Yg798AfcV4rGHto37ftxnrd9rPw-uD2s7l9mY4LxFWjxYZrczlnnfLmmK7ao2vZ1pvrc5xu07arq6QUduTaZxuVWOp3AmrhQrdrnmu5tqfxYHyfOfns2peGK2qMXs3+7X809ekcvRlGWuabbwL6d1vJl83prL-755Gvn-XXCAA 1000,400 noborder}}
</WRAP>

~~PAGEBREAK~~~CLEARFIX~~~

<imgref BildNr40> shows on the left a symmetrical construction of a network of equal resistors $R$. For better understanding, in the middle of the same circuit additional switches and test points (TP) are installed, which indicate the voltage to ground.

The switches can be used to check whether a current flows if the respective nodes are connected. In the simulation it can be seen that this is not the case. In the symmetrical setup, these nodes are each at the same potential.

This also allows the circuit to take the form shown in <imgref FigNo40> on the right. This circuit is again easy to calculate:

\begin{align*}
R_g = R || R + R || R || R + R || R || R + R || R = {{1}\over{2}}\cdot R + {{1}\over{4}}\cdot R + {{1}\over{4}}\cdot R + {{1}\over{2}}\cdot R = 1.5\cdot R
  \{{align}}


~~PAGEBREAK~~~CLEARFIX~~~

<panel type="info" title="Task 2.7.1 Circuit Simplification Task I"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>

{{youtube>8nhzwwRYaUI}}

</WRAP></WRAP></panel>

<panel type="info" title="Task 2.7.2 Circuit Simplification Task II + III"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>

{{youtube>QqUQF3ky7gk}}
{{youtube>UmvFJbS21co}}

</WRAP></WRAP></panel>

<panel type="info" title="Task 2.7.3 Circuit Simplification Task IV"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>

{{youtube>SzXWWrPRsDU}}

</WRAP></WRAP></panel>

<panel type="info" title="Task 2.7.4 Circuit Simplification Task IV"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>

{{youtube>MhaO6kiB4dk}}

</WRAP></WRAP></panel>


<panel type="info" title="Task 2.7.5 Circuit Simplification Task V"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>

{{youtube>9eIRRUNba4A}}

</WRAP></WRAP></panel>


<panel type="info" title="Task 2.7.6 Circuit Simplification VI Task"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>

{{youtube>glzTvhIW-nk}}

</WRAP></WRAP></panel>

{{page>task_2.7.7_with_invoice&nofooter}}
{{page>task_2.7.8_with_invoice&nofooter}}
{{page>aufgabe_2.7.9&nofooter}}
{{page>aufgabe_2.7.10&nofooter}}


<panel type="info" title="other tasks"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>

More Anfgaben can be found online on the pages of [[https://www.eit.hs-karlsruhe.de/hertz/teil-b-gleichstromtechnik/zusammenschaltung-von-widerstaenden-und-idealen-quellen/uebungsaufgaben-zusammenschaltung-von-widerstaenden/berechnung-von-ersatzwiderstaenden.html|HErTZ]] (selection on the left in the menu).
</WRAP></WRAP></panel>
\\ \\ \\ \\
<WRAP group><WRAP half column>~REVEAL theme=dokuwiki&fade=convex&controls=1&show_progress_bar=1&build_all_lists=1&show_image_borders=0&horizontal_slide_level=2&enlarge_vertical_slide_headers=0&show_slide_details=0&open_in_new_window=1&size=1024x768~~
</WRAP><WRAP half column>

</WRAP></WRAP>
{{background>:capacitor-1835729_1280.jpg 30% 50%,90%}}
====== 0.Introduction to Electrical Engineering ======
\\ \\ \\ \\

---->>
===== 0.0 my imagination =====

---->
==== Resume ====
<WRAP hide>
{{drawio>Me01}}
<WRAP>

---->
==== Resume ====
<WRAP hide>
{{drawio>Me02}}
<WRAP>

---->
==== Resume ====
<WRAP hide>
{{drawio>Me03}}
<WRAP>

---->
==== Resume ====
<WRAP hide>
{{drawio>Me04}}
<WRAP>

---->
==== Resume ====
<WRAP hide>
{{drawio>Me05}}
<WRAP>

---->>
==== my-subjects ====
  * Electrical Engineering I/II
  * Basics Digital Technology
  * Electronic Circuitry
  * Electronics Lab
  * Microcontroller Technology
  * Electronic Systems

---->
==== other support services ====
  * laboratory work
  * Bachelor seminar papers
  * Bachelor Thesis
  * Master seminar papers
  * Master's thesis
  * Doctoral thesis

---->>
===== 0.0 Your presentation =====

---->
==== View in the round ====

<WRAP hide>
{{url>https://padlet.com/timffischer/Bookmarks 1000, 600 noborder}}
</WRAP>

---->
==== Coordinate origin ====

<WRAP hide>
{{url>https://padlet.com/timffischer/fzxz8rwofph6o00k 1000, 600 noborder}}
</WRAP>

---->>
===== 0.1 What does your future look like? =====

---->
{{drawio>Future01}}
{{drawio>Future01}}

---->
{{drawio>Subjects01}} (MR) ====
{{drawio>Subjects01}}

---->
{{drawio>Subjects02}} (MR) ====
{{drawio>Fan02}}

---->>
===== 0.2 What should you bring with you? =====

---->
==== General ====
<WRAP right>{{url>https://pixabay.com/get/5ee2dd454e54b10ff3d89960c62e377c1237d8ed5254_640.jpg 640, 352 noborder}}</WRAP>

  * Ability to engage in abstract questions
  * Motivation for lecture-based learning
  * The secret of ability lies in the will

~PAGEBREAK~ ~CLEARFIX~~
---->
==== math/physics =====
<WRAP right>{{url>https://pixnio.com/free-images/custom-size/pixnio-252485-450x300.jpg 450, 300 noborder}}</WRAP>

  * Understanding of physics
  * Vectors
  * Linear systems of equations / matrices
  * Differential and integral calculus
  * In ET2: complex numbers

~PAGEBREAK~ ~CLEARFIX~~
---->>
===== 0.3 Rework Sources =====

|Hagmann, Gert|**Fundamentals of Electrical Engineering**, \\ AULA-Verlag \\roughly on the same level as the course; covers ET1 and ET2||
| - |Online Simulator by [[http://www.falstad.com/circuit/circuitjs.html|Falstad]]** <notes>It is a good idea to illustrate electrical engineering principles via simulations. A nice possibility is offered by the Falstad Simulator. There, under the menu item "Circuits", a wide variety of setups can be found.</notes>|
|- |**MEXLE-Wiki**|

---->>
===== 0.4 Scared by the topics in the first week? =====

  * Use the [[https://www.hs-heilbronn.de/mr|Math Learning Center]] (further down the link). <notes> Contrary to what you might first think from the name, you can also go there for questions about "related subjects". There you will find students from higher semesters who can help you with homework and problems with lecture material. They can also give you tips on how to study. </notes>
  * Try to do as many assignments as possible.
  * Try to stick with it and study and read in a timely manner. The semester picks up quickly
  * Form study groups / join study groups. \ <fragment-block fade-in>BUT: first try the assignments yourself and get creative, then ask fellow students!</fragment-block>

---->>
===== 0.5 More about ET1 =====

---->
==== ILIAS course ====

  * The course for Electrical Engineering I can be found in [[https://ilias.hs-heilbronn.de|ILIAS]]: <wrap lo>\ Faculty of Mechanics and Electronics >> Mechatronics and Robotics (Bachelor) >> SPO 1 >> Basic Studies >> (134040) Electrical Engineering >> \\ [[https://ilias.hs-heilbronn.de/ilias.php?ref_id=103611&cmdClass=ilrepositorygui&cmdNode=sl&baseClass=ilrepositorygui|(134041) Electrical Engineering 1 - Prof. Dr. Tim Fischer]]</wrap>.

---->
==== Tutorials ====

  * 2 tutors
  * Discord
  * please give prompt feedback

---->
==== exam ====

  * Processing time: 60 minutes
  * Aids:
    * Calculator
    * 2 sheets of DIN A4 handwritten formulary
  * Note: There must be a legible and comprehensible calculation for each result.

---->>
===== 0.6 More about ET2 =====

---->
==== ILIAS course ====

  * The course for Electrical Engineering II can be found in [[https://ilias.hs-heilbronn.de|ILIAS]]: <wrap lo>\ Faculty of Mechanics and Electronics >> Mechatronics and Robotics (Bachelor) >> SPO 1 >> Grundstudium >> (134040) Elektrotechnik >> \\ [[https://ilias.hs-heilbronn.de/ilias.php?ref_id=22188&cmd=frameset&cmdClass=ilrepositorygui&cmdNode=sl&baseClass=ilrepositorygui|(134042) Elektrotechnik 2 - Prof. Dr. Tim Fischer]]</wrap>.
  * Joining the course is required!

---->
==== exam ====

  * Processing time: 120 minutes
  * Aids:
    * Calculator
    * 2 sheets of DIN-A4 handwritten formulary
  * Note: There must be a legible and comprehensible calculation for each result====== 1. basics and basic concepts ======

===== 1.1 Physical quantities =====

<callout>
=== Targets ===

After this lesson, you should:

  - know the basic physical quantities and the corresponding SI units.
  - know the most important prefixes. They can assign a power of ten to the respective abbreviation (G, M, k, d, c, m, µ, n).
  - be able to insert given numerical values and units into an existing quantity equation. From this you should be able to calculate the correct result using a calculator.
  - be able to assign the Greek letters.
  - always calculate with numerical value and unit.
  - Know that a referred quantity equation is dimensionless!
</callout>

<callout>
The KIT bridge course offers a similar introduction to [[https://lx3.mint-kolleg.kit.edu/onlinekursphysik/html/1.1.2/modstart.html|physical quantities]]
</callout>

==== Basic Sizes ====
<WRAP 50%>
Brief presentation of SI units
{{youtube>Fq0J-V4PUoc}}

<tabcaption tab01| SI units>

^ base quantity ^ name ^ unit sign ^ definition ^
| time | second | s | oscillation of a $Cä$ atom |
| length | meter | m | over s and speed of light |
| current | ampere | A | via s and elementary charge |
| mass | kilogram | kg | still via kg prototype |
| temperature | Kelvin | K | via triple point of water |
| amount of substance | mole | via number of $^{12}C$-nuclide |
| luminous intensity | candela | cd | via given radiant intensity |
</tabcaption>
</WRAP>
  * For the practical application of physical laws of nature, **physical quantities** are put into mathematical relationships.
  * There are basic quantities based on the SI system of units (French for Système International d'Unités), see below.
  * In order to determine the basic quantities quantitatively (quantum = lat. "how big"), **physical units** are defined, e.g. $metre$ for length.
  * In electrical engineering, the first three basic quantities (cf. <tabref tab01>) are particularly important. \ Mass is important for representing energy and power.
  * Any physical quantity is given by a product of **number value** and **unit**: \ e.g. $I = 2 A$
    * This is the short form of $I = 2\cdot 1A$.
    * $I$ is the physical quantity, here: electric current intensity
    * $\{I\} = 2 $ is the numerical value
    * $ [I] = 1 A$ is the (measurement) unit, here: Ampere

~~PAGEBREAK~~ ~~CLEARFIX~~
==== derived quantities, SI units and prefixes ====

<WRAP 50%>
<WRAP><WRAP half>
<tabcaption tab02| Prefixes I>
^ Prefix ^ Prefix character ^ Meaning ^
| Yotta | Y | $10^{24}$ |
| Zetta | Z | $10^{21}$ |
| Exa | E | $10^{18}$ |
| Peta | P | $10^{15}$ |
| Tera | T | $10^{12}$ |
| Giga | G | $10^{9}$ |
| Mega | M | $10^{6}$ |
| kilo | k | $10^{3}$ |
| hecto | h | $10^{2}$ |
| deca | de | $10^{1}$ |
</tabcaption>
</WRAP><WRAP half >
<tabcaption tab02| Prefixes II>
^ prefix ^ prefix character ^ meaning ^
| Dezi | d | $10^{-1}$ |
| centi | c | $10^{-2}$ |
| Milli | m | $10^{-3}$ |
| Micro | u, $\mu$ | $10^{-6}$ |
| nano | n | $10^{-9}$ |
| Piko | p | $10^{-12}$ |
| Femto | f | $10^{-15}$ |
| Atto | a | $10^{-18}$ |
| Zeppto | z | $10^{-21}$ |
| Yokto | y | $10^{-24}$ |
</tabcaption>
</WRAP></WRAP>
</WRAP>
  * In addition to basic quantities, there are also quantities derived from them, e.g. $1{{m}\over{s}}$.
  * SI units should be preferred for calculations. These can be derived **without a number factor** from the basic quantities.
    * The pressure unit bar ($bar$) is an SI unit.
    * BUT: The obsolete pressure unit atmospheric ($=1.013 bar$) is **__not__** an SI unit.
  * To prevent the numerical value from becoming too large or too small, it is possible to replace a decimal factor with a prefix (prefix). These are listed in the <tabref tab02>.


Example of power calculation

{{youtube>fwUyMBtdrvw}}

~PAGEBREAK~ ~CLEARFIX~~
==== physics-equations ====

  * Physical equations allow a connection of physical quantities.
  * There are two types of physical equations to distinguish:
    * quantity equations
    * normalized quantity equations (also called related quantity equations)

<WRAP><WRAP half>
<callout color="gray">

=== Equations of quantity ===
The vast majority of physical equations result in a physical unit that is not equal to $1$.
\\ \\

Example: Force $F = m \cdot a$ with $[F] = kg \cdot {{m}\over{s^2}}$
\\ \\

  * Unit control should **always** be used with quantity equations.
  * Equations of magnitude should generally be preferred

</callout>
</WRAP><WRAP half>
<callout color="gray">
=== Normalized size equations ===

With normalized quantity equations, the measured value or calculated value of a quantity equation is divided by a reference value.
The result is a dimensionless quantity relative to the reference value.

Example: Efficiency $\eta = {{P_{down}}\over{P_{up}}$

The reference value is often:
  * nominal values (maximum permissible value in continuous operation) or
  * maximum values (maximum value achievable in the short term)
are used.  

  * For normalized quantity equations, the units should **always** cancel out.

</callout>
</WRAP></WRAP>

<callout title="Example Calculation for a Size Equation">

Let a body with mass $m = 100kg$ be given. The body is lifted by the distance $s=2m$. \\
What work is done?
\\ \\
Physical equation:
<WRAP indent><WRAP indent>
Work = Force $\cdot$ Displacement
\W = F \cdot s \quad\quad\quad\;$ where $F=m \cdot g$
\\ $W = m \cdot g \cdot s \quad\quad$ with $m=100kg$, $s=2m$ and $g=9.81{{m}\over{s^2}}$
\cdot $W=100kg \cdot 9.81{{m}\over{s^2}} \cdot 2m $
\cdot $W = 100\cdot 9.81 \cdot 2 \;\; \cdot \;\; kg \cdot {{m}\over{s^2}} \cdot m$
\\\ $W = 1962 \quad\quad \cdot \quad\quad\; \left( kg \cdot {{m}\over{s^2}} \) \cdot m $
\cdot $W = 1962 Nm = 1962 J $
</WRAP></WRAP>

</callout>

==== Letters for physical quantities ====
 
<WRAP 50%>
<WRAP><WRAP half>
<tabcaption tab03| Greek letters>
^ uppercase-\ letters ^ lowercase-\ letters^ name ^
| $A$ | $\alpha$ | Alpha |
| $B$ | $\beta$ | Beta |
| $\Gamma$ | $\gamma$ | Gamma |
| $\Delta$ | $\delta$ | Delta |
| $E$ | $\epsilon$, $\varepsilon$ | Epsilon |
| $Z$ | $\zeta$ | Zeta |
| $H$ | $\eta$ | Eta |
| $\Theta$ | $\theta$, $\vartheta$ | Theta |
| $I$ | $\iota$ | Iota |
| $K$ | $\kappa$ | Kappa |
| $\Lambda$ | $\lambda$ | Lambda |
| $M$ | $\mu$ | My |
</tabcaption>
</WRAP><WRAP half >
<tabcaption tab03| Greek letters>
^ uppercase-\ letters ^ lowercase-\ letters^ Name ^
| $N$ | $\nu$ | Ny |
| $\Xi$ | $\xi$ | Xi |
| $O$ | $\omicron$ | Omicron |
| $\Pi$ | $\pi$ | Pi |
| $R$ | $\rho$, $\varrho$ | Rho |
| $\Sigma$ | $\sigma$ | Sigma |
| $T$ | $\tau$ | Tau |
| $\Upsilon$ | $\upsilon$ | Ypsilon |
| $\Phi$ | $\phi$, $\varphi$ | Phi |
| $X$ | $\chi$ | Chi |
| $\Psi$ | $\psi$ | Psi |
| $\Omega$ | $\omega$ | Omega |
</tabcaption>
</WRAP></WRAP>
{{youtube>UwNCixgrVzY}}

</WRAP>

In physics and electrical engineering, attempts have often been made to find letters for physical quantities that are close to the (English) term. \\
Thus $C$ for //**__C__**apacity//, $Q$ for //**__Q__**uantity// and $\varepsilon_0$ for the //**__E__**lectical Field Constant// and others can be explained.
Here it can already be seen that $C$ is used for the thermal capacity as well as for the electric capacity.

The Latin alphabet does not have enough letters for the scope of physics to avoid conflicts.
Therefore, for various physical quantities, Greek letters are used (see <tabref tab03>).

Especially in electrical engineering, upper/lower case is used to distinguish between.
  * is a constant (time-independent) quantity, \\ e.g. the period $T$
  * or a time-dependent quantity, \ e.g. the instantaneous voltage $u(t)$

The relevant Greek letters for electrical engineering are described in the following video.


~PAGEBREAK~ ~CLEARFIX~~
==== exercises ====

<panel type="info" title="Task 1.1.1 Conversions I - pre-calculated unit conversion example"> <WRAP><WRAP 2%>{{fa>pencil?32}}</WRAP><WRAP 92%>
{{youtube>xGyAw8MvxSA}}
</WRAP></WRAP></panel>

<panel type="info" title="Task 1.1.2 Conversions II"> <WRAP><WRAP 2%>{{fa>pencil?32}}</WRAP><WRAP 92%>
Convert the following values step by step:
  - A vehicle speed of 80 km/h in m/s
  - An energy of 60 joules into kWh (1 joule = 1 watt*second).
  - The number of electrolytically deposited single positively charged copper ions of 1.2 coulombs (a copper ion has the charge of about $1.6 \cdot 10^{-19} C$)
  - Absorbed energy of a smallest consumer if it consumes 1 µW uniformly in 10 days.
</WRAP></WRAP></panel>

<panel type="info" title="Task 1.1.3 Conversions III"> <WRAP><WRAP 2%>{{fa>pencil?32}}</WRAP><WRAP 92%>
Convert the following values step by step:
How many minutes could an ideal 10 kWh battery run a 3W load?
</WRAP></WRAP></panel>

<panel type="info" title="Task 1.1.4 Conversions IV"> <WRAP><WRAP 2%>{{fa>pencil?32}}</WRAP><WRAP 92%>
Convert the following values step by step:
How much energy does an average household use in a day if it consumes an average power of 500 W? How many candy bars (2000 kJ each) does that equal?
</WRAP></WRAP></panel>
====== 1. Basics and Basic Concepts ======

===== 1.1 Physical quantities =====

<callout>
=== Targets ===

After this lesson, you should:

  - know the basic physical quantities and the corresponding SI units.
  - know the most important prefixes. They can assign a power of ten to the respective abbreviation (G, M, k, d, c, m, µ, n).
  - be able to insert given numerical values and units into an existing quantity equation. From this you should be able to calculate the correct result using a calculator.
  - be able to assign the Greek letters.
  - always calculate with numerical value and unit.
  - Know that a referred quantity equation is dimensionless!
</callout>

<callout>
The KIT bridge course offers a similar introduction to [[https://lx3.mint-kolleg.kit.edu/onlinekursphysik/html/1.1.2/modstart.html|physical quantities]]
</callout>

==== Basic Sizes ====
<WRAP right 50%>
Short presentation of the SI units
{{youtube>Fq0J-V4PUoc}}

<tabcaption tab01| SI units>

^ base quantity ^ name ^ unit sign ^ definition ^
| time | second | s | oscillation of a $Cä$ atom |
| length | meter | m | over s and speed of light |
| current | ampere | A | via s and elementary charge |
| mass | kilogram | kg | still via kg prototype |
| temperature | Kelvin | K | via triple point of water |
| amount of substance | mole | via number of $^{12}C$-nuclide |
| luminous intensity | candela | cd | via given radiant intensity |
</tabcaption>
</WRAP>
  * For the practical application of physical laws of nature, **physical quantities** are put into mathematical relationships.
  * There are basic quantities based on the SI system of units (French for Système International d'Unités), see below.
  * In order to determine the basic quantities quantitatively (quantum = lat. "how big"), **physical units** are defined, e.g. $metre$ for length.
  * In electrical engineering, the first three basic quantities (cf. <tabref tab01>) are particularly important. \ Mass is important for representing energy and power.
  * Any physical quantity is given by a product of **number value** and **unit**: \ e.g. $I = 2 A$
    * This is the short form of $I = 2\cdot 1A$.
    * $I$ is the physical quantity, here: electric current intensity
    * $\{I\} = 2 $ is the numerical value
    * $ [I] = 1 A$ is the (measurement) unit, here: Ampere

~~PAGEBREAK~~ ~~CLEARFIX~~
==== derived quantities, SI units and prefixes ====

<WRAP right 50%>
<WRAP group><WRAP half column>
<tabcaption tab02| Prefixes I>
^ prefix ^ prefix character ^ meaning ^
| Yotta | Y | $10^{24}$ |
| Zetta | Z | $10^{21}$ |
| Exa | E | $10^{18}$ |
| Peta | P | $10^{15}$ |
| Tera | T | $10^{12}$ |
| Giga | G | $10^{9}$ |
| Mega | M | $10^{6}$ |
| kilo | k | $10^{3}$ |
| hecto | h | $10^{2}$ |
| deca | de | $10^{1}$ |
</tabcaption>
</WRAP><WRAP half column>
<tabcaption tab02| prefixes II>
^ prefix ^ prefix character ^ meaning ^
| Dezi | d | $10^{-1}$ |
| centi | c | $10^{-2}$ |
| Milli | m | $10^{-3}$ |
| Micro | u, $\mu$ | $10^{-6}$ |
| nano | n | $10^{-9}$ |
| Piko | p | $10^{-12}$ |
| Femto | f | $10^{-15}$ |
| Atto | a | $10^{-18}$ |
| Zeppto | z | $10^{-21}$ |
| Yokto | y | $10^{-24}$ |
</tabcaption>
</WRAP></WRAP>
</WRAP>
  * In addition to basic quantities, there are also quantities derived from them, e.g. $1{{m}\over{s}}$.
  * SI units should be preferred for calculations. These can be derived **without a number factor** from the basic quantities.
    * The pressure unit bar ($bar$) is an SI unit.
    * BUT: The obsolete pressure unit atmospheric ($=1.013 bar$) is **__not__** an SI unit.
  * To prevent the numerical value from becoming too large or too small, it is possible to replace a decimal factor with a prefix (prefix). These are listed in the <tabref tab02>.


Example of power calculation

{{youtube>fwUyMBtdrvw}}

~PAGEBREAK~ ~CLEARFIX~~
==== physics-equations ====

  * Physical equations allow a connection of physical quantities.
  * There are two types of physical equations to distinguish:
    * quantity equations
    * normalized quantity equations (also called related quantity equations)

<WRAP group><WRAP half column>
<callout color="gray">

=== Equations of quantity ===
The vast majority of physical equations result in a physical unit that is not equal to $1$.
\\ \\

Example: Force $F = m \cdot a$ with $[F] = kg \cdot {{m}\over{s^2}}$
\\ \\

  * Unit control should **always** be used with quantity equations.
  * Equations of magnitude should generally be preferred

</callout>
</WRAP><WRAP half column>
<callout color="gray">
=== Normalized size equations ===

With normalized quantity equations, the measured value or calculated value of a quantity equation is divided by a reference value.
The result is a dimensionless quantity relative to the reference value.

Example: Efficiency $\eta = {{P_{down}}\over{P_{up}}$

The reference value is often:
  * nominal values (maximum permissible value in continuous operation) or
  * maximum values (maximum value achievable in the short term)
are used.  

  * For normalized quantity equations, the units should **always** cancel out.

</callout>
</WRAP></WRAP>

<callout title="Example calculation for a size equation">

Let a body with mass $m = 100kg$ be given. The body is lifted by the distance $s=2m$. \\
What work is done?
\\ \\
Physical equation:
<WRAP indent><WRAP indent>
Work = Force $\cdot$ Displacement
\W = F \cdot s \quad\quad\quad\;$ where $F=m \cdot g$
\\ $W = m \cdot g \cdot s \quad\quad$ with $m=100kg$, $s=2m$ and $g=9.81{{m}\over{s^2}}$
\cdot $W=100kg \cdot 9.81{{m}\over{s^2}} \cdot 2m $
\cdot $W = 100\cdot 9.81 \cdot 2 \;\; \cdot \;\; kg \cdot {{m}\over{s^2}} \cdot m$
\\ $W = 1962 \quad\quad \cdot \quad\quad\; \left( kg \cdot {{m}\over{s^2}} \right) \cdot m $
\cdot $W = 1962 Nm = 1962 J $
</WRAP></WRAP>

</callout>

==== Letters for physical quantities ====
 
<WRAP right 50%>
<WRAP group><WRAP half column>
<tabcaption tab03| greek letters>
^ uppercase-\ letters ^ lowercase-\ letters^ name ^
| $A$ | $\alpha$ | Alpha |
| $B$ | $\beta$ | Beta |
| $\Gamma$ | $\gamma$ | Gamma |
| $\Delta$ | $\delta$ | Delta |
| $E$ | $\epsilon$, $\varepsilon$ | Epsilon |
| $Z$ | $\zeta$ | Zeta |
| $H$ | $\eta$ | Eta |
| $\Theta$ | $\theta$, $\vartheta$ | Theta |
| $I$ | $\iota$ | Iota |
| $K$ | $\kappa$ | Kappa |
| $\Lambda$ | $\lambda$ | Lambda |
| $M$ | $\mu$ | My |
</tabcaption>
</WRAP><WRAP half column>
<tabcaption tab03| Greek letters>
^ uppercase-\ letters ^ lowercase-\ letters^ Name ^
| $N$ | $\nu$ | Ny |
| $\Xi$ | $\xi$ | Xi |
| $O$ | $\omicron$ | Omicron |
| $\Pi$ | $\pi$ | Pi |
| $R$ | $\rho$, $\varrho$ | Rho |
| $\Sigma$ | $\sigma$ | Sigma |
| $T$ | $\tau$ | Tau |
| $\Upsilon$ | $\upsilon$ | Ypsilon |
| $\Phi$ | $\phi$, $\varphi$ | Phi |
| $X$ | $\chi$ | Chi |
| $\Psi$ | $\psi$ | Psi |
| $\Omega$ | $\omega$ | Omega |
</tabcaption>
</WRAP></WRAP>
{{youtube>UwNCixgrVzY}}

</WRAP>

In physics and electrical engineering, attempts have often been made to find letters for physical quantities that are close to the (English) term. \\
Thus $C$ for //**__C__**apacity//, $Q$ for //**__Q__**uantity// and $\varepsilon_0$ for the //**__E__**lectical Field Constant// and others can be explained.
Here it can already be seen that $C$ is used for the thermal capacity as well as for the electric capacity.

The Latin alphabet does not have enough letters for the scope of physics to avoid conflicts.
Therefore, for various physical quantities, Greek letters are used (see <tabref tab03>).

Especially in electrical engineering, upper/lower case is used to distinguish between.
  * is a constant (time-independent) quantity, \\ e.g. the period $T$
  * or a time-dependent quantity, \ e.g. the instantaneous voltage $u(t)$

The relevant Greek letters for electrical engineering are described in the following video.


~PAGEBREAK~ ~CLEARFIX~~
==== exercises ====

<panel type="info" title="Task 1.1.1 Conversions I - pre-calculated example of converting units"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>
{{youtube>xGyAw8MvxSA}}
</WRAP></WRAP></panel>

<panel type="info" title="Task 1.1.2 Conversions II"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>
Convert the following values step by step:
  - A vehicle speed of 80 km/h in m/s
  - An energy of 60 joules into kWh (1 joule = 1 watt*second).
  - The number of electrolytically deposited single positively charged copper ions of 1.2 coulombs (a copper ion has the charge of about $1.6 \cdot 10^{-19} C$)
  - Absorbed energy of a smallest consumer if it consumes 1 µW uniformly in 10 days.
</WRAP></WRAP></panel>

<panel type="info" title="Task 1.1.3 Conversions III"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>
Convert the following values step by step:
How many minutes could an ideal 10 kWh battery run a 3W load?
</WRAP></WRAP></panel>

<panel type="info" title="Task 1.1.4 Conversions IV"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>
Convert the following values step by step:
How much energy does an average household use in a day when consuming an average power of 500 W? How many candy bars (2000 kJ each) does that equal?
</WRAP></WRAP></panel>

===== 1.2 Introduction to the structure of matter =====

<WRAP><callout>
=== Objectives ===
After this lesson, you should:
  - know the magnitude of the elementary charge
</callout></WRAP>

==== Elementary charge ====

<WRAP right>
<imgcaption imageNo0 | atomic model after Bohr / Sommerfeld>
</imgcaption>
{{drawio>atom model }}
</WRAP>
 
  * Explanation of charge using the Bohr and Sommerfeld atomic models (see <imgref imageNo0>).
  * Atoms consist of
    * atomic nucleus (with protons and neutrons)
    * electron shell
  * electrons are carriers of the elementary charge $|e|$
  * elementary charge $|e| = 1.6022\cdot 10^{-19} C$
  * proton is the antagonist, i.e. has opposite charge
  * Sign is arbitrarily chosen:
    * electron charge: $-e$
    * proton charge: $+e$
  * all charges on/in bodies can only occur as integer multiples of the elementary charge
  * Due to the small numerical value of $e$, the charge is considered as a continuum when viewed macroscopically.

==== Conductivity ====
<WRAP group><WRAP column third>
<callout color="grey">
=== Conductors ===

Charge carriers are free to move in the conductor.
\\ \\ \\ \\
Examples:
  * Metals
  * Plasma

</callout>
</WRAP><WRAP column third>
<callout color="grey">
=== Semiconductor ===

In the semiconductor, charge carriers can be generated by heat and light irradiation. Often a small movement of the electrons is already possible by the room temperature.

Examples:
  * Silicon, diamond

</callout>
</WRAP><WRAP column third>
<callout color="grey">
=== Isolator ===

In the insulator, charge carriers are tightly bound to the atomic shells.
\\ \\ \\ \\

Examples:
  * many plastics and salts


</callout>
</WRAP></WRAP>

==== Exercises ====

<panel type="info" title="Task 1.2.1 Charges I"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>
How many electrons make up the charge of one coulomb?
</WRAP></WRAP></panel>

<panel type="info" title="Task 1.2.2 Charges II"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>
A balloon has a charge of $Q=7nC$ on its surface. How many additional electrons are on the balloon?
</WRAP></WRAP></panel>

===== 1.3 Effects of Electric Current =====
<WRAP><callout>
=== Objectives ===
After this lesson, you should:
  - Know that forces act between charges.
  - Know and be able to apply Coulomb's law.
</callout></WRAP>

  * What effects of electric current do you know?

==== first approximation to the el. charge ====

<WRAP right>
<imgcaption imageNo1 | Experiment 1 with two suspended charges >
</imgcaption>
{{drawio>Experiment1_Charges}}
<WRAP>
 
  * first experiment (see <imgref FigNo1>):
    * two charges ($Q_1$ and $Q_2$) are suspended at a distance $r$.
    * charges are generated by high voltage source and transferred to the two samples
  * Result
    * Test specimens with equal charges $\rightarrow$ repulsion
    * test specimens with charges of different sign $\rightarrow$ attraction
  * Findings
    * The forces cannot be explained mechanically
    * Two different types of charges seem to exist $\rightarrow$ positive (+) and negative (-) charge

==== Coulomb force ====

<WRAP right 50%>
Structure for your own experiments \\
{{url>https://phet.colorado.edu/sims/html/charges-and-fields/latest/charges-and-fields_de.html 500,400 noborder}} \\
Take a charge ($+1nC$) and position it. Measure the field across a sample charge (a sensor).

Experiment on Coulomb's law
{{youtube>mBYlnkm3gbE}}
</WRAP>

  * [[https://lx3.mint-kolleg.kit.edu/onlinekursphysik/html/1.4.1/modstart.html|Chapter 4.1.1 in KIT Bridge Course]]
  * Qualitative investigation using second experiment.
    * two charges ($Q_1$ and $Q_2$) at a distance $r$
    * additional measurement of force $F_C$ (e.g. via spring balance).
  * experiment results:
    * force increases linearly with larger charge $Q_1$ or $Q_2$ \ $ F_C \sim Q_1$ and $ F_C \sim Q_2$
    * force falls quadratically at greater distance $r$ \ $ F_C \sim {1 \over {r^2}}
    * with a proportionality factor $a$: \ $ F_C = a \cdot {{Q_1 \cdot Q_2} \over {r^2}}$
  * Proportionality factor $a$
    * The proportionality factor $a$ is defined in such a way that simpler relations arise in electrodynamics.
    * $a$ thus becomes: \\ $a = {{1} \over {4\pi\cdot\varepsilon}$
    * $\varepsilon_0$ is the {{wpde>electric field constant}}. In a vacuum, $\varepsilon_0 = \varepsilon$
  * The formula is similar to that of the gravitational force: $F_G = {\gamma \cdot {{m_1 \cdot m_2} \over {r^2}}$


<callout icon="fa fa-exclamation" color="red" title="Merke:">
The Coulomb force (in a vacuum) can be calculated by \\ $\boxed{ F_C = {{1} \over {4\pi\cdot\varepsilon_0}} \cdot {{Q_1 \cdot Q_2} \over {r^2}}}}$ \
Where $\varepsilon_0 = 8.85 \cdot 10^{-12} \cdot {{C^2 \over {m^2\cdot N}} = 8.85 \cdot 10^{-12} \cdot {{As} \over {Vm}}$
</callout>

~~PAGEBREAK~~ ~CLEARFIX~~
===== 1.4 Charge and Current =====

<WRAP><callout>
=== Objectives ===
After this lesson, you should:
  - Be able to distinguish the technical direction of current and the flow of electrons.
  - Be able to determine cathode and anode for components
  - Be able to apply the definition of current
</callout></WRAP>


<WRAP right>
The electric charge
{{youtube>JnYrmCaQfcM}}
</WRAP>

  * From previous experiments it is clear that there are two types of charge. In matter, these are:
    * (+) $\rightarrow$ excess of positive charges.
    * (-) $\rightarrow$ excess negative charges
  * further experiment:
    * (+) and (-) are connected by a conductor.
    * $\rightarrow$ electrons move from (-) pole to (+) pole
    * $\rightarrow$ electric current

~~PAGEBREAK~~ ~~CLEARFIX~~~
==== qualitative view ====
<WRAP right>
<imgcaption imageNo2 | part of a ladder >
</imgcaption>
{{drawio>charges_in_the_ladder}}
</WRAP>

  * In the thought experiment, let the following be given (see <imgref FigNo2>):
    * the above conductor with a control cross section $A$ perpendicular to the conductor.
    * the quantity of charges $\Delta Q = n \cdot e$, which in a certain period of time $\Delta t$, pass through the area $A$.
  * In the case of a uniform charge transport over a longer period of time, i.e. direct current (DC), the following applies:
    * The amount of charges per time flowing through the surface is constant: \{\Delta Q} \over {\Delta t}} = const. = I$
    * $I$ denotes here the strength of the direct current.
    * The unit of $I$ is the SI unit ampere: $1 A = {{1 C}\over{1 s}}$ . Thus, for the unit coulomb $1 C = 1 A\cdot s$.

<callout icon="fa fa-comment" color="blue" title="Definition of current intensity">.
The current of $1 A$ flows when an amount of charge of $1 C$ is transported through the cross-section of the conductor in $1 s$.

Before 2019: The current of $1 A$ flows when two parallel conductors, each $1m$ long and $1m$ apart, exert a force of $F_C = 0.2\cdot 10^{-6} N$ on each other.
 </callout>

<callout icon="fa fa-exclamation" color="red" title="Merke:">
 An electric current is the directed motion of free, electric charge carriers.
</callout>

==== Specifying the direction of the current ====
<WRAP right>
<imgcaption imageNo4 | Part of a conductor with different charged charges >
</imgcaption>
{{drawio>pos_neg_charges_in_conductor}}
</WRAP>

Charge transport can take place by (<imgref FigNo4>):
  * negative charge carriers $\color{midnightBlue}{\Delta Q_n}$ (e.g. electrons in a metallic conductor).
  * positive charge carriers $\color{brown}{\Delta Q_p}$ (e.g. certain semiconductor materials or in electrochemical cells)
  * positive and negative charge carriers (e.g. certain semiconductor materials, plasma)

The total cargo transported is
$\Delta Q = \color{brown}{\Delta Q_p} - \color{midnightBlue}{\Delta Q_n} = n_p \cdot e - n_n \cdot (-e)$

$\rightarrow$ The direction of the current must be determined independently of the direction of motion of the electric charge carriers.

<callout icon="fa fa-comment" color="blue" title="Definition of current direction (according to DIN5489)">
The current in a conductor from a cross-sectional area $A_1$ to a cross-sectional area $A_2$ is calculated to be positive if:
  * positive charge carriers move from $A_1$ to $A_2$, resp.
  * negative charge carriers move from $A_2$ to $A_1$.

The direction of current (or technical direction of current) is the sense of direction of the positive current, that is, the positive charge carriers.
</callout>

<WRAP right>
<imgcaption imageNo8 | electrodes on diode>
</imgcaption>
{{drawio>Diode_Electrodes}}
</WRAP>

<callout icon="fa fa-comment" color="blue" title="Definition of electrodes (in accordance with DIN5489)">
An electrode is a connection of an electrical component.
Usually electrodes are characterized by the fact that a change of material takes place (e.g. metal->semiconductor, metal->liquid).
  * **Anode**: Electrode at which the current enters the component.
  * **Cathode**: Electrode at which the current exits into the component.

As a mnemonic, you can remember the structure, shape, and electrodes of the diode (see <imgref imageNo8>).
</callout>

~PAGEBREAK~ ~CLEARFIX~~
==== exercises ====

<panel type="info" title="Task 1.4.1 Determining current from charge per time"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>

<WRAP right>
<imgcaption imageNo3 | time course of charge>
</imgcaption>
{{drawio>Time_course_of_charge}}
</WRAP>

Let the charge gain per time on an object be given.
  * From the diagram opposite (see <imgref Fig.No.3>), determine the currents and plot them on the diagram.
  * How could you proceed if the amount of charge on the object changes non-linearly?

</WRAP></WRAP></panel>

<panel type="info" title="Task 1.4.2 Electron flow"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>

How many electrons pass through a control cross section of a metallic conductor when the current of $40mA$ flows for $4.5s$?

</WRAP></WRAP></panel>

===== 1.5 Voltage, potential and energy =====

<WRAP><callout>
=== Objectives ===
After this lesson you should:
  - Be able to determine the energy gain of a charge when a voltage difference is overcome.
</callout></WRAP>

==== energetic approach ====

<WRAP right>
<imgcaption imageNo9 | Symbol image of a circuit >
</imgcaption>
{{drawio>symbol_image_circuit}}
</WRAP>

Given an electrical conductor ("load") on a battery (see <imgref FigNo9>).
  * $\rightarrow$ current flows
  * Similar to the transport of a mass in the gravitational field, energy is needed to transport the charge in the "voltage field"
  * We will look at the specific electric field later in the semester.
  * A point charge $q$ is moved from electrode ① to electrode ② \ Which is similar to a moving point of mass in the gravitational field.
  * $\rightarrow$ there is a turnover of energy.
  * The energy turnover is proportional to the amount of charge $q$ transported.
  * In many cases, characterize the "energetic path" from ① to ② **charge-independent**: \ $\boxed{{W_{1,2}}\over{q}} = U_{1,2}}$
  * In English, $V$ (for Voltage) is often used to denote the quantity: * E.g.
    * $VCC = 5V$ : voltage supply of an IC (__V__oltage __C__ommon __C__ollector),
    * $V_{S+} = 15V$ : voltage supply of an operational amplifier (__V__oltage __S__upply plus).

~PAGEBREAK~ ~CLEARFIX~~
==== CompareMechanicstoElectrics====

<WRAP group><WRAP half column>
<callout color="grey">
<WRAP right>
<imgcaption imageNo5 | mechanical potential>
</imgcaption>
{{drawio>mechanical_potential}}
<WRAP>
=== Mechanics ===

**Potential energy**

Potential energy is always related to a reference level.

The energy required to move $m$ from $h_1$ to $h_2$ is independent of the reference level.

$\Delta W = W_1 - W_2 = m \cdot g \cdot (h_1 - h_2)$
</callout>
</WRAP><WRAP half column>
<callout color="grey">
<WRAP right>
<imgcaption imageNo6 | electrical potential>
</imgcaption>
{{drawio>electric_potential}}
<WRAP>
=== Electrical===

**Potential**

The potential $\varphi$ is always fixed relative to a reference point.

Common is:
  * ground potential (earth, ground, ground).
  * infinitely distant point

To shift the charge, the potential difference must be overcome. The potential difference is independent of the reference potential.
$\boxed{\delta W_{1,2} = W_1 - W_2 = Q \cdot \varphi_1 - Q \cdot \varphi_2 = Q \cdot (\varphi_1 - \varphi_2)}$
</callout>
</WRAP>
\\ \\
From this we get: \\
$\boxed{{\Delta W_{1,2} \over {Q}} = \varphi_1 - \varphi_2 = U_{1,2}}$

<callout icon="fa fa-exclamation" color="red" title="Merke:">
  * Voltage is always a potential difference.
  * The unit of voltage is volts: $1 V$.
</callout>

<callout icon="fa fa-comment" color="blue" title="Definition of Voltage">.
A voltage of $1 V$ is present between two points if a charge of $1 C$ undergoes a change in energy of $1J = 1Nm$ between those two points.

From $W=U \cdot Q$ we also get the unit: $1Nm = 1V\cdot As \rightarrow 1V = 1{{Nm}\over{As}}$
</callout>

==== Voltage between two points ====

For the tension between two points the following definition results with the previous knowledge:

$U_{12} = \varphi_1 - \varphi_2 = -U_{21} = - (\varphi_2 - \varphi_1)$

Therefore, the order of the indices must always be observed in the following.

<callout icon="fa fa-comment" color="blue" title="Definition of the conventional direction of voltage (nacg DIN5489) ">
The voltage of $U_{12}$ along a path from point ① to ② becomes positive if the potential at ① is greater than the potential at ②.
</callout>

==== Exercises ====

<panel type="info" title="Task 1.5.1 Direction of stress"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>

<WRAP right>
<imgcaption imageNo21 | example of conventional voltage specification>
</imgcaption>
{{drawio>ExampleConventionalVoltageSpecification}}
</WRAP>

For the voltages $U_{Batt}$, $U_{12}$, and $U_{21}$ in <imgref FigNo21>, indicate whether they are positive or negative according to the voltage definition.
~~PAGEBREAK~~ ~CLEARFIX~~
</WRAP></WRAP></panel>


===== 1.6 Resistance and conductance =====

<WRAP><callout>
=== Objectives ===
After this lesson you should:

  - know and be able to apply Ohm's law.
  - be able to calculate the resistance from the specific resistance.
  - be able to determine the conductance from the resistance or the specific conductivity.
  - know which cases of temperature dependence are distinguished and how they are named.
  - be able to calculate the resistance at different temperatures.
  - know that there are different types of construction and that the physical value of the resistance does not depend on the geometric value.

</callout></WRAP>

<WRAP right 60%>

Illustrative explanation of ohmic resistance
{{youtube>IyFJqQa_Dfw}}

<WRAP group><WRAP column>
<imgcaption imageNo11 | Resistance as Second Gate>
</imgcaption>
{{drawio>Resistance_second_port}}

</WRAP><WRAP column>
{{url>https://www.falstad.com/circuit/circuitjs.html?running=false&ctz=CQAgjCAMB0l3BWK0AsKAcYBs6CckB2SAZmPUixSxAUhpsgCgAnEAJgWuzo+rfXRRwcRmAJtwuCd3AEuWOnQL0ki5EgBqAewA2AFwCGAcwCmjI+0iD+1qyGKQUURsSKTpC9mycyIixgDultYC7JzsoUwARuyk4ChIbLjUDtRMQWze4J682f5AA 350,300 noborder}}
</WRAP></WRAP></WRAP>

Current flow generally requires energy input. This energy is extracted from the electric circuit and is usually converted into heat.
The reason for this is the resistance of the conductor.

A resistor is an electrical component with two terminals. Components with two terminals are called two-terminal or one-terminal (<imgref FigNo11>). In the second semester, four-terminal or two-terminal components will also be added.

In general, the cause-and-effect relationship is such that an applied voltage across the resistor produces the current flow. However, the reverse relationship also applies: as soon as an electric current flows across a resistor, a voltage drop is generated across the resistor.

In electrical engineering, circuit diagrams work with idealized components. The resistance of the supply lines is either neglected - if it is very small compared to all other resistance values - or drawn in by a separate resistor.

~~PAGEBREAK~~ ~CLEARFIX~~
==== Linearity of resistors ====
<WRAP group><WRAP half column>
<callout color="grey">
=== Linear resistors ===
<imgcaption imageNo13 | linear resistors in U-I diagram>
</imgcaption>
{{drawio>linear_resistor_UI}}

  * For linear resistors, the resistance value $R={{U_R}\over{I_R}}=const.$ and thus is independent of $U_R$.
  * This results in **Ohm's law**: \ $\boxed{R={{U_R}\over{I_R}}$ with unit $[R]={[U_R]}\over{[I_R]}}= 1{{V}\over{A}}= 1\Omega$
  * In <imgref FigNo13>, the value $R$ can be read from the course of the straight line $R={{\Delta U_R}\over{\Delta I_R}}$
  * The reciprocal value of the resistance is called conductance: $G={{1}\over{R}}$ with the unit $1 S$ (Siemens). This value can be seen as a slope in the $U$-$I$ diagram.

</callout>

</WRAP><WRAP half column>
<callout color="grey">
=== Nonlinear resistors ===
<imgcaption imageNo14 | nonlinear resistors in U-I diagram>
</imgcaption>
{{drawio>nonlinear_resistor_UI}}

  * The point in the $U$-$I$-diagram, which is taken in a system, is called **working point** or **operating point**. In <imgref FigNo14>, a **working point** is marked with a circle in the left-hand diagram.
  * For non-linear resistors, the resistance value is $R={{U_R}\over{I_R(U_R)}}=f(U_R)$. This resistance value depends on the operating point.
  * Often small changes around the operating point are interesting (e.g. for small disturbances of load machines). For this, the **differential resistance** $r$ (also dynamic resistance) is determined: \ $\boxed{r={{dU_R}\over{dI_R}}\approx{{\delta U_R}\over{\delta I_R}}$ with unit $[R]=1\Omega$.
  * As with the resistor $R$, the reciprocal of the differential resistance $r$ is the differential conductance $g$.
  * In <imgref FigNo14>, the differential conductance $g$ can be read from the slope of the straight line at each point $g={{dI_R}\over{dU_R}}$
</callout>

</WRAP></WRAP>

==== Resistance as a material property ====

<WRAP right 50%>
Illustrative explanation of resistivity
{{youtube>_F-Y_I8DJEE}}
</WRAP>

The resistance value can also be derived from the geometry of the resistor. For this purpose, an experiment can be carried out with resistors of different shapes. Thereby it can be stated:
  * the resistance value $R$ increases proportionally with the length $l$ that the current has to travel: $R \sim l$
  * the resistance value $R$ decreases inversely proportional with the cross-sectional area $A$ through which the current passes: $R \sim {{1}\over{A}}$
  * the resistance value $R$ depends on the material (<tabref tab04>)
  * this gives: \\ $R \sim {{l}\over{A}}$
 
~~PAGEBREAK~~~CLEARFIX~~~
<WRAP right 30%>
<tabcaption tab04| resistivity for different materials>
^ material ^ $\rho$ in ${{\Omega\cdot {mm^2}}\over{m}}$ ^
| silver | $1.59\cdot 10^{-2}$ |
| copper | $1.79\cdot 10^{-2}$ |
| aluminum | $2.78\cdot 10^{-2}$ |
| gold | $2.2\cdot 10^{-2}$ |
| lead | $2.1\cdot 10^{-1}$ |
| graphite | $8\cdot 10^{0}$ |
| battery acid (lead acid battery) | $1,5\cdot 10^4$ |
| blood | $1.6\cdot 10^{6}$ |
| (tap) water | $2 \cdot 10^{7}$ |
| paper | $1\cdot 10^{15} ... 1\cdot 10^{17}$ |

</tabcaption>
</WRAP>

<callout icon="fa fa-exclamation" color="red" title="Merke:">
The resistance can be calculated using \ $\boxed{R = \rho \cdot {{l}\over{A}} } $
  * $\rho$ is the material dependent, **specific resistance** with the unit: $[\rho]={[R]\cdot[A]}\over{l}}=1{{\Omega\cdot m^{\not{2}}\over{\not{m}}=1 \Omega\cdot m$
  * Often, instead of $1 \Omega\cdot m$, the unit $1 {{\Omega\cdot {mm^2}}\over{m}}$ is used. It holds that $1 {{\Omega\cdot {mm^2}}\over{m}}= 10^{-6} \Omega m$
</callout>

  * There exists also a specific conductance $\kappa$, given by the conductance $G$ : $G= \kappa \cdot {{A}\over{l}}$
    * The specific conductance $\kappa$ is the reciprocal of the specific resistance $\rho$: $\kappa$

==== Temperature dependence of resistors ====

<WRAP right 50%>
Explanation of the temperature dependence of resistors
{{youtube>Xw7QXJ9sV6s}}

<WRAP group><WRAP column>
<imgcaption imageNo15 | Circuit for measuring the effect of temperature on a resistor>
</imgcaption>
{{drawio>Resistor_Temperature_Circuit}}
</WRAP></WRAP>

<WRAP group><WRAP column>
<imgcaption imageNo16 | Effect of temperature on resistance>
</imgcaption>
{{drawio>Resistance_Temperature}}
</WRAP></WRAP>
</WRAP>

The resistance value is - apart from the influences of geometry and material mentioned so far - also influenced by other effects. These are e.g.:
  * Heat (thermoresistive effect, e.g. in the resistance thermometer).
  * Light (photoresistive effect, e.g. in the component photo resistor)
  * Magnetic field (magnetoresistive effect, e.g. in hard disks)
  * pressure (piezoresistive effect e.g. tire pressure sensor)
  * chemical environment (chemoresistive effect e.g. chemical analysis of breathing air)

In order to summarize these influences in formulas, the mathematical tool of the {{wpde>Taylor series}} is often used. This will be used here practically on the basis of the thermoresistive effect. The thermoresistive effect, or the {{wpde>temperature coefficient|temperature dependence of resistors}} is one of the most common (disturbance) influences in components.

The starting point for this should again be an experiment. The ohmic resistance is to be determined as a function of temperature. For this purpose, the resistance is determined by means of a voltage source, a voltmeter (voltmeter) and an ammeter (ammeter) and the temperature is changed (<imgref BildNr15>).

This results in a plot of resistance $R$ versus temperature $\vartheta$ as shown in <imgref FigNo16>. These are approximated to a first approximation by a linear curve about an operating point. This results in:

$R(\vartheta) = R_0 + c\cdot (\vartheta - \vartheta_0)$

  * The constant is replaced here by $c = R_0 \cdot \alpha$
  * $\alpha$ is here the **linear resistance-temperature coefficient** with the unit: $ [\alpha] = {{1}\over{[\vartheta]}} = {{1}\over{K}} $
  * Besides the linear term, it is also possible to increase the accuracy of the calculation of $R(\vartheta)$ with higher exponent of the temperature influence. This approach is considered again in more detail in mathematics under {{wpde>potency series}}.
  * These resistance-temperature coefficients are described by Greek letters: $\alpha$, $\beta$, $\gamma$, ...

~~PAGEBREAK~~~CLEARFIX~~~
<callout icon="fa fa-exclamation" color="red" title="Merke:">

<WRAP right 30%>
<imgcaption imageNo22 | Effect of temperature on resistance>
</imgcaption>
{{drawio>resistance_temperature_coefficient}}
<WRAP>

The temperature dependence of a resistor is described by the following equation:
$\boxed{ R(\vartheta) = R_0 (1 + \alpha \cdot (\vartheta - \vartheta_0) + \beta \cdot (\vartheta - \vartheta_0)^2 + \gamma \cdot (\vartheta - \vartheta_0)^3 + ...}$

Where:
  * $\alpha$ the **(linear) resistance-temperature coefficient** with unit: $ [\alpha] = {{1}\over{K}} $
  * $\beta$ the **(quadratic) resistance-temperature coefficient** with unit: $ [\beta] = {{1}\over{K^2}} $
  * $\gamma$ the **resistance-temperature coefficient** with unit: $ [\gamma] = {{1}\over{K^3}} $
  * $\vartheta_0$ the given reference temperature (or reference temperature), usually $0°C$ or $25°C$.

The further the temperature range deviates from the reference temperature, the more temperature coefficients are required to replicate the actual curve (<imgref FigNo22>).
~~PAGEBREAK~~ ~CLEARFIX~~
</callout>

<callout icon="fa fa-info" color="grey" title="Outlook">
In addition to specifying the parameters $\alpha$,$\beta$, ... the specification of $R_{25}$ and $B_{25}$ can occasionally be found. This is a different variant of approximation, which refers to the temperature of $25°C$. This is based on the {{wpde>Arrhenius equation}}, which in chemistry links reaction kinetics to temperature. For NTCs, the Arrhenius equation links the inhibition of charge carrier motion by lattice vibrations to temperature $R(T) \sim e^{{B}\over{T}} $

Again, a series expansion can be applied: $R(T) \sim e^{A + {{B}\over{T}} + {{C}\over{T^2}} + ...}$

However, often only $B$ is given. \\ By taking the ratio of any temperature $T$ and $T_{25}=298.15 K$ ($\hat{=} 25°C$) we get:
${R(T)}\over{R_{25}} = {{exp \left({{B}\over{T}}\right)} \over {exp \left({{B}\over{298.15 K}}\right)}} $ with $R_{25}=R(T_{25})$

This allows the final formula to be determined:

$R(T) = R_{25} \cdot exp \left( B_{25} \cdot \left({{1}\over{T}} - {{1}\over{298.15 K}} \right) $

</callout>
=== Types of temperature dependent resistors ===

In addition to temperature dependence as a disturbance influence, there are also components which have been deliberately bred for a specific temperature influence. These are called thermistors (a portmanteau of __therm__ally-sensitive res__istor__). Thermistors are basically divided into hot conductors and cold conductors.

A special form are materials which have been explicitly optimized for minimum temperature dependence (e.g. Constantan or Isaohm).

<WRAP group><WRAP half column><callout color="grey">
=== Thermistors===

<imgcaption imageNo17 | Hot conductor in U-I diagram>
</imgcaption>
{{drawio>Hot_conductor_UI}}

  * As the name suggests, the {{wpde>thermistor}} has a lower resistance at higher temperatures.
  * A thermistor is also called an NTC (for __n__egative __t__emperature __c__oefficient).
  * Examples are semiconductors
  * Applications are inrush current limiters and temperature sensors. For the desired operating point, a strongly non-linear curve is selected there (e.g. fever thermometer).

</callout></WRAP><WRAP half column><callout color="grey">
=== PTC thermistors===

<imgcaption imageNo18 | PTC thermistor in U-I diagram>
</imgcaption>
{{drawio>PTC thermistor_UI}}

  * The {{wpde>pTC thermistor}} shows a higher resistance at higher temperatures.
  * A PTC thermistor is also called a PTC (__p__positive __t__temperature __c__oefficient).
  * Examples are doped semiconductors or metals.
  * Applications are temperature sensors. For this they often offer a wide temperature range and good linearity (e.g. PT100 in the range $-100°C$ to $200°C$).
  * [[https://www.geogebra.org/m/VVA2YUJQ#material/EQQm5kbT|Interactive example]] on PTC thermistors.
</callout>

</WRAP></WRAP>

==== Resistor designs ====

<WRAP right 50%>
{{youtube>m3jagiGFcoY?start=156}}
</WRAP>

The designs are not explained in detail here. Please refer to the video on the right.


~PAGEBREAK~ ~CLEARFIX~~
==== exercises ====

<panel type="info" title="Task 1.6.1 pre-calculated resistivity example"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>
{{youtube>UbWsbwqQc-E}}
~~PAGEBREAK~~ ~CLEARFIX~~
</WRAP></WRAP></panel>

<panel type="info" title="Task 1.6.2 Resistance of a pencil stroke"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>

Assume that a soft pencil lead is 100% graphite. What is the resistance of a $5cm$ long and $0.2mm$ wide line if it has a height of $0.2\mu m$?

The spec. resistance is given by <tabref tab04>.

</WRAP></WRAP></panel>

<panel type="info" title="Task 1.6.3 Resistance of a cylindrical coil"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>

Let a cylindrical coil in the form of a multi-layer winding be given, as can also occur in motors, for example. The cylindrical coil has an inner diameter of $d_i=70mm$ and an outer diameter of $d_a = 120mm$. The number of turns is $n_W=1350$ turns, the wire diameter $d=2.0mm$ and the specific conductivity of the wire $\kappa_{Cu}=56 \cdot 10^6 {{S}\over{m}}$.

First calculate the wound wire length and then calculate the ohmic resistance of the entire coil.
</WRAP></WRAP></panel>

<panel type="info" title="Task 1.6.4 Resistance of a Lead Wire"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>

The supply line to a consumer is to be replaced. Due to the application, the conductor resistance must remain the same.
  * The old aluminium supply line had a specific conductivity $\kappa_{Al}=33\cdot 10^6 {{S}\over{m}}$ and a cross-section $A_{Al}=115mm^2$.
  * The new copper feed line has a specific conductivity $\kappa_{Cu}=56\cdot 10^6 {{S}\over{m}}$

Which wire cross-section $A_{Cu}$ must be selected ?

</WRAP></WRAP></panel>

<panel type="info" title="Task 1.6.5 Strain gauges"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>

t.b.d.

</WRAP></WRAP></panel>

{{page>task_1.7.6_with_calculation&nofooter}}


===== 1.7 Power and efficiency =====

<WRAP><callout>
=== Objectives ===
After this lesson, you should:
  - Be able to calculate the electrical power and energy across a resistor.
</callout></WRAP>

==== Determining the electrical power in a direct current circuit ====
From chapter [[#1.5 Voltage, Potential and Energy]] it is known that a movement of a charge across a potential difference corresponds to a change in energy. So charge transport automatically means energy expenditure. Often, however, the energy expenditure per time unit is of interest.

<WRAP right 30%>
<imgcaption imageNo19 | History of power and energy>
</imgcaption>
{{drawio>PowerEnergy}}

<imgcaption imageNo20 | source and consumers>
</imgcaption>
{{drawio>SourceConsumer}}

</WRAP>

The energy expenditure per unit of time represents the power: \\
$\boxed{P={\Delta W}\over{\Delta t}}$ with the unit $[P]={[W]}\over{[t]}}=1 {{J}\over{s}} = 1 {{Nm}\over{s}} = 1 V\cdot A = 1 W$

For a constant power $P$ and an initial energy $W(t=0)=0$ holds: \\
$\boxed{W=P \cdot t}$ \
If the above restrictions do not apply, the generated/needed energy must be calculated via an integral.

In addition to current flowing from the source to the load (and back), power also flows from the source to the load.

Considering only __a DC circuit__, the following energy is converted between the terminals (see also <imgref FigNo19> and <imgref FigNo20>): \\
$W=U_{12}\cdot Q = U_{12} \cdot I \cdot t$

This gives the power (i.e. energy converted per unit time): \\
$\boxed{P=U_{12} \cdot I}$ with the unit $[P]= 1 V\cdot A = 1W \quad$ ... $W$ here stands for watt.

For ohmic resistors applies:

$\boxed{P=R\cdot I^2 = {{U_{12}^2}\over{R}}$

==== Nominal values of ohmic loads ====

^ Name of the nominal quantity ^ Physical quantity ^ Description ^
| rated power | $P_N$ | $P_N$ is the power output of a device (consumer or generator) permissible in continuous operation |
| Rated current | $I_N$ | $I_N$ is the current occurring during operation at rated power |
| rated voltage | $U_N$ | $U_N$ is the voltage occurring in operation with rated power |


==== Efficiency ====

<WRAP right 30%>
<imgcaption imageNo23 | power flow diagram>
</imgcaption>
{{drawio>Power Flow}}

</WRAP>

The usable (= outgoing) $P_A$ power is always smaller than the supplied (incoming) power $P_E$. The difference is called power loss $P_V$. It is thus valid:

$P_E = P_A + P_V$

Instead of the power loss $P_V$ the efficiency $\eta$ is often given:

$\boxed{\eta = {{P_{A}}\over{P_{E}}\overset{!}{<} 1}$

For systems connected in series (see <imgref FigNo23>), the total resistance is given by:

$\boxed{\eta = {{P_{A}}\over{P_{E}} = {\not{P_{1}}\over{P_{E}}\cdot {\not{P_{2}}\over \not{P_{1}}\cdot {{P_{A}}\over \not{P_{2}} = \eta_1 \cdot \eta_3 \cdot \eta_3}$

~PAGEBREAK~ ~CLEARFIX~~
==== exercises ====

<panel type="info" title="Task 1.7.1 pre-calculated example on electrical power and energy"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>
{{youtube>c31qvyXKpNc}}
~~PAGEBREAK~~ ~~CLEARFIX~~
</WRAP></WRAP></panel>

<panel type="info" title="Task 1.7.2 Performance"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>

A SMD resistor is used on a circuit board for current measurement. The resistance value should be $R=0,2\Omega$, the maximum power $P_N=250 mW$.

What is the maximum current that can be measured?

</WRAP></WRAP></panel>

<panel type="info" title="Task 1.7.3 Power dissipation and efficiency I"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>

<WRAP right>
<imgcaption imageNo29 | sketch of the structure>
</imgcaption>
{{drawio>sketchbatterymonitor}}
<WRAP>

  * The battery monitor BQ769x0 measures the charge and discharge currents of a lithium-ion battery by means of the voltage across a measuring resistor (shunt). In <imgref FigNo29>, the analog-to-digital converter ($ADC$) of this chip is connected to shunt $R\_S$ via the circuit board. Through the shunt, the discharge current flows from the battery terminal $BAT+$ to $OUT+$ and back to $BAT-$ via $OUT-$. The shunt shall be designed so that the bipolar measurement signals have a voltage level in the range of $-0.20 V$ to $+0.20 V$. The analog-to-digital converter has a resolution of $15uV$. The currents can be used to count the charge in the battery to determine the state of charge (SOC).
    * Draw an equivalent circuit with voltage source (battery), measuring resistor and load resistor $R_L$. Also draw the measurement voltage and load voltage.
    * The shunt should have a resistance value of $1m\Omega$. What maximum charge/discharge currents are still measurable? What minimum current change is measurable?
    * What is the power dissipation at the shunt in the extreme case?
    * Now the efficiency is to be calculated
      * Find the efficiency as a function of $R\_S$ and $R_L$. Note that the same current flows through both resistors.
      * Special task: let the battery have a nominal voltage of $10V$ (3 cells) and let the maximum discharge current flow. What are the efficiencies obtained from the measurement alone?
</WRAP></WRAP></panel>

<panel type="info" title="Task 1.7.4 Power Dissipation and Efficiency II"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>

A water pump ($\eta_P = 60\%$) has an electric motor drive ($\eta_M=90\%$).
The pump is to pump up $500l$ of water $12m$ per minute.

  * What is the power rating of the motor?
  * What current does the motor draw on the $230V$ mains?
</WRAP></WRAP></panel>

~PAGEBREAK~ ~CLEARFIX~~
===== CONTINUED =====

  - [[http://omegataupodcast.net/303-das-si-system-der-einheiten/|Omega Tau No. 303]] : Podcast with a BTP researcher on the evolution of the SI unit system.

====== 3. Linear sources and dipoles ======


It is known from everyday life that battery voltages drop under heavy load. This can be seen, for example, when turning the ignition key in winter: The load from the starter motor is sometimes so great that the low beam or radio briefly cuts out. \\
Another example are $1.5V$ batteries: If such a battery is short-circuited by a piece of wire, not so much current flows that the piece of wire glows, but noticeably less.

So it makes sense here to develop the concept of the ideal voltage source further.
In addition, we will see that this also opens up a possibility to convert and simplify more complicated circuits.

<WRAP right>
<imgcaption imageNo1 | passive two-pole>
</imgcaption>
{{drawio>PassiveTwoPole}}
</WRAP>

First, the concept of the two-terminal from the chapter [[basics_and_basic_concepts?#resistance_and_conductance|basics and basic concepts]] is to be expanded (<imgref BildNr1>).
  - As **passive two-terminal** in the following a two-terminal is called, which acts exclusively as a consumer. Thus it is valid for the passive two-terminal that the current-voltage-characteristic always runs through the origin (see also chapter [[simple_dc-circuits#consumer|simple circuits]]).
  - **Active two-terminal circuits**, on the other hand, also act as generators of electrical energy. Thus, the current-voltage characteristic there __does not__ pass through the origin. Active dipoles always contain at least one source (i.e. at least one current or voltage source).

~~PAGEBREAK~~~CLEARFIX~~~

===== 3.1 linear sources =====

<callout>

=== Goals ===

After this lesson, you should:

  - Be able to describe the difference between an ideal and a linear voltage or current source.
  - Know and be able to apply the relationship between output voltage, source voltage $U_q$ and internal resistance $R_i$.
  - know and be able to apply the relationship between the current supplied, the source current $I_q$ and the internal conductance $G_i$.
  - be able to represent the voltage curve of the linear (voltage/current) source using open-circuit voltage and short-circuit current.
  - be able to determine the open-circuit voltage and the short-circuit current using two current/voltage measuring points.
  - Be able to explain the reason for the duality of current and voltage sources.
  - Be able to convert a linear current source into a linear voltage source and vice versa.

</callout>

==== practical example of a realistic source ====

<WRAP right>
<imgcaption imageNo2 | Battery model with load resistor>
</imgcaption> <WRAP>.
{{url>https://www.falstad.com/circuit/circuitjs.html?running=false&ctz=CQAgjCAMB0l3AWAnC1b0DYrQOwGYAOJSAVg0iSL0gxIjyzGhJATwChoQAhAQwBd+AUwBOATyjgQAJhkgAOgAcpkSQFpJqiACUhAZwCWe-gHsRAUQA2AW3l6wd6fJEA1E5f68A5kKu29kI5Q8nYkdoHhIXoIEfZhkQkBzOwIYFrkMpAE4DhY0jiyEHyCohKBSSzl5TF2TGEBkRXsykiMuSCtMgUqUOwA7uBIsvmFYMPdkP3g0tlg7WBDOViTIiB447LrMtIImtBgU51zWFvHvQOdI2sb5x153WBjXbKTXoM3C8ME2ZN4OFozJZ3IEQVSTPTXIFbaQ7TQgABmvEseiE7AhajAbSwGLysK0CKRKKmp3aOKBkwGZJhuypEymtM2N0mAA8QAQ8DlVEgWHMObswLsAJLyACO7FZaWGAumqmkc1Y4F2AFVReLWHk8BAyDI8Lt+bttGqSNlpIQQLQZKkFdKADLsExSUGsTEcjDQaQkIRqJBrd1en5yH0cM27MH7cCqXSGYy8AB2AGNUUA 600,500 noborder}}

</WRAP>


For the ideal voltage source it was defined that it always supplies the same voltage independent of the load. In <imgref FigNo2>, in contrast, an example of a "realistic" voltage source is shown as an active two-terminal. \\
  - This active two-terminal generates a voltage of $1.5V$ and a current of $0A$ when the circuit is open.
  - If a resistor is added, the voltage decreases and the current increases. For example, a voltage of $1.2V$ is applied to the resistor of $2 \Omega$ and a current of $0.6A$ flows.
  - The terminals of the active two-terminal can be directly connected via the outer switch. Then a current of $3A$ flows at a voltage of $0V$.

This realization shall now be described with some technical terms:
  * One speaks of **open circuit** when no current is drawn from an active two-terminal: $I_{LL}=0$. \\ The voltage corresponds to the **open circuit voltage** $U=U_{LL}$ (English: OCV for Open Circuit Voltage). \ The open circuit power is $P_{LL}=U_{LL} \cdot I_{LL} = 0$.
  * The term **short circuit** is used when the terminals of the two-pole are bridged without resistance. The current then flowing is called the **short-circuit current** $I=I_{KS}$. \ The short-circuit voltage is $U_{KS}=0V$. \ Also, the short-circuit power is $P_{KS}=U_{KS} \cdot I_{KS} = 0$.
  * In the region between no-load and short-circuit, the active two-terminal outputs power to a connected load.

Important: As will be seen in the following, the short-circuit current inside the two-terminal can cause considerable power loss and thus a lot of waste heat. Not every real two-terminal is designed for this.

~~PAGEBREAK~~~CLEARFIX~~~
<WRAP right>
<imgcaption imageNo3 | Current-voltage characteristic of a linear voltage source>
</imgcaption>
{{drawio>IU_CharacteristicLinearVoltageSource}}
</WRAP>

What is interesting now is the current-voltage characteristic of the circuit in <imgref FigNo2>. This can be seen in the simulation below. The result is a linear curve (see <imgref imageNo3>).

From a purely mathematical point of view, the course can be represented by the basic equation of linear graphs with the y-axis intercept $I_{KS}$ and a slope of $-{{I_{KS}}\over{U_{LL}}$:

\begin{align*}
I = I_{KS} - {{I_{KS}}\over{U_{LL}}\cdot U \tag{3.1.1} \end{align*}

On the other hand, the formula can also be resolved to $U$:

\begin{align*}
U = U_{LL} - {{U_{LL}}\over{I_{KS}}\cdot I \tag{3.1.2} \end{align*}

~~PAGEBREAK~~~CLEARFIX~~~
<callout icon="fa fa-exclamation" color="red" title="Remember:">
If a two-terminal results in a linear curve between $U_{LL}$ and $I_{KS}$, it is called a **linear source**. This curve describes in good approximation the behavior of many real sources. Often one finds synonymous to the term 'linear source' also the term 'real (voltage) source'. However, this is somewhat misleading as it is a simplified model for reality.
</callout>

~~PAGEBREAK~~~CLEARFIX~~~
<WRAP right>
<imgcaption imageNo4 | equivalent circuit images of linear sources>
</imgcaption> <WRAP>.
{{url>https://www.falstad.com/circuit/circuitjs.html?running=false&ctz=CQAgjCAMB0l3AWAnC1b0DYqzAVgMz67IYYAcCyATBWSPlmNLiAvgFAAOIZk4A7Fl4gq-.KuChR2AdxFUEIsSCRZR4yDPBhxYQXIW6sGgE7hVSsKvmSYYTcMNmBR+3zVP3G2cPcrF69gBzEABafHF3MIiyOg18fj4wbWceRL0IPg0AZydHS31JCAAzAEMAGyyAU3YcqJS6qmtEkFKK6tl8xzrHL1DwgoalXsGdcwCANwL3RoUZm1ZE5nmYXHZTOY3rHuxVgCNwSjM3JAgqJAUNfbBDhjoziFxIANNHBAc9fAobZk06t8SsP8pLI-v0-sJevlPgo6tCpLV+nDQeoJK0qjUnHD8mwUcVyuj4olknDhFibEE+uIgX8-L1hEC-EC6XwgdiIa56F8oV8TE5WYD+ploHYOsSuR8eZpXsJGezuKSvvSHOSAJYiBAsjCzDULTL0TTyFlwdV8OG9Q2cgwSi4GnUILUm1js2QW1l6JnsAAenPAnxEDAOEAMdDKKoAdpUSsZKgAdLIAZU4JTDYYArmHAlkAI6pyplMrVb0YcT4ZAiXT0XBU8Ah8OR6Nx+MAF2MAHsALY5vMFr0gcgCcRkfACFjBkAASRjWd7GH49H4SB4JbI1bAdEn0+9ZBYVGu4EgdwrY4AqlPe2RFwgtskr3OT2fvWBziIni-VJBRzWQAAlXtUSBzv+h5uEkrBfiqvZrnQCCgf+LJ7mOv7ev+VL-i+o5UMOY4QY+K7qsOT7aoCX7HpBz7yIucEiFWYHJBoyE0TBpy4HeCFfuOf64ICT4iMWrCNLRATIZRAYwf6l5fkh9Asmw-GsPatF0FJsnyA8w7yAoY4ADK9rJCA0QQrC4J+a4gDpraFJIMEYMOuCVCEdBINAghcVQdkhGAw5uDw+oWf0GSsJYtn2Y5zkYK57meRIEQKBwfpedgEBgHw36VFkKpZE2yYAMbVH6ZzfElKVpRlWVhrl7BAA 600,700 noborder}}

</WRAP>
So what does the inside of the linear source look like? In <imgref imageNo4> two possible linear sources are shown, which will be considered in the following.

==== Linear voltage source ====

The linear voltage source consists of a series connection of an ideal voltage source with the source voltage $U_0$ (English: EMF for Electro-Magnetic-Force) and the internal resistance $R_i$.
To determine the voltage outside the active two-terminal, the system can be considered as a voltage divider. The following applies:

\U = U_0 - R_i \cdot I \end{align*}

The source voltage $U_0$ of the ideal voltage source is to be measured at the terminals of the two-terminal, if this is unloaded. Then no current flows through the internal resistor $R_i$ and there is no voltage drop there. Therefore: The source voltage is equal to the open circuit voltage $U_0 = U_{LL}$.

\U = U_{LL} - R_i \cdot I \end{align*}

When the external voltage $U=0$, it is the short circuit case. In this case, $0 = U_{LL} - R_i \cdot I_{KS}$ and transform $R_i = {{U_{LL}}\over{I_{KS}}$. Thus, equation $(3.1.2)$ is obtained:
\begin{align*} U = U_{LL} - {{U_{LL}}\over{I_{KS}} \cdot I \end{align*}

Is this the structure of the linear source we are looking for? To verify this, we will now look at the second linear source.

==== Linear current source ====

The linear current source now consists of a __parallel circuit__ of an ideal current source with source current $I_0$ and internal resistance $R_i$, or internal conductance $G_i = {{1}\over{R_i}}$. To determine the voltage outside the active two-terminal, the system can be considered as a current divider. Here, the following holds:

\begin{align*} I = I_0 - G_i \cdot U \end{align*}

Here, the source current can be measured at the terminals in the event of a short circuit. The following therefore applies: $I_{KS}= I_0$

\begin{align*} I = I_{KS} - G_i \cdot U \end{align*}

When the external current $I=0$, it is the no-load case. In this case, $0 = I_{KS} - G_i \cdot U_{LL}$ and transform $G_i = {{I_{KS}}\over{U_{LL}}$. \\ Thus, equation $(3.1.1)$ is obtained:
\begin{align*} I = I_{KS} - {{I_{KS}}\over{U_{LL}} \cdot U \end{align*}

So it seems that the two linear sources describe the same thing.

==== Duality of linear sources ====

<WRAP right>
<imgcaption imageNo7 | duality of linear sources>
</imgcaption>
{{drawio>DualitySources}}
</WRAP>

Through the previous calculations, we came to the interesting realization that both the linear voltage source and the linear current source provide the same result. It is true: For a linear source, both a linear voltage source and a linear current source can be specified as an equivalent circuit! As already in the case of the star-delta transformation, this not only provides two explanations for a black box. Also here linear voltage sources can be transformed into linear linear current sources and vice versa.

The <imgref Fig.No.7> compares again the two linear sources and their characteristics:
  - The linear voltage source is given by the source voltage $U_0$, or the open circuit voltage $U_{LL}$ and the internal resistance $R_i$.
  - The linear current source is given by the source current $I_0$, or the short-circuit current $I_{KS}$ and the internal conductance $G_i$.

The conversion is now done in such a way that the same characteristic curve is obtained:

  - __From linear voltage source to linear current source__: \\ __Given: Source voltage $U_0$, or open circuit voltage $U_{LL}$, internal resistance $R_i$ \Looked for: source current $I_0$, or short circuit current $I_{KS}$ , internal conductance $G_i$ \ $\boxed{I_{KS} = {{U_{LL}}\over{R_i}}$ , $\boxed{G_i = {{1}\over{R_i}}$
  - __From linear current source to linear voltage source__: \\ __Given: Source current $I_0$, resp. short-circuit current $I_{KS}$ , internal resistance $G_i$ __Sought: Source voltage $U_0$, resp. open-circuit voltage $U_{LL}$, internal resistance $R_i$ __$ $\boxed{U_{LL} = {{I_{KS}}\over{G_i}}$ , $\boxed{R_i = {{1}\over{G_i}}$

~PAGEBREAK~ ~CLEARFIX~~
==== Operating point of a real voltage source ====

<WRAP right>
<imgcaption imageNo5 | Source and consumer characteristics>
</imgcaption> <WRAP>.
{{url>https://www.falstad.com/circuit/circuitjs.html?running=false&ctz=CQAgjCAMB0lwrFaYwGZVgCwA5P1ZAGyHYBMhqImAnCIpqgFABuIAtCqSKfIe5+FLYoIzJDojxMeIwBO-MF0XC2qJUJFg4kRgHcFXHn1WHeUPVWFH2a7mZ0BzS3b45u2YToBG-THy1cbJiY3CTm8gQhyiCRLppwFqSkURqx0Tr6SSHWsdY6rFncYtzJ4IRS4FTiEDXQqIRovNhghPBt8ADsHUgyPuRRkLSo8OKkHYgZOnlyCbumMToXWix7mmaXJo+v2iaUtmymejAAe-MQg2OLUENiU2SEAKgCmAM4ALs8AxgAWAIYANq8AK4AOwcAB1ngAzAA-sghAEVAY8-n9HscyuIOrRMB1hGQqEUQE83p9fgCQeCobCIQA1R6yLyyH6A7709E8cDYWgdGrULgpEDMH6yACWPy8qNk6LAiFIcBA424kFughAAHURQATelvH7AzWMaCC4UAfQASiJKJQuGCAA4gfwiNjcSoQGkAYRpAGUAKJ-AC2ELAENIENQEMwYNkAHFZAB7EGav2B57wKMe73JsMQtPPDoQwhBqNml4it5x2RZ57B56hmPxxNVyOyM0-EV-KvdSBgkM92sAKgAFD82F4AJQQ7vPbvZiHD0dj-uPAB6g4+E+nfZGk53s+eYj7+d3m8n0BkmC0pwxV6EwggQtk5qgvYH8-Hkmzzzfi5Xa7Hz9TCop3Dfcp3zTcIOkRg43ACBCBCcQL0ILh4EeNhdjQjlRlgmJoJAWgIEQloULQu80IIl0hkYIA 700,400 noborder}}
</WRAP>

<imgref FigNo5> shows the characteristics of the linear voltage source (left) and a resistive resistor (right). For this purpose, both are connected to a test system in the simulation: In the case of the source with a variable ohmic resistor, in the case of the load with a variable source. The characteristic curves formed in this way were described in the previous chapter.

~~PAGEBREAK~~~CLEARFIX~~~

<WRAP right>
<imgcaption imageNo6 | Determining the operating point>
</imgcaption>
{{drawio>WorkingPointLinear}}
</WRAP>

The operating point can be determined from both characteristic curves. This is assumed when both the linear voltage source is connected to the ohmic resistor (without the respective test systems). In <imgref FigNo6> both characteristic curves are drawn in a current-voltage diagram. The point of intersection is just the operating point that sets in. If the load resistance is varied, the slope changes in inverse proportion and a new operating point is established (light grey in the figure).

The derivation of the working point is also [[https://www.youtube.com/watch?v=Okt6oy4sF_A|hier]] explained again in a video.

~~PAGEBREAK~~~CLEARFIX~~~

<WRAP right>
<imgcaption imageNo9 | Straight line arrays for source parameter variation>
</imgcaption>
{{drawio>StraightSharesSourceParameters}}
<WRAP>

The variation of the different source parameters will be briefly discussed.
\\ \\
For the linear current source, the source current $I_0$ and the internal conductance $G_i$ can be varied. This results in the straight line arrays in <imgref FigNo9> above. The source current shifts the straight lines while keeping the slope constant. The internal conductance changes only the slope; this results in a straight line array around the intersection $I_0 = I_{KS}$. \\ Since an ideal current source should always supply the source current, its internal conductance $G_i=0$.

\\ \\
For the linear voltage source, the source voltage $U_0$ and the internal resistance $R_i$ can be varied. This results in the straight line arcs in <imgref FigNo9> below. The source voltage shifts the straight lines while keeping the slope constant. The internal resistance changes only the slope; this results in a straight line array around the intersection $U_0 = U_{LL}$. \\ Since an ideal voltage source should always supply the source voltage, its internal resistance $R_i=0$.

~~PAGEBREAK~~~CLEARFIX~~~

<panel type="info" title="Task 3.1.1 Convert current source to voltage source"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>

{{youtube>eQ60i0ZIfw8}}

</WRAP></WRAP></panel>

<panel type="info" title="Task 3.1.2 Convert voltage source to current source"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>

{{youtube>GsNY0_Rsk7I}}

</WRAP></WRAP></panel>

{{page>task_3.1.3_with_calculation&nofooter}}


===== 3.2 Conversion of any linear two-terminal =====

<callout>
=== Objectives ===
After this lesson you should:

  - know that any linear circuit with two connections of ohmic resistors and sources can be understood as a linear current source or linear voltage source.
  - Be able to apply source conversion to more complicated circuits with multiple current sources or voltage sources.
  - know how to determine the open circuit voltage $U_{LL}$ and the short circuit current $I_{KS}$).
  - be able to calculate the parameters of the equivalent voltage source (internal resistance $R_i$ and source voltage $U_q$) of any linear circuit.
  - understand and be able to draw the graphical interpretation of voltage and current at the linear two-terminal in the form of a characteristic curve.

</callout>

 
<WRAP right>
<imgcaption imageNo7 | Resistance of linear sources>
</imgcaption> <WRAP>.
{{url>https://www.falstad.com/circuit/circuitjs.html?running=false&ctz=CQAgDOB0YzCsICMZICYaoOyYMxgByoBsAnCZkiAgCw4BQAbiCUSKnKy0pqxBNRAR8ocOgCdmrRD25SZEZDDoB3SWw6z1vOqkStUqalraHNClSaPs5+jWAvSbls3QCWIHOmOeF8qPAsfTRx8I0dwcQ9Q4OjrcCRYQNiNILj7VRCwmWoiLO0M5NYcqzsdPRBi40rw8wBzZziDK3x8CPrKuI6WiIAjECJEJDgwohw2TFRetnIh1upEQaxJ+wAPabCcGmoSJBCKpFaAG1cAOwBTAEMxM4AdAGcAZQAXMQB7AFsARwBXM8PDs50Na5Xb6IiTRA4HZhI6nS7Xe4PAAOFxOJ2+J1qdx+fwBdCAA 700,400 noborder}}
</WRAP>

In <imgref FigNo7>, it can be seen that the internal resistance of the linear current source measured by the ohmmeter (resistance meter) is exactly equal to that of the linear voltage source.

If you look at the properties of the ohmmeter in the simulation, you will see that a measuring current is used there to determine the resistance value. This concept will still be part of the electrical engineering lab experiment on [[electrical_engineering_lab:1. resistors|resistors]]s in the 2nd semester.
However, a very large measuring current of $1A$ is used here. This could lead to high voltages or destruction of components in real setups.
\\ \\
Why is this nevertheless chosen so high in the simulation? Set the measuring current for both linear sources to (more realistic) $1mA$. What do you notice?

~~PAGEBREAK~~~CLEARFIX~~~
<WRAP right>
<imgcaption imageNo8 | circuit with two current sources>
</imgcaption>
{{drawio>CircuitTwoCurrentSources}}
</WRAP>

The circuit in <imgref FigNo8> shows this circuit again. The ohmmeter is replaced by a current source and a voltmeter, since only the electrical properties are important in the following. In this setup, it can be seen that the current through $G_i$ is just given by $I_i = I_0 + I_\Omega$ (node theorem). Thus, the two sources in the circuit can be reduced.

This should make the situation clear with a measuring current of $1mA$. The voltage at the resistor is now given by $U_\Omega = R \cdot (I_0 + I_\Omega)$. Only when $I_\Omega$ is very large does $I_0$ become negligible. The current of a conventional ohmmeter cannot guarantee this for every measurement.

<callout icon="fa fa-exclamation" color="red" title="Merke:">
If resistors are to be measured in a circuit, at least one terminal of the resistor must be disconnected from the circuit. Otherwise, other sources or resistors may falsify the measurement result.
</callout>

==== More complex example ====
<WRAP right>
<imgcaption imageNo9 | circuit with multiple sources>
</imgcaption>
{{drawio>CircuitMultipleCurrentSources}}
</WRAP>

But this knowledge can now be used for more complicated circuits. In <imgref fig.no.9> such a circuit is drawn. This is to be converted into a searched equivalent conductance $G_g$ and a searched equivalent current source with $I_g$.

**Important here**: Only two-terminals can be converted via source duality. This means that only 2 nodes may act as output terminals for selected sections of the circuit. If there are more nodes the conversion is not possible.
\\ \\ \\
  - As a first step, sources are to be converted in such a way that resistors can be combined after the conversion. In this example this is done by:
    - converting the linear voltage source $R_1$ and $U_1$ into a linear current source with $I_1={{U_1}\over{R_1}}$ and $R_1$ (or $G_1={{1}\over{R_1}}$)\
    - converting the linear current source $R_4$ and $I_4$ into a linear voltage source with $U_4={{I_4}\cdot{R_4}}$ and $R_4$ \\ \\ \\\ \\\ \\\\
  - In the second step, the linear voltage source $U_4$ formed in 1. with $R_4$ can be connected to the resistor $R_3$. From this again a linear current source can be created. This now has a resistance of $R_5 = R_3+R_4$ and an ideal current source with $I_5 = {{U_4}\over{R_3+R_4}}= {{I_4}\cdot{R_4}\over{R_3+R_4}} $.  \\ \\ \\ \\ \\ \\
  - The circuit diagram that now emerges is a parallel circuit of ideal current sources and resistors. This can be used to determine the values of the ideal equivalent current source and the equivalent resistance:
    - ideal equivalent current source $I_g$: \begin{align*} I_g = I_1 + I_3 + I_5 = I_1 + I_3 + I_4\cdot{{R_4}\over{R_3+R_4}} \end{align*}
    - Substitute conductance $G_g$: \begin{align*} G_g = \Sigma G_i = {{1}\over{R_1}}+{{1}\over{R_2}}+{{1}\over{R_5}}={{1}\over{R_1}}+{{1}\over{R_2}}+{{1}\over{R_3+R_4}} \end{align*}

~~PAGEBREAK~~~CLEARFIX~~~

<callout icon="fa fa-exclamation" color="red" title="Notice:">

<WRAP right>
<imgcaption imageNo63 | Spare voltage and current source>
</imgcaption> <WRAP>.
{{url>https://www.falstad.com/circuit/circuitjs.html?running=false&ctz=CQAgjA7CAMB00IQVhrAnAFgBwQgZiSzwDYCskAmYkJaGkDPAKAHMQBaCiC8MHz7uApYYrBtnBoeGCRSTVoTAG4c8JXvy48wfGHox0KsLNAiE9dOEiYB3DlpDEM9wTp6K77DGGpuO36idRACdVdT8vHw09HQhYCCwMKghbcRF2KJkRMClROyDhESDLJlC-.E2jIBXAEUpA8Cjoq+sShESba0KzJHjxWhw7FFTxoZxGx-sFLBjoUaatlR2dCkAoKZax2-UMLVGsVcsNobOFdg3p52GtQtecgw5jEOobDeRblqZrFUL7nPhFujl3F86v9VucCpttt9VutVlDxvVRnpGkNYcs4bd4Vs6Oc5hYrotERUscVthcCftKhBfKd1vMZhTLITho1wDT6my5NVcbNdgsVCswDJ3m0znzLtckc4SVDhc5BnVGicREKRYqbpi4S92dVUUrjmKsViNdTqMqxYrPM00NQMhyStaObaOM0wR5XdQXRlae1Uj7OZ6QC7FABLV3Ou1REOoaAUPokUh1e2+oPu3hxBJJYgpNgBsH59VMAAerrQTTQeFWkmoy2cAFFggBnACGABcAF5NtvBAD2AFsAI4AVwApgAbcejkv+Vy5TjxyQoOsgRutztNgAOLYAdjvhzuWE2RxOp8mSCIGi5esC8GBYMQ0DP2Go0OB0hQ0CgKGAGOCQAASgA+gAMosL7EOkV4CPwV5NDMEqUkwADK-hRHQEE4uAIAAGYtuOTajiiz4kBhpCqGAKDsMQuL-iBLbds+YDQFROYcDR7HOCu9GMZ4DhBDBgYeoJ0EOC+7j+gESxoTy-oXkGmGqBJnjyVEilwf6fhqX4GnWroAYZPpulBtBakaahKZ6IZwLwXhBFEYY4ERKZdpwdhBiIcy1iluwRDvqob4-n+cLAWBoTWYGAZuXeD5PnYJrouACipFi4mJV4fp2DqUVshloioc0Aapeh2F2YRxGCnCeXZRI8EeUyewznIVa9Mx9ArqFYBKnCUQ6iVMWPqkhVRAJ6GpKNgQKnknIYZkEgBh62W9WyY0AEb+EEFScHgUA+BJ62cMKIAVGAVZ7aIB0YJBHBIM4nDeI4JQHT+6RIL+b7kYopbCqxvjluxVbcQxbZML2DC+NsaBvvecixvi0jUFWa7tl2257geR6DrAoOrNDkPQ7AsNwCx2E8J+9Srs2KPdn2Q7Y2Du34+AhNzPA37VlAVYtsOTY4bAAA6TbIQAxgAFvhbYY0wQA 600,400 noborder}}
</WRAP>


Any interconnection of __linear__ voltage sources, current sources, and __ohmic__ resistors can be.
  * as a single, linear voltage source \\ ({{wpde>Thévenin theorem}}) or
  * as a single, linear current source \\ ({{wpde>Norton theorem}})
{{wpde>Norton theorem}}

In <imgref FigNo63> it can be seen that the three circuits give the same result (voltage / current) with the same load. This is also true when an (AC) source is used instead of the load.
</callout>


~~PAGEBREAK~~ ~CLEARFIX~~
==== Simplified determination of the internal resistance ====

<callout icon="fa fa-exclamation" color="red" title="Notice:">

<WRAP right>
<imgcaption imageNo10 | equivalent resistance of ideal sources>
</imgcaption>
{{drawio>SubstituteResistorIdealSources}}
<WRAP>

If only the equivalent resistance of a more complex circuit is sought, the following approach can be used:
  - Replace all ideal voltage sources by a short circuit \\ (= internal resistance of the ideal voltage source).
  - Replace all ideal current sources by an open contact \\ (= internal resistance of the ideal current source)
  - Add the remaining resistors to an equivalent resistance using the rules for parallel and series connection.

The equivalent circuits for the ideal sources can be seen via the circuit diagrams (see <imgref FigNo10>).

</callout>

Thus also the equivalent resistance of the complex circuit above can be derived quickly. \\
For the source current $I_0$ ideal equivalent current source resp. the source voltage $U_0$ ideal equivalent voltage source this derivation can not be used.

The reason that the internal resistance can be determined in this simple way will be explained in the next chapter [[analysis_of_dc_networks#superposition_methodsuperposition_principle|analysis_of_dc_networks: Superposition method]] is explained.

<WRAP right>
<imgcaption imageNo11 | Simplified determination of internal resistance>
</imgcaption>
{{drawio>SimplifiedDeterminationOfInternalResistance}}
</WRAP>


~~PAGEBREAK~~~CLEARFIX~~~

<panel type="info" title="Task 3.2.1 Solving a circuit simplification I"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>

{{youtube>Jb0GIMbZwjA}}

</WRAP></WRAP></panel>

<panel type="info" title="Task 3.2.2 Solving a circuit simplification II"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>

{{youtube>9e1o5QNkzf8}}

</WRAP></WRAP></panel>

<panel type="info" title="Task 3.2.3 Solution sketch for a more difficult circuit simplification"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>

{{youtube>eYLGsmgAzSY}}

</WRAP></WRAP></panel>

<panel type="info" title="Task 3.2.4 3 short but interesting circuit tasks"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>

{{youtube>IyMY7bAQtGY}}

</WRAP></WRAP></panel>


===== 3.3 Power at two poles and reference values =====

<callout>

=== Goals ===

After this lesson, you should:

  - be able to calculate the source power and consumer power.
  - be able to distinguish between the optimisation objectives for power engineering and communications engineering.
  - be able to calculate the efficiency and utilization factor.

</callout>

Power and efficiency have already been considered in [[basics_and_basic_concepts#power_and_efficiency|1st chapter]] and [[simple_dc_circuits#generator_and_consumer_reference_arrow_systems|2nd chapter]] for a simple dc circuit. In the following, this will be analyzed again with the knowledge of two-terminals.
This is especially important for the fields of communications and power engineering. The goals here are different:
  - In __power engineering__, power transmission is the goal. Power is thus to be delivered without losses as far as possible.
  - In __communications engineering__, the focus is on information transmission. So that, for example, the best possible signal can be extracted from an antenna, the maximum power must be extracted here.

These two goals seem similar at first, but they are quite different, as will be seen in a moment.

==== Power measurement ====

<WRAP right>
<imgcaption imageNo12 | Power measurement on linear voltage source>
</imgcaption>
{{drawio>PowerMeasurementLinearVoltageSource}}
<WRAP>

First, it is necessary to consider how to determine the power. The power meter (or wattmeter) consists of a combined ammeter and voltmeter.

In <imgref Fig.-No.12> the wattmeter with the circuit symbol can be seen as a round element with crossed measuring inputs.
The circuit also shows one wattmeter each for the (not externally measurable) output power of the ideal source $P_Q$ and the input power of the load $P_R$.

~~PAGEBREAK~~ ~CLEARFIX~~
==== Power and reference variables in the diagram ====

<WRAP right>
<imgcaption imageNo13 | power adjustment>
</imgcaption> <WRAP right>
{{url>https://www.falstad.com/circuit/circuitjs.html?running=false&ctz=CQAgjCAMB0mQrFaB2AzADgJwIGyU1qpDvBPOiIgCyoBQAbiFRQExWRMVhVVR-uU+HGPFoB3TuB4hMOKb0i0ATjLndeYPPL5g4iiS0hdpbDuqjKQqQ+GRzrHUzrjirNlshYhDrTxYmajn7M3n76kh5eIU6KKizoxrxO5sKwkGC0LGBR1iA4vFT5eQpWIAAUAIYAtABGAJQA9DBwYABUZTVVAMZ1rk5FNF5F4Q7aGBrSI+5+46Fe4YPebEy52RSKAOaU8arbXNkWhsh7xZRaAyAACgD6ADJ9AmAHyZMPHEXJByPovLJWP9oFrkWMtFpELDVwGoDmBkJhvAQId4jG5KJgIJoFLQAB5LcCQLyaDEIJjgCgAZQADhUAHY0gCuNI2AGcAI70gCmABsuRycUskhxUMgkqhSWAKLcOQBLZkAF0ZLIAthzmczFX0UeZAoDNb5CVpwYpKf8JrxZuCIMJXKNzLa7BZhULIGa3GZpFb-G7vAJRkabQC-os-t8ksFgWF+SCQOg5A6sOKKFzpTSORUlHzcTg5IY46x8ImQABVfnoLxZLwJbxgajgeb8p4aPzZNTIWsSkAAJQbRFsXFW6AgGi80qjserEBYiKyBTJXf5REcmisxGr8OHIHuuJY8HhTxzyAxYHXc+727Qdccg7rYo35L6VbtLt1uNQu5jXmFEHQ7YoJdfh4xvYcIxieBz3ri7DHMeFBUGAYqGEOdabgumBiset4ksecgaH+C66HWn7Ltk7ZeFuK7ocuRD7GBFAAJK0AA9uAiCetkU6pHAuY4E8x5ULuejMHIzQtHwhJYsxGj8Jo6HQDuHJVFwcnwApBGTqSdAYKSZhwEweibhU8piNKAAmHJKPKtImQAOsynZ3LQQA 700,400 noborder}}
</WRAP>

The simulation in <imgref FigNo13> shows the following:
  * The circuit with linear voltage source ($U_0$ and $R_i$), and a resistive load $R_L$.
  * A simulated wattmeter, where the ammeter is implemented by a measuring resistor $R_S$ (English: shunt) and a voltage measurement for $U_S$. The power is then: $P_L = {{1}\over{R_S}}\cdot U_S \cdot U_L$.
  * in the oscilloscope section (below).
    * On the left is the power ''P_L'' plotted against time in a graph.
    * On the right is the already known current-voltage diagram of the current values.
  * The slider ''Load resistance R_L'', with which the value of the load resistance $R_L$ can be changed.

Now try to vary the value of the load resistance $R_L$ (slider) in the simulation so that the maximum power is achieved. Which resistance value is set?

~~PAGEBREAK~~~CLEARFIX~~~

<WRAP right>
<imgcaption imageNo14 | current-voltage diagram, power-voltage diagram and efficiency-voltage diagram>
</imgcaption>
{{drawio>CurrentVoltageDiagramWithPower}}
</WRAP>

<imgref imageNo14> shows three diagrams:
  * Diagram top: current-voltage diagram of linear voltage source.
  * Diagram in the middle: source power $P_Q$ and consumer power $P_L$ versus delivered voltage $U_L$.
  * Diagram below: Reference quantities over delivered voltage $U_L$.

The two powers are defined as follows:
  * source power: $\, \, \large{ P_Q = U_0 \cdot I_L} $
  * consumer power: $\large{ P_L = U_L \cdot I_L} $

  - Both power $P_Q$ and $P_L$ are equal to 0 without current flow. The source power becomes maximum, at maximum current flow, that is, when the load resistance $R_L=0$. In this case, all the power flows out through the internal resistor. The efficiency drops to $0\%$. This is the case, for example, with a battery shorted by a wire.
  - If the load resistance becomes just as large as the internal resistance $R_L=R_i$, the result is a voltage divider where the load voltage becomes just half the open circuit voltage: $U_L = {{1}\over{2}}\cdot U_{LL}$. On the other hand, the current is also half the short-circuit current $I_L=I_{KS}$, since the resistance at the ideal voltage source is twice that in the short-circuit case.
  - If the load resistance becomes high impedance $R_L\rightarrow\infty$, less and less current flows, but more and more voltage drops across the load. Thus, the efficiency increases and approaches $100\%$ for $R_L\rightarrow\infty$.

The whole context can be seen in a [[https://www.falstad.com/circuit/circuitjs.html?running=false&ctz=CQAgjCAMB0mQrFaB2AzADgJwIGyU1qpDvBPOiIgCyoBQAbiALSY4gBMVkzr4VVUQV0qDuMeLQDuIKhTD8QveQMi0ATorbLwePivBxV09pDkLO3bao2oT4ZG1vcLgsHCkgnHZOw6nvvkY6zj46AVAeLGzsoVEcwtYc6GYCLlZIhmC07GC+NL44AvwUYp5+hnAAVAAUAIZMAEYAlDUNTADGTR4uhTK2IL1BXtoYAlYeXjG+o+FB+Rycfb65JbQA5pTsFLzwW+C5ESbImxS9uxSyMiAACgD6ADLdwmAHaQpBPanPB0PoArwzcbSebsRYg0KqBrMFb7ZbITAcAgRKEmCj9eCYCBgQa0AAeCwMyxwWIQVzAFAAygAHWoAO1pAFdaWsAM4ARwZAFMADbczl4hapbjwSDHdioMkUe6cgCWLIALkzWQBbTksllKgA6LNqyu1ADVOWoGmpagz2gALI3dfzabGWd42ihTYKzWhUzx-PSe1KhCBiCZ2EZBhwRVDIbhEMYKYKf0RaSTYSTCETL28ea8X6+vL9F2qfGgkDoNihrCSkDcmW0zm1NT8-E4aLEEDIZ34CsAVQF6F8OV8yQ4YGo4ECApeYz97C0yBH5JAACVx0R7HJ+uSIGNfDKBVtoi9Ec75BWl-iiM5sZ4WzkEVuQI9Cxj9tFkFiwLfwBRT94JblnOgsXFMlfApbpBxGSBoxUAVUCfXtPFfYs5wobsz0QksEIRcs71A-EuGOd8LjACUTE3Ud7xgzBfyomRSXfNgxhQmC3FHaZL1yOdfAfK9f0vIg5HfCsAEk0wEOJ5jiVRw24Jh13eb14wLQVoVCeAcF8JhsQ-edpTlRVmRZVV1S1HVaW1AATI1tWqGVLNqPlaSabUAEUuV5fkUX8TTdAxCAmD2VQjhOaE9nOaEFAEO5nPWYK4jCph0ADaRLxdS8mBxaRZI0gEvUkyJEpCi5+k0gLsgOCSYWKaFMECMoTAqSAanqZomo6LoNCy70EodfQYEySIlAUbrvSCYaSooYagWYOSxOKrMPBS2JeHzDwJN6NKMtcCKUkOA4cFiF4CkHTTQ2cEAGgAejqTUqT6sAug2BwJoKp7mEuIJ4phfaNJhIJvsKgYDp+Dwwt6f7NvBgR-uGwLRVbHAJsO1tcqRgROXlWpaChQoZPtVtMARfzVkbKg-LAdDUDJhGKwAdRlNQAGslRZNZTUstRlRleV5W5JVaAAezKQYZGxPJoAJiXJalzBOW88BZfJwRpk8AXWyxIR8BKaB1MQGB1d8RC6EF3ZwA1pFdegCA+3AXXaAwK5LDgWinfuWoFUkWyjQVOlzO1BcHloIA|extensive Simulation]] will be analyzed again.

==== The reference values efficiency and utilisation rate ====

In order to understand the lower diagram in <imgref FigNo14>, the definition equations of the two reference quantities shall be described here again:

The **efficiency** $\eta$ describes the delivered power (consumer power) in relation to the supplied power (power of the ideal source):
\begin{align*}\eta = {{P_{down}}\over{P_{up}} = {{R_L\cdot I_L^2}\over{(R_L+R_i) \cdot I_L^2}}
\quad \Rightarrow \quad \boxed{ \eta = {{R_L}\over{R_L+R_i}} }
\end{align*}

The **efficiency** $\varepsilon$ describes the delivered power in relation to the maximum possible power of the ideal source. Here, the currently supplied power is not assumed (as in the case of efficiency), but the best possible power of the ideal source, i.e. in the short-circuit case:
\begin{align*}\varepsilon = {{P_{down}}\over{P_{to,max}} = {{R_L\cdot I_L^2}\over{{U_0^2}\over{R_i}} = {{R_L\cdot R_i \cdot I_L^2} \over {U_0^2}} = {{R_L\cdot R_i \cdot \left({{U_0}\over{R_L+R_i}}\right)^2} \over {U_0^2}}
\quad \Rightarrow \quad \boxed{\varepsilon = {{R_L\cdot R_i } \over {(R_L+R_i)^2}} = {{R_L} \over {(R_L+R_i)}}\cdot {{R_i} \over {(R_L+R_i)}}
\end{align*}

In __power engineering__ a situation close to (1.) in <imgref FigNo14> is desired: maximum power output with lowest losses at the internal resistance of the source. Thus, the internal resistance of the source should be low compared to the load $R_L \gg R_i $. The efficiency should go towards $\eta \rightarrow 100\%$.

In __communications engineering__, one situation is different and corresponds to situation (2.): The maximum power is to be taken from the source, without consideration of the losses via the internal resistance. For this purpose, the internal resistance of the source (e.g., a receiver) and the load (e.g., the downstream evaluation) are matched. This case is called **{wpde>power matching| power matching or resistance matching}}**. The utilization factor here becomes maximum: $\varepsilon = 25\%$

The power adjustment is also [[https://www.youtube.com/watch?v=BJLlXUD6CsM|hier]] explained again in a video.====== sample solution winter semester 2020/2021 ======

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====== 7. Switching operations on RC combinations ======

<WRAP onlyprint>
  - Capacitor in IC's --> MOSFET
  - Charge / discharge FET capacitor.
</WRAP>

<callout>

<WRAP right>
<imgcaption imageNo01 | Capacitor in electrical circuit>
</imgcaption> <WRAP>.
{{drawio>CapacitorInElectricCircuit}} \\
</WRAP>


At the previous chapter the capacitor was already described. It consists of two insulated conductors separated by an insulator (cf. <imgref BildNr01>). \\
They serve as energy storage. This is done in the following manner:
  - An external source draws charge carriers from one of the electrodes and carries them to the other electrode.
  - If the external source is a voltage source with the voltage $U$, a stationary state is reached after a certain time. \\ In this state there is a fixed number of $+Q$ on the positive electrode and $-Q$ on the negative electrode.
  - These charges form an electric field in the space between the electrodes. This field stores the supplied energy.

It is true that the larger the voltage $U$, the more charges $Q$ are stored on the electrode.
This relationship is directly proportional to the proportionality constant $C$:

\begin{align*}
C = {{Q}\over{U}} \quad \text{with:} \quad [C]=1 {{As}\over{V}}= 1 F = 1\; Farad
\end{align}
 
But it is not always directly recognizable that a structure contains a capacitor. \\ So the following examples are also capacitors:
  * **open switch**: If there is voltage between the two metal parts, charges can also accumulate there. \ Since the distances are usually large and air is used as the dielectric, the capacitance of the capacitor formed in this way is very small.
  * **Overhead line**: An overhead line also represents a capacitor against the ground potential of the earth. The charging and discharging by the alternating current leads to the fact that polarizable molecules can align themselves. For example, the water drops near the line are rolled through the field and hum with $100Hz$ and many times that (harmonics). Peak discharge results in the high frequency crackle.
  * **Conductor Trace**: A trace on a PCB can also be a capacitor against a nearby ground plane. This can be a problem for digital signals (see charge and discharge curves below).
  * **Human Body**: The human body can likewise pick up charge. The charge thus absorbed forms a capacitor with respect to other objects. This can be charged up to some $kV$. This is a particular problem in electrical laboratories, as the mere touching of components can destroy them.
  * **Membrane of nerve cells**: Nerve cells also result in a capacitor due to the lipid bilayer (membrane of the nerve cell) and the two cellular fluids with different electrolytes (ions). The nerve cells are surrounded by a thick layer (myelin layer) for faster transmission. This lowers the capacitance and thus increases the successive charging of successive parts of the nerve cell. In diseases such as Creutzfeldt-Jakob or multiple sclerosis, this layer thins out. This leads to delayed signal transmission which characterizes the disease patterns.

<WRAP right>
<imgcaption imageNo02 | Circuit for viewing charge and discharge curve>
</imgcaption> <WRAP right>
{{drawio>CircuitDischargeCurve}} \\
</WRAP>

In the following, the charging process of a capacitor is to be considered in more detail. For this purpose, one has to realize, that during charging of the capacitor, besides the voltage source $U_q$ and the capacitor $C$, there is always a resistance $R$ in the circuit. This is composed of the internal resistance of the (non-ideal) voltage source, the internal resistance of the capacitor and the parasitic (=interfering) resistance of the line. In practical applications it is often desired that capacitors charge in a certain time range. For this purpose, another real resistor is inserted into the circuit. The resulting series of resistor and capacitor is called a **RC element**. It resembles a voltage divider in which a resistor has been replaced by a capacitor. \\

To start the loading, an (ideal) switch $S$ is inserted.
The circuit to be considered then looks like shown in <imgref FigNo02>. \\
An ideal switch is characterized by:
  * infinitely fast switching
  * resistance of $0\Omega$ in closed state ("short circuit")
  * resistance $\rightarrow \infty$ in open state ("open line")
  * no capacitive effect

</callout>

<callout>

In this chapter also time-varying quantities are considered. These are generally marked with small letters. Examples of time-varying quantities are:
  * A **time-varying voltage $u_C(t)$ across a capacitor** or the **voltage $u$ of an ac voltage source** as opposed to a constant voltage $U_q$ across a constant voltage source.
  * A **time-varying current $i_L(t)$ across a coil** or **time-varying current $i_L(t)$ across a capacitor**.

Since the time dependence is already clear from the small letter, these quantities are occasionally not indicated by the trailing $(t)$. So it is $u = u(t)$.

</callout>

===== 7.1 Time course of the charging and discharging process =====

<callout>

=== Goals ===

After this lesson, you should:

  - know the time constant $\tau$ and in particular be able to calculate it.
  - Be able to determine the time characteristic of the currents and voltages at the RC element for a given resistance and capacitance.
  - know the continuity conditions of electrical quantities.
  - know when (=according to which measure) the capacitor is considered to be fully charged / discharged, i.e. a steady state can be considered to have been reached.

</callout>

<WRAP right>{{url>https://www.falstad.com/circuit/circuitjs.html?running=false&ctz=CQAgjCAMB0l3BOJyWoVaYEBYDsuBmbSbMSAVmwDZcqRyJyRsCAoAN3G2xACZiuPMNyiiBTSKJjlWAJxAECdYT2KSVospFYBjZnEH7J-SadjxIvKebCsA7oZN8BT7Q6dO1z0-e9+nYLRQrADKfAAc4Y7kyiKmIABmAIYANgDOAKaivL68MeC8URpghcEOxaXFcbn5YFTGAnU+7o31foF02gAezrjgkOi8VATgJczOICGsPXUjYOFW8+rhTDz8IACq0+Dh6ghzCHt0QlEAjtsEuEJYCggQJVZrPABKF228Awr4fGNPIADCrEuxkihkUsSEUlYAAdmNgiiJuFEAlCHEi-F5XL50RovBpunwEMoxgRIPcwKsJgBLC5kgqSAhgOa8EYnAHbPBCb7w5S7cbzdk9bDkdCBSR4RaEcbrACurAA9nxwGIBugYBZaLx9jhyIy8hgLAhaDF9cYlSMqQB9f4AGgAOmkZdagcjrBY6erDWhvUgAUloUkdFSAC5JAB2OgyDsBBCKmnU8BAzwyaSpaVDEajaVeQA 600,400 noborder}}
</WRAP>

In the simulation on the right you can see the circuit mentioned above in a slightly modified form:
  * The capacitance $C$ can be charged via the resistor $R$ if the toggle switch $S$ connects the DC voltage source $U_q$ to the two.
  * But it is also possible to short-circuit the rich circuit of $R$ and $C$ via the switch $S$.
  * Furthermore the current $i_C$ and the voltage $u_C$ are displayed in the oscilloscope as data points over time and in the circuit as numerical values.
  * Additionally it is possible to change the capacitance value $C$ and resistance value $R$ with the sliders ''Capacitance C'' and ''Resistance R''.

Tasks:
  - Become familiar with how the capacitor current $i_C$ and capacitor voltage $u_C$ depend on the given capacitance $C$ and resistance $R$. \\ To do this, use for $R=\{ 10\Omega, 100\Omega, 1k\Omega\}$ and $C=\{ 1\mu F, 10 \mu F\}$. How fast does the capacitor voltage $u_C$ increase in each case n?
  - Which quantity ($i_C$ or $u_C$) is continuous here? Why must this one be continuous? Why must the other quantity be discontinuous?

\\

At the following, this circuit is divided into two separate circuits, which consider only charging and only discharging.

~~PAGEBREAK~~~CLEARFIX~~~

<WRAP right>
<imgcaption imageNo02 | circuit for viewing the charge curve>
</imgcaption> <WRAP right>
{{drawio>CircuitDischargeCurve2}} \\
</WRAP>

To understand the charging process of a capacitor, an initially uncharged capacitor with capacitance $C$ is to be charged by a DC voltage source $U_q$ via a resistor $R$.
  * In order that the voltage $U_q$ acts at a certain time $t_0 = 0 s$ the switch $S$ is closed at this time.
  * Directly after the time $t_0$ the maximum current ("charging current") flows in the circuit. This is only limited by the resistor $R$. The uncharged capacitor has a voltage $u_C(t_0)=0V$ at that time. The maximum voltage $u_R(t_0)=U_q$ is applied to the resistor. The current is $i_C(t_0)={{U_q}\over{R}}$.
  * The current causes charge carriers to flow from one electrode to the other. Thus the capacitor is charged and its voltage increases $u_C$.
  * Thus the voltage $u_R$ across the resistor is reduced and so is the current $i_R$.
  * With the current thus reduced, less charge flows on the capacitor.
  * Ideally, the capacitor is not fully charged to the specified voltage $U_q$ until $t \rightarrow \infty$. It then carries the charge: $q(t \rightarrow \infty)=Q = C \cdot U_q$

The process is now to be summarized in detail in formulas. \\
Linear components are used in the circuit, i.e. the component values for the resistor $R$ and the capacitance $C$ are independent of the current or the voltage. \\
Then definition equations for the resistor $R$ and the capacitance $C$ are also valid for time-varying or infinitesimal quantities:

\begin{align*}
R = {{u_R(t)}\over{i_R(t)}} = {{du_R}\over{di_R}} = const. \\
C = {{q(t)}\over{u_C(t)}} = {{dq}\over{du_C}} = const.  \tag{7.1.1}
\end{align*}

The following explanations are also well explained in these two videos on [[https://www.youtube.com/watch?v=Ml3d3_WnuRE|loading]] and [[https://www.youtube.com/watch?v=GAFeYmi4WUk|unloading]].

==== Charging a capacitor at time t=0 ====

By considering the mesh, the general result is: the voltage of the source is equal to the sum of the two voltages across the resistor and capacitor.

\begin{align*}
U_q =u_R + u_C = R \cdot i_C + u_C \tag{7.1.2}
\end{align*}

At the first instant $dt$, an infinitesimally small charge "chunk" $dq$ flows through the circuit driven by the current $i_C$ from the voltage source. \\
For this, $(7.1.1)$ gives:

\begin{align*}
i_C = {{dq}\over{dt}} \quad \text{and} \quad dq = C \cdot du_C
\end{align*}

The charging current $i_C$ can be determined from the two formulas:

\begin{align*}
i_C = C \cdot {{du_C}\over{dt}} \tag{7.1.3}
\end{align*}

Thus $(7.1.2)$ becomes:

\begin{align*}
U_q &=u_R + u_C \\
&= R \cdot C \cdot {{du_C}\over{dt}} + u_C
\end{align*}

--> here follows some mathematics: #

This result represents a 1st order differential equation. \\
This should generally be rewritten so that the part that depends (on the variable) is on one side and the rest is on the other. \\
This is already present here. The appropriate approach to such a problem is:

\begin{align*}
u_C(t) = \mathcal{A} \cdot e^{\mathcal{B}\cdot t} + \mathcal{C}
\end{align*}

\begin{align*}
U_q &= R \cdot C \cdot {{d}\over{dt}}(\mathcal{A} \cdot e^{\mathcal{B}\cdot t} + \mathcal{C}) + \mathcal{A} \cdot e^{\mathcal{B}\cdot t} + \mathcal{C} \\
    &= R \cdot C \cdot \mathcal{AB} \cdot e^{\mathcal{B}\cdot t} + \mathcal{A} \cdot e^{\mathcal{B}\cdot t} + \mathcal{C} \\
U_q - \mathcal{C} &= ( R \cdot C \cdot \mathcal{AB} + \mathcal{A} ) \cdot e^{\mathcal{B}\cdot t} \\
\end{align*}

This equation must hold for every $t$. This is only possible if the left as well as the right term become equal to 0. \\ Thus:

\begin{align*}
\mathcal{C} = U_q \

R \cdot C \cdot \mathcal{AB} + \mathcal{A} &= 0 \quad \quad | : \mathcal{A} \quad | -1 \
R \cdot C \cdot \mathcal{B} &= - 1 \\
\mathcal{B} &= - {{1}\over{R C}} \\
\end{align*}

So it follows:

\begin{align*}
u_C(t) = \mathcal{A} \cdot e^{\large{- {{t}\over{R C}} }} + U_q
\end{align*}

For the solution it must still hold that at time $t_0=0$ $u_C(t_0) = 0$ just holds:

\begin{align*}
0 &= \mathcal{A} \cdot e^{\large{0}} + U_q \\
0 &= \mathcal{A} + U_q \
\mathcal{A} &= - U_q
\end{align*}

So the solution is:

\begin{align*}
u_C(t) &= - U_q \cdot e^{\large{- {{t}\over{R C}}}} + U_q
\end{align*}

<--


And this results in:
\begin{align*}
u_C(t) &= U_q \cdot (1 - e^{\large{- {{t}\over{R C}}}})
\end{align*}

And with $(7.1.3)$, $i_C$ becomes:
\begin{align*}
i_C(t) &= {{U_q}\over{R}} \cdot e^{\large{- {{t}\over{R C}} } }
\end{align*}

~~PAGEBREAK~~~CLEARFIX~~~

In <imgref FigNo04>, the two time courses for the charging voltage $u_C(t)$ and the charging current $i_C(t)$ of the capacitor are shown.

<WRAP right>
<imgcaption imageNo04 | loading curve>
</imgcaption> <WRAP right>
{{drawio>Charge Curve}} \\
</WRAP>

<callout icon="fa fa-exclamation" color="red" title="Notice:">
  * There must be a unitless term in the exponent. So $RC$ must also represent a time. \ This time is called **time constant** $\tau =R \cdot C$. \\ \\
  * At time $t=\tau$, we get: $u_C(t) = U_q \cdot (1 - e^{- 1}) = U_q \cdot (1 - {{1}\over{e}}) = U_q \cdot ({{e-1}\over{e}}) = 0.63 \cdot U_q = 63\% \cdot U_q $ \ **So the capacitor is charged to $63$% after one $\tau$. \\ \\
  * At time $t=2 \cdot \tau$ we get: $u_C(t) = U_q \cdot (1 - e^{- 2}) = 86 \% \cdot U_q = (63 \% + (1-63 \%) \cdot 63 \% ) \cdot U_q$ \\ **So after each additional $\tau$, the uncharged remainder ($1-63 \%$) is recharged to $63\%$**. \\ \\
  * **After about $t=5 \cdot \tau$, the result is a capacitor** charged to over $99\%$. \ In real circuits, a charged capacitor can be assumed after $5 \cdot \tau$.
  * The time constant $\tau$ can be determined graphically in several ways:
    * Plotting the voltage value corresponding to $63\%$ on the y-axis. Finding the point of intersection with the graph. Reading the time (see green lines in <imgref FigNo04>).
    * Plotting the tangent to the (voltage) charge curve at the time of the discharged capacitor. \\ This intersects a horizontal line at the level of the charging voltage at the point $t=\tau$ (see black and light blue lines in <imgref imageNo04>).
</callout>

==== Discharging a capacitor at time t=0 ====

<WRAP right>
<imgcaption imageNo03 | circuit for viewing discharge curve>
</imgcaption> <WRAP right>
{{drawio>CircuitDischargeCurve3}} \\
</WRAP>

The following situation is considered for the discharge:
  * A capacitor charged to voltage $U_q$ with capacitance $C$ is short-circuited across a resistor $R$ at time $t=t_0$.
  * As a result, the full voltage $U_q$ is initially applied to the resistor: $u_R(t_0)=U_q$
  * The initial discharge current is thus defined by the resistance: $i_C ={u_R}\over{R}$
  * The discharging charges lower the voltage of the capacitor $u_C$, since: $u_C = {{q(t)}{C}}$
  * Ideally, the capacitor is not fully discharged until $t \rightarrow \infty$.

Also this process now is to put into formula in detail.
By looking at the mesh, the general result is: the sum of the two voltages across the resistor and capacitor add up to zero.

\begin{align*}
0 =u_R + u_C = R \cdot i_C + u_C
\end{align*}

This gives $(7.1.3)$:

\begin{align*}
0 =u_R + u_C = R \cdot C \cdot {{du_C}\over{dt}} + u_C
\end{align*}

--> also here uses some mathematics: #

This result again represents a 1st order differential equation. \\
The appropriate approach to such a problem is:

\begin{align*}
u_C(t) = \mathcal{A} \cdot e^{\mathcal{B}\cdot t} + \mathcal{C}
\end{align*}

\begin{align*}
0 &= R \cdot C \cdot {{d}\over{dt}}(\mathcal{A} \cdot e^{\mathcal{B}\cdot t} + \mathcal{C}) + \mathcal{A} \cdot e^{\mathcal{B}\cdot t} + \mathcal{C} \\
  &= R \cdot C \cdot \mathcal{AB} \cdot e^{\mathcal{B}\cdot t} + \mathcal{A} \cdot e^{\mathcal{B}\cdot t} + \mathcal{C} \\
0 - \mathcal{C} &= ( R \cdot C \cdot \mathcal{AB} + \mathcal{A} ) \cdot e^{\mathcal{B}\cdot t} \\
\end{align*}

This equation must hold for every $t$. This is only possible if the left as well as the right term become equal to 0. \\ Thus:

\begin{align*}
\mathcal{C} = 0 \\ \

R \cdot C \cdot \mathcal{AB} + \mathcal{A} &= 0 \quad \quad | : \mathcal{A} \quad | -1 \
R \cdot C \cdot \mathcal{B} &= - 1 \\
\mathcal{B} &= - {{1}\over{R C}} \\
\end{align*}

So it follows:

\begin{align*}
u_C(t) = \mathcal{A} \cdot e^{\large{- {{t}\over{R C}} }}
\end{align*}

For the solution it must still hold that at time $t_0=0$ $u_C(t_0) = U_q$ just holds:

\begin{align*}
U_q &= \mathcal{A} \cdot e^{\large{0}}  \\
U_q &= \mathcal{A}  \\
\mathcal{A} &= U_q
\end{align*}

<--

<WRAP right>
<imgcaption imageNo05 | discharge curve>
</imgcaption> <WRAP>.
{{drawio>Discharge Curve}} \\
<WRAP>


And this results in:
\begin{align*}
u_C(t) &= U_q \cdot e^{\large{- {{t}\over{\tau}}}} \quad \text{with} \quad \tau = R C
\end{align}

And with $(7.1.3)$, $i_C$ becomes:
\begin{align*}
i_C(t) &= - {{U_q}\over{R}} \cdot e^{\large{- {{t}\over{R C}} } }
\end{align*}

In <imgref FigNo05> the two time histories are again shown; this time for the discharge voltage $u_C(t)$ and the discharge current $i_C(t)$ of the capacitor. \\
Since the current now flows out of the capacitor, the sign of $i_C$ is negative.

~~PAGEBREAK~~~CLEARFIX~~~

==== Periodic switching operations ====

<WRAP right>{{url>https://www.falstad.com/circuit/circuitjs.html?running=false&ctz=CQAgjCAMB0l3BOJyWoVaYEBYDsuBmbSbMSAVmwDZcqRyJyRsCAoAN3G2xACZiuPMNyiiBTSKJjlWAJxAECdYT2KSVosqwDGzOIL2T+kk7HiReC6AhLFsOMPjCKwrAO4HjfAV8jvvRgJqAVD+XuECjnR+HrzkyrwAHAZgSaEeGqnJmSIxfPHgVIHqRekhYKVeUaEAHt644JDovFQE4KnM3iAAyqx1FW1giZZD6olMPPwgAKp94InqCIM24Ah0Qsl+dQS4QlgKCBCplpM8AEpzSkZNCvh8HacgAMKsO0aJ2SKKyiIQJqwAB2Y2E+qhB5Sk-m4yS8wV8UPBGmCGi2fDW7UsBEgRzAEy6AEtLtjwLxJARnCS2htnnM8EI7iDlAtOkMaXVsOR0I5JHgRoROlMAK6sCrvUH5H48a6Ncyy1hnPi4EaVJWrEZ8GVGaBMXjaqTa1gAew1f2YYH2EmgcR2jl2vAqThsbSMGra+IA+k8ADQAHQAzoLPa8YfrZX8zPA0FHkM8AIYA2PafEAF1jADttABTf0vAjZTTqeAgM6Zv34v2pjPZv0XPNdQb6MhFgBiskzAEdBZmMwBPf0AM1YQA 600,450 noborder}}
</WRAP>

In the simulation on the right, a periodic switching operation can be seen. The capacitor is periodically charged and discharged via the switch.
Three sliders are given in the simulation to change the resistance $R$ (%%Resistance R%%), the capacity $C$ (%%Capacity C%%) and the frequency $f$ (%%Frequency f%%).  \\
In the simulation below, the voltage $u_C$ across the capacitor is shown in green and the current $i_C$ is shown in yellow.

Tasks:
  - Increase the the frequency to $f=10kHz$ using the appropriate slider. What is the change for $u_C$ and $i_C$?
  - Now increase the capacitance to $C=10 \mu F$ using the corresponding slider. What is the change for $u_C$ and $i_C$?
  - Now increase the resistance to $R= 1 k\Omega$ using the corresponding slider. What is the change for $u_C$ and $i_C$?

~~PAGEBREAK~~~CLEARFIX~~~

===== 7.2 Energy of a capacitor =====

<callout>

=== Goals ===

After this lesson, you should:

  - Be able to calculate the energy content in a capacitor.
  - Be able to calculate the change in energy of a capacitor resulting from a change in voltage between the capacitor terminals.
  - Be able to calculate (initial) current, (final) voltage and charge when balancing the charge of several capacitors (also via resistors).
 
</callout>

<WRAP right>
<imgcaption imageNo02 | circuit for viewing the charge curve>
</imgcaption> <WRAP right>
{{drawio>CircuitDischargeCurve2}} \\
</WRAP>

Now the capacitor as energy storage is to be looked at more closely. This derivation is also explained in [[https://www.youtube.com/watch?v=je40ruFNKig|this youtube video]]. For this we consider again the circuit in <imgref BildNr02> an.
According to the chapter [[Basics and Basic Concepts#Determination_of_electrical_power_in_the_DC_circuit|Basics and Basic Concepts]], the power for constant values (DC) is defined as:

\begin{align*}
P={{{delta W}\over{{{delta t}} = U \cdot I
\end{align*}

For variable signals, the instantaneous power is given as:

\begin{align*}
p={dw}\over{dt}} = u \cdot i
\end{align*}

=== Energy consideration of the capacitor ===

Charging the capacitor at time $t_0 = 0$ results in $\Delta W = \Delta W_C$ for the stored energy at a later time $t_1 =t$:

\begin{align*}
\delta W_C = \int_{t_0}^{t_1} dw = \int_{0}^t u \cdot i \cdot dt = \int_{0}^t u_C \cdot i_C dt \tag{7.2.1}
\end{align*}

During the charging process

\begin{align*}
u_C(t) = U_q\cdot (1 - e^{ - {{t}\over{\tau}} }) \\
i_C(t) = {{U_q}\over{R}} \cdot e^{ -{t}\over{\tau}} } \tag{7.2.2}
\end{align}}

In particular:

\begin{align*}
C = {{q(t)}\over{u_C(t)}} \quad &\rightarrow \quad &q(t) &= {u_C(t)}\cdot{C}  \\
i_C(t) = {{d q(t)}\over{dt}} \quad &\xrightarrow{C=const} \quad &i_C(t) &= C \cdot {{d u_C(t)}\over{dt}}
\end{align*}

Thus, the stored energy from formula $(7.2.1)$:

\begin{align*}
\delta W_C &= \int_{0}^t u_C(t) \cdot C \cdot {{d u_C(t)}\over{dt}} dt \quad & | \text{ substitution of integration variable: } t \rightarrow u_C\\\
  &= \int_{U_0}^{U_1} u_C(t) \cdot C \cdot du_C \quad & | \text{ Since the capacity is constant, it can be written before the integral}\
  &= C \cdot \int_{U_0}^{U_1} u_C \, d u_C \
  &= C \cdot \left[{{1}\over{2}} u_C^2 \right] _{U_0}^{U_1} \\
\end{align*}
\begin{align}
\boxed{\delta W_C= {{1}\over{2}} C \cdot (U_1^2-U_0^2)} \tag{7.2.3}
\end{align*}

Thus, for a fully discharged capacitor ($U_q=0V$), the energy stored when charging to voltage $U_q$ is $\delta W_C={{1}\over{2}} C \cdot U_q^2$.

=== Energy consideration of resistance ===

The converted energy can also be determined for the resistance:

\begin{align*}
\delta W_R = \int_{0}^t u_R \cdot i_R dt = \int_{0}^t R \cdot i_R \cdot i_R dt = R \cdot \int_{0}^t i_R^2 dt
\end{align*}

Since the current through the capacitor $i_C$ is equal to that through the resistor $i_R$, it follows via $(7.2.2)$:

\begin{align*}
\Delta W_R &= R \cdot \int_{0}^t \left( { {U_q}\over{R}} \cdot e^ { -{t}\over{\tau}} \right)^2 dt \
   &= { {U_q^2}\over{R}} \cdot \int_{0}^t e^ { -{2 \cdot t}\over{\tau}} dt \\
   &= { {U_q^2}\over{R}} \cdot \left[ -{{\tau }\over{2}} \cdot e^ { -{2 \cdot t}\over{\tau}} \right]_{0}^t \quad & | \text{with } \tau = R \cdot C \.
   &= -{{1}\over{2}} \cdot {U_q^2}\cdot{C} \cdot \left[ e^ { -{{2 \cdot t}\over{\tau}} \right]_{0}^t \\
\end{align}}

For $t \rightarrow \infty$ we get:

\begin{align*}
\delta W_R &= -{{1}\over{2}} \cdot {U_q^2}\cdot{C} \cdot \left[ e^ { -{{2 \cdot t}\over{\tau}} \right]_{0}^{\infty} \\
   &= -{{1}\over{2}} \cdot {U_q^2}\cdot{C} \cdot \left[ 0 - 1 \right] \.
\end{align}
\begin{align}
\boxed{ \Delta W_R = {{1}\over{2}} \cdot{U_q^2}\cdot{C}} \tag{7.2.4}
\end{align}}

This means that the energy converted at the resistor is independent of the resistance value (for an ideal constant voltage source $U_q$ and given capacitor $C$)! At first, this doesn't really sound comprehensible. No matter if there is a very large resistor $R_1$ or a tiny small resistor $R_2$: The same waste heat is always produced. \\
Graphically, this apparent contradiction can be resolved like this: A higher resistor $R_2$ slows down the small charge packets $\Delta q_1$, $\Delta q_2$, ...  $\Delta q_n$ more strongly. But a considered single charge packet $\Delta q_k$ will nevertheless pass the same voltage across the resistor $R_1$ or $R_2$, since this is given only by the accumulated packets in the capacitor: $u_r = U_q - u_C = U_q - {{q}\over{C}}$.

In real applications, as mentioned in previous chapters, ideal voltage sources are not possible. Thus, without a real resistor, the waste heat is dissipated proportionally to the internal resistance of the source and the internal resistance of the capacitor. The internal resistance of the capacitor depends on the frequency, but is usually smaller than the internal resistance of the source.

=== Consideration of total energy expenditure ===

In the previous considerations, the energy conversion during the complete charging process was also considered. It was found that the capacitor stores the energy $W_C= {{1}\over{2}} \cdot {U_q^2}\cdot{C} $ (see $(7.2.3)$) and at the resistor the energy $W_R= {{1}\over{2}} \cdot {U_q^2}\cdot{C} $ (see $(7.2.4)$) into heat. So, in total, the voltage source injects the following energy:

\begin{align*}
\delta W_0 &=\delta W_R + \delta W_C = {U_q^2}\cdot{C}
\end{align*}

This also follows via $(7.2.1)$:

\begin{align*}
\Delta W_0 &= \int_{0}^{\infty} u_0 \cdot i_0 \cdot dt \quad | \quad u_0 = U_q \text{ is constant because constant voltage source!} \\
&= U_q \cdot \int_{0}^{\infty} i_C dt \
&= U_q \cdot \int_{0}^{\infty} {{dq}\over{dt}} dt \
&= U_q \cdot \int_{0}^Q dq = U_q \cdot Q \quad | \quad \text{where } Q= C \cdot U_q \\
&= U_q^2 \cdot C \
\end{align*}

This means that only half of the energy emitted by the source is stored in the capacitor! Again, this doesn't really sound comprehensible at first. Again, it helps to look at small packets of charge that have to be transferred from the ideal source to the capacitor. \ <imgref FigNo06> shows current and voltage waveforms across the capacitor and the stored energy for different resistance values. There, too, it can be seen that the maximum stored energy (dashed line in the figure at right) is given by $\Delta W= {{1}\over{2}} alone. U_q^2 \cdot C = {{1}\over{2}} \cdot (5V)^2 \cdot 1 \mu F = 12.5 \mu Ws$ is given.


<WRAP right>
<imgcaption imageNo06 | Current, voltage and energy during charging and discharging>
</imgcaption> <WRAP right>
{{drawio>ChargeCurrentVoltageEnergy}} \\
</WRAP>

~~PAGEBREAK~~~CLEARFIX~~~

This can also be tested in the following simulation. In addition to the RC element shown so far, a power meter and an integrator are also drawn in here. It is possible to display the instantaneous power and the stored energy. Via the slider %%Resistance R%% the resistance value can be varied. The following values are shown in the oscilloscopes:
  * left: Current $u_C$ and voltage $i_C$ at the capacitor.
  * middle: Instantaneous power $p_C = u_C \cdot i_C$ of the capacitor.
  * right: stored energy $w_C = \int u_C \cdot i_C \; dt$ of the capacitor

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</WRAP>


~~PAGEBREAK~~~CLEARFIX~~~

==== Tasks ====

<panel type="info" title="Task 7.2.1 Capacitor charging/discharging practice task"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>
{{youtube>QJAK2OpLqe0}}
</WRAP></WRAP></panel>

<panel type="info" title="Task 7.2.2 further capacitor charging/discharging practice task"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>
{{youtube>6QGnGR47eaA}}
</WRAP></WRAP></panel>

<panel type="info" title="Task 7.2.3 further practice charging the capacitor"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>
{{youtube>PpNmdPFNs90}}
</WRAP></WRAP></panel>

<panel type="info" title="Exercise 7.2.4 Exercise on charge balance of two capacitors"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>
{{youtube>ww7qLsWyVk0}}
</WRAP></WRAP></panel>

<panel type="info" title="Exercise 7.2.5 Exercise on charge balance of two capacitors"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>

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</WRAP>

On the right you see a simulation containing the two capacitors $C_1$ and $C_2$.
At the beginning $C_1$ is charged to $10V$ and $C_2$ to $0V$. With the switches $S_1$ and $S_2$ you can choose whether
  - the capacitances $C_1$ and $C_2$ are shorted, or
  - the capacitors $C_1$ and $C_2$ are connected via resistor $R$.

On the right side of the simulation there are some additional "measuring devices" to calculate the stored potential energy from the voltages across the capacitors. \\

In the following, the charging and discharging of a capacitor is to be explained with this construction.

~~PAGEBREAK~~ ~~CLEARFIX~~
Under the electrical structure, the following quantities are shown over time:

^ Voltage $u_1(C_1)$ of the first capacitor ^ Voltage $u_2(C_2)$ of the second capacitor ^ Stored energy $w_1(C_1)$ ^ Stored energy $w_2(C_2)$ ^ Total energy $\sum w$ ^
| Initially charged to $10V$ | Initially neutrally charged ($0V$) | Initially holds: \w_1(C_1)= {1 \over 2} \cdot C \cdot U^2 = {1 \over 2} \cdot 10\mu F \cdot (10V)^2 = 500\mu W$ \In the oscilloscope, $1V \sim equals 1W$ | Initially, $w_2(C_2)=0$ , since the capacitor is not charged. | The total energy is $w_1 + w_2 = w_1$|

The capacitor $C_1$ has thus initially stored the full energy and via a closing of the switch $S_2$ one would expect a balancing of the voltages and an equal distribution of the energy $w_1 + w_2 = 500\mu W$.

  - Close the switch $S_2$ (the toggle switch $S_1$ should point to the switch $S_2$). What do you find?
    - What do the voltages $u_1$ and $u_2$ do?
    - What the energies and the total energy? \ How is this understandable with the previous total energy?
  - Open $S_2$ - the changeover switch $S_1$ should not be changed. What do you find?
    - What do the voltages $u_1$ and $u_2$ do?
    - What the energies and the total energy? \\ How is this understandable with the previous total energy?
  - Repeat 1. and 2. several times. Can anything be deduced regarding the distribution of the energy?
  - Change the switch $S_2$ to the resistor. What do you observe?
    - What do the voltages $u_1$ and $u_2$ do?
    - What the energies and the total energy? \\ How is this understandable with the previous total energy?

</WRAP></WRAP></panel>

{{page>task_7.2.6_with_calculation&nofooter}}


~~NOCACHE~~
<affix offset-top="0" position-top="0" position-left="20">
<button size="xs" type="link" collapse="sidebar">{{icon>list}} Course Content</button>

<collapse id="sidebar" collapsed="false"><well>
\ [[start|{icon>home?18}}]]<fs large> [[start|Electrical Engineering 1]]</fs>

\ [[introduction_to_electrical_engineering_1|introduction_to_electrical_engineering 1]]\\

__Direct Current Technology__
\\ __ [[basics_and_basic_concepts|1. basics_and_basic_concepts]]
\\ ___ [[simple_dc_circuits|2. simple_circuits]]
\ [[linear_sources_and_bipoles|3. Linear sources and bipoles]]
\ [[analysis_of_dc_networks|4. Analysis of power systems]]
\\
\\ __Stationary fields and capacitances__]
\ [[the_electrostatic_field|5. The electrostatic field]]
\ [[the_stationary_electric_current_field|6. The_stationary_electric_current_field]]
\ [[switching_operations_on_rc-combinations|7. switching_operations_on_rc_members]]
</well>
~~BARCODE~url=https://wiki.mexle.org/elektrotechnik_1/start~size=S~~
[[https://creativecommons.org/licenses/by-sa/4.0/deed.de|{electronic_circuit_engineering:by-sa.png?100}}]
</collapse></affix>
====== Electrical Engineering 1 ======

You already know U-R-I and you don't only think of music when you think of AC/DC? \\
Great, then we should take it a step further. \\
In this course, we will look at,
  * How to understand the electric field and *
  * what resistors and capacitors do individually and in networks.
 
\\ \\
<WRAP indent><WRAP indent>
[[electrical_engineering_1:introduction_to_electrical_engineering_1|introduction_to_electrical_engineering 1]]\.
Or: What is the best way to handle the event?
</WRAP></WRAP>

__Direct Current Technology__
  - [[Fundamentals and Basic Concepts]] \\ ___ [[watt is current]]
  - [[Simple DC Circuits]] \\ ___ [[Branches and Stars]]
  - [[Linear sources and dipoles]] \\ or: something with 2 terminals and why a short circuit can be useful.
  - [[DC network analysis]] \\ [[Recipes for Networking]] \ <WRAP outdent><WRAP outdent> (nbsp) __Stationary Fields and Capacitances__ </WRAP></WRAP>.
  - [[The electrostatic field]] \\ __or: from attraction and repulsion to capacitor
  - [[The steady-state electric flow field]] \\ [[The steady-state electric flow field]] or: what's flowing?
  - [[Switching operations on RC combinations]] \\ or: charging to infinity

<WRAP indent><WRAP indent>
[[electrical engineering_1:tips for the electrical engineering 1 exam]]
</WRAP></WRAP>

====== Further links ======

==== Reading material ====

  * [[https://lx3.mint-kolleg.kit.edu/onlinekursphysik/html/sectionx1.4.0.html|Online Bridge Course of KIT/Uni Stuttgart]]: Nice, partially animated online script covering chapters 1, 2, 3 and 5.
  * [[https://www.eit.hs-karlsruhe.de/hertz/quicklink/startseite.html|H.Er.T.Z of HS Karlsruhe]]: The __**H**__university-open __**E**__lekt__**r**__o__**t**__echnik __**Z**__entrum of the Karlsruhe HS has a nice [[https://www.eit.hs-karlsruhe.de/hertz/nc/downloads/downloads-teil-b-gleichstromtechnik/downloadstatistik.html?download=Skript_Gleichstromechnik.pdf&did=9|Online-Skript]]
  * [[https://www.leifiphysik.de/elektrizitaetslehre|LeifiPhysics]]: Here you can find further explanations of our chapters on vocational school/high school level.
  * [[https://phys.libretexts.org/Bookshelves/University_Physics/Book%3A_University_Physics_(OpenStax)/Map%3A_University_Physics_II_-_Thermodynamics%2C_Electricity%2C_and_Magnetism_(OpenStax)|Electricity and Magnetism]]: beautiful online course covering most of the parts of my course.

==== Exercises ====

  * Matching the H.Er.T.Z script (see above), there are further [[https://www.eit.hs-karlsruhe.de/hertz/nc/downloads/downloads-teil-b-gleichstromtechnik/downloadstatistik.html?download=GET_Uebungsaufgaben.pdf&did=8|exercises]]
  * Further exercises will be handed out via ILIAS

==== Youtube recommendations ====

  * [[https://www.youtube.com/user/TheSimplePhysics/playlists|simple club]]: Explanatory videos on electrical engineering in the physics section ; subscription not necessary!
  * [[https://www.youtube.com/channel/UCyFX1F6hpkyXeEbLFURd8PA/playlists|Electrical engineering simply explained]]: still few, but well developed videos
  * [[https://www.youtube.com/channel/UCKC9gZw2CxA8y9sgBA58Wnw/featured|Electrical engineering in 5 minutes]]: good fund of short videos~~REVEAL theme=dokuwiki&fade=convex&controls=1&show_progress_bar=1&build_all_lists=1&show_image_borders=0&horizontal_slide_level=2&enlarge_vertical_slide_headers=0&show_slide_details=1&open_in_new_window=1&size=1024x768~~

<wrap footer>Electrical Engineering 1</wrap>

{{no-footer}}
====== Electrical Engineering Exam Tips ======

<notes>
This page is optimized for presentation mode.
</notes>

===== Specifications from SPO =====

  * 60 minutes
  * LK - course-accompanying by written examination
    * <wrap lo> usually in conjunction with a course and related to its content</wrap>.
    * <wrap lo>Subject examination related to exactly one course</wrap>.
    * <wrap lo>Grades: 1.0; 1.3; 1.7; ... 3.7; 4.0 and 5.0 (= anything worse than 4.0)</wrap>
  * Electrical Engineering Lab (2nd Sem):
    * <wrap lo>Electrical Engineering 1 with ≤ 4.0 </wrap>.

===== Coordinated specifications in T1 =====

  * Auxiliaries
    * <wrap lo>own collection of formulas (2 A4 pages)</wrap>
    * <wrap lo>non-programmable calculator</wrap>
  * Grade 4.0 ≙ 40%.
  * Total score: 50 points

===== General tips =====

  * Do not forget the derivation of results.
  * Read the task well. \\ Work on the task in the required way.
  * **Don't Panic** \ if a task is too difficult / takes too long \ → there are still others
  
  

-------------------------------------------------




====== 1. The magnetostatic field ======

===== 1.1 Magnetic phenomena =====

<callout>
=== Targets ===

After this lesson you should:
  - Know that forces act between magnetic poles and know the direction of the forces.
  - Know that a magnetic field is formed around a current-carrying conductor.
  - be able to sketch the field lines of the magnetic field. Know the direction of the field and where the field is densest.

</callout>

==== Effects around permanent magnets ====

<WRAP right>
<imgcaption imageNo01 | First approximation to magnetism></imgcaption>{{drawio>electrical engineering_2:iron ore}}
</WRAP>

First permanent magnets made of magnetic ironstone ($Fe_{3} O_{4}$) were found in Greece in the region around Magnesia. Besides the iron materials, other elements also show a similar "strong and permanent magnetic force effect", which is also called ferromagnetism after iron: Cobalt and nickel, as well as many of their alloys, also exhibit such an effect. Chapter [[#1.5 Matter in the magnetic field]] describes the subdivision of magnetic materials in detail.

Here now the "magnetic force effect" is to be looked at more near. For this purpose, a few thought experiments are carried out with a magnetic iron stone <imgref BildNr01> ([[https://www.youtube.com/watch?v=pmmmwLuc2ns|This video]] gives a similar introduction).

  - From the iron ore should now first be separated a handy elongated part. If one is lucky, the found iron ore is already magnetic by itself. This case will be considered in the following. The elongated piece is now to be cut into two small pieces.
  - As soon as the two pieces are removed from each other, one notices that the two pieces attract each other again directly at the cut surface.
  - If one of the two parts is turned (the upper part in the picture on the right), a repulsive force acts on the two parts.

So it seems that there is a directed force around each of the two parts. If you dig a little deeper you will find that this force is focused on one part of the outer surface.

Of course you already know magnets and also know that there are poles. The considered thought-experiment shall clarify, how one could have proceeded at an unknown appearance. In further thought experiments, such magnet iron stones can also be cut into other directions and the forces analysed.

~~PAGEBREAK~~~CLEARFIX~~~

The result here is:
  - There are two poles. These are called the north pole and the south pole. The north pole is coloured red, the south pole green.
  - Poles with the same name repel each other. Unequal poles attract each other. This is similar to the electric field (opposite charges attract).
  - So magnets experience a force in the vicinity of other magnets.
  - A compass is a small rotating "sample" magnet and is also called a magnetic needle. This sample magnet can thus represent the effect of a magnet. This is also similar to the sample charge of the electric field.
  - The naming of the magnetic poles was done by the part of the compass which points to the geographic north pole. The reason for this is that the magnetic south pole is found at the geographic north pole.
  - Magnetic poles are not isolatable. Even the smallest fraction of a magnet shows either no magnetism, or both north and south poles.

<WRAP right>
<imgcaption imageNo02 | Magnetic field becomes visible through iron filings></imgcaption>{{drawio>electrotechnology_2:Iron filings}}
</WRAP>

An interesting aspect is that even non-magnetised, ferromagnetic materials experience a force effect in the magnetic field. A non-magnetic nail is attracted by a permanent magnet. This even happens independently of the magnetic pole. This also explains the visualization about iron filings (= small ferromagnetic parts), see <imgref FigNo02>. Also here there is a force effect and a torque, which aligns the iron filings. The visible field seems to form field lines here.

<callout icon="fa fa-exclamation" color="red" title="Notice:">
  * Field line images can be visualized by iron filings. Conceptually, these can be understood as a string of sample magnets.
  * The **direction of the magnetic field** defined via the sample magnet: The north pole of the sample magnet points in the direction of the magnetic field.
  * The **amount of magnetic field** is given by the torque experienced by a sample magnet oriented perpendicular to the field.
  * Field lines seem to repel each other (transverse pressure). e.g. visible when the field exits the permanent magnet.
  * Field lines attempt to travel as short a path as possible (longitudinal pull).
</callout>

~~PAGEBREAK~~~CLEARFIX~~~

<WRAP right>
{{youtube>lM_ogtchwNc?start=79}}
</WRAP>

~PAGEBREAK~ ~CLEARFIX~~
==== Effects around current-carrying wires ====

<WRAP right>
<imgcaption imageNo03 | Magnetic field around a current-carrying conductor></imgcaption>{{drawio>electrotechnology_2:magnetic_field_around_conductor}}
</WRAP>


In 1820, Christian Ørsted discovered by chance during a lecture that current-carrying conductors also have an effect on a compass. This experiment is illustrated in <imgref FigNo03>. A long, straight conductor with a circular cross-section has current $I$ flowing through it. Due to symmetry considerations, the field line pattern must be radially symmetric with respect to the conductor axis. By an experiment with a magnetic needle it can be shown that the field lines form concentric circles.

<callout icon="fa fa-exclamation" color="red" title="Merke:">
  * If the technical direction of current is considered, the magnetic field lines surround the current in the sense of a right-hand screw. ("right screw rule")
  * This rule can also be remembered in another way: If the thumb of the **__r__**right hand points in the (technical) st**__r__**om direction, the fingers of the hand surround the conductor like the magnetic field lines. Likewise, if the thumb of the **__l__**left hand points in the E**__l__**ectron flow direction, the fingers of the hand surround the conductor like the magnetic field lines.
</callout>


~~PAGEBREAK~~~CLEARFIX~~~

==== Comparison of electrostatics and magnetostatics ====

^Property ^ Electrostatic ^ Magnetostatic ^
| Field line images | <imgcaption imageNo21 | Electrostatic field lines></imgcaption>{{drawio>electrotechnology_2:Electrostaticfieldlines}} | <imgcaption imageNo22 | Magnetostatic field lines></imgcaption>{{drawio>electrotechnology_2:Magnetostaticfieldlines}} |
| sample for the field | positive sample charge | compass |
| field lines | - start on a positive charge | - end on a negative charge | - are closed | - have no start and no end |
| field line ends | there are sources and sinks | there are __no__ sources and sinks |
| field type | vortex-free **__source field__** | source-free **__vortex field__** |

~~PAGEBREAK~~~CLEARFIX~~~

===== 1.2 Magnetic field strength =====

<callout>
=== Targets ===

After this lesson, you should:

  - know the two field-describing quantities of the magnetic field.
  - be able to describe and apply the relationship between these two quantities.

</callout>

==== Simulation and superposition of the magnetostatic field ====

<WRAP right>
Superposition of magnetic fields (only until 04:08).
{{youtube>qAOfVXJMxk8?end=248}}
</WRAP>

Before the magnetic field strength will be considered in more detail, the simulation and superposition of the magnetic field will be discussed in more detail here.

Magnetostatic fields can be superposed, just like electrostatic fields. This allows the fields of several current-carrying lines to be combined into a single one.
This trick is used in the following chapter to examine the magnetic field in more detail.

~~PAGEBREAK~~~CLEARFIX~~~

<WRAP right>
{{url>https://www.falstad.com/vector3dm/vector3dm.html?f=InverseRotational&d=streamlines&sl=none&st=19&ld=12&rx=75&ry=0&rz=0&zm=1.8 700,350 noborder}}
</WRAP>

On the right side the magnetic field of a single current-carrying conductor is shown. This was already derived at the previous chapter by symmetry considerations. The representation in the simulation can be simplified a bit here to see the conditions more clearly: Currently, the field lines are displayed in 3D, which is done by selecting "Display: Field Lines" and "No Slicing". If you change the selection to "Show Z Slice" instead of "No Slicing", you can switch to a 2D display. In this display, small compass needles can also show the magnetic field. To do this, select "Display: Field Vectors" instead of "Display: Field Lines". In addition, a "magnetic sample", i.e. a moving compass, can be found at the mouse pointer in the 2D display.

~~PAGEBREAK~~~CLEARFIX~~~

<WRAP right>
{{url>https://www.falstad.com/vector3dm/vector3dm.html?f=InverseRotationalDouble&d=streamlines&sl=z&st=20&ld=8&a1=51&rx=33&ry=0&rz=0&zm=1.2 700,450 noborder}}
</WRAP>

If there is another current-carrying conductor near the first conductor, the fields overlap. In the simulation on the right, the current of both conductors is directed in the same direction. The field between the conductors overlaps just enough to weaken. This can also be deduced by previous knowledge, if just the middle point between both conductors is considered: There the right hand rule results for the left conductor a vector directed towards the observer. For the right conductor results a vector directed away from the observer. These just cancel each other out. Further outward field lines go around both conductors. North- and south-poles here are not fix localized towards outside.

~~PAGEBREAK~~~CLEARFIX~~~
<WRAP right>
{{url>https://www.falstad.com/vector3dm/vector3dm.html?f=InverseRotationalDipole&d=streamlines&sl=z&st=20&ld=8&a1=51&rx=33&ry=0&rz=0&zm=1.2 700,450 noborder}}
<WRAP>

If, on the other hand, the current in the second conductor is directed in the opposite direction to the current in the first conductor, the picture changes: Here there is a reinforcing superposition between the two conductors.
Using the nomenclature from the previous chapter, it is also possible to assign north and south poles locally. Towards the outside, one pole appears to be located in front of the two conductors and another one behind.

in both simulations, the distances between the conductors can also be changed using the "Line Separation" slider. What do you notice in each case when the two lines are brought close together?

~~PAGEBREAK~~ ~CLEARFIX~~
==== Derivation of magnetic field strength (part 1, via toroidal coil) ====

<WRAP right>
<imgcaption imageNo04 | Magnetic field in a toroidal coil></imgcaption> {{url>https://www.falstad.com/vector3dm/vector3dm.html?f=ToroidalSolenoidField&d=streamlines&sl=none&st=1&ld=8&a1=77&a2=26&a3=100&rx=0&ry=0&rz=0&zm=1.8 700,450 noborder}}
<WRAP>

So far the magnetic field was defined quite pragmatically by the effect on a compass.
For a deeper analysis of the magnetic field, the field is now to be considered again - as with the electric field - from __two_ directions.
The magnetic field will also be considered as a "causer field" (a field produced by magnets) and an "acting field" (field acting on a magnet).
This chapter will first discuss the effective magnetic field. For this, it is convenient to consider the effects inside a toroidal coil (= toroidal coil).
This can be seen in <imgref FigNo04>. For reasons of symmetry, it is also clear here that the field lines form concentric circles.

In an experiment, a magnetic needle inside the toroidal coil is now to be aligned at right angles to the field lines.
Then the magnetic field will generate a torque which tries to align the magnetic needle in the field direction.

It now follows:
  - $M \neq M(\phi)$ : For the same distance from the axis of symmetry, the torque $M$ is independent of the angle $\phi$.
  - $M \sim I$ : The stronger the current flowing through a winding, the stronger the effect, i.e. the stronger the torque.
  - $M \sim N$ : The greater the number $N$ of windings, the stronger the torque $M$.
  - $M \sim {1 \over l}$ : The smaller the mean coil circumference $l$ the greater the torque. The mean coil circumference $l$ is equal to the field line length.

To summarize:
\begin{align}}
M \sim {{I \cdot N}\over{l}}
\end{align*}


The **magnetic field strength** $H$ inside the toroidal coil is given as:
\begin{align}}
\boxed{H ={{I \cdot N}\over{l}} \quad \quad | \quad \text{applies to toroidal coil only}}
\end{align}}

As unit of the **magnetic field strength** $H$ we get $[H] = {{[I]}\over{[l]}}= 1{{A}\over{m}}$

==== Magnetic flux ====

The cause of the magnetic field is the current in the winding of the coil. If this current $I$ and/or the number $N$ of windings is increased, the effect is amplified. To make this easier to handle, we introduce the **magnetic flux leakage**. The magnetic flux $\theta$ is defined as

\begin{align}
\boxed{\theta = N \cdot I}
\end{align*}
The unit of $\theta$ is: $[\theta]= 1A$ (obsoletely called ampere-turn).

Thus the magnetic field strength $H$ of the toroidal coil is given by: $H= {{\theta}\over{l}}$

==== Derivation of the magnetic field strength (part 2, straight conductor) ====

The previous derivation from the toroidal coil is now to be used to derive the field strength around a long, straight conductor. The flux $\theta$ for a single conductor is given by $\theta = N \cdot = 1 \cdot I = I$. For the toroidal coil, the magnetic field strength was given by the flux $\theta$ divided by the (mean) field line length. Because of the (same rotational) symmetry, this is also true for the single conductor.

The length of a field line around the conductor is given by the distance $r$ of the field line from the conductor: $l = l(r) = 2 \cdot \pi \cdot r$. \ For the magnetic field strength of the single conductor we then get:
\begin{align*}
\boxed{H ={{theta}\over{l}} = {{I}\over{2 \cdot \pi \cdot r}} \quad \quad \text{applies only to the long, straight ladder}}
\end{align}

<panel type="info" title="Task 1.2.1 Magnetic field strength around a horizontal straight conductor"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>

The current $I = 100A$ flows in a long straight conductor with a round cross-section. The radius of the conductor is $r_{L}= 4mm$.

  * What is the magnetic field strength $H_1$ at a point $P_1$, which is __outside__ the conductor at a distance of $r_1 = 10cm$ from the conductor axis?
  * What is the magnetic field strength $H_2$ at a point $P_2$, which is __inside__ the conductor at a distance of $r_2 = 3mm$ from the conductor axis?

</WRAP></WRAP></panel>

<panel type="info" title="Task 1.2.2 Superposition"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>

<WRAP right>
<imgcaption imageNo01 | ladder arrangement>
</imgcaption> <WRAP right>
{{drawio>Task1LadderArrangement}} \\
<WRAP>

Three long straight conductors are arranged in a vacuum so that they lie at the vertices of an equilateral triangle (see <imgref FigNo01>). The radius of the circumcircle is $r = 2 cm$.

What is the magnetic field strength $H$ at the center of the equilateral triangle?
</WRAP></WRAP></panel>

~~PAGEBREAK~~~CLEARFIX~~~

In the electric field, the field line density was a measure of the strength of the field. This is also used for the magnetic field. Looking at the simulations in Falstad (below left) with this understanding, one notices an inconsistency: contrary to the relationship just given, the field line density in the Falstad simulation __**not**__ indicates the strength of the field. Below right, a realistic simulation is shown for comparison, which makes the difference clear: the field is stronger near the conductor. Thus the field line density must also be stronger there.

<WRAP right>
{{url>https://www.geogebra.org/material/iframe/id/fy0yrGYK/width/500/height/500/border/888888/rc/false/ai/false/sdz/false/smb/false/stb/false/stbh/true/ld/false/sri/true/at/preferhtml5 700,350 noborder}}
</WRAP>

<WRAP right>
{{url>https://www.falstad.com/vector3dm/vector3dm.html?f=InverseRotational&d=streamlines&sl=z&st=19&ld=12&rx=0&ry=0&rz=0&zm=1.8 700,350 noborder}}
</WRAP>

~~PAGEBREAK~~ ~CLEARFIX~~
<callout icon="fa fa-exclamation" color="red" title="Attention:">
  * The density of the field lines is a measure of the field strength.
  * The simulation in Falstad cannot represent this in this way. Here the field strength is coded by the color intensity (dark green = low field strength, light green to white = high field strength).
</callout>

==== Derivation of the magnetic field strength (part 3, generalization) ====

So far, only rotationally symmetric problems could be solved. Now this shall be generalized. For this purpose we will have a look back to the electric field. For the electric field strength $E$ of a capacitor with two plates at a distance of $s$ and the potential difference $U$ holds:

\begin{align*}
U = E \cdot s \quad \quad | \text{applies to capacitor only}
\end{align}

This was extended to $U = \int_s E ds$. If we transform the formula for the magnetic field strength $H$ of a toroidal coil with the mean field line length $l$ for comparison, we get

\begin{align}
\theta = H \cdot l \quad \quad | \quad \text{applies to toroidal coil only}
\end{align*}

Can you see the similarities? Again, the magnitude of the field strength is multiplied by the length to arrive at another field-describing quantity (here, the flux $\theta$). Because of the similarity - which continues below - the so-called **magnetic (circulating) voltage $V_m$** is introduced:

\begin{align*}
V_m = H \cdot s \quad \quad | \text{applies to toroidal coil only}
\end{align*}

Now what is the difference between the magnetic voltage $V_m$ and the flux $\theta$?
  - The first equation of the toroidal coil ($\theta = H \cdot l$) is valid for one turn along a field line. In addition, the flux is fixed by current and number of windings: $\theta = N \cdot I$.
  - The second equation ($V_m = H \cdot s$) holds independently of the path length $s$ along the field line. If just $s = l$ is chosen, the magnetic voltage equals the magnetic flux.

Thus, for each infinitesimally small path $ds$ along a field line, the resulting infinitesimally small magnetic voltage $dV_m = H \cdot ds$ can be determined. If now along the field line the magnetic field strength $H = H(\vec{s})$ changes, then the magnetic voltage from point $\vec{s_1}$ to point $\vec{s_2}$ results to:

\begin{align*}
V_{m12} = V_m(\vec{s_1}, \vec{s_2}) = \int_\vec{s_1}^\vec{s_2} H(\vec{s}) ds
\end{align*}

Up to now only the situation was considered that one always walks along the same field line. $\vec{s}$ therefore always arrived at the same field line. If one wants to extend this to arbitrary directions (also transverse to field lines), then only that part of the magnetic field strength $\vec{H}$ may be used in the formula, which is parallel to the path $d \vec{s}$. This is made possible by scalar multiplication. Thus, it is generally valid:

\begin{align*}
\boxed{V_{m12} = \int_\vec{s_1}^\vec{s_2} \vec{H} \cdot d \vec{s}}
\end{align*}


  * closed ring integral so that $V_m = \theta$

==== Application of the generalized form ====

=== one or more current-carrying conductors ===

  * Verification of the equation for single conductor
  * For multiple, $\theta = \sum I$
  * Node theorem
  * graphical examples for magn. voltage per circuit
 
=== spatially extended flow ===

  * Right-hand screw between $d \vec{s}$ and $d \vec{a}$

~~PAGEBREAK~~~CLEARFIX~~~

<panel type="info" title="Task 1.2.3 magnetic tension"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>

<WRAP right>
<imgcaption imageNo05 | different trajectories around current-carrying conductors>
</imgcaption> <WRAP>.
{{drawio>Task3MagneticFieldCurrentFlowingConductor}} \\
</WRAP>

Given are the adjacent closed trajectories in the magnetic field of current-carrying conductors (see <imgref Fig.No.05>). Let $I_1 = 2A$ and $I_2 = 4.5A$ be valid.

In each case, the magnetic voltage $V_m$ along the drawn path is sought.

</WRAP></WRAP></panel>

~~PAGEBREAK~~~CLEARFIX~~~

===== 1.3 Ampere's power law, magnetic flux density =====
 
<callout>

=== Goals ===

After this lesson, you should:

  - know the force law for current-carrying conductors.
  - Be able to determine the direction of the forces using given current directions and, if applicable, flux density.
  - be able to represent the acting force vectors in a sketch.
  - be able to determine a force vector by superimposing several force vectors using vector calculus.
  - be able to state the following quantities for a force vector:
      - Force vector in coordinate representation
      - magnitude of the force vector
      - Angle of the force vector

</callout>


{{youtube>cU6Pbb71dyo}}

Please have a look at the contents (text, videos, exercises) on the page of the [[https://lx3.mint-kolleg.kit.edu/onlinekursphysik/html/1.4.2/xcontent1.html|KIT-Bridge Course >> 4.2.2 Magnetic Field]]. Make sure that "Total" is selected in the selection bar at the top. The last part on "Magnetic field with matter" can be skipped - it will come in 2-3 terms.

\begin{align*}
F = {{\mu _0}\over{2 \pi}} \cdot {{I_1 \cdot I_2 }\over{r}} \cdot l
\end{align}}

With:
  * Conductor length $l$
  * distance between conductors $r$
  * currents through the conductors $I_1$ and $I_2$
  * vacuum permeability $\mu _0 = 4 \pi \cdot 10^{-7} {{Vs}\over{Am}}$

===== 1.4 Lorentz force =====

<callout>
=== Targets ===

After this lesson, you should:

  - be able to represent the vectors of the magnetic flux density in a sketch when several current-carrying conductors are specified.
  - be able to determine the resulting vector of magnetic flux density by superimposing several vectors using vector calculus.
  - Be able to determine the force on a current-carrying conductor in a magnetostatic field by applying the force law for current-carrying conductors in a magnetic field:
    - Force vector in coordinate representation
    - magnitude of the force vector
    - Angle of the force vector
  
</callout>

==== Video ====

Please have a look at the contents (text, videos, exercises) on the page of the [[https://lx3.mint-kolleg.kit.edu/onlinekursphysik/html/1.4.2/xcontent2.html|KIT-Brückenkurs >> 4.2.3 Lorentz-Kraft]]. Make sure that "Total" is selected in the selection bar at the top. The last part on "Magnetic field with matter" can be skipped.


===== 1.5 Matter in the magnetic field =====

<callout>

=== Goals ===

After this lesson, you should:

  - know the two field-describing quantities of the magnetostatic field.
  - be able to describe and apply the relationship between these two quantities via the material law.
  - know the classification of magnetic materials.
  - be able to read the relevant data from a magnetisation characteristic curve.

</callout>


===== 1.6 Poynting Vector (not part of the curriculum) =====

  * Clear picture of the Poynting vector along an electric circuit: https://de.cleanpng.com/png-jyy1vj/
  * Good explanation of theEnergy flow via a current model: https://www.forphys.de/Website/elekt/stromodl.html
  * Very detailed view of the energy flow in an electric circuit: http://www.didaktik.physik.uni-duisburg-essen.de/~backhaus/publicat/Energie.pdf

Force effect on dia- and paramagnetic materials in the magnetic field.
{{youtube>jeLzdmTnrHM}}

A living frog ("diamagnet") floats in a very strong magnetic field
{{youtube>KlJsVqc0ywM?start=45}}

Explanation of the hysteresis curve
{{youtube>WKCrchkTXv0}}

Nice illustration of magnetization and demagnetization of soft magnetic material.
{{youtube>cEGP50lopYA?start=7&end=128}}

Wandering magnetic domains in a ferromagnetic material (from [[https://en.wikipedia.org/wiki/Ferromagnetism#/media/File:Moving_magnetic_domains_by_Zureks.gif|Zureks@en.wikipedia.org]] at [[https://creativecommons.org/licenses/by-sa/3.0|CC-BY-SA 3.0]]).

{{https://upload.wikimedia.org/wikipedia/commons/0/06/Moving_magnetic_domains_by_Zureks.gif|}}

===== Tasks =====

 **<fs large>Task 1</fs>**

<WRAP group> <WRAP half column>

<quizlib id="quiz" rightanswers="[['a0'],['a2'], ['a2'], ['a0'], ['a1'], ['a2']]" submit="Check Answers">
    <question title="1. Which hand can be used to infer magnetic field direction from currents?" type="radio">The right hand|The left hand</question>.
    <question title="2. In the derivation from 1. how are the fingers to be assigned?" type="radio"> Thumb for current direction, remaining fingers for magnetic field direction | Thumb for magnetic field direction, remaining fingers for current direction| both possibilities are correct </question>
    <question title="3. Two conductors carrying current are parallel and close to each other. The current in both is flowing in the same direction. What force effect is seen?" type="radio"> none | The conductors attract | The conductors repel</question>.
    <question title="4. Two conductors carrying current are at right angles to each other. Current flows through both of them. What force effect can be seen?" type="radio"> none | The conductors attract | The conductors repel</question>.
    <question title="5. What is the magnetic field inside the earth or a permanent magnet?" type="radio"> from the magnetic north pole to the south pole | from the magnetic south pole to the north pole | the inside is free of field</question>.
    <question title="6. At which location of a current-carrying coil are the field lines densest?" type="radio"> at the magnetic north pole | at the magnetic south pole | inside the coil | at both poles </question>

</quizlib>

</WRAP> <WRAP half column>
++++Tip to 1|
For St<fs large>__**R**__</fs>omfluss, you use which hand?
++++

++++Tip for 2|
  * Imagine a coil with a winding pictorially, or draw it on.
  * Now think of a generated field through this to it. What direction must the current causing it be? Does this fit the rule of thumb?
  * Then try it the other way round: If a current is given, where do the field lines go in and where out? What poles are created there?
++++

++++Tip for 3| See 3rd video.
  * Picture the two wires, or draw it on.
  * In which direction would the outer field run in each case?
  * The field is a linear vector field. So the total field can be created from several individual fields by adding them together. Does adding the field in between make it larger, or smaller?
++++

++++Tip to 4|
  * First imagine the parallel wires again. What happens when the current flows in the same direction and what happens when the current flows in opposite directions? Are the resulting forces equal in magnitude?
  * The reversal of the direction of the current can now also be produced by turning the wire instead of changing the current - just so that the wires are perpendicular to each other in the meantime when turning.
  * If now with parallel wires and different current-direction the amount-wise same force arises. so this is also with every angle in between (detailed about integration of the force over single wire-pieces).
  * But then there must be a point at which the force becomes 0.
++++

++++Tip for 5|
  * The magnetic field lines must be closed.
  * Compare the field curve between coil and permanent magnet.
++++
++++Tip for 6|
  * In video 1 you can see the course outside and inside the coil.
++++

</WRAP> </WRAP>


--> References to the media used #

^ Element ^ License ^ Link ^
| <imgref ImageNo01> | [[https://creativecommons.org/licenses/by-sa/3.0/|CC-BY-SA 3.0]]  | https://de.wikipedia.org/wiki/Magnetit#/media/Datei:Chalcopyrite-Magnetite-cktsr-10c.jpg |
| <imgref imageNo02> | Public Domain | https://commons.wikimedia.org/wiki/File:Magnetic_field_of_bar_magnets_attracting.png |
| <imgref imageNo22> | [[https://creativecommons.org/licenses/by-sa/3.0/|CC-BY-SA 3.0]]  | https://commons.wikimedia.org/wiki/File:VFPt_Solenoid_correct.svg |

<--~NOTOC~~

====== 2. The time-varying magnetic field ======


===== 2.1 Magnetic flux and induction (motion induction, rest induction) =====

--> Goals and Video #

<callout>

=== Goals ===

After this lesson, you should:

  - know how magnetic flux is defined.
  - be able to state the magnetic flux of an arrangement for a given area from the magnetic flux density.
  - know the general meaning of "Gauss's theorem for the magnetic field".
  - be able to select a closed enveloping surface appropriately and apply the Gaussian theorem.
  - be able to apply Lenz's rule.
  - be able to calculate the induced voltage in the case of a variable magnetic field and/or variable, flowed-through area.  

</callout>

==== Video ====

Please have a look at the contents (text, videos, exercises) on the page of the [[https://lx3.mint-kolleg.kit.edu/onlinekursphysik/html/1.4.2/xcontent3.html|KIT-Brückenkurs >> 4.2.4 Elektromagnetische Induktion]]. Make sure that "Total" is selected in the selection bar at the top.

Simple online experiments on induction are available at [[https://phet.colorado.edu/sims/html/faradays-law/latest/faradays-law_en.html|PhET - Induction]] and [[https://phet.colorado.edu/sims/cheerpj/faraday/latest/faraday.html?simulation=faraday|PhET - Induction Construction Kit]]. These can help to visualize the effects.

Comparison electric vs magnetic flux
{{youtube>vActzYz_Q2E?start=560}}

==== Tasks ====
see Ilias Course: "3.1 Induction Effects"
  * For this, you should already know the unit [[https://de.wikipedia.org/wiki/Magnetischer_Fluss#Ma%C3%9Feinheit|Weber]] from the video above.


<WRAP pagebreak />

===== 2.2 The chained flow =====

see PPT slides in ILIAS


<WRAP pagebreak />
===== 2.3 Self Induction =====

<callout>

==== Targets ====

After this lesson, you should:

  - know how magnetic flux is defined.
  - be able to state the magnetic flux of an arrangement for a given area from the magnetic flux density.
  - know the general meaning of "Gauss's theorem for the magnetic field".
  - be able to select a closed enveloping surface appropriately and apply the Gaussian theorem.

</callout>

=== Video ===

Please have a look at the contents (text, videos, exercises) on the page of the [[https://lx3.mint-kolleg.kit.edu/onlinekursphysik/html/1.4.2/xcontent4.html|KIT-Bridge Course >> 4.2.5 Self Induction and Coil (*)]]. Make sure that "Total" is selected in the selection bar at the top.

==== Tasks ====

see Ilias course: "3.2 Self-inductance and self-induction".



<WRAP pagebreak />
===== 2.4 Interconnection of inductors =====

   - Excitation field
   - the magnetic voltage is also sometimes called MMK (magnetomotive force) or flux (since it is related to the flowing current). More memorable for the consideration of magnetic effects, however, is the magnetic voltage.

<callout>

==== Targets ====

After this lesson, you should:

  - know the two field-describing quantities of the magnetic field.
  - be able to describe and apply the relationship between these two quantities.
  - Know the definition of the magnetic voltage and be able to calculate it in a magnetic field.
  - Understand why the calculation of the magnetic voltage is path-independent.
  - know the law of flux leakage.
  - Be able to apply the flow law to a given arrangement of current-carrying conductors.

</callout>

==== Video ====

Please have a look at the contents (text, videos, exercises) on the page of the [[https://lx3.mint-kolleg.kit.edu/onlinekursphysik/html/1.4.3/xcontent3.html|KIT-Bridge Course >> 4.3.4 Rewind (*)]]. Make sure that "Total" is selected in the selection bar at the top.


==== Assignments ====
See KIT Bridge Course


<WRAP pagebreak />
====== Related Links ======

  * Slides of the Children's University 2019: [[:slides_and_explanation_for_the_children's_uni|From Duplo engine to aluminium foil brake]]
  * [[https://www.ipes.ethz.ch/mod/lesson/view.php?id=21&pageid=64| IPES ETHZ]]: interactive illustration of flow change with permeability and geometry of a flowed object
  * [[https://www.ipes.ethz.ch/mod/lesson/view.php?id=21&pageid=63| IPES ETHZ]]: interactive representation for magnetic shielding
  * [[https://www.ipes.ethz.ch/mod/lesson/view.php?id=22| IPES ETHZ]]: interactive representation for the field of a coil ~~NOTOC~~

====== The magnetic circuit ======

--> 7.1 The magnetic voltage #

--> Goals and Video #

<WRAP group> <WRAP half column><callout>

==== Targets ====

After this lesson, you should:

  - know under which assumptions the calculation can be performed on a linear magnetic circuit.
  - Know the definition of magnetic potential and be able to calculate it in a magnetic field.
  - Understand why the calculation of the magnetic potential is path-independent.
  - Be able to apply the law of flux to a magnetic circuit.
  - Know Ohm's law of the magnetic circuit.
  - be able to construct an equivalent circuit for a magnetic circuit.
  - be able to calculate the magnetic resistance of a linear magnetic circuit.
  - be able to calculate all relevant quantities of the linear magnetic circuit.

</callout></WRAP> <WRAP half column>

==== Video ====

The electric charge

{{youtube>JnYrmCaQfcM}}

</WRAP> </WRAP>

<--

--> Tasks #
=== Task 1 ===

<WRAP group> <WRAP half column> text

</WRAP> <WRAP half column>

[[https://phet.colorado.edu/sims/html/charges-and-fields/latest/charges-and-fields_de.html|PHET: Charges and Fields]] </WRAP> </WRAP>

<--

<--

--> 7.2 The electrical flow #

   - the electric flux (= related to the flowing current) denotes the magnetic voltage of a closed circuit. It is also occasionally referred to as MMK (magnetomotive force). More memorable for the consideration of magnetic effects, however, is the magnetic voltage.

--> Goals and Video #

<WRAP group> <WRAP half column><callout>

==== Targets ====

After this lesson, you should:

  - know under which assumptions the calculation can be done on a linear magnetic circuit.
  - be able to apply the flux law to a magnetic circuit.
  - know the Ohm's law of the magnetic circuit.
  - be able to construct an equivalent circuit for a magnetic circuit.
  - be able to calculate the magnetic resistance of a linear magnetic circuit.
  - be able to calculate all relevant quantities of the linear magnetic circuit.

</callout></WRAP> <WRAP half column>

==== Video ====

The Flood Law
{{youtube>Mv-tmCDmOj8}}

Flow, resistance, remanence.
{{youtube>m7FdpjHfQT4?start=67}}

</WRAP> </WRAP>

<--

--> Tasks #
=== Task 1 ===

<--

<--

--> 7.3 The linear magnetic circuit #

--> Goals and Video #

<WRAP group> <WRAP half column><callout>

==== Targets ====

After this lesson, you should:

  - know under which assumptions the calculation can be done on a linear magnetic circuit.
  - be able to apply the flux law to a magnetic circuit.
  - know the Ohm's law of the magnetic circuit.
  - be able to construct an equivalent circuit for a magnetic circuit.
  - be able to calculate the magnetic resistance of a linear magnetic circuit.
  - be able to calculate all relevant quantities of the linear magnetic circuit.

</callout></WRAP> <WRAP half column>

==== Video ====

Example: Magnetic circle with air gap
{{youtube>R1oQFfstFlQ}}

</WRAP> </WRAP>

<--

--> Tasks #
=== Task 1 ===

<--

<--

--> 7.4 The nonlinear magnetic circuit #

--> Goals and Video #

<WRAP group> <WRAP half column><callout>

==== Targets ====

After this lesson, you should:

  - know the limits of the linearised calculation of a magnetic circuit.
  - be able to solve simple non-linear problems with the aid of a magnetisation characteristic.

</callout></WRAP> <WRAP half column>

==== Video ====

The electric charge

{{youtube>JnYrmCaQfcM}}

</WRAP> </WRAP>

<--

--> Tasks #
=== Task 1 ===

<WRAP group> <WRAP half column> text

</WRAP> <WRAP half column>

[[https://phet.colorado.edu/sims/html/charges-and-fields/latest/charges-and-fields_de.html|PHET: Charges and Fields]] </WRAP> </WRAP>

<--

<--

--> 7.5 The counter induction#

--> Goals and Video #

<WRAP group> <WRAP half column><callout>

==== Targets ====

After this lesson, you should:

  - know under which assumptions the calculation can be done on a linear magnetic circuit.
  - be able to apply the flux law to a magnetic circuit.
  - know the Ohm's law of the magnetic circuit.
  - be able to construct an equivalent circuit for a magnetic circuit.
  - be able to calculate the magnetic resistance of a linear magnetic circuit.
  - be able to calculate all relevant quantities of the linear magnetic circuit.

</callout></WRAP> <WRAP half column>

==== Video ====

The electric charge

{{youtube>JnYrmCaQfcM}}

</WRAP> </WRAP>

<--

--> Tasks #
=== Task 1 ===

<WRAP group> <WRAP half column> text

</WRAP> <WRAP half column>

[[https://phet.colorado.edu/sims/html/charges-and-fields/latest/charges-and-fields_de.html|PHET: Charges and Fields]] </WRAP> </WRAP>

<--

<--

--> 7.6 Coupling of coils#

--> Goals and Video #

<WRAP group> <WRAP half column><callout>

==== Targets ====

After this lesson, you should:

  - know under which assumptions the calculation can be done on a linear magnetic circuit.
  - be able to apply the flux law to a magnetic circuit.
  - know the Ohm's law of the magnetic circuit.
  - be able to construct an equivalent circuit for a magnetic circuit.
  - be able to calculate the magnetic resistance of a linear magnetic circuit.
  - be able to calculate all relevant quantities of the linear magnetic circuit.

</callout></WRAP> <WRAP half column>

==== Video ====

The electric charge

{{youtube>JnYrmCaQfcM}}

</WRAP> </WRAP>

<--

--> Tasks #
=== Task 1 ===

<WRAP group> <WRAP half column> text

</WRAP> <WRAP half column>

[[https://phet.colorado.edu/sims/html/charges-and-fields/latest/charges-and-fields_de.html|PHET: Charges and Fields]] </WRAP> </WRAP>

<--

<--

--> 7.7 The magnetic energy #

--> Goals and Video #

<WRAP group> <WRAP half column><callout>

==== Targets ====

After this lesson, you should:

  - know under which assumptions the calculation can be done on a linear magnetic circuit.
  - be able to apply the flux law to a magnetic circuit.
  - know the Ohm's law of the magnetic circuit.
  - be able to construct an equivalent circuit for a magnetic circuit.
  - be able to calculate the magnetic resistance of a linear magnetic circuit.
  - be able to calculate all relevant quantities of the linear magnetic circuit.

</callout></WRAP> <WRAP half column>

==== Video ====

Energy in the magnetic field

{{youtube>O-pmfVgh6Do}}

</WRAP> </WRAP>

<--

--> Tasks #
=== Task 1 ===

<WRAP group> <WRAP half column> text

</WRAP> <WRAP half column>

[[https://phet.colorado.edu/sims/html/charges-and-fields/latest/charges-and-fields_de.html|PHET: Charges and Fields]] </WRAP> </WRAP>

<--

<--

--> 7.8 Application examples#

--> Goals and Video #

<WRAP group> <WRAP half column><callout>

==== Targets ====

After this lesson, you should:

  - know what an elementary charge is and that forces act between charges.
  - know Coulomb's law.
  - Be able to determine the direction of the forces on the basis of given charges.
  - be able to represent the acting force vectors in a sketch.
  - Be able to determine a force vector by superimposing several force vectors using vector calculus.
  - be able to state the following quantities for a force vector:
      - Force vector in coordinate representation
      - magnitude of the force vector
      - Angle of the force vector

</callout></WRAP> <WRAP half column>

==== Video ====

The electric charge

{{youtube>JnYrmCaQfcM}}

</WRAP> </WRAP>

<--

--> Tasks #
=== Task 1 ===

<WRAP group> <WRAP half column> text

</WRAP> <WRAP half column>

[[https://phet.colorado.edu/sims/html/charges-and-fields/latest/charges-and-fields_de.html|PHET: Charges and Fields]] </WRAP> </WRAP>

<--

<--

===== Related Links =====
  * [[https://www.ipes.ethz.ch/mod/lesson/view.php?id=22&pageid=66| IPES ETHZ]]: interactive representation of the field in a soft iron core
  * [[https://www.ipes.ethz.ch/mod/lesson/view.php?id=22&pageid=67| IPES ETHZ]]: interactive representation of the flux in a soft iron core====== Information on the electrical engineering exam ======

The 2-page DIN A4 formularies for the exams "Electrical Engineering I" and "Electrical Engineering II" have to be handwritten.

_2-sided DIN A4__ means either one DIN A4 page with writing on the front and back, or two DIN A4 pages, each with writing on one side.

The aim of the __handwritten preparation__ is that you think about what should be in your formulary and how. Tip: It is advisable to write the formulary with a soft pencil and to keep it in a transparent envelope. Then changes are easily possible and contamination during exercises is avoided.

Please avoid photocopied or printed formularies, or a formularies with more than 2 pages DIN A4. Since these boundary conditions are examination requirements (which are consistently handled in this way in the faculty), a deviation must be evaluated as an "unauthorized aid" - i.e. the examination must be graded with 5.0.====== 5. Multiphase systems ======

===== 5.0 Preliminary consideration =====
So far, AC current, AC voltage and their effects have been considered on a circuit that had only included a source AC voltage. \\
These circuits can be understood as.
  * the sinusoidal alternating voltage is produced by the rotation of a coil in a homogeneous magnetic field, and
  * the sinusoidal alternating current is formed by a connected load (or complex impedance).
are formed.

<WRAP right>
<imgcaption imageNo01 | voltage generation in generator></imgcaption>
{{drawio>electrotechnology_2:voltagegenerationgenerator}}
</WRAP>
This will be briefly illustrated here. In <imgref imageNo01> a coil with $w$ windings is seen in a magnetic field with a magnetic flux density $\vec{B}$. The coil rotates - starting from $\alpha_0$ with angular velocity $\omega$. The rotation changes the chained flux $\Psi$ through the coil and thus a voltage $u(t)$ is induced.
For the rotation angle $\alpha$ holds:
\begin{align*}
\alpha(t)=\omega t + \alpha_0 \quad \text{with} \quad \alpha_0=\alpha(t=0) \
\end{align}

Thus, the induced voltage $u(t)$ is given by:
\begin{align*}
u(t) &= \frac{d\Psi}{dt} \\
     &= w\cdot\frac{d\Phi}{dt} \\
     &= wBA\cdot\frac{d cos \alpha}{dt} \\
     &= \hat{\Psi}\cdot\frac{d cos (\omega t + \alpha_0)}{dt} \\
     &= -\omega \hat{\Psi}\cdot sin (\omega t + \alpha_0) \
     &= -\hat{u}\cdot sin (\omega t + \alpha_0) \
\end{align*}

Such single-phase systems are therefore alternating current systems, which use one outgoing line and one return line each for the current conduction. \\
At a connected load the power was considered. A distinction was made between:
  * an active power: $P = UI cos \varphi$, which permanently represents an energy drain from the electrical system.
  * a reactive power: $Q = UI sin \varphi$, which describes the "sloshing back and forth" of the energy into the electric and magnetic fields.

In the following, the way to multiphase systems will be described.

===== 5.1 Technical terms of the multiphase system =====

==== General ====

Various general technical terms in the multiphase system will now be briefly discussed.

  - A **$m$-phase system** describes a circuit in which $m$ sinusoidal voltages transport the power. The voltages are generated by $m$ coils of the same shape, which are arranged offset to each other by an angle of $\alpha = 2\pi/m$. \ Example: 3-phase systems.
  - An $m$ phase system is **symmetrical** if the voltages of the individual phases are offset at the same angle to each other and exhibit the same amplitude. Thus, the voltage pointers $\underline{U}_1 ... \underline{U}_m$ form a symmetrical star.   \ Example: 3-phase systems with $\alpha = 120°$.
  - First, each coil represents a string which provides 2 potentials. The notion of concatenation now represents the following circumstance:
    - If all strings are independently connected to a load, the phase system is called **not chained**.
    - If all strands are connected to each other, then the phase system is called **chained**. \ <WRAP outdent> <WRAP outdent>(nbsp) With concatenation, fewer wires are needed. Star or ring circuits can be used for daisy chaining.</WRAP></WRAP>
  - The instantaneous power of a phase is initially variable in time. However, if the instantaneous power $p$ does not change with time, this system is called **balanced**. \\ If a balanced load is used, then $m$ phase systems are balanced with $m\geq3$. Balanced multiphase systems have the lowest conduction cost \\ \\ For $m\geq3$, the following is obtained for the instantaneous voltage: $\quad \quad p = m \cdot U \cdot I \cdot cos\varphi = P$

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</WRAP>

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</WRAP>

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</WRAP>
~~PAGEBREAK~~~CLEARFIX~~~

This gives the three-phase system as advantages over a DC or AC system :
  - Simple three-phase machines can be used for generation.
  - If a symmetrical load can be assumed, the energy flow is constant in time.
  - For energy transport, the voltage can be up-transformed and thus the AC current, as well as the associated waste heat power, can be reduced.
  - Rotary field machines can also be simply connected to the load, which converts the electrical energy into mechanical energy.

<callout icon="fa fa-exclamation" color="red" title="Note:">
  * If no load is connected, there is no three-phase network or three-phase system, but a three-phase voltage network. The colloquial term of three-phase current is thus not clean.
  * The windings are called $U$, $V$, $W$; the winding connections correspondingly: $U1$, $U2$, $V1$, $V2$, $W1$, $W2$.
</callout>

~~PAGEBREAK~~~CLEARFIX~~~

==== Targets ====

After this lesson, you should:

  - know what a symmetrical polyphase system is
  - be able to assign the terms neutral point, neutral conductor, neutral conductor


==== Text and video ====

{{:3phasemotorstrangeschema.jpg?nolink&200|}}



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===== Related Links=====
  * [[https://www.ipes.ethz.ch/mod/lesson/view.php?id=23| IPES ETHZ]]: interactive display on the field in the engine
  
====== 4. Networks at variable frequency ======

Further content can be found at {{https://www.elektroniktutor.de/analogtechnik/rei_swkr.html|elektroniktutor}}

==== Introduction ====

At the previous chapters it was explained how the "influence of a sinusoidal current flow" of capacitor and inductors look like. To describe this, the impedance was introduced. This can be understood as a complex resistance for sinusoidal excitation.

It applies to the capacitor:

\begin{align*}
\underline{U}_C = \frac{1}{j\omega \cdot C} \cdot \underline{I}_C \quad \rightarrow \quad \underline{Z}_C = \frac{1}{j\omega \cdot C}
\end{align*}

and for the inductance

\begin{align}
\underline{U}_L = j\omega \cdot L \cdot \underline{I}_L \quad \rightarrow \quad \underline{Z}_L = j\omega \cdot L
\end{align*}

Complex impedances can be dealt with in much the same way as ohmic resistances in Electrical Engineering 1 (see: [[electrical_engineering_1:simple_dc_circuits]], [[electrical_engineering_1:linear_sources_and_bipoles]], [[electrical_engineering_1:analysis_of_dc_networks]]). In these transformations, the fraction $ j\omega \cdot$ is preserved.
Circuits with impedances such as inductors and capacitors will show a frequency dependence accordingly.


===== 4.1 Frequency-dependent voltage divider =====

<callout>
=== Targets ===

After this lesson, you should:
  - know that ...
  - know that ... is formed.
  - be able to ...  can ...
</callout>

==== From two-pole to four-pole ====

<WRAP right>
<imgcaption imageNo01 | two-pole and four-pole>
</imgcaption> <WRAP>.
{{drawio>TwoPoleandFourPole}} \\
</WRAP>

Until now, components such as resistors, capacitors and inductors have been understood as two-terminal. This is also obvious, since there are only two connections. In the following however circuits are considered, which behave similar to a voltage divider: On one side a voltage $U_E$ is applied, on the other side $U_A$ is formed with it. This results in 4 terminals. The circuit can and will be considered as a four-terminal circuit in the following. But the input and output values will be complex.

For quadripoles, the relation of "what goes out" (e.g. $\underline{U}_A$ or $\underline{U}_2$) to "what goes in" (e.g. voltage $\underline{U}_E$ or $\underline{U}_1$) is important.
Thus, the output and input variables ($\underline{U}_A$) and ($\underline{U}_E$) give the quotient:

\begin{align}
\underline{A} & = \frac {\underline{U}_A}{\underline{U}_E} \\
& \text{with} \; \underline{U}_E = U_E \cdot e^{j \varphi_{uE}} \\
& \text{and} \; \underline{U}_A = U_A \cdot e^{j \varphi_{uA}} \\ \\

\underline{A}& = \frac {\underline{U}_A}{\underline{U}_E} = \frac {U_A \cdot e^{j \varphi_{uA}}{U_E \cdot e^{j \varphi_{uE}} \\
& = \frac {U_A}{U_E}\cdot \cdot e^{j (\varphi_{uA}-\varphi_{uE})} \\
\end{align}
\begin{align}
\boxed{\underline{A} = \dfrac {\underline{U}_A}{\underline{U}_E} = \frac {U_A}{U_E}\cdot e^{j \Delta\varphi_{u}}
\end{align*}


<callout icon="fa fa-exclamation" color="red" title="Remember:">
  * The complex-valued quotient ${\underline{U}_A}/{\underline{U}_E}$ is called the **transfer function**.
  * The frequency-dependent magnitude of the quotient $A(\omega)={{U}_A)}/{U}_E}$ is called **amplitude response** and the angular difference $\delta\varphi_{u}(\omega)$ is called **phase response**.
</callout>

The frequency behaviour of the amplitude response and the frequency response is not only important in electrical engineering and electronics, but will also play a central role in control engineering.

~~PAGEBREAK~~~CLEARFIX~~~

==== RL series connection ====

<WRAP right>
<imgcaption imageNo02 | RL-series>
</imgcaption> <WRAP>.
{{drawio>RLRow}} \\
<WRAP>

First, a series connection of a resistor $R$ and an inductor $L$ shall be considered (see <imgref FigNo02>). This structure is also called RL-element. \\
Here, $\underline{U}_E= \underline{X_E} \cdot \underline{I}_E$ with $\underline{X}_E = R + j\omega \cdot L$ and corresponding for $\underline{U}_A$:
\begin{align*}
\underline{A} = \dfrac {\underline{U}_A}{\underline{U}_E} = \frac {\omega L}{\sqrt{R^2 + (\omega L)^2}}\cdot e^{j\left(\frac{\pi}{2} - arctan \frac{\omega L}{R} \right)}
\end{align*}

This results in the following for
  * the amplitude response: $A = \frac {\omega L}{\sqrt{R^2 + (\omega L)^2}}$ and
  * the phase response: $\Delta\varphi_{u} = \frac{\pi}{2} - arctan \frac{\omega L}{R}$

The main focus should first be on the amplitude response. Its frequency response can be derived from the equation in various ways.
  - Limit value consideration of the RL arrangement (in the equation and in the system)
  - Plotting amplitude and frequency response
  - Determination of prominent frequencies

These three points are now to be gone through.
\\ \\
=== Limit value consideration of the RL arrangement ===

For the limit consideration we look at what, happens when the frequency $\omega$ runs to the definition range limits, i.e. $\omega \rightarrow 0$ and $\omega \rightarrow \infty$:
  * For $\omega \rightarrow 0$, $A = \frac {\omega L}{\sqrt{R^2 + (\omega L)^2}} \rightarrow 0$ as the numerator approaches zero and the denominator remains greater than zero.
  * For $\omega \rightarrow \infty$, $A \rightarrow runs 1$, because in the root in the denominator $(\omega L)^2$ becomes larger and larger in the ratio $R^2$ to . So the root tends to $\omega L$ and thus to the numerator.

It can thus be seen that:
  * at small frequencies there is no voltage $U_2$ at the output.
  * at high frequencies $A = \frac {U_A}{U_E} = \rightarrow 1$, so the voltage at the output is equal to the voltage at the input.

The RL element shown here therefore only allows large frequencies to pass (= pass through) and small ones are filtered out. The circuit corresponds to a **high pass**. \\
This can also be derived from understanding the components: At small frequencies, the current in the coil and thus the magnetic field changes only slowly. So only a negligibly small reverse voltage is induced. The coil acts like a short circuit at low frequencies. At higher frequencies, the current generated by $U_E$ through the coil changes faster, the induced voltage $U_i = - dI / dt$ becomes large. As a result, the coil inhibits the current flow and a voltage drops across the coil. If the frequency becomes very high, only a negligible current flows through the coil - and hence through the resistor. The voltage drop at $R$ thus approaches zero and the output voltage $U_A$ tends towards $U_E$.

For further consideration, the equation of the transfer function $\underline{A} = \dfrac {\underline{U}_A}{\underline{U}_E}$ is to be rewritten so that it becomes independent of component values. This allows for a generalized representation. This representation is called **normalization**:

\begin{align*}
\underline{A} = \dfrac {\underline{U}_A}{\underline{U}_E} = \frac {\omega L}{\sqrt{R^2 + (\omega L)^2}}\cdot e^{j\left(\frac{\pi}{2} - arctan \frac{\omega L}{R} \right)}
\quad \xrightarrow{\text{normalization}} \quad \quad
\underline{A}_{norm} = \frac {\omega L / R}{\sqrt{1 + (\omega L / R)^2}}\cdot e^{j\left(\frac{\pi}{2} - arctan \frac{\omega L}{R} \right)} = \frac {x}{\sqrt{1 + x^2}} \cdot e^{j\left(\frac{\pi}{2} - arctan x \right)}
\end{align*}

This equation behaves quite the same as the one considered so far.

~~PAGEBREAK~~~CLEARFIX~~~
\\
=== Plotting amplitude and frequency response ===

The transfer function can also be decomposed into amplitude response and frequency response. This can be done by
  * the amplitude response double logarithmic and
  * the phase response single logarithmic
logarithmically. <imgref FigNo03> shows the two plots. On the x-axis, $x = \omega L / R$ has been plotted as the normalization variable. This represents a weighted frequency.

<WRAP>
<imgcaption imageNo03 | Amplitude and phase response of the RL high-pass filter>.
</imgcaption> <WRAP>.
{{drawio>amplitudephaseresponseRLhighpass}} \\
</WRAP>


Here, too, the behavior determined in the limit value observation can be seen: at small frequencies $\omega$ (corresponds to small $x$), the amplitude response tends toward zero. At high frequencies, the ratio $U_A / U_E = 1 $ is established.

Interesting in the phase response is the point $x = 1$.
  * Further to the left of this point (i.e. at smaller frequencies) a tenfold increase of the frequency $\omega$ produces a tenfold increase of $U_A / U_E$.
  * Further to the right of this point (i.e. at higher frequencies) $U_A / U_E = 1$ remains.  

So this point marks a limit. Far to the left, the ohmic resistance is significantly greater the amount of impedance of the coil: $R >> \omega L$. \ far to the right is just the opposite.

The point $x=1$ just marks the cutoff frequency.
It holds

\begin{align}
\underline{A}_{norm} = \frac {x}{\sqrt{1 + x^2}} \cdot e^{j\left(\frac{\pi}{2} - arctan x \right)} = \frac {U_A}{U_E} \cdot e^{j\varphi}\quad \quad

  \left\{
    \begin{array}{l}
      x \ll 1 & \widehat{=} \omega L \ll R &: \quad\quad \frac{U_A}{U_E}=x &, \varphi = \frac{\pi}{2} & \widehat{=} 90°\\
      x \gg 1 & \widehat{=} \omega L \gg R &: \quad\quad \frac{U_A}{U_E}=1 &, \varphi = 0 & \widehat{=} 0°\\
      x = 1 & \widehat{=} \omega L = R &: \quad\quad \frac{U_A}{U_E}=\frac{1}{\sqrt{2}} &, \varphi = \frac{\pi}{4} &, \widehat{=} 45{°}
    \end{array}
  \right.
  
\end{align*}

<callout icon="fa fa-exclamation" color="red" title="Merke:">
  * The **cutoff frequency** for high-pass and low-pass filters is the frequency at which the ohmic resistance just equals the value of the impedance.
  * The cutoff frequency separates a range in which the filter allows signals through from one in which they are suppressed (=blocked).
  * At the cutoff frequency, the phase $\varphi = 45°$ and the amplitude $A = \frac{1}{\sqrt{2}}$.

These statements apply to single-stage passive filters, i.e. one RL or one RC element. Multistage filters are considered in circuit engineering.
</callout>

The cut-off frequency in this case is given by:

\begin{align*}
R &= \omega L \\
\omega_{Gr} &= \frac{R}{L} \\
2 \pi f_{Gr} &= \frac{R}{L}  \quad \rightarrow \quad
\boxed{f_{Gr} = \frac{R}{2 \pi \cdot L}}
\end{align*}

== SYNC, CORRECTED BY ELDERMAN ==
=== low pass ===

<WRAP right>
<imgcaption imageNo04 | Circuit, pointer diagram, and amplitude and phase response of RL low-pass filter>.
</imgcaption> <WRAP>.
{{drawio>amplitudephaseresponseRLlowpass}} \\
</WRAP>

So far, only one variant of the RL element has been considered, namely the one where the output voltage $\underline{U}_A$ is tapped at the inductance. Here we will briefly discuss what happens when the two components are swapped.

In this case, the normalized transfer function is given by:

\begin{align*}
\underline{A}_{norm} = \frac {1}{\sqrt{1 + (\omega L / R)^2}}\cdot e^{-j arctan \frac{\omega L}{R} }
\end{align*}

The cutoff frequency is again given by $f_{Gr} = \frac{R}{2 \pi \cdot L}$.

~PAGEBREAK~ ~CLEARFIX~~
==== RC Series ====

=== RC high pass ===

<WRAP right>
<imgcaption imageNo05 | Circuit, pointer diagram, and amplitude and phase response of the RC high-pass filter>
</imgcaption> \\
{{drawio>amplitudePhaseRChighpass}} \\
</WRAP>

Now a voltage divider is to be constructed by a resistor $R$ and a capacity $C$. Quite similar to the previous chapters, the transfer function can also be determined here.  

Here results as normalized transfer function:

\begin{align*}
\underline{A}_{norm} = \frac {\omega RC}{\sqrt{1 + (\omega RC)^2}}\cdot e^{\frac{\pi}{2}-j arctan \omega RC }
\end{align*}

In this case, the normalization variable $x = \omega RC$. Again, the cutoff frequency is determined by equating $R$ and the magnitude of the impedance of the capacitance:

\begin{align*}
R &= \frac{1}{\omega_{Gr} C} \\
\omega_{Gr} &= \frac{1}{RC} \\
2 \pi f_{Gr} &= \frac{1}{RC}  \quad \rightarrow \quad
\boxed{f_{Gr} =\frac{1}{2 \pi\cdot RC} }
\end{align}}


==PAGEBREAK ==CLEARFIX ==
=== rc low pass ===

<WRAP right>
<imgcaption imageNo06 | Circuit, pointer diagram, and amplitude and phase response of RC low-pass filter>
</imgcaption> <WRAP>.
{{drawio>amplitudephaseresponseRClowpass}} \\
</WRAP>

Again, the voltage at the impedance is to be used as the output voltage. This results in a low-pass filter.  

Here results as normalized transfer function:

\begin{align*}
\underline{A}_{norm} = \frac {1}{\sqrt{1 + (\omega RC)^2}}\cdot e^{-j arctan \omega RC }
\end{align*}

 Also, the cutoff frequency is given by $f_{Gr} =\frac{1}{2 \pi\cdot RC}$

~PAGEBREAK~ ~CLEARFIX~~
===== 4.2 Resonance phenomena =====



~PAGEBREAK~ ~CLEARFIX~~
==== RLC - Series Resonant Circuit ====

<WRAP right>
<imgcaption imageNo06 | circuit of the series resonant circuit>
</imgcaption> <WRAP>.
{{drawio>circuit of the series resonant circuit}} \\
</WRAP>

If a resistor $R$, a capacitor $C$ and an inductance $L$ are connected in series, the result is a series resonant circuit. In this case the output voltage is not clearly defined. It must be considered in the following how the voltages behave across the individual components.
The total voltage (= input voltage $U_E$) results to:

\begin{align}
\underline{U} = \underline{U}_R + \underline{U}_L + \underline{U}_C
\end{align}

Since the current in the circuit must be constant, the total impedance can be determined here in a simple way:

\begin{align}
\underline{U} &= R \cdot \underline{I} + j \omega L \cdot \underline{I} + \frac {1}{j\omega C } \cdot \underline{I} \\
\underline{U} &= \left( R + j \omega L - j \cdot \frac {1}{\omega C } \right) \cdot \underline{I} \\
\underline{Z}_{ges} &= R + j \omega L - j \cdot \frac {1}{\omega C }
\end{align*}

As the magnitude of the (input) voltage $U$ or the (input or total) impedance $Z$ and the phase result to:

\begin{align*}
U &= \sqrt{U_R^2 + (U_Z)^2} = \sqrt{U_R^2 + (U_L - U_C)^2}  \\
\end{align*}

\begin{align*}
Z &= \sqrt{R^2 + (Z)^2} = \sqrt{R^2 + (\omega L - \frac{1}{\omega C})^2}  \\
\end{align*}

\begin{align*}
\varphi_u = \varphi_Z &= arctan \frac{\omega L - \frac{1}{\omega C}}{R}
\end{align*}

There are now 3 different situations to distinguish:
  * If $U_L > U_C$ the whole setup behaves like an ohmic-inductive load. This is the case at high frequencies.
  * If $U_L$ equals $U_C$, the total input voltage $U$ is applied to the resistor. In this case, the total resistance $Z$ is minimal and only ohmic. \\Thus, the current $I$ is then maximal. If the current is maximum, then the responses of the capacitance and inductance - their voltages - are also maximum. This situation is the **resonance case**.
  * If $U_L < U_C$ then the whole setup behaves like a resistive-capacitive load. This is the case at low frequencies.

Again, there seems to be an excellent frequency, namely when $U_L = U_C$ or $Z_C = Z_L$ holds:

\begin{align*}
\frac{1}{\omega_0 C} & = \omega L \\
\omega_0 & = \frac{1}{\sqrt{LC}} \\
2\pi f_0 & = \frac{1}{\sqrt{LC}}
\rightarrow
\boxed{ f_0 = \frac{1}{2\pi \sqrt{LC}} }
\end{align*}

The frequency $f_0$ is called **resonance frequency**.



^ ^ $\quad$ ^ $f \rightarrow 0$ ^ $\quad$ ^ $f = f_0$ ^ $\quad$ ^ $f \rightarrow \infty$ ^
| voltage $U_R$ \ at the resistor | | $\boldsymbol{0}$ | | $\boldsymbol{U}$ \ since the impedances just cancel | | $ \boldsymbol{0}$ |
| voltage $U_L$ \at the inductor | | $\boldsymbol{0}$ \because $\omega L$ becomes very small | | $\boldsymbol{\omega_0 L \cdot I = \omega_0 L \cdot \frac{U}{R} = \color{blue}{\frac{1}{R}\sqrt{\frac{L}{C}}\cdot U}$ | | $\boldsymbol{U}$ \ since $\omega L$ becomes very large |
| $\boldsymbol{U}$ \voltage $U_C$ \at the capacitor | | $\boldsymbol{U}$ \because $\frac{1}{\omega C}$ becomes very large | | $\boldsymbol{\frac{1}{\omega_0 C} \cdot I = \frac{1}{\omega_0 C} \cdot \frac{U}{R} = \color{blue}{\frac{1}{R}\sqrt{\frac{L}{C}}\cdot U}$ | | $\boldsymbol{0}$ \because $\frac{1}{\omega C}$ becomes very small |

The calculation in the table shows that in the resonance case, the voltage across the capacitor or inductor deviates from the input voltage by a factor $\color{blue}{\frac{1}{R}\sqrt{\frac{L}{C}}$. This quantity is called **quality** $Q_S$:

\begin{align*}
\boxed{ Q_S = \frac{U_C}{U} |_{\omega = \omega_0} = \frac{U_L}{U} |_{\omega = \omega_0} = \color{blue}{\frac{1}{R}\sqrt{\frac{L}{C}} }
\end{align*}

The quality can be greater than, less than or equal to 1.
  * If the quality is very high, the overshoot of the voltages at the impedances becomes very large in the resonance case. This is useful and necessary in various applications, e.g. in an RLC element as an antenna.
  * If the Q is very small, overshoot is no longer seen. Depending on the impedance at which the output voltage is measured, a high-pass or low-pass is formed similar to the RC or RL element. However, this has a steeper slope in the blocking range. This means that the filter effect is better.

The reciprocal of the Q is called **attenuation** $d_S$. This is specified when using the circuit as a non-overshooting filter.

\begin{align*}
\boxed{ d_S = \frac{1}{Q_S} = R \sqrt{\frac{C}{L}} }
\end{align*}


~~PAGEBREAK~~~CLEARFIX~~~

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~~PAGEBREAK~~~CLEARFIX~~~

===== Decoupling capacitor on the microcontroller =====

[[http://www.falstad.com/circuit/circuitjs.html?cct=$+1+3.125e-9+0.7389056098930651+50+5+50%0AR+-256+144+-256+80+0+0+40000+0+5+0+0.5%0Ar+624+160+624+192+0+0.1%0Aw+624+144+624+160+0%0Aw+624+144+672+144+0%0Ag+672+208+672+224+0%0Ap+672+144+672+208+1+0%0Ap+-256+144+-256+208+1+0%0Ag+-256+208+-256+224+0%0Ar+176+144+224+144+0+0.02%0Al+112+144+176+144+0+3.0000000000000004e-7+0.0012646053584079516%0Aw+224+144+288+144+0%0Aw+-256+144+-208+144+0%0Al+-160+144+-96+144+0+0.000003+0.001714943880265035%0Ar+-96+144+-48+144+0+0.2%0Af+576+224+624+224+33+1.5+0.02%0Aw+624+192+624+208+0%0Ag+624+240+624+272+0%0AR+576+224+560+224+5+5+10000000+2.5+2.5+0+0.5%0Ar+288+272+288+208+0+0.01%0Al+288+208+288+144+0+1e-8+-3.95516952522712e-14%0Aw+336+144+288+144+0%0Aw+-160+144+-208+144+0%0Aw+624+144+560+144+0%0Ab+-192+96+-35+301+0%0Ax+-180+65+-54+68+4+18+Ersatzschaltbild%0Ac+-96+144+-96+208+0+1e-10+5.6099395909659755%0Ag+-96+208+-96+224+0%0Ax+-188+87+-34+90+4+18+5..50cm%5CsLeiterbahn%0Ax+83+87+242+90+4+18+0.5..5cm%5CsLeiterbahn%0Ax+91+65+217+68+4+18+Ersatzschaltbild%0Ag+176+208+176+224+0%0Ac+176+144+176+208+0+1e-10+5.828890665512928%0Ab+80+96+237+301+0%0Aw+80+144+112+144+0%0Ab+248+96+341+413+0%0Ax+232+444+358+447+4+18+Ersatzschaltbild%0Ax+231+470+337+473+4+18+100nF%5CsKerKo%0Aw+560+144+496+144+0%0Ax+555+87+673+90+4+18+Mikrocontroller%0Ax+552+65+678+68+4+18+Ersatzschaltbild%0Ab+521+96+726+301+0%0Aw+352+144+384+144+0%0Ab+352+96+509+301+0%0Ac+448+144+448+208+0+1e-11+5.8885188471740335%0Ag+448+208+448+224+0%0Ax+341+87+525+90+4+18+0.05..0.5cm%5CsLeiterbahn%0Al+384+144+448+144+0+3.0000000000000004e-8+0.0009675223894950857%0Ar+448+144+496+144+0+0.002%0Ax+-41+470+65+473+4+18+100nF%5CsKerKo%0Ax+-40+444+86+447+4+18+Ersatzschaltbild%0Ab+-24+96+69+413+0%0Aw+64+144+16+144+0%0Al+16+208+16+144+0+1e-8+1.5154544286133387e-13%0Ar+16+272+16+208+0+0.01%0Aw+-48+144+16+144+0%0Aw+64+144+80+144+0%0Ax+366+65+492+68+4+18+Ersatzschaltbild%0Aw+336+144+352+144+0%0Ax+186+352+225+355+4+32+S2%0As+288+320+288+352+0+1+false%0Ag+288+352+288+384+0%0Ac+288+272+288+320+2+1.0000000000000001e-7+5.013963438724142%0Ac+16+272+16+320+2+1.0000000000000001e-7+4.9684040165331345%0Ag+16+352+16+384+0%0As+16+320+16+352+0+1+false%0Ax+-79+352+-40+355+4+32+S1%0Ax+201+493+406+496+4+18+%22nahe%5Csam%5CsMikrocontroller%22%0Ao+6+8+0+4106+24.87604116742552+0.0001+0+2+5+0%0A|Simulation in Falstad]]\.
Note: The simulation gives a highly simplified picture. The response of the microcontroller is shown reduced to a triangular signal, since the slope of the voltages cannot be represented. A real simulation requires a powerful SPICE program in which the [[https://de.wikipedia.org/wiki/Leitungstheorie|conduction theory ]]can be represented.

Further details can be found [[https://rn-wissen.de/wiki/index.php/Abblockkondensator|here (practice)]] or [[http://www.lothar-miller.de/s9y/categories/14-Entkopplung|here (board layout)]].

~~NOCACHE~~
<affix offset-top="0" position-top="0" position-left="20">
<button size="xs" type="link" collapse="sidebar">{{icon>list}} Course Content</button>

<collapse id="sidebar" collapsed="false"><well>
\ [[start|{icon>home?18}}]]<fs large> [[start|Electrical Engineering 2]]</fs>


\ [[electrical_engineering_1:introduction_to_electrical_engineering_1|introduction_to_electrical_engineering 2]]
\\
\ [[alternating current technology
\ [[The magnetostatic field|1. The magnetostatic field]]
\ [[The time-varying magnetic field|2. The time-varying magnetic field]]
\ [[Alternating current technology|3. Alternating current technology]]
\ [[Variable frequency networks|4. Variable frequency networks]]
\ [[Polyphase systems|5. Polyphase systems]]
\\
\ Magnetics
\ [[The magnetic circuit|6. The magnetic circuit]]
\ [[Transformer|7. The transformer]]
\\
\ [[ET exam information|]]
</well>
~~BARCODE~url=https://wiki.mexle.org/elektrotechnik_2/start~size=S~~
</collapse></affix> ====== Electrical Engineering II ======



<WRAP column 20%>
\\ \\
{{::permanent_magnet_sketch.jpg?200|}}
{{::spule_sketch.jpg?150|}}
</WRAP><WRAP column 30%>
[[electrical_engineering_1:introduction_to_electrical_engineering_1|introduction_to_electrical_engineering 2]]_.
Or: What is the best way to handle the event?

\ [[The magnetostatic field]]\
or: known names in an unknown field

[[The time-varying magnetic field]]\
or: there comes movement into play

</WRAP>
<WRAP column 30%>
{{::AdvancedOrganizer_Magnetism.jpg?400|}}
</WRAP>
<WRAP clear></WRAP>

<WRAP column 20%>
{{::alternating_current_sketch.jpg?120|}}
</WRAP><WRAP column 30%>
\\ {{::alternating current technology}}
Or: what works, what shines?
</WRAP><WRAP clear></WRAP>

<WRAP column 20%>
{{::networks_veraenderl_frq_sketch.jpg?600|}}
</WRAP><WRAP column 30%>
[[Networks at varying frequency]]\
[[or: it oscillates and dampens
</WRAP><WRAP clear></WRAP>

<WRAP column 20%>
{{::multiphase_sys1_sketch.jpg?100|}}
{{::multiphase_sys2_sketch.jpg?75|}}
</WRAP><WRAP column 30%>
\\ {{::multiphase-sys2_sketch}}
{[[strong! Power!
</WRAP><WRAP clear></WRAP>

<WRAP column 20%>
{{::mangetic_circle_sketch.jpg?150|}}
</WRAP><WRAP column 30%>
\\ {{::mangeticcircle_sketch}}
Or: what's flowing?
</WRAP><WRAP clear></WRAP>

<WRAP column 20%>
{{::transformator_sketch.jpg?300|}}
</WRAP><WRAP column 30%>
\\ {{::transformer_sketch}}
or: the transformation of change (!?)
</WRAP><WRAP clear></WRAP>


[[electrical_engineering_2:information_on_et_exam]]
~~NOTOC~~

====== 3. AC technology ======

<WRAP right>
<imgcaption imageNo5 | example of a circuit>
</imgcaption> \\
{{url>https://www.falstad.com/circuit/circuitjs.html?ctz=CQAgjCAMB0l3BWcMBMcUHYMGZIA4UA2ATmIxAUgpABZsKBTAWjDACgAncYlcMX3DT68qYeGwDGIQcOmQhYHlCix4EGHHWr2AczkKlMlDSqm2acjNYC8C-tQDyAVwAu5yOUW9r4QqPtIAJIAdu7kNHiyeP68QaEAbiBM0bIxyhB+0MRCSKYqCGx6yWnFIMZ5kGwAzun2EangIABmAIYANlUMbAD20iCEQlQmxJG50Opl-VBsQA 800,500 noborder}}


</WRAP>


Comparison of electrics with fluid mechanics
{{youtube>6Dlva2_XOvI}}


===== 3.1 Classification of time-dependent signals =====


==== Targets ====

After this lesson, you should:

  - know which types of time-dependent waveforms there are and be able to assign them


==== Video ====


<WRAP pagebreak />

===== 3.2 Descriptive variables of alternating variables =====


==== Targets ====

After this lesson, you should:

  - Know the relationship between amplitude and peak-to-peak value.
  - Know the relationship between period, frequency and angular frequency.
  - Know the difference between zero phase angle and phase shift angle.
  - Know the direction of the phase shift angle.
  - know the formula symbols of the above-mentioned quantities.


==== Video ====



<WRAP pagebreak />

===== 3.3 Mean values of alternating variables =====

==== Targets ====

After this lesson, you should:

  - be able to calculate the arithmetic mean, the rectified value and the effective value.
  - know these mean values for sinusoidal quantities.
  - know the reason for using the effective value.
  

==== Video ====

Mean values of alternating variables

{{youtube>p4XP77d59ak}}

RMS value (using the example of a sine function)

{{youtube>nyfwyf_sQc0}}

Rectification value (using the example of a sine function)

{{youtube>fcharnFGmtk}}

<WRAP pagebreak />

===== 3.4 Two-pole for alternating variables =====


==== Targets ====

After this lesson, you should:

  - know that real, lossy components are described by equivalent circuits of ideal components.
  - know and be able to apply the definition of apparent resistance, apparent conductance, impedance, and admittance.


==== Video ====


<WRAP pagebreak />

===== 3.5 AC resistors =====


==== Targets ====

After this lesson, you should:

  - know how sine variables can be symbolized by a vector.
  - know which parameters can determine a sinusoidal quantity.
  - be able to graphically derive a pointer diagram for several existing sine variables.
  - be able to plot the phase shift on the vector and time plots.
  - Be able to add sinusoidal quantities in vector and time representation.
  - know and be able to apply the impedance of components.
  - know the frequency dependence of the impedance of the components. In particular, you should know the effect of the ideal components at very high and very low frequencies and be able to apply it for plausibility checks.


==== Video ====

Please have a look at the contents (text, videos, exercises) on the page of the [[https://lx3.mint-kolleg.kit.edu/onlinekursphysik/html/1.4.3/xcontent4.html|KIT-Bridge Course >> 4.3.5 AC Resistors]]. Make sure that "Total" is selected in the selection bar at the top.


<WRAP pagebreak />

===== 3.6 Insertion: Complex numbers =====


==== Targets ====

After this lesson you should be able to:
  - be able to convert complex-valued numbers from polar to Cartesian coordinates and vice versa.
  - know how to add, subtract, multiply, and divide two complex-valued numbers using a formula and complex plane.
  - know what multiplication by / division by j means graphically.
  - know how to find the absolute value of a complex number.


==== Video ====

What's the point of complex numbers?
{{youtube>8su2_dLFGt8}}

Calculating with complex numbers, sum, difference, product

{{youtube>Ea4BjgzMIno}}

Geometric interpretation of the complex multiplication \\.
Or: Why is the amount and angle to be added in the multiplication?

{{youtube>fbExg5Cs2C0}}


<WRAP pagebreak />

===== 3.7 Complex AC resistors =====


==== Targets ====

After this lesson you should:
  - Be able to draw and read pointer diagrams.
  - Know and be able to apply the complex value formulas of impedance, reactance, resistance.
   


==== Video ====

Pointer diagrams; complex alternating current calculus.
{{youtube>Csc79xym6Os}}

Complex alternating current calculation - basic terms: Impedance, Reactance, Resistance

{{youtube>c962sw5zchQ}}

Capacitor and inductance as complex resistors; pointer diagram

{{youtube>1c1cWT550sM}}

detailed explanation of impedances
{{youtube>jcAm6kjzLAg}}

<WRAP pagebreak />

===== 3.8 Power in AC technology =====


==== Targets ====

After this lesson you should:
  - Know the formula of the instantaneous power of the resistor, coil and capacitor and be able to determine its values.


==== Video ====

detailed explanation of power in alternating current technology
{{youtube>ZNwBW6NYqOc}}

[[https://www.geogebra.org/m/yfwfhtn7|Simulation of instantaneous power as a function of phase]]

<WRAP pagebreak />


===== 3.9 Generation of alternating current =====

==== Targets ====

After this lesson, you should:
  -


==== Video ====

Please have a look at the contents (text, videos, exercises) on the page of the [[https://lx3.mint-kolleg.kit.edu/onlinekursphysik/html/1.4.2/xcontent5.html|KIT-Bridge Course >> 4.2.6 AC]]. Make sure that "Total" is selected in the selection bar at the top.

Please work through the [[http://katgym.by.lo-net2.de/c.wolfseher/web/Generator/generator.html|How does the sine wave get into the socket?]] page.

<WRAP pagebreak />


===== 3.10 Exercises =====


==== Video ====

Parallel connection of complex resistors / impedances \\
Why can a circuit with impedances only act purely ohmic?
{{youtube>WdkZD90DTYo}}

more complex exam task: complex circuit I
{{youtube>rHg_37qYdNw}}

more complex exam task: complex circuit II
{{youtube>sjji0w9h-ag}}
{{youtube>0JSLd_15Kyw}}
{{youtube>1YSKzvaeEbk}}
{{youtube>5c4Ehwb9lJ0}}

Examination task: complex circuit III
{{youtube>EjLG0WbP1Do}}




-------------------------------------------------


====== 0. Resources ======

===== 0.1 Digital =====

The free simulation program [[https://github.com/hneemann/Digital|Digital]] helps to practically implement the various concepts of digital technology. The program offers the possibility ...
  * ... Calculations with binary and hexadecimal numbers. (Chapter 1.)
  * ... build logic circuits (chapter 3.)
  * ... synthesize logic circuits directly from KV diagrams or logical expressions (Chapter 4.-7.)
  * ... analyze logic circuits

<WRAP info>
--> Introduction to Digital 0 - Explanation of digital, download and installation #
<WRAP group>
<WRAP>

The program can be downloaded directly from [[https://github.com/hneemann/Digital/releases/latest|Github]] as "Digital.zip".

=== German instructions ===

On the above mentioned page you can also find manuals in different languages. All functions are explained in them. But the docu is also already available in "Digital.zip" under the docu folder, so you don't have to download it separately.
</WRAP>

<WRAP help>
=== Tasks ===
Please install Digital. \\ ___ The following tips:
  * There is no separate "installer". I.e. you should unzip the zip file to a sensible location. A folder like ''C:\Program Files\Digital'' is recommended.
  * The program opens by double-clicking on ''Digital.exe''. If the menu and the icons are too small, it is recommended to open the application via ''Digital_noD3D.exe''.
  * Furthermore it is a good idea to put a link on the desktop. This can be done e.g. via drag-and-drop with the right mouse button.
</WRAP>
</WRAP>
<--

--> Introduction to Digital 1 - Structure of Digital, First Circuit#.
<WRAP group>
<WRAP>
=== Objectives ===
After this lesson you should:
  - Know the different component strips in Digital,
  - insert components,
  - edit the values of components,
  - draw connections.

</WRAP>
<WRAP>
=== Step-by-step 1: The first look ===
<WRAP right>
<imgcaption image_number1 | Elements of the Digital program>
{{fundamentals_of_digital:digital_empty.jpg?600}}
</imgcaption>
</WRAP>

  - Open the program by double-clicking on ''Digital.exe''.
  - You should now see an "empty circuit" in Digital (see <imgref imageNo1>).
  - There are several icon areas in the menu bar, these will be used below:
    - File selection: New, Open, Save
    - Zoom: Auxiliary dialog, Zoom in, Zoom out, Fit.
    - Undo, Redo, Delete
    - Start simulation, start until stop signal, stop
    - Gate step modes
    - Test execution
  - Below the menu bar is the work area, which is a grid of gray dots.

==PAGEBREAK ==CLEARFIX ==
=== Step-by-step 2: Create input and output ===

The first circuit to be created is an input and an output
    - <WRAP right><imgcaption imageNo2 | Input and output connected>{{basics_of_digitaltechnology:digital_simplecircuit.jpg?600}}</imgcaption></WRAP> To do this, go to ''Components''. There you will find all the necessary components. For now, we only need inputs/outputs. These are located under ''IO''. First select ''Input''
    - After that you will see the component marked with a mint-colored circle. Mint-colored circled elements are always marked for moving in the Digital program. Click on any position in the workspace
    - Then add an ''Output'' to the right of the input. Here we will present another variant. Click on ''View >> Tree view of components'' (or ''F5''). A column with the different components is then displayed on the left. In this column the output is located under ''IO''. This can be inserted into the work area by drag and drop.
    - Both components can be connected with a line. A line can be started from any grid point by left clicking and ended by ''Esc''. A line can also start or end at an output (red dot on a component: <fc #ff0000>-</fc> ) or an input (red dot on a component: <fc #0000ff>-</fc> ).
    - This already creates the first simple circuit (see <imgref imageNo2>). The circuit can be activated by pressing the start icon ''►''. If you have forgotten the connection, or there is another error, an error message appears. After confirming this error, the faulty component is highlighted in red.
    - When the circuit is started, input components can be actuated. Existing other components will then also become active. You should test this by pressing the input. Then stop the simulation by pressing the stop icon ''■'' (stopping by ''►'' is also possible). The circuit can now be edited again.

==PAGEBREAK ==CLEARFIX ==
=== Step-by-step 3: Change properties and extend circuit ===

    - <WRAP right><imgcaption picNo3 | Circuit with 8 bit bus>{{basics_of_digital_technology:digital_properties.jpg?600}}</imgcaption></WRAP> The previous circuit will now be further developed. The goal now is to have a circuit input and output of 8 bits. How a bit/byte is defined is shown in the [[number systems]] chapter. The trivial solution would be to multiply the existing circuit with ''<Ctrl>+<A>'', ''<Ctrl>+<C>'', ''<Ctrl>+<V>''. But here we will present another variant, which transports the 8 bits through a data bus. A data bus is defined as lines that belong together.
    - For this purpose, the properties of the input can be changed by right-clicking on it. Here the number of ''Data-Bits'' should be increased to 8 and the ''Designation'' to ''A''. The changes have to be confirmed with ''OK''.
    - If now the circuit is started the somewhat cryptic error message appears: ''1 bits are needed, but 8 bits were found. Affected are: Line out''. The connection line and the output are marked. We forgot to set the output to 8 bits...
    - To solve this, again not the trivial variant (right click >> change) shall be described, but a slightly different possibility:
      - mark everything with ''<Ctrl>+<A>''.
      - right click
      - ''Change data bits''. This option is grayed out though, because the components have different values there. By clicking on the box ''□'' next to data bits, the option can be changed.
    - If the circuit is now started, a 0 appears over input, line and output (see <imgref FigNo3>).
    - A (left)click on the input now no longer directly changes the value, but opens an input window. In this window numbers can be entered as decimal value (''10''), as binary value (''0b10''), as hexadecimal value (''0x10'') or as octal value (''010''). Pressing Line break (''<↵>'') accepts the value and closes the input window. Clicking Shift + Newline (''<↑><↵>'') accepts only the value. The cursor keys up/down allow the displayed value to be counted up/down.

</WRAP>

<WRAP help>
=== Tasks ===
  - Determine the hexadecimal values for 070, 64, 0b01100110.
  - Change the output to decimal value display and determine the decimal values for 064, 0x70, and 0b10011001.
</WRAP>
</WRAP>
<--



====== 2. Binary Logic ======

==== Motivation ====

<WRAP group> <WRAP third column> In the first chapter we got to know how to understand the number systems relevant for computers and microcontrollers. However, this does not allow us to understand how, for example, a program is executed. What we still need to know for this:

How can you build logical functions with the binary numbers?

==== Tasks ====

The assignments can be found in ILIAS under: (2) Introduction, Lecture Notes and Assignments >> Assignments

For details, see [[fundamentals_of_digital_technology:introduction|introduction to fundamentals of digital technology]]

</WRAP> <WRAP third column> {{:breaknes_mos_6502.jpg?200}}

(Image: [[https://commons.wikimedia.org/wiki/File:BreakNES_MOS_6502.jpg|Sbp@Wikimedia]], [[https://creativecommons.org/licenses/by-sa/4.0/deed.en|CC BY-SA 4.0]])

</WRAP> </WRAP>

----

===== 2.1 Binary logic =====

<WRAP group> <WRAP half column><callout>

==== Targets ====

After this lesson, you should:

  - Know the Boolean functions, their notation and function tables.
  - Be able to apply the Boolean calculation rules.
  - be able to simplify Boolean expressions.
  - know the following technical terms: (logic) gates, names of arithmetic rules.

</callout></WRAP> <WRAP half column>

==== Video ====

{{youtube>-QdOJh0X8Lw}}

</WRAP> </WRAP>

----

===== 2.2 Binary logic circuit symbols =====

<WRAP group> <WRAP half column><callout>

==== Targets ====

After this lesson, you should:

  - Be able to understand the two types of circuit symbols.
  - Know timing diagrams.
  - To know the representation of negation of inputs/outputs.
  - Be able to trace gates to NAND and NOR gates.

</callout></WRAP> <WRAP half column>

==== Video ====

{{youtube>9W78OV7blYo}}

</WRAP> </WRAP>

----

===== 2.3 Applications of binary logic =====

<WRAP group> <WRAP half column><callout>

==== Targets ====

After this lesson, you should:

  - Understand the purpose of the Tri-State Gate and State "Z".
  - Understand the use of the "Don't care" state.
  - Be able to convert interconnections from a few logic gates to function tables and vice versa.
  - Be able to trace gates to NAND and NOR gates.

</callout></WRAP> <WRAP half column>

==== Video ====

{{youtube>rCsrxsIXtgs}}

</WRAP> </WRAP>

----

===== further links =====

<WRAP hide> * [[https://web.archive.org/web/20170630154405/http://www.elektroniker-bu.de/boolesche.htm#ergebnis|solver for boolean functions]]: The solver specifies which axioms can be used to simplify Boolean equations. (Unfortunately only available via internet archive) </WRAP>

  * [[http://siliconzoo.org/nxp.html|Silicon Zoo]]: Here is a practical implementation of logic gates in silicon.

==== Applications ====

  * [[https://github.com/djrtwo/evm-opcode-gas-costs/blob/master/README.md|Etherium]]: With this cryptocurrency, computations are possible on the blockchain. Here, the basic logical functions are the most favorable - a program should therefore be feasible with as few boolean operators as possible
  * [[https://onlinegdb.com/r1A7GJGvv|Example in C]] for the use of boolean algebra: by clicking on the ''Fork this'' button, the code is changeable.<WRAP group><WRAP half column>~REVEAL theme=dokuwiki&fade=convex&controls=1&show_progress_bar=1&build_all_lists=1&show_image_borders=0&horizontal_slide_level=2&enlarge_vertical_slide_headers=0&show_slide_details=0&open_in_new_window=1&size=1024x768~~
</WRAP><WRAP half column>

</WRAP></WRAP>
====== Introduction ======

---->>
===== Online lecture and presence plenary =====

<WRAP left round box 35%>
"What do you mean there is only online lecture?"
<WRAP>
<WRAP center round box 60%>
"Why isn't the lecture still offered alongside the videos?"
</WRAP>
<notes>Maybe you have asked yourself similar questions too. I'd like to give you a brief introduction here as to why the lecture is offered in this manner</notes>.

---->>
===== essence-of-study =====
<WRAP right>
{{::study_aigledore_cc-by-2.0jpg.jpg?300}}<notes>\ <fs 70%>Source: [[https://commons.wikimedia.org/wiki/File:Study_(16840395246).jpg|Moyan Brenn @ Wikimedia.org]] \\
 ([[https://creativecommons.org/licenses/by/2.0/deed.en|CC BY 2.0 license]])</fs></notes>
</WRAP>
<WRAP hide>
\begin{align} study =\int_{1.semester}^{7.semester+\varepsilon}literature + interaction \: dt \end{align}
</WRAP>
<notes>
The aim of the course is to be able to handle scientific and technical documents independently and to familiarize yourself with unfamiliar topics by means of textbooks, bibliographical references and (engineering) scientific material. To make this easier for you in the first semester, I offer you the complete German script and additionally compact videos on the content of the script. In the course of your (Bachelor or Master) studies, it is quite possible that you will meet professors who require you to familiarize yourself independently without a script, using only the literature references provided.

However, not everything is eaten as hot as it is cooked. Take advantage of the opportunities that arise. Look for like-minded people, get organised and support each other. Chances are, the person you help in the first semester can in turn support you in higher semesters.
</notes>

---->>
===== Time Management =====
<WRAP right>
{{::clock_alexas_photos_cc0.jpg?200|}}<notes>
\fs 70%>Source: [[https://pixabay.com/en/the-eleventh-hour-disaster-1364075/|Alexas_Photos @ Pixabay]] \\
 ([[https://creativecommons.org/publicdomain/zero/1.0/deed.en|CC0 1.0]])</fs></notes>
</WRAP>

<notes>
Another reason for missing a lecture is that attendance time is too precious to me. I don't see any advantage in filling the attendance time with the same material you can already find in the script or videos as well.
According to the current SPO, we have 2SWS - that is 16.5h in the current semester - of attendance time available for the fundamentals of digital technology.
If we also wanted to completely cover the material in the context of a blackboard during the attendance time, there would be hardly any time left for your concrete questions and the calculation of tasks.

On the other hand, the given [[https://de.wikipedia.org/wiki/European_Credit_Transfer_System#ECTS-Punkte_.28Credits.29|ECTS points]] result in about 50h to 60h of effort for one student, for the whole course. You are welcome to prepare and follow up the exercises in the lab. However, the exercises are essentially planned as practice and consolidation of the material and accordingly do not have the classical character of a laboratory at the university.

This means that for **1 hour of presence plenary about 3-4 hours of independent preparation and follow-up**. Independent does not have to mean that you do this alone. Use the possibilities to organize yourself in groups and again: **Help others.

</notes>
<WRAP hide>
  - <wrap lo>''Presence time = unique;'' </wrap> ''videos = replayable;''
  - <wrap lo>''if (all == anDerTafel) { timeForQuestions = 0;}''</wrap>
  - <wrap lo>''EffortPerWeek = (50h - 16.5h)/ 11 weeks''</wrap>
  - <wrap lo> Help others </wrap>
</WRAP>

---->>
===== Interaction =====
<notes>
Here's what emerged from a 2017 survey:
  - slightly more than half of the students come from Gymnasium. A third come from vocational college and were previously in junior high school or an apprenticeship.
  - The breakdown of prior experience in electronics is about 1/3 high, 1/3 medium, and 1/3 low.
  - About 10% rated their programming experience as high, 60% as low or very low.
</notes>
<WRAP hide>
Previously:
  * about 50% from high school, about 30% vocational college.
  * Prior experience:
    * approx. 33% high, 33% medium, 33% low.
    * 10% high programming experience, 60% low

</WRAP>
---->
<WRAP right>
{{::pigdog_pixabay_suju_cc0.jpg?200|}}<notes>
\fs 70%>Source: [[https://pixabay.com/en/son-of-a-bitch-inner-pig-dog-dog-2613125/|Suju@Bixabay]] \\
 ([[https://creativecommons.org/publicdomain/zero/1.0/deed.en|CC0 license]])</fs></notes>
</WRAP>
<notes>
This is roughly in line with my assumptions with which I laid out the course. There is no such thing as the standard student who comes into the course with exactly the prior knowledge. On the other hand, there are those among you who already have experience. Some may have already been able and required to show initiative and responsibility. Be aware: The university lives from you! Whether in eRacing, VDI working group, student council or even the lectures / events. Have courage and leave the pig in the basket.
</notes>
<WRAP hide>
  * help others
  * show initiative + responsibility
  * college lives on you
  * put the pig in the basket
</WRAP>

---->>
===== Learning resources and exercises =====

<notes>
<WRAP right>
{{::logo-ilias_240.png?100|}}
\fs 70%>Source: [[https://www.ilias.de/docu/goto_docu_cat_2818.html?lang=de|ilias.com]] \\
 ([[http://www.gnu.org/licenses/gpl|GNU GPLv3]])</fs>
</WRAP>

The [[https://ilias.hs-heilbronn.de/goto.php?target=file_23491_download&client_id=iliashhn|Skript]] can be found in ILIAS in the course "Fundamentals of Digital Technology". There are also exercises given for each chapter.

I offer to correct the exercises. For this you have to upload the exercise in time in the ILIAS instance of the course under [[https://ilias.hs-heilbronn.de/goto.php?target=exc_23534&client_id=iliashhn|Homework]]. The dates are given there directly. If you want to take advantage of the offer, you should go through the assignments together with your exercise partner and also participate in the reviews afterwards. You will receive the documents one day after the deadline, which you should review.

In addition, I have prepared compact videos for you on the content of the script and attached them in the following chapters. As described above, I ask you to use them as you would have used a lecture.

Furthermore, I recommend working through the chapter [[0. Tools]], as the tools mentioned there can be used to directly implement what you have learned.
</notes>
<WRAP hide>
{{::logo-ilias_240.png?100|}}{{wiki:dokuwiki-128.png?100|}}
  * Script and exams in ILIAS.
  * Exercises
  * Videos in the wiki
</WRAP>

---->>
===== tips =====
<notes>
The following may help you with the Fundamentals of Digital Technology lecture:  

<callout type="tip" icon="true" title="Bonus Feature: Tutorial">.
In addition to assigning you to lab and practice groups, I can also offer you a student-led tutorial. For this I need motivated and interested students who plan the dates and topics independently with the course. As a thank you, I can offer tutors a tutoring contract, which means they would also get a financial reward. Again, the call: **Help others.
</callout>

<callout type="tip" icon="true" title="Bonus Feature: Instructor Feedback">

In addition, I offer you to send me your important questions already before the plenary session. This gives me an overview of your problems in the current topic and allows me to better prepare the plenary. You are also welcome to give me praise and criticism. If you want to give feedback anonymously, just drop a few lines of text into my inbox (mailboxes between D- and A-building).
If you want direct feedback, it's easier to write me a [[tim.fischer@hs-heilbronn.de|Mail]] directly.
</callout>

<callout type="tip" icon="true" title="Actively engaging with the presence plenary">
<WRAP right>
{{::talk.png?100|}}
<fs 70%>Source: [[https://de.wikipedia.org/wiki/Datei:Internet-talk-error.svg|Public Domain]]</fs>
<WRAP>
Take the opportunity to ask their questions in plenary. As already written, I would be pleased if you could write me the questions already **before**. Then the question-searching in the first minutes is reduced and we have more time for the answers. Also, take the opportunity to shorten the open questions if you know the answer or an easier way - other fellow students who don't know the answer will be grateful for your word. Have courage and speak up. There are no grades in plenary and heads have not been bitten off yet.
</callout>

<callout type="tip" icon="true" title="Actively Preparing">
<WRAP right>
{{::notebook_monicore_pixabay_cc0jpg.jpg?200|}}
\ <fs 70%>Source: [[https://pixabay.com/en/notebook-pen-paper-office-business-1042595/|monicore@pixabay.com]] \fs> ([[https://creativecommons.org/publicdomain/zero/1.0/deed.en|CC0 license]])</fs>
</WRAP>
Whether you use the videos, internet research on the topic, or the script to prepare: In either case, it is a good idea to have a **pen and pad** handy and take notes. Some people like to watch the videos / the script on the bus, while eating, cooking or on the toilet. But: would you do the same in a lecture? Or think of it another way: Your washing machine broke down and as a mechatronics engineer you don't miss the chance to fix it yourself. Unfortunately, you only have a desktop PC with Internet at your disposal. Wouldn't you write down one or two things before you go down to the basement?

The goal of my course should be the consolidation of knowledge. For many, this is done through the application of the material, i.e. through (the) exercise(s). For others, it is through discussion and controversy, i.e. group work and the plenary. Writing down can also support. And for the exams, it won't hurt to have summarized the things that are important to you yourself.
</callout>
</notes>
<WRAP hide>
{{::talk.png?100|}}{::notebook_monicore_pixabay_cc0jpg.jpg?200|}}
  * tutorial
  * Tutor feedback
  * Active intervention in the tutorial
  * Active preparation
</WRAP>

---->>
<WRAP hide>
{{https://cdn.pixabay.com/photo/2016/11/07/19/58/calendar-1806776_960_720.jpg}}
</WRAP>
====== At the heart of a computer ======

==== Introduction ====

<imgcaption pic1|A microcontroller topless>{{ :Decapped_IC_Travis_Goodspeed.jpg?300|View inside a microcontroller}}</imgcaption>

The core of a computer is the processor in which the instructions are executed. This central processing unit (CPU) is also used in microcontrollers, which can be found around us in almost every device: Mobile phones, cars, bank cards, washing machines... Often there are even several microcontrollers built into the devices.

In addition to the command-executing microprocessor (more precisely the {{wpde>Arithmetic-logic unit|arithmetic-logic unit}}), the microcontroller also contains other peripherals such as memory, clock generation, analog-digital converters and much more. This makes it a compact tool for many applications. If you look at the microcontroller under the optical microscope, you will see the picture on the right (<imgref pic1>). The various peripheral components can be seen in it.

~~CLEARFIX~~ <imgcaption pic2|Microcontroller under the microscope>{{ :atmega8-hd_wiki.jpg?300|atmega8-hd_wiki.jpg}}</imgcaption>
<imgcaption pic3|Microcontroller schematic>{{ :atmega8-hd_interior.jpg?300|atmega8-hd_interior.jpg}}</imgcaption>

However, let us now look at the structure of the processor. The chip shown in <imgref pic2> and <imgref pic3> was developed in 1990 by [[https://spectrum.ieee.org/tech-history/silicon-revolution/chip-hall-of-fame-atmel-atmega8|two students]] and consists of several tens of thousands of transistors. This chip paved the way to cheap, fast, yet easily programmable controllers, from the fax machine to the hobbyist's basement, and can be found on Arduinoboards, among others. You will get to know and program the ATmega328 - a distant successor with several hundred thousand transistors - in higher semesters.

In the following we will now look at how a simple calculation like $y=a+b$ runs in the calculator.

~~PAGEBREAK~~~CLEARFIX~~~

<WRAP right>
{{youtube>Fxv3JoS1uY8?size=543x300}}
</WRAP>

~~PAGEBREAK~~~CLEARFIX~~~

==== two transistor(types) will do ====
<WRAP right><well>
<imgcaption pic4|n- and p-channel MOSFETs></imgcaption> \\

{{url>https://www.falstad.com/circuit/circuitjs.html?running=false&ctz=CQAgjCAMB0l3BOJyWoVaYDsYAcuA2BAJgFZdjcEBmSgkAFlJGYeoCgAzES3cBhiGrV6YAS2IZmMSMXYAlISJ74l9MvUhRGW6dpil2AczXgs9YaJJQFZ0eLCV+giJLDwe0ZsS-7oDBHYAdztncGsxQTkQy3ACLVjsTXYGSGo4rTBHFT5I7WoAfTACyCLiYgQGAtIS2Dhi0uICwo5uXiFIQVjqTpZ0rRlonlJ7KNU8yGNTWgTlalJJScVEp1iETW1Uln0pYOHR-Y7BSZCNI5zzyYAPIQJ1HBZIPmIsQRc+AAcAWgBpAEMAHZ-AA2AB0AM4AWQA8gBlABiAFEACrsG7ULDpSosMDMBBYRjgPgA36AkEQmEIlFolTMMB3HhIOLod4gADiNPatGYPUktD4rI56KYcR5pAJ9IFRJAsJp1BFFQS4sZ6VZsuFdPlQmVYAxhLwIAAInLxLQCfMLD19XxjZDQnkkhltD5DCEwBFxGcJuwgA 600,500 noborder}}

</well></WRAP>

Ok, now we know at least that the chip consists of many transistors. But how do they work and how can you build something as complex as a processor from them? The exact function is the content of a course in the 2nd and 3rd semester. For digital applications it is sufficient to have a simple picture of a certain type of transistor - the MOSFET - in your head. This has the three connections:
  * **S**ource ("source"), the inflow of charge carriers.
  **D**rain ("sink"), the drain of the charge carriers
  * **G**ate ("gate"), the gatekeeper that controls the passage between source and drain: Is there the correct voltage at the Gate terminal((in fact, it is the voltage between Gate and Source that matters, not just the voltage from the Gate. But more about this in [[elektronische_schaltungstechnik:2_transistoren|Elektronische Schaltungstechnik]]), the connections source and drain are short-circuited, that means a current can flow and the voltage drop between them becomes small.
Two types of MOSFET are important for the following digital circuits:
  * one that is __non-conducting__ at the gate at __low__ voltages ($0V$, logic $0$, low, $L$ or false) (__n__-channel MOSFET) and.
  * one that is non-conducting at the gate at high voltages ($5V$, logic $1$, High, $H$ or True) (p-channel MOSFET).
In the picture on the right (<imgref pic4>), you can see the two variants in action when the voltage at the gate just takes the digital voltage values (cf. [[fundamentals_of_digital_engineering:number_systems|video 1 on number_systems]]).

~~PAGEBREAK~~ ~~CLEARFIX~~
Now, it is interesting to note that these two types of transistors are sufficient to build all variants of logical functions. Logical functions combine one or more inputs ($X_1, ... X_n$) in such a way, that every kind of input uniquely leads to an output ($Y_1, ... Y_n$). In circuits, so-called gates correspond to logical functions. They need not be only such simple functions as an AND-gate (logical AND, conjunction, "only if both inputs are true, the output becomes true."). Mathematical operations can also be mapped in this way. To do this, gates must be cleverly combined with each other. In an exercise on [[fundamentals_of_digital_engineering:binaer_logic|binary_logic]], it is shown that all gates can be constructed using NAND or NOR gates. So if we could figure out how to build a NAND or NOR gate from transistors, we could in turn build all gates from that up to addition.
~~PAGEBREAK~~ ~CLEARFIX~~

==== gate logical? ====

<WRAP right><well>
<imgcaption pic20|Simulation of an inverter></imgcaption> \\

{{url>https://www.falstad.com/circuit/circuitjs.html?running=false&ctz=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-MFgDCGwA8kWGgAVesVxsFrMFnMAe3rtebQA 600,500 noborder}}

</well></WRAP>

Before the NAND gate is considered, a very simple circuit is to build, which generates the output value $Y(0)=1$ from the digital input value $X=0$ and for $X=1$ correspondingly $Y(1)=0$. So this circuit always negates the input value and is called inverter. For this purpose the two types of transistors are combined similar to a voltage divider or a half bridge. Thus always only one transistor becomes conductive, the other one correspondingly high impedance.

In <imgref pic20> is also drawn the corresponding circuit with normally open and normally closed switches. It is important for the coming consideration that the logic voltages ($0V$, $5V$) are just switched complementary via the switches. For this reason this technique is also called CMOS-technique: __C__omplementary __MOS__FET. In today's electronics, this technology is used throughout and has completely replaced older variants (e.g. TTL).

~~PAGEBREAK~~~CLEARFIX~~~
<WRAP right><well>
<imgcaption imageNo1 | inverter gate in CMOS on chip>
</imgcaption>
{{drawio>InverterGateMicroscope}}
</well></WRAP>

From this example, it can be seen how a logical $1$ can become $0$ in mobile phones, vehicle control units and television sets. The <imgref imageNo1> shows the realization of this gate in silicon:
  * Figure (1) shows the image of a scanning electron microscope, which shows several layers simultaneously. The three circular elements are electrical vias through several layers. In green are the two structures of the MOSFETs, which can act as a "valve" to open the connection up ($5V$) or down ($0V$).
  * Picture (2) is a false color picture. In beige is the top conductive layer and in blue is the non-conductive area of the top layer.
  * In figure (3), the different signals have been highlighted. \\ The output $A$ is taken out via a via with the right connection.


~PAGEBREAK~ ~CLEARFIX~~
==== non-UND, right? ====

<WRAP right><well>
<imgcaption pic21|Simulation of a CMOS NAND gate></imgcaption> \\

{{url>https://www.falstad.com/circuit/circuitjs.html?running=false&ctz=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-6IAG95PncuD6G+nwujBBhuM2fyRqiQjITUsRoEqTQ0C06SZNkwp5AURQoCUiETCBCCoVCU61Mu2EjqkSYdNk3S9P0gxQEE0yUZGVzoU6SFTgJU59q2glIV+M7SsJWgSUi07DPOF5EJ+548J+jz3puKk6rRJjkJp-4XnehknpkMQCXMUGfoosQ0EEQA 600,500 noborder}}

</well></WRAP>

So how does the NAND gate work, on which any logic can be built? The concept behind this gate is that the output will only output logic false ($Y=0$) if both inputs are set to logic true ($A=1$ and $B=1$). The gate is shown in <imgref pic21> above. This must be implemented using the two transistor types addressed. So this structure must be such that:
  - only if both inputs switch transistors with $A=1$ and $B=1$ at the same time, $0V$ shall be applied,
  - if one of the inputs $A$ or $B$, or both equal 0, $5V$ shall be applied.
The former is possible by connecting the transistors in series at $0V$. These must short the source and drain terminals at a gate voltage of $5V$ (input to logic $1$), so n-channel MOSFETs are required. \ The latter requires a parallel connection of transistors against $5V$. These must short the source and drain with a gate voltage of $0V$. Here, p-channel MOSFETs are used (<imgref pic21> below).

The conversion to silicon (<imgref ImageNo2>) again appears somewhat opaque at first glance. Also in this illustration again three pictures are to be seen. In the first picture the transistors are marked green again and also the vias are to be recognized again over white circles. If you take a closer look, you will notice that the via at $5V$ can be reached via the left or right MOSFET. The via at $0V$ can only be reached if both lower MOSFETs are short-circuited. Thus, the setup is consistent with the circuit determined so far. The <imgref pic7> and <imgref pic22> are to clarify this again.


~~PAGEBREAK~~~CLEARFIX~~~

<WRAP right>
<WRAP group><WRAP column>
<WRAP right><well>
<imgcaption image #2 | NAND gate in CMOS on chip>
</imgcaption>
{{drawio>NANDGateMicroscope}}
</well></WRAP>

<imgcaption pic7|>{{ :cmos_nand_schematic.gif|NAND gate in CMOS on chip (schematic)}}</imgcaption>

</WRAP><WRAP column>
<well>
<imgcaption pic22|Simulation of a CMOS NAND gate (structure similar to Si-Die)></imgcaption> \\

{{url>https://www.falstad.com/circuit/circuitjs.html?running=false&ctz=CQAgDOB0YzCMI5kgJhig7BgzGAHCgGwCcxGIKArCNZWAFADmFeeI2hbKrIALHIXD0ASnzaVu7FDUkJpaCCkjUl1bJDj9KxIpWyU8ZYtii88IsX0K8pVm3MQwKy52o0ZrpFNjyEUxQgxiNmRKegB3Swk2XmsZEPpeMBM8CF4Ainx48EQAfWtcsFy4FH9eXMpC6HhC3JRi3OxE5JBUvmx5LNibRXzyov4wesrkWDAyQko-EbrZpoBZCiSpNLYOiEVlZpTFHjQ1tBzCPtrB4aq4WrgI9g5EXhsONk0em6f7mySIF6FI7Dv3oDOL92MD3l9QQlIrwOpCxBB3gxItw1mCwI9gQxRLhPmt0TkNnwINRCaE3vizO1pJSkVTLDDpD9aTjLCymW9DnAdCsKChXtCBBQpnxBesQfxBFzqYL9uKZV0ZXzxTE4vpqXFmYcqIIxWLmcCfu8SlClgjDmYzRsbhaeTbEfRFpTolESjlVA76XFKQ8Qq56AAzHn-HWHd4SRAuUYoAMikNpUXA8NwSPQaOBsVSnk-PQRkmpjmWoN3BiBiWIbll7MmZN5ob0AAeIAAtLQ4NQ4AhKHAYksQABBBvN6i8JMIEfD3sAITeBskfO+klp88yXGWepuy9ly+N4rizu6pt3NmXDMPzLuy7VZ7e4cv4YPDEbTdDVHYZp7SoA8gBVAAq1plYVT2XWlgOWMtSn5HlINuQQQLeC81y1YUGDbBAIDaTDCWkYlBybPBaGkJsSBoCclWnJ8CNaExiOIVpBGPGwB0WJtYTiVjqV9d0WNhDIONabCtkifiNRADUbn4jIICkiwRJsJtKTiBwkE2FQXHUOAPHSHQfD8AIghCLZREkwQFLYDIHAUVxrI0rQdEmfRDCCExkDMG4sMQV02lpDtpDaXzm0FHzXSbQUzMCwRH0CylQoQJtgmbD8bB-f8WJisK2lCyK-SAA 600,600 noborder}}

</well>

</WRAP></WRAP></WRAP>

~PAGEBREAK~ ~CLEARFIX~~
==== a simple bill ====

<WRAP right><well>
<imgcaption pic24|Simulation of an adder></imgcaption> \\

{{url>https://www.falstad.com/circuit/circuitjs.html?running=false&ctz=CQAgjCAMB0l3BWcMBMcUHYMGZIA4UA2ATmIxAUgpABZsAoMBCFGqsDQkbNcTqECmoJ6AGT5d2-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 600,500 noborder}}

</well></WRAP>

How can the NAND gates now be connected in such a way that the arithmetic unit can perform operations such as $y=a+b$? For this purpose, the operation is first considered only for binary values. In order to distinguish binary values from decimal values, they are preceded by $0b$. So the following combinations are possible:
  * $0 + 0 = 0$ resp. $0b0 + 0b0 = 0b00$
  * $0 + 1 = 1$ or $0b0 + 0b1 = 0b01$
  * $1 + 0 = 1$ resp. $0b1 + 0b0 = 0b01$
  * $1 + 1 = 2$ resp. $0b1 + 0b1 = 0b10$

It can be seen that only if both inputs are even $1$ the second digit of the bit value is set. This just corresponds to an AND. But as everything is to build out of NAND-gates, a clever interconnection of these gates is needed. For this an inverter-gate is connected to a NAND-gate. The inverter-gate in turn is obtained via a NAND-gate, if both inputs are connected. In <imgref pic24> this circuit is shown below by the two lower NAND-gates. \\ A circuit can also be found for the first digit of the bit value. How can one come up with these circuits? This is explained in the chapter [[switching networks]].

~~PAGEBREAK~~ ~~CLEARFIX~~
If many inputs or outputs are combined, larger numerical values can be converted. That is, the calculation $3+3$ or in binary $0b\color{green}1\color{violet}{1} + 0b\color{blue}{1}\color{red}{1}$ is broken down to several individual calculations. This is similar to adding by hand by writing the numerical values one below the other and calculating step by step. In this example, $0b\color{violet}{1} + 0b\color{red}{1}$ would need to be calculated first, resulting in $0b\boldsymbol{1}0$. In the next step $0b\color{green}1 + 0b\color{blue}{1}$, the overflow from the previous calculation $0b\boldsymbol{1}$ must also be taken into account. Thus, in principle, numbers of any length can be linked together.

In order to make the numbers more manageable for man and machine, a sensible grouping was set up: 8 binary number values are combined into one byte. This can be read by humans as decimal value $0...255$ or hexadecimal value $0x00 ... 0xFF$ in programs. In the microprocessor these numerical values are always handled as binary values.

~PAGEBREAK~ ~CLEARFIX~~
==== Nano-Lego ====

<imgcaption pic10|>{{ https://upload.wikimedia.org/wikipedia/commons/a/aa/Silicon_chip_3d.png?300|three-dimensional structure of a silicon chip}}</imgcaption>



~PAGEBREAK~ ~CLEARFIX~~
-----


===== further links =====

  * A nice overview of core ideas for [[https://informatik.mygymer.ch/ef2019/rechnen-mit-strom|calculating with electricity]] has been compiled by Gymnasium Kirchenfeld (CH)
  * Various microscope images at [[http://www.siliconzoo.org/megamos.html|SiliconZoo.org]]
  * [[http://www.righto.com/2016/12/die-photos-and-analysis-of_24.html|http://www.righto.com/2016/12/die-photos-and-analysis-of_24.html]]
  * explanation of [[https://ca.wikipedia.org/wiki/CMOS|CMOS]] in the English Wikipedia
  * Wiki page on [[https://de.wikipedia.org/wiki/Integrierter_Schaltkreis|Integrated Circuits]]

~PAGEBREAK~ ~CLEARFIX~~
----

===== image references =====
<imgref pic1>: [[https://www.flickr.com/photos/travisgoodspeed/2818745933|TravisGoodspeed@Flickr]]
[[https://creativecommons.org/licenses/by/2.0/|CC BY 2.0]]

<imgref pic2>: [[https://commons.wikimedia.org/wiki/File:Atmega8-HD.jpg|ZeptoBars@Wikimedia]],[[https://creativecommons.org/licenses/by/3.0/deed.en|CC BY 3.0]]

<imgref pic1>, <imgref pic6>: [[http://www.siliconzoo.org/megamos.html|SiliconZoo.org]], license unknown.

<imgref pic7>: [[https://commons.wikimedia.org/wiki/File:Silicon_chip_3d.png|David Carron@Wikimedia]],public domain====== 4. Realization of Switching Networks ======

==== Motivation ====

==== Assignments ====
The assignments can be found in ILIAS under:
(2) Introduction, Lecture Notes and Assignments >> Assignments.

For details, see the [[introduction]]

----
===== 4.1 Realization of switching networks I =====

<WRAP group>
<WRAP half column>
==== Targets ====
After this lesson, you should:
  - Have heard of the different logic families.
  - Understand why any logic function can be implemented with a two-level combinational circuit.
  - Know the graphical simplification for logic operations.
  - Have heard about the different types of programmable devices.

<WRAP>
<WRAP half column>
==== video ====

{{youtube>-Mrg5uSy35Q}}

</WRAP>
</WRAP>
----
===== 4.2 Realization of switching networks II =====
<WRAP group>
<WRAP half column>
==== Objectives ====
After this lesson, you should:
  - Know the structure of nonvolatile data memories.
  - Know the types of different data stores and their characteristics.
  - Know the graphical simplification for links.
  - Have heard of the applications of these data stores.

<WRAP>
<WRAP half column>
==== video ====

{{youtube>Ti_zNM1d61Q}}

</WRAP>
</WRAP>
----
===== 4.3 More complex switching networks =====
<WRAP group>
<WRAP half column>
==== targets ====
After this lesson, you should:
  - Know the structure of more complex logic devices.
  - Be able to classify the terms gate, macrocell, function block.
  - Know applications of CPLD and FPGA.
</WRAP>
<WRAP half column>
==== Video ====

{{youtube>gxem36C0KgQ}}

</WRAP>
</WRAP>
----

[[===== applications]]
[[https://imgur.com/a/Hn5wC#0|Nibbler]]: A few chips in the 7400 family can even be used to build a functional CPU.====== 3. Switching networks ======

==== Motivation ====

==== Assignments ====
The assignments can be found in ILIAS under:
(2) Introduction, Lecture Notes and Assignments >> Assignments.

For details, see the [[introduction]]
----

===== 3.1 Synthesis of switching networks =====
<WRAP group>
<WRAP half column><callout>
==== targets ====
After this lesson, you should:
  - Know the following technical terms and their properties: Switching network, function table, min/max terms, KNF, KF, DNF, DF.
  - Be able to determine min terms and max terms from the function table.
  - Be able to create normal forms from the min and max terms.

</callout></WRAP>
<WRAP half column>
==== video ====

{{youtube>hAU4yrkKe6c}}

</WRAP>
</WRAP>
----
===== 3.2 Switching function optimization I =====
<WRAP group>
<WRAP half column><callout>
==== targets ====
After this lesson, you should:
  - Know the derivation of the KV diagram.
</callout></WRAP>
<WRAP half column>
==== Video ====

{{youtube>Mi-iTdn1_cc}}

</WRAP>
</WRAP>
----
===== 3.3 Switching function optimization II =====
<WRAP group>
<WRAP half column><callout>
==== targets ====
After this lesson, you should:
  - Be able to apply the KV diagram (fill, summarize, form logic function).
  - be able to exploit "Don't Care" states for switching purposes.
  - understand the different positions of the KV diagram.
  - know the allowed summaries.

</callout></WRAP>
<WRAP half column>
==== video ====

{{youtube>mvSDUvY87n4}}

</WRAP>
</WRAP>
----
===== 3.4 Switching function optimization III =====
<WRAP group>
<WRAP half column><callout>
==== targets ====
After this lesson, you should:
  - Know the following technical terms and their properties: Full conjunction, full disjunction, (core prime) implicant, (non) eliminable prime implicant.
- be able to optimize the switching function using the KV diagram.
- have heard the limits and alternatives to the KV diagram.

</callout></WRAP>
<WRAP half column>
==== video ====

{{youtube>BUnPVh3uLmQ}}

</WRAP>
</WRAP>
----
===== Related Links =====
[[https://web.archive.org/web/20170630164134/http://www.elektroniker-bu.de/kvdiagramm.htm|solver for KV diagrams]]: This can be used to plot KV diagrams and the minimized form directly from an entered logical formula. The solver can handle up to 8 variables. (Unfortunately only available in the internet archive)
\ [[http://www.mathematik.uni-marburg.de/~thormae/lectures/ti1/code/karnaughmap/|interactive KV diagram]]
\ [[http://www.mathematik.uni-marburg.de/~thormae/lectures/ti1/code/normalform/index.html| generator for KNF and DNF]]
\ [[https://play.google.com/store/apps/details?id=karnagh.ammsoft.karnagh|kmap solution app]]: free Android app for solving KV diagrams and displaying function table and gate logic====== 6. Switches ======

==== Motivation ====

State machines facilitate structuring in many applications. In digital technology, state machines are often called switching circuits.

==== Exercises ====
Some exercises on switchgears can be found {{ ::uebungen_zu_schaltwerken.pdf |here}}.

==== videos ====
There are no videos from me at this time. However, you can already find the important concepts explained in the videos of the flipflops and on the take notes in ILIAS.
\\ In addition, the following short videos can serve as a basis.

{{youtube>ULY9dMyHQ1Q}}
\\ {{youtube>mS2VxqL6TV0}}
\\ {{youtube>MFnRF07SoFo}}
\\ {{youtube>PQMtZkkFFv0}}
\{youtube>Pi-IV4don_Q}}

===== related-links =====
  * [[http://madebyevan.com/fsm|Finite State Machine Designer]]: the FSM Designer allows you to graphically create a state transition diagram.
  * An application of state machines can also be found in [[https://codeincomplete.com/posts/game-state-in-breakout/|games]] and sequences in software.
  * Another nice example of state machines in protocols can be found in the access of several processors to a memory location in RAM. This usually uses a "[[https://de.wikipedia.org/wiki/MESI|MESI]]-State Machine" (states **M**odified - **E**xclusive - **S**hared - **I**nvalid).~~NOCACHE~~
<affix offset-top="0" position-top="0" position-left="20">
<button size="xs" type="link" collapse="sidebar">{{icon>list}} Course Content</button>

<collapse id="sidebar" collapsed="false"><well>
<WRAP group><WRAP column 10>[[start|{fa>power-off?32}}]]</WRAP><WRAP column><fs large> [[start|Fundamentals of]] \ [[start|Digital technology]]</fs></WRAP></WRAP>
\ [[introduction]] \\
\ [[0. Aids]]
\ [[number systems|1st number systems]]
\ [[binary_logic|2. Binary logic]]
\ [[switching_networks|3. Switching_networks]]
\ [[realization_of_switching_networks|4. realization of ]] \ [[realization_of_switching_networks|Switching_networks]]
\ [[storage_elements|5th storage_elements]]
\ [[switchgears|6. switchgears]]

</well>
~~BARCODE~url=https://wiki.mexle.org/grundlagen_der_digitaltechnik/start~size=S~~ [[https://creativecommons.org/licenses/by-sa/4.0/deed.de|{electronic_circuit_engineering:by-sa.png?100}}]]
</collapse></affix>
====== 5. Memory Elements ======

==== Motivation ====

<WRAP group> <WRAP third column>

So far we only looked at expressions and gates, which depended exclusively on input values. However, it is not uncommon to consider intermediate memories for internal states. This is used extensively in programmable logic controllers in control cabinets, among other things.

==== Tasks ====

The assignments can be found in ILIAS under: (2) Introduction, Lecture Notes and Assignments >> Assignments

For details, see [[fundamentals_of_digital_technology:introduction|introduction to fundamentals of digital technology]]

</WRAP> <WRAP third column> {{:sps_s71500.jpg|SPS}}

(Image: [[https://de.wikipedia.org/wiki/Steuerungstechnik#/media/File:S71500.JPG|Palatinatian@Wikimedia]], [[https://creativecommons.org/licenses/by/3.0|CC BY 4.0]])

</WRAP> </WRAP>

----

===== 5.1 Evolution of flip-flops =====

<WRAP group> <WRAP half column><callout>

==== Targets ====

After this lesson, you should:

  - know the structure of RS flip-flop, D latch and clock edge controlled D flip-flop.
  - know what is the disallowed state and transparency, what are the problems with it and how to fix it.
  - Know the reason for the delay time.
  - know what times to consider when dealing with clock edge controlled flip-flops.

</callout></WRAP> <WRAP half column>

==== Video ====

{{youtube>UN_E1Jt5DE8}}

</WRAP> </WRAP>

----

===== 5.2 other flip-flop types =====

<WRAP group> <WRAP half column><callout>

==== Targets ====

After this lesson, you should:

  - Know the derivation and structure of JK flip-flop and T flip-flop.
  - know what asynchronous inputs are for clock edge controlled flip-flops.

</callout></WRAP> <WRAP half column>

==== Video ====

{{youtube>HCngft2Byxc}}

</WRAP> </WRAP>

----

===== 5.3 Conversion of flip-flops =====

<WRAP group> <WRAP half column><callout>

==== Targets ====

After this lesson, you should:

  - know how to convert flipflops.
  - understand the procedure that is necessary for the conversion of flipflops. This will be described in more detail in the next chapter.

</callout></WRAP> <WRAP half column>

==== Video ====

{{youtube>dPmFvrIyr08}}

</WRAP> </WRAP>

===== 5.4 Applications with Flipflops =====

<WRAP group> <WRAP half column><callout>

==== Targets ====

After this lesson, you should:

  - know the structure of registers, shift registers and frequency dividers.
  - be able to distinguish between synchronous and asynchronous coupling.
  - know the difference between (up, down) counter and frequency divider.
  - Be able to read the timing diagram.

</callout></WRAP> <WRAP half column>

==== Video ====

{{youtube>8iMnGJ2klwY}}

</WRAP> </WRAP>

===== further links =====

[[https://tams.informatik.uni-hamburg.de/applets/hades/webdemos/16-flipflops/10-srff/chapter.html|Hamburg Design System of the University of Hamburg]]: Java applets depicting flip-flops and their applications (caution requires Java-enabled browser, e.g. Internet Explorer).
\ [[http://www.falstad.com/circuit/|Simulation tool Falstad]]: In this simulation tool there is a nice application of flipflops to recreate a sine wave. In the menu: Circuits >> Analog/Digital >> Digital Sine Signal====== Basics of Digital Technology ======

No matter if you have to deal with a programmable logic controller (PLC) in the future, or if you just want to know how a mobile phone controls a display (or a microcontroller controls a light emitting diode) - the following chapters are indispensable for this. The knowledge gained can also be transferred to pure application programming.

<wrap indent>[[introduction|introduction to fundamentals of digital technology]]</wrap>.
<wrap indent>or: how best to handle the event? </wrap>
  - [[payment systems|payment systems]] \\ or: when does 1 and 1 = 1?
  - [[binary_logic|binary logic]] \\ [[binaeric logic|Binary logic]] or: 2B + ¬2B?
  - [[switching_networks|Switching_networks]] \ or: logic puzzles in a different way
  - [[realization_of_switching_networks|realization of switching networks]] \\ or: logic on the refectory card
  - [[storage_elements|storage_elements]] \\ [[switchgear|memory elements]] or: short term memory of logic
  - [[switchgear|switchgear]] \\ [[coke machine]] or: [[coke machine

[[===== insertion =====
  - [[basics_of_digital_technology:At the heart of a computer]]

===== Tips for the exam =====

  - [[basics_of_digital_engineering:tips_for_the_exam|tips_for_the_exam ]]

===== Further links =====

  - [[https://github.com/hneemann/Digital/releases/latest|Digital]]: This tool is used for simulating digital circuits. From Github, only the zipfile: Digital.zip is needed.
  - [[http://www.iris.uni-stuttgart.de/lehre/eggenberger/eti/|Extensive online script from the University of Stuttgart]]. \\ [[This covers a broader scope than we have in the lecture HHN in the MM and MR department.
  - [[http://elektroniktutor.de/digital1.html|Digitaltechnik auf elektroniktutor.de]] and [[http://www.inf-schule.de/|INF-Schule]]: \\ [[|Digital Technology on elektroniktutor.de]] and [[|INF-School]] for further explanations of our chapters on vocational school/gymnasium level.

====== General ======
  * No electronic aids are allowed on the exam (no calculator, mobile phone, laptop)**. Be prepared to do the calculation by hand.
  * **Results without derivation will not be marked**.
  * Please read the assignment carefully. Work on the assignment in the manner asked (e.g., as gate circuits, with function table, stating the rule of arithmetic).
  * **Take the opportunity to prepare documents**. A collection of formulas (e.g. with laws of Boolean algebra) shortens the search in the documents. Furthermore, it is a good idea to equip the script with sticky notes to make searching easier. \ <fc #ff0000>But keep in mind that neither formulary nor script are allowed in the exam.</fc>

====== Conversion of the number systems ======
  * Please always indicate the number system (e.g. hexadecimal numbers: 0x10, 10H, 10h).
  * Conversion from hex to binary or back (1101b <-> Dh) is easiest for one nibble at a time via a table.
  * Conversion from hex to binary or back is usually easier than from decimal to either system. Converting from binary to decimal is a bit more involved. If the conversion from decimal to hex and binary is required, then it is almost always easier to first convert the decimal number to a hex number and then to a binary number (55d = 37h = 0011 0111b).
  * The decimal places are also combined into nibbles starting from the decimal point.
  * Note that the decimal places must also be converted, of course (0.5h ≠ 0.5d !).
====== Simplifying Boolean Expressions ======
  * In function tables for gate circuits, for example, columns for intermediate results are useful. Otherwise, if the result is wrong, I cannot trace the intermediate steps and thus cannot award points.
  * Shorten logical expressions so far that as few as possible logical connections are needed (often a shortening is still possible over the De Morgan's law, distributive law or the definition of the XOR connection).
  * Acquire [[https://youtu.be/2UwxdCLFW70?t=2m43s|this method]] to be able to write binary numbers quickly among each other.

====== Setting up the KNF and DNF======
  * Careless errors in setting up the DNF and KNF can be avoided if the same sorting of the inputs is used as in the function table.
  * If a minimization of the normal form is required, then you should apply the Boolean laws (see Simplifying Boolean Expressions).

====== KV diagrams ======
===== Creating a KV Diagram =====
  * If you are to minimize the Boolean function for many outputs and are faced with a KV diagram for which you have no template, you can proceed as follows: First, create a KV diagram in which you enter only the decimal numbers of the positions. Two coordinates are quickly entered in. The 0 is always in the cell where all inputs are negated (usually top left). The maximum value is always in the cell where all inputs are non-negated (usually in the middle).

===== Reading the prime implicants of a KV diagram =====
  - First, determine what is being looked for:
    * minterms = conjunctive form = "groups of **0**es" or
    * Maxterms = disjunctive form = "groups of **1**en".
  - Find the implicants you are looking for (**0**en or **1**en).
  - Starting from the implicants, try to form a prime implicant (= area / group with the same binary value) as large as possible. "Don't care" states (marked with "d.c.", "-" or "x") may be considered as a suitable implicant. Note that only certain types of groups are allowed as prime implicants:
    * The size of prime implicants can only be a power to the base 2 (1, 2, 4, 8, ...).
    * Only straight implicants can be connected, or implicants that are not just adjacent across corners (e.g. ◨, ▣).
  - Once you have found all the largest possible prime implicants, then it is necessary to check what dependencies they have on each other. Often there is a constellation of prime implicants that has the fewest terms.
  - On the basis of the plotted prime implicants the logical function can be derived in equation form. In general, this can still be simplified (e.g. via definition of the XOR or De Morgan's law).


====== Switching mechanisms / state machines ======
==== Rear derailleur synthesis ====
The following procedure is recommended for every rear derailleur synthesis:
  - Careful reading of the specifications
  - Determine which specifications already exist, e.g.
    * Type of automaton (asynchronous = Mealy automaton, synchronous = Medvedev automaton or Moore automaton)
    * type of flip-flops to be used (D-FF, JK-FF, SR-FF, ...)
  - Determining the input, state and output variables
  - Determine the number of flip-flops required
    * for Mealy/Moore automata via the number of different states
    * for Medvedev automata via the maximum value to be output
  - Drawing up a state transition diagram
    * For Mealy and Moore automata, the assignment of state values to output values can be arbitrary.
    * For Medvedev automata, the assignment of state to output values is directly specified.
  - Creating the state transition table from the state transition diagram
    * The column arrangement below is suitable. \\ This puts the inputs and outputs for the different switching networks (transition and output networks) right next to each other. \ <WRAP center round box 90%>
^State- \ transition- \ table ^ <fc #800000>Current state</fc> ^^^ <fc #008000>Input value \ of the vending machine</fc> ^ <fc #ff00ff>Next state</fc> ^^^ <fc #6495ed>Output value \ of the vending machine</fc> ^^
| <fc #800000>**Z2**</fc> |<fc #800000>**Z1**</fc> | <fc #800000>**Z0**</fc> | <fc #008000>**X0**</fc> | <fc #ff00ff>**Z2'**</fc> | <fc #ff00ff>**Z1'**</fc> | <fc #ff00ff>**Z0'**</fc> | <fc #6495ed>**Y1**</fc> | <fc #6495ed>**Y0**</fc> |
| ::: | 0 | 0 | 0 | 0 | 0 | 0 | 1 | 0 | 0 |
| ::: | 0 | 0 | 0 | 1 | 0 | 1 | 0 | 1 | 0 |
| ::: | 0 | 0 | 1 | 0 | 1 | 1 | 0 | 1 | 0 |
| ::: | ... | ... | ... | ... | ... | ... | ... | ... | ... |
| ::: | |||||||||
| \ **Transition Network (ÜNW)** ||||||||||
|input value \ of the ÜNW | <fc #800000>current state</fc> ||| <fc #008000>input value \ of the automat</fc> | |||||
|| Output value \ of the OB</fc>) | |||| <fc #ff00ff>Next state</fc> || ||
| \ **output network (ANW)** ||||||||||
|Input value \ of the ANW | <fc #800000>Current state</fc> || only for Mealy: | <fc #008000>Input Value</fc> | |||||
|output value</fc> of the ANW | |||||||<fc #6495ed>output value</fc>| of the automaton</fc>|

</WRAP>
    * *) The output value of the transition network is only the <fc #ff00ff>next state</fc> if the switch mechanism is implemented on DD flip-flops. If the switch network is not implemented on D-FF, a conversion to FF input values must be considered.
  - From the transition network and output network columns, the KV diagrams for the respective switching networks can be created.  ====== 1. Number systems ======

==== Motivation ====
<WRAP right>
{{:chapter2_codeexample.jpg?200|}}
{{::chapter2_hexample.jpg?150|}}
</WRAP>

You may already know how to represent code in various programming languages. In your studies you will get to know programming in C. You can see an excerpt of it in the adjacent picture.

But what does this look like in the memory of the microprocessor? You can see this in the 2nd picture next to it. In the memory only bits are stored. These are combined for a better representation as hexadecimal number in so-called [[https://de.wikipedia.org/wiki/Intel_HEX|hex files]]. But how can you get back numerical values in the decimal system from them?

This is the aim of the following chapter.

Another goal is to understand basic arithmetic at the level of bits and bytes.

===== 1.1 Bits and Bytes =====
<WRAP group>
<WRAP half column><callout>
==== targets ====
After this lesson, you should:
  - Be able to distinguish between positive and negative logic.
  - Know the situations in the level diagram. Which of these are valid states?
  - Know the names for different groupings of bits.
  - know the powers of two from 2^0 to 2^10, and 2^12 and 2^16.
  - know the powers of two from 2^-1 to 2^-12.

</callout></WRAP>
<WRAP half column>
==== video ====

{{youtube>78kq0hOIJpk}}

</WRAP>
</WRAP>
----
===== 1.2 The binary number system =====
<WRAP group>
<WRAP half column><callout>
==== targets ====
After this lesson you should:
  - Be able to distinguish between the terms base, digit, eigenvalue, place value.
  - Be able to convert numbers from dual to decimal and vice versa.
  - know the basic arithmetic operations in the dual system.

</callout></WRAP>
<WRAP half column>
==== video ====

{{youtube>rvRB7kffVJ0}}

</WRAP>
</WRAP>
----
===== 1.3 Other number systems =====
<WRAP group>
<WRAP half column><callout>
==== targets ====
After this lesson, you should:
  - Be able to convert hexadecimal values to dual and decimal, and vice versa.
  - Be able to convert numbers with decimal places.
  - Be able to use the BCD code.
  - Be able to label different number systems.

</callout></WRAP>
<WRAP half column>
==== video ====

{{youtube>r3M9TFRur_8}}

</WRAP>
</WRAP>
----

===== 1.4 Two's Complement =====
<WRAP group>
<WRAP half column><callout>
==== targets ====
After this lesson, you should:
  - Be able to represent negative numbers as two's complement.
  - Be able to do addition and subtraction with two's complements.
</callout></WRAP>
<WRAP half column>
==== video ====

{{youtube>NG6zc1j0XbQ}}

</WRAP>
</WRAP>
----

===== Tasks =====
==== Tasks with review ====

The assignments can be found in ILIAS under:
[[2]] introduction, [[lecture notes]] and [[assignments]] >> [[assignments]]
\ For details, see the [[Introduction]]

==== Online quizzes ====

In addition, the following quizzes give you the opportunity to test what you have learned:

--> Quizz 1#
{{url>https://LearningApps.org/watch?app=985512 800,400 noborder}}

<--

--> Quizz 2#

{{url>https://LearningApps.org/watch?app=1478110 800,400 noborder}}

<--

--> Quizz 3#

{{url>https://LearningApps.org/watch?app=pfs1eduq517 800,600 noborder}}

<--

===== Further links =====

  * [[http://www.arndt-bruenner.de/mathe/scripts/Zahlensysteme.htm | Illustrative description of converting numbers with decimal places]]
  * An example of adding two BCD numbers can be found [[https://www.youtube.com/watch?v=qZ6wvs4xGWs|here]]



-------------------------------------------------




====== 0.Aid ======

In the following you will get to know two simulation tools. With these you can reconstruct circuits and display their voltages and currents. This is particularly important for understanding circuits with operational amplifiers and transistors.

<callout>
==Why two simulation tools?==

  * For learning and understanding, on the one hand, it is important that you get a feeling for how currents and voltages are set in circuits.
  * On the other hand, you should practice using professional tools that enable high-quality simulations.  
</callout>
The former tools should allow for easy visualization. The latter tools tend to be a bit harder to use, but allow more complex circuits and more detailed measurements.
For high-quality simulation, you will use the free TINA TI program below. The illustration of the concepts allows the online simulation of Falstad.

----
===== 0.1 First steps in TINA TI =====

TINA is a SPICE (Simulation Program with Integrated Circuit Emphasis) program from Texas Instruments. With this you can reconstruct electronic circuits and map the time sequence. This is very useful for learning and testing.

<WRAP info>
--> Introduction to TINA TI 0 - Explanation of TINA, download and installation #
<WRAP group>
<WRAP>

Already the free version of the program is very comprehensive. You can find it as follows (recommended is link 1!):
  - [[mexle:TINA TI Download]] on this wiki (login via university account)
  - in the [[https://ilias.hs-heilbronn.de/goto.php?target=fold_24013&client_id=iliashhn|ILIAS course]]
  - Download (a bit awkward) via the [[http://www.ti.com/tool/TINA-TI|TI page]]
\\
=== German manual of the full version ===

An English manual can be found {{https://wiki.mexle.org/_media/elektronische_schaltungstechnik/tina_9.0_manual.pdf#page=53|here}} on the wiki. \\
A German manual is only available from the [[http://designsoftware.com/home/German/tina/TINA_9.0_manual.pdf#page=53|full version]].
</WRAP>

<WRAP help>
=== Tasks ===
Please install TINA TI. \ The following tips for this:
  * For "User Name" and "Company Name" pseudonyms (''Hinz&Kunz'', ''HHN'') can be entered.
  * at "Select schematic symbol set you want to use" select ''European (DIN)''.
  * All paths should fit as recommended.
</WRAP>
</WRAP>
<--

--> Introduction to TINA TI 1 - Structure of TINA TI, first circuit#.
<WRAP group>
<WRAP half column>
=== Objectives ===
After this lesson you should:
  - Be familiar with the various component bars in TINA TI,
  - be able to insert and rotate components and their descriptions,
  - edit component values,
  - be able to draw connections.

<WRAP>
<WRAP half column>
{{youtube>c-08I51qbbA?size=543x392}}

<WRAP>
<WRAP help>
=== Tasks ===
  - Rebuild the circuit from the video in TINA TI.
  - Additionally, change the following values:
    - Voltage source output value: 10 V
    - Size of resistor R1: 20k
    - Size of the resistor R1: 30k
</WRAP>
</WRAP>
<--

--> Introduction to TINA TI 2 - Output of values and debugging #
<WRAP group>
<WRAP half column>
=== Objectives ===
After this lesson, you should:
  - Be able to perform the "Electrical Rule Check" and use it for debugging,
  - be able to measure output values such as voltages and currents,
  - be able to use multimeters in TINA TI to measure voltages and currents,
  - be able to insert current measuring points correctly,


<WRAP>
<WRAP half column>
{{youtube>L9eF64ZuLS0?size=543x392}}
</WRAP>
</WRAP>
<WRAP help>
=== Tasks ===
<WRAP group>
<WRAP column 70%>
- Rebuild the circuit from the video in TINA TI.
  - Use the Ammeter instead of the Current Arrow and the Voltage Pin instead of the Voltmeter. \\ Are there any differences when you use them?
  - Imagine you want to build a small circuit with 3 light emitting diodes and wonder how much the voltage of the two 1.5V batteries will drop.
    - Build the simplified parallel circuit 1 (see picture). \\ The internal resistance of the battery should be **500mOhm** (property "Internal Resistance").
      - What voltage/current is measured?
      - What is the current through a string?
    - **__Optional__**: Rebuild parallel circuit 2 for this purpose (see picture). The internal resistance of the battery is to be maintained.
      - What voltage/current is measured?
      - What is the current through one string?
Solution: <wrap spoiler>Voltage of the battery 2.97V, battery current for variant 1: 74.07mA, battery current for variant 2: 67.76mA </wrap>.
</WRAP><WRAP column 25%>
<imgcaption pic1|parallel circuit 1>{{electronic_circuitry:tina_parallel_circuit2.jpg?200}}</imgcaption>
<imgcaption pic2|parallel circuit 2>{{electronic_circuitry:tina_parallel_circuit1.jpg?200}}</imgcaption>
</WRAP></WRAP>

</WRAP>
<--

--> Introduction to TINA TI 3 - Even more bugs and lots of diagrams#
<WRAP group>
<WRAP half column>
=== Objectives ===
After this lesson, you should:
  - Be able to fix the most common errors in the simulation itself,
  - plot time histories of signals,
  - work with the cursor in diagrams, separate curves and insert a legend,
  - be able to create and evaluate curves over temperature and other variables,
  - be able to create a structured simulation with title.

The template file can be found at <wrap tip>Tips for TINA TI</wrap>.
</WRAP>
<WRAP half column>
{{youtube>PJs9rDz5Dbs?size=543x392}}
</WRAP>
</WRAP>
<WRAP help>
=== Tasks ===

- You will find exercises for this video in the following chapter

</WRAP>
<--
</WRAP>



<WRAP tip>
-->Tips for TINA TI #
  * Avoid nodes directly on the output of a component.
  * The following keyboard shortcuts make Tina easier to use:
    * ''<Ctrl>+<R>'': Right-rotate a selected component.
    * ''<Ctrl>+<L>'': Left-rotation of a selected component
    * ''<Ctrl>+<Space>'': Switch to connection mode (wire).
    * ''<Ctrl>+<C>'', ''<Ctrl>+<V>'': Copy, Paste.
    * ''<Ctrl>+<Z>'', ''<Ctrl>+<Y>'': Undo, Redo
  * Please use the template file {{ :template_est_v01.tsc |template_EST.TSC}} when starting a simulation.
  * [[advanced tips for TINA TI]]

<--

-->General tips #
  * Use rule checks available with simulation tools, such as the "Electrical Rule Check" (ERC). \ Rule Checks show errors and warnings. If there are errors, the simulation will not run. For warnings, it will run, but there are unclear areas in the circuit.
  * Avoid unclean identifiers and text. I.e. try to write text so that it is legible (not overlapping, aligned the same).
  * Always specify a reference potential (ground).

<--

</WRAP>

----
===== 0.2 Online Circuit Simulator =====
<WRAP info>
--> Introduction to Online Circuit Simulator - explanation, example #.
<WRAP group>
<WRAP third column>
In addition to TINA TI, another simulation tool will be used in this course: The [[http://www.falstad.com/circuit/|Online Circuit Simulator]]. \\ This can help to better understand the currents and voltages in different circuits. The program animates the current flow and the applied voltage. Under "Circuits" >> "Operational Amplifiers (OPVs)" you will find various circuits that are useful for this course. \\ The Landesberufsschule Salzburg has created a [[https://www.lbs4.salzburg.at/fileadmin/user_upload/lbs4/Typo3_7.6/News/20170418_Falstad-Schaltungssimulator/Falstad-Schaltungssimulator_Shortcard1_v2.1.5js_LBS4.pdf|Short manual for the Online Circuit Simulator]]. The [[https://github.com/hausen/circuit-simulator|Source Code]] of the simulator can be found on GitHub.
</WRAP><WRAP twothirds column>
{{url>https://www.falstad.com/circuit/circuitjs.html?running=false&startCircuit=resistors.txt 600,400 noborder}}
</WRAP>
 
<WRAP help>
=== Tasks ===

Familiarize yourself with the Online Circuit Simulator by Falstad
  * first with the embedded example on the right
    * If the circuit is too small, click on ''Edit'' >> ''Center Circuit'' \\ If only part of the available space is taken up by the circuit after that, refreshing the browser window (''<Ctrl>+R'') or editing directly on the [[http://www.falstad.com/circuit/|Online Circuit Simulator homepage]] will help.
    * Most circuits in the wiki are initially stopped. The simulation can be started by pressing ''RUN/Stop''.
    * Check what change occurs in the current flow over different switch settings.
    * Change the resistor values by double clicking on the respective resistor
  * With further examples of equivalent voltage source (Thévenin's theorem) and equivalent current source (Norton's theorem).
    * Search under ''Circuits'' >> ''Basics'' >> ''Thévenin-Theorem''
    * At the bottom left of the window See two running graphs of the currents and voltages of the two circuits. \\ Click with the right mouse button on one of the diagrams and select ''Combine''. The curves should now be directly on top of each other
    * Move the mouse over one of the two voltage sources. \ The respective voltage/current curve is highlighted. \ Are both curves on top of each other?
    * Do the same for the equivalent current source (Norton theorem).
</WRAP>
</WRAP>
<--
</WRAP>

===== 0.3 Further simulation tools =====

In addition to the tools explained so far, there are other simulation tools. Some of them are briefly summarized here with their limitations.

{{tablelayout?colwidth=",298px,557px"}}
^ name ^ focus ^ limitations ^ commercial product ^
^ [[http://www.ti.com/tool/TINA-TI|Tina TI]]                                                               | Simulation of analog circuits \ (e.g. amplifier circuits) | Function limited \ in the commercial product the simulation of mixed digital-analog circuits is also possible | [[https://www.tina.com/circuit-simulator/|Designsoft Tina]]  |
^ [[https://www.microchip.com/mplab/mplab-mindi|MPLab Mindi]]                                              | Simulation of transient transitions | (e.g. DCDC converters) | Number of nodes < 150 (approx.) | [[https://www.simplistechnologies.com/|SIMETRIX / SIMPLIS]]  |
^ [[https://www.analog.com/en/design-center/design-tools-and-calculators/ltspice-simulator.html|LTSpice]]  | broad setup | flat learning curve (unusual key mapping) | - |

===== 0.4 Literature recommendations =====

For self-study I recommend the following literature.

^ Title ^ Author ^ Short description ^
| Operational amplifiers | J. Federau | Textbook with illustrative approaches. Accessible via university network or VPN [[https://rd.springer.com/book/10.1007/978-3-8348-2146-1|viewable]].                                                                                                                       |
| OP Amp Applications Handbook || very nice and detailed textbook by the manufacturer Analog Devices, "Freeware", Online [[http://www.analog.com/en/education/education-library/op-amp-applications-handbook.html|viewable]]                                                              |
| Handbook of Operational Amplifiers Application || A similar nice handbook is available from the competitor [[https://www.ti.com/lit/an/sboa092b/sboa092b.pdf|Texas Instruments]]                                                                                                                             |
| Semiconductor Circuit Technology | U. Tietze, Chr. Schenk, E. Gamm | very detailed reference book. Available via university network or VPN [[https://link.springer.com/book/10.1007/978-3-662-56563-6|viewable]]. | Additionally, there is a [[https://link.springer.com/book/10.1007/978-3-662-56569-8|Collection of Exercises]]  |
====== 1. Basics of amplifiers ======

<WRAP right><panel type="default">
<imgcaption tablelabel|Overview of the different areas of electronics></imgcaption>
{{drawio>OverviewElectronics}}
</panel></WRAP>


===== 1.0 What is electronic circuitry =====

Electronic circuit technology comprises various subareas of electronics.
In <imgref tablelabel> the basics of electronics, its specializations and the interfaces addressed are presented.
\\ \\ \\
In [[Fundamentals of Digital Technology:start]], digital technology with gates and various simple programmable circuits have already been described. In [[Electrical Engineering 1:start]] and [[Electrical Engineering 2:start]] the basics of simple electrical components and circuits were laid.
\\ \\ \\
Electronic circuitry now focuses on __electr**on**ic components and their circuits__, which pertains to analog electronics.
\\ \\ \\
These components and circuits often connect the digital world with the analog world or adjust voltages and currents so that they can be reused. In addition, the "transistor" and "diode" components form the basis for both digital, power and high frequency electronics. Without exaggerating, these components are the basis of the modern world; today they form the [[basics_of_digital_technology:at_the_heart_of_a_computer|heart of every computer]] and every calculator.

~~PAGEBREAK~~~~CLEARFIX~~
<WRAP right noprint>
{{youtube>hCyR7a6szFc?size=543x342}}
<WRAP>
<WRAP noprint>
\\ \\ \\
But now what distinguishes electronics from electrical engineering? And what does "electronic circuitry" mean? The video opposite aims to answer these questions.
</WRAP>

<WRAP onlyprint>
But what is the difference between electronics and electrical engineering? And what is the meaning of electronic circuitry?\
For this purpose, it is useful to take a closer look at the individual parts of the term "electronic circuit technology": \\
The word __**technology**__ is derived from the Greek "τέχνη" (téchne) and means something like art or craft. At first glance, this seems a bit surprising, since technique is rarely associated with an artistic activity such as painting. In the higher semesters, however, you will learn that understanding, for example, the arrangement of electronics (circuit design) and the application of the individual components requires trained skill and creativity. \\ \\

A __**circuit**__ is an arrangement of electrical or electronic components to form a functioning whole or an electric circuit. We have already learned the term circuit in the first semester in [[:Electrical Engineering_1:start|Electrical Engineering 1]]. \\ \\

__**Electronics**__ is a portmanteau of electrons and technology. "Electron engineering" considers circuits in which an electric current or voltage is controlled by other electric signals. This means, for example, that a voltage $U_{in}$ controls an output voltage $U_{out}$. Physically, the two signals do not have to be electrically connected. \\ \\

First, the term electronics will be examined in more detail using various examples. First of all, a transformer will be considered. Is this an electrical or electronic component? In a transformer, the output voltage is generated by the change in the magnetic field. The magnetic field change in turn is caused by the current change on the primary side. So there is a direct transformation (a conversion) of the signals. Thus the transformer is not an electronic component. If this is still a bit unclear, it helps to work intensively on this chapter and to compare the following examples. \\ \\
The second component to be considered is the so-called contactor. A contactor is an electrically controlled switch. Here a coil, if energized, closes a secondary or output circuit. There is no direct electrical connection here. The contactor is often not yet regarded as an electronic, but as an electromechanical component. \\ \\
The last thing to be examined is the electron tube in the light of electronics and electrical engineering. An electron tube is a vacuum vessel, with several connections. Two of the terminals lead internally to one electrode each, which face each other. These can be brought to a potential difference against each other and heated. This allows electrons to escape from the electrode and generate a current to the other electrode through the vacuum. A grid is placed between the two electrodes. If this is set to an opposite potential, the current flow can be stopped. Here, the grid potential can be used to change the current flow. The electron tube is already considered as an electronic component. Nowadays, the electron tube has been replaced by semiconductor components.
In this course we deal only with semiconductor electronic components and essentially with silicon as a semiconductor.

</WRAP>

~PAGEBREAK~~~~CLEARFIX~~

===== 1.1 Why amplifiers? =====
<WRAP right noprint>
{{youtube>KbrHOcgmRkI?size=543x342}}
</WRAP>

The amplifier is one - if not the - central component in electronics. To understand what it is used for, various examples are shown in the three short videos opposite.

In example 1, the amplifier is briefly explained from the point of view of acoustics.
<WRAP RIGHT NOPRINT>
<WRAP right noprint>
{{youtube>HC4NXIigD98?size=543x342}}
</WRAP>
In example 2, a sensor with variable resistance is to be evaluated. This case is very common with passive sensors (i.e., sensors without further signal conditioning). Many of these sensors produce only a small voltage difference or a very small current. If these sensors were connected directly to a load or another unit with a low input resistance, the measurement voltage would collapse and the measurement signal determined from it would be distorted.
<WRAP RIGHT NOPRINT~~~PAGEBREAK~~~CLEARFIX~~~
<WRAP right noprint>
{{youtube>Ki_uU418-xQ?size=543x342}}
</WRAP>
Example 3 shows various amplifiers interacting with a microcontroller. In addition to evaluating sensors, many applications require signal output. If powerful motors, LEDs, antennas or loudspeakers are to be controlled, an amplifier must convert the microcontroller signal appropriately. In addition, an amplifier can take over control tasks analogously and usually with little electronic effort.

~~PAGEBREAK~~~CLEARFIX~~~

===== 1.2 Amplifier - a black box is specified =====

<WRAP right noprint>
{{youtube>wB498oinin4?size=543x302}} ~~PAGEBREAK~~ ~CLEARFIX~~
</WRAP>

Before the amplifier is examined in more detail in the application, the interfaces and essential characteristics are to be dealt with first.

<WRAP onlyprint>
The switching symbol of the amplifier is a rectangle with an inset triangle. The input terminals on the left side are labeled $IN+$ and $IN-$. The output terminals on the right side are correspondingly labeled $OUT+$ and $OUT-$. The input voltage $U_E$, or $U_{IN}$ is applied between the input terminals and the output voltage $U_A$, or $U_{OUT}$ is applied between the output terminals. \\ \\
On the left-hand side, the signal to be amplified comes from any source. Often this can be thought of as an ideal (voltage) source - i.e. with internal resistance. The amplified signal is fed to a load on the right-hand side. In the simplest case, this load is an ohmic resistor. \\ \\
</WRAP>

<WRAP column 40%>
<panel type="danger" title="Notice">
<WRAP group><WRAP column 7%>{{fa>exclamation?32}}</WRAP><WRAP column 80%>
An amplifier is a system that uses a low-power input signal to control a much higher-power output signal. \\ \\
The necessary power is taken from the power supply!
</WRAP></WRAP></panel>
</WRAP>

~PAGEBREAK~ ~CLEARFIX~~
==== parameters ====

<WRAP right><panel type="default">
<imgcaption pic1|amplifier with source and load>
</imgcaption>
{{drawio>replica_of_an_amplifier_blackbox}}
</panel></WRAP>



A voltage amplifier is required for most applications. Accordingly, this is the basis for the following explanation. However, the determined values apply accordingly to other amplifiers. In <imgref pic1> a voltage amplifier is shown as a black box. The voltage amplifier always tries to output a given multiple of the input voltage $U_E$ at the output as output voltage $U_A$. This "multiple" can be determined as a ratio. \\ \\

On the right is a **simulation of an ideal amplifier**. The input source specifies the voltage to be amplified. The amplifier with amplification factor 100 has the connections for input and output voltage drawn in. On the right side a resistor is provided as load; this can be varied by a switch.
\\ \\
~~PAGEBREAK~~ ~~CLEARFIX~~~

<WRAP right>{{url>https://www.falstad.com/circuit/circuitjs.html?running=false&cct=$+1+0.001+0.04723665527410147+50+5+43%0A368+288+192+288+160+0+0%0A368+-112+192+-112+160+0+0%0Aw+224+192+288+192+0%0Ax+260+101+305+104+4+24+Last%0Ab+252+115+377+367+0%0Ab+-206+114+-49+366+0%0Ax+-198+100+-129+103+4+24+Quelle%0Aw+192+288+224+288+0%0Ag+224+288+224+320+0%0Ar+288+288+288+192+0+100%0Aw+224+288+288+288+0%0A370+192+192+224+192+1+0%0Ag+-16+288+-16+320+0%0Ar+-112+192+-112+240+0+10000%0Aw+-16+288+-112+288+0%0Av+-112+288+-112+240+4+5+40+0.05+-0.025+0+0.5%0A370+-32+192+0+192+1+0%0Aw+-112+192+-32+192+0%0Aw+288+192+320+192+0%0Aw+288+288+336+288+0%0As+336+192+336+240+0+1+false%0Ar+336+240+336+288+0+100%0Aw+320+192+336+192+0%0Aw+144+224+144+192+0%0Aw+144+256+144+288+0%0Aw+48+256+48+288+0%0Aw+48+224+48+192+0%0A212+48+224+80+224+0+2+100*(a-b)%0Aw+16+288+-16+288+0%0Ab+16+114+173+366+0%0Ax+1+99+189+102+4+24+idealer%5CsVerst%C3%A4rker%0Aw+48+192+0+192+0%0Aw+48+288+16+288+0%0Aw+192+192+144+192+0%0Aw+192+288+144+288+0%0Ax+24+185+43+188+4+12+IN%5Cp%0Ax+26+303+41+306+4+12+IN-%0Ax+135+184+167+187+4+12+OUT%5Cp%0Ax+136+303+164+306+4+12+OUT-%0A 730,400 noborder}}
</WRAP>

<WRAP right column 60em>
<panel type="default" title="Characteristics" no-body="true">
{{tablelayout?colwidth=""}}
^ Characteristic groups ^ # ^ Characteristic (german) ^ Characteristic (english) ^ Formula ^
| Ratio ^ 1 | Voltage Amplification $A_V$ | $\large{A_V =\frac{U_A}{U_E}}$ |
| :::            ^ 2 | Current Amplification $A_C$ | Current Amplification $A_C$ | $\large{A_C =\frac{I_A}{I_E}}$ |
| :::            ^ 3 | Transmission Resistance $R_ü$ | Transmission Resistance, \\ Transimpedance $R_T$ | $\large{R_ü =\frac{U_A}{I_E}}$ |
| :::            ^ 4 | Transmission Conductance (Slope) $G, S$ | Transmission Conductance, \\ Transconductance (Slope) $S$ | $\large{G = S = \frac{I_A}{U_E}}$ |
| Input/Output Resistance ^ 5 | Input Resistance $R_E$ | Input Resistance $R_I$ | $\large{R_E = \frac{U_E}{I_E}}$ |
| :::                           ^ 6 | Output Resistance $R_A$ | Output Resistance $R_A$ | $\large{R_A =-\frac{\Delta U_A}{\Delta I_A}}$ |
| Reaction ^ 7 | Voltage Reaction $A_{rV}$ | - | $\large{A_{rV} =\frac{U_E}{U_A}}$ |
| :::         ^ 8 | current feedback $A_{rC}$ | - | $\large{A_{rC} =\frac{I_E}{I_A}}$ |
</panel>
</WRAP>
~~PAGEBREAK~~ ~~CLEARFIX~~

In the simulation some characteristics of an amplifier can be seen:
  - Ideally, no current flows into the amplifier on the input side.
  - The current on the output side depends on the connected load. If the load resistance is reduced with the help of the switch, the current increases. The amplifier thus tries to maintain the desired voltage.
  - On the output side of the amplifier, the current can flow in either direction. \\ The amplifier adjusts the current so that the amplified voltage $U_A=\pm 2.5V$ can be measured at the output.
\\
The ratios of the input and output quantities of a black box are called **characteristics**. For example, a well-known characteristic is the efficiency $\eta = \frac{P_A}{P_E}$. In the amplifier, only the voltages and currents are considered as input and output quantities. Various amplifier characteristics are shown in the table. \\
Depending on the desired input variable, which is to be used as input, and the output variable, which is to be controlled, different **translation ratios** result. It is important to note that transfer resistance $R_ü$ and transfer conductance $S$ do not correspond to any electrical component, since current and voltage are not measured at the same terminals. \\
If current and voltage are related at the same terminals, the **input resistance** $\bold symbol{R_E}$ and output resistance $R_A$ are obtained. From [[Electrical engineering_1:start|Electrical engineering 1]] it is known that for an equivalent resistance of a system the quotient of open circuit voltage $U_{LL}$ and short circuit current $I_{KS}$ can be used. Important: A resistor must always be considered in the load arrow system: For a positive resistance value, the current and voltage arrows must be plotted in the same direction. For input resistance, this is already the case (see also <imgref pic1>). \\ However, for **output resistance** $\boldsymbol{R_A}$ (__in__ amplifier), the arrow of $U_A$ and $I_A$ is antiparallel. The quotient is a negative value and accordingly must be negated. However, for the output resistance, the measurement of the short-circuit current $I_{A,KS}$ is also problematic. At the beginning of this subchapter it was already described that the voltage amplifier always tries to output a given multiple of the input voltage $U_E$ at the output as input voltage $U_A$. It would also attempt to do this in the event of a short circuit. The current $I_A$ would accordingly become very large. When measuring the short-circuit current $I_{A,KS}$, the disproportionately increasing losses $P=R_A\cdot I_A^2$ via the output resistance could then destroy the amplifier. Thus, determining the output resistance via the quotient of $U_A$ and $I_A$ is not possible. Since the output resistance is assumed to be ohmic resistance - i.e. slope in the $U$-$I$ diagram - any two (load) points on the straight line can be used to form the quotient. Specifically, to measure the output resistance, consider the amplifier with two different loads $R_{L,1}$ and $R_{L,2}$ and find the differences $\Delta U_A=U_{A,1}-U_{A,2}$ and $\Delta I_A=I_{A,1}-I_{A,2}$. \\
The most important characteristics of the voltage amplifier are voltage gain $A_V$, input and output resistance $R_E$ and $R_A$.

~PAGEBREAK~ ~CLEARFIX~~
==== equivalent-circuit diagram ====

<WRAP right><panel type="default">
<imgcaption pic2|Amplifier with source and load (with real voltage sources)>
</imgcaption>
{{drawio>replica_of_an_amplifier}}
</panel></WRAP>

After the characteristics, the first attempt will be made to understand the internal structure of the amplifier. In the previous section, current-voltage relationships were determined from the external view. There, an input resistance $R_E$ was described on the input side of the amplifier. This is now considered in the equivalent circuit (see <imgref pic2>). A resistance was also determined on the output side by considering it as a black box. Here, however, it must additionally be noted that the amplifier - according to its name - is supposed to amplify the input signal. So here, besides the output resistor $R_A$, also an element has to be used, which outputs the amplified voltage. This is the voltage source [(Note1>The voltage source used in the amplifier is a [[https://de.wikipedia.org/wiki/Gesteuerte_Quelle|controlled source]], this term will not be discussed further in this course)]. The voltage of the voltage source is based on the voltage applied to the input resistor $R_E$. It is more accurately larger by the factor of the voltage gain $A_V$.

~~PAGEBREAK~~~CLEARFIX~~~

<WRAP right>{{url>https://www.falstad.com/circuit/circuitjs.html?running=false&ctz=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-1ktlisu8S9hgAHM-3UX9gXENASwvBg4gkT0uw9A8R3AOAv1dBD4N-dtvn-RtB3Aps1DAetINA58AOA-8sO7E921glNhyQBiQMvf56yomhmIAgA3OMMBrX8uNdKQ6FdGBIDoTRxIQOhL2vHC4ynJtGIZCD62Ywd4y9PsJHvL00FNbSYIkQyCIMntqKeBgwhMwIA3MliGQAM14MJC2dRCyNsrydM8wc0D0wdsJ0M9d2IA8wtpbxwAYBTDxo29dJbIiGBUcUVPSrzoqdPiMubIznS4wT8oAlQrTPANyqylL+z5BDTKwjs+X8wygoYAAjbsQxbX80EazrIX4kNk1dPqMG9Tryz0VN9wcchvSAA 730,400 noborder}}
</WRAP>

On the right you can see a **simulation of a (simulated) real amplifier**. The input source has a high internal resistance. This means it has a high impedance and can only supply a small amount of current. The amplifier with amplification factor 100 has - besides the connections for input and output voltage - the connections for the supply voltage drawn in. On the right side a resistor is provided as load; this can be varied by a switch.
~~PAGEBREAK~~ ~CLEARFIX~~

In the simulation some characteristics of an amplifier can be seen:
  - On the input side, a small current flows into the amplifier.
  - The current on the output side depends on the connected load. If the load resistance is reduced with the help of the switch, the current increases. The amplifier thus tries to maintain the desired voltage.
  - The amplifier can output current as well as draw current. \\ The current on the output side flows in and out of the amplifier through the supply voltage terminals.
  - The simulation is based on a real amplifier. This has a small deviation from the expected value $U_A=\pm 2.5V$ at the output voltage. Part of the deviation will be described later in this chapter.

~PAGEBREAK~ ~CLEARFIX~~
==== idealized amplifier basic types ====

<WRAP right column 40em>
<panel type="default" title="idealized amplifier basic types" no-body="true">
{{tablelayout?colwidth=""}}
^ # ^ amplifier ^ graph ^ $\boldsymbol{R_E}$ ^ $\boldsymbol{R_A}$ ^ gain ^
^ 1 | voltage amplifier | {{voltage-amplifier_bb.png?200}}(nbsp)(nbsp)(nbsp) | \ \large{\rightarrow \infty}$(nbsp)(nbsp)(nbsp) | \ \ $ \large{\rightarrow 0}$ | \ \ $\large{A_V =\frac{U_A}{U_E}}$ |
^ 2 | current amplifier | {{current amplifier_bb.png?200}}          | \large{\rightarrow 0}$ | \large{\rightarrow \infty}$(nbsp)(nbsp)(nbsp) | \large{A_C =\frac{I_A}{I_E}}$ |
^ 3 | current-voltage-converter | {{current-voltage-amplifier_bb.png?200}} | \large{\rightarrow 0}$ | \large{\rightarrow 0}$ | \large{R_ü =\frac{U_A}{I_E}}$ |
^ 4 | voltage-to-current converter | {{voltage-current-amplifier_bb.png?200}} | \large{\rightarrow \infty}$ | \large{\rightarrow \infty}$ | \large{S =\frac{I_A}{U_E}}$ |
</panel>
</WRAP>

Depending on which input variable is to change which output variable, different basic amplifier types result. These are listed in the table opposite. As **graphic** in [[https://de.wikipedia.org/wiki/Blockschaltbild#Blockschaltbilder_in_der_Elektrotechnik|Block diagrams]] a square with diagonal is drawn for the respective amplifier basic type, which contains the corresponding formula symbol on the input and output side. \\ The **input resistors** $\boldsymbol{R_E}$ **and output resistors** $\boldsymbol{R_A}$ **for __ideal voltage amplifiers__** shall now be considered in more detail.
If a voltage is read in, the input resistance should load the source as little as possible so that the voltage to be measured does not collapse (cf. <imgref pic2>). This can also be easily checked in the simulation of the real amplifier (see above). If the resistance of the load is increased there (double-click), it approaches the input resistance of the amplifier. If the value is set to $1 M \Omega $, the voltage collapses to half. The source resistance is then equal to the input resistance of the amplifier. So it is important that the input resistance is as high as possible, or ideally tends to infinity. \\
A similar consideration can be made for the **output resistance** $\boldsymbol{R_A}$. If a voltage is to be output, the output resistance must be dimensioned so that the voltage at the load does not collapse at the output either. The output resistance should be as small as possible, so that the voltage dropping there becomes low. \\

~~PAGEBREAK~~ ~CLEARFIX~~
<WRAP right><panel type="default">
<imgcaption pic3|amplifier with source and load (with real current sources)>
</imgcaption>
{{drawio>replica_of_an_amplifier_current_source}}
</panel></WRAP>

If now the **input resistors** $\boldsymbol{R_E}$ **and output resistors** $\boldsymbol{R_A}$ **are considered for __ideal current amplifiers__**, a different view of the amplifier is convenient. In <imgref pic3>, the same amplifier considered so far can be seen. In this case, however, all real voltage sources have been replaced by real current sources. This transformation has already been described in [[Electrical engineering_1:start|Electrical engineering 1]]. Depending on the considered, electrical quantity, one or the other real source can be advantageous. \\ With this knowledge, the input and output resistance of the current amplifier shall now be considered. On the input side, most of the input current $I_E$ should flow into the amplifier. The input resistance $R_E$ must accordingly tend towards zero. The maximum current should also flow out of the amplifier at the amplifier output. Here, the output resistance $R_A$ has to strive towards infinity accordingly, so that the lowest possible current flows through it. \\
For **__current-voltage and voltage-current converters__** applies:
  - The respective name is composed of input variable-output variable.
  - For the input and output resistances, the corresponding consideration of the ideal current amplifier or voltage amplifier can be used.


~PAGEBREAK~ ~CLEARFIX~~
===== 1.3 Feedback =====

<WRAP right><panel type="default">
<imgcaption pic4|Block diagram of a feedback amplifier>
</imgcaption>
{{drawio>block diagramfeedback}}
</panel></WRAP>


One of the fundamental principles of control engineering, digital technology and electronics is **feedback**. Thus, in [[Fundamentals of Digital Technology:start]], the output value of a NOR gate was already fed back to its input via detours for the development of a flip-flop. Similarly, here the output value of the ideal amplifier is to be fed back to the input. In contrast to digital technology, in control engineering and electronics a fraction (in rare cases: a multiple) of the output value is fed back. \\
Additionally, in control engineering, digital engineering, and electronics, there is another tool: the **block diagram**, or signal flow diagram. In Electrical Engineering 1, circuit diagrams were previously used. With circuit diagrams there is an interaction of all components due to Kirchhoff's rules, furthermore voltage differences everywhere or current over all components can be measured. \\
In contrast to this is the block diagram. This shows single blocks (also called links) which connect a cause with an effect. In general, no feedback from the effect to the cause is assumed. Causes and effects can be voltages or currents, which are then written on the respective connecting arrow. The block diagram does not claim to be a conservation of energy or charge, but serves as an overview of the effects and relationships. Therefore Kirchhoff's rules are usually not applicable there. \\
<imgref pic4> shows a **block diagram of a feedback amplifier** with an ideal voltage amplifier with gain $A_D$ drawn in the center. Via a feedback element, the output voltage $U_A$, reduced by a factor $k$, is fed back. The circle symbol with the arithmetic symbols (in the block diagram on the left) shows how the incoming values must be offset against each other. Thus, the value $k \cdot U_A$ is subtracted from the input value $U_E$ in the indicated block diagram.
\\ \\
The advantage of a real amplifier in negative feedback is that the gain $A_V$ of the whole system depends only negligibly on the gain factor $A_D$ of the real amplifier if $A_D$ is very large (see also task 1.3.2). In this case, the gain $A_V=\frac{1}{k}$. To avoid oscillation of the whole system, it must contain a delay element. This is present in the real amplifier in such a way that the output voltage $U_A$ cannot change infinitely fast [(Note2>The fact that a voltage change can only occur in finitely long time is also true for the input voltage. However, this cannot be influenced by the amplifier, but is externally specified.)] (see also task 1.3.1).

<WRAP column 80%>
<panel type="danger" title="Notice: feedback, positive feedback, negative feedback">
<WRAP group><WRAP column 3%>{{fa>exclamation?32}}</WRAP><WRAP column 85%>
**__Feedback__** refers to the feedback of a portion of the output signal of an amplifier. \\
In **__feedback__**, the part of the output signal with positive sign is fed back. \\
Thus, the output value always increases in magnitude to the input value.
\\ \\
With **__feedback__** the part of the output signal with negative sign is fed back. \\
So the output value always attenuates the input value.
</WRAP></WRAP></panel>
</WRAP>
~~PAGEBREAK~~~CLEARFIX~~




<WRAP column 80%>
<panel type="danger" title="Remember: amplification factors">
<WRAP group><WRAP column 3%>{{fa>exclamation?32}}</WRAP><WRAP column 85%>
The **difference gain** $\boldsymbol{A_D}$ refers only to input and output voltage of the inner amplifier: $A_D=\frac{U_A}{U_D}$ \\ This acts only without external feedback. It is also called open-loop gain. \\ \\

The **voltage gain** $\boldsymbol{A_V}$ refers to input and output voltage of the whole circuit __with feedback__: $A_V=\frac{U_A}{U_E}$ \ It is called closed-loop gain in English. \\ \\
</WRAP></WRAP></panel>
</WRAP>

~PAGEBREAK~ ~CLEARFIX~~
#REFNOTES #

====== Tasks ======

{{page>Exercise Sheet1&nofooter}}
{{page>ExerciseSheet2&nofooter}}

======== study questions ==
=== for self study ===
  * What is the definition of a reinforcer?
  * Explain with an example what is the essence of an amplifier.
  * How do you determine the input and output resistance of an operational amplifier?
  * How to choose the input and output resistance of a current amplifier? Why?
  * Name 2 basic types of amplifiers.
  * When is it called positive feedback and when is it called negative feedback?
  * Explain the principle of negative feedback.
  * What is the difference between voltage gain and differential gain? Briefly describe the difference between $A_V$ and $A_D$.
  * How does $A_D$ affect the output voltage $U_A$ when there is no feedback in an operational amplifier circuit?
  * What is the effect of $A_D$ on the output voltage $U_A$ when feedback is present in an op-amp circuit and $A_D$ is increased from 100,000 to 200,000?
  * At what value for k does the feedback become maximum?
  * What values can k assume for a passively feedback amplifier?
  * What effect does k have on the amplifier?
  * What happens if you feed back the complete voltage?

=== with answers ===

<quizlib id="quiz" rightanswers="[['a3'],['a2'], ['a3'], ['a1'], ['a3'], ['a0', 'a3']]" submit="check answers">
    <question title="How is the output resistance of an amplifier defined?" type="checkbox">
$R_A = \Delta U_E / \Delta I_A$|
$R_A = U_E / I_A$|
$R_A = \Delta U_A / \Delta I_A$|
$R_A = -\Delta U_A / \Delta I_A$|
$R_A = U_A / I_A$
</question>
    <question title="When do we talk about negative feedback, when about feedback?" type="checkbox">
Feedback = negative feedback|
Feedback = neg. feedback, negative feedback = feedback general|.
negative feedback, negative feedback = feedback in general |
negative feedback, negative feedback = positive feedback
</question>
    <question title="Ideal resistances of a voltage-to-current converter" type="checkbox">
$R_E → 0$, $R_A → ∞$|
$R_E → 0$, $R_A → 0$|
$R_E → ∞$, $R_A → 0$|
$R_E → ∞$, $R_A → ∞$
</question>
    <question title="What can you use a linear controller for?" type="checkbox">
To control linear motors|
To output fixed voltage values|
To control linear circuits|
For the output of fixed current values
</question>
    <question title="What type of amplifier produces an output current $I_A$ from an input voltage $U_E$ in such a way as to produce an output voltage $U_A = C \cdot U_E$ with constant $C$?" type="checkbox">
current-voltage converter|
current-amplifier|
voltage-to-current converter|
Voltage-Amplifier
</question>
    <question title="The transfer resistor ..." type="checkbox">
Cannot be measured using a resistance meter|
can be used for voltage dividers|
is given by ${U_E} \over {I_A}$, with input voltage $U_E$ and output current $I_A$ |
gives a gain
</question>
</quizlib>

====== 2. Diodes and transistors ======
<WRAP><callout type="info" icon="true">
=== Introductory example===
Microcontrollers have many digital inputs that evaluate signals between $0...5V$ as a digital signal. However, the input signal can be disturbed during transmission by small coupled pulses. This interference can cause the signal to leave the permitted voltage range of approx. $-0.5...5.5V$ and thus destroy the logical unit.

To prevent such destruction, an overvoltage protection circuit consisting of diodes is installed (see e.g. [[https://ww1.microchip.com/downloads/en/DeviceDoc/Atmel-7810-Automotive-Microcontrollers-ATmega328P_Datasheet.pdf#page=58|ATmega 328]]). In case of an over-/undervoltage one of the diodes becomes conductive and lowers the input voltage by the resulting current. In the simulation it can be seen that the interference on the input side can be reduced to an acceptable, low level by the protection circuit.

This chapter explains why a diode becomes conductive at a certain voltage, what has to be considered when using diodes and which different types of diodes are available.

<WRAP>{{url>https://www.falstad.com/circuit/circuitjs.html?running=false&cct=$+17+0.000001+45.7+82+10+50%0Ad+224+208+224+160+2+1N4148%0Ad+224+256+224+208+2+1N4148%0Ag+224+256+224+288+0%0AR+224+160+224+128+0+0+40+5+0+0+0.5%0A207+224+208+320+208+4+zur%0Ax+335+211+425+214+4+12+logischen%5CsEinheit%0Aw+224+208+16+208+0%0Ar+-48+208+16+208+0+10%0Ap+-48+208+-48+272+1+0%0Ag+-48+272+-48+288+0%0AR+-272+208+-288+208+1+2+20+2.5+2.5+0+0.5%0Aw+-144+208+-112+208+0%0Aw+-112+208+-48+208+0%0A159+-112+208+-112+272+0+20+10000000000%0AR+-128+240+-144+240+5+5+137+1+1.5001+0+0.01%0Ar+-224+208+-272+208+0+1000%0AR+-112+272+-112+288+0+2+315+20+0+0+0.5%0A403+32+208+160+384+0+8_1024_0_4098_20_0.1_1_1%0A403+448+240+576+336+0+4_1024_0_4098_10_0.1_0_2_4_3%0Aw+-224+208+-144+208+0%0Ab+-320+112+-197+314+0%0Ab+-171+112+16+314+0%0Ab+167+111+434+313+0%0Ax+-321+336+-186+362+4+20+Digitales%5Cs%5C%5CnEingangssignal%0Ax+-135+338+10+341+4+20+St%C3%B6reinstrahlung%0Ax+226+337+446+363+4+20+Microcontroller%5Csmit%5Cs%5C%5Cninterner%5CsSchutzschaltung%0A 900,400 noborder}}

For the protection of digital interfaces that leave the device housing (e.g. USB), additional separate ICs are used that support this protection of the data processing chips. These protection diode ICs suppress the short-time voltages and are called __T__ransient __V__oltage __S__uppressor or TVS diodes. Typical TVS ICs are [[https://www.onsemi.com/pdf/datasheet/nup2301mw6t1-d.pdf|NUP2301]] or for USB [[https://www.onsemi.com/pdf/datasheet/nup4201mr6-d.pdf|NUP4201]].


</WRAP></callout></WRAP>

<callout type="info"><WRAP group><WRAP column 7%>{{fa>leanpub?32}}</WRAP><WRAP column 80%>
=== Further reading ===

  * With a depth beyond this course can be found the topic [[https://link.springer.com/chapter/10.1007/978-3-662-56563-6_2|Diodes in Tietze Schenk]]
  * For a deeper look at the level of this course and in pleasant morsels, see [[https://www.youtube.com/watch?v=Wsf6Ks-w3LE&list=PLjkreDBz1mhDrDO590zUy3eM1_4EtO5aW&index=1|Electrical Engineering in 5 Minutes - Diode Topic]]. \\ Here, the considerations of interconnecting diodes with nonlinear components are beyond the subject matter of this course.
  * A nice introduction on with less depth can be found in [[https://lx3.mint-kolleg.kit.edu/onlinekursphysik/html/1.4.3/xcontent5.html|KIT Bridge Course - 4.3.6 Diodes and Transistors (*)]]. \\ Some of the following passages, videos and pictures are taken from this introduction.
  * One with similar introductory character is available at [[https://www.leifiphysik.de/elektronik|LEIFIphysik]].

  * **as a podcast**: Several of the aspects of the silicon PN transition presented here are explained in [[http://minkorrekt.de/minkorrekt-folge-164-corona-unterhaltungsshow/?t=1%3A55%3A30|Methodically Incorrect Episode 164 - "Small Grain of Sand Hope"]].
</WRAP></WRAP></callout>.

<pagebreak>

<WRAP><callout>
=== Objectives ===
After this lesson, you should:
  - Know how to distinguish electron mobility in metals, semiconductors, and insulators,
  - know what the intrinsic conductivity of a semiconductor is,
  - distinguish between electron and hole conduction and relate them to p- and n-doping,
  - know what doping is and what it is used for...
  - know the difference between real and ideal diode,
  - be able to show the course in forward and reverse direction,
  - be able to choose the correct diode from different diode types,
  - be able to explain physical quantities such as reverse/residual current, reverse/residual voltage, breakdown voltage. 

</callout></WRAP>

===== 2.1 Current conduction in semiconductors =====

<WRAP noprint>
<WRAP right>
{{youtube>YB3pQ7P8SNg?size=543x392}}
<WRAP>

++++ Video transcript (Alternative to the explanation in the video) |
In metals, electrons are free to move. If an external voltage is applied, they follow the potential difference to the positive electrode: current flows. In insulators, on the other hand, the electrons are firmly bound to the atomic trunks. If a voltage is applied, they can at best be polarized. No current flows.

A semiconductor is a material whose conductivity lies between that of metals and that of insulators. The technologically most important example of a semiconductor is silicon. In the silicon crystal, the electrons are not freely movable as in a metal, because they are bound to the atomic trunks. But a small supply of energy (e.g. thermal energy) is sufficient to release the electrons from the atoms. Then, when a voltage is applied, an electric current flows. This is called the **intrinsic conduction** (intrinsic conduction) of the semiconductor. When the electrons move around in the semiconductor, this is called **electron conduction**.  

A hole with a positive electrical charge is created at the silicon atom from which the electron was removed. This is also called a defect electron. These holes can also move through the crystal lattice and thus generate an electric current. This is called **hole conduction**. Hole conduction can be thought of as a hole being filled by an electron from the neighboring atom. However, this creates a hole in the neighboring atom. Effectively, such a hole has migrated from one atom to another, carrying with it a positive electric charge.  

<imgcaption pic2|p-doping with aluminum>{{ :electronic_circuitry:mfile6xpsilicon.png?nolink&200|p-doping with aluminum}}</imgcaption>
<imgcaption pic1|n-doping with phosphorus>{{ :electronic_circuitry:mfile6xnsilicon.png?nolink&200|n-doping with phosphorus}}</imgcaption>

Most semiconductors are elements of the fourth main group, i.e. they have four electrons in the outer shell. This also applies to the element silicon. In the silicon lattice, each silicon atom is therefore connected to four neighbouring atoms via a bond. If foreign atoms are added to this semiconductor material, the electrical conductivity can be modified. This is called **doping**.  

Atoms of the fifth main group (e.g. phosphorus) have five electrons in the outer shell. If these are added to the silicon crystal lattice, one electron is surplus at these points, as it is not needed for the four bonds in the crystal lattice. This electron is much more mobile than the electrons that contribute to the bond and therefore greatly increases conductivity by electron conduction. This addition of free negative charge carriers is called **n-doping** (see <imgref pic1>).

On the other hand, by adding atoms of the third main group (e.g. aluminium), a so-called hole can be created at these points, as these atoms only have three electrons in the outer shell. This leads to an increase in conductivity by hole conduction. This addition of free positive charge carriers is called **p-doping** (see <imgref pic2>).
++++
</WRAP>

<WRAP onlyprint>
In metals, electrons are free to move. If an external voltage is applied, they follow the potential difference to the positive electrode: current flows. In insulators, on the other hand, the electrons are tightly bound to the atomic trunks. If a voltage is applied, they can at best be polarized. No current flows.

A semiconductor is a material whose conductivity lies between that of metals and that of insulators. The technologically most important example of a semiconductor is silicon. In the silicon crystal, the electrons are not freely movable as in a metal, because they are bound to the atomic trunks. But a small supply of energy (e.g. thermal energy) is sufficient to release the electrons from the atoms. Then, when a voltage is applied, an electric current flows. This is called the **intrinsic conduction** (intrinsic conduction) of the semiconductor. When the electrons move around in the semiconductor, this is called **electron conduction**.  

A hole with a positive electrical charge is created at the silicon atom from which the electron was removed. This is also called a defect electron. These holes can also move through the crystal lattice and thus generate an electric current. This is called **hole conduction**. Hole conduction can be thought of as a hole being filled by an electron from the neighboring atom. However, this creates a hole in the neighboring atom. Effectively, such a hole has migrated from one atom to another, carrying with it a positive electric charge.  

<imgcaption pic2|p-doping with aluminum>{{ :electronic_circuitry:mfile6xpsilicon.png?nolink&200|p-doping with aluminum}}</imgcaption>
<imgcaption pic1|n-doping with phosphorus>{{ :electronic_circuitry:mfile6xnsilicon.png?nolink&200|n-doping with phosphorus}}</imgcaption>

Most semiconductors are elements of the fourth main group, i.e. they have four electrons in the outer shell. This also applies to the element silicon. In the silicon lattice, each silicon atom is therefore connected to four neighbouring atoms via a bond. If foreign atoms are added to this semiconductor material, the electrical conductivity can be modified. This is called **doping**.  

Atoms of the fifth main group (e.g. phosphorus) have five electrons in the outer shell. If these are added to the silicon crystal lattice, one electron is surplus at these points, as it is not needed for the four bonds in the crystal lattice. This electron is much more mobile than the electrons that contribute to the bond and therefore greatly increases conductivity by electron conduction. This addition of free negative charge carriers is called **n-doping** (see <imgref pic1>).

On the other hand, by adding atoms of the third main group (e.g. aluminium), a so-called hole can be created at these points, as these atoms only have three electrons in the outer shell. This leads to an increase in conductivity by hole conduction. This addition of free positive charge carriers is called **p-doping** (see <imgref pic2>).
</WRAP>

~PAGEBREAK~ ~CLEARFIX~~
==== A Quantum Mechanical View ====

<imgcaption picz|Bohr's atomic model and ribbon model>{{electronic_circuitry:shell_bendermodel.png?420|shell_bendermodel.jpg?420|Bohr's atomic model and ribbon model}}</imgcaption>

The above model of conductivity in semiconductors will now be considered in a little more depth. In the Bohr atomic model (<imgref picz>, 1), it is assumed that the electrons in the **atom** move in certain circular orbits around the nucleus - similar to the planets in the planetary system. Here, more strongly bound electrons are in closer orbits and weaker ones are in orbits further out. This also behaves similarly to satellites in the gravitational field, which, when farther from the center, are more weakly attracted. Bohr postulated 3 axioms to make the model and measurement results fit together plausibly:
  - Circular orbits are discrete. There are only certain paths on which the electrons may move \\ (and thus discrete energies for the electrons).
  - Each jump of an electron from one orbit to another is accompanied by an energy absorption or release.
  - The exact energy amounts of the orbits result from quantum physics.
Unfortunately, this representation produces quite a few physical contradictions - but the model is sufficient for explaining conductivity in semiconductors[(Note3>The contradictions of Bohr's atomic model were only resolved by quantum physics and [[https://www.youtube.com/watch?v=A4Ip1p6o9TU|orbital theory]])]. The Bohr atomic model and the [[https://de.wikipedia.org/wiki/Oktettregel|Octet rule]] (tendency of higher orbits to be saturated with 8 electrons) are enough to gain a deeper insight into semiconductor physics. \\ \\
<imgref picz> 1a shows the electrons in the discrete circular orbits, i.e., in a $x$-$y$ coordinate system. More strongly bound electrons are shown in black on inner orbits; on the outermost noncompletely occupied green orbit, electrons are shown in blue. In addition to the occupied orbits, other, outer, nonoccupied orbits are also present (blue in <imgref picz> 1a).  \\
The same electrons can also be sorted into an $x$-$W$ coordinate system (see <imgref picz> 1b). Here $W$ is the binding energy, or work released when an unbound electron jumps into the orbit under consideration. The origin of the binding energy (i.e., the binding energy of an unbound electron: $W=0$) is above the unoccupied levels. Thus, as expected, the magnitude of the binding energy of the fully occupied level is the highest. The discrete orbits also result in discrete energy levels on the energy axis. \\
~~PAGEBREAK~~ ~~CLEARFIX~~

<WRAP right noprint>
{{youtube>_pE-ZX79wX4?size=543x392}}
</WRAP>

If we consider a **section of a solid** instead of a single atom, the electron configuration changes. In <imgref picz> 2a, the situation is again shown in the $x$-$y$ coordinate system. Here, the inner electrons and the nucleus are now reduced to a single, yellow circle with the resulting charges. The electrons from the (in the example atom) partially occupied levels now satisfy the octet rule. \\ However, depending on the element, there are different properties of the electrons here. In metals the electrons are freely movable - thus a good conductivity is measurable, but in semiconductors initially not. This statement cannot be explained by the Bohr atomic model, but by the band model and some quantum physics very well. As already for the atom, the electrons of the solid are now entered into a $x$-$W$-coordinate system.
Here are now many electrons from the same atomic levels close to each other. The laws of quantum physics forbid that electrons occupy exactly the same energy level at the same location. This results in a broadening of the discrete levels into energy bands (<imgref picz> 2b).
In the example, a semiconductor is drawn. In the semiconductor, the energetically highest-lying band is completely occupied. The energetically highest-lying and occupied band is called the **valence band**, and the next highest non-occupied (or not fully occupied) band is called the **conduction band**. The energetic gap between the conduction and valence bands is called the band gap. The conduction band of the semiconductor just corresponds to the electrons strongly bound in the $x$-$y$ coordinate system. Thus, there are initially no mobile electrons in the semiconductor (the conduction band is unoccupied, and the valence band is fully occupied). The band gap in semiconductors is approximately in the range of $0.1 ... 4eV$ [(Note4> The electron volt (eV) corresponds to the energy absorbed by an electron when it passes through into a potential difference of one volt. One electron volt is equal to $1.602\cdot 10^{-19} J$. Since energy in joules is unwieldy and not easily understood, this is converted to the energy gain of an electron in volts. For this purpose, the [[https://de.wikipedia.org/wiki/Elementarladung|elementary charge]] $e_0=1.602\cdot 10^{-19} C$ is used.)] \\
Electrons can be released from bonds with **addition of energy**. An electron can get the energy it needs in two ways: Either by an excitation of the electromagnetic field, i.e. a quantum of light, or by an excitation of the elastic field, i.e. lattice vibrations of the crystal. Light quanta are also called photons, quantized lattice vibrations are also called phonons. In <imgref picz> 2a, top left, a photon is absorbed by an electron, thus breaking the bond. The electron absorbs the energy of the photon. It is excited and raised by that amount on the $W$ axis. It also follows that only quanta of energy can be absorbed that allow it to be lifted to an existing and free level. The energy absorption results in an electron in the conduction band that is mobile in the crystal. In addition, the electron leaves a positively charged hole in the valence band. This process is called **generation of electron-hole pairs**. Both electron and hole conduction contribute to conductivity in the undoped semiconductor. The reverse process - the **recombination** of electrons with holes, occurs in silicon after a few tens of microseconds, or a few tens of micrometers. In this process, the amount of energy in the bandgap is released again. \\

<imgcaption pic0|band model and doping>{{ https://upload.wikimedia.org/wikipedia/commons/c/c0/Halbleiter1.PNG?420|band model and doping}}</imgcaption>

Since the crystal lattice already contains thermal energy at room temperature (the atomic trunks move), phonons are also present in the crystal. The phonons have a broad, energetic distribution. At room temperatures, the average energy of a phonon is $k_B\cdot T = 26 meV$ ($k_B$ is the [[https://de.wikipedia.org/wiki/Boltzmann-Konstante|Boltzmann constant]]).  
In silicon, about 0.000 000 01% (about one in $10^{13}$) of phonons have sufficient energy to lift an electron from the valence band to the conduction band. However, this is sufficient to provide about 10 billion charge carriers ($10^{10}$) to pure silicon at room temperature and a volume of $1 cm^3$ (about $5\cdot 10^{22}$ atoms). These charge carriers enable the intrinsic conduction described above.

The previous subchapter also described another way of increasing the number of charge carriers: doping with impurity atoms. This requires that the semiconductor material used is very pure and crystalline. Impurities and crystalline impurities can also produce conductive charge carriers. The semiconductor material should have less than one defect per $10^{10}$ atoms (equivalent to about one person to humanity). In this case, intrinsic conduction would predominate in it. For doping, one impurity atom is added to $10^5...10^{10}$ semiconductor atoms. In the band model, n-doping results in additional electrons in the conduction band and additional positively charged fixed recombination centers due to the fixed positive atomic hulls, so-called (electron)**donors** (<imgref pic0>: red marking for n-doping in b,c,d). A p-doping creates additional holes in the valence band and fixed negatively charged recombination centers, so-called (electron)**acceptors**.

~~PAGEBREAK~~ ~CLEARFIX~~
===== 2.2 PN junction and operating principle of a diode =====

<WRAP right noprint>
{{youtube>HdXaTn-JRCo?size=543x392&start=52}}
</WRAP>

<imgcaption pic3|Switching sign of a diode, with the designations of the doping and electrodes>{{ electronic_circuitry:switching-sign-diode.png?nolink&200|Switching sign of a diode}}</imgcaption>

<WRAP noprint>
++++ video-transcript |
A diode is a semiconductor device that allows current to pass in only one direction. So it can be thought of as a valve for the current. The circuit symbol is shown in <imgref pic3>.

The arrowhead indicates the direction in which the diode allows current to pass, here meaning the technical direction of current, i.e. the movement of positive charge carriers. This means that the diode conducts the current when the positive pole on the left and the negative pole on the right of a DC voltage source are applied ("dash" of the diode is connected to the negative pole). If you connect the diode with the opposite polarity, it will not conduct the current. If the diode conducts the current, it is connected in forward direction, if it does not conduct the current, it is connected in reverse direction.  
++++
</WRAP>

<WRAP onlyprint>
A diode is a semiconductor device that allows current to pass in only one direction. So it can be considered as a valve for the current. The circuit symbol is shown in <imgref pic3>.

The arrowhead indicates the direction in which the diode allows current to pass, here meaning the technical direction of current, i.e. the movement of positive charge carriers. This means that the diode conducts the current when the positive pole on the left and the negative pole on the right of a DC voltage source are applied ("dash" of the diode is connected to the negative pole). If you connect the diode with the opposite polarity, it will not conduct the current. If the diode conducts the current, it is connected in forward direction, if it does not conduct the current, it is connected in reverse direction.  
</WRAP>

For the circuit symbol there are the following mnemonics: Viewed from the cathode side, the circuit symbol resembles a "K". From the anode side, the circuit symbol resembles a horizontal "A". Mnemonic for sorting: **K**athode - **N**negative - **A**node - **P**positive.

In the simulation shown below, three examples of diodes in circuits are considered. \\ In the **first example on the left**, the voltage source is polarized so that the diode is forward biased. The light bulb is on. \\
In the first example on the right, the diode is reverse biased. The light bulb remains dark. \\
In the second example (middle), an **ideal diode** - i.e. a directional current valve - can be seen. Next to it is the transfer characteristic or current-voltage characteristic (in this case also called diode characteristic). The voltage at the diode is plotted on the x-axis, the current through the diode on the y-axis. The diode is non-conducting at all voltages below 0V, and conducts current at all voltages above 0V.  \\
In the last example (right) a **real diode** is connected. The real diode differs from the ideal diode in the following ways:
  - The real diode does not have such a steep slope.
  - The real diode has a non-linear resistance; it is not an ohmic resistor.
  - The real diode seems to require a minimum voltage to allow a current to flow.

The details of the real diode are described below.

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<imgcaption pic5|Evolution of the p-n junction>{{ https://upload.wikimedia.org/wikipedia/commons/d/d5/Halbleiter2.PNG?400|Evolution of the p-n junction}}</imgcaption>

In a diode, two differently doped layers of silicon collide: p-doped silicon ("p-crystal") on one side and n-doped silicon ("n-crystal") on the other.

The situation __without external voltage__ will be considered first (compare <imgref pic5>). On the n-doped side, many free-moving electrons will dissolve at room temperature, leaving acceptors stationary. The same can be seen on the p-doped side: the free-moving holes leave behind donors. in the middle, at the **pn-junction**, both moving charge carriers, electrons and holes, meet. When they meet directly, the two charge carriers will cancel each other out, they **recombine**. This creates a photon (electromagnetic vibration) and/or a phonon (lattice vibration). The recombination forms a layer, the **barrier layer**, which is largely free of free moving charge carriers. The barrier layer initially acts as an insulator.


__With external voltage $U_D$__ on the diode, two cases are to be distinguished (<imgref pic6>):
  - Applying a positive voltage from p-doped side to n-doped side \\ (diode voltage = forward voltage $U_D = U_F$, $U_F>0$).
  - Applying a negative voltage from p-doped side to n-doped side \\ (diode voltage = reverse voltage $U_D = -U_R$, $U_R>0$).

~~PAGEBREAK~~~CLEARFIX~~
<imgcaption pic6|Functionality of a semiconductor diode>{{ https://upload.wikimedia.org/wikipedia/commons/7/7b/Halbleiter3.PNG?400|Functionality of a semiconductor diode}}</imgcaption>

A triangular or sawtooth signal can be applied to create the diode characteristic (see Falstad simulations).

=== Forward voltage $U_F>0$ ===
If a positive potential__ is applied to the __p-doped side, the freely moving holes there are driven towards the pn-junction. Negative potential is then applied to the n-doped side, which also drives the freely moving electrons towards the pn-junction. At the pn-junction, holes and electrons can neutralize each other. Thus, holes from the positive terminal and electrons from the negative terminal can continue to move in, and an electric current flows through the diode. The diode is connected in the **conducting direction**. In common diodes, about $0.7 V$ is dropped in the forward direction. This means, of course, that the current does not pass the diode completely without resistance, but that the forward voltage $U_S$ of about $0.7 V$ must be applied from the outside.[(Note2>In the literature, the forward voltage can be found under other names: Forward voltage, Threshold voltage, Forward voltage, Buckling voltage, Forward voltage.)]. This voltage results from the energy difference of the band gap related to one electron, which is about $1.1eV$ for silicon, but is reduced by thermal energy (phonons).
On closer inspection, the curve resembles an exponential function. This can be described by the [[https://de.wikipedia.org/wiki/Shockley-Gleichung|Shockley equation]]:

^ $\boxed{\large{I_F = I_S(T)\cdot (e^{\frac{U_F}{m\cdot U_T}}-1)}$ ||
| $\small{I_F}$ | <fs small> forward current at the diode (Forward Current)</fs> | "positive current at the diode" |
| $\small{U_F}$ | <fs small> forward voltage (Forward Voltage)</fs> | "positive voltage at the diode" |
| $\small{I_S(T)}$ | <fs small> Reverse Current (Saturation or Leakage Current)</fs> | "current present when connected in reverse direction" |
| $\small{m}$ | <fs small> Emission Coefficient (1...2) </fs> | "Trickle factor, only part of the energy of $U_F$ acts on the charge carriers" |
| $\small{U_T}$ | <fs small> temperature voltage ($26mV$ at room temp.)</fs> | "energy due to temperature related to charge" |

Several consequences can be derived from the exponential function:
  - The forward voltage $U_S$ of about $0.7 V$ depends on which current (/voltage) range is considered. $0.6...0.7V$ is a suitable value for currents in the range of $5...100mA$. This range is used in most circuits. For smaller currents, the forward voltage $U_S$ also decreases (e.g., for $5...100mA$ \rightarrow about 0.4V$,(nbsp)(nbsp) $0.1...1mA$ \rightarrow about 0.2V$, see the following Falstad simulation).
  - The forward voltage and the voltage response are temperature dependent. The higher the temperature, the more current flows for the same voltage. So if a diode is connected directly to a voltage source, at currents higher than about $50mA$ the current would increase directly via self-heating [(Note1>The self-heating $Q$, or temperature increase $\Delta \vartheta$ results directly via the power dissipation $P_{loss}=U_D \cdot I_D = \dot{Q} = C\cdot \Delta \vartheta$.)] up to / above the maximum current.

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A diode behaves like an NTC resistor, that is, the warmer it gets, the lower the resistance, the more current flows ($I\sim \frac{1}{R}$), the more power dissipation there is ($P_{loss}\sim I$), the warmer it gets ($\vartheta\sim P_{loss}$). This relationship can lead to the disturbance of the diode.

If a diode is used, it should therefore be noted that it must be thermally stabilized. A frequently used method is the use of a resistor, e.g. load resistor or series resistor for an LED.

Correspondingly, when diodes are connected in parallel, they must either be measured beforehand and compared for similar characteristics or a series resistor must also be provided.

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=== Blocking voltage $U_R>0$ ===

If the __diode is contacted in the opposite direction__, i.e. positive pole on the n-doped side and negative pole on the p-doped side, no current can flow. This is because the free electrons from the positive pole are sucked out of the n-doped side, and the free holes are removed from the p-doped side. In between, a so-called depletion zone without free charge carriers is created. No electric charge can be transported through this depletion zone and the diode cannot conduct "any current". The diode is connected in the **blocking direction**. In fact, the diode can still conduct a very small **blocking current** $\boldsymbol{I_S}$: Thermal energy keeps creating free electrons and free holes, which allow it to conduct. Thus it is also clear that the reverse current is temperature dependent. $I_S(T)$ lies in the range $100nA...10\mu A$.

<imgcaption pic8|Switching sign of a Z-diode>{{ electronic_circuitry:switching-sign-z-diode.png?200|Switching sign of a Z-diode}}</imgcaption>

If the reverse voltage is increased further, the free charge carriers are sucked out more and more. Above a certain negative voltage, the energy of the free charge carriers becomes so great that they knock out more charge carriers, which in turn knock out more charge carriers. This results in an avalanche of free moving charge carriers and the diode becomes abruptly conductive. This situation is called **breakthrough**. The voltage is denoted $U_Z$, after the discoverer [[https://de.wikipedia.org/wiki/Zener-Effekt|Clarence Zener]]. For an ordinary diode, breakdown is problematic because it does not occur at a defined voltage. Without precise knowledge of this voltage, the rapid current rise will quickly destroy the diode.
In addition to avalanche breakdown at high negative voltages, there is Zener breakdown at low voltages in highly doped materials due to quantum mechanical processes. By suitable structuring it is possible to combine both effects in the so-called **Z-diode** (formerly called Zener diode). This has two major advantages: on the one hand, it makes it possible to create diodes that let through at arbitrary (negative) voltages. On the other hand, the temperature dependence of the characteristic can be compensated.

In the older circuit symbols - which should no longer be used - the Z of the Z-diode can still be seen (<imgref pic8>). In the current symbol for the Z diode, only a dash is drawn on the top or bottom. The is to avoid accidental confusion with other diode circuit symbols.

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Z-diodes are wired for operation in the opposite direction. Z-diodes are available prefabricated for various breakdown voltages $U_Z$.
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In the diode characteristic $I_D(U_D)$, the individual voltage ranges are designated according to their respective effects: Breakdown range ($U_D = -U_R < U_Z$), Rejection range ($U_Z < U_D < U_S$), Forward range ($U_D = U_F > U_S$).

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~~PAGEBREAK~~ ~CLEARFIX~~

===== 2.3 Special diodes =====

<imgcaption pic12|Absorption of photons in the intrinsic layer of a photodiode>{{ https://upload.wikimedia.org/wikipedia/commons/f/f0/Pin-Photodiode.png?nolink&300|Absorption of photons in the intrinsic layer of a photodiode}}</imgcaption>

In addition to the silicon PN diode and the Z diode discussed so far, other diodes are available for various applications. In the following, the most important ones will be briefly described.

=== PIN diode ===

In the PIN diode, there is an undoped region (**i**ntrically non-conducting) between the **p**-doped and **n**-doped regions. The name is therefore derived from the existing layers of the diode. In all diodes, the carrier-free junction results in a capacitor. The capacitance of this capacitor is reciprocally proportional to the distance $d$ between the conducting regions: $C\sim \frac{1}{d}$. With the additional undoped region inserted, $d$ becomes larger and thus the capacitance becomes smaller. This capacitance is alternately charged and discharged in AC applications. A smaller capacitance improves the blocking performance at high frequencies. The broadened junction also increases the dielectric strength of the diode. The same circuit symbol is used for the PIN diode as for the classic PN diode.

==PAGEBREAK~ ==CLEARFIX~~
=== Photodiode (solar cell) ===

<imgcaption pic11|Switching sign of a photodiode>{{ electronic_circuitry:switching-sign-photodiode.png?nolink&150|Switching sign of a photodiode}}</imgcaption>

A photodiode is a PIN diode which is constructed in such a way that the cross-section of the junction occupies a very large area. The structure of a photodiode is: n-doped layer, intrinsically conductive layer, p-doped layer. When a photon hits the diode, an electron-hole pair is generated, which is separated by the electric field in the PN junction: the electrons accumulate in the n-doped layer, the holes in the p-doped layer (see <imgref pic12>). In a photodiode, the charge carriers are dissipated in a voltage-free manner. The number of charge carriers is proportional to the absorbed photons. The circuit symbol (<imgref pic11>) shows the incoming photons with arrows.

If the voltage generated by a diode is used as the source voltage, it is referred to as a solar cell. Silicon is often used as the material for solar cells for cost and processing reasons. Solar cells can never convert the complete irradiated energy: Photons with energies below the band gap $W_g$ cannot generate an electron-hole pair. For photon energies $W_{ph}$ above the band gap $W_g$, part of the energy ($W_{ph}-E_g$) is emitted as heat or phonon. In addition, there are other technical reasons for losses. With silicon, up to 26% of the irradiated energy can be converted into electrical energy. The efficiency can be increased with several layers of different materials.

==PAGEBREAK~ ==CLEARFIX~~
=== germanium diode ===

In the germanium diode, germanium is used as the semiconductor instead of silicon. Germanium has a smaller band gap and thus also a lower forward voltage of about $U_D=0.3V$. Thus, the germanium diode is already closer to an ideal diode at low currents and voltages. A disadvantage compared to silicon PN diodes is that the reverse current is higher and the forward current is lower. In other words, the diode does not "block" quite as well and the "kink" in the characteristic curve is less pronounced. The same circuit symbol is used for the germanium diode as for the PN diode. In contrast to silicon, germanium has an additional electron shell, so the core is more strongly shielded. The easier detachment of the outer electrons in the crystal leads on the one hand to a smaller band gap, but also to a higher intrinsic conductivity (density). A consequence of this is that germanium diodes have lower operating temperatures (approx. 70..90°C) than silicon.

~~PAGEBREAK~~ ~CLEARFIX~~
<imgcaption pic12|Switching sign of a Schottky diode>{{electronic_circuitry:switching-sign-schottky-diode.png?nolink&150|Switching sign of a Schottky diode}}</imgcaption>

=== Schottky diode ===

The Schottky diode also uses a different material. In the silicon Schottky diode, however, a metal is used instead of silicon only on the previously p-doped side. This eliminates the holes as charge carriers, which allows for "faster switching": available Schottky diodes have a forward voltage in the range of $U_S=0.15...0.45V$. Although silicon Schottky diodes have similar advantages and disadvantages to the germanium diode, the disadvantages such as higher reverse current (compared to silicon PN diode) and lower slope are less pronounced.
In most applications, the silicon Schottky diode is superior to the germanium diode. The circuit symbol (<imgref pic12>) stylizes an S for Schottky, which is especially visible with the non-normalized symbol.

=== Power diodes ===

For power applications, silicon__carbide__ Schottky diodes (SiC Schottky) with a forward voltage of $U_S=0.8V$ or gallium nitride diodes (GaN) are used. The large band gap of these two materials results in lower intrinsic conductivity. This leads to the fact that these diodes can be operated at high temperatures (and thus high losses and voltages). The reverse voltage is noticeably higher than that of silicon diodes (SiC: up to approx. 1,600V, GaN approx. 200V).

Both materials are also used in power transistors for higher power (kW to MW range).

== SYNC, CORRECTED BY ELDERMAN ==
=== (O)LED ===

<imgcaption pic10|Switching sign of a LED>{{ electronic_circuitry:switching-sign-led.png?nolink&150|Switching sign of a LED}}</imgcaption>

In the (organic) light-emitting diode, other (also organic) semiconductor materials are used instead of silicon. These are optimized in such a way that photons of a certain wavelength are essentially formed by recombination in the forward direction. This requires a large band gap, which also produces a high forward voltage $U_S>2V$. As with all diodes, care must be taken with LEDs to ensure that the negative temperature coefficient does not cause the diode to fail at high currents: Each LED requires a series resistor, which should be designed to limit the current. Usually LEDs are rated at $I_D=20mA$. A nice overview of the various voltages and maximum currents can be found at [[https://www.reichelt.de/reicheltpedia/index.php/LED#Genutzte_Halbleiter|Reicheltpedia]]; for specific use of an LED, the datasheet should be inspected. The circuit symbol (<imgref pic10>) shows with arrows the outgoing photons.

The following simulation allows a comparison of different diodes in the voltage range $U=\pm0.05V$. This reveals the different reverse currents $I_S$ in the negative voltage range and the early rise of germanium and Schottky diodes.

<WRAP>{{url>https://www.falstad.com/circuit/circuitjs.html?running=false&cct=$+17+0.0005+0.2031+87+2+50%0Ag+192+336+192+352+0%0AR+48+192+48+160+0+3+50+0.1+0+0+0.5%0A403+288+160+576+384+0+1_64_0_4163_1.25e-72_8e-73_1_2_1_3%0Ax+438+167+443+170+4+18+I%0Ax+557+288+569+291+4+18+U%0Ax+483+300+530+303+4+18+0,05V%0Ax+501+279+505+282+4+18+%7C%0Ad+64+288+64+336+2+1N5711%0Ad+112+288+112+336+2+1N4148%0Ar+112+336+192+336+0+1000%0Ax+38+359+111+404+4+18+Schottky%5C%5CnDiode%5C%5Cn(1N5711)%0Ax+131+274+206+319+4+18+Silizium%5C%5CnDiode%5C%5Cn(1N4148)%0A34+1N34_+0+0.0000026+6.5+1.6+75%0Ad+16+288+16+336+2+1N34_%0Aw+64+224+96+224+0%0Aw+112+336+64+336+0%0Aw+64+336+16+336+0%0Aw+112+256+112+288+0%0Aw+80+256+64+256+0%0Aw+64+256+64+288+0%0Aw+16+256+16+288+0%0Aw+32+224+16+224+0%0Aw+16+224+16+256+0%0Ax+-109+278+-2+302+4+18+Germanium%5C%5CnDiode%5Cs(1N34)%0AS+48+192+48+224+0+0+false+0+2%0AS+96+224+96+256+0+0+false+0+2%0A 1000,400 noborder}}
</WRAP>
~~PAGEBREAK~~ ~CLEARFIX~~

===== 2.4 Calculations with diodes =====

<WRAP right>{{url>https://www.falstad.com/circuit/circuitjs.html?running=false&cct=$+17+0.00005+1.4391916095149893+50+1+50%0Ab+656+128+956+360+0%0Ax+688+110+810+113+4+24+reale%5CsDiode%0Ag+688+320+688+336+0%0Ad+688+192+688+320+2+1N4004%0AR+688+192+688+160+0+3+50+0.8+0+0+0.5%0A403+704+160+944+336+0+4_64_0_4163_1.25e-32_8e-33_1_2_4_3%0Ax+928+272+940+275+4+18+U%0Ax+830+168+835+171+4+18+I%0Ax+877+272+914+275+4+18+0,7V%0Av+1008+272+1008+224+0+0+40+0.7+0+0+0.5%0Ar+1008+192+1008+224+0+1%0Ax+1008+110+1192+113+4+24+idealisierte%5CsDiode%0Ag+1008+320+1008+336+0%0A34+fwdrop%5Cq0.01+0+1e-7+0+0.024+0%0Ad+1008+272+1008+320+2+fwdrop%5Cq0.01%0AR+1008+192+1008+160+0+3+50+0.8+0+0+0.5%0A403+1040+160+1280+336+0+14_1024_0_4162_5e-88_1e-89_0_2_14_3%0Ab+982+128+1298+360+0%0Ax+1266+268+1278+271+4+18+U%0Ax+1166+166+1171+169+4+18+I%0Ax+1214+270+1251+273+4+18+0,7V%0A 600,400 noborder}}
</WRAP>

The idealized diode consists of:
  * an ideal diode, which is maximally conductive when $U>0$,
  * a voltage source providing the forward voltage: $U_Q = U_S$
  * an ohmic resistor with a resistance value such that for the desired voltage range (=working point) the real diode characteristic is approximated. This resistance is called differential resistance $r_D$.

If one is only interested in the behavior of the diode in at voltages above the forward voltage ($U>U_S$), then the ideal diode can also be removed.
In the videos below two of the essential calculations in dealing with diodes are shown.

~~PAGEBREAK~~~CLEARFIX~~~

<WRAP right noprint>
{{youtube>iNZj91TSRUg?size=500x360}}
</WRAP>

<WRAP noprint>
{{youtube>ckDSBLjS1SM?size=500x360}}
</WRAP>



~~PAGEBREAK~~~CLEARFIX~~~



===== 2.5 Applications =====

In the following, the most important applications for diodes will be presented. The respective simulations can be found under the links.

A frequently used application are rectifiers, which convert an AC voltage into a DC voltage. A distinction is made between the [[http://www.falstad.com/circuit/e-rectify.htmll|half-wave rectifier]] and the [[http://www.falstad.com/circuit/e-fullrect.html|bridge rectifier or full-wave rectifier]]. Rectifiers are used on the one hand in the power supply. On the other hand, a rectifier (using amplifiers) can be used to easily determine the peak value of a voltage waveform.

A similar concept also makes it possible to generate [[http://www.falstad.com/circuit/e-voltdouble.html|double]], [[http://www.falstad.com/circuit/e-volttriple.html|triple]] or [[http://www.falstad.com/circuit/e-voltquad.html|multiple voltage]] from an AC voltage.

As overvoltage protection, for example, the sparks when switching inductances can be avoided by [[http://www.falstad.com/circuit/e-inductkick-block.html|"free running" of the current]]. Also too large excursions of alternating quantities can be limited by an [[http://www.falstad.com/circuit/e-diodelimit.html|antiparallel interconnection]] in an analogous way (e.g. for actuators like loudspeakers). This also includes the suppressor diodes shown in the initial example, which protect electronic circuits from short-term overvoltages.

The relatively stable voltage drop across diodes makes it possible to use them as a voltage reference. For this purpose, a voltage divider consisting of a resistor and a diode is used, at which a constant voltage is dropped over a wide input voltage range. Especially [[http://www.falstad.com/circuit/e-zenerref.html|Z diodes]] are predestined for this application due to the wide range of different breakdown voltages. In a further step, the voltage reference can be further developed in an amplifier circuit to a [[http://www.falstad.com/circuit/e-opamp-regulator.html|precision voltage source]].

====== Tasks ======

{{page>uebung_2.1.1&nofooter}}
{{page>uebung_2.1.2&nofooter}}
{{page>uebung_2.1.3&nofooter}}
{{page>uebung_2.1.4&nofooter}}
{{page>uebung_2.1.5&nofooter}}

====== Study Questions ======
=== For self-study ===
  * On a U-I diagram, draw the characteristic of an ideal diode and a real silicon diode and explain the differences.
  * What is meant by n-doped and p-doped?
  * How does a junction form inside the diode?
  * What is meant by a threshold voltage?
  * Why does voltage drop across a diode?
    * Sketch the layered structure of a diode and label the three layers formed.
    * What kind of free charge carriers are available in each of these layers?
    * Draw the electric fields formed in the diode when no external field is applied.
    * Explain how an external voltage can bring the diode into a conducting state.
  * Explain the working of a diode with the help of a sketch. Draw the following areas: p-doped part, n-doped part, junction.
  * Given is a layered structure of a diode (n-doping and p-doping can be seen). How would the diode have to be connected to pass current?
  * Typical diode characteristic for silicon diodes.
    * Draw a characteristic curve for silicon diodes.
    * What is the characteristic voltage? (Technical term of voltage, magnitude of voltage, relevant current range).
    * Insert the technical terms for the three relevant voltage ranges.
  * Z-diode
    * Explain the operating principle of a Z-diode using its characteristic curve.
    * Draw the current waveform of a 6V8 diode on a voltage-current diagram.
  * What needs to be considered while choosing a type of diode?
  * State three applications of diodes and sketch the construction.
  * Rectifier circuits
    * Draw a half-wave rectifier. Draw a bridge rectifier.
    * Given a sinusoidal input voltage of 3V. Draw the waveform of the input voltage and the output voltage of the two rectifiers over 2 periods for 50 Hz in a graph.
    * How can the output voltage be smoothed? How can the output current be smoothed?
    * Given a sinusoidal input voltage of 3V. What should be considered if very high frequencies are to be rectified? Draw a possible signal waveform of the input voltage and the output voltage of the two rectifiers over 2 periods for 50 GHz in a diagram.
  * Draw a circuit with which the __positive__ half-wave of a sinusoidal voltage can be cut off.

=== with answers ===
<quizlib id="quiz" rightanswers="[['a1'],['a0', 'a1', 'a2', 'a3'], ['a1', 'a3', 'a5'], ['a1', 'a2', 'a3'], ['a0', 'a1', 'a2', 'a3'], ['a0', 'a1']" submit="Check answers">

<question title="Which of the following statement(s) is/are correct?" type="checkbox">
p-doping produces quasi-free electrons|
Conductivity in semiconductor happens via conduction and valence band|
The diode blocks at any negative voltage (reverse voltage|).
The diode can be modeled as a voltage source and capacitor
</question>

<question title="On which physical properties does the forward voltage $U_S$ depend?" type="checkbox">.
temperature|
current range considered|
(semiconductor) material|
LED color|
breakdown voltage of the Z-diode
</question>

<question title="Which statement(s) about the junction is/are correct?" type="checkbox">
There is no electric field in the junction|
The junction does not contain free charge carriers|
The junction becomes larger when current is passed through it|
Electron-hole pairs are created in the junction by photons|
The junction is enlarged in the Schottky diode compared to the PN diode|
The junction forms a capacitor
</question>

<question title="Statements about the conduction/valence band" type="checkbox">.
Photon capture can move electrons from the conduction band to the valence band|
"Recombination" removes an electron from the valence band and a hole from the conduction band|
A donor creates one or more quasi-free electrons|
The band gap indicates the maximum energetic distance between the conduction and valence bands
</question>

<question title="The forward voltage ..." type="checkbox">
... for silicon is about 0.6 ... 0,7V|
... serves to allow electrons to cross the bandgap|
... ...depends on the current range under consideration|
... is smaller for germanium diodes than for silicon diodes.
</question>

<question title="The pass current ..." type="checkbox">
... Is dependent on the temperature|
... depends on the forward voltage|
... is logarithmic with respect to the forward voltage|
... depends on the reverse voltage
</question>
</quizlib>

--> References to the media used #

^ Element ^ License ^ Link ^
| Video: Circuit Elements - Diodes and Transistors - Part 1 | [[https://creativecommons.org/licenses/by/3.0/legalcode|CC-BY (Youtube)]]  | https://www.youtube.com/watch?v=YB3pQ7P8SNg |
| Video: Circuit Elements - Diodes and Transistors - Part 2 | [[https://creativecommons.org/licenses/by/3.0/legalcode|CC-BY (Youtube)]]  | https://www.youtube.com/watch?v=HdXaTn-JRCo |
| <imgref pic1> | [[https://creativecommons.org/licenses/by-sa/3.0/de/|CC-BY-SA 3.0]]        | https://commons.wikimedia.org/wiki/File:Schema_-_n-dotiertes_Silicium.svg |
| <imgref pic3> | [[https://creativecommons.org/licenses/by-sa/3.0/de/|CC-BY-SA 3.0]]        | https://commons.wikimedia.org/wiki/File:Schema_-_n-dotiertes_Silicium.svg |
| <imgref pic2> | [[https://creativecommons.org/licenses/by-sa/3.0/de/|CC-BY-SA 3.0]]        | https://de.wikipedia.org/wiki/Datei:Schema_-_p-dotiertes_Silicium.svg |

<--====== 2. Diodes and Transistors ======

<callout>
A nice introduction can be found in [[https://lx3.mint-kolleg.kit.edu/onlinekursphysik/html/1.4.3/xcontent5.html|KIT Bridge Course - 4.3.6 Diodes and Transistors (*)]]. \\ Some of the following passages, videos and pictures are taken from this introduction.
\\ \\
For another introduction, see [[https://www.leifiphysik.de/elektronik|LEIFIphysics]].
</callout>

<WRAP>
<callout type="info" icon="true">
=== Introductory example===
The electronics in personal computers, mobile phones, electric toothbrushes, and like all other digital companions, are based on transistor circuits (see [[fundamentals_of_digital_engineering:at_the_heart_of_a_computer]]). In [[:fundamentals_of_digital_technology:start]] it has already been explained that all logic circuits can be traced back to NAND and NOR gates, respectively, via conjunctive and disjunctive normal forms. These in turn consist of transistors. In the simulation below, the structure of a NAND gate is shown in the current CMOS structure. CMOS here indicates the structure of the circuit and semiconductor structure: **C**omplementary **m**etal-**o**xide-**s**emiconductor - an oppositely complementary circuit of semiconductors of the metal-oxide-semiconductor structure.
The complementary structure is shown by the fact that.
  * from the digital output ($OUT2$) to ground two transistors of one kind are connected in series and
  * from the digital output ($OUT2$) to the 5V supply, two transistors of a different type are connected in parallel.
These two different kinds of MOS-transistors and further used kinds shall be explained in this chapter.

<WRAP>{{url>https://www.falstad.com/circuit/circuitjs.html?running=false&cct=$+0+0.0002+0.134+1+5+50%0Af+368+128+416+128+5+1.5+0.02%0Af+368+224+416+224+4+1.5+0.02%0Aw+416+144+416+176+0%0Aw+416+176+416+208+0%0AM+448+176+448+192+0+2.5%0Af+352+288+416+288+4+1.5+0.02%0Aw+416+240+416+272+0%0Ag+416+304+416+320+0%0Aw+368+128+368+224+0%0Af+288+128+336+128+5+1.5+0.02%0Aw+336+80+336+112+0%0Aw+336+80+416+80+0%0Aw+416+80+416+112+0%0AR+416+80+416+48+0+0+40+5+0+0+0.5%0Aw+288+128+288+288+0%0Aw+288+288+352+288+0%0Aw+336+144+336+176+0%0Aw+336+176+416+176+0%0AR+-112+272+-128+272+1+2+100+2.5+2.5+0+0.5%0AR+-112+288+-128+288+1+2+50+2.5+2.5+0+0.5%0A207+-112+272+-96+272+4+IN1%0A207+-112+288+-96+288+4+IN2%0A207+240+224+224+224+4+IN1%0A207+240+288+224+288+4+IN2%0AM+240+224+240+240+0+2.5%0Aw+240+224+368+224+0%0Aw+240+288+288+288+0%0AM+240+288+240+304+0+2.5%0Aw+416+176+448+176+0%0A207+448+176+464+176+4+OUT2%0A151+-144+128+-80+128+0+2+5+5%0A207+-160+112+-176+112+4+IN1%0A207+-160+144+-176+144+4+IN2%0A207+-64+128+-48+128+4+OUT1%0AM+-64+128+-64+144+0+2.5%0Aw+-80+128+-64+128+0%0Ab+192+32+507+341+0%0Ab+-209+32+-6+341+0%0Ax+-366+49+-224+79+4+24+NAND-Gatter%5C%5Cnals%5CsBlackbox%0Ax+36+50+178+80+4+24+NAND-Gatter%5C%5Cn%C3%BCber%5CsCMOS%0AM+-160+144+-160+160+0+2.5%0Aw+-160+144+-144+144+0%0Aw+-144+112+-160+112+0%0AM+-160+112+-160+128+0+2.5%0A403+32+224+160+256+0+33_1_0_4102_5_0.1_0_2_33_3%0A403+32+112+160+144+0+31_1_0_4102_5_0.1_0_2_31_3%0A403+32+144+160+176+0+32_1_0_4102_5_0.1_0_2_32_3%0A403+32+256+160+288+0+29_1_0_4102_5_0.1_0_2_29_3%0A 900,400 noborder}}
</WRAP>

</callout></WRAP>


<WRAP><callout>
=== Targets for the bipolar transistor ===
After this lesson, you should:
  - Know what types of bipolar transistors there are, what their layer structure and circuit symbol looks like.
  - Know how the two types of bipolar transistors are controlled.
  - Know what are the main characteristics of the bipolar transistor and what they look like.
</callout></WRAP>

<WRAP><callout>
=== Targets for the field-effect transistor ===
After this lesson, you should:
  - Know what types of MOSFETs there are, what their layer structure and circuit symbol looks like.
  - Know what the output characteristic field of the MOSFET looks like.
  - know what the body diode is and where it comes from.
  - know what to look for in the output characteristic field when designing a semiconductor element.
</callout></WRAP>

===== 2.6 Functional principle of a (bipolar) transistor =====

<WRAP right>
{{youtube>KjyHta5p9WE?size=543x392}}

</WRAP>

A variable resistor can be developed from the diode or PN junction. With this controlled transition resistor ("__tran__sfer re__sistor__" or better transistor) the resistance can be changed by a current and thus the current let through can be adjusted.

++++ Video-Transcript (Alternative to the explanation in the video) |
A transistor consists of two diodes connected against each other, which have a common n- or p-layer, e.g. a thin p-doped layer is placed between two n-doped layers. This is an npn transistor, the more common design. However, pnp transistors are also used for special applications. All three layers are electrically contacted, so the transistor has three terminals. The contact to the middle layer is called base (B), the contacts to the two outer layers collector (C) and emitter (E). The circuit symbol of a transistor is shown in <imgref pic5>.

A transistor is usually operated as a switch or as a current amplifier. To explain how it works, a typical transistor circuit is shown in the figure below. The circuit containing the consumer, here the incandescent lamp, is called the working circuit. Here, the voltage source must be connected in such a way that the technical direction of current through the transistor runs from the collector to the emitter, i.e. in the direction of the arrow indicated on the emitter.

The second circuit, in which a positive control voltage is applied to the base, is the control circuit. Holes are pumped from the p-doped base into the n-type emitter by the positive control voltage, since a negative voltage is applied to it. Thus, the base-emitter diode is forward biased. On the other hand, a positive voltage is applied to the collector, so that this diode actually blocks.

If the voltage $U_{BE}$ in the control circuit exceeds a certain threshold, a current $I_C$ can now flow in the operating circuit. In this respect, the transistor acts as a switch. The small current $I_B$ in the control circuit can therefore be used to control the large current $I_C$ in the operating circuit.  

In a certain range, the current $I_B$ in the working circuit is proportional to the current $I_C$ in the control circuit. This ratio is called the current gain $\beta={I_C}/{I_B}$ of the transistor. This behavior can be understood by considering that the p-type base layer is very thin compared to the n-type layers. Electrons supplied through the control circuit diffuse through it rapidly, reaching $99\% $ in the collector connected to the positive terminal, and are pumped back through the working circuit into the emitter. Only a few pass through the emitter directly back into the control circuit. Therefore, the current in the control circuit is much less than the current in the working circuit.  
++++

~~PAGEBREAK~~~CLEARFIX~~~

=== Switching characters ===
<WRAP right><panel type="default">
<imgcaption pic1|Switching sign and simplified structure of npn and pnp bipolar transistors>
</imgcaption>
{{drawio>bipolar_transistor_npn_or_pnp_structure}}
</panel></WRAP>

As just described, the bipolar transistor is built by a three-layer alternately doped layer structure, which corresponds to two diodes opposite and connected in series. Depending on the layer sequence (or "direction of the diodes"), pnp or npn transistors result, represented by different circuit symbols with three terminals (see <imgref pic1>).
In both transistor variants, charge carriers are emitted from the emitter terminal (E) toward the collector terminal (C) if a suitable current flows through the base terminal (B). In simplified terms, the negative charge carriers of the n-doped sides could represent a current through an NPN structure if negative charge carriers were also present in the p-doped layer. The current $I_C$ flowing with it in the technical current direction is illustrated in the circuit symbol by the arrow direction at the emitter. In the NPN transistor, the current $I_C$ flows from the collector to the emitter. Since positive charge carriers enable conductivity in the PNP transistor, the technical current direction here points from the emitter to the collector and the arrow on the emitter points towards the collector. The direction of the arrow is similar to the direction of the diode or the PN junction.
Other mnemonic devices for the direction of the arrow are:
  * "If the arrow hurts the base, it′s **pnp**"
  * "If the arrow wants to separate from the base, it is **npn**".

~~PAGEBREAK~~~CLEARFIX~~~

=== Correct wiring of the transistors ===

<WRAP right>{{url>https://www.falstad.com/circuit/circuitjs.html?running=false&cct=$+17+0.00005+1.56+61+1+50%0Ag+320+288+320+304+0%0A181+320+144+320+192+0+312286+0.0001+0.1+0.0001+0.0001%0AR+320+144+320+112+0+0+50+5+0+0+0.5%0Ax+329+228+341+231+4+18+C%0Ax+289+230+301+233+4+18+B%0At+288+240+320+240+0+1+-2.62+0.66+100%0Ar+160+240+240+240+0+1000%0Aw+320+288+320+256+0%0Aw+320+224+320+192+0%0Av+144+288+144+240+0+0+40+0.8+0+0+0.5%0Aw+144+288+320+288+0%0Aw+144+240+160+240+0%0Ax+328+267+340+270+4+18+E%0Ax+180+220+215+223+4+18+R_B%0Ax+260+248+274+251+4+24+%3E%0Aw+240+240+288+240+0%0Ax+260+231+272+234+4+18+B%0Ax+255+223+260+226+4+18+I%0Ax+314+209+326+212+4+24+v%0Ax+298+199+310+202+4+18+C%0Ax+294+191+299+194+4+18+I%0Ax+279+273+299+276+4+40+%E2%86%93%0Ax+243+270+255+273+4+18+U%0Ax+255+276+279+279+4+18+BE%0At+560+192+592+192+0+-1+2.6167525896333887+-0.6632208486391651+100%0Ax+516+175+540+178+4+18+BE%0Ax+504+169+516+172+4+18+U%0Ax+538+176+558+179+4+40+%E2%86%91%0Ax+564+247+569+250+4+18+I%0Ax+568+255+580+258+4+18+C%0Ax+586+240+598+243+4+24+v%0Ax+529+213+534+216+4+18+I%0Ax+534+221+546+224+4+18+B%0Aw+512+192+560+192+0%0Ax+532+200+546+203+4+24+%3E%0Ax+455+219+490+222+4+18+R_B%0Ax+602+175+614+178+4+18+E%0Aw+416+192+432+192+0%0Aw+416+144+592+144+0%0Av+416+192+416+144+0+0+40+0.8+0+0+0.5%0Aw+592+176+592+144+0%0Aw+592+240+592+208+0%0Ar+432+192+512+192+0+1000%0Ax+561+182+573+185+4+18+B%0Ax+604+211+616+214+4+18+C%0AR+592+144+592+112+0+0+50+5+0+0+0.5%0A181+592+240+592+288+0+312286.73671908857+0.0001+0.1+0.0001+0.0001%0Ag+592+288+592+304+0%0Ab+112+96+370+320+0%0Ab+387+96+645+320+0%0Ax+124+86+356+89+4+24+NPN-Bipolartransistor%0Ax+397+86+627+89+4+24+PNP-Bipolartransistor%0Ao+5+512+6+4097+5+0.8+0+1%0A 600,400 noborder}}
</WRAP>

The simulation on the right shows the correct connection of the transistors. In general, the arrow of the symbol of the technical current direction must point at the correct interconnection. The base current $I_C$ is almost always generated in the circuits by a voltage source between base and emitter with a voltage $U_{BE}$. In this case, a positive voltage with respect to the emitter is required for the NPN transistor and a negative voltage with respect to the emitter is required for the PNP transistor. In practical applications the NPN transistors predominate, among other things because the negative charge carriers used there produce a higher conductivity. For the following explanations only NPN transistors are considered.

A central question that arises from a closer look at the simulation on the right is: Why does a technical current __flow__ into the base, i.e. positive charge carriers into the P-layer, have to be supplied for the NPN transistor? Wouldn't it be more plausible if the negative charge carriers, which are not present and needed for transport, had to be supplied?

~~PAGEBREAK~~~CLEARFIX~~~
<wrap #Transistor_in_Baendermodel />

=== Transistor in band model ===
<imgcaption pic5|Transistor in ribbon model>{{ https://upload.wikimedia.org/wikipedia/commons/2/28/Bipolartransistor2.PNG?400|Transistor in ribbon model}}</imgcaption>

To understand this, the knowledge of the PN junction is needed. In the figure <imgref pic5>, the structure of the NPN transistor is shown in the band model. In the n-doped collector and emitter, the free-moving negative charge carriers (blue) and stationary positive charge carriers (red) are drawn, and in the base, correspondingly, the free-moving positive charge carriers (red) and stationary negative charge carriers (blue). Both PN-junctions have formed a junction. A positive voltage $U_{CE}$ is applied to the transistor, which cannot generate any current flow in the situation shown. Due to the positive voltage $U_{CE}$ and the missing potential at the base, the voltage $U_{BE}$ decreases, which leads to a reduction of the junction. In contrast, the voltage $U_{CB} = U_{CE} -U_{BE}$ increases. Thus, the junction between the base and the collector becomes larger. When the external voltage $U_{CE}$ is varied, there will always be at least one PN junction that is reverse biased, i.e. the transistor will block.

~~PAGEBREAK~~~CLEARFIX~~~

<WRAP right><panel type="default">
<imgcaption pic9|function of npn bipolar transistor>
</imgcaption>
{{drawio>function_of_the_bipolar_transistor2}}
</panel>
</WRAP>

To remove the junction between the collector and the base, the latter must be connected in the forward direction. Until the transistor is switched through, this is several steps, which are described below via <imgref pic9>:

  - Figure: The physics of this takes place in the narrow p-layer. The following images refer to the highlighted section.
  - Image - Situation $U_{CE}=0V, U_{BE}=0V$: In this picture the unpowered transistor is shown. In it the free charge carriers (electrons in blue, holes in red) and the junction layers between base and emitter, and base and collector in yellow. Only the junction layer shows the stationary charge carriers with their sign. As shown in the band model, the stationary charge carriers are present everywhere in both doped regions.
  - Figure - Situation $U_{CE}=0V, 0V<U_{BE}<0.6V$: First, consider a small, positive voltage $U_{BE}$. This provides holes in the base with current $I_B$. This operates the PN junction between the base and emitter in the forward direction. In the figure, it is indicated with black circles that the injected holes compensate some stationary negative charge carriers in both junction layers. Electrons also flow through the emitter into the n-region, which attenuate the junction on the other side.
  - Figure - Situation $U_{CE}=0V, U_{BE}>0.6V$: When the forward voltage of the PN junction between base and emitter is exceeded, the injected holes and electrons cancel the bottom junction. In the simulation below, it can be seen that the circuitry of the transistor is such that in the diode circuit (which is not physically correct), the diode between the base and emitter becomes conductive.
  - Figure - Situation $U_{CE}>0V, U_{BE}>0,6V$: Now with this voltage at the base, the working circuit, i.e. a voltage $U_{BE}>0$ should be present at the output. In the real system, the base is very small compared to the mean free path length of the electrons ("path to recombination with a hole"). This changes the situation at the upper PN junction. In a classical diode, no electrons are present in the p-doped region. However, the electrons present here can cross the base and compensate for the stationary positive charge carriers in the upper junction. The holes injected into the base in turn compensate for the stationary negative charge carriers. Thus, this junction layer is also removed. This is possible as long as enough holes are injected into the base.
  - Figure - Situation $U_{CE}>0V, U_{BE}>0.6V$: Thus, in the NPN bipolar transistor, both holes (to remove the junction layers) and electrons (as the "main agents" responsible for charge transport, the so-called majority carrier charges) contribute to the conductivity. This is where the name __bipolar__transistor comes from.

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The simulation on the right shows the simplified model of the opposing diodes. The necessary input current $I_C$ and the corresponding input voltage $U_{BE}$ resemble the ratios of the diode between base and emitter. In <imgref picJ> the principle of operation is shown. The current $I_B$ across the diode between base and emitter regulates the current $I_C$ in the working circuit. This regulation is done by the variable resistor $R_{CE}$.

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<WRAP right><imgcaption picJ|function of npn bipolar transistor>
{{electronic_circuitry:jbt_function.jpg?400}}
</imgcaption></WRAP>

=== Characteristic, characteristic curves, characteristic diagrams ===

In the previous chapter [[1 Fundamentals of Amplifiers]] the characteristics of a black box have already been discussed, there especially for an amplifier. The methodology can also be applied here. In the video above, the first parameter has already been described: The **current gain** $\beta=\frac{d I_C}{d I_B}$, or in the form of a graph, the **current control characteristic** $I_C(I_B)$.[(Note1>In practice, a distinction is still made between small-signal current gain $\beta = h_{fe}$ and large-signal current gain $B = h_{FE}$. In small-signal behavior, a relatively small change around a fixed operating point (e.g., around certain values $I_C$ and $U_{CE}$) is considered. In large-signal behavior, a change between 0 and a given value is considered. For nonlinear characteristics, the two quantities may differ. In this course, only the small-signal behavior is described. The large-signal behavior and the distinction between the two considerations are not considered in this course)].

Another characteristic is the **input characteristic field** $U_{BE}({I_B})$ or as differential characteristic (=slope in the characteristic) the **differential input resistance** $r_{BE}=\frac{d U_{BE}}{d I_B}$. As described earlier, the structure between the base and emitter resembles a diode. Accordingly, the input characteristic field resembles that of a diode. Since the current flow $I_B$ is very small (a few microamps or smaller), the input resistance $r_{BE}$ is large.

The following simulation shows the current control characteristic $I_C(I_B)$ and input characteristic(nfield) $U_{BE}({I_B})$ by varying $U_{BE}$ (or $I_B$).
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For the description of the transistor, the **output characteristic field** $U_{CE}({I_C})$ and the **differential collector-emitter resistance** $r_{CE}=\frac{U_{CE}}{I_C}$ present in it as a slope is particularly important. This can be seen in the following simulation for different input voltages $U_{BE}$ (and thus different control currents $I_B$). The output characteristic field can be divided into different ranges:
  - __junction__: at low input voltages $U_{BE}< 600mV$, the junction is not degraded. Accordingly, the entire transistor becomes non-conducting. In the output characteristic field, this can be seen by the fact that when the output voltage $U_{CE}$ is positive, the output current $I_C$ becomes very small. In this case, the transistor on the output side corresponds to a high-impedance resistor, or an open switch.
  - __Gain region__ (or active region): at larger input voltages $U_{BE}> 600mV$, the junction is degraded. In the gain region, the output characteristic behaves as a straight line. The output current $I_C$ is thus only dependent on $I_B$, as defined by the current gain $\beta=I_C/I_B$.
  - __Saturation region__: The saturation region is found at larger input voltages $U_{BE}> 600mV$ and only small output voltage $U_{CE}$. At constant input voltage $U_{BE}$ the output voltage behaves to the output current like a high non-linear resistor. in this case the transistor on the output side corresponds to a low impedance resistor, or a conducting switch.

In the datasheet, a different nomenclature is occasionally found, resulting from the so-called [[https://de.wikipedia.org/wiki/Zweitor|H-characteristic of quadrupole theory]][(Note2>Quadpole theory is not considered in this course, but is derived in more detail in the Power Electronics subject.)]:
  * current gain $h_{fe}=\beta(I_C, U_{CE})=\frac{I_C}{I_B}$
  * input resistance $h_{ie}=r_{BE}(I_C, U_{CE})=\frac{U_{BE}}{I_B}$
  * output resistance $h_{oe}=r_{CE}(I_B, U_{BE})=\frac{U_{CE}}{I_C}$

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The bipolar transistor is used where a low threshold voltage or current amplifier is required. This is advantageous in various amplifier circuits, for example.
Bipolar transistors are also found in some simple power supplies.
The most common bipolar transistor circuit is the so-called collector circuit. This is characterized by the fact that a constant voltage - the supply voltage - is applied to the collector. Several collector circuits can be operated by a common voltage supply. This means that the same voltage is applied to all collector connections. Because of the wide use that bipolar transistors have had, even today the common voltage supply of electronic circuits is called $V_{CC}$, where $CC$ stands for **Common Collector**. This is often seen even when bipolar transistors are no longer used.

A major disadvantage of the bipolar transistor is that a control current is required for switching. Especially in digital circuits, but also in power electronics, this results in a non-negligible input power $P=U_{BE}\cdot I_B$. This leads to losses and waste heat, which must be taken into account in the power supply and thermal design. For this reason, bipolar transistors are no longer used in current microcontrollers. In these fields, the bipolar transistor has been displaced by the field-effect transistor.

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<WRAP> <panel type="danger" title="Note: Bipolar transistors"> <WRAP group><WRAP column 7%>{{fa>exclamation?32}}</WRAP><WRAP column 80%>

There are 2 different types of bipolar transistors.
These differ in the type of layer structure, or majority carrier charges:
  * **npn bipolar transistors**: Major conduction occurs via electrons. These cannot pass through the p-doped region without current $I_B$ across the base. By $I_B>0$ holes are introduced into the base, which remove junction layers.
  * **pnp bipolar transistors**: The main conduction occurs through holes. These cannot pass through the n-doped region without current $I_B$ across the base. By $I_B<0$ electrons are introduced into the base, which degrade junction layers.

In the __bipolar__transistor, __both types of charge carriers are involved in the transport__.

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===== 2.7 Operating principle of a field-effect transistor =====

<WRAP right><imgcaption picP|Function of the MOSFET>
{{electronic_circuitry:mosfet-front-final-test_jpg_project-body.jpg?400}}
</imgcaption></WRAP>

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<imgcaption pic6|FET circuit symbols>
</imgcaption>
{{drawio>FET_Principle}}
</panel></WRAP>


A field effect transistor (FET) also consists of two diodes connected against each other, which have a common n- or p-layer. However, the conductivity of the field-effect transistor is not generated by applying a control__current__, but solely by a control__voltage__. In the case of the bipolar transistor, the control current was also generated by a control voltage. However, the control current must flow continuously to drive the bipolar transistor, since the charge carriers introduced via the base recombine internally.

In <imgref pic6> a special field-effect transistor is drawn, the so-called "metal-oxide-semiconductor field-effect transistor". This will be explained in more detail below. The <imgref picP> outlines the principle of operation: the control voltage $U_{GS}$ (in English $V_{GS}$) regulates the current $I_D$ in the working circuit. This is done by the resistance between $R_{DS}$.

To distinguish the transistor types, and to emphasize the physics behind them, the terminals are labeled differently for the field-effect transistor:

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  * **(S) Source**: Terminal from which the charge carriers pass through the transistor (roughly corresponds to the emitter).
  * **(G) Gate**: Terminal at which a voltage can be used to change the conductivity (roughly corresponds to the base, with control currents being injected there).
  * **(D) Drain**: Terminal at which the charge carriers arrive and leave the transistor (corresponds approximately to the collector).
In addition, there is the **"Bulk" (B)** in the structure, which refers to the basic substrate of the transistor. This is usually not led out separately, but shorted to the source terminal.
In some FETs, the bulk is represented by the middle connection.

In the simulation on the right, you can see that the field-effect transistor behaves much like a switch, which is controlled by a voltage. No current seems to flow on the gate, but when the voltage on the gate changes, the behavior changes from "conductive" to "open".

==== Metal Oxide Semiconductor Field Effect Transistor (MOSFET) ====

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<imgcaption pic3|MOSFET layering>
</imgcaption>
{{drawio>MOSFET_Pinch_off}}
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The structure of the metal oxide semiconductor field effect transistor (**Metal Oxide Semiconductor Field Effect Transistor: MOSFET)** resembles the bipolar transistor at first glance. In <imgref pic3>, the individual figures (1)...(3) show the layering of an n-channel (English n-channel) MOSFET, and in (4) the circuit symbol is shown again. In contrast to the npn-bipolar transistor, the middle p-doped layer (bulk) is not directly connected to the control electrode. Rather, the metal layer of the gate (<imgref pic3>, Fig. (5), gray), the insulating layer of the oxide (shown in purple), and the conductive p-doped layer of the bulk (shown in red) form a capacitor. It should be noted that the bulk is at the potential of the source connection (dotted line in the picture).

Without voltage difference $U_{GS}$ between gate and source, a (small) junction is formed at the p-n junctions. If the voltage difference $U_{GS}$ is increased, the capacitor between gate and bulk is charged. This accumulates electrons opposite the gate electrode (<imgref pic3>, Fig. (2), dark blue "wedge"). If the voltage difference $U_{GS}$ exceeds a certain threshold voltage, the enriched electrons form a channel between source and gate. This allows a current $I_D \gg 0$ to flow through the MOSFET (<imgref pic3> Fig. (3)).

The switching symbol (<imgref pic3>, figure (4)) can also be described as follows: Capacitors form between gate and source, between gate and base, and between gate and drain, respectively, in the off state because of the oxide layer (purple in Fig. (1))[(Note2>In field-effect transistors, an additional capacitor forms between source and drain, which can lead to overvoltages at the MOSFET, especially during fast switching of inductors)]. To drive the MOSFET, the voltage at the gate $U_{GS}$ must be such that a PN junction forms in the bulk, indicated by the white filled triangle in figure (4). Since the apex of the triangle (or the diode symbol sketched with it) points toward the gate, it is clear that we are dealing with an n-channel MOSFET.

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In the simulation on the right, the same voltage ratios are shown as in <imgref pic3> (1)...(3). The toggle switch on the left makes it possible to invert the voltage $U_{DS}$ across the transistor. If this becomes negative, a slightly different situation arises: The MOSFET appears to become conductive regardless of what voltage $U_{GS}$ assumes. This is due to the fact that another diode has been hidden in the layer structure: a junction has formed between the bulk (p) and drain (n), which is operated at $U_{DS}<0$ and with the bulk and source connected in the forward direction. This so-called body diode is explicitly built into the simulation at (3b).

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==== Output characteristics of the MOSFET ====

The **output characteristic field** $U_{DS}({I_D})$ is also to be considered for the MOSFET. This is also similar to the bipolar transistor, but now the different characteristics are adjustable by different control voltages $U_{GS}$ and not by a control current.

Unfortunately, the naming of the different operating ranges of a MOSFET differs from that of the bipolar transistor:

~~PAGEBREAK~~~CLEARFIX~~~

  - __Blocking range__: at low input voltages $U_{GS}$, no channel can be formed. Accordingly, the entire transistor becomes nonconducting. In the output characteristic field this can be seen by the fact that at positive output voltage $U_{DS}$ the output current $I_D$ becomes very small. In this case, the transistor corresponds to a high-impedance resistor, or an open switch, on the output side.
  - __Saturation region__: for larger input voltages $U_{GS}> U_{th}$ above a threshold, a conductive channel is formed. In the saturation region, the output characteristic behaves like a straight line. The output current $I_D$ is thus only dependent on $U_{GS}$.
  - __linear region__ (active region) : The linear region is found at larger input voltages $U_{GS}> U_{th}$ and only small output voltage $U_{DS}$. At constant input voltage $U_{GS}$, the output voltage to output current behaves like a high non-linear resistor. In this case, the transistor on the output side corresponds to a low-impedance resistor, or a conducting switch.

It should be noted that the saturation region for MOSFET and bipolar transistor characterizes different operating ranges.

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==== Variants of MOSFETs ====

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<panel type="default">
<imgcaption pic2|FET circuit symbols>
</imgcaption>
{{drawio>FET_Switch_Icon}}
</panel>

</WRAP>

The so far considered (and also most frequently used) field effect transistor is the so-called "**n-channel enhancement type MOSFET**". The part "n-channel" comes from the type of the current-forming charge carrier and was already given above. The part "enhancement type" represents, that the charge carriers are not present at first and have to be accumulated in the bulk by means of the voltage $U_{GS}$ for conductivity.

Some circuits (especially digital circuits) also use "**p-channel enhancement type MOSFET**", where holes are the current-forming charge carriers. In the simulation on the right, this type of MOSFET is shown. Most clearly, when the p-channel enhancement type MOSFET is connected, the drain and source are generally reversed. Thus, the numerical values of $U_{DS}$ and $I_D$ in the output characteristic field become negative. To enrich holes in the p-channel, a negative voltage must be applied to the gate $U_{DS}<0$.

In the <imgref pic2> the circuit symbols of different variants of MOSFETs are shown. In the MOSFETs in the top row, an n-channel is formed for charge transport, and in the bottom row, a p-channel is formed.

Three variations of an **n-channel enhancement type MOSFET** are shown in <imgref pic2> in the upper left. In the first circuit symbol, the circle represents that it is a discrete device, i.e., a single MOSFET not integrated with others in a chip. The second circuit symbol has already been used in the previous chapters. The third circuit symbol of the same n-channel enhancement type MOSFET is the reduced version (i.e., without bulk). This representation is used for simplification in digital circuits.

In <imgref pic2> on the lower left, three variations of a **p-channel enhancement type MOSFET** are shown. Again, the circle on the first circuit symbol indicates that it is a discrete device, but now the direction of the arrow on the bulk is rotated. The second switching symbol is used in the same way as for the n-channel MOSFET - in integrated circuits. The third symbol is again the reduced version (without bulk). For the digital circuit it is only important whether the switch closes or opens at high signal ($= 5V$). Since the p-channel enhancement type MOSFET opens, this is drawn with a negation sign (small circle) at the gate.

~~PAGEBREAK~~~CLEARFIX~~~

In <imgref pic2> on the right, so-called **n-channel and p-channel depletion-type MOSFET** are shown. The MOSFETs considered so far were not conductive in the off state (i.e. $U_{DS}=0$). However, in some applications, it would be good if the MOSFET resembled a conductive switch when off. Looking at the layer structure (<imgref pic3>, Figure (1)...(3)), this is possible via selective redoping of the region opposite the gate. The doping can be used to dislocate a conductive channel. The charge carriers of this channel can be displaced or depleted by a suitable field - and thus suitable gate voltage $U_{GS}$. Thus, the MOSFET becomes non-conducting in the presence of a reverse voltage $U_{GS}$. In the circuit symbol, the "short circuit" between source and drain is also drawn in pictorially.

<WRAP group>
<WRAP> <panel type="danger" title="Remember: MOSFETs"> <WRAP group><WRAP column 7%>{{fa>exclamation?32}}</WRAP><WRAP column 80%>

There are 4 different types of MOSFETs.
On the one hand, these differ in the type of current-forming charge carriers:
  * **n-channel**: The current-forming charge carriers are electrons.
  * **p-channel**: The current-forming charge carriers are holes.

The second distinguishing feature is the off-state conductivity ($U_{GS}=0$):
  * **enhancement type**: When the gate voltage is $U_{GS}=0$, no conductive channel is present. Only by charging the gate bulk capacitor, the channel is formed or the carriers are enriched.
  * **depletion type**: At a gate voltage of $U_{GS}=0$, a conductive channel is present. By charging the gate bulk capacitor, the channel is reduced or the charge carriers are displaced ("depleted").

In the field-effect transistor, the electric field of the gate-bulk capacitor enriches or depletes only those charge carriers that contribute to charge transport.

</WRAP></WRAP></panel> </WRAP>


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<WRAP right><panel type="default">
<imgcaption pic8|workspace of semiconductor elements>
</imgcaption>
{{drawio>Working_space_of_semiconductor_elements}}
</panel>
</WRAP>

=== Semiconductor element design ===

For all transistors and diodes, various limit values must be observed for the circuit design. These can be entered directly in the output characteristic field (<imgref pic8>, above).
Due to the heating of the component and the resulting increase in intrinsic conduction, two limit values result:
  * In the conducting state, the power dissipation $P_{loss}=R(T)\cdot I^2$ forms a direct reference to the current through the semiconductor element $I_C, I_D, I_D$ (bipolar transistor, MOSFET, diode). This results in current $I_{max}$, which should not be exceeded.
  * In the state where there is both a noticeable current and voltage, there is a maximum allowed power $P_{tot}=const.=U\cdot I$. This is a hyperbola in the output characteristic. If the output current exceeds this hyperbola, the semiconductor element heats up to such an extent that, due to the increasing intrinsic conductivity, the conductivity drops, which in turn leads to an increasing current. This effect leads to the thermal destruction of the component.
In addition, a maximum voltage $U_{max}$ must not be exceeded. This is usually due to the (internal) dielectric strength of the component.

These limits are especially important if, for example, a MOSFET is to be used as a switch (example: <imgref pic8>, below). In this case, there are two states:
  * Switch is conductive: a low voltage $U_{DS}$ is applied, at which a large current $I_D<I_{max}$ flows.
  * Switch is non-conductive: A high voltage $U_{DS}<U_{max}$ is applied, at which no current flows.
When switching from "conductive" to "non-conductive", even if the individual current and voltage limits are taken into account, this can destroy the switch. In <imgref pic8>, this case can be seen in the diagram below. Current flow $I_D$ is initially maintained (or are only small), although voltage $U_{DS}$ increases (blue line). In this case, $P_{tot}$ may be exceeded and the MOSFET is destroyed due to thermal overload.

To speed up the switching process (especially for power MOSFETs, e.g. for motor drivers), so-called **driver circuits** generate the voltage $U_{GS}$. With these driver circuits, the control voltage can be made available and reset very quickly. For this purpose, currents in the range of several amperes must be provided for a short time for charging and discharging the gate capacitor.

<WRAP group>
<WRAP> <panel type="danger" title="Remember: Maximum output values of a semiconductor element"> <WRAP group><WRAP column 7%>{{fa>exclamation?32}}</WRAP><WRAP column 80%>

For each semiconductor element, there are three maximum values to consider at the output:
  * a maximum voltage limit $U_{max}$,
  * a maximum current limit $I_{max}$,
  * a maximum power limit $P_{tot}=U \cdot I$

</WRAP></WRAP></panel> </WRAP>
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===== 2.8 Applications for bipolar transistors =====

====Darlington-Transistor====

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The Darlington circuit or the Darlington transistor (as a discrete element) is a simple construction, which makes it possible to control the output voltage $U_{BE}$ with a considerably lower base current $I_B$. On the right is the Darlington circuit compared to a simple bipolar transistor. Details can be found in [[https://de.wikipedia.org/wiki/Darlington-Schaltung|Wikipedia under Darlington circuit]].

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==== Internal life of an operational amplifier ====

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The operational amplifier as an "almost ideal" differential voltage amplifier represents a central component of electronic circuit technology from the next chapter on. In the chapter [[electronic_circuit_engineering:1_basics_to_amplifiers#feedback|basics to amplifiers - feedback]] an ideal differential voltage amplifier was already used. In the simulation on the right, the core of the differential voltage amplifier is simplified. Accordingly, there is no differential voltage at the input, but a small sinusoidal voltage. This is first applied to the base of the first bipolar transistor, which is a high impedance input amplifier stage. The current $I_C$ regulated by this in turn leads to a base of another bipolar transistor and then to the output amplifier stage. In simulation, this setup achieves a differential gain of about $A_D=10,000,000$. In real differential amplifiers, this is more in the range $A_D ≈100'000$. Details can be found in [[https://de.wikipedia.org/wiki/Operationsverst%C3%A4rker#Herk%C3%B6mmlicher_Operationsverst%C3%A4rker_(VV-OP)|Wikipedia under operational amplifier]].

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===== 2.8 Applications for field effect transistors =====

==== NOT gate ====
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Just about all consumer electronics products have field-effect transistors at their core. In detail, this is based on [[https://de.wikipedia.org/wiki/Complementary_metal-oxide-semiconductor|CMOS technology]] (CMOS: Complementary metal-oxide-semiconductor) is used. The MOSFETs on the ground side and the MOSFETs on the power supply side behave in opposite ways, i.e. complementary. The simulation on the right shows the simplest gate, the NOT gate. Another gate was considered in an introductory way.

~~PAGEBREAK~~~CLEARFIX~~~

==== Reverse polarity protection ====

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Many chips (such as microcontrollers) can be destroyed by an incorrectly polarized power supply. Battery powered electronics should have an active protection circuit for this. A diode is not practical for the power supply (why?). Instead, a MOSFET can be used, which does not pass negative voltages. Details are well explained on the [[http://www.lothar-miller.de/s9y/categories/39-Verpolschutz|page of Lothar Miller]].

~~PAGEBREAK~~~CLEARFIX~~~

==== Level converter ====

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During electronics development it can happen that several integrated circuits (e.g. intelligent light sensor, microcontroller, intelligent LED) require different voltage levels. This can lead to problems especially during data exchange, if logic High has to be in a certain voltage range. This problem can be solved by a level converter.
The level converter (also logic level converter, level shifter) enables the bidirectional connection of digital connections of different voltage levels, e.g. 5 V to 3.3 V.

For the level converter, any n-channel enhancement MOSFET whose threshold voltage is below $1.8...2.0 V$ can be used. This limit is due to the minimum logic level of $2.0 V$ for logic high. For simplicity, "logic level enhancement mode MOSFET" are used, which are just optimized for the logic voltage of $3.3V$.

The way it works is well explained on [[https://de.wikipedia.org/wiki/Pegelumsetzer|Wikipedia]] and can be derived with simulation.

~~PAGEBREAK~~~CLEARFIX~~~


==== Voltage doubler/inverter ====

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As a power supply for electronics, $5V$ or $3.3V$ is often used. In the following chapter, we will see that a bipolar power supply is often used for operational amplifier circuits. To be able to generate $-5V$ at low currents from a $5V$ supply, [[https://de.wikipedia.org/wiki/Ladungspumpe#Spannungsverdopplung|charge pumps]] are often used. One such can be seen on the right in the simulation. In the oscilloscope (in the simulation below), the voltage $U_{C1}$ is displayed at the input capacitor C1 and $U_{C2}$ at the storage capacitor C1. This circuit can be found, for example, in IC [[https://www.renesas.com/eu/en/www/doc/datasheet/icl7660.pdf|ICL7660]] (Renesas), [[https://www.ti.com/lit/ds/symlink/lmc7660.pdf|LMC7660]] (TI), [[http://ww1.microchip.com/downloads/en/DeviceDoc/21465C.pdf|TC7660]] (Microchip) integrated. Details on how it works can be found in [[https://www.youtube.com/watch?v=85CCafHIbA4|this video]], for example.

Study Questions:

  * In which state is the voltage $U_{C1}$ equal to $1 V$? * In which state is the difference between the voltages $U_{C2}-U_{C1}$ across the two capacitors equal to $1V$?
  * What happens if the voltage sources for $0V$ and $1V$ are reversed?
  * How can this circuit be implemented with diodes instead of changeover switches?

~~PAGEBREAK~~~CLEARFIX~~~

==== Voltage inverter in the microcontroller ====

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</WRAP>

In some microcontrollers a negative voltage is required internally (e.g. for operational amplifiers).
Since this voltage is not supplied externally, the microcontroller must provide it via an internal circuit.
The simulation on the right shows a circuit that can be integrated into a microcontroller in this way. The ring oscillator generates a high frequency clock signal, which drives an inverter stage (logical NOT gate). The charge can then be shoveled down via the two capacitors in such a way that the capacitor provides a negative voltage at the output. For more information, see [[https://de.wikipedia.org/wiki/Ladungspumpe#Spannungsinvertierung|Wikipedia under charge pump]] and "[[http://www.righto.com/2018/08/inside-die-of-intels-8087-coprocessor.html|Inside the 8087's substrate bias circuit]]".


~PAGEBREAK~ ~CLEARFIX~~
==== four-quadrant ====

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In many applications, current and voltage must be controlled independently of each other. This is the case, for example, with a motor (= ohmic-inductive load). There, the current is essentially proportional to the torque and the voltage to the speed. If voltage and current are to be output bipolar (or in the application: Torque and speed are to be controlled in both directions), a four-quadrant controller made of transistors is suitable. In modern integrated circuits, these are made of MOSFETs, directly equipped with the MOSFET driver, and several four-quadrant controllers can be found next to each other (e.g. the stepper motor driver [[http://www.ti.com/lit/ds/symlink/drv8835.pdf#page=9|DRV8835]]).
Details can be found on [[https://de.wikipedia.org/wiki/Vierquadrantensteller|Wikipedia under four-quadrant actuators]].


==== Other MOSFET Applications====

MOSFETs are not only used for pure switching of currents. Further applications are also:
  - as a display element in TFT screens ({{wpde>Thin Film Transistor|TFT ... Thin Film Transistor}}).
  - as memory element e.g. in SD cards ({{wpde>Floating Gate Transistor}}, or also new approaches, like {{wpde>Ferroelectric_Random_Access_Memory}})
  - as an integrated "upstream" element for power bipolar transistors, especially in the {{wpde>Bipolar_transistor_with_insulated_gate_electrode}} (IGBT)
  - as a chemical sensor for various materials (see {{wpde>Chemical_sensitive_field_effect_transistor}})
  - as a link between photonics/optoelectronics and classical electronics

~~PAGEBREAK~~~CLEARFIX~~~



====== Learning questions ======

=== for self-study ===
  * Describe the function of a transistor.
    * Sketch the layered structure of a bipolar transistor. Explain the switching through of a PNP bipolar transistor with the help of the sketch drawn.
    * Draw the simplified diode equivalent circuit of an NPN transistor describe the working.
  * Explain the difference between a PNP and NPN transistor.
    * Draw a circuit each with the respective switch connected to U+ = 5V and ground in such a way that switching through is possible with a voltage between U+ and ground at the base.
    * Name the respective connections of the transistors in the drawing.
    * What voltage must be applied to the base in each case for the transistor to switch through?
    * How should the sign of the control current be chosen in each case?
    * In what size range is a typical current gain?
  * Current-controlled and voltage-controlled transistors
    * Explain the difference between a current controlled transistor and a voltage controlled transistor.
    * Which type of transistor is current controlled and which is voltage controlled?
    * Draw a circuit diagram each for a current controlled transistor and a voltage controlled transistor.
    * What is the doping order of the transistors drawn?
  * What are the two basic types of transistors?
  * MOSFET
    * What are the advantages of a MOSFET over a bipolar transistor?
    * How is a MOSFET constructed? (layer structure, connections)
  * H-bridge
    * Draw an H-bridge with switches (ideal switch), a resistive/inductive load and an external voltage source with V+ and GND.
    * How can the various switches be controlled to have any voltage between V+ and V- applied to the load? What is the technical term for the method of control?
  * Draw the PWM signal necessary to generate a sinusoidal output when a full bridge is used.
  * What are the uses for transistors
    * What are some uses for transistors?
    * Draw a voltage doubler.
    * What is a level converter?
    * Why is it preferred to use field effect transistors rather than bipolar transistors nowadays?

=== with answers ===

{{electronic_circuitry:transistor_question.png}}

<quizlib id="quiz" rightanswers="[['a2'],['a1','a2'], ['a1']]" submit="Check answers">

<question title="Looking at the picture above, which of the following statement(s) is/are correct?" type="checkbox">
The transistor has an npn structure internally|
The collector terminal is at the bottom|
It is a bipolar transistor|
In order to make I_C flow, the voltage U_BE must become positive
</question>

<question title="Which statement(s) about bipolar transistors is/are correct?" type="checkbox">
The current I_C or the voltage U_BC controls the current flow I_B|.
The input characteristic of a bipolar transistor corresponds to that of a diode.
The disadvantage of the bipolar transistor is the continuous current flow required in the conductive state.
VCC stands for Voltage Common Connector
</question>

<question title="Which statement(s) about MOSFETs is/are correct?" type="checkbox">.
MOSFET stands for the structure of the field-effect transistor made of metal oxide and semiconductor|.
Due to the body diode the MOSFET acts in one direction like a diode|
Enrichment type MOSFET are conductive with $U_{GS} =0V$|
In n-channel MOSFETs, holes are the current-carrying charge carriers.
</question>
</quizlib>

====== Image references ======

--> References to the media used #

^ Element ^ License ^ Link ^
| Video: Circuit Elements - Diodes and Transistors - Part 4 | [[https://creativecommons.org/licenses/by/3.0/legalcode|CC-BY (Youtube)]]  | https://www.youtube.com/watch?v=KjyHta5p9WE |
| <imgref picJ>: Function of the npn bipolar transistor | (c) Open Music Lab, with permission for further use | Source: Mail of the illustrator |
| <imgref picP>: function of the MOSFET | (c) Open Music Lab, with permission to reuse | [[https://www.crowdsupply.com/open-music-labs/mosfet-girl|CrowdSupply]]                               |

<--
====== 3. Basic circuits of operational amplifiers I ======

<callout>.
  * [[https://www.mikrocontroller.net/articles/Operationsverst%C3%A4rker-Grundschaltungen|Operational amplifier basic circuits on Microcontroller.net]] provide a nice introduction.
  * [[https://rd.springer.com/book/10.1007/978-3-8348-2146-1|Textbook and Workbook Operational Amplifiers (Joachim Federau)]] (viewable via the university network).
  * On Youtube the following two videos by Prof. Griesbauer are recommended: [[https://www.youtube.com/watch?v=nl3JZbj9kHs|Operation amplifier properties]], [[https://www.youtube.com/watch?v=wiq93KjXwao|Operation amplifier circuits]]
  * interactive animation of an inverted amplifier can be viewed in the [[https://www.ipes.ethz.ch/mod/lesson/view.php?id=24&pageid=70|iPES course of ETH Zurich]] (Login as guest!)

</callout>

<WRAP>
<callout type="info" icon="true">
<WRAP right><panel type="default">
<imgcaption pic11|undistorted signal>
</imgcaption>
\ {{electronic_circuitry:hello.mp3}}
\\ {{drawio>hello}}
</panel></WRAP>

<WRAP right><panel type="default">
<imgcaption pic12| overdriven signal>
</imgcaption>
\ {{electronic_circuitry:hello_distorted.mp3}}
\ {{drawio>hello_distorted}}
</panel></WRAP>

=== Introductory example===

Acoustic amplifiers, such as those found in mobile phones, laptops or hi-fi systems, often exhibit an unpleasant characteristic when heavily amplified: the previously undistorted signal is no longer passed on as usual, but [[https://de.wikipedia.org/wiki/Klirrfaktor#Das_Klirren|clatters]]. It is distorted in such a way that it no longer sounds pleasant.

For this purpose, you will find an acoustic example with pictures in <imgref pic11> and <imgref pic12> respectively. The bottom of each image shows the time course of the voltage output to a loudspeaker (x-axis: time, y-axis: frequency). The upper picture has three dimensions: It shows in color intensity which frequencies are used at which time. The frequencies in grey areas are not used. If a frequency is shown in red at one point in time, it has a relatively large amplitude.

It can be seen that the distorted signal has large amplitudes in the time course of the voltage as well as a wide distribution of frequencies (= a broad spectrum). The high frequencies in particular can promote wear of the diaphragm in loudspeakers.

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</WRAP>

The signal distortion is due to the design of the amplifier, which can only output a maximum possible voltage and otherwise [[https://de.wikipedia.org/wiki/%C3%9Cbersteuern_(signal processing)|overdrives]]. The structure of an acoustic amplifier is similar to that of a feedback operational amplifier, as seen in the simulation on the right.

Acoustic amplifiers are usually constructed like operational amplifiers, which will be considered in this chapter.
</callout></WRAP>

<WRAP><callout>
=== Objectives for basic circuits of operational amplifiers ===
After this lesson you should:
  - Have learned the basic equation of operational amplifiers and the golden rules.
  - Know how the basic circuits (inverting and non-inverting amplifiers) are constructed, what their voltage gain is, and how their input and output resistances behave.
  - Be able to calculate the voltage gain of simple amplifier circuits.
  - Understand the concept of virtual ground.
</callout></WRAP>

~~PAGEBREAK~~~CLEARFIX~~~

===== 3.0 Introduction =====

<WRAP right><panel type="default">
<imgcaption pic0|operatingamplifier>
</imgcaption>
{{drawio>OP_amplifier}}
</panel></WRAP>

=== Historical outline ===
To understand the origin of the operational amplifier, a brief outline of microprocessor history will be presented here.
The first microprocessor was commercially distributed in 1971 by INTEL as [[https://de.wikipedia.org/wiki/Intel_4004|Intel 4004]], which could only represent a 4-bit number space with one word and had a clock frequency of 500 kHz. This was already extended after half a year as [[https://de.wikipedia.org/wiki/Intel_8008|8008]] on a 8-Bit Word, so that calculations could be accomplished substantially faster. From this second generation the [[https://de.wikipedia.org/wiki/X86-Prozessor|8086]] was developed in 1978, which had been the basis of many PCs and other electronic components with a 16-bit word for a very long time. In 1985 this processor was extended as 80386 to a 32-Bit Word. From this time also the perhaps well-known question originates whether software should be installed as x86 or x64 code. Processors with 64-bit Word have been available since 1999, but they have not yet penetrated all niches.

With this historical outline in mind, the question arises as to how Apollo 11 was able to control both the first lunar mission and the lunar module, and how the orbit could be calculated so accurately that this would not pose a risk. Development of the [[https://de.wikipedia.org/wiki/Apollo_Guidance_Computer|Apollo Gudance Computer]] was started in 1961, 10 years before the first commercial microprocessor. This processor, with about 1MHz frequency and 16-bit word width, could provide performance as early as 1966 that was not commercially available until 15 years later. Nevertheless, these computers could not solve the differential equation systems necessary for trajectory planning in a reasonable amount of time. This was only possible by combining the well-known digital processor and the [[https://de.wikipedia.org/wiki/Analogrechner|analog computers]] that had been in use since the 1920s. Analog computers are only slightly reminiscent of the classical image of the computer. Although they also consist of a network of components, they do not have any software. Rather, various properties of electronic components are used for computation (e.g., the exponential behavior of the Shockley equation of the diode). For the Apollo missions, the [[http://www.analogmuseum.org/library/eai380_short.pdf|EAI 380 hybrid computer]] was used. This and similar hybrid computers continued to be used into the 1990s to solve systems of differential equations for aerospace, construction, and (under a slightly different guise) as synthesizers in the music industry.
The interested reader is referred to the lecture by [[https://youtu.be/UnKCCpRFrrk|Hr. Ulmann on analog computers]], which explains why they are superior to even the most modern computers.

<WRAP right>
<imgcaption picZ|Operational Amplifier Module (OPA) in a PIC microcontroller manufactured by Microchip (c) Microchip>.
{{ electronic_circuitry:opa_8bitpic.jpg?200|}}
</imgcaption>
</WRAP>

The core component of analog computers is the operational amplifier. With this it is possible to build and combine adders, subtractors, multipliers, exponentials, integrators and differentiators. The "calculation" happens almost instantaneously after setting the desired input voltages.

The operational amplifier has outlasted this time and nowadays offers a starting point for a wide variety of signal conditioning, such as active filters, amplifier stages, input/output resistance matching and various (integrated) control loops. Several microcontrollers have op-amps built right in and can be configured by suitable code. In <imgref picZ> a PIC microprocessor is shown, which contains an integrated operational amplifier.

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===== 3.1 Circuit symbols and basic circuitry =====

This chapter deals with operational amplifiers or in general with measuring amplifiers. One application for this is the measurement of voltages, currents and resistances. These must be determined very precisely in some applications, for example for an accurate temperature measurement. In this case, amplification of the measurement signals is useful and necessary.

This amplification is done by measuring amplifiers. Measuring amplifiers have to fulfil some characteristics:
  * Measurement amplifiers should not have any feedback effect on the measured variable. An operational amplifier should have the highest possible input resistance. This prevents the voltage to be amplified from collapsing.
  * Measuring amplifiers should have a high sensitivity. An operational amplifier should have a large differential gain $A_D$.
  * Measuring amplifiers should show a defined transmission behaviour, i.e. the output signal should be clearly related to the input signal. An operational amplifier concretely should show a linear relationship.
  * Measuring amplifiers should show a good dynamic behaviour. The output signal of an operational amplifier should follow the input signal without any time delay.
  * Measurement amplifiers should produce an "impressed output signal". This means that the components at the amplifier output cannot change the output signal. An operational amplifier, specifically, should be able to maintain the desired output signal with the necessary current to do so. Since the current $I_A$ can become very large (by electronic standards), this means that an operational amplifier must have a low output resistance $R_A =\frac{U_A}{I_A}$.

<WRAP right><panel type="default">
<imgcaption pic1|Equivalent circuit diagram of an amplifier>
</imgcaption>
{{drawio>replacement_circuit_diagram_of_an_amplifier}}
</panel></WRAP>

In general, a measurement amplifier is constructed as in <imgref pic1>. This has already been described in chapter [[1_basics_to_amplifiers#amplifiers_-_a_blackbox_will_be_specified|1 Basics to amplifiers]]. In the following, only operational amplifiers will be considered. An operational amplifier is a measurement amplifier, which is often used in electrical engineering.
\\ \\ \\ \\ \\ \\ \\
The circuit symbol of the amplifier is an isosceles triangle, at the apex of which the output signal originates and in the base of which the input signal enters. In <imgref pic3> you can see different switching symbols:

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<WRAP right><panel type="default">
<imgcaption pic3|Switching symbols of amplifiers>
</imgcaption>
{{drawio>Switching_symbols_OPV}}
</panel></WRAP>

  * Switching symbol (1): In block diagrams (not to be confused with circuit diagrams, see [[1_basics_to_amplifiers#feedback|chapter 1]]) this switching symbol is used for general amplifiers. The input signal enters an input and exits through an input. This symbol will not be found again until Chapter 5.
  * Switching symbol (2): According to DIN EN 60617, this circuit diagram is to be used for operational amplifiers. It indicates with the infinity sign the ideally infinitely high amplification. in the following this symbol is not used, because it is not used in all international circuits and tools.
  * Switching symbol (3): The switching symbol (3) is the most commonly used symbol for an operational amplifier. On the left is the **inverting input** with voltage $U_m$ (__m__inus) and the **non-inverting input** with $U_p$ (__p__lus). The output with voltage $U_A$ is shown on the right.
  * Switching symbol (4): The switching symbol (4) is additionally drawn with the supply voltages $U_{sp}$ (__s__upply __p__lus) and $U_{sm}$ (__s__upply __m__inus). Power is provided from the supply for the output voltage of the operational amplifier.
  * Switching symbol (5) and (6): these symbols show __**no**__ operational amplifier. These symbols show the NOT gate and the tri-state gate. Both of these components have already been discussed in [[fundamentals_of_digital_engineering:binaer_logic#applications_of_binaer_logic|fundamentals_of_digital_engineering]]. Unfortunately, the representation of these digital components in various circuits is not unlike the operational amplifier. An example of this is the transceivers[("__trans__mitter and re__ceiver__", meaning transmitter-receiver, or interface adapter)] {{electronic_circuitry:sp3485cn-ltr.pdf|SP3481 or SP3485}}. If digital input values are considered, assume that the circuit symbol does not represent an operational amplifier.

<WRAP column 100%> <panel type="danger" title="Notice: opamp input"> <WRAP group><WRAP column 7%>{{fa>exclamation?32}}</WRAP><WRAP column 80%>

<WRAP right><panel type="default">
<imgcaption pic2|Voltages at the operational amplifier>
</imgcaption>
{{drawio>Voltages_at_OPV}}
</panel></WRAP>

The inputs of the operational amplifier are designated as **inverting input** $U_m$ and **non-inverting input** $U_p$.

The voltage $U_D = U_p - U_m$ is called the differential voltage \ (see <imgref pic2>).

</WRAP></WRAP></panel> </WRAP>

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===== 3.2 Basic equation / golden rules =====

<WRAP right><panel type="default">
<imgcaption picA|Operational amplifier equivalent circuit>
</imgcaption>
{{drawio>OPV_replacement circuit}}
</panel></WRAP>

The operational amplifier is a voltage amplifier. Thus it follows from chapter [[1_basics_to_amplifiers#idealized amplifier basic types]] that for the ideal case the input resistance must be infinite and the output resistance $R_A=0$.

The <imgref picA> shows an ideal voltage amplifier. This is characterized by the following properties:
  * **input resistance**: The input side is drawn in the figure. The resistance given here is the differential resistance $R_D$ where the differential voltage $U_D$ drops. Therefore $R_D \rightarrow \infty$ is valid. Thus, the input currents $I_p \rightarrow 0$ and $I_m \rightarrow 0$.
  * **Output resistance**, **Basic equation**: On the output side, $R_A=0$ gives that $U_A = A_D\cdot U_D$. This is the basic equation of the amplifier circuit. Ideally, the op-amp amplifies linearly, as indicated in the equation. Specifically, for a differential voltage of $0V$, the output voltage is $0V$.
  * **Voltage Gain**: From the chapter [[1_basics_to_amplify#feedback]] it is known that $A_D$ must be very large. Ideally, the following applies: $A_D \rightarrow \infty$

<WRAP column 100%> <panel type="danger" title="Remember: basic equation and golden rules"> <WRAP group><WRAP column 7%>{{fa>exclamation?32}}</WRAP><WRAP column 80%>

  - The output voltage depends on the differential voltage via the differential gain: $U_A = A_D \cdot U_D$ \\ This is the **basic equation** of the amplifier circuit.
  - The **golden rules** of the ideal amplifier are:
    - The differential gain goes to infinity: $A_D \rightarrow \infty$
    - The input resistance goes to infinity: $R_D \rightarrow \infty$
    - The output resistance is 0: $R_A = 0$

</WRAP></WRAP></panel> </WRAP>
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These rules have different limits in the real amplifier:
  * $\boldsymbol{U_A = A_D \cdot U_D}$:
    * The output voltage can only follow the input voltage as far as the power supply allows. In real operational amplifiers, only so-called **rail-to-rail** operational amplifiers can exploit the range down to a few $100mV$ to $U_S$. Other operational amplifiers have an **output limit**, which is $1...2V$ below the supply voltage.
    * If the supply voltages are not symmetrical ($U_{sm} \neq -U_{sp}$), then the characteristic also shifts.
    * The ideal operational amplifier produces the same output voltage $U_A=A_D \cdot U_D$ as long as $U_D = U_p - U_m$ is the same. For the real operational amplifier with fixed $A_D$, output voltage $U_{A1}$ for $U_{D1}=5V - 4.9V$ is different from $U_{A2}$ for $U_{D1}=0.1V - 0V$.
  * $\bold symbol{A_D}$: The differential gain is usually between $A_D = 20'000 ... 400'000$.
  * $\boldsymbol{R_D}$: For real operational amplifiers, the input resistance $R_E > 1 M\Omega$ and the input current $|I_p|$ or $|I_m|$ is less than $1 \mu A$.
  * $\boldsymbol{R_A}$: In real operational amplifiers, the output resistance $R_A$ is usually a few $\Omega$ and limited by a maximum current (in the range of a few tens of $mA$ to a few $A$)

<WRAP right>{{url>https://www.falstad.com/circuit/circuitjs.html?running=false&ctz=CQAgjA7CAMB00OgFgJxvRzA2GsUA4UwloUsBmcpAVmomp3KjGhGteoFMBaMMAKABM0KFiQhBDcGEESp4gKoB9AIJCRIXjkElN+VjtaKlAB34B3EEnxycY2zH4AnCTom7Bb8shjho-.AHNXcW9xTxDhX38ScnAkcWIwiFlwqKV8JWglEgIlYUzYMEy84vJ1KFTDcBQU3WM1ACNNYRtU3kjKMEcAD2byFBByKV5IQeoBhJsAUQBLADsAgEMFgGcVmYC5xYAbfibiCsEBwSwoShx-XrBvMe0IViGJ8BsAGUWVgBcAHRXFgFsfioAK4rJYLcqaMB3WTcfQSZJWEDKOb8RaQ7TyVpSAZgaiQvEsRAIXBE2h8LB8aj4QTJejUGi+QmICFaB4+VkSfA2Ywo4RQDmSHAChHGMwAN3RbPEAqkrC6ujgiH6mBV6AuUVg1H4MWabkSmjIwUZ1HSxRyGUVCAp0CGuKwkgKRSygiUuKUZR13EN52ksiQ5FkcpNLB0ZqOWSQeFVqp4UNdsYgxRdWUufiMpF9VhkiLANgU-Cu1yzzEkVkikxAABEC+A8ZQkoMmDmbGpenW9Wcac2kTWulQcX3qH7niAlDXYjlwBPsxWx1cJrm2FAkIJ5SPeNQAGo17gBsY2B4U7u47dAA 700,400 noborder}}
</WRAP>

The op-amp In the simulation on the right replicates a real op-amp in some respects: The voltage gain is $A_D = 100'000$. The transfer characteristic $U_A(U_D)$ shows proportional behavior only when the output value is smaller in magnitude than the supply voltage $|U_{sp}|=|U_{sm}|=15V$ (not shown). The modulation limits and the voltage gain can be changed in the simulation via "Edit component" (double click).

~~PAGEBREAK~~~CLEARFIX~~~

<WRAP right><panel type="default">
<imgcaption pic10|unipolar and bipolar care>
</imgcaption>
{{drawio>unipolar_and_bipolar_care}}
</panel></WRAP>

=== Power supply of the operational amplifier ===

For the voltage supply of the operational amplifier, a distinction is made between unipolar and bipolar:

With the **bipolar voltage supply**, the same voltage with different sign is applied to both supply terminals in terms of magnitude (<imgref pic10> (1)). This allows the output voltage $U_A$ of the amplifier to point in either direction (<imgref pic10> (2)). The power supply must be designed in such a way that it can provide both voltages. A differential voltage of $U_D=0$ also results in an output voltage of $U_A=0$.

With the **unipolar power supply**, the negative supply terminal is connected to ground (<imgref pic10> (3)). As a result, the output voltage $U_A$ of the amplifier can only take on non-negative values (<imgref pic10> (4)). In this case, the supply can be provided by a single voltage source (e.g., a battery). With a differential voltage of $U_D=0$, the output voltage is $U_A=\frac{1}{2}\cdot U_{sp}$.


~PAGEBREAK~ ~CLEARFIX~~
===== 3.3 Voltage follower =====

<WRAP right><panel type="default">
<imgcaption pic4|tension follower>
</imgcaption>
{{drawio>voltage_follower_circuit}}
</panel></WRAP>

In the chapter [[1_basics_to_amplify#feedback]] it was described that an amplifier with high open-loop gain can be "tamed" by feeding back a part of the output signal with a negative sign. In the simplest case, the output signal could be fed directly to the negative input of the operational amplifier. The input signal $U_E$ of the entire circuit is applied to the positive input.
In <imgref pic4> this circuit is shown.

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Using this circuit, the procedure for solving amplifier circuits is now to be illustrated.
  - The aim is always to create a relation between output voltage $U_A$ and input voltage $U_E$. \ <fc #800080>Thus, the goal here is the voltage gain $A_V=\frac{U_A}{U_E}$.</fc>
  - Before calculating, it should be checked how many equations describe the system and thus have to be set up. This can be determined by the **number of variables**. This is done by counting through the currents and voltages of the circuit. \\ _fc #800080>In this case, there are 3 currents and 3 voltages. So the **number of equations** needed is 6.</fc>
  - Now **equations are set up** that can be used. These are:
    - **Basic equation**: (1) $U_A = U_D \cdot A_D$
    - **Golden rules**: $R_D \rightarrow \infty$ so that (2+3) $I_p = I_m = 0$, $A_D \rightarrow \infty$, $R_A = 0$
    - Consideration of the existing **meshes**: \ <fc #800080>in this example, there is only one mesh (4) $-U_E + U_D + U_A =0$.</fc> \ **__Caution__: Meshes can __not__ enter the amplifier through an input and exit through the output! Also to be noted is the direction of $U_D$.
    - Consideration of the existing **nodes**:  \\ __fc #800080>in this example, there is only one node (5) $I_o = I_m$</fc>.
  - <fc #800080>There appears to be a missing equation. However, this is not correct, because there is still an equation hidden in the objective: (0) $A_V=\frac{U_A}{U_E}$</fc>
  - Now, to **solve the equations**, the equations must be cleverly inserted into each other in such a way that there are no dependencies on the variables left at the end.

The calculation is done here once in detail (clicking on arrow to the right "►" leads to the next step, [[rechnung_spannungsfolger|alternative Darstellung]]):
{{url>https://wiki.mexle.org/_export/revealjs/elektronische_schaltungstechnik/rechnung_spannungsfolger?theme=dokuwiki&fade=fade&controls=1&show_progress_bar=1&build_all_lists=1&show_image_borders=0&horizontal_slide_level=2&enlarge_vertical_slide_headers=0&size=2024x128#/ 1024,108 left noborder}}
~~PAGEBREAK~~ ~CLEARFIX~~

<WRAP right>{{url>https://www.falstad.com/circuit/circuitjs.html?running=false&ctz=CQAgjAnCAMB00IKw1gJldA7FsAOMiAbNIdgCyEiLJjRV2ICmAtGGAFAaYiaoipEQESgMpkQAVQD6AQXYBDEMzK5+ggMx9RIVQSV7aCIyDJpUBCJjJtEmdetzdT6MIXXDUmRGS-QB4IwR2AHcQTTVKcIxVaBCwrWhVZVVomDjk-kSlFX4yOljQ5l5+YuZcOk8+WK59TBFSyC1i8WkAUU4sJXKSvmZCcUqTSSl2gCVa+t7UwbowEwYYRbhEdjz1JVQ88FQksB3MmJBCKTApaCk8iFwpRDPYU-PUKWP1Veh1vdn99AGs-JOzhdoFcbncHlInudXgAjDZbH76DBhPSxAAeG3cYXU3FYYG46i24jwIFaAEsAHYAc3kVIAzrTSZTyfIADbsWFsET+VDqSKuNLoyxY7g8-GE8CqAAy8lpABcADq0+QAW0VMgArrTqVT2HiiRAtFtXBVinNDPk0Lh0CBRowGXKaQBjRgdbjGiLgRBNPgtWTsSngA09QNaXCHWLYug5bTFbRzfJxaMaLSCCPYJSx0rR01LdhAA 700,400 noborder}}
</WRAP>

So the voltage gain is $A_V=1$. This would also have been seen from chapter [[1 Fundamentals of amplifiers#feedback]]. There it was derived that for $A_D\rightarrow\infty$ the voltage gain just results from $k$: $A_V=\frac{1}{k}$. Since the entire output voltage is fed back here, $k=1$ and thus also $A_V=1$.

The output voltage $U_A$ is therefore equal to the input voltage $U_E$. This is where the name "voltage follower" comes from. Now one could assume, that this amplifier is of little help, because also a direct connection would deliver $U_A=U_E$. But the important thing here is: because of the operational amplifier, there is __no feedback__ from U_A to U_E. This means that a resistor on the output side will not load the input side. In the simulation on the right, the "Resistance" slider (on the right) can be used to change the load resistance. This changes the current flow, but not the voltage.

This behaviour can also be explained in another way: The input signal usually come from a voltage source, which can only produce low currents. That means the input signals are high impedance ($\text{high impedance}=\frac{\text{voltage}}{\text{low current}}$). However, a load of arbitrary impedance can be applied to the output. That is, to keep the output signal constant, a large current must be provided depending on the load. As the output resistance of the amplifier approaches 0, the signal is actually low impedance ($\text{low impedance}=\frac{\text{voltage}}{\text{u.U. large current}}$). This is where the second name of the circuit "**impedance converter**" comes from.

#~PAGEBREAK~ #~CLEARFIX~~
<wrap #steps_to_goal />

<WRAP column 100%> <panel type="danger" title="Remember: steps to the goal"> <WRAP group><WRAP column 7%>{{fa>exclamation?32}}</WRAP><WRAP column 80%>

To solve tasks, the following procedure helps:
  - Where to? Clarification of the goal (here: always the relation between output and input signal)
  - What to? Clarification of what is needed (here: always equations. Number of equations can be determined by number of variables)
  - With what? Clarification of what is already available (here: known equations: Stress amplification equation, basic equation, golden rules, mesh/node theorem, relationships of stresses and currents of components).
  - Go. Work out the solution (here:inserting the equations) \\ It helps to rearrange the equation so that $1/A_D$ appears without a prefactor. It is valid: $1/A_D \xrightarrow{A_D \rightarrow \infty} 0$

</WRAP></WRAP></panel> </WRAP>
~~PAGEBREAK~~ ~CLEARFIX~~
===== 3.4 Non-inverting amplifier =====
So far, the entire output voltage has been negative-feedback. Now only a part of the voltage is to be fed back. To do this, the output voltage can be reduced using a voltage divider $R_1+R_2$. The circuit for this can be seen in <imgref pic5>.

By considering the feedback, the result can be quickly derived here as well: only $\frac{R_2}{R_1+R_2}\cdot U_A$ is fed back from the output voltage $U_A$. So the feedback factor is $k=\frac{R_2}{R_1+R_2}$ and thus the voltage gain becomes $A_V=\frac{R_1+R_2}{R_2}$.

This "trick" via $A_V=\frac{1}{k}$ is no longer possible for some of the following circuits. Accordingly, a possible solution via network analysis is to be derived here as well.

{{page>uebung_3.4.1&nofooter}}

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<WRAP right><panel type="default">
<imgcaption pic5|non-inverting amplifier>
</imgcaption>
<button size="xs" type="link" collapse="Non-inverting_amplifier_circuit">{{icon>undo}}</button>
 
<collapse id="Non-inverting_amplifier_circuit" collapsed="true">{{drawio>Non-inverting_amplifier_circuit}}</collapse>
<collapse id="Non-inverting_amplifier_circuit" collapsed="false">{drawio>Non-inverting_amplifier_circuit2}}</collapse>
</panel></WRAP>

^Step^Description^Implementation^
|1|What is wanted?|<button size="xs" type="link" collapse="Step_1_1">{icon>eye}}</button><collapse id="Step_1_1" collapsed="true">$A_V = \frac{U_A}{U_E}=?$</collapse>|
|2|Counting the variables \ $->$ Number of equations needed|<button size="xs" type="link" collapse="Step_1_2">{{icon>eye}}</button><collapse id="Step_1_2" collapsed="true">5 voltages + 5 currents \ $->$ Number of equations needed: 10</collapse>|
|3|Setting up the equations|<button size="xs" type="link" collapse="Step_1_3">{{icon>eye}} always usable equations</button><collapse id="Step_1_3" collapsed="true">(1) Basic equation: $U_A = A_D \cdot U_D$ \ <fc #ffa500> Golden rules: \\ $R_D \rightarrow \infty$ so that (2+3) $I_p \rightarrow 0$ and $I_m \rightarrow 0$ \ $R_A = 0$ \ $A_D \rightarrow \infty$ </fc></collapse><button size="xs" type="link" collapse="Step_1_4">{{icon>eye}} Meshes and nodes</button> (see <button size="xs" type="link" collapse="Non-inverting_amplifier_circuit">{{icon>undo}}</button>)<collapse id="Step_1_4" collapsed="true">(4) Mesh I: $-U_E + U_D + U_2 = 0$ \ (5) Mesh II: $-U_2 -U_1 + U_A = 0$ \ (6) Node I: $I_o = I_1$ \ (7) Node II / voltage divider: $I_1 - I_2 - I_m = 0$</collapse><button size="xs" type="link" collapse="Step_1_5">{{icon>eye}} $U, I$ relationships across components</button><collapse id="Step_1_5" collapsed="true">(8) Resistor $R_1= \frac{U_1}{I_1}$ \ (9) Resistor $R_2= \frac{U_2}{I_2}$</collapse> |

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The calculation is done here again in detail (clicking right arrow "►" leads to the next step, [[calculation_non-inverting_amplifier|alternative representation]]):
{{url>https://wiki.mexle.org/_export/revealjs/elektronische_schaltungstechnik/rechnung_nichtinvertierender_verstaerker?theme=dokuwiki&fade=fade&controls=1&show_progress_bar=1&build_all_lists=1&show_image_borders=0&horizontal_slide_level=2&enlarge_vertical_slide_headers=0&size=2024x128#/ 1024,108 noborder}}
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<WRAP right>{{url>https://www.falstad.com/circuit/circuitjs.html?running=false&ctz=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-uBILjuJTI6WgA 700,400 noborder}}
</WRAP>

So the voltage gain of the non-inverting amplifier is $A_V=\frac{R_1+R_2}{R_2}$ or $A_V=1+\frac{R_1}{R_2}$. Thus, the numerical value $A_V$ can only become larger than 1. This is shown again in the simulation on the right. In real circuits, the resistors $R_1$ and $R_2$ will be in the range between a few $100 \Omega$ and a few $M\Omega$. If the sum of the resistors is too small, the operational amplifier will be heavily loaded. However, the output current must not exceed the maximum current. If the sum of the resistors is too large, the current $I_1=I_2$ can come into the range of the current $I_m$, which is present in the real operational amplifier.

The __**input and output resistance of the entire circuit**__ should also be considered here. Both resistors are marked here with a superscript 0 to distinguish them from the input and output resistance of the operational amplifier. The input resistance $R_{E}^0$ is given by $R_{E}^0=\frac{U_E}{I_E}$ with $I_E=I_p$. Thus, for the ideal operational amplifier, it is also true that the input resistance $R_{E}^0=\frac{U_E}{I_p} \rightarrow \infty$ becomes when $I_p \rightarrow 0$.

In the **real case** it is important in how far the total input resistance depends on the input resistance of the operational amplifier $R_{E}^0(R_D)$. This can be derived as follows: (clicking on right arrow "►" leads to the next step, [[calculation_non-inverting_amplifier_input_resistance|alternative representation]]):
{{url>https://wiki.mexle.org/_export/revealjs/elektronische_schaltungstechnik/rechnung_nichtinvertierender_verstaerker_eingangswiderstand?theme=dokuwiki&fade=fade&controls=1&show_progress_bar=1&build_all_lists=1&show_image_borders=0&horizontal_slide_level=2&enlarge_vertical_slide_headers=0&size=2024x128#/ 1024,108 noborder}}
~~PAGEBREAK~~ ~CLEARFIX~~

So it can be assumed simplistically, that the input resistance of the whole circuit is many times higher than the input resistance of the operational amplifier.
The output resistance $R_A^0$ of the whole circuit with real operational amplifiers shall only be sketched: In this case, the output resistance $R_A$ of the operational amplifier is in parallel with $R_1 + R_2$. Thus the output resistance $R_A^0$ will be somewhat smaller than $R_A$.

<WRAP column 100%>
<panel type="danger" title="Notice: non-inverting amplifier"> <WRAP group><WRAP column 7%>{{fa>exclamation?32}}</WRAP><WRAP column 80%>

For the __non-inverting amplifier__, the following holds:
  * The input voltage $U_E$ is at the __non-inverting input__ of the operational amplifier.
  * The feedback is done by a voltage divider $R_1 + R_2$
  * The voltage gain is $A_V=\frac{R_1+R_2}{R_2}$ or $A_V=1+\frac{R_1}{R_2}$ and is always greater than 1.
  * Both input and output resistances of the overall circuit are smaller than these for the (real) operational amplifier used.
</WRAP></WRAP></panel>
 </WRAP>

~~PAGEBREAK~~ ~CLEARFIX~~
===== 3.5 Inverting Amplifier =====
<WRAP right><panel type="default">
<imgcaption pic6|Inverting Amplifier>
</imgcaption>
{{drawio>Inverting_amplifier_circuit}}
</panel></WRAP>

<WRAP right><imgcaption pic8|Converting non-inverting amplifier to inverting amplifier>
{{electronic_circuitry:inv_2_ninv.gif}}
</imgcaption></WRAP>

The circuit of the inverting amplifier can be derived from that of the non-inverting amplifier (see <imgref pic8>). To do this, first consider the noninverting amplifier as a system with 3 connections (or as a quadripole): $U_E$, $GND$ and $U_A$. These terminals can be rearranged - while keeping the output terminal $U_A$. Thus the voltage divider $R_1 + R_2$ is no longer between $U_A$ and $GND$, but between $U_A$ and $U_E$, see <imgref pic6>.

In this circuit, the resistor $R_2$ is also called the negative feedback resistor.

~~PAGEBREAK~~~CLEARFIX~~~

<WRAP right>{{url>https://www.falstad.com/circuit/circuitjs.html?running=false&ctz=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-A0M1SB54KQWVVA3lIAVKmASkaKqbcGs6o0aKrWntQ6NDsd+jgpGdYBdelcpgBpdh3Y2lVWPNTiJtsIA 700,400 noborder}}
</WRAP>

Before the voltage gain is determined, the node $K1$ in <imgref pic6> is to be considered first. This is just larger than the ground potential by the voltage $U_D$; thus, it lies on the potential difference $U_D$. For a feedback amplifier with finite voltage supply, $U_A$ can only be finite and thus $U_D= U_A / A_D \rightarrow 0$ (cf. [[3_basic_circuits_i#basic_equationgolden_rules|basic_equation of the operational amplifier]]), since $A_D \rightarrow \infty$ holds. Thus it can be seen that the node $K1$ is __always__ at ground potential in the ideal operational amplifier. This property is called **virtual ground** because there is no direct short to ground. The op-amp regulates its output voltage $U_A$ in such a way that the voltage divider sets a potential of $0V$ at node $K1$. This can also be seen in the simulation by the voltage curve at $K1$.

<WRAP column 100%> <panel type="danger" title="Notice: virtual ground"> <WRAP group><WRAP column 7%>{{fa>exclamation?32}}</WRAP><WRAP column 80%>
 For the ideal feedback amplifier, $U_D \rightarrow 0$ holds. This means that the same voltage is always present at both inputs. if one of the two voltages is fixed, for example by connecting ground potential or even by a fixed voltage source, this property is called **virtual ground**.
</WRAP></WRAP></panel> </WRAP>
~~PAGEBREAK~~ ~CLEARFIX~~

<WRAP right><panel type="default">
<imgcaption pic9|Voltage divider in inverting amplifier>
</imgcaption>
{{drawio>voltage_divider_in_inverting_amplifier}}
</panel></WRAP>

For the determination of the voltage gain, the consideration of the feedback $A_V=\frac{1}{k}$ seems to be of little use at first. Instead, however, the determination via network analysis is possible. <imgref pic6> shows a possible variant to choose the meshes for this purpose. However, network analysis is not to be done here, but is given as Task 3.5.1 below.

Instead, two other ways of derivation will be shown here to bring further approaches closer.
For the first derivation, the **voltage divider** $\boldsymbol{R_1 + R_2}$ is considered.
For the unloaded voltage divider, the general rule is:

\begin{align*}
U_2 = U_G \cdot \frac{R_2}{R_1 + R_2}
\end{align*}

This equation is now to be adapted for concrete use. First <imgref pic6>, the voltages of the voltage divider can be read as given in <imgref pic9>. From this, using the general voltage divider formula:

\begin{align*}
U_2 = ( U_E - U_A ) \cdot \frac{R_2}{R_1 + R_2}
\end{align*}


With the virtual masses at node $K1$ in <imgref pic9>, it holds that $U_2$ points away from the (virtual) mass and thus equals $U_A$ in magnitude. By the same reasoning, $U_E = U_1$ holds. Thus it follows:

\begin{align*}
 - U_A = ( U_E - U_A ) \cdot \frac{R_2}{R_1 + R_2}
\end{align*}

And from that:

<WRAP left 70%>
\begin{align*}
 - U_A &= U_E \cdot \frac{R_2}{R_1 + R_2} - U_A \cdot \frac{R_2}{R_1 + R_2}  \\
 - U_A + U_A \cdot \frac{R_2}{R_1 + R_2} &= U_E \cdot \frac{R_2}{R_1 + R_2} \\
   U_A \cdot (1 - \frac{R_2}{R_1 + R_2}) &= U_E \cdot \frac{R_2}{R_1 + R_2} \\
   \frac{U_A}{U_E} &= \frac{R_2}{R_1 + R_2}}{1 - \frac{R_2}{R_1 + R_2}} \\
   \frac{U_A}{U_E} &= \frac{R_2}{R_1 + R_2 - R_2} = \frac{-R_2}{R_1} \\ \\
   \boxed{A_V = - \frac{R_2}{R_1}}
\end{align*}
</WRAP>

~~PAGEBREAK~~~CLEARFIX~~~

<WRAP right><panel type="default">
<imgcaption pic7|Inverting Amplifier - Animation>
</imgcaption>

{{url>https://www.geogebra.org/material/iframe/id/hhxhcqbp/width/600/height/700/border/888888/smb/false/stb/false/stbh/false/ai/false/asb/false/sri/false/rc/false/ld/false/sdz/false/ctl/false 400,500 noborder}}
</panel></WRAP>

For the second derivation, the current flow through the resistors $R_1$ and $R_2$ of the unloaded voltage divider is to be considered. These two currents $I_1$ and $I_2$ are just equal. Thus:

\begin{align*}
I_\boxed{}=\frac{U_\boxed{}}{R_\boxed{}}=const. \quad \text{with} \: \boxed{}=\{1,2\}
\end{align}

respectively

\begin{align*}
\frac{U_1}{R_1}=\frac{U_2}{R_2}
\end{align*}

This can also be converted into a "seesaw" or mechanical analogue via **like triangles**.
In the mechanical analog, the potentials are given by height. As in the electrical case with the ground potential, a height reference plane must be chosen in the mechanical picture. The electric currents correspond to forces (i.e., a momentum flux) - but the consideration of forces is not necessary here. [(Note1> To complete the mechanical analogue of the setup, one can assume that there is an external "force source". This always acts in such a way that it always lands on the height reference surface at the point corresponding to the virtual mass)].

Now, if a certain height (voltage $U_E$) is set, a certain height on the right side (voltage $U_A$) is obtained via force arm (resistor $R_1$) and load arm (resistor $R_2$). This is shown in <imgref pic7> above. In the figure, all points marked in red (<fc #ff0000>{{fa>circle?10}}</fc>) can be manipulated. Accordingly, the input voltage $U_E = U_{in}$ is adjustable and automatically results in a voltage $U_A=U_{out}$.
In the circuit (figure below), the resistors $R_1$ and $R_2$ can be changed.

The **input resistance of the entire circuit** $R_E^0=\frac{U_E}{I_E}$ is easily obtained by considering the input side: since $K1$ is at $0V$, $U_1 = U_E$. The complete current flowing into the input passes through resistor $R_1$. So it is then true that he input resistance is $R_E = R_1$. At the **output resistance of the whole circuit** $R_A^0$ there is again a parallel connection between the output resistance of the operational amplifier $R_A$ and the resistor $R_2$. So the output resistance will be slightly smaller than the output resistance of the operational amplifier $R_A$.

~~PAGEBREAK~~~CLEARFIX~~~

<WRAP column 100%>
<panel type="danger" title="Note: Inverting Amplifier"> <WRAP group><WRAP column 7%>{{fa>exclamation?32}}</WRAP><WRAP column 80%>

In the case of the __inverting amplifier__:
  * The input voltage $U_E$ is at the __inverting input__ of the operational amplifier.
  * The feedback is done by a voltage divider of $R_1$ and $R_2$.
  * The voltage gain is $A_V= - \frac{R_2}{R_2}$ and is always less than or greater than 0. However, the magnitude of the voltage gain can be greater than or less than 1.
  * The input resistance of the whole circuit is defined by $R_1$ and is usually smaller than the input resistance of the used (real) operational amplifier. \\ The output resistance is smaller than that of the used (real) operational amplifier.
</WRAP></WRAP></panel>
 </WRAP>

~PAGEBREAK~ ~CLEARFIX~~
====== tasks ======

{{page>uebung_3.3.1&nofooter}}
{{page>uebung_3.3.2&nofooter}}
{{page>uebung_3.3.3&nofooter}}
{{page>uebung_3.4.2&nofooter}}
{{page>uebung_3.5.1&nofooter}}
{{page>uebung_3.5.2&nofooter}}
{{page>uebung_3.5.3&nofooter}}
{{page>uebung_3.5.4&nofooter}}

~~PAGEBREAK~~ ~~CLEARFIX~~~
====== Learning Questions ======
  * Explain the difference between unipolar and bipolar power supply of an OPV.
  * Draw a sketch for bipolar and one for unipolar power supply.
  * What are the advantages and disadvantages of unipolar and bipolar supply in OPV?
  * What is the meaning of "virtual ground"? Under what circumstances is it found in opamp circuits?
  * What are the golden rules in negative feedback amplifier circuits?
  * What is the basic equation in negative feedback amplifier circuits?
  * What are impedance transformers used for?
  * How does the input voltage relate to the output voltage in an impedance converter and why?
  * What does "voltage follower" mean in the context of operational amplifiers and what are the characteristics of the voltage follower?

--> References to the media used #

^ Element ^ License ^ Link ^
| <imgref picZ>: Image with internal operational amplifier | (c) Microchip | http://ww1.microchip.com/downloads/en/appnotes/90003132a.pdf |
<--
====== 4. Basic Circuits II ======

<callout>
  * Also recommended for basic circuits II is the [[https://www.mikrocontroller.net/articles/Operationsverst%C3%A4rker-Grundschaltungen|Operational amplifier basic circuits on Microcontroller.net]]
  * [[https://rd.springer.com/book/10.1007/978-3-8348-2146-1|Teaching and Workbook Operational Amplifiers (Joachim Federau)]] [viewable via the university network]]
</callout>

<WRAP>
<callout type="info" icon="true">
=== Introductory example===

In various applications, currents must be measured. In an electric motor, for example, the torque is caused by the current flowing through the motor. A motor control and also a simple overcurrent shutdown is based on the knowledge of the current. For further processing, a voltage must be generated from the current. The simplest current to voltage converter is the ohmic resistor. A sufficiently large voltage as required by a microcontroller, for example, cannot be achieved with this. So not only the current has to be converted, but also the generated potential difference has to be amplified.

One such current sense amplifier is the [[http://www.ti.com/lit/ds/symlink/ina240.pdf|INA 240]] device. This is installed as shown below. In the simulation, a real current source feeds the electrotechnical image of a DC motor on the left (in the example: inductance with $L_L=10mH$ and internal resistance $R_L=1\Omega$). The current flowing from the motor is conducted through a measuring resistor ($R_M=0.01\Omega$) which is noticeably smaller than the internal resistance of the motor. Thus, most of the power acts in the motor and the current is only marginally affected by the sense resistor. The simulation on the right shows the inner workings of the current measuring amplifier.

The following explains ways in which such circuits can be understood.

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</WRAP>

</callout></WRAP>

<WRAP><callout>
=== Objectives for Basic Circuits II ===
After this lesson, you should:
  - Be able to apply the superpostition method to operational amplifier circuits.
  - Know how differential amplifiers and instrumentation amplifiers differ (circuit, applications, advantages and disadvantages).
  - Know what the circuit and transfer function of a voltage-to-current converter and current-to-voltage converter look like.
  - Be able to name applications for the inverse-radiator, voltage-to-current converter, and current-to-voltage converter.
</callout></WRAP>

===== 4.1 Reverse shredder =====

<WRAP right><panel type="default">
<imgcaption pic1| Reverse eraser>
</imgcaption>
\\ {{drawio>reverse-eraser}}
</panel></WRAP>

From the [[3_basic_circuits_i#inverting_amplifier|inverting amplifier]] another circuit can be derived, which can be seen in <imgref pic1>. Here, both the green part of the circuit and the purple part correspond to an inverting amplifier.

How can $U_A$ be calculated in this circuit? To do this, it is first important to understand what is being sought (compare [[electronic_circuit_engineering:3_basic_circuits_i#steps_to_the_goal|steps to the goal]]). The goal is to find the relationship between output and input signals: $U_A(U_{E1}, U_{E2})$. Different ways to get there were explained in [[electrical_engineering_1:analysis_of_dc_networks|Electrical_engineering 1: Analysis of dc_networks]]. Here we will now outline a different way.

In the case of a circuit with several sources, superposition is a suitable method, in particular the superposition of the effect of all sources in the circuit. For superposition, it must be ensured that the system behaves linearly. The circuit consists of ohmic resistors and the operational amplifier. These two components give twice the output value when the input value is doubled - they behave linearly. For superposition, the effect of the two visible voltage sources $U_{E1}$ and $U_{E2}$ must be analyzed in the present circuit. \\
In **case 1** the voltage source $U_{E1}$ must be considered - the voltage source $U_{E2}$ must be short-circuited for this purpose. The equivalent circuit formed corresponds to an inverting amplifier across $R_2$ and $R_0$. However, there is an additional resistor $R_1$ between the inputs of the operational amplifier. What is the influence of this resistor? The differential voltage $U_D$ between the inputs of the operational amplifier approaches 0. Thus, the following also applies to the current through $R_1$: $I_1^{(1)} \rightarrow 0$. Thus the circuit in case 1 is exactly an inverting amplifier. For case 1, $A_V^{(1)} = \frac{U_A^{(1)}}{U_{E1}} = - \frac{R_0}{R_1}$ and thus: $U_A^{(1)}= - \frac{R_0}{R_1} \cdot U_{E1}$. \\
Using the same procedure, **case 2** for considering the voltage source $U_2$ gives: $U_A^{(2)}= - \frac{R_0}{R_2} \cdot U_{E2}$. \\
In superposition, the effect results from the **addition of partial effects**:

$\boxed{U_A = \sum U_A^{(i)} = - (\frac{R_0}{R_2} \cdot U_{E2} + \frac{R_0}{R_1} \cdot U_{E1})}$.

Also, considering the node set for $K1$ in <imgref pic1> gives the same result.

<WRAP right>{{url>https://www.falstad.com/circuit/circuitjs.html?running=false&cct=$+1+0.00019999999999999998+0.17188692582893286+57+5+50%0Aa+384+224+480+224+8+15+-15+1000000+-0.00006516129982611548+0+100000%0Ag+384+240+384+272+0%0Ar+400+160+480+160+0+1000%0Aw+480+160+480+224+0%0Aw+384+160+384+208+0%0A368+480+160+528+160+0+0%0Aw+384+160+400+160+0%0Ar+288+208+352+208+0+1000%0AR+288+208+240+208+0+1+40+5+0+0+0.5%0AR+288+144+240+144+0+1+80+5+0+0+0.5%0Ar+288+144+352+144+0+2000%0AR+288+80+240+80+0+1+120+5+0+0+0.5%0Ar+288+80+352+80+0+4000%0Aw+384+80+384+144+0%0Aw+384+144+384+160+0%0A403+448+32+576+128+0+5_2_0_4098_10_0.1_0_2_5_3%0Aw+352+80+384+80+0%0Aw+352+208+384+208+0%0Aw+384+16+384+80+0%0Ar+288+16+352+16+0+8000%0AR+288+16+240+16+0+1+160+5+0+0+0.5%0A403+64+0+192+64+0+20_2_0_4098_5_0.1_0_2_20_3%0A403+64+64+192+128+0+11_2_0_4098_5_0.1_0_2_11_3%0A403+64+128+192+192+0+9_2_0_4098_5_0.1_0_2_9_3%0A403+64+192+192+256+0+8_2_0_4098_5_0.1_0_2_8_3%0AR+288+-48+240+-48+0+1+200+5+0+0+0.5%0Ar+288+-48+352+-48+0+16000%0Aw+352+-48+384+-48+0%0A403+64+-64+192+0+0+25_2_0_4098_5_0.1_0_2_25_3%0Aw+384+-48+384+16+0%0Aw+352+144+384+144+0%0Aw+352+16+384+16+0%0A38+7+0+1000+20000+R1%0A38+10+0+1000+20000+R2%0A38+12+0+1000+20000+R3%0A38+19+0+1000+20000+R4%0A38+26+0+1000+20000+R5%0A 800,600 noborder}}
</WRAP>

The reverse adder can be extended to any number of inputs. The simulation on the right shows the superposition of several inputs. Depending on the resistances at the different inputs, a different current flows into the circuit.

This circuit was used in analog [[https://en.wikipedia.org/wiki/Electronic_mixer|audio mixers]]. This allows to combine several signals with different gain (by the input resistors $R_i$ with $i=1, ..., n$). Furthermore, the overall gain can be changed by $R_0$. A big advantage of this circuit is also that the summation at node $K1$ is done on potential $U_D$. This means that capacitive interference with respect to the ground potential is virtually non-existent.

A very similar concept allows the construction of a [[8_continuing#digital-analog-wander_dac|digital-to-analog converter]] [[digital analog converter, DAC]].

~~PAGEBREAK~~~CLEARFIX~~~

{{page>uebung_4.1.1&nofooter}}

~~PAGEBREAK~~ ~CLEARFIX~~
===== 4.2 Differential amplifier / subtractor =====

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</WRAP>

In addition to the (reverse) adder, there is also a circuit for subtracting two input values. This circuit became the core of the introductory example. But also in the simulation below this circuit is shown in another example: In this case, a [[https://de.wikipedia.org/wiki/Symmetrische_Signal%C3%BCbertragung|differential input signal]] is shown on the left. Differential means that the signal on one line is __not__ transmitted with respect to a reference voltage (usually ground potential) on a second line. Instead, the signal is transmitted to both lines in opposite directions. If a disturbance acts equally on both lines (which is often the case when lines are close to each other), the effect of the disturbance can be eliminated by forming the difference.

How can the relationship $U_A(U_{E1}, U_{E2})$ between output and input signals be determined for this circuit?

<WRAP right><panel type="default">
<imgcaption pic2| differential amplifier>
</imgcaption>
\\ {{drawio>Difference Amplifier}}
</panel></WRAP>

Again, various network analysis concepts could be used to look at the circuit (e.g. superposition or mesh and node sets). Again, another possibility is to split the circuit as color-coded in the <imgref pic2>. \\
The __green part__ shows a voltage divider $R2 + R4$. Since the input resistance of the operational amplifier is very large, this voltage divider is unloaded. The voltage at node $K2$ or at the noninverting input $U_p$ is just given by the voltage divider: $U_p = U_{E2}\cdot \frac{R_4}{R_2+R_4}$. \\
The __violet part__ corresponds to an inverting amplifier, but the voltage at the node $K1$ or at the inverting input $U_m$ is just equal to $U_p$ due to the feedback, since $U_D \rightarrow \infty$. Thus, the current flowing into node $K1$ via $R_1$ results from $I_1=\frac{U_{E1} - U_p}{R_1}$. The output voltage is given by $U_A = U_p - U_3$, where the voltage $U_3$ is given by the resistance $R_3$ and the current through $R_3$. The current through $R_3$ is just the same as the current through $R_1$, i.e. $I_1$.

The result is: \\
$U_A = U_{E2}\cdot \frac{R_4}{R_2+R_4} - R_3 \cdot \frac{U_{E1} - U_p}{R_1} $ \\
$U_A= U_{E2}\cdot \frac{R_4}{R_2+R_4} - U_{E1} \cdot \frac{R_3}{R_1} + U_{E2} \cdot (\frac{R_3}{R_1}\cdot \frac{R_4}{R_2+R_4})$
$\boxed{U_A= U_{E2}\cdot \frac{R_4}{R_2+R_4}  \frac{R_1+R_3}{R_1} - U_{E1} \cdot \frac{R_3}{R_1}$


~~PAGEBREAK~~~CLEARFIX~~~

<WRAP right><panel type="default">
<imgcaption pic3|differential amplifier - animation>
</imgcaption>

<collapse id="foo" collapsed="true"><well>
{{url>https://www.geogebra.org/material/iframe/id/msxjcgz4/width/450/height/300/border/888888/smb/false/stb/false/stbh/false/ai/false/asb/false/sri/false/rc/false/ld/false/sdz/false/ctl/false 450,300 noborder}}
</well></collapse>

<collapse id="foo" collapsed="false">
<button type="warning" collapse="foo">To view the animation: click here!</button>
</collapse>
 


</panel></WRAP>

Two simplifications should be considered here:
  - If $R_1 = R_2$ and $R_3 = R_4$ are chosen, the equation further simplifies to: \boxed{U_A= U_{E2}\cdot \frac{R_3}{R_1} - U_{E1} \cdot \frac{R_3}{R_1} = \frac{R_3}{R_1}\cdot(U_{E2}-U_{E1})}$. \ This variant can be found in various measurement circuits. \\ \\
  - Alternatively, if $R_1 = R_3$ and $R_2 = R_4$ is chosen, the result is: \ (nbsp) $\boxed{U_A= U_{E2}-U_{E1}}$ \ This would also result in case 1. if $R_1 = R_2 =R_3 = R_4$ is chosen.

The animation on the right shows how the 2nd case would result with similar triangles. The connection of the two rockers at the point $K_1 K_2$ is caused by the operational amplifier, through which the voltage $U_p$ and $U_m$ converge to $U_D \rightarrow 0$.

A big advantage of this circuit is that even very large voltages can be used as input voltage if $R_1 \gg R_3$ and $R_2 \gg R_4$ are chosen. This would divide the input voltages down and display a fraction of the difference as the result. The main drawback of the circuit is that the gain/attenuation depends on more than one resistor. This makes a quick choice of gain difficult.

~PAGEBREAK~ ~CLEARFIX~~
===== 4.3 Instrumentation Amplifier =====

<WRAP right><panel type="default">
<imgcaption pic3| instrumentation amplifier>
</imgcaption>
\\ {{drawio>Instrument Amplifier }}
</panel></WRAP>

~~PAGEBREAK~~~CLEARFIX~~~

{{page>uebung_4.3.1&nofooter}}

<WRAP right>{{url>https://www.falstad.com/circuit/circuitjs.html?running=false&cct=$+17+0.00019999999999999998+0.26073472713092677+56+5+50%0Aa+288+192+432+192+8+15+-15+1000000+1.3742754673027664+1.3743254638029614+100000%0Aw+288+64+288+128+0%0Aw+432+144+432+64+0%0Ar+288+64+432+64+0+1000%0Ar+192+64+288+64+0+1000%0Ar+320+320+400+320+0+1000%0Ag+432+320+432+336+0%0Ar+288+320+192+320+0+1000%0AR+96+48+32+48+0+1+60+5+0+0+0.5%0AO+432+192+496+192+0%0Av+64+336+0+336+0+2+60+1+0+0+0.5%0AR+0+336+-32+336+0+1+60+5+0+0+0.5%0Aa+96+320+192+320+8+15+-15+1000000+0.748750922592489+0.748778409101765+100000%0Aw+288+320+288+256+0%0Aw+320+320+288+320+0%0Aw+432+320+400+320+0%0Ar+192+320+192+272+0+1000%0Ar+192+64+192+112+0+1000%0Aa+96+64+192+64+9+15+-15+1000000+-0.25119907990738605+-0.2512215908982351+100000%0Aw+96+80+96+112+0%0Aw+96+112+192+112+0%0Aw+192+272+96+272+0%0Aw+96+272+96+304+0%0A174+192+160+160+208+0+1000+0.5+Resistance%0Aw+432+176+432+192+0%0Aw+288+160+288+176+0%0Aw+288+208+288+224+0%0Aw+192+112+192+128+0%0Aw+192+128+160+128+0%0Aw+160+128+160+176+0%0Aw+96+336+64+336+0%0Aw+160+176+160+192+0%0Aw+192+224+192+272+0%0Aw+288+224+288+256+0%0Aw+288+160+288+128+0%0Aw+432+144+432+176+0%0A403+416+208+544+272+0+9_1_0_4098_5_0.0000244140625_1_1%0A403+-64+240+64+304+0+30_1_0_4102_5_0.1_1_2_30_3%0A403+-64+80+64+144+0+8_1_0_4098_7_0.025_0_2_8_3%0A 700,400 noborder}}
</WRAP>


~~PAGEBREAK~~ ~CLEARFIX~~
===== 4.4 Current-Voltage-Converter =====

<WRAP right><panel type="default">
<imgcaption pic4| current-voltage converter>
</imgcaption>
\\ {{drawio>current-voltage converter}}
</panel></WRAP>

<WRAP right>{{url>https://www.falstad.com/circuit/circuitjs.html?running=false&cct=$+1+0.0005+0.03528660814588489+57+5+50%0Aa+384+224+480+224+8+15+-15+1000000+0.00009999900000999991+0+100000%0Ag+384+240+384+272+0%0Ar+384+160+480+160+0+1000%0Aw+480+160+480+224+0%0Aw+384+160+384+208+0%0A370+320+208+384+208+1+0%0Ai+272+208+320+208+0+0.01%0Aw+272+208+272+256+0%0Ag+272+256+272+272+0%0Ax+258+167+332+170+4+14+Stromquelle%0A368+480+224+528+224+0+0%0A38+6+0+-0.01+0.01+Strom%5Csder%5CsStromquelle%0A 500,400 noborder}}
</WRAP>

~~PAGEBREAK~~~CLEARFIX~~~

In <imgref pic4> you can see the circuit of a current-voltage converter. The current-to-voltage converter changes its __output voltage__ based on an __input current__. This circuit is also called a [[https://de.wikipedia.org/wiki/Transimpedanzverst%C3%A4rker|transimpedance amplifier]] because here the transfer resistance - that is, the transimpedance - represents the gain. Generally, the gain was expressed as $A={ {output} \{input} }$. In the case of the current-to-voltage converter, the gain is defined as.

$$ R = {{U_{out}} \over I_{in}} = - R_1 $$

$R_1$ is the resistor used in the circuit.

In the simulation, the slider on the right ("Current of current source") can be varied. This changes the input current and thus also the output voltage.
 
This circuit can be used, for example, to read a [[https://de.wikipedia.org/wiki/Photodiode#Betrieb_im_Quasi-Kurzschluss|photodiode in volt-free circuit]] ([[https://de.wikipedia.org/wiki/Transimpedanzverst%C3%A4rker#Anwendung|further explanation]] and integrated circuit {{electronic_circuit_technology:tsl250r.pdf}}).


~~PAGEBREAK~~ ~CLEARFIX~~
===== 4.5 Voltage-to-Current Converter =====

<WRAP right><panel type="default">
<imgcaption pic5| current-voltage converter>
</imgcaption>
\\ {{drawio>voltage-to-current converter}}
</panel></WRAP>

<WRAP right>{{url>https://www.falstad.com/circuit/circuitjs.html?running=false&cct=$+1+0.0005+0.20306040966347483+58+5+50%0Ag+416+368+416+384+0%0Ag+256+304+256+336+0%0Aa+304+256+416+256+9+15+-15+1000000+1.2434121333518957+1.2434494357158963+100000%0A370+416+256+448+256+1+0%0Ar+416+304+416+368+0+500%0Aw+416+304+304+304+0%0Aw+304+304+304+272+0%0Av+256+240+256+304+0+1+40+5+0+0+0.5%0Aw+304+240+256+240+0%0Ar+512+256+512+304+0+1000%0Aw+416+304+512+304+0%0Aw+512+256+448+256+0%0Ab+480+224+544+335+0%0Ax+470+351+564+354+4+14+Lastwiderstand%0A38+4+0+500+10000+Widerstand%0A 500,400 noborder}}
</WRAP>

~~PAGEBREAK~~~CLEARFIX~~~

Next, consider the voltage-to-current converter. With this, an output__current__is set proportional to an input__voltage__.

Here, the general gain $A={ {output} \{input} }$ to

$$ S ={{I_{out}} \over U_{in}} $$

The quantity $S$ is called the transmission slope or the transmission conductance.

This circuit can be used, for example, to generate a voltage-regulated current source.

~~PAGEBREAK~~~CLEARFIX~~~

====== Tasks ======

{{page>exercise sheet6&nofooter}}


====== Further Reading ======
  * [[https://en.wikipedia.org/wiki/Low-dropout_regulator| Low Dropout Controller]]

====== Learning Questions ======
  * State applications for the reverse adder.
  * Explain the working of a current to voltage converter.
  * Name 3 applications for an operational amplifier.
====== 5 Filter Circuits ======

<callout>
  * Also recommended for basic circuits II is the [[https://www.mikrocontroller.net/articles/Operationsverst%C3%A4rker-Grundschaltungen|Operational amplifier basic circuits on Microcontroller.net]]
  * [[https://rd.springer.com/book/10.1007/978-3-8348-2146-1|Teaching and Workbook Operational Amplifiers (Joachim Federau)]] [viewable via the university network]]
</callout>

<WRAP>
<callout type="info" icon="true">
=== Introductory example===

<WRAP right>{{url>https://www.falstad.com/circuit/circuitjs.html?running=false&cct=$+1+0.0000999999999999+2.008553692318767+41+5+43%0Aa+288+128+384+128+8+15+-15+1000000+-0.000058679015510918+0+100000%0Aw+384+112+384+64+0%0Ac+384+64+288+64+0+5.000000000000001e-7+5.867960230107311%0Aw+288+64+288+112+0%0Ar+208+112+288+112+0+1000%0Ar+384+32+288+32+0+1000%0Aw+288+32+288+64+0%0Aw+384+32+384+64+0%0Ag+288+144+288+176+0%0AR+96+128+64+128+0+2+40+5+0+0+0.5%0A207+384+128+384+176+4+U_A1%0A403+304+208+448+272+0+10_4_0_12294_7.438926379638319_0.0001_0_2_10_3_U%5CsA1%0A170+144+96+144+64+2+20+1000+5+0.1%0AS+176+112+144+112+0+0+false+0+2%0Aw+176+112+208+112+0%0A403+160+208+288+272+0+4_4_0_12294_9.983191019672175_0.0001_0_2_4_3_U%5CsE%5Csverrauscht%0A207+96+128+96+160+4+U_E%0Av+144+128+96+128+0+6+4000+5+0+0+0.5%0A207+560+128+560+176+4+U_A2%0Ag+464+144+464+176+0%0Aw+560+32+560+64+0%0Aw+464+32+464+64+0%0Ar+560+32+464+32+0+1000%0Ar+384+112+464+112+0+1000%0Aw+464+64+464+112+0%0Ac+560+64+464+64+0+0.0000015+-5.260573781691551%0Aw+560+128+560+64+0%0Aa+464+128+560+128+8+15+-15+1000000+0.00005260521176479787+0+100000%0Aw+384+112+384+128+0%0A403+480+208+640+272+0+18_4_0_4102_8.4590010899438_0.0001_0_2_18_3_U%5CsA2%0A403+0+208+144+272+0+16_4_0_12294_5.000000000000001_0.0001_0_2_16_3%0A 900,400 noborder}}
</WRAP>

In various applications in harsh environments (e.g. sensors in the engine compartment or in industrial environments, communication with satellites), the clear digital transmit signals become noisy signals at the receiver. In the simulation on the right, the left oscilloscope image shows the original signal. The second oscilloscope image shows the noisy signal.

One possibility to process such signals is the use of filters. Filters have already been described in [[:electrical_engineering_2:networks_at_variable_frequency|electrical_engineering 2]]. The classical $R C$-filters are passive. This means that although the voltage value is filtered, the output current of the filter is always lower than the current measured at the input. To enable better filtering and subsequent use of the signal, active filters can be used. These can be built up by operational amplifiers.
 
Two low-pass filters are shown in the simulation. These attenuate the high-frequency components in the signal. The signal $U_{A1}$ after the first filter stage already shows significantly less noise. In the signal $U_{A2}$, even less noise is visible, but the rising and falling edges are also no longer displayed sharply.

With the switch (left in the simulation) a frequency-variable test signal ([[https://de.wikipedia.org/wiki/Sweep_(Signal Processing)|Sweep]]) can be fed in. This clearly shows that the filter produces a smaller amplitude at the filter output from high-frequency oscillations at the input.

This chapter will explain the basics of active high and low pass filters from operational amplifiers.

</callout></WRAP>

<WRAP><callout>
=== Objectives for Basic Circuits II ===
After this lesson, you should:
  - Be able to apply the superpostition method to operational amplifier circuits.
  - Know how differential amplifiers and instrumentation amplifiers differ (circuit, applications, advantages and disadvantages).
  - Know what the circuit and transfer function of a voltage-to-current converter and current-to-voltage converter look like.
  - Be able to name applications for the inverse-radiator, voltage-to-current converter, and current-to-voltage converter.
</callout></WRAP>

===== 5.0 Representation of numerical values =====

In order to be able to analyse filter circuits, various possibilities for representing the numerical values should be explained beforehand.

==== 5.0.1 The dB measure ====

The decibel measure is an auxiliary unit of measure that facilitates handling with ratios (e.g. $U_2/U_1$). These ratios are called **level** in engineering. The level makes it possible to refer to a reference quantity. In electronic circuitry, the decibel is used as a dimensionless unit for current or voltage ratios. In the future, this will be particularly interesting for the amplification $A_V = \frac{U_A}{U_E}$ and factors.

The conversion to a level in $dB$ is defined for current or voltage ratios by the following equation:


$\boxed{A_V^{dB}=20 dB \cdot log_{10}\left(\frac{U_2}{U_1}\right)=20 dB \cdot log_{10}A_V}$(nbsp)(nbsp)(nbsp) resp. (nbsp)(nbsp)(nbsp) $A_C^{dB}=20 dB \cdot log_{10}\left(\frac{I_2}{I_1}\right)$

\\
=== Technical levels in dB ===
<WRAP right>

^ Name ^ Symbol ^ Formula ^ Reference value for 0dB ^
| [[wpde>voltage level]]         | $dBV$ | $20dB \cdot log_{10}(V/V_{ref})$ | $0dBV \widehat{=} 1V$ |
| [[wpde>power level]]         | $dBm$ | $10dB \cdot log_{10}(P/P_{ref})$ | $0dBm \widehat{=} 1mW$ |
| [[wpde>power level]]         | $dBW$ | $10dB \cdot log_{10}(P/P_{ref})$ | $0dBW \widehat{=} 1W$ |
| [[wpde>dBFS|full-scale level]]  | $dBFS$ | $20dB \cdot log_{10}(V/V_{max})$ | $0dBFS \widehat{=} V_{max}$ |
| [[wpde>sound pressure level]]       | $dBA$ | $20dB \cdot log_{10}(p/p_{ref})$ | $0dBA \widehat{=} 20\mu Pa$ |
</WRAP>

Note that this equation changes somewhat for __power__ quantities, i.e. ratios of $P$. If $P \sim U^2$ or $U \sim P^\frac{1}{2}$ is considered, then we get: \cdot $A_P^{dB}=20 dB \cdot log_{10}\left(\frac{P_2^\frac{1}{2}}{P_1^\frac{1}{2}}\right)$\ (nbsp)(nbsp)(nbsp)(nbsp)(nbsp)$ =\color{blue}{20 dB \cdot \frac{1}{2}} \cdot log_{10}\left(\frac{P_2}{P_1}\right) =\color{blue}{10 dB} \cdot log_{10}\left(\frac{P_2}{P_1}\right) $

To the right of the table are various levels frequently used in engineering. In the following, only the voltage level is used and indicated with the symbol $dB$.

\\
=== Simple examples of voltage levels in dB===
<WRAP right>
^ linear factor ^ level [$dB$] ^
| $\times 0.01$ | $-40dB$ |
| $\times 0.1$ | $-20dB$ |
| $\times 1$ | $0dB$ |
| $\times 2$ | $\approx +6dB$ |
| $\times 10$ | $+20dB$ |
| $\times 100$ | $+40dB$ |
</WRAP>

By this equation, various linear factors and ratios $A = \boxed{}_2 / \boxed{}_1$ can be converted into a level $A^{dB}$ in $dB$. Occasionally, the superscript $\boxed{}^{dB}$ is omitted below, in which case the level is denoted by the unit after the numerical value.

Examples:
  - For $A_V= \color{green}{1} $ we get $ A_V^{dB}(\color{green}{1}) = 20 dB \cdot \underbrace{log_{10}\left(\color{green}{1}\right)}_\color{blue}{x} \qquad \qquad \rightarrow \boldsymbol{A_V^{dB}(\color{green}{1})} \quad = 20 dB \cdot 0 \quad \ \boldsymbol{= 0 dB}$ \ Here $\color{blue}{x}$ has the value that gives $10^\color{blue}{x} = \color{green}{1}$, so just $\color{blue}{x}=0$. \\ \\ \\
  - For $A_V= \color{green}{0,01} $ we get $ A_V^{dB}(\color{green}{0,01}) = 20 dB \cdot \underbrace{log_{10}\left(\color{green}{0,01}\right)}_\color{blue}{x} \qquad \rightarrow \boldsymbol{A_V^{dB}(\color{green}{0,01})} = 20 dB \cdot (-2) \boldsymbol{= -40 dB}$ \ Here $\color{blue}{x}$ has the value, that gives $10^\color{blue}{x} = \color{green}{0.01}$, so even $\color{blue}{x}=-2$. \\ \\ \\
  - For $A_V= \color{green}{2} $, we get $ A_V^{dB}(\color{green}{2}) = 20 dB \cdot \underbrace{log_{10}\left(\color{green}{2}\right)}_\color{blue}{x} \qquad \qquad \rightarrow \boldsymbol{A_V^{dB}(\color{green}{2})} \quad\approx 20 dB \cdot 0.30103 \boldsymbol{\approx 6 dB}$ \here $\color{blue}{x}$ has the value that gives $10^\color{blue}{x} = \color{green}{2}$. Thus, $\color{blue}{x}\approx equals 0.30103$. \\ \\ \\

=== Use of the dB measure ===

The decibel offers some advantages, which are used in the filter elements considered below:
  * **handier numerical values**: If very large or very small linear values are needed, the number of the resulting level has fewer digits. Example: $A_V = 10000000 \rightarrow A_V^{dB}= 140dB$. This also results in less "zero counting".
  * **Relationship to Sensory Perceptions**: Sensory perceptions such as brightness and loudness have an almost exponential effect. That is, any tenfold increase in the underlying physical quantity (number of photons or sound pressure) does not have ten times the effect, but seems to have an additive effect.
  * **easier math**: The logarithm in the defining equation turns any multiplication of linear factors into an addition of levels: \A_V^{dB}(A_1 \cdot A_2) = 20dB \cdot log_{10}(A_1 \cdot A_2) = 20dB \cdot log_{10}(A_1) + 20dB \cdot log_{10}(A_2) = A_V^{dB}(A_1) + A_V^{dB}(A_2) $

~~PAGEBREAK~~~CLEARFIX~~~

<WRAP right><panel type="default">
<imgcaption pic0| example: filter in row>
</imgcaption>
\\ {{drawio>example_filter_in_row}}
</panel></WRAP>

Especially the last point of the calculation should be considered again. In <imgref pic0> several amplifiers connected in series can be seen with exemplary voltage gain values. \\ The total gain here is the product of the individual gains: $A_{V,ges}=\prod A_i = A_1 \cdot A_2 \cdot A_3$. \\ The determination of the total gain was rather laborious before the times of the pocket calculator due to the multiplications. For the levels, the result is an addition: $A_{V,ges}^{dB}=\sum A_i^{dB} = A_1^{dB} + A_2^{dB} + A_3^{dB}$. \\ Here this would be: $A_{V,ges}^{dB}=\sum A^{dB} = 88dB + (-58dB) + 14dB = 44dB$.


<WRAP column 100%> <panel type="danger" title="Remember: dB measure"> <WRAP group><WRAP column 7%>{{fa>exclamation?32}}</WRAP><WRAP column 80%>

For current and voltage levels:
  - A linear factor of $\color{green}{\times 10}$ results in level $+ 20dB$.
  - A linear factor of $\color{green}{\times 2}$ results in a level of $+ 6dB$.
  - The linear value $A_V = 1$ corresponds to $0 dB$.

For systems connected in series, to determine the amplification
  - multiply the linear measure $A_V$ and
  - add the level $A_V^{dB}$.

</WRAP></WRAP></panel> </WRAP>

~~PAGEBREAK~~~CLEARFIX~~~

\\
=== more difficult examples of voltage level in dB===

With this knowledge, the interpolation points $\color{green}{\times 10} \rightarrow + 20dB$ and $\color{green}{\times 2} \rightarrow + 6dB$ the linear values can easily be determined from a level in $dB$ without a calculator.

Examples:
  - $A_V^{dB}=58dB$ \color{blue}{2}\cdot: $A_V^{dB}=58dB = 40dB + 18dB = \color{blue}{2}\cdot 20dB + \color{magenta}{3}\cdot 6dB$ \ This becomes linear to $ \qquad \qquad \qquad \qquad \ A_V = 10^\color{blue}{2} \qquad \cdot \qquad 2^\color{magenta}{3} \qquad = 100 \cdot 8 = 800$ \\ \
  - $A_V^{dB}=56dB$ \ with interpolation points: $A_V^{dB}=56dB = 80dB - 24dB = \color{blue}{4}\cdot 20dB + \color{magenta}{-4}\cdot 6dB$ \ This becomes linear to $ \qquad \qquad \qquad \qquad \ A_V = 10^\color{blue}{4} \qquad \cdot \qquad 2^\color{magenta}{-4} \qquad = 10000 \cdot \frac{1}{16} = 625$ \ or $A_V^{dB} = 20dB + 36dB \rightarrow A_V= 10^\color{blue}{1} \cdot 2^\color{magenta}{6} = 10 \cdot 64 = 640$ \rightarrow
  - $A_V^{dB}=55dB$ \ with interpolation points: $A_V^{dB}=56dB = 40dB + 18dB - 3dB = \color{blue}{2}\cdot 20dB + \color{magenta}{3}\cdot 6dB + \color{teal}{-\frac{1}{2}}\cdot 6dB$ \ This becomes linear to $ \qquad \qquad \qquad \qquad \qquad \ A_V = 10^\color{blue}{2} \qquad \cdot \qquad 2^\color{magenta}{3}  \qquad \cdot \qquad 2^\color{teal}{-\frac{1}{2}} \approx 100 \cdot 8 \cdot 0,707 = 560$

\\
The value $-3dB$ will still be used in the following.

~~PAGEBREAK~~~CLEARFIX~~~

==== 5.0.2 The Bode diagram ====

<WRAP right><panel type="default">
<imgcaption picA| Representation of complex numbers>
</imgcaption>
\\ {{drawio>representation_of_complex_numbers}}
</panel></WRAP>

The aim of the floor diagram is to show the transmission behaviour of systems clearly and concisely.

=== Preliminary consideration: complex numbers ===

A complex number can always be reduced to two real number values. For the exact definition of these numerical values there are different possibilities (<imgref picA>):
  - Definition over real part $\Re(\underline{A}_V)=A_V \cdot cos(\varphi)$ and imaginary part $\Im(\underline{A}_V)=A_V \cdot sin(\varphi)$ in $\underline{A}_V= \Re(\underline{A}_V) + j \cdot \Im(\underline{A}_V)$
  - Definition over magnitude $A_V = |\underline{A}_V|$ and phase $\varphi = arctan \left( \frac{\Im(\underline{A}_V)}{\Re(\underline{A}_V)} \right)$ in $\underline{A}_V=A_V \cdot e^{j \varphi}$

The 2nd definition is more appropriate when considering frequency-dependent voltage gain, since it allows the "time shift" (phase) to be separated from the gain.

~~PAGEBREAK~~~CLEARFIX~~~

=== path to the floor diagram ===

<WRAP right>{{url>https://www.falstad.com/afilter/circuitjs.html?running=false&cct=$+1+0.000005+5+44+0.5+50%0A%25+1+8761050.782000184%0Ac+240+160+240+48+0+1e-9+0%0Ar+112+48+240+48+0+16000%0AO+240+48+352+48+1%0Ag+240+160+240+192+0%0A170+112+48+80+48+3+20+1000+50+0.1%0Ax+326+77+343+80+0+24+U%0Ax+339+86+355+89+0+24+A%0Ax+114+88+130+91+0+24+E%0Ax+101+79+118+82+0+24+U%0Ao+4+16+0+34+5+0.00009765625+0+-1+in%0Ao+2+16+0+34+2.5+0.00009765625+1+-1+out%0A 600,400 noborder}}
</WRAP>

To better understand the frequency dependence of the voltage gain, it can be plotted as $|A_V(f)|$ as a function of frequency. It is useful to represent the voltage gain as level $|A_V^{dB}(f)|$. The simulation on the right shows the $|A_V^{dB}(f)|$ curve for a lowpass filter in the lower half of the image. The curve starts at $0dB$ on the left and drops to about $-45dB$. If the mouse is dragged over the diagram, the gain value in $dB$ can be displayed for each frequency. Clicking on the curve displays current flow and voltage ratios. Only clicking on the leftmost frequency range results in a situation where $U_A$ assumes noticeably high voltages (visible via the color of the line).  

In this type of display, it is difficult to determine values such as the cut-off frequency from the frequency curve.
However, if the frequency axis is also logarithmized, a different picture results. This is done via ''Options'' >> ''Linear Scale''. Now the initially flat course of the voltage gain for low frequencies and the drop for higher frequencies becomes very clear. Also the cut-off frequency can be read in the "kink".

The phase can be made visible via ''Options'' >> ''Show Phase''.

== SYNC, CORRECTED BY ELDERMAN ==
=== Description of the floor diagram ===

<WRAP right><panel type="default">
<imgcaption picB| Principle image of the floor diagram>
</imgcaption>
\\ {{drawio>principle_image_of_the_body_diagram}}
</panel></WRAP>

The bottom diagram allows the **representation of a complex-valued, frequency-dependent quantity in logarithmic form**. It is also referred to as the **frequency response** and is divided into (cf. <imgref picB>):
  * the **amplitude response** which represents amplitude in double logarithmic form (dB level is a logarithmic representation); and
  * the **phase response** which represents the phase single logarithmically.

A jump in frequency by a factor of $\times 10$ is called a **decade** (abbreviated Dec.). \\
The amplitude response of various functions will be briefly discussed ($\mathcal{C}$ is an arbitrary frequency-independent factor):
  - $|A_V(f)| = \mathcal{C} \cdot f$: If a function is considered that increases linearly with $f$, then a tenfold increase in frequency results in a tenfold increase in voltage gain. This results in an increase of $+20dB$ per decade. \\ \\
  - $|A_V(f)| = \mathcal{C} / f$: If a function is considered that is reciprocal to $f$, then a tenfold increase in frequency results in a decrease in voltage gain to one-tenth. This results in a drop of $-20dB$ per decade (cf. <imgref picB> at high frequencies). \\ \\
  - $|A_V(f)| = \mathcal{C} / f^n$: Considering a function reciprocal to $f^n$, a tenfold increase in frequency results in a drop in voltage gain to one $1/10^n$th. This results in a drop of $-20dB \cdot n$ per decade.

As an alternative to the actual course, $|A_V(f)|$ and $\varphi(f)$ are occasionally represented idealized with straight line segments.

~~PAGEBREAK~~~CLEARFIX~~
<WRAP column 100%> <panel type="danger" title="Remember: floor diagram"> <WRAP group><WRAP column 7%>{{fa>exclamation?32}}</WRAP><WRAP column 80%>

The bottom diagram (= frequency response) consists of:
  - Amplitude response: amplitude in dB over logarithmic frequency (i.e. double logarithmic representation).
  - Phase response: linear phase over logarithmic frequency (i.e. single logarithmic representation)

This gives a straight line in the amplitude response for functions of the form $A(\omega) \sim \omega^n$. \\
In particular, this is true for $A(\omega) \sim \omega$, i.e., a slope of +20dB/decade, and for $A(\omega) \sim \frac{1}{\omega}$, i.e., a slope of -20dB/decade
</WRAP></WRAP></panel> </WRAP>
~~PAGEBREAK~~ ~CLEARFIX~~


===== 5.1 Reverse integrator =====

<WRAP right><panel type="default">
<imgcaption pic1| filter circuits>
</imgcaption>
\\ {{drawio>general_filter_circuits}}
</panel></WRAP>

Up to now, operational amplifier circuits have been considered in which negative feedback took place via ohmic resistors. In the following, operational amplifier circuits with storing components ($C$, $L$) will be analyzed.

In electronics, inductors are rarely used for this purpose. This has several reasons:
  - Inductors are possible in integrated circuits, but are somewhat more difficult to calculate as such an element.
  - Inductors require a current source as current storage. The internal resistance results in a continuous power loss.

Instead of inductors, capacitors are used in microelectronics and filter technology for sensor signals. The passive circuits formed are correspondingly also called $R$-$C$-links and were analyzed in [[:elektrotechnik_2:start|Elektrotechnik 2]].

The following basic circuit is a modified, [[3_basic_circuits_i#inverting_amplifier|inverted amplifier]] (<imgref pic1>) in which one or both ohmic resistors are replaced by (complex-valued) impedances.

For first circuit only the part between output voltage $U_A$ and virtual ground should be replaced by a capacitor (<imgref pic2>).

~~PAGEBREAK~~~CLEARFIX~~~

==== 5.1.1 Circuit analysis with differential equations ====

<WRAP right>{{url>https://www.falstad.com/circuit/circuitjs.html?running=false&cct=$+1+0.00009999999999999999+0.32112705431535615+57+5+50%0Ag+96+224+96+240+0%0Aw+336+112+336+160+0%0Aw+192+112+192+144+0%0Aa+192+160+336+160+8+15+-15+1000000+-0.00007695554889407606+0+100000%0Ac+192+112+336+112+0+0.0000058+-7.695631844956501%0AO+336+160+400+160+0%0Av+96+224+96+160+0+2+40+2+0+3.141592653589793+0.5%0Av+96+160+96+112+0+2+80+1+0+0+0.5%0Aw+96+112+128+112+0%0Ar+128+112+192+112+0+1000%0Ag+192+176+192+192+0%0A403+304+16+432+80+0+5_4_0_12290_8.523156660430175_0.0001_1_1%0A403+16+16+144+80+0+8_4_0_12290_3_0.0001_0_2_8_3%0Ax+194+97+210+100+4+24+K%0Ax+209+105+222+108+4+24+1%0A 600,400 noborder}}
</WRAP>

The first active filter circuit can be seen on the right side of the simulation. The superimposed square-wave voltage sources result in a step function as input voltage. This generates a current via the RC element. If we now look at the output voltage in comparison to the input voltage, we can see that:
  - for each constant input value $U_E \neq 0$ an output value with a fixed slope results and
  - for each positive input value $U_E > 0$ a negative slope results, for a negative input value a positive slope results.

The circuit thus created is called a **reverse integrator**.

If you look at the circuit, you can see that the node $K_1$ is at virtual ground. With a constant input voltage, the input current is therefore constant and defined only by the resistance. Thus, the capacitor charges with a constant current; the charge increases linearly. The voltage across the capacitor therefore also increases linearly.

~~PAGEBREAK~~ ~CLEARFIX~~
<WRAP right><panel type="default">
<imgcaption pic2| reverse integrator>
</imgcaption>
\\ {{drawio>reverse integrator}}
</panel></WRAP>

Just as in the basic circuits, the relationship between the output value and the input value is now to be determined mathematically. The meshes are drawn in <imgref pic2> for this purpose. The **transfer function $U_A = f(U_E)$** is now to be determined.

$A_V = ? \quad -> \quad U_A = f(U_E) $

=== given equations ===

Given the following equations:

|I.|Basic equation|$U_A = A_D \cdot U_D$|
|II.|mesh 1|$ -U_E+U_R-U_D=0 $|
|III.|mesh 2|$U_D+U_C+U_A=0$|
|IV.|Mesh|$I_R=I_C$|
|V.|Capacity C|$C= { Q \over U_C } = { 1 \over U_C }\cdot(\int_{t_0}^{t_1} I_C dt+ Q_0(t_0)) $|
|VI.|Resistance R|$R = { U_R \over I_R }$|

=== Derivation ===

The calculation is performed once in detail here (clicking on right arrow "►" leads to the next step, [[calculation_reverse_integrator|alternative representation]]):
{{url>https://wiki.mexle.org/_export/revealjs/elektronische_schaltungstechnik/rechnung_umkehrintegrator?theme=dokuwiki&fade=fade&controls=1&show_progress_bar=1&build_all_lists=1&show_image_borders=0&horizontal_slide_level=2&enlarge_vertical_slide_headers=0&size=2024x128#/ 1024,108 left noborder}}

~~PAGEBREAK~~~CLEARFIX~~~
\\

==== 5.1.2 Signal Time History ====

<WRAP right><panel type="default">
<imgcaption pic3| Example of a signal-time history of the reverse integrator>
</imgcaption>
\\ {{drawio>Signal_Time_Progression_of_the_Reverse_Integrator}}
</panel></WRAP>

By means of an example, the signal-time curve at the reverse integrator shall be explained. \\
  - Let $R=5 c\Omega$, $C=1 \mu F$, and the input voltage waveform $U_E$ shown in <imgref pic3> be given.
  - We are looking for the output voltage $U_A$.

Solution: \\
  - Over the given values of $R$ and $C$, the time constant $\tau$ is determined.
  - With the inverse integrator the input value is integrated and inverted. For the given course of the input voltage the calculation of interpolation points is sufficient.
  - With the formula derived in 5.1.1 $U_A$ can be composed section by section:


The calculation is performed once in detail here (clicking on right arrow "►" leads to the next step, [[calculation_signal_time_course_reverse_integrator|alternative representation]]):
{{url>https://wiki.mexle.org/_export/revealjs/elektronische_schaltungstechnik/rechnung_signalzeitverlauf_umkehrintegrator?theme=dokuwiki&fade=fade&controls=1&show_progress_bar=1&build_all_lists=1&show_image_borders=0&horizontal_slide_level=2&enlarge_vertical_slide_headers=0&size=1624x128#/ 624,108 left noborder}}

<WRAP column 100%> <panel type="danger" title="Notice: signal-time history of the reverse integrator"> <WRAP group><WRAP column 7%>{{fa>exclamation?32}}</WRAP><WRAP column 80%>

If a constant input voltage $U_E$ is applied to the reverse integrator, the output voltage $U_A$ just equals $-U_E$ after the time constant $\tau$ (see <imgref pic3>, light gray arrow).

</WRAP></WRAP></panel> </WRAP>

~~PAGEBREAK~~~CLEARFIX~~~

==== 5.1.3 Determination of amount and phase ====

In order to be able to determine magnitude and phase, purely sinusoidal input and output variables are to be considered first. \\ The following function is used as input voltage $U_E$:

$ U_E(t)= \hat{U}_E \cdot sin(\omega \cdot t)$

This definition of the input voltage can now be substituted into the above equation for $U_A$:

The calculation is performed once in detail here (clicking on right arrow "►" leads to the next step, [[calculation_amountand_phase_reverse_integrator|alternative representation]]):
{{url>https://wiki.mexle.org/_export/revealjs/elektronische_schaltungstechnik/rechnung_betragundphase_umkehrintegrator?theme=dokuwiki&fade=fade&controls=1&show_progress_bar=1&build_all_lists=1&show_image_borders=0&horizontal_slide_level=2&enlarge_vertical_slide_headers=0&size=2524x128#/ 1024,108 left noborder}}


<WRAP right><panel type="default">
<imgcaption pic4| Sketch of the bottom diagram from the reverse integrator>
</imgcaption>
\\ {{drawio>sketch_of_the_bottom_diagram_from_the_reverse_integrator}}
</panel></WRAP>
~~PAGEBREAK~~ ~CLEARFIX~~

The **amount** $|A_V|$ is given by the amplitude ratio of $\hat{U}_A \over \hat{U}_E$: $$|A_V|={\hat{U}_A \over \hat{U}_E} = {1 \over {\omega \cdot R\cdot C}} $$

The **phase** can be determined from the "time offset" of the peak input voltage $U_E = \hat{U}_E \cdot sin(\omega \cdot t)$ and output voltage $U_A = { {\hat{U}_E } \over {\omega \cdot R\cdot C} } \cdot cos(\omega \cdot t)$ can be determined. The phase is given by considering the trigonometric functions and the sign:

$U_E = + \hat{U}_E \cdot sin(\omega \cdot t)$ \
$U_A = + \hat{U}_A \cdot cos(\omega \cdot t) = + \hat{U}_A \cdot sin(\omega \cdot t + 90°)$ \
$ \rightarrow \ \varphi = 90°$

\\
=== Extreme value consideration ===

In order to be able to sketch the course in the bottom diagram, the **behavior of the transfer function $U_A=f(U_E)$ in the extreme cases for low (${\omega}\rightarrow 0$) and high frequencies (${\omega}\rightarrow \infty$)** shall be considered. For magnitude $|A_V|$ and phase $\varphi$ we obtain:

$ |A_V({\omega}\rightarrow 0 \ \; )| \quad=\quad{1 \over {\color{blue}{\omega} \cdot R\cdot C}} \quad\xrightarrow{\color{blue}{\omega}\rightarrow 0}\quad \infty$ \.
$ |A_V({\omega}\rightarrow\infty)| \quad=\quad{1 \over {\color{blue}{\omega} \cdot R\cdot C}} \quad\xrightarrow{\color{blue}{\omega}\rightarrow\infty}\quad 0$ \\.

$\varphi = +90° \qquad \forall \ \omega$

From these boundary conditions the frequency response can already be sketched, see <imgref pic4>.

~~PAGEBREAK~~~CLEARFIX~~~

==== 5.1.4 Circuit analysis with complex calculation====

In the previous chapters it could be seen that the circuit analysis with differential equations is already very tough and computationally intensive for a simple circuit like the reverse integrator. Now the complex calculation will be considered as a method which simplifies the analysis of such circuits. For the complex calculation, the resistances and capacitances are replaced by complex impedances:

$U_R=R\cdot I \qquad \qquad \qquad \rightarrow \underline{U}_R = R \cdot \underline{I}$ \\
$U_C={ 1 \over C }\cdot(\int_{t_0}^{t_1} I_C \ dt+ Q_0(t_0)) \qquad \rightarrow \underline{U}_C = \underline{Z}_C \cdot \underline{I} \quad$ with $\quad \underline{Z}_C= \frac{1}{j \cdot \omega \cdot C}$ \.

However, this consideration can only be implemented under certain boundary conditions:
  - **sinusoidal quantities**: complex current or voltage pointers (cf. [[:elektrotechnik_2:wechselstromtechnik|ET2 Wechselstromtechnik]]) can only represent sinusoidal quantities.
  - **Swinged state**: The systems of equations consider only sinusoidal oscillations that have already existed for infinite time. This corresponds to a long time since switch-on. This takes out disturbances that are generated by switching on.

<WRAP right><panel type="default">
<imgcaption pic5| Circuit of the reverse integrator with complex impedances>
</imgcaption>
\\ {{drawio>circuit_of_the_reverse_integrator_with_complex_impedances}}
</panel></WRAP>

This can now be used to calculate the circuit (<imgref pic5>):

$\underline{Z}_1=R$ \\\
$\underline{Z}_2=\frac{1}{j \cdot \omega \cdot C} = \frac{-j}{\omega \cdot C}$

From the basic circuit of the [[3_basic_circuits_i#inverting_amplifier|inverting amplifier]], its voltage gain is known:

$A_V = \frac{U_A}{U_E}=-\frac{R_2}{R_1}$

This results in the complex:

$\underline{A}_V = \frac{\underline{U}_A}{\underline{U}_E}=-\frac{\underline{Z}_2}{\underline{Z}_1} = \frac{j}{\omega \cdot R \cdot C} $


==PAGEBREAK ==CLEARFIX ==
=== Plausibility check via extreme value consideration ===
^From the formula^From the circuit^
|$\underline{A}_V \xrightarrow{\omega \rightarrow 0} \infty$ | For $\omega \rightarrow 0$, the capacitor becomes a high impedance resistor \\ The opamp must output an output voltage of $\underline{U}_D \rightarrow 0$ for $\underline{U}_A \rightarrow \infty$.  |
|$\underline{A}_V \xrightarrow{\omega \rightarrow \infty} 0$ | For $\omega \rightarrow \infty$ the capacitor becomes a short circuit \ The operational amplifier must output an output voltage of $\underline{U}_D \rightarrow 0$ for $\underline{U}_A \rightarrow 0$. |

==PAGEBREAK ==CLEARFIX ==
=== magnitude and phase ===

<WRAP right><panel type="default">
<imgcaption pic6| course of the arc tangent>
</imgcaption>
\\ {{drawio>course_arc tangent}}
</panel></WRAP>

The amount $A_V$ is given by:

$|\underline{A}_V|= \frac{1}{\omega \cdot R \cdot C}  \sim \frac{1}{f}$

Specifically, for a magnitude of $|\underline{A}_V(0dB)|$ at $0dB$, the result is:

$|\underline{A}_V(0dB)|\overset{!}{=} 1 \widehat{=} 0dB \rightarrow \omega(0dB) = \frac{1}{R \cdot C}$

The phase $\varphi$ is calculated via

$\varphi = arctan \left( \frac{\Im(\underline{A}_V)}{\Re(\underline{A}_V)} \right) = arctan \left( \frac{\omega \cdot R \cdot C}{0} \right) = arctan \left( \infty \right) = +90°$

The phase $\varphi=+90°$ for $arctan(x)|_{x\rightarrow \infty}$ is also evident from the course of the arctangent (<imgref pic6>) for $x \rightarrow \infty$.

~~PAGEBREAK~~~CLEARFIX~~~

==== 5.1.5 Frequency response ====

<WRAP right><panel type="default">
<imgcaption pic7| Bottom diagram of the reverse integrator>
</imgcaption>
\\ {{drawio>Body diagram_of_the_reverse_integrator}}
</panel></WRAP>

The frequency response is to be illustrated by means of an example.
  - Let $R=1 k\Omega$, $C=16 nF$ be given.
  - We are looking for the Bode diagram

Solution
  - Determining the time constant: \\ $\tau = R \cdot C = 16 \mu s$ \\ \\.
  - Determining the frequency $f$ for $|\underline{A}_V(0dB)|$: $\omega(0dB)= \frac{1}{\tau} = 2\pi \cdot f(0dB)$ \ This gives $f(0dB)$ via: \f(0dB)=\frac{1}{2\pi} \cdot \frac{1}{16 \mu s} \approx 10kHz$ \\ \
  - Consideration of the slope: \ $|\underline{A}_V|= \frac{1}{\omega \cdot R \cdot C}  \sim \frac{1}{f}$ \ From this, a tenfold increase in $f$ results in one-tenth the amount $|\underline{A}_V|$, i.e., a slope of $-20dB$ per decade \ \.
  - From this information, the full Bode diagram can be determined (<imgref pic7>)


~~PAGEBREAK~~~CLEARFIX~~~

===== 5.2 Lowpass =====

<WRAP right><panel type="default">
<imgcaption pic7_1| Circuit of the active low-pass filter>
</imgcaption>
\\ {{drawio>circuit_of_the_active_low_pass}}
</panel></WRAP>

Another circuit can be derived from the reverse integrator. For this purpose, the hitherto purely capacitive value of the impedance between the output side and virtual ground is to be supplemented by a resistive component. This circuit can be seen in <imgref pic7_1>.
In the following, this circuit
  * first be considered practically with a simulation,
  * then a picture of the system's effect will be drawn up without detailed calculation, and finally *
  * be checked by a circuit analysis with complex calculation.

==PAGEBREAK ==CLEARFIX ==
=== Lowpass in simulation ===

<WRAP right>{{url>https://www.falstad.com/afilter/circuitjs.html?running=false&cct=$+1+0.000005+5+63+5+50%0A%25+4+984968.4014609919%0Ag+208+208+208+256+0%0Aw+352+112+352+192+0%0Aw+208+112+208+176+0%0Aa+208+192+352+192+4+15+-15+100000000%0Ac+288+112+336+112+0+1.5915000000000002e-7+0.16559840149986407%0Ar+112+112+208+112+0+10000%0AO+352+192+416+192+0%0A170+112+112+64+112+2+20+4000+5+0.1%0Ar+288+64+336+64+0+1000%0Aw+352+64+352+112+0%0Aw+208+64+208+112+0%0AB+224+32+336+144+0+Box%0As+288+64+224+64+0+0+false%0As+288+112+224+112+0+0+false%0Aw+208+64+224+64+0%0Aw+208+112+224+112+0%0Aw+336+112+352+112+0%0Aw+336+64+352+64+0%0Ax+229+53+251+56+0+18+S1%0Ax+229+98+251+101+0+18+S2%0Ao+0+32+0+34+10+0.0125+0+-1%0A 600,500 noborder}}
</WRAP>

In the simulation on the right, the circuit from <imref pic7_1> is shown again. In addition, two switches $S1$ and $S2$ are built into the circuit by which the various feedback paths can be disabled:
  - If only switch $S1$ is closed, the circuit is an inverting amplifier.
  - If only switch $S2$ is closed, the circuit is an inverting integrator.
In the simulation, the Bode diagram is sketched below. By clicking on the Bode diagram, the distribution of the current in the circuit corresponding to the frequency is displayed and - in addition to the Bode diagram - also the gain in $dB$, or the phase in degrees.

~~PAGEBREAK~~~CLEARFIX~~~

<panel type="info" title="Task 5.2.1. circuit analysis in simulation"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>

  - First set up an inverting amplifier and read off the gain and phase (click on the Bode diagram).
  - Now change the circuit to an inverting integrator and read off the gain and phase there as well.
  - Now both switches should be closed.
    - In which frequency ranges does the inverting amplifier or the reverse integrator work approximately?
    - Consider the distribution of the currents in the circuit at different frequencies. At what frequency does the current divide approximately equally between the $1k\Omega$ resistor and the capacitor? What is the value of the gain and phase here?
    - After reading through the following analyses, the gain and phase at the "kink point" can be determined. Do these deviate from your measurement?

</WRAP></WRAP></panel>
~~PAGEBREAK~~ ~CLEARFIX~~

==== 5.2.1 Consideration without detailed account ====


==PAGEBREAK=- ==CLEARFIX=-
===Extreme value consideration ===

From the observation of the capacitor for $\omega \rightarrow 0$ and $\omega \rightarrow \infty$, the variation of magnitude of voltage gain and phase can be analyzed.
  - $\omega \rightarrow 0$:  
    - The capacitor acts like an open circuit
    - Thus the impedance of the capacitor is greater than that of the resistor: $R_2 \ll |\underline{X}_C|$
    - Thus, in the parallel circuit of $R_2$ and $C$, $R_2$ acts essentially
    - Thus the circuit behaves like an inverting amplifier: $|\underline{A}_V|=|-\frac{R_2}{R_1}|$
    - For the inverting amplifier, a sinusoidal input signal results in a negated sinusoidal signal. This corresponds to a phase of $\varphi=\pm 180°$. The sign is not defined by the inverting amplifier.
  - $\omega \rightarrow \infty$:  
    - The capacitor acts like a short circuit
    - Thus the impedance of the capacitor is smaller than that of the resistor: $R_2 \gg |\underline{X}_C|$
    - Thus, in the parallel circuit of $R_2$ and $C$, essentially $C$ acts
    - Thus the circuit behaves like a reverse integrator: $|\underline{A}_V|=|-\frac{1}{\omega \cdot R_1 \cdot C}|$ and $\varphi=+ 90°$

From this it can be seen that
  * for low frequencies a constant gain is expected and
  * for high frequencies a drop as known from the reverse integrator.

== SYNC, CORRECTED BY ELDERMAN ==
=== expected floor diagram ===

<WRAP right><panel type="default">
<imgcaption pic10|expected bottom diagram of low-pass filter>
</imgcaption>
\\ {{drawio>expected_ground_diagram_of_the_low_pass_filter}}
</panel></WRAP>

A floor diagram can be estimated from the extreme value consideration.

**Frequency Response:**
  * For low frequencies, the filter behaves like an inverting amplifier with $|\underline{A}_V|=|-\frac{R_2}{R_1}|$
  * For higher frequencies, the filter behaves like an inverting integrator with $|\underline{A}_V|=|-\frac{1}{\omega \cdot R_1 \cdot C}|$
  * There is a frequency where both situations seem to occur simultaneously

**Phase response:**
  * For low frequencies, the filter behaves like an inverting amplifier with $\varphi=\pm 180°$.
  * For higher frequencies the filter behaves like an inverting integrator with $\varphi=+90°$
  * There is a frequency where both situations seem to occur simultaneously

For the intermediate area, there must be a transition between the two extremal situations.

One problem still seems to be that for the inverting amplifier it is not clear whether the phase is now $+180°$ or $-180°$.
In the mathematical consideration of the inverting integrator, it turned out that for a capacitor an integration step ($U=\frac{1}{C} \int I_C \ dt$) must be performed. In the inverting amplifier, no integration step was necessary. Thus, a sinusoidal input signal is shifted by 90° at most. So there must be just $90°$ phase shift from inverting amplifier to high frequencies at low frequencies.

From this knowledge, we get an expected floor diagram as seen in <imgref pic10>.

<WRAP column 100%> <panel type="danger" title="Notice: filters in the floor diagram"> <WRAP group><WRAP column 7%>{{fa>exclamation?32}}</WRAP><WRAP column 80%>

The following rules apply to filters:
  * for each energy store present, the order of the filter increases by 1.
  * for each energy store present ($C$, $L$) there is an amplitude change of $-20dB/dec$
  * each energy store present ($C$, $L$) results in a phase change of $-90°$
  * The phase response is monotonically decreasing.

</WRAP></WRAP></panel> </WRAP>

==PAGEBREAK ==CLEARFIX ==
=== RC element and cutoff frequency ===

In the circuit, the parallel circuit $R_2$ and $C$ behaves like a passive RC element. That is, it affects the frequency response. At a certain frequency, the circuit behaves just in such a way that the current runs half over $R_2$ and $C$, thus acting "half" like an inverting amplifier and "half" like a reverse integrator. This frequency is the cutoff frequency $f_{Gr}$:

For this: \\
$\underline{X}_C|=R_2$ \
$\frac{1}{\omega_{Gr} \cdot C}=R_2 \rightarrow \omega_{Gr} = \frac{1}{R_2 \cdot C} = 2\pi \cdot f_{Gr}$\\
$\boxed{f_{Gr} = \frac{1}{2\pi \cdot R_2 \cdot C}} $\

~~PAGEBREAK~~ ~CLEARFIX~~
==== 5.2.2 Circuit Analysis with Complex Calculus ====

Now the circuit is to be analyzed again by means of complex calculation. The impedances are again understood as complex numbers. The starting point is again the voltage gain of the (complex) inverting amplifier. The impedances from <imgref pic7_1> are taken into account:

$\underline{A}_V=\frac{-\underline{Z}_2}{\underline{Z}_1}=\frac{-R_2||C}{R_1}=\frac{-\frac{R_2 \cdot \frac{1}{j \cdot \omega \cdot C}}{R_2 + \frac{1}{j \cdot \omega \cdot C}}{R_1}=\frac{-R_2}{R_1 \cdot R_2 \cdot \omega \cdot C + R_1}$ \\
$\boxed{\underline{A}_V= - \frac{R_2}{R_1}\cdot \frac{1}{1 + j \omega \cdot R_2 \cdot C}$

==PAGEBREAK ==CLEARFIX ==
=== Calculation of magnitude and phase ===

For the calculation of the amount $A_V$ a "trick" can be used. In principle, the amount can always be determined by multiplication with the conjugate complex value. But here it is easier to calculate the amount of the fraction via the amount of numerator and amount of denominator:

$|\underline{A}_V| = |\mathcal{a} \cdot \frac{\mathcal{b}}{\mathcal{c}}| = |\mathcal{a}| \cdot \frac{|\mathcal{b}|}{\mathcal{c}|}$

This results in for the amount:

$\boxed{|\underline{A}_V| = \frac{R_2}{R_1}\cdot \frac{1}{\sqrt{1^2 + (\omega \cdot R_2 \cdot C)^2}}$

For the phase $\varphi$ real value $\Re(\underline{A}_V)$ and imaginary value $\Im(\underline{A}_V)$ must be determined by multiplication with the conjugate complex value.

$\varphi = arctan(\frac{\Im(\underline{A}_V)}{\Re(\underline{A}_V)})$

But here, too, there is a "trick":

$\underline{A}_V= \color{blue}{- \frac{R_2}{R_1}\cdot \frac{1}{1 + j \omega \cdot R_2 \cdot C}} \cdot \frac{1 - j \omega \cdot R_2 \cdot C}{\color{blue}{1 - j \omega \cdot R_2 \cdot C}}$

After all, it is just the conjugate complex value that is multiplied to get a real denominator. Thus the blue marked part is a real constant $\mathcal{C}$, because all factors of the constant are real:

$\underline{A}_V= \color{blue}{\mathcal{C}} \cdot (1 - j \omega \cdot R_2 \cdot C)$

Thus, the phase $\varphi = arctan\left(\frac{\color{teal}{\Im(\underline{A}_V)}}{\color{brown}{\Re(\underline{A}_V)}}\right)$ is obtained as.

$\underline{A}_V= \mathcal{C} \cdot (\color{brown}{1} + j \cdot (\color{teal}{-\omega \cdot R_2 \cdot C}))$

$\boxed{\varphi = arctan\left(\frac{\color{teal}{-\omega \cdot R_2 \cdot C}}{\color{brown}{1}}\right)}$

\\
=== Extermal value consideration ===

For the __amount__ we get
  - for $\omega \rightarrow 0$:  \$|\underline{A}_V| = \frac{R_2}{R_1}\cdot \frac{1}{\sqrt{1^2 + \color{blue}{(\omega \cdot R_2 \cdot C)^2}} \rightarrow \frac{R_2}{R_1}\cdot \frac{1}{\sqrt{1^2 }}$, (nbsp) (nbsp) since $\color{blue}{(\omega \cdot R_2 \cdot C)^2} \gg 1$ \\ So the magnitude of the gain goes towards $|\underline{A}_V| = \frac{R_2}{R_1}$. \\ The effect is similar to the inverting amplifier \\
  - at $\omega \rightarrow \infty$: \ $|\underline{A}_V| = \frac{R_2}{R_1}\cdot \frac{1}{\sqrt{1^2 + \color{blue}{(\omega \cdot R_2 \cdot C)^2}} \rightarrow \frac{R_2}{R_1}\cdot \frac{1}{\sqrt{\color{blue}{(\omega \cdot R_2 \cdot C)^2}} $, (nbsp) (nbsp) da $\color{blue}{(\omega \cdot R_2 \cdot C)^2} \ll 1$ \\ So the amount of gain goes towards $|\underline{A}_V| = \frac{1}{\color{blue}{\omega \cdot} R_1 \color{blue}{\cdot C}}$. \ The action resembles the inverse integrator

<WRAP right>
<panel type="default">
<imgcaption pic20|arc tangent>
</imgcaption> <panel type=default>

<collapse id="openAni1" collapsed="true"><well>
{{url>https://www.geogebra.org/material/iframe/id/t2abuwjr/width/730/height/400/border/888888/smb/false/stb/false/stbh/false/ai/false/asb/false/sri/false/rc/false/ld/false/sdz/false/ctl/false 450,250 noborder}}
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<collapse id="openAni1" collapsed="false">
<button type="warning" collapse="openAni1">To view the animation: click here!</button>
</collapse>
 
</panel>
</WRAP>


For finding the phase $\color{red}{\varphi} = arctan\left(\color{teal}{-\omega \cdot R_2 \cdot C}\right)$ it helps to look at the arctangent in the diagram. To do this, the argument $Arg = \color{teal}{-\omega \cdot R_2 \cdot C}$ is plotted against the phase $\color{red}{\varphi}$ (<imgref pic20>). For the extremal values $\omega$ of results in:
  - at $\omega \rightarrow 0$:  \the argument $Arg = \color{teal}{-\omega \cdot R_2 \cdot C}$ goes towards $-0$.
  - at $\omega \rightarrow \infty$:  \Arg = \color{teal}{-\omega \cdot R_2 \cdot C}$ argument goes towards $-\infty$.

In the diagram the argument $Arg$ is changeable by the slider in the upper left corner. The course in the diagram must be continuous, because also the argument $Arg = \color{teal}{-\omega \cdot R_2 \cdot C}$ is continuous between $-0$ and $-\infty$. This is only possible on the __upper branch__: the point $Arg \rightarrow -0$ then just corresponds to approaching the y-axis (here $\color{red}{\varphi}$-axis) from the left in <imgref pic20>. The point $Arg \rightarrow -\infty$ corresponds to the path to the left in the diagram.

This results in a progression of the phase $\color{red}{\varphi}$ from $\color{red}{\varphi}(\omega \rightarrow -0)=\pi = 180°$ to $\color{red}{\varphi}(\omega \rightarrow -\infty)=\frac{\pi}{2} = 90°$.

==PAGEBREAK ==CLEARFIX ==
=== Calculation of the cutoff frequency ===

The cut-off frequency can also be understood as the transition from the inverting amplifier to the reverse integrator. In the Bode diagram (<imgref pic10>), the cut-off frequency can be found at the intersection of the straight lines for the inverting amplifier and the reverse integrator.

Thus, for the cut-off frequency $f_{Gr}$ we get

$\frac{R_2}{R_1} = \frac{1}{\omega_{Gr} R_1 \cdot C}$ \
$\omega_{Gr} = \frac{1}{ R_2 \cdot C} = 2 \pi \cdot f_{Gr}$

At the cut-off frequency, the result is an amount of:

$|\underline{A}_{V,Gr}| = \frac{R_2}{R_1}\cdot \frac{1}{\sqrt{1^2 + (\omega_{Gr} \cdot R_2 \cdot C)^2}}= \frac{R_2}{R_1}\cdot \frac{1}{\sqrt{1^2 + (\frac{1}{ R_2 \cdot C} \cdot R_2 \cdot C)^2}} = \frac{R_2}{R_1}\cdot \frac{1}{\sqrt{1^2 + 1^2}} = \frac{R_2}{R_1}\cdot \frac{1}{\sqrt{2}}$

$\boxed{|\underline{A}_{V,Gr}| = \frac{1}{2} \sqrt{2} \cdot \frac{R_2}{R_1} = -3dB + |\underline{A}^{dB}_V(\omega \rightarrow 0)|}$

The phase at the cutoff frequency is:

$\varphi_{Gr} = arctan\left(-\omega_{Gr} \cdot R_2 \cdot C\right) = arctan\left(-\frac{1}{ R_2 \cdot C} \cdot R_2 \cdot C\right) = arctan\left(-1 \right)$

$\boxed{\varphi_{Gr} = \frac{3}{4} \pi =135°}$

Because of the $-3dB$ attenuation of the low-frequency gain at the cutoff frequency, it is also called the **$-3dB$ cutoff frequency**.


~PAGEBREAK~ ~CLEARFIX~~
===== 5.3 Reverse Differential =====

<WRAP right><panel type="default">
<imgcaption pic11_1| Circuit of the reversal differentiator>
</imgcaption>
\\ {{drawio>circuit_of_the_reversing_differentiator}}
</panel></WRAP>

<WRAP right>{{url>https://www.falstad.com/circuit/circuitjs.html?running=false&cct=$+1+0.000049999999999999996+1.9265835257097934+41+5+43%0Aa+288+128+384+128+8+15+-15+1000000+-0.000039999600019248555+0+100000%0Aw+384+112+384+80+0%0Ac+208+112+160+112+0+5.000000000000001e-7+-3.42003999865076%0Aw+288+80+288+112+0%0Ar+384+80+288+80+0+10000%0Ag+288+144+288+176+0%0AR+128+112+96+112+0+3+40+5+0+0+0.5%0A207+384+128+432+128+4+U_A%0A403+320+208+464+272+0+7_8_0_12294_4.008796818347498_0.0001_0_2_7_3_U%5CsA%0A207+128+112+128+144+4+U_E%0Aw+384+112+384+128+0%0A403+112+208+256+272+0+9_8_0_12294_4.9840000009474466_0.0001_0_2_9_3_U%5CsE%0Aw+128+112+160+112+0%0Aw+272+112+288+112+0%0Ar+208+112+272+112+0+10%0A 600,400 noborder}}
</WRAP>

~~PAGEBREAK~~~CLEARFIX~~~

<WRAP right><panel type="default">
<imgcaption pic11| Bode diagram of the reversal differentiator>
</imgcaption>
\\ {{drawio>Bode_diagram_of_the_reversal_differentiator}}
</panel></WRAP>


In <imgref pic11_1> an inverse differentiator is shown. Compared to the integrator here just the resistor and the capacitor are swapped.

In the simulation next to it, the effect of the circuit can be seen: the derivative of the inverted input signal is output at the output. The derivative at the reversal points ("peaks") of the signal cannot be determined (see [[wpde>Differentiability#Sum function|Differentiability of the sum function]]). This leads to problems in the calculation during the simulation and can be seen as overshoot or "deflection" at $U_A$. To reduce this, a small resistor (relative to the feedback resistor) is inserted after the capacitor.

In the following, only the results will be discussed without derivation.

Circuit analysis via differential equation yields:

$\boxed{U_A = - R \cdot C \frac{d}{dt}U_E}$

With complex calculation, the transfer function becomes:

$\boxed{\underline{A}_V=-j \cdot \omega \cdot R \cdot C}$

From this, the Bode diagram shown in <imgref pic11> can be determined.

~~PAGEBREAK~~~CLEARFIX~~
<panel type="info" title="Task 5.3.1 Reverse Differential"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>

<WRAP right>
{{drawio>default_timeline_reverse_integrator}}</WRAP>

For the inverse differentiator shown in <imgref pic11_1>, derive the complex voltage gain, and its magnitude and phase using complex calculus as shown for the [[#inverse integrator]]. In doing so, implement the following steps:

  - Circuit analysis by means of differential equation
  - Determination of magnitude and phase from differential equation (incl. consideration of extreme cases)
  - Example of a signal-time-curve with: $R = 10 k\Omega$ and $C = 2µF$ and $U_E$ as right shown
  - Circuit analysis by means of complex calculation
  - Consideration of magnitude and phase for $\omega \rightarrow 0$ and $\omega \rightarrow \infty$
  - Frequency response (Bode plot) for circuit with: $R = 10 k\Omega$ and $C = 16nF$

</WRAP></WRAP></panel>
~~PAGEBREAK~~ ~CLEARFIX~~
===== 5.4 High Pass =====

<WRAP left>{{url>https://www.falstad.com/afilter/circuitjs.html?running=false&cct=$+1+0.000005+5+63+5+50%0A%25+4+984968.4014609919%0Ag+240+176+240+224+0%0Aw+384+80+384+160+0%0Aw+240+80+240+144+0%0Aa+240+160+384+160+4+15+-15+10000000000%0Ac+144+80+208+80+0+1.5915e-9+0.16559840149986407%0Ar+80+80+144+80+0+10000%0AO+384+160+448+160+0%0A170+80+80+32+80+2+2+400000+0.005+0.1%0Ar+256+80+368+80+0+1000%0AB+64+48+224+160+0+Box%0As+144+112+80+112+0+1+false%0As+208+112+144+112+0+1+false%0Aw+240+80+256+80+0%0Aw+368+80+384+80+0%0Ax+78+147+100+150+0+18+S1%0Ax+149+149+171+152+0+18+S2%0Aw+144+80+144+112+0%0Aw+80+80+80+112+0%0Aw+208+80+208+112+0%0Aw+208+80+240+80+0%0Ao+0+32+0+34+10+0.0125+0+-1%0A 600,500 noborder}}
</WRAP>

<WRAP right><panel type="default">
<imgcaption pic12_1| Circuit of the high-pass filter>
</imgcaption>
\\ {{drawio>circuit_of_the_high_pass_filter}}
</panel></WRAP>


~~PAGEBREAK~~~CLEARFIX~~~

<WRAP right><panel type="default">
<imgcaption pic12| Bode diagram of the high-pass filter>
</imgcaption>
\\ {{drawio>Bode_diagram_of_the_high_pass_filter}}
</panel></WRAP>

A high-pass filter can be created from the reversal differentiator if the purely capacitive impedance for $Z_1$ is suitably extended via a resistive component (<imgref pic12_1>). The simulation above shows this high pass. A reverse integrator forms from this with switch $S1$ closed and switch $S2$ open. When the switch is inverted, an inverting amplifier is formed. In the simulation, clicking on a frequency point again shows the distribution of the currents.

With complex calculation this results in: $\boxed{\underline{A}_V = - \frac{R_2}{R_1} \cdot \frac{j \cdot \omega \cdot R_1 \cdot C}{1 + j \cdot \omega \cdot R_1 \cdot C}} $

From this, the Bode diagram shown in <imgref pic12> can be determined.

~~PAGEBREAK~~~CLEARFIX~~
<panel type="info" title="Task 5.4.1 1st order high pass"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>

In [[#lowpass|previous chapter]] the gain $A_V$ of the 1st order lowpass was derived based on its circuit. In the same way now the gain for a high pass (cf. <imgref pic12_1>) shall be derived.

  - Behavior of magnitude and phase for $\omega \rightarrow 0$ and $\omega \rightarrow \infty$.
  - Expected Bode diagram
  - RC element and cutoff frequency
  - Circuit analysis with complex calculation
  - Calculation of magnitude and phase

</WRAP></WRAP></panel>

~~PAGEBREAK~~ ~CLEARFIX~~
===== 5.5 Overview high pass / low pass =====

<WRAP right><panel type="default">
<imgcaption pic13| Overview high pass / low pass>
</imgcaption>
\\ {{drawio>overview_highpass/lowpass}}
</panel></WRAP>

~PAGEBREAK~ ~CLEARFIX~~
====== tasks ======

<panel type="info" title="Task 5.0.1. converting linear factors to dB"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>

Derive the linear factor for the following levels in dB. State how this factor can be determined in each case using the interpolation points 20 dB ≙ factor 10 and 6 dB ≙ factor 2.

Solve without a calculator \ (note: $\sqrt{2}\approx 1.414$, ${1\over\sqrt{2}}\approx 0.707$).

As an example, the derivation is sketched for the value $10 dB$.

^level^over interpolation points in dB^over interpolation points linear^linear factor|
|$10 dB$|$5 \cdot 6 dB - 20 dB$|$2^5 \cdot {1\over 10}$|$3,2$|
|$2 dB$| |
|$4 dB$| | |
|$6 dB$| | |
|$8 dB$| | |
|$12 dB$| | |
|$14 dB$| | |
|$16 dB$| | |
|$18 dB$| | |
|$15 dB$| | |
|$79 dB$| | |
|$128 dB$| | |

</WRAP></WRAP></panel>

<panel type="info" title="Task 5.1.1 Analysis of Circuits in Tina TI"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>

<WRAP right>
{{drawio>Reverse Integrator}}</WRAP>

Let the circuit shown opposite with $R= 10 k\Omega$, $C = 1.6 uF$ and a sinusoidal input voltage $U_E = 1 V $ with $f = 1 kHz$ be given. As described in the course, the Bode diagram can be displayed in Tina TI via Analysis > AC Analysis > AC Transfer Characteristic. In the following, frequencies from 100 Hz to 1 GHz are relevant.

  - Simulate this circuit in Tina TI \\ - with an ideal operational amplifier \\ - using the operational amplifiers uA776, LM301A and LM318.
      - Attach the Bode diagram.
      - Briefly describe the differences in the amplitude response of the gain $A_V$.
  - What happens if instead of $R= 10 k\Omega$, $C = 1.6 uF$ the same time constant is implemented with $R= 10 M\Omega$, $C = 1.6 nF$? Verify this by comparing the Bode plot of the LM318 operational amplifier.
  - Simulate an inverting amplifier in Tina TI. To do this, replace the capacitor in the circuit with a resistor $R_2 = 10 k\Omega$ and use the LM318 op-amp.
      - Attach the Bode diagram.
      - What should the Bode diagram (amplitude and phase response) look like for an ideal inverting amplifier? What deviation do you notice in the real setup?
      - Up to what frequency can the circuit be operated as an inverting amplifier (maximum deviation of 1dB)? Use zoom and/or cursor to determine.
  - Simulate an open-loop operational amplifier LM318 (i.e. without feedback network). The non-inverting input should be connected to ground. The inverting input should have the above sinusoidal input voltage.
      - Attach the Bode diagram.
      - What is the cut-off frequency?
      - How many dB per decade does the amplitude response drop at high frequencies?

</WRAP></WRAP></panel>

====== Learning questions ======

  * What affects the rise time of an amplifier circuit?
  * What is the difference between an ideal and a real operational amplifier?

\\

====== References ======

--> References to the media used #

^ Element ^ License ^ Link ^
| <imgref pic6>: Arc tangent course | [[https://creativecommons.org/licenses/by-sa/4.0/|CC-BY-SA 4.0]]  | https://de.m.wikipedia.org/wiki/Datei:Arctangent.svg |

<--
====== 6. Filter Circuits II - Higher Order Filters ======

===== 6.1 Bandpass filter =====

<WRAP right><panel type="default">
<imgcaption pic0| Example: WLAN channels>
</imgcaption>
\\ {{drawio>WLAN_Channels}}
</panel></WRAP>

When analyzing different signals, only a part of the entire frequency spectrum is desired. In <imgref pic0>, the channels of the WLAN standard 802.11 are shown as an example; these are used alternately for data transmission. Another example arises with vibration spectra of a motor in a machine, which contains not only the vibrations (usable for diagnostics), but also interference from other machine parts. Other examples are cabled data transmission or [[wpde>electroencephalography#EEG-frequencyb%C3%A4nder_and_Graphoelements|bands of brain waves]].

To separate the desired frequencies, a filter can be used which only passes a given band between two frequencies (//frequency band//). This is possible with a **bandpass filter**.

~~PAGEBREAK~~~CLEARFIX~~~

<WRAP right><panel type="default">
<imgcaption pic1b| Tolerance scheme for a bandpass filter>
</imgcaption>
\\ {{drawio>tolerance_scheme_bandpass_filter}}
</panel></WRAP>

=== Frequency response ranges ===

The range between the two frequencies is called the **passband**, or bandwidth.
Outside the passband, the gain drops off. A real filter cannot attenuate infinitely.
Also, there are various ideal filters where outside the passband, gain does not approach zero, but just falls below a threshold.
Often the sloping region is called the **transition region** and the region below the threshold is called the **blocking region**.
The threshold itself is called **blocking area**. In <imgref pic1b>, the ranges are drawn. However, the terms are not clearly defined; in various textbooks, the transition region is already called the blocking region.


~~PAGEBREAK~~~CLEARFIX~~~

<WRAP right><panel type="default">
<imgcaption pic1| block diagram bandpass>
</imgcaption>
\\ {{drawio>Block diagram_bandpass}}
</panel></WRAP>

=== Assembling the bandpass filter ===

This filter can be composed over by basic low-pass and high-pass filters. If the signal is first filtered through a low-pass filter and then through a high-pass filter, the desired filter is created. The order of the filters can be reversed. <imgref pic1> shows this in the block diagram - where in (1) is a commonly used and in (2) with the circuit symbols to be used according to EN 60617. Thus the transfer function $\underline{A}_{BP}$ of the bandpass filter simply results from the transfer function of the lowpass and highpass filters $\underline{A}_{TP}$ and $\underline{A}_{HP}$, since the signal passes through the filter stages one after the other:

$$\underline{A}_{BP}= {{\underline{U}_A}\over{\underline{U}_E}} = {{\underline{U}_A}\over{\underline{U}_1}} \cdot {{\underline{U}_1}\over{\underline{U}_E}} = \underline{A}_{TP} \cdot \underline{A}_{HP}$$

~~PAGEBREAK~~~CLEARFIX~~~

<WRAP right><panel type="default">
<imgcaption pic2| amplitude response bandpass>
</imgcaption>
\\ {{drawio>amplitude_response_bandpass}}
</panel></WRAP>

=== Amplitude response of the bandpass filter ===

In <imgref pic2>, the amplitude response of the bandpass filter can be seen. Since in the amplitude response the transfer function is represented in $dB$ ($\underline{A}^{dB}$), multiplying the transfer functions of the low-pass and high-pass filters $\underline{A}_{TP}$ and $\underline{A}_{HP}$ results in an addition of $\underline{A}_{TP}^{dB}$ and $\underline{A}_{HP}^{dB}$. In the amplitude response, we can see that it results in a $20 dB/dec$ change twice: once at $f_{Gr,HP}$ and once at $f_{Gr,TP}$. So the filter has an order of 2.

Important: The cutoff frequency of the low-pass filter $f_{Gr,TP}$ must be larger than the cutoff frequency of the high-pass filter $f_{Gr,HP}$ (see <imgref pic2>).

But what does the frequency response look like? This is to be derived in the following.

~~PAGEBREAK~~~CLEARFIX~~~

<WRAP right><panel type="default">
<imgcaption pic3| Circuit of the bandpass filter based on the inverting amplifier >
</imgcaption>
\\ {{drawio>circuit_bandpass_filter_inverting_amplifier}}
</panel></WRAP>

==== 6.1.1 Bandpass based on the inverting amplifier ====
== Realization ==
From [[5_filter_circuits_i#lowpass|chapter 5]], the circuits of highpass and lowpass filters are known. From this, the circuit shown in <imgref pic3> can be derived.
This will be considered in some detail.

The extremal value consideration yields:
  * for $ \boldsymbol{\omega \rightarrow 0} $:\ The magnitude of the impedance of the capacitances becomes large \ and thus $|\underline{X}_{C_1}| \gg R_1$ , as well as $|\underline{X}_{C_2}| \gg R_2$ \ Thus $\underline{X}_{C_1}$ prevails at $\underline{Z}_1$ and $\underline{R}_2$ prevails at $\underline{Z}_2$. \rightarrow$ **A reverse differentiator results at low frequencies.
  * for $ \boldsymbol{\omega \rightarrow \infty} $:\\ The magnitude of the impedance of the capacitances becomes small and thus $|\underline{X}_{C_1}| \ll R_1$ , as well as $|\underline{X}_{C_2}| \ll R_2$ \\ Thus $\underline{R}_1$ predominates at $\underline{Z}_1$ and $\underline{X}_{C_2}$ predominates at $\underline{Z}_2$. \rightarrow$ **A reverse integrator results at high frequencies.**

\\
== complex-valued consideration of the transfer function ==

<wrap #transmissionfunction1 />

The transfer function is again to be derived from a complex-valued inverting amplifier:

$\underline{A}_V = {{\underline{U}_A}\over{\underline{U}_E}} = - {\underline{Z}_2}\over{\underline{Z}_1}} = - {\underline{Z}_2}\cdot {1\over{\underline{Z}_1}} = - \Large{{R_2\cdot {1\over{j\omega C_2}}\over{R_2 + {1\over{j\omega C_2}}}}}}\cdot{1 \over{R_1 + {1\over{j\omega C_1}}}}}= - \Large{{R_2}\over{j\omega C_2 R_2 + 1}}}}\cdot{j\omega C_1 \over{j\omega C_1 R_1 + 1}} \Bigg| {{\cdot R_1}\over{\cdot R_1}}$

$\boxed{\underline{A}_V = - \color{blue}{R_2 \over R_1 } \cdot \large\color{teal}{1 \over {1+ j\omega \cdot C_2 R_2}} \cdot \large\color{brown}{{j\omega \cdot C_1 R_1} \over {1+ j\omega \cdot C_1 R_1}}$

\\
Clever reshaping yields an interesting result of the following parts:
  - $- \color{blue}{R_2 \over R_1 }$: This corresponds to a [[3_basic_circuits_i#inverting_amplifier|inverting amplifier]]
  - $\large\color{teal}{1 \over {1+ j\omega \cdot C_2 R_2}$: This corresponds to a [[5_filter circuits_i#lowpass|lowpass 1st order]] with a cutoff frequency of $\color{teal}{\omega_{Gr, TP}= {1 \over {C_2 R_2}}$
  - $\large\color{brown}{{j\omega \cdot C_1 R_1} \over {1+ j\omega \cdot C_1 R_1}}$: This corresponds to a [[5_filter circuits_i#highpass|highpass 1st order]] with a cutoff frequency of $\color{brown}{\omega_{Gr, HP}= {1 \over {C_1 R_1}}$

\\
This results in a function via the extremal value consideration:

  * for $ \boldsymbol{\omega \rightarrow 0 } $:\ $\underline{A}_V = - \Large{R_2 \over R_1 } \cdot \Large\color{teal}{1 \over {1+ \color{black}{\underbrace{\color{teal}{j\omega \cdot C_2 R_2}}_{\rightarrow 0}}}} \cdot \large\color{brown}{{j\omega \cdot C_1 R_1} \over {1+ \color{black}{\underbrace{\color{brown}{j\omega \cdot C_1 R_1}}_{\rightarrow 0}}}} \rightarrow - {R_2 \over R_1 } \cdot \color{teal}{ 1 \over 1} \cdot \large\color{brown}{{j\omega \cdot C_1 R_1} \over 1} \rightarrow - \color{brown}{\normalsize{j\omega \cdot C_1 \color{black}{R_2}}$ \ The equation is the same as that of a reverse differentiator \cdot
  * for $ \omega \rightarrow \infty $:\ $\underline{A}_V = - \Large{R_2 \over R_1 } \cdot \Large\color{teal}{1 \over {1+ j\omega \cdot C_2 R_2}} \cdot \Large\color{brown}{{j\omega \cdot C_1 R_1} \over \color{brown}{1+ {j\omega \cdot C_1 R_1}} \rightarrow - {R_2 \over R_1 } \cdot \color{teal}{ 1 \over {j\omega \cdot C_2 R_2}} \cdot \large\color{brown}{1 \over 1} \rightarrow - \color{teal}{1 \over {j\omega \cdot C_2 \color{black}{R_1}}$ \ The equation is equivalent to that of an inverse integrator

\\
== Determination of magnitude and phase from complex-valued observation ==

For the magnitude $|\underline{A}_V|$ of the transfer function, the following hint can be used: $|a\cdot b\cdot c| = |a| \cdot |b| \cdot |c| $. \\
Thus, for the magnitude $|\underline{A}_V|$, we get: \\
$ |\underline{A}_V| = {R_2 \over R_1 } \cdot \large{1 \over \sqrt{1+ \omega^2 C_2^2 R_2^2}} \cdot \large{{\omega \cdot C_1 R_1} \over \sqrt{1+ \omega^2 C_1^2 R_1^2}}
$\xrightarrow{\color{teal}{\omega_{Gr, TP}}, \ \color{brown}{\omega_{Gr, HP}}$
$\boxed{|\underline{A}_V| = {R_2 \over R_1 } \cdot \large{1 \over \sqrt{1+ \omega^2 / \color{teal}{\omega_{Gr, TP}^2}} \cdot \large{{\omega / \color{brown}{\omega_{Gr, HP}} \over \sqrt{1+ \omega^2 \color{brown} / {\omega_{Gr, HP}}^2}}$

For the phase $\varphi$ must be conjugate complexly extended again. \\
At first this produces an unwieldy equation - but a real-valued constant can be separated from it.

$\underline{A}_V = - \color{blue}\large{R_2 \over R_1 } $
$\cdot \large\color{teal }{ 1 \over \color{lightgray}{\boxed{\color{teal }{\small{1+ j\omega \cdot C_2 R_2}}}}}$
$\cdot \large\color{teal }{{1- j\omega \cdot C_2 R_2} \over \color{lightgray}{\boxed{\color{teal }{\small{1- j\omega \cdot C_2 R_2}}}}}$
$\cdot \large\color{brown}{{ j\omega \cdot C_1 R_1} \over \color{ pink }{\boxed{\color{brown}{\small{1+ j\omega \cdot C_1 R_1}}}}}$
$\cdot \large\color{brown}{{1- j\omega \cdot C_1 R_1} \over \color{ pink }{\boxed{\color{brown}{\small{1- j\omega \cdot C_1 R_1}}}}}$

$\underline{A}_V = \quad \quad \mathcal{C} \quad \quad \quad \quad$
$ \cdot \color{teal }{(1- j\omega \cdot C_2 R_2)}$
$\ \cdot \color{brown}{ j\omega \cdot C_1 R_1 }$
$\ \cdot \ \color{brown}{(1- j\omega \cdot C_1 R_1)}$

$\underline{A}_V = \quad \quad \mathcal{C} \quad \quad \quad$
$ \cdot (j + \omega R_2 C_2 + \omega R_1 C_1 - j \omega R_1 C_1 \omega R_2 C_2)$

From this equation it is easy to read the proportions for real part $\Re(\underline{A}_V)$ and imaginary part $\Im(\underline{A}_V)$. \\
This gives for the phase $\varphi$ :

$ \varphi = arctan \left( \frac{\Im(\underline{A}_V)}{\Re(\underline{A}_V)} \right) = arctan \left( \frac{1 - \omega R_1 C_1 \omega R_2 C_2}{\omega R_2 C_2 + \omega R_1 C_1} \right)$
$\xrightarrow{\color{teal}{\omega_{Gr, TP}}, \ \color{brown}{\omega_{Gr, HP}}$
$\boxed{\varphi = arctan \left( \frac{\color{teal}{\omega_{Gr, TP}} \color{brown}{\omega_{Gr, HP}} - \omega^2 }{\omega (\color{teal}{\omega_{Gr, TP}}+\color{brown}{\omega_{Gr, HP}})} \right)}$

\\
The formal for the phase $\varphi$ says
The extremal consideration can now be carried out for some salient frequencies:
  * for $ \boldsymbol{\omega \rightarrow 0} $:\\ $\varphi(0) = arctan \left( \frac{\mathcal{C}_1 - \omega^2}{\omega \mathcal{C}_2} \right) \rightarrow arctan \left( \frac{\mathcal{C}_1 - "0"}{"0"} \right) = arctan \left( "+\infty" \right)$ \\ \
  * for $ \boldsymbol{\omega \rightarrow \infty} $: \varphi(\infty) = arctan \left( \frac{\mathcal{C}_1 - \omega^2}{\omega \mathcal{C}_2} \right) \rightarrow arctan \left( \frac{\mathcal{C}_1 - "\infty"^2}{"\infty"} \right) = arctan \left("-\infty" \right)$ \\ \
  * for a **(circular) frequency** $\boldsymbol{\omega= \omega_0}$ **for which the argument of the** $\boldsymbol{arctan}$** function becomes zero**. \ Thus the phase: \varphi(\omega_0) = arctan \left( 0 \right)$. \ The corresponding frequency is given by: \large\frac{\color{teal}{\omega_{Gr, TP}} \color{brown}{\omega_{Gr, HP}} - \omega^2 }{\omega (\color{teal}{\omega_{Gr, TP}}+\color{brown}{\omega_{Gr, HP}})} = 0 \quadrightarrow\quad \omega_0^2 = \color{teal}{\omega_{Gr, TP}} \color{brown}{\omega_{Gr, HP}} \quad\rightarrow\quad \omega_0 = \large\sqrt{\color{teal}{\omega_{Gr, TP}} \color{brown}{\omega_{Gr, HP}}$ \
  * for the cutoff frequency of the high pass filter $\boldsymbol{\omega = \color{brown}{\omega_{Gr, HP} = {1 \over {R_1 C_1}}}}$. \ For this, if the passband is sufficiently large, $\color{brown}{\omega_{Gr, HP}} \color{teal}\omega_{Gr, TP}$ can be assumed. \ Thus we get: \varphi(\color{brown}{\omega_{Gr, HP}}) = arctan \left( \large\frac{\color{teal}{\omega_{Gr, TP}} \color{brown}{\omega_{Gr, HP}} - \color{brown}{\omega_{Gr, HP}}^2 }{\color{brown}{\omega_{Gr, HP}} (\color{teal}{\omega_{Gr, TP}}+\color{brown}{\omega_{Gr, HP}})} \right) = arctan \left( \large\frac{\color{teal}{\omega_{Gr, TP}} - \color{brown}{\omega_{Gr, HP}} }{ (\color{teal}{\omega_{Gr, TP}}+\color{brown}{\omega_{Gr, HP}})} \right) \xrightarrow{\color{brown}{\omega_{Gr, HP}} \ll \color{teal}{\omega_{Gr, TP}} \varphi(\color{brown}{\omega_{Gr, HP}}) = arctan (1)$ \color{brown}{\omega_{Gr, HP}}
  * for the cutoff frequency of the lowpass filter $\boldsymbol{\omega = \color{teal}{\omega_{Gr, TP} = {1 \over {R_2 C_2}}}}$. \For this, if the passband is sufficiently large, $\color{brown}{\omega_{Gr, HP}} \color{teal}{\omega_{Gr, TP}}$ can be assumed. \ Thus we have: \varphi(\color{teal}{\omega_{Gr, TP}}) = arctan \left( \large\frac{\color{teal}{\omega_{Gr, TP}} \color{brown}{\omega_{Gr, HP}} - \color{teal}{\omega_{Gr, TP}}^2 }{\color{teal}{\omega_{Gr, TP}} (\color{teal}{\omega_{Gr, TP}}+\color{brown}{\omega_{Gr, HP}})} \right) = arctan \left( \large\frac{ \color{brown}{\omega_{Gr, HP}} - \color{teal}{\omega_{Gr, TP}}{ (\color{teal}{\omega_{Gr, TP}}+\color{brown}{\omega_{Gr, HP}})} \right) \xrightarrow{\color{brown}{\omega_{Gr, HP}} \gg \color{teal}{\omega_{Gr, TP}} \varphi(\color{teal}{\omega_{Gr, TP}}) = arctan (-1)$

<WRAP right><panel type="default">
<imgcaption pic7| Arc tangent for bandpass>
</imgcaption>
\\ {{drawio>ArcusTangens_Bandpass}}
</panel></WRAP>

\\
This results in the following for individual points:

^ $\boldsymbol{\omega}\quad$ | $\rightarrow 0$ | $\color{brown}{\omega_{Gr, HP}}$ | $\sqrt{\color{teal}{\omega_{Gr, TP}} \color{brown}{\omega_{Gr, HP}}$ | $\color{teal}{\omega_{Gr, TP}}$ | $\rightarrow \infty$ |
^ $\boldsymbol{\varphi}$ | $arctan \left( "+\infty" \right)$ | $arctan ( +1)$ | $arctan ( 0)$ | $arctan ( -1)$ | $arctan \left( "-\infty" \right)$ |
^ | $+90°$ | $+45°$ | $0°$ | $-45°$ | $-90°$ |

The results also seem plausible with the course of the arc tangent (<fc #ff0000>red curve</fc> in <imgref pic7>): for low frequencies the argument of the arc tangent goes towards $+\infty$ and thus the phase $\varphi$ seems to go towards $+90°$, for high frequencies towards $-90°$.

<fs x-large>ABER:</fs> Looking at the phase progression in the simulation below, it shows more of a progression that goes along with the black line.

~~PAGEBREAK~~~CLEARFIX~~~

<panel type="info" title="Task 6.1.1 Bandpass based on the inverting amplifier"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>

  - Consider again the [[#uebertragungsfunktion1| transfer function]] and find the complex gain for $\omega_0 = \large\sqrt{\color{teal}{\omega_{Gr, TP}} \color{brown}{\omega_{Gr, HP}}}$. \ Is this value positive (= no phase shift) or negative (= phase shift by $\pm 180°$)?
  - Consider the circuit in the simulation below at the following points:
    - Increase of +20dB/dec at low frequencies.
    - Middle of the passband ("plateau")
    - Drop of -20dB/dec at high frequencies \ <WRAP outdent> <WRAP outdent> Which capacitor behaves like a short circuit at each point? \ Knowing the behavior of the capacitors: What equivalent circuit describes the system in the forward region? </WRAP> </WRAP>
</WRAP></WRAP></panel>

<WRAP right>{{url>https://falstad.com/afilter/circuitjs.html?cct=$+1+0.000005+5+50+5+50%0A%25+4+33623165.424224265%0Ac+256+128+304+128+0+0.000006799999999999999+0%0Ar+192+128+256+128+0+33%0AO+400+144+464+144+0%0Ag+304+160+304+192+0%0A170+192+128+160+128+3+20+1000+5+0.1%0Aa+304+144+400+144+0+15+-15+100000000%0Ar+304+80+400+80+0+100%0Ac+304+32+400+32+0+6.8000000000000005e-9+0%0Aw+400+32+400+80+0%0Aw+400+80+400+144+0%0Aw+304+128+304+80+0%0Aw+304+80+304+32+0%0Ao+4+16+0+34+5+0.00009765625+0+-1+in%0Ao+2+16+0+34+2.5+0.00009765625+1+-1+out%0A 700,400 noborder}}
</WRAP>

This can be used to determine the floor diagram.

~~PAGEBREAK~~ ~CLEARFIX~~
==== 6.1.2 Multi-Feedback Bandpass ====

<WRAP right><panel type="default">
<imgcaption pic7| Circuit of the multi-feedback bandpass filter>
</imgcaption>
\\ {{drawio>Circuit_MultiFeedbackBandpassFilter}}
</panel></WRAP>


<WRAP right><panel type="default">
<imgcaption pic6| bottom diagram bandpass>
</imgcaption>
\\ {{drawio>Body diagram_bandpass}}
</panel></WRAP>

~~PAGEBREAK~~~CLEARFIX~~~

====== 6.2 Tape lock ======

Electrical Engineering 2 and Electrical Engineering Lab have already given insights into oscillating circuits. Within these circuits, at certain frequencies come up swinging motions, which can take up energies of system

~~PAGEBREAK~~~CLEARFIX~~~

<WRAP onlyprint>
{{url>https://www.geogebra.org/material/iframe/id/zhvkeaa8/width/1000/height/700/border/888888/smb/false/stb/false/stbh/false/ai/false/asb/false/sri/false/rc/false/ld/false/sdz/false/ctl/false 1000,700 noborder}}

From page [[https://www.geogebra.org/m/zhvkeaa8|www.geogebra.org/m/zhvkeaa8]], author: Tim Fischer.
</WRAP>

~~PAGEBREAK~~~CLEARFIX~~~

==== Homework ====

Example: Evaluation of an infrared sensor:
  * Nodes are missing in the circuit from the manufacturer --> correct circuit is to be drawn.
  * to which basic circuits do OPV 1 and 2 correspond? What filtenn do both correspond to?
{{electronic_circuitry:murata_example_opv_circuit.jpg?600}}

====== References ======

--> References to the media used #

^ Element ^ License ^ Link ^
| <imgref pic5>: Superposition of sinusoidal oscillations | Public Domain | https://en.wikipedia.org/wiki/File:Fourier_transform_time_and_frequency_domains.gif |
| <imgref pic6>: Superposition of sinusoidal oscillations | [[https://creativecommons.org/licenses/by-sa/3.0/deed.en|CC-BY SA 3.0]]  | https://en.wikipedia.org/wiki/File:Fourier_series_square_wave_circles_animation.gif |
| <imgref pic7>: Superposition of sinusoidal oscillations | [[https://creativecommons.org/licenses/by-sa/4.0/deed.en|CC-BY SA 4.0]]  | https://commons.wikimedia.org/wiki/File:Example_of_Fourier_Convergence.gif |
<--
====== 7. Non-linear applications ======

===== 7.1 Schmitt Trigger=====

<callout>
  * interactive animation of a trigger circuit: [[https://www.ipes.ethz.ch/mod/lesson/view.php?id=24&pageid=71|iPES Zurich]]
</callout>

===== 7.3 Timer 555 =====

The 555 timer is sometimes referred to as "the time machine". It has historically been of great importance in the generation and detection of temporal signal sequences. \\
Before discussing the device itself, the various modes should be briefly considered:

Mode^other name^explanation^application^examples^
|astable|oscillator, multivibrator|output signal changes value periodically|creation of a periodic, rectangular signal, with adjustable pulse width and frequency|motor control, dimming of LEDs, generation of tones|
|monostable|"flip-flop", monoflop|Output signal is triggered by a trigger and goes back to 0V after a defined time|Extending pulses that are too short, setting a fixed pulse length|Conditioning sensor signals of a proximity sensor|
|bistable|flipflop|set/reset when thresholds are exceeded/fallen short|bounce-free switch, circuits with hysteresis|two-point controller, e.g. for heating elements|

==== 7.3.1 Comparison 555 and Microcontroller ====

It can be seen that there are a wide variety of applications for this component. The question can be asked what differentiates the application of this component from a microcontroller:

^Property^Microcontroller^Timer 555^
|Cost (single unit, 2018)|from 3ct (chin. µC, [[https://lcsc.com/product-detail/PADAUK_PADAUK-Tech-PMS150C_C129127.html|PADAUK PMS150C]])|from 29ct (western µC[[https://www.mouser.de/ProductDetail/Microchip-Technology-Atmel/ATTINY9-TS8R?qs=sGAEpiMZZMvqv2n3s2xjsRLJ5ROJ0gc0Te0wtn3axF4%3d|ATtiny]]) | from 30ct
|cost (>10'000, 2018)|from approx. 2..3ct (chin. µC)|from 20ct (westl. µC) |from 2..3ct |
|other components|interference suppression|interference suppression|transistors, resistors and other capacitors |depending on the application|
|Complexity|in software|hardly available|
|Flexibility|Updates possible|if potentiometers were used, limited|
|advantages for small quantities|easily changeable by programming, \no component scattering|easily changeable by reassembling on the board, \no software bugs, \no tools necessary|
For a long time, the Timer 555 was the most cost-effective solution for the tasks mentioned above. Currently (2018), Timer 555 and microcontroller prices are about the same in large quantities. Nevertheless, due to its simplicity, the Timer 555 is still found in various consumer electronic products.

==== 7.3.2 Pinning and principle circuit====
====== Advanced simulations ======

<WRAP right><panel type="default">
<imgcaption pic3|JFET Pinch Off>
</imgcaption>
{{drawio>>JFET}}
</panel>
</WRAP>

===== Junction Field Effect Transistor (JFET) =====

The structure of the junction field effect transistor (JFET)** resembles the bipolar transistor at first glance. In <imgref pic3>, the individual images (1)...(3) show the layering of an n-channel (English n-channel) JFET and the circuit symbol in the upper left. In contrast to the pnp bipolar transistor, however, here the p-doped layers are jointly supplied with voltage and the n-doped layer is transversely fluxed.

Without voltage difference $U_{GS}$ between gate and source, a (small) junction is formed at the p-n junctions. Electrons can flow unhindered through the n-doped layer: a current $I_G \gg 0$ flows through the FET (<imgref pic3> Fig. (1)).  The "path" between the two junction layers is called the **n-channel**.

<WRAP right><panel type="default">
<imgcaption pic3|JFET Pinch Off>
</imgcaption>
{{drawio>Pinch_off}}
</panel>
</WRAP>

When the voltage difference $U_{GS}$ becomes smaller than zero, the junction layers increase, and the diode between G and S is operated in the reverse direction. The n-channel is constricted and geometrically decreases the electron or current flow $I_G$ (<imgref pic3> Fig. (2)). At a certain voltage $U_{GS}=U_p$ (**pinch-off or pinch-off voltage**), the two junction layers are so large that no n-channel is left - the channel is **pinched-off** (<imgref pic3> figure (3)). Above this voltage, there can be no current flow.
The principle is thus similar to the situation when the flow from a water hose is regulated by squeezing the hose.

In the simulation on the right, the same voltage ratios are shown. The toggle switch on the left makes it possible to invert the voltage $U_{DS}$ via the transistor. If this becomes negative, a slightly different situation arises: The JFET seems to become conductive in all, regardless of what voltage $U_{GS}$ assumes.


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</WRAP>
~~PAGEBREAK~~ ~CLEARFIX~~

===== Digital Analogue Converter (DAC) =====

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</WRAP>

In chapter 3, a digital-to-analog converter was described in [[3_basic_circuits_i#tasks|task 3.5.3]]. The R-2R ladder described there allows a pure digital value to be output as an analog voltage as an integrated circuit. In the simulation on the right a simplified version can be seen. However, the simplified version requires many very precisely tuned resistors. In contrast, the R-2R ladder requires only 2 resistor values and this is easier to manufacture in terms of microsystem technology.

~PAGEBREAK~ ~CLEARFIX~~
===== Time Domain - Frequency Domain =====

=== arbitrary periodic signals ===

<WRAP right><imgcaption pic5|Superposition of sinusoidal oscillations>
{{electronic_circuitry:fourier_transform_time_and_frequency_domains.gif}}
</imgcaption> </WRAP>

At the previous chapter a sinusoidal input voltage was used for the analysis. How do the filters work Here we want to focus on it again.

In <imgref pic5> it can be seen that a square wave signal can be approximated by several sinusoidal signals. If the amplitudes of these signals are plotted versus frequency, an image of the signal in the frequency domain is obtained. This transformation is done computationally via the [[https://de.wikipedia.org/wiki/Fourier-Transformation|Fourier transform]] and is treated in detail in advanced subjects such as control engineering, signals and systems and digital signal processing.

For very fast changes, high-frequency components are required.

Already with the [[4_grundschaltungen_ii#umkehraddierer|Umkehraddierer]] it became apparent that a periodic sawtooth signal can be assembled from several sinusoidal voltages.

~~PAGEBREAK~~~CLEARFIX~~~

<WRAP right><imgcaption pic6|Superposition of sinusoidal oscillations>
{{electronic_circuitry:fourier_series_square_wave_circles_animation.gif}}
</imgcaption> </WRAP>

<WRAP right><imgcaption pic7|Approximation of any signal>
{{electronic_circuitry:example_of_fourier_convergence.gif}}
</imgcaption> </WRAP>

From the sinus

The video [[https://www.youtube.com/watch?v=r6sGWTCMz2k|But what is a Fourier series?]] explains clearly how even vector images can be generated by superimposing sine functions.
~~REVEAL theme=white&fade=fade&controls=1&show_progress_bar=1&build_all_lists=1&show_image_borders=1&horizontal_slide_level=2&enlarge_vertical_slide_headers=0&show_slide_details=0&open_in_new_window=1&size=2400x168~~

---->
|$U_A = -{ 1 \over {R\cdot C} }\cdot\int_{t_0}^{t_1} \color{blue}{U_E(t)} \ dt + U_{A0}$|insert sine function|$ \color{blue}{U_E(t)}= \hat{U}_E \cdot sin(\omega \cdot t)$|
|$\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad$|$\qquad\qquad\qquad\qquad$|$\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad$|
<----

---->
|$U_A = -{ 1 \over {R\cdot C} }\cdot\color{blue}{\int_{t_0}^{t_1} \hat{U}_E \cdot sin(\omega \cdot t) \ dt} + U_{A0}$|insert root function with \ limits|$\color{blue}{\int_{x_0}^{x_1} sin(a \cdot x) \ dx} = [- {1 \over a} \cdot cos(a \cdot x) ]_{x_0}^{x_1}$|
|$\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad$|$\qquad\qquad\qquad\qquad$|$\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad$|
<----

---->
|$U_A = -{ 1 \over {R\cdot C} }\cdot [- \color{blue}{\hat{U}_E \over \omega} \cdot cos(\omega \cdot t) ]_{t_0}^{t_1} + U_{A0}$ |put constant before \ integral| |
|$\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad$|$\qquad\qquad\qquad\qquad$|$\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad$|
<----

---->
|$U_A = { 1 \over {R\cdot C} }\cdot {\hat{U}_E \over \omega} \cdot \color{blue}{[ cos(\omega \cdot t) ]_{t_0}^{t_1}} + U_{A0}$ |insert limits|$t_0=0$, $t_1=t$|
|$\qquad\qquad\qquad\qquad\qquad\qquad\qquad$|$\qquad\qquad\qquad\qquad\qquad$|$\qquad\qquad\qquad\qquad$|$\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad$|
<----

---->
|$U_A = {{\hat{U}_E } \over {\omega \cdot R\cdot C} } \cdot (} cos(\omega \cdot t) - \color{blue}{cos(0)} ) + U_{A0}$ |$\color{blue}{cos(0)}=1$|
|$\qquad\qquad\qquad\qquad\qquad\qquad\qquad$|$\qquad\qquad\qquad\qquad\qquad$|$\qquad\qquad\qquad\qquad$|$\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad$|
<----

---->
|$U_A = \color{blue}{{ \hat{U}_E } \over {\omega \cdot R\cdot C} } \cdot (} cos(\omega \cdot t) - 1 \color{blue}{)} + U_{A0}$ |multiply| |
|$\qquad\qquad\qquad\qquad\qquad\qquad\qquad$|$\qquad\qquad\qquad\qquad\qquad$|$\qquad\qquad\qquad\qquad$|$\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad$|
<----

---->
|$U_A = { {\hat{U}_E } \over {\omega \cdot R\cdot C} } \cdot cos(\omega \cdot t) \color{blue}{-{ {\hat{U}_E } \over {\omega \cdot R\cdot C}} + U_{A0}}$ |consider the \non-cosine terms| |
|$\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad$|$\qquad\qquad\qquad\qquad\qquad$|$\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad$|
<----

---->
|$U_A = { {\hat{U}_E } \over {\omega \cdot R\cdot C} } \cdot cos(\omega \cdot t) \color{blue}{-{ {\hat{U}_E } \over {\omega \cdot R\cdot C}} + U_{A0}}$ |This part is independent in time. Since we assume purely sinusoidal quantities, \the for the initial voltage of the capacitor must be: $U_{C0} = U_{A0}={{\hat{U}_E} \over {\omega \cdot R\cdot C}}$||
|$\qquad\qquad\qquad\qquad\qquad\qquad\qquad$|$\qquad\qquad\qquad\qquad\qquad$|$\qquad\qquad\qquad\qquad$|$\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad$|
<----

---->
|$U_A = { {\hat{U}_E } \over {\omega \cdot R\cdot C} } \cdot cos(\omega \cdot t)$| |
|$\qquad\qquad\qquad\qquad\qquad\qquad\qquad$|$\qquad\qquad\qquad\qquad\qquad$|$\qquad\qquad\qquad\qquad$|$\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad$|
<----
~~REVEAL theme=whide&fade=fade&controls=1&show_progress_bar=1&build_all_lists=1&show_image_borders=1&horizontal_slide_level=2&enlarge_vertical_slide_headers=0&show_slide_details=0&open_in_new_window=1&size=1324x168~~


---->
|$\quad$|$R_{E}^0=\frac{U_E}{I_p}$ |$\quad$|
|$\quad$|$\quad$quad$|$quad$quad$quad$|$quad$quad$quad$quad$quad$quad$|$quad$quad$quad$quad$|
<----

---->
|$\quad$|$R_{E}^0=\frac{U_E}{\color{blue}{I_p}}$ |with $I_p$ from $R_D=\frac{U_D}{I_p}$|
|$\quad\quad\quad\quad\quad\quad$|$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad$|$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad$|
<----

---->
|$\quad$|$R_{E}^0=\frac{U_E\cdot \color{blue}{R_D}}{\color{blue}{U_D}}$ |$\quad$|
|$\quad$quad$quad$quad$|$\quad$quad$quad$quad$quad$quad$quad$|$quad$quad$quad$quad$quad$quad$quad$quad$quad$quad$quad$quad$quad$quad$quad$quad$quad$quad$
<----

---->
|$\quad$|$R_{E}^0=\frac{U_E\cdot R_D}{\color{blue}{U_D}}$ |with $U_D = \frac{U_A}{A_D}$|
|$\quad\quad\quad\quad\quad\quad$|$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad$|$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad$|
<----

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|$\quad$|$R_{E}^0=\frac{U_E\cdot R_D \cdot \color{blue}{A_D}}{\color{blue}{U_A}}$ |transformed|
|$\quad\quad\quad\quad\quad\quad$|$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad$|$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad$|
<----

---->
|$\quad$|$R_{E}^0=\frac{U_E}{U_A}\cdot R_D \cdot A_D$ |$\quad$|
|$\quad$quad$quad$quad$|$quad$quad$quad$quad$quad$quad$quad$|$quad$quad$quad$quad$quad$quad$quad$quad$quad$quad$quad$quad$quad$quad$quad$quad$quad$quad$
<----

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|$\quad$|$R_{E}^0=\color{blue}{\frac{U_E}{U_A}}\cdot R_D \cdot A_D$ |where $A_V = \frac{U_A}{U_E} =\frac{R_2}{R_1+R_2}$|
|$\quad\quad\quad\quad\quad\quad$|$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad$|$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad$|
<----

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|$\quad$|$R_{E}^0={{1}\over{A_V}}\cdot R_D \cdot A_D$ | |
|$\quad$quad$quad$quad$|$\quad$quad$quad$quad$quad$quad$quad$quad$|$\quad$quad$quad$quad$quad$quad$quad$quad$quad$quad$quad$quad$quad$quad$quad$quad$
<----

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|$\quad$|$R_{E}^0={{A_D}\over{A_V}}\cdot R_D $ |
|$\quad$|$\quad$$$quad$$$$quad$$$quad$$$quad$$$$quad$$$quad$$$$quad$$$$quad$$$$quad$$$quad$$$quad$$$quad$$quad$$$quad$$$
<----~~REVEAL theme=whide&fade=fade&controls=1&show_progress_bar=1&build_all_lists=1&show_image_borders=1&horizontal_slide_level=2&enlarge_vertical_slide_headers=0&show_slide_details=0&open_in_new_window=1&size=1324x168~~



---->
$I.\quad$ consideration of flows
<----

---->
|from (2+3)|$\color{blue}{I_p} = \color{blue}{I_m} = 0$ |$I_p$ and $I_m$ are thus defined|
|$\quad\quad\quad\quad\quad\quad$|$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad$|$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad$|
<----

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|f (6)|$\color{blue}{I_o} = I_1 $ |$I_o$ is thus known if $I_1$ is known|
|$\quad\quad\quad\quad\quad\quad$|$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad$|$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad$|
<----

---->
|from (7) and (3)|$I_1 - I_2 -\color{blue}{0} = 0 $ |$\quad$|
|$\quad$quad$quad$quad$|$quad$quad$quad$quad$quad$quad$quad$|$quad$quad$quad$quad$quad$quad$quad$quad$quad$quad$quad$quad$quad$quad$quad$quad$quad$
<----

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|$\quad$|$I_1 = I_2 = I_o$ |$\quad$|
|$\quad$|$\quad\quad\quad\quad$|$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad$|$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad$|
<----

---->
|$\quad$|$\color{blue}{I_1} = \color{blue}{I_2} = \color{blue}{I_o} $ |with (8) and (9): $I_\boxed{}=\frac{U_\boxed{}}{R_\boxed{}$ and (5)|
|$\quad\quad\quad\quad\quad\quad$|$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad$|$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad$|
<----

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| $\quad$ |$\frac{U_1}{R_1}= \frac{U_2}{R_2} = \frac{U_A}{R_1 + R_2}$ |voltage divider formula, $I=const.$ |
|$\quad\quad\quad\quad\quad\quad$|$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad$|$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad$|
<----

---->
| (10)|$U_2= U_A\cdot\frac{R_2}{R_1+R_2}$ |voltage divider formula|
|$\quad\quad\quad\quad\quad\quad$|$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad$|$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad$|
<----

---->
$II.\quad$ consideration of voltage gain
<----

---->>
|off (0) |$\color{blue}{A_V}=\frac{U_A}{U_E}$ | $\quad$|
|$\quad$quad$quad$quad$|$quad$quad$quad$quad$quad$quad$quad$|$quad$quad$quad$quad$quad$quad$quad$quad$quad$quad$quad$quad$quad$quad$quad$quad$quad$quad$quad$|
<<----

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| $\quad$ |$A_V=\frac{U_A}{\color{blue}{U_E}}$ |with (4): $U_E=U_2+U_D$|
|$\quad\quad\quad\quad\quad\quad$|$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad$|$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad$|
<<----

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| $\quad$ |$A_V=\frac{U_A}{\color{blue}{U_2+U_D}}$ | $\quad$ |
|$\quad$$$$$quad$$|$\quad$$$quad$$$quad$$$$quad$$$$quad$$$$quad$$$$$quad$$$$quad$$$$quad$$$$quad$$$$quad$$$$
<<----

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| $\quad$ |$A_V=\frac{U_A}{\color{blue}{U_2}+U_D}$ | with (10): $U_2= U_A\cdot\frac{R_2}{R_1+R_2}$ |
|$\quad\quad\quad\quad\quad\quad$|$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad$|$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad$|
<<----

---->>
| $\quad$ |$A_V=\frac{U_A}{\color{blue}{U_A\cdot\frac{R_2}{R_1+R_2}}+U_D}$ | $\quad$ |
|$\quad\quad\quad\quad\quad\quad$|$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad$|$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad$|
<<----

---->>
| $\quad$ |$A_V=\frac{U_A}{U_A\cdot\frac{R_2}{R_1+R_2}+U_D}$ | $\quad$ |
|$\quad\quad\quad\quad\quad\quad$|$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad$|$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad$|
<<----

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| $\quad$ |$A_V=\frac{U_A}{U_A\cdot\frac{R_2}{R_1+R_2}+\color{blue}{U_D}}$ | with (1)|
|$\quad\quad\quad\quad\quad\quad$|$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad$|$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad$|
<<----

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| $\quad$ |$A_V=\frac{U_A}{U_A\cdot\frac{R_2}{R_1+R_2}+\color{blue}{\frac{U_A}{A_D}}$ | $\quad$ |
|$\quad\quad\quad\quad\quad\quad$|$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad$|$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad$|
<<----

---->>
| $\quad$ |$A_V=\frac{U_A}{U_A\cdot\frac{R_2}{R_1+R_2}+\frac{U_A}{A_D}}$ | $\quad$ |
|$\quad\quad\quad\quad\quad\quad$|$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad$|$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad$|
<<----

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| $\quad$ |$A_V=\frac{\color{blue}{U_A}}{\color{blue}{U_A}\cdot\frac{R_2}{R_1+R_2}+\frac{\color{blue}{U_A}}{A_D}$ | Extend with $\frac{1}{U_A}$ |
|$\quad\quad\quad\quad\quad\quad$|$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad$|$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad$|
<<----

---->>
| $\quad$ |$A_V=\frac{1}{\frac{R_2}{R_1+R_2}+\frac{1}{A_D}}$ | $\quad$ |
|$\quad\quad\quad\quad\quad\quad$|$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad$|$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad$|
<<----

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| $\quad$ |$A_V=\frac{1}{\frac{R_2}{R_1+R_2}+\color{blue}{\frac{1}{A_D}}$ | with $\frac{1}{A_D} \xrightarrow{A_D \rightarrow \infty} 0$ |
|$\quad\quad\quad\quad\quad\quad$|$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad$|$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad$|
<<----

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| $\quad$ |$A_V=\frac{1}{\frac{R_2}{R_1+R_2}}$ | Transform Fraction |
|$\quad$$$quad$$|$\quad$$$quad$$$quad$$$$quad$$$$quad$$$quad$$$quad$$$$quad$$$$quad$$$quad$$$$quad$$$$quad$$$$quad$$$
<<----

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| $\quad$ |$A_V=\frac{R_1+R_2}{R_2}$ | $\quad$ |
|$\quad$|$\quad\quad\quad\quad\quad$|$\quad\quad\quad\quad\quad\quad\quad\quad\quad$|$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad$|
<<----~~REVEAL theme=white&fade=fade&controls=1&show_progress_bar=1&build_all_lists=1&show_image_borders=1&horizontal_slide_level=2&enlarge_vertical_slide_headers=0&show_slide_details=0&open_in_new_window=1&size=1324x168~~

---->
$I.\quad$ At point $t_1$
<----

---->
|$U_{A}(t_1) \ \ = -\quad { 1 \over {\tau} } \quad \ \cdot \int_{t_0}^{t_1} \quad U_E \ dt \ + \ U_{A}(t_0)$ | |
|$\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad$|$\qquad\qquad$|
<----

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|$U_{A}(t_1) \ \ = -{ 1 \over {5 k\Omega \cdot 1 \mu F} }\cdot\int_{0}^{10ms} 1V \ dt + 0V$ | |
|$\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad$|$\qquad\qquad$|
<----

---->
|$U_{A}(t_1) \ \ = - \quad { 1 \over {5 ms} } \quad \cdot 1V \ \cdot \int_{0}^{10ms} \ dt\quad\quad$ | |
|$\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad$|$\qquad\qquad$|
<----

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|$U_{A}(t_1) \ \ = - \quad { 1 \over {5 ms} } \quad \cdot 1V \ \cdot [t]_{0}^{10ms} = \quad -2V$ | |
|$\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad$|$\qquad\qquad$|
<----

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$I.\quad$ At point $t_2$
<----

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|$U_{A}(t_1) \ \ = -\quad { 1 \over {\tau} } \quad \ \cdot \int_{t_0}^{t_1} U_E \ dt \ + \ U_{A}(t_0)$ | |
|$\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad$|$\qquad\qquad$|
<----

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|$U_{A}(t_1) \ \ = -{ 1 \over {5 ms} } \quad \cdot (-1V) \ \cdot [t]_{10ms}^{20ms} + 2V = 0V$ | |
|$\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad$|$\qquad\qquad$|
<----

---->
$I.\quad$ At point $t_3$
<----

---->
|$U_{A}(t_1) \ \ = -\quad { 1 \over {\tau} } \quad \ \cdot \int_{t_0}^{t_1} U_E \ dt \ + \ U_{A}(t_0)$ | |
|$\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad$|$\qquad\qquad$|
<----

---->
|$U_{A}(t_1) \ \ = -{ 1 \over {5 ms} } \quad \cdot (-2V) \ \cdot [t]_{10ms}^{20ms} + 0V = -2V$ | |
|$\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad$|$\qquad\qquad$|
<----
~~REVEAL theme=whide&fade=fade&controls=1&show_progress_bar=1&build_all_lists=1&show_image_borders=1&horizontal_slide_level=2&enlarge_vertical_slide_headers=0&show_slide_details=0&open_in_new_window=1&size=322x104~~



---->
$I.\quad$ consideration of flows
<----

---->
|from (2+3)|$\color{blue}{I_p} = \color{blue}{I_m} = 0$ |$I_p$ and $I_m$ are thus defined|
|$\quad\quad\quad\quad\quad\quad$|$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad$|$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad$|
<----

---->
|from (3) and (5)|$\color{blue}{I_o} = I_m = 0$ |$I_o$ is thus defined|
|$\quad\quad\quad\quad\quad\quad$|$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad$|$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad$|
<----

---->
$II.\quad$ consideration of voltage gain
<----

---->>
|off (0) |$\color{blue}{A_V}=\frac{U_A}{U_E}$ | $\quad$|
|$\quad$quad$quad$quad$|$quad$quad$quad$quad$quad$quad$quad$|$quad$quad$quad$quad$quad$quad$quad$quad$quad$quad$quad$quad$quad$quad$quad$quad$quad$quad$quad$|
<<----

---->>
| $\quad$ |$A_V=\frac{U_A}{\color{blue}{U_E}}$ |with (4)|
|$\quad$|$\quad\quad\quad\quad\quad$|$\quad\quad\quad\quad\quad\quad\quad\quad\quad$|$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad$|
<<----

---->>
| $\quad$ |$A_V=\frac{U_A}{\color{blue}{U_A+U_D}}$ | $\quad$ |
|$\quad\quad\quad\quad\quad\quad$|$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad$|$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad$|
<<----

---->>
| $\quad$ |$A_V=\frac{\color{blue}{U_A}}{\color{blue}{U_A}+U_D}$ |with (1) |
|$\quad\quad\quad\quad\quad\quad$|$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad$|$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad$|
<<----

---->>
| $\quad$ |$A_V=\frac{\color{blue}{A_D\cdot U_D}}{\color{blue}{A_D\cdot U_D}+U_D}$ | $\quad$ |
|$\quad\quad\quad\quad\quad\quad$|$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad$|$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad$|
<<----

---->>
| $\quad$ |$A_V=\frac{A_D\cdot U_D}{A_D\cdot U_D + U_D}$ | $\quad$ |
|$\quad\quad\quad\quad\quad\quad$|$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad$|$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad$|
<<----

---->>
| $\quad$ |$A_V=\color{blue}{\frac{A_D\cdot U_D}{A_D\cdot U_D + U_D}}$ |Expand with $\frac{1}{A_D\cdot U_D}$ |
|$\quad\quad\quad\quad\quad\quad$|$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad$|$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad$|
<<----

---->>
| $\quad$ |$A_V=\color{blue}{\frac{A_D\cdot U_D\cdot\frac{1}{A_D\cdot U_D}}{(A_D\cdot U_D + U_D)\cdot \frac{1}{A_D\cdot U_D}}$ | $\quad$ |
|$\quad\quad\quad\quad\quad\quad$|$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad$|$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad$|
<<----

---->>
| $\quad$ |$A_V=\color{blue}{\frac{1}{1 + \frac{1}{A_D}}$ | $\quad$ |
|$\quad\quad\quad\quad\quad\quad$|$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad$|$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad$|
<<----

---->>
| $\quad$ |$A_V=\frac{1}{1 + \frac{1}{A_D}}$ | $\quad$ |
|$\quad$$$$$quad$$|$\quad$$$$quad$$$$quad$$$$quad$$$$quad$$$$$quad$$$$$$$quad$$$$quad$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$
<<----

---->>
| $\quad$ |$A_V=\frac{1}{1 + \color{blue}{\frac{1}{A_D}}$ |with $\frac{1}{A_D} \xrightarrow{A_D \rightarrow \infty} 0$ |
|$\quad\quad\quad\quad\quad\quad$|$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad$|$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad$|
<<----

---->>
| $\quad$ |$A_V=\frac{1}{1 + \color{blue}{0}}$ | $\quad$ |
|$\quad$$$$$quad$$$$$quad$$$$$quad$$$$quad$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$
<<----

---->>
| $\quad$ |$A_V=\frac{1}{1}=1$ | $\quad$ |
|$\quad$|$\quad\quad\quad\quad\quad$|$\quad\quad\quad\quad\quad\quad\quad\quad\quad$|$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad$|
<<----~~REVEAL theme=white&fade=fade&controls=1&show_progress_bar=1&build_all_lists=1&show_image_borders=1&horizontal_slide_level=2&enlarge_vertical_slide_headers=0&show_slide_details=0&open_in_new_window=1&size=624x158~~


---->
|$U_A = f(U_E)$ |with III.|
|$\qquad\qquad\qquad\qquad\qquad$|$\qquad\qquad\qquad\qquad$|$\qquad\qquad\qquad\qquad\qquad\qquad\qquad$|
<----

---->
|$U_A=\color{blue}{-U_D}-U_C$ |with II. and I.|$ \color{blue}{U_D} = { 1 \over A_D } \cdot U_A \overset{A_D -> \infty}\longrightarrow 0$|
|$\qquad\qquad\qquad\qquad\qquad$|$\qquad\qquad\qquad\qquad\qquad$|$\qquad\qquad\qquad\qquad\qquad\qquad\qquad$|
<----

---->
|$U_A= \quad 0 \quad -\color{blue}{U_C}$|with V.|$\color{blue}{U_C}={ 1 \over C }\cdot(\int_{t_0}^{t_1} I_C \ dt+ Q_0(t_0))$|
|$\qquad\qquad\qquad\qquad\qquad$|$\qquad\qquad\qquad\qquad\qquad$|$\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad$|
<----

---->
|$U_A = {-{ 1 \over C }\cdot}(\int_{t_0}^{t_1} \color{blue}{I_C} \ dt+ Q_0(t_0)) $|with IV.|$\color{blue}{I_C}=I_R$|
|$\qquad\qquad\qquad\qquad\qquad\qquad$|$\qquad\qquad\qquad\qquad\qquad$|$\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad$|
<----

---->
|$U_A = \color{blue}{-{ 1 \over C }\cdot(}\int_{t_0}^{t_1} I_R \ dt+ Q_0(t_0)\color{blue}{)} $|brackets|
|$\qquad\qquad\qquad\qquad\qquad\qquad$|$\qquad\qquad\qquad\qquad\qquad$|$\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad$|
<----

---->
|$U_A = -{ 1 \over C }\cdot\int_{t_0}^{t_1} I_R \ dt - \color{blue}{ Q_0(t_0) \over C } $|consider integration constant \\$\color{blue}{ Q_0(t_0) \over C }= U_C(t_0) = -U_{A0}$|
|$\qquad\qquad\qquad\qquad\qquad\qquad$|$\qquad\qquad\qquad\qquad\qquad$|$\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad$|
<----

---->
|$U_A = -{ 1 \over C }\cdot\int_{t_0}^{t_1} \color{blue}{I_R} \ dt + U_{A0}$|with VI. and II.|$\color{blue}{I_R}={ U_R \over R}={ U_E \over R} $|
|$\qquad\qquad\qquad\qquad\qquad\qquad$|$\qquad\qquad\qquad\qquad\qquad$|$\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad$|
<----

---->
|$U_A = -{ 1 \over C }\cdot\int_{t_0}^{t_1} \color{blue}{1 \over R} \cdot U_E \ dt + U_{A0}$|prefer constant|
|$\qquad\qquad\qquad\qquad\qquad\qquad$|$\qquad\qquad\qquad\qquad\qquad$|$\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad$|
<----

---->
|$U_A = -{ 1 \over {R\cdot C} }\cdot\int_{t_0}^{t_1} U_E \ dt + U_{A0}$| insert time constant \ $\tau = R \cdot C$ |
|$\qquad\qquad\qquad\qquad\qquad\qquad$|$\qquad\qquad\qquad\qquad\qquad$|$\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad$|
<----

---->
|$U_A = -{ 1 \over {\tau} }\cdot\int_{t_0}^{t_1} U_E \ dt + U_{A0}$| |
|$\qquad\qquad\qquad\qquad\qquad\qquad$|$\qquad\qquad\qquad\qquad\qquad$|$\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad$|
<----

~~NOCACHE~~
<affix offset-top="0" position-top="0" position-left="20">
<button size="xs" type="link" collapse="sidebar">{{icon>list}} Course Content</button>
 
<collapse id="sidebar" collapsed="false"><well>
\ [[start|{icon>home?18}}]]<fs large> [[start|Electronic]] \(nbsp)(nbsp)(nbsp)(nbsp)(nbsp)(nbsp)[[start|Circuitry]]</fs>

\ [[0_aids|0. Aids]]
\ [[1_basics_to_amplifiers|1. Basics to amplifiers]]
\ [[2_diodes|2. diodes]] \ [[2_transistors|Transistors]]
\ [[3_basic_circuits_i|3. basic_circuits I]]
\ [[4_basic_circuits_ii|4. basic_circuits II]]
\ [[5_filter circuits_i#filter circuits|5. filter circuits I]]
\ [[6_filter circuits_ii|6. Filter circuits II]]
\ [[7_non-linear_applications|7. Non-linear applications]]
</well>
~~BARCODE~url=https://wiki.mexle.org/elektronische_schaltungstechnik/start~size=S~~ [[https://creativecommons.org/licenses/by-sa/4.0/deed.de|{electronic_circuit_engineering:by-sa.png?100}}]]
</collapse></affix>
====== Electronic Circuitry ======

<WRAP right> {{elektronische_systeme:tech-1137487_640.jpg?200|tech-1137487_640.jpg}}\\
<fs 70%>Source: [[https://pixabay.com/en/bitcoin-currency-crypto-cyber-2057405/|Pixabay]] ([[https://creativecommons.org/publicdomain/zero/1.0/deed.en|CC 0 license]])</fs> </WRAP>

The course Electronic Circuit Technology expands your knowledge of switching and amplifying elements. The knowledge gained from this course is used in a wide range of products. Microcontrollers in mobile phones and automobiles, data processing in sensors (e.g. motion detectors) and much more are based on the components and circuits presented in this course. These are simple amplifier circuits, as well as diodes and transistors. The course is based on Electrical Engineering I and is linked to Electronics II (Filters), as well as ET1 Lab.

===== Teaching material for the individual parts =====


<WRAP column 15%>
{{chapter_image_return.jpg?100}}
</WRAP><WRAP column 70%>
\ [[0_resources|0 block - resources]]
 \ [[1_basics_to_amplifiers|1 block - basics to amplifiers]]
</WRAP><WRAP clear></WRAP>

<WRAP column 15%>
{{chapter_image_transistor.jpg?80}}{{chapter_image_diode.jpg?70}}
</WRAP><WRAP column 70%>
\ [[2_diodes|2 block (I) - diodes]]
\ [[2_transistors|2 block (II) - transistors]]
</WRAP><WRAP clear></WRAP>

<WRAP column 15%>
{{chapter_image_opv_circuits.jpg?100}}
</WRAP><WRAP column 70%>
[[3_basic_circuits_i|3 block - basic_circuits I]] \\
[[4_basic_circuits_ii|4 block - basic circuits II]]  \\
</WRAP><WRAP clear></WRAP>

<WRAP column 15%>
{{chapter_image_filter_circuits.jpg?100}}
</WRAP><WRAP column 70%>
\ [[5_filter_circuits_i|5 block - filter_circuits I]] \\ \\
[[6_filter circuits_ii|6 block - filter circuits II]] \\
</WRAP><WRAP clear></WRAP>

<WRAP column 15%>
{{chapter_image_trigger_circuits.jpg?70}}
</WRAP><WRAP column 70%>
\ [[7_non-linear_applications|7 block - non-linear applications]]
</WRAP><WRAP clear></WRAP>

\ [[8_continuing|8 block - continuing simulations]]

<WRAP pagebreak></WRAP>
<panel type="info" title="Task 2.1.1 Diode at higher frequencies"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>
{{electronic_circuit_technology:single-ended_rectifier_with_sensor_and_oszi_sketch.png?300}}

In your company "HHN Mechatronics & Robotics" you have built a single-ended rectifier to rectify a sinusoidal measuring signal of ($f=200kHz$, amplitude $\hat{U} = 5.0V$, output resistance of the sensor $R_q = 10k$). For this purpose you built a simple circuit with the "Si rectifier diode" $D=1N5400$ and a smoothing capacitor with $C=10pF$. As measuring instrument you used an oscilloscope (Rigol DS1000E). The circuit is drawn in the adjacent sketch.

Your colleague has already pointed out to you that at high frequencies some diodes get a problem with rectification. You also noticed this when measuring the setup and looking at the oscilloscope...

Write down the expected signal curve before the respective simulation. Note that you must consider a steady-state system in the simulation!

  - Find in the {{::ds1000en_anleitung_en.pdf|Instruction of the oscilloscope}} the values of the input impedance, which are needed in the circuit for the input resistance $R_E$ and the input capacitance $C_E$. \\ Replicate the circuit in using the information from TINA TI above (**Circuit 1**). Take into account the input impedance of the oscilloscope, as shown in the sketch. \\ Simulate circuit 1 with the specified signal. Briefly describe the expected and measured signal waveforms.<WRAP onlyprint> \
  - Try tuning the capacitance of capacitor $C$ to get the expected rectified value. What do you find? <WRAP onlyprint> \WWRAP>
  - Since something seems to be funny, you want to debug the circuit, that is, determine the error. To do this, you use a [[elektronik_labor:tipps_for_debugging#general|generic approach to debugging]] and want to break down the unclear system to a minimum. Specifically, you build a modified circuit (**Circuit 2**):
    - the sensor is replaced by a function generator (same frequency and amplitude, but $R_q = 50 \Omega$),
    - the smoothing capacitor $C$ is replaced by an open lead (so it is no longer present)
  - Simulate circuit 2 with the signal given so far. Briefly describe the expected and measured signal characteristics. 
  - Now take another step back and try to get a little more current flowing across the diode. In circuit 2, the current was limited by $R_E$ and thus the diode was not yet operating above $U_S=0.7V$. The idea now in **Circuit 3** is to also switch the input resistor to $R_E = 50 \Omega$ (this is possible on some oscilloscopes). The rest of circuit 3 is the same as circuit 2. Simulate circuit 3 with the signal given so far. 
  - Now you seem to be getting closer to the problem. You vary input resistance to $R_E = 500 \Omega$ (**Circuit 4**) \ Simulate circuit 4 with the given signal. Briefly describe the expected and measured signal waveforms.<WRAP onlyprint> \
  - Your colleague tips you that the progression (see diagram on the right) is typical of <wrap right>{{electronic_circuit_engineering:task_2.1.1._circuit4_diagram.png?300}}</wrap>
    - A reverse recovery time $t_{rr}$ that is too large. This is reproduced in Tina via the transit time $TT$.
    - an excessive junction capacity (junction capacity $C_j$ or diode capacity $C_D$).
  - These values can be changed in Tina TI by the following procedure: Double-click on the diode >> click on ''...'' at Type >> search for the mentioned values. \ You now want to analyze how the reverse bias and the junction capacitance affect the voltage curve (for circuit 4). \\ Simulate and __describe the voltage curve__ if
    - on the one hand the reverse bias is reset to $0s$ or
    - on the other hand, the junction capacitance is reset to $0F$. <WRAP onlyprint> __describe __the voltage waveform__.
  - Instead of diode $D=1N5400$, choose diode $D=1N4148$ and simulate again circuit 3 and circuit 1. \ Now how does the voltage waveform behave and why? <WRAP onlyprint> </WRAP></WRAP></WRAP></WRAP></WRAP></WRAP></WRAP></WRAP></WRAP></WRAP></WRAP></WRAP></WRAP></WRAP></WRAP></WRAP>.
</WRAP></WRAP></panel><WRAP pagebreak></WRAP>
<panel type="info" title="Task 2.1.2 Calculating a series resistor for a diode"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>
You want to drive a red light-emitting diode on a voltage source of $U_q = 5.0 V$ with a current of $I_D = 20 mA$. On the Internet, you have found a voltage drop of $U_D = 2.3 V$ for red LEDs. Now you want to know which resistor from the E12 series (available to you) is the correct one.

  - Draw the circuit of the LED with series resistor and plot the voltages. 
  - What must be the value of the series resistor $R_V$ to give the voltage drop $U_S$ above? 
  - In the red LED datasheet {{electronic_circuitry:tlur640.pdf|TLUR6400}}, find the relationship between forward voltage ("Forward Voltage") and forward current ("Forward Current"). What voltage drop $U_D$ can be determined from this for $I_D = 20 mA$ and what is the correct series resistance? <WRAP onlyprint> \\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ </WRAP>.
  - Briefly compare the different results considering the [[https://de.wikipedia.org/wiki/E-Reihe#Werte|E12 series]]. <WRAP onlyprint> \\\ \\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ </WRAP>

</WRAP></WRAP></panel><WRAP pagebreak></WRAP>
<panel type="info" title="Task 2.1.3 Z-diode as voltage reference"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>

{{ electronic_circuitry:z-diode_as_voltage_reference.png?nolink&300|z-diode-as-voltage-reference}}

In a voltage reference circuit, a Z-diode enables - despite a fluctuating input voltage - an output voltage to be kept relatively constant. In the simplest case, a circuit as shown on the right can be used for this purpose.
The following quantities are to be used for the task:
  * Source voltage $U_q=7.0 ... 13,0V$ (e.g. via sinusoidal input voltage with $f = 50 Hz$),
  * series resistor $R_V=1,0k\Omega $
  * load resistor $R_L=10k\Omega $
  * Z-diode $D_Z$ as $BZX84C6V2$ ($U_Z = 6,2 V$)

An ideal (Z) diode would be assumed to hold the breakdown voltage $U_Z$ at all reverse currents $I_S$. \\
This is to be checked for the real diode.

  - Model the circuit in Tina TI and insert a picture of the circuit.
  - Compare the progression of $U_L$ to $U_q$. In particular, measure the maximum and minimum values of $U_L$. 
  - Change the load resistance to $R_L=1.0k\Omega $ and perform the same comparison of $U_L$ to $U_q$ again. <WRAP onlyprint> \\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ </WRAP>.
  - How can the difference be explained? <WRAP onlyprint> \\ \ \ \ \ \ \ \ \ \ \ \ \ \ </WRAP>

</WRAP></WRAP></panel>
<WRAP pagebreak></WRAP>
<panel type="info" title="Task 2.1.4 Calculating the differential resistance of a diode"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>

<WRAP right>
<imgcaption imageIdealizedDiode | Idealized Diode>
</imgcaption>
{{drawio>ImageIdealizedDiode}}
<WRAP>

The differential resistance $r_D$ of a diode was already described in the chapter. This is necessary if a diode is to be simulated via a simplified diode model (voltage source + resistor + ideal diode, if applicable). In <imgref imageIdealizedDiode>, see the differential conductance $g_D={{1}\over{r_D}}$ as the local slope at the desired operating point.
Calculate the differential resistance $r_D$ at forward current $I_D=15 mA$ for room temperature ($T=293K$) and $m=1$ from Shockley's equation: ${I_F = I_S(T)\cdot (e^{\frac{U_F}{m\cdot U_T}}-1)}$ with $U_T = \frac{k_B \cdot T}{e}$.
To do this, first calculate the general formula for the differential resistance $r_D$.

Steps:
  - First, simplify Shockley's equation for $U_F >> U_T$ <WRAP onlyprint> \frac {d I_F}{d U_F}$.<WRAP onlyprint> \frac {d I_F}{d U_T$.
  - Find a formula for $\frac {d I_F}{d U_F}$.<WRAP onlyprint> \\ \ \ \ \ \ \ \ \ \ \ \ \ \ </WRAP>
  - Again, replace part of the result with $I_F$ and rotate the fraction to calculate the differential resistance by $r_D = \frac {d U_F}{d I_F}$. \ As a result, you should now have $r_D = \frac {d U_F}{d I_F} = \frac {m \cdot U_T}{I_F} $ <WRAP onlyprint> \\ \ \ \ \ \ \ \ \ \ \ \ \ \ </WRAP>.
  - Calculate $r_D$. <WRAP onlyprint> \\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \WRAP>

</WRAP></WRAP></panel><WRAP pagebreak></WRAP>
<panel type="info" title="Task 2.1.5 two LEDs on one digital output"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>
For a circuit you need to control two LEDs, but unfortunately only one digital output is free. The supply voltage of the microcontroller and the board is 5V. LED1 should have a forward voltage $U_{f,1}=2,3V$ (red), LED2 should have a forward voltage $U_{f,2}=2,1V$ (green).

  - Design a circuit from the existing voltage levels so that both LEDs can be driven. \\ Existing voltage levels:
    - Supply: 5V
    - Digital output: 5V, 0V or high impedance
    
  - Calculate the values for the other necessary components <WRAP onlyprint> \\ \ \ \ \ \ \ \ \ \ \ \ \ </WRAP>.
</WRAP></WRAP></panel><panel type="info" title="Task 3.3.1 Analysis of impedance converter circuit with different operational amplifiers"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>

<WRAP right><panel type="default">
<imgcaption pic3_3_1 | passive and active shielding>
</imgcaption>
{{drawio>pic3_3_1 }}
</panel></WRAP>

Imagine that you work in the company "HHN Mechatronics & Robotics", which wants to build a cheap mobile [[https://de.wikipedia.org/wiki/Elektrokardiogramm|EKG]] - that is a measuring device for the electrocardiogram, or the heart tension curve - for athletes and people in need.
The measurement signal here has only a few millivolts and microamperes. To protect the signal from electromagnetic radiation on its way from the glued-on electrode to the evaluation electronics, a [[https://de.wikipedia.org/wiki/Abschirmung_(Electrical engineering)#Electromagnetic_Fields|Shielding]] is placed around the lead (see <imgref pic3_3_1>, above). However, this sets up a parasitic capacitor, so a colleague suggested active shielding to you. In this case, the shield is always held at the measurement voltage applied to the line via a voltage follower (see <imgref pic3_3_1>, below). The parasitic capacitor is never charged by this setup, since there is the same voltage on both its sides - there is no distortion of the signal.
**It is important for the application that the voltage follower reacts quickly.

You are supervised with the design of this voltage follower and shall analyze the available operational amplifiers $LM318$, $uA741$ and $uA776$ in the voltage follower circuit (see script page).

A short report (problem description, circuit from Tina, results, discussion) is to be prepared; Tina TI is to be used as the analysis tool.

  - Model the circuit described above for a realistic operational amplifier in Tina. Use a voltage generator as the source as a [[https://de.wikipedia.org/wiki/Heaviside-Funktion|jump function]] ("Unit step") with the amplitude $U_A = 1.0 V$ .
  - Use "Analysis">>"Transient..." to simulate the time response for the specified operational amplifiers. \\ Determine the time that elapses until the output value of $0.1 V$ reaches $0.9 V$ for the first time (10% to 90% of the amplitude, also called [[https://de.wikipedia.org/wiki/Anstiegs-_und_Abfallzeit|rise time]]).
  - Describe the time course in each case. Are there any other differences besides the rise time?
  - Which of the three operational amplifiers would you choose - based on the given information - for the problem?

In-depth information (not relevant for homework):

  * Master's thesis on [[https://www.uni-muenster.de/imperia/md/content/fachbereich_physik/technik_didaktik/entwicklung_und_bau_eines_demonstrationsmessg.pdf#page=42|Development and Construction of a Demonstration Measuring Device]]
  * Paper [[https://pdfs.semanticscholar.org/f2b9/2ee5f3e3adb034e18ddb85bd3770dc8c2c29.pdf|on the Stability of Shield-Driver Circuits]]
  * Detailed description of a [[https://www.elektronik-kompendium.de/public/schaerer/diffemg.htm|EMG/EKG preamplifier circuit]]
</WRAP></WRAP></panel><panel type="info" title="Task 3.3.2 Additional Task to 3.3.1"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>

<WRAP right>
{{ electronic_circuitry:idealeropv.jpg?300|}}\ <fs 70%>figure 1 real OPV</fs>
<WRAP>

This task should also be based on the same problem as in task 3.3.1. This time, however, the influence of the impedance converter on the sensor signal is to be analysed. The signal curve of the ECG voltage must be simulated by the voltage source. To do this, select "Piecewise linear" as the voltage source (3rd from the right) and use the curve shown below. \\ In addition, the influence on the current (=output resistance of the sensor) must be taken into account. To do this, insert a resistor with 20 MOhm directly before the non-inverting inputs.

++++current|
<code>
120ms, 0
160ms, 0.1m
200ms, 0
260ms, 0
270ms, -0.2m
320m,1m
340m,-0.3m
350m,0
440ms,0
520ms,0.1m
600ms,0
820ms, 0
860ms, 0.1m
900ms, 0
960ms, 0
970ms, -0.2m
1020m,1m
1040m,-0.3m
1050m,0
1140ms,0
1220ms,0.1m
1300ms,0
1400ms,0
</code>
++++

  - Now simulate the time course for 0 s to 2 s with "Use initial conditions".  
  - What is the only difference between the output signals? (note y-axis)
  - For the real operational amplifier, 3 additional current sources and one additional voltage source must be taken into account in the equivalent circuit diagram (cf. figure on the right). In Tina, the values of these current sources can be viewed in the specifications of the simulated operational amplifier (double click on the OPV in Circuit >> Type ... >> Model Parameters). The analysis of the whole circuit (incl. equivalent circuit) will not be discussed here. It is assumed that the "input bias current" of the operational amplifier is the decisive value. \\ Compare the values of the different operational amplifiers.
  - For simplicity, consider the additional input voltage resulting from the current of a single bias current source at the sense resistor.
</WRAP></WRAP></panel><panel type="info" title="Task 3.3.3 Using Amplifiers in Microcontrollers"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>
In the application notes [[http://ww1.microchip.com/downloads/en/appnotes/90003132a.pdf|MTP3132]] of the manufacturer Microchip, a block diagram is given on the 2nd page, which explains the use of the operational amplifier module of various PIC controllers.
  - In the block diagram the logic input signal "UG" is drawn. What happens in the logic circuit when this takes the value TRUE?
  - Which operational amplifier circuit is set by this?
  - For what purpose is this set operational amplifier circuit used?
</WRAP></WRAP></panel><panel type="info" title="Task 3.4.1 Noninverting Amplifier"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>

Derive the voltage gain for the non-inverting amplifier. Use the procedure that was used for the voltage follower.
</WRAP></WRAP></panel>

<WRAP pagebreak></WRAP>
<panel type="info" title="Task 3.4.2 Input resistance of feedback systems"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>
In Section 3.4.2, the output resistance from the noninverting amplifier was calculated. Analogously, determine the relationship between the input resistance $R_{e,cL}$ of a closed-loop amplifier circuit and the open-loop input resistance $R_{e,oL}$.
</WRAP></WRAP></panel><WRAP pagebreak></WRAP><panel type="info" title="Task 3.5.1 inverting amplifier"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>

Derive the voltage gain for the inverting amplifier. Use the procedure that was used for the non-inverting amplifier. \\ Consider that for the differential gain $A_D$ of the ideal OPV applies: $A_D \rightarrow \infty$. \ Thus, it also holds: $1/A_D \rightarrow 0$ , **but** it does not always hold ${{C}\over{U_x \cdot A_D}} \rightarrow 0$, for an unknown contant $C$ and a voltage $U_x$!
    * What is sought?
    * Number of variables?
    * Number of equations needed?
    * Formulation of the known equations.
    * Deriving the voltage amplification <wrap onlyprint> \\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ </wrap>.
  - Which of the amplifiers (inverting or non-inverting) has a lower input impedance? Why?
</WRAP></WRAP></panel><WRAP pagebreak></WRAP>
<panel type="info" title="Task 3.5.2. variations of the non-inverting amplifier"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>

<wrap noprint>Bottom</wrap><wrap onlyprint>On the following pages</wrap> you will find circuits with an ideal operational amplifier, which are similar to the non-inverting amplifier and whose voltage gain $A_V$ is to be determined.

__assumptions__
  * $R_1 = R_3 = R_4 = R$
  * $R_2 = 2 \cdot R$
  * $U_E$ comes from a low impedance source
  * $U_A$ is connected to a high impedance load

__Tasks__
  - For each circuit, give the voltage gain $A_V$. Detailed calculation as before is not necessary.
  - For Figure 7, state how the voltage gain can be determined.
  - Generalize with justification how
    - a short circuit of the two OPV inputs has to be considered,
    - resistors are to be taken into account if they are
      - with one terminal each ("on one side") directly and exclusively connected to one OPV input,
      - with one terminal each connected directly to an OPV input.
  - In which circuits do resistors $R_3$ and $R_4$ represent an unloaded voltage divider?

To approach the problems, try to use the knowledge from the inverting amplifier. It may be a good idea to simulate the circuits using [[http://www.falstad.com/circuit/|Falstad-Circuit]] or Tina TI. As a support you can see tips below the figure in the first two circuits.  
\\ \\
**__Important__**: As always in your studies, you should try to generalize the knowledge from the assignment.

<WRAP onlyprint>
**Tips**.
  * What is the current flow into the inverting and non-inverting inputs for an ideal op amp? So what would be the voltage drop across a resistor with one terminal going to only one input of the op amp?
  * The operational amplifier always tries to put out as much current at the output so that the required minimum voltage $U_D$ results between inverting and non-inverting input. How large can $U_D$ be assumed to be? Can this voltage also be built up via a resistor?
  * Can different resistors (e.g. because they are between the same nodes) be combined?
</WRAP>

<WRAP pagebreak></WRAP><WRAP group><WRAP half column>

<WRAP right><panel type="default">
Fig. 1 {{drawio>pic3_5_2_task1 }}
{{drawio>pic3_5_2_Task1 }}
</panel></WRAP>
\\ \\ \\ \\ \\ \\ \\ \\
<WRAP noprint>
++++ tips|
  * What is the current flow into the inverting and non-inverting input for an ideal op amp? Thus, what would be the voltage drop across a resistor with one terminal leading to only one input of the op amp ($R_3$)?
  * The operational amplifier always tries to put out as much current at the output, so that between inverting and non-inverting input the required minimum voltage $U_D$ results. How large can $U_D$ be assumed to be? Can this voltage also be built up via a resistor ($R_4$)?
++++
</WRAP>

</WRAP><WRAP half column>

<WRAP right><panel type="default">
Fig. 2 {{drawio>pic3_5_2_task2 }}
{{drawio>pic3_5_2_Task2 }}
</panel></WRAP>

\\ \\ \\ \\ \\ \\ \\ \\
<WRAP noprint>
++++ tips|
  * How much current must flow across $R_4 = R$ to result in the expected voltage $U_4$?
  * Correspondingly, how much current must flow across $R_2 = 2 \cdot R$?
  * So how much current flows through $R_1 = R$? so what voltage results at $R_1$?
++++

</WRAP></WRAP></WRAP>

<WRAP onlyprint><WRAP half column>
<WRAP group><WRAP half column>

<WRAP right><panel type="default">
Fig. 3 {{drawio>pic3_5_2_task3 }}
{{drawio>pic3_5_2_Task3 }}
</panel></WRAP>
</WRAP><WRAP half column>

<WRAP right><panel type="default">
Fig. 4 {{drawio>pic3_5_2_task4 }}
{{drawio>pic3_5_2_Task4 }}
</panel></WRAP>
</WRAP></WRAP>

<WRAP pagebreak></WRAP>
<WRAP group><WRAP half column>

<WRAP right><panel type="default">
Fig. 5 {{drawio>pic3_5_2_task5 }}
{{drawio>pic3_5_2_Task5 }}
</panel></WRAP>
</WRAP><WRAP half column>

<WRAP right><panel type="default">
Fig. 6 {{drawio>pic3_5_2_task6 }}
{{drawio>pic3_5_2_Task6 }}
</panel></WRAP>
</WRAP></WRAP>

<WRAP onlyprint><WRAP half column>
<WRAP group><WRAP half column>

<WRAP right><panel type="default">
Fig. 7 {{drawio>pic3_5_2_task7 }}
{{drawio>pic3_5_2_Task7 }}
</panel></WRAP>
</WRAP><WRAP half column>

<WRAP right><panel type="default">
Fig. 8 __.
{{drawio>pic3_5_2_Task8 }}
</panel></WRAP>
</WRAP></WRAP>

<WRAP pagebreak></WRAP>
<WRAP group><WRAP half column>

<WRAP right><panel type="default">
Fig. 9 {{drawio>pic3_5_2_task9 }}
{{drawio>pic3_5_2_Task9 }}
</panel></WRAP>
</WRAP><WRAP half column>
</WRAP></WRAP>

</WRAP></WRAP></panel><WRAP pagebreak></WRAP><panel type="info" title="Task 3.5.3. R-2R Ladder"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>

<WRAP right>
{{ electronic_circuitry:dac.jpg?400|}}
<fs 70%>figure 1 </fs>
</WRAP>
You work in a company called "HHN Mechatronics & Robotics", which is to build a battery model for a customer. This model is to emulate a real battery. For this purpose, a voltage is to be output, which is specified by a software model of the battery. A digital-to-analog converter (DAC) is therefore required.

You have found the DAC7741 for this purpose. In the data sheet on [[http://www.ti.com/lit/ds/symlink/dac7741.pdf#page=12|page 12]] you can see an image of the internal structure - this is similar to the figure on the right. For an error analysis you now want to understand this structure more closely.

The drawing on the right shows the current switch position for 000b, i.e. all switches $SW_1$ ... $SW_3$ are connected to ground. It is a good idea to rebuild the circuit in [[http://www.falstad.com/circuit/|Falstad-Circuit]] for a better understanding. In this case it is useful to measure the single node voltages $K_1$ ... $K_3$ with to measure.

  - Let $SW_3$ = 1, $SW_2$ = 0 and $SW_1$ = 0 - only the switch $SW_3$ is connected to $U_{logic}$.
    - To do this, draw the equivalent circuit without the switch.
    - Simplify this equivalent circuit by an equivalent resistor.
    - This results in a resistor which is between the inverting and non-inverting input. The operational amplifier always tries to feed as much current into the resistor net surrounding it, so that a small differential voltage $U_D$ results. This is also possible with a (not too small) resistor between inverting and non-inverting input. \\ So what is the resulting gain?
  - Let $SW_3$ = 0, $SW_2$ = 1 and $SW_1$ = 0 - so only the switch $SW_3$ is switched to $U_{logic}$.
    - Again, draw the equivalent circuit without the switch.
    - Simplify this equivalent circuit using equivalent resistors as well.
    - Again, the statement about the above resistance between inverting and non-inverting input applies. Furthermore the voltage of the node $K_3$ should be clear to you. \ Now draw an equivalent circuit of the left-hand side, assuming the voltage at node $K_3$ of the ideal amplifier.
    - Now determine the voltage at node $K_2$.
    - This voltage at node $K_2$ is the input voltage of an inverting amplifier, which starts from node $K_2$ to the right. Now calculate the gain of the resulting network.
  - By now the concept should be understood. Now indicate which input/switch specifies the [[https://en.wikipedia.org/wiki/Bit_numbering#Least_significant_bit|LSB]].

</WRAP></WRAP></panel><WRAP pagebreak></WRAP><panel type="info" title="Task 3.5.4. Converting a unipolar signal to a bipolar signal"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>

<WRAP right 60%>{{url>https://www.falstad.com/circuit/circuitjs.html?ctz=CQAgjCAMB0l5YCcyWrQDhAJmuyBmMANiwBZ1FSBWLIsUkKyRkU-AKAENxEtwB2IiHxZmYQSETgqIALRgZYePDkwi6dEnRMN4qmBGro69KNr48kbUTqNlkdgCVwpBsWZ5wRZmNa+fUNBU7ABOwqICQiLMWFRCMcqh4TFxycL48eCJAOZp+Bl5NFDsAO5psUJIfBXFZVU8fPU1DmXRkazk7Q6i-B2Y4kI0-RIMAKoA+gCCpS5u3g1ePuw9s4ur9Axj4wCi7ADO5amkcNipvgBmnAA2ewCmSccpQqREDM3Yic4vb6lDp5nMY4sAIwYIAD2wniwSGwiF6WARrGwDEc7Ah+BOCN6hBiGKRYH6aNh2MgUUgJKkbkwWCJWH48MQ1MowgQSLIIFREO+2HwlP4b1IfCpwiJrjehFYtiw+E2yI5RIsQjIlMg1SKApAAGV2AB7LJeKB+RnA6AQGL6jh6-qZY7GpSBCAQPiYDhAA noborder}}
</WRAP>

You are working in the company "HHN Mechatronics & Robotics" and should generate a bipolar signal (-10V..+10V) from a unipolar signal of a digital-to-analog converter (0..5V). A colleague has recommended the circuit shown on the right.

  - First, analyze what change occurs when the S switch is pressed. How does the output signal change?
  - Try to mathematically determine the relationship of $U_A$ and $U_E$ as $U_A(U_E)$ by superposition.
  - The circuit still has the problem that the positive half wave is output negative. What other circuit must be provided behind it to solve this problem?

</WRAP></WRAP></panel><WRAP>
<panel type="info" title="Task 4.1.1 Transfer Task - Signals and Systems"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>

What form of oscillation appears to result from the superposition of the sinusoidal signals if they are each doubled in frequency and halved in effective amplitude? (Initial state of the simulation above)

What mode of oscillation appears to result when the $80Hz$ and $160Hz$ voltage sources are removed?
</WRAP></WRAP></panel></WRAP><WRAP pagebreak></WRAP><panel type="info" title="Task 4.3.1 Instrumentation Amplifier"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>
Find out how the instrumentation amplifier works by doing an internet search. \\

    - To do this, look at the <wrap noprint>below</wrap><wrap onlyprint>on the wiki at [[https://wiki.mexle.org/elektronische_schaltungstechnik/4_grundschaltungen_ii#instrumentenverstaerker|4 Basic Circuits II : Instrumentation Amplifiers]]</wrap> and change the variable resistor. In particular, analyze the minimum and maximum values of the output voltage. This can be done by hovering the mouse pointer over the output signal.
    - What happens if you remove the variable resistor and leave the connections open? What is the circuitry of the OPVs at the input now?<wrap onlyprint> \\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \/wrap>
    - Compare the following situations. What can be determined?
      - variable resistor is replaced by resistor with 2 kOhm.
      - variable resistor is replaced by 1 kOhm to ground at the top terminal and 1 kOhm to ground at the bottom terminal.
      - variable resistor is replaced at the top terminal by 1 kOhm to a voltage source of 1 V and at the bottom by 1 kOhm to 1 V. 
  - What is the transfer equation of the initial setup when all resistors (except the variable resistor $R_g$) have the same value $R$? $U_A = f(U_2,U_1, R_g, R) = ?$ <wrap onlyprint> \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ </wrap>
  - What are the advantages of the instrumentation amplifier over the differential amplifier? <wrap onlyprint> \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ </wrap>
</WRAP></WRAP></panel><WRAP onlyprint>
<fs xx-large>exercise sheet 1</fs>
\\ \\ \\ \\
Please upload the padded PDF to ILIAS.

For details, tips, and tools for filling in and inserting images, see: \\ [[:tools_for_teaching_learning_events|Tools for Teaching/Learning Events]]
\\ \\ \\ \\

^ Name ^ First name ^ Matriculation number ^
| $\quad\quad\quad\quad\quad\quad\quad\quad$ \\ (nbsp) \\ (nbsp)| $\quad\quad\quad\quad\quad\quad\quad\quad$| $\quad\quad\quad\quad\quad\quad\quad\quad$ |
| (nbsp) \\ (nbsp) \ (nbsp)| | |
\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\
</WRAP>
<panel type="info" title="Task 1.1.1 Microphone Amplifier I">
<WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>
Given an amplifier circuit which is to amplify a microphone signal so that a loudspeaker ($R_{LS}= 8.0 \Omega$) can be driven. Let the [[https://de.wikipedia.org/wiki/Effektivwert#Sinusf%C3%B6rmige_Spannung|effective value]] of the desired voltage across the speaker be $U_{eff,LS} = 10 V$. It is assumed that a sinusoidal signal is to be output. The voltage supply is done by two voltage sources with $V_{S+} = 15 V$ and $V_{S-} = - 15 V$.

For understanding (especially for task 2. and 3.) look at the simulation under the [[https://wiki.mexle.org/elektronische_schaltungstechnik/1_grundlagen_zu_verstaerkern?do=edit#ersatzschaltbild|Subchapter Equivalent Circuit in chapter "1. Basics"]]. There, a realistic amplifier is already built, but the current flow can already be guessed.

  - Draw a labeled sketch of the circuit with the amplifier as the black box. <WRAP onlyprint> \\ \ \ \ \ \ \ \ \ \ \ \ \ </WRAP>.
  - What power $P$ does the speaker draw? <WRAP onlyprint> \ \ \ \ \ \ \ \ \ \ \ \ \ </WRAP>
  - From this, how can we determine the rms current $I_{eff,S}$ of the power supply at which the above desired voltage $U_{eff,LS}$ is output at the speaker?  
  - From the previous task, determine the minimum maximum current $I_{max,S}$ for which the two power supplies must be designed. \ (Note that for simple amplifiers, the output current $I_A$ is always less than or equal to the current $I_S$ of the power supply.)<WRAP onlyprint> \ \ \ \ \ \ \ \ \ \ \ \ \ \/WRAP>
</WRAP></WRAP></panel>

<WRAP pagebreak></WRAP>
<panel type="info" title="Task 1.1.2 Microphone Amplifier II">
<WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>
Given a voltage amplifier circuit, which is to amplify a microphone signal so that a loudspeaker ($R_{LS}= 8.0 \Omega$) can be driven. Neither amplification nor the desired voltage at the loudspeaker is known. This amplifier circuit is internally protected against overcurrents above $I_{max,amplifier}= 5,0 A$ by a [[https://de.wikipedia.org/wiki/Schmelzsicherung#Ger%C3%A4teschutzsicherungen_(fine-wire fuses)|fuse]]. It is known that no overcurrents occur in the allowed voltage operation of $8.0 \Omega$ loudspeakers.
  - By what factor does the current change if a $4.0 \Omega$ speaker is used instead of an $8.0 \Omega$ speaker? <WRAP onlyprint> \WRAP>.
  - What effect will this have on the backup? <WRAP onlyprint> \\ \ \ \ \ \ \ \ \ \ \ \ \ \ \/WRAP>
</WRAP></WRAP></panel>

<WRAP pagebreak></WRAP>
<panel type="info" title="Task 1.1.3 Wheatstone bridge circuit">
<WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>

<WRAP right><panel type="default">
<imgcaption pic1_1_3_Wheat|Wheatstone bridge circuit with a temperature sensor>
</imgcaption>
{{drawio>wheatstone_bridge_circuit_tsensor}}
</panel></WRAP>

Imagine that you work in the company "HHN Mechatronics & Robotics". You are developing an IoT system that will be used in a harsh environment and will contain a rechargeable battery. The temperature of the battery must be monitored during operation and charging. If the temperature is too high, charging must be aborted or a warning issued. For the temperature measurement at the housing of the used lithium-ion cell {{electronic_circuit_engineering:ncr18650b.pdf|NCR18650}} a measuring circuit is to be built.
A suggested circuit is as follows:
  - Wheatstone bridge circuit with $R_1 = R_2 = R_3 = R_4 = 1.0k \Omega $.
  - Let the resistor $R_4$ be a PT1000 with a temperature coefficient $\alpha = 3850 \frac{ppm}{K}$.
  - For the other resistors, resort to components that have an unknown temperature coefficient according to the data sheet, but within $\alpha = \pm 100 \frac{ppm}{K}$.
  - The voltage source of the system generates a voltage of 5V with sufficient accuracy.
  - The determined voltage $\Delta U$ is amplified by a factor of 20 by another amplifier circuit, output as $U_{A}$, and further used by an analog-to-digital converter in a microcontroller [(Note3>In real systems, an analog-to-digital converter would most likely not be used, as it has a relatively large power consumption for IoT applications. For Atmel chips, this is a few $10$\mu A$, which adds up to a rapid battery drain over time)].

~~PAGEBREAK~~~~CLEARFIX~~
A short report is to be created; Tina TI is to be used as the analysis tool.
  - Create a problem description. <WRAP onlyprint> <WRAP pagebreak></WRAP>.
  - Rebuild the circuit in TINA TI and include it <wrap noprint>in your description </wrap><wrap onlyprint>here</wrap>. Consider the following note when doing so. <panel type="info" title="Note">Use a simple resistor for in the simulation for the PT1000. For Tina TI, 27°C (room temperature) is selected as the reference temperature for the temperature profile. For the PT1000, the reference temperature is often 0°C (in practical applications, this should be checked in the data sheet). With Tina TI, the reference temperature can be changed by entering the value 27 under ''Temperature [C]'' in the properties (double-click on Resistance).</panel><WRAP onlyprint> \\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \WRAP>
  - From the datasheet linked above, determine in what range from $T_{min}$ to $T_{max}$ may be charged and what temperature $T_{lim}$ may not be exceeded in any of the states.  <WRAP onlyprint> \\\ \\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ </WRAP>.
  - First, for temperature invariant $R_1 = R_2 = R_3 = 1.0k \Omega$ and a temperature invariant $R_4$, determine the voltage change $\Delta U$ over the temperature of $-30...70°C$ in TINA TI. To do this, create a plot with $\Delta U$ as a function of temperature.  \\ Read $\Delta U^0 (T_{min})$, $\Delta U^0 (T_{max})$, $\Delta U^0 (T_{lim})$, from the graph and plausibilize the values by calculation. <WRAP onlyprint> \Delta U^$, \Delta U^$, \Delta U^$.
  - Determine $\Delta U$ when the temperature dependence of $R_1$, $R_2$ and $R_3$ is considered. To do this, create a suitable diagram with $\Delta U$ as a function of temperature in TINA TI. \ At what voltages $U_A (T_{min})$, $U_A (T_{max})$ must the microcontroller intervene and disable charging? \ At what value $U_A (T_{lim})$ must a warning be issued? 
  - Discuss the results <WRAP onlyprint> \\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ WRAP>

</WRAP></WRAP></panel>
<WRAP onlyprint>
<fs xx-large>exercise sheet 2</fs>
\\ \\ \\ \\
Please upload the padded PDF to ILIAS.

For details, tips, and tools for filling in and inserting images, see: \\ [[:tools_for_teaching_learning_events|Tools for Teaching/Learning Events]]
\\ \\ \\ \\

^ Name ^ First name ^ Matriculation number ^
| $\quad\quad\quad\quad\quad\quad\quad\quad$ \\ (nbsp) \\ (nbsp)| $\quad\quad\quad\quad\quad\quad\quad\quad$| $\quad\quad\quad\quad\quad\quad\quad\quad$ |
| (nbsp) \\ (nbsp) \ (nbsp)| | |
\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\
</WRAP>

<panel type="info" title="Task 1.2.1 Converting Amplifiers">
<WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>

<WRAP right><panel type="default">
<imgcaption pic1_1_3_Wheat|black box of a voltage amplifier>
</imgcaption>
{{drawio>blackboxamplifier}}
</panel></WRAP>

In the lecture the different amplifier types were presented as black boxes. Thereby the adjacent picture was used for the input and output quantities and the ideal resistance values were derived.
In the following, you are to consider how these can be converted into each other by interconnecting them with further passive, electrical components.

How can the following amplifiers be converted into each other? To do this, draw a circuit with the amplifier as shown opposite as a black box in each case.
  
  - Current amplifier in voltage current amplifier <WRAP onlyprint> \\ \ \ \ \ \ \ \ \ \ \ </WRAP>
</WRAP></WRAP></panel>

<WRAP pagebreak></WRAP><panel type="info" title="Task 1.3.1 Back coupling in TINA">
<WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>

<WRAP right>
{{ electronic_circuitry:feedback_tina.jpg?400|}}
</WRAP>

The simulation [[https://wiki.mexle.org/doku.php?id=simulationstools_fuer_elektronische_schaltungstechnik#tina_-_ti|TINA]] is generally used to simulate circuit diagrams. In the following a block diagram of the feedback (see picture) is to be examined. Please download the following file and work on the task contained therein.

File: [[https://wiki.mexle.org/_media/elektronische_schaltungstechnik/aufgabe_1.3.1.tsc|Task 1.3.1.tsc]]

Compare the results with the findings from the chapter [[electronic_circuit_engineering:1_basics_to_amplifiers#feedback|feedback]]!

</WRAP></WRAP></panel>

<WRAP pagebreak></WRAP>
<panel type="info" title="Task 1.3.2 Back coupling calculations">
<WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>


<WRAP right><panel type="default">
<imgcaption pic1_3_2_feedback2|block diagram of a feedback amplifier>
</imgcaption>
{{drawio>block diagramfeedback}}
</panel></WRAP>

For the principle of negative feedback the adjacent block diagram was given in the script. Here $A_D$ is the so-called differential gain, i.e. the gain of the difference between the input voltage $U_E$ and the feedback voltage.
  - Derive the voltage gain $A_V$ as a function of the differential gain $A_D$ and the feedback factor $k$. Note that $A_V = {{U_A}\over{U_E}} = f(A_D, k)$ and give the derivation.<WRAP onlyprint> \\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ </WRAP>
  - What is the voltage gain $A_V$ for an ideal differential gain ($A_D \rightarrow \infty $)? <WRAP onlyprint> \WRAP>
  - Find the voltage gain $A_V$ for feedback $k = 0.001$ with differential gain $A_{D1} = 100,000$ and $A_{D2} = 200,000$. \ Real differential amplifiers, more precisely operational amplifiers, are considered in more detail in Chapter 3. Two operational amplifiers of the same type can have noticeably different values in the differential gain, e.g., due to specimen scattering, aging, or temperature drift. \ Looking at the result from $A_{D1}$ and $A_{D2}$, what can be said about such a variation of a large differential gain value by, say, 50%? 
  - State how the voltage gain behaves for the following feedbacks $k$ with an ideal differential gain and correctly assign the following statements (some are not needed, some are needed more than once): \\ (A) positive feedback, \ (B) negative feedback, \ (C) damping, \ (D) gain, \ (E) voltage gain equals open-loop gain, \ (F) $U_A = U_E$, \ (G) $U_A = - U_E$, \ (H) gain equals 0. \.
    - $k < -0$ <WRAP onlyprint> \\ \\ </WRAP>
    - $k = 0$ <WRAP onlyprint> \\ \ </WRAP>
    - $0 < k < 1$ <WRAP onlyprint> \\ \ </WRAP>
    - $k = 1$ <WRAP onlyprint> \\ \ </WRAP>
    - $k > 1$ <WRAP onlyprint> \\ \\ </WRAP>

</WRAP></WRAP></panel>
<WRAP pagebreak></WRAP><WRAP onlyprint>
<fs xx-large>exercise sheet 3</fs>
\\ \\ \\ \\
Please upload the padded PDF to ILIAS.

For details, tips, and tools for filling in and inserting images, see: \\ [[:tools_for_teaching_learning_events|Tools for Teaching/Learning Events]]
\\ \\ \\ \\

^ Name ^ First name ^ Matriculation number ^
| $\quad\quad\quad\quad\quad\quad\quad\quad$ \\ (nbsp) \\ (nbsp)| $\quad\quad\quad\quad\quad\quad\quad\quad$| $\quad\quad\quad\quad\quad\quad\quad\quad$ |
| (nbsp) \\ (nbsp) \ (nbsp)| | |
\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\
</WRAP>

{{page>uebung_2.1.1&nofooter}}
{{page>uebung_2.1.2&nofooter}}
{{page>uebung_2.1.3&nofooter}}
{{page>exercise_2.1.4&nofooter}}<WRAP onlyprint>
<fs xx-large>exercise sheet 4</fs>
\\ \\ \\ \\
Please upload the padded PDF to ILIAS.

For details, tips, and tools for filling in and inserting images, see: \\ [[:tools_for_teaching_learning_events|Tools for Teaching/Learning Events]]
\\ \\ \\ \\

^ Name ^ First name ^ Matriculation number ^
| $\quad\quad\quad\quad\quad\quad\quad\quad$ \\ (nbsp) \\ (nbsp)| $\quad\quad\quad\quad\quad\quad\quad\quad$| $\quad\quad\quad\quad\quad\quad\quad\quad$ |
| (nbsp) \\ (nbsp) \ (nbsp)| | |
\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\
</WRAP>

{{page>uebung_3.3.1&nofooter}}
{{page>uebung_3.5.1&nofooter}}
{{page>uebung_3.5.2&nofooter}}
<WRAP onlyprint>
<fs xx-large>exercise sheet 6</fs>
\\ \\ \\ \\
Please upload the padded PDF to ILIAS.

For details, tips, and tools for filling in and inserting images, see: \\ [[:tools_for_teaching_learning_events|Tools for Teaching/Learning Events]]
\\ \\ \\ \\

^ Name ^ First name ^ Matriculation number ^
| $\quad\quad\quad\quad\quad\quad\quad\quad$ \\ (nbsp) \\ (nbsp)| $\quad\quad\quad\quad\quad\quad\quad\quad$| $\quad\quad\quad\quad\quad\quad\quad\quad$ |
| (nbsp) \\ (nbsp) \ (nbsp)| | |
</WRAP>

<WRAP pagebreak></WRAP><panel type="info" title="Task 4.2.1 Determining the transfer response of the differential amplifier"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>

In the following you have to calculate the transfer function of the differential amplifier. To do this, you are to follow a few steps.
  - Derive the function $U_A = f(U_{E1}, U_{E2})$ using superposition.
    - To do this, first draw an equivalent circuit for each.
    - Briefly describe the resulting circuit. Which amplification circuit results in each case?
    - Then calculate the voltages $U_A1$ and $U_A2$, and from them $U_A$.
  - Find the function $U_A = f(U_{E1}, U_{E2})$ for the resistance values of the circuit shown.

</WRAP></WRAP></panel>

<WRAP pagebreak></WRAP><panel type="info" title="Task 4.4.1 Transfer resistance of the current-voltage transformer"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>
 Derive the gain, that is, the transfer resistance, for the current-voltage converter. Use the procedure we used for the other amplifiers.

  - Draw a circuit with the relevant voltages, currents, and resistors and the OPV.
  - What are you looking for?
  - Number of variables?
  - Number of equations needed?
  - Set up the known equations
  - Derive the transfer resistance
</WRAP></WRAP></panel>

<WRAP pagebreak></WRAP><panel type="info" title="Task 4.5.1 Voltage-to-current transformer transfer slope"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>

Derive the amplification, i.e. the transmission slope, for the voltage-current transformer here as well. Use the same procedure as we used for the other amplifiers.

  - Draw a circuit with the relevant voltages, currents, and resistors and the OPV.
  - What are you looking for?
  - Number of variables?
  - Number of equations needed?
  - Set up the known equations
  - Derive the transfer slope
</WRAP></WRAP></panel>

<WRAP pagebreak></WRAP><panel type="info" title="Task 4.5.2 Earth-related load"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>

{{ electronic_circuit:earth_relatedelast.jpg?200|}}

If the voltage-current transformer is used as a current source, care must be taken that the load has no contact to ground.

  - Draw the voltage-current transformer with a load which has a contact to ground.
  - Why does the transfer slope derived above no longer apply as a gain factor in this case?
  - Does the output current become higher or lower in this case?

</WRAP></WRAP></panel>

<WRAP pagebreak></WRAP><panel type="info" title="Task 4.6 Meta Question"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>

In preparation for the exam, you should also consider the material and possible questions about it. For this purpose, develop **two individual questions** on the material so far that
  * either concern an aspect that you do not yet understand or
  either a part of the material that you do not understand yet, or * a part that you have learned in the last lesson.
Please note that the level of difficulty of the questions should be such that they can also be included in the exam. Too general formulations ("Explain an operational amplifier") should be avoided.
\The submitted questions will be played back to you via ILIAS and will be included in the exam.
\\
\\
You are welcome to write me other content questions beyond this assignment that you would like an answer to.
 
However, you should then label these separately.
</WRAP></WRAP></panel>====== further tips for TINA TI ======

===== 1. Parameterization of circuits =====

An interesting possibility to play through several identical circuit configurations with different component values is offered by "control objects": This allows component values to be varied automatically. To use them, proceed as follows:
  * **Select Control Objects**: ''Analysis >> Select Control Object'' and click on the desired component. \\ A menu opens in which the properties can be changed. This is similar to the classic properties menu.
  * **Change property values**: Note that the component values cannot be entered directly here. Configuration is done by pressing the ''...'' button to the right of the value entry. Here the "Parameter Stepping" can be entered by start and end value, as well as number of values (sum of intermediate values plus start and end value). \ Example: If the resistance values of 50 k$\Omega$, 100 k$\Omega$, 150 k$\Omega$, 200 k$\Omega$ are to be simulated through, the start value would be: 50 k$\Omega$, the end value: 200 k$\Omega$ and the number of values: 4.
  * **Simulation**: The simulation happens as before. The simulation mode is automatically set to use the parameters. Manually this can be reset, or changed. This is done by ''Analysis >> Mode''. Here the use of the parameterization can be deactivated again by the selection ''Single''.

More tips on this:
  - In the diagram window, note the following:
    * Via ''Legend'' the variation can be displayed as text for a maximum of 20 functions.
    * For a single, selected graph, the individual variation can be displayed via right mouse button >> ''Modified Components...''.
  - If you want to delete some variations that are not needed, the following procedure is recommended:
    * Maximize graph window
    * Show legend (so it is clear which curve name is assigned to which variation)
    * ''View >> Show/Hide Curve'' (un)select the curve name(s) that are not needed.
  - If you have chosen a lot of component values and / or a lot of control objects, the simulation and also a deletion of variations can take longer. Against this it helps to combine **component values cleverly**. \ Example: You want to simulate the 4 variations $\{(R_1 = 10k\Omega, R_2 = 20k\Omega, C_1 = 10nF),$ $(R_1 = 20k\Omega, R_2 = 10k\Omega, C_1 = 20nF),$ $(R_1 = 30k\Omega, R_2 = 30k\Omega, C_1 = 30nF),$ $(R_1 = 40k\Omega, R_2 = 20k\Omega, C_1 = 10nF)\}$. A simulation with all variations then leads to $R_1 \in \{10k\Omega, 20k\Omega, 30k\Omega, 40k\Omega \}$, $R_2 \in \{10k\Omega, 20k\Omega, 30k\Omega \}$, $C_1 \in \{10nF, 20nF, 30nF \} $, thus $ 4 \cdot 3 \cdot 3 = 36 $ results.
      * Here it would be useful to split into several simulations. in this case 2 simulations: ($R_1 \in \{10k\Omega, 40k\Omega \}$, $R_2 = 10k\Omega$, $C_1 = 10nF$ and $R_1 \in \{20k\Omega, 30k\Omega\}$, $R_2 \in \{10k\Omega, 30k\Omega \}$, $C_1 \in \{20nF, 30nF \}$) would be appropriate. These give $2 + 2 \cdot 2 \cdot 2 = 10$ Results.
      * In the diagram window the not needed variations can be deleted with right click ''Delete'' or as shown above under point 2.
      * If the different variations should still appear in one diagram, all graphs can be marked (''<CTRL>+<A>''), copied (''<CTRL>+<C>'') and pasted (''<CTRL>+<V>'') into the second diagram. \ **Caution**: This often creates problems with bodediagrams. In this case, the output into individual amplitude and phase diagrams helps.
      * A representation in a combined bodiagram is possible, but cumbersome:
        * All results of the various simulations - except the last one - must first be output into individual amplitude and phase diagrams.
        * These are then transferred via copy and paste into only one amplitude and phase diagram each.
        * For the last simulation, a combined ground diagram is then output.
        * The amplitude response of the other can be copied directly into this.
        * For the phase response the graphs can be copied. Before pasting, however, a graph must be marked in the phase diagram of the ground diagram. Only then the graphs may be pasted via ''<CTRL>+<V>''. Then the graphs are also displayed in the correct diagram.

===== 2. Importing a SPICE Model =====
==== 2.1 Creating the SPICE Model ====
 
For many components [[wpde>SPICE_(Software)|SPICE models]] are available on the net. These are mostly available as ''*.cir'' files.

The code in this file starts by defining the subcircuit using ''.SUBCKT <name of circuit> <pin_list> PARAMS: <parameter list>'' and ends with ''.ENDS <name of circuit>''.

All comments are prefixed with ''*'' or '';''. For lines beginning with ''.'', comments can lead to errors.

The individual components have a similar designation:

^ Component ^ General nomenclature ^ Example ^ Description ^
| resistor | ''Rxxxx <node1> <node2> <resistance value>'' | ''Rtest 1 Node2 10k'' | resistor named Rtest between node \ ''1'' and ''Node2'' with value $10k\Omega$ |
| capacitor | ''Cxxxx <node1> <node2> <capacitance value>'' | ''C_1 Node2 Node3 10p'' | capacitor named C_1 between node \\ ''Node2'' and ''Node3'' with value $10pF$ |
| inductance | ''Lxxxx <node1> <node2> <inductance value>'' | ''Lpar Node2 1 {L}''           | inductance named Lpar between node \ ''Node1'' and ''1'' with value $L$; \ this value must be given externally |
| diode | ''Dxxxx <node1> <node2> <name of model>'' | ''Dfw Node2 1 D_1N1183_temp'' | diode named Dfw between node \ ''Node2'' and ''1'' with model ''D_1N1183_temp''; \ this function must be provided externally |

Component names can be freely chosen.

Nodes can be names (e.g., ''Node_One'') or numbers (e.g., ''1'').
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<imgcaption image_number1 | Illustration of the CIR export in TINA TI>.
{{electronic_circuitry:tina_cirexport.jpg?400}}
</imgcaption>
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For the description of further details (e.g. models) please refer to the [[http://bwrcs.eecs.berkeley.edu/Classes/IcBook/SPICE/UserGuide/elements_fr.html|SPICE UserGuide by Berkeley]]. A simpler way to create a TINA TI circuit is to export it as a ''*.CIR'' file. This must then be modified in a text editor so that the first lines incl. ''.TRAN ...'' are removed and replaced by ''.SUBCKT <name of circuit> <pin_list> PARAMS: <parameter list>''. At the end, instead of ''.END'', there should be the line ''.ENDS <name of circuit>''.

~~PAGEBREAK~~ ~CLEARFIX~~
==== 2.2 Example of a SPICE Model ====

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<imgcaption imageNo2 | equivalent circuit diagram for electrolytic capacitors>
</imgcaption>
{{drawio>equivalent_circuit_diagram_for_electrolytic_capacitors}}
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In this manual, a replacement model for electrolytic capacitors is to be illustrated (cf. <imgref FigNo2>) and integrated into TINA TI.

In parallel with the actual capacitance $C_{pri}$ is a high impedance resistor $R_{par}$, across which a small leakage current can flow, and a diode $D_{dmy}$. In front of this circuit is the equivalent series resistor $R_{ESR}$ and the equivalent series inductor $L_{ESL}$. This is a simple equivalent model; more complicated models include other capacitors for physical effects of the dielectric.

The code below shows the SPICE model, which must be saved to a file ''C_electrolytic.CIR'' by a text editor.

<sxh text>
.SUBCKT C_el Pin_pos Pin_neg PARAMS: C = 10U

* Author: Tim Fischer (03.06.2020)
* Explanation: simple electrolytic capacitor

* Circuit:
* +-- R_par ---+
* | |
* +-- C_pri ---+
* | |
* ------ L_ESL ---+--- R_ESR ---+-- D_dmy ---+----
* Pin_pos Node1 Node2 Pin_neg
*
*
*

* Code:

	L_ESL Pin_pos Node1 1n * Equivalent Series Inductance of the capacitor
	R_ESR Node1 Node2 1m * Equivalent Series Resistance of the capacitor
	R_par Node2 Pin_neg 1G * Resistance parallel to capacitance
	C_pri Node2 Pin_neg {C} * primary capacitance
	D_dmy Pin_neg Node2 dummyDiode * diode

* Model for Diode
.MODEL dummyDiode D VJ = 0.2

.ENDS C_el
</sxh>

==== 2.3 Import into TINA TI ====

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By following the steps below, the file can be imported:
  - Menu: ''Tools'' >> ''New Macro Wizard...'' (or ''<Ctrl> + <Shift> + M'')
  - In the Macro Wizard - file selection (<imgref BildNr3>).
    - Inserting the macro name. In the example: ''C_electrolytic'' or ''Elko''
    - Opening a ''*.CIR'' file. In the example: The file created in 1.2.
  - In the Macro Wizard - Symbol selection (<imgref BildNr4>).
    - This dialog can be used to automatically create the symbol. However, this is usually a bit clunky.
    - Better is the search in the library via ''Load shape from library''.
    - Additionally ''Show suggested shapes only'' should be deactivated, so that all symbols are displayed.
    - Use ''Number of pins'' to limit the symbols to those with the correct number of pins.
  - In the Macro Wizard - Pin Selection (<imgref ImageNo5>).
    - In the pin selection, the ''unconnected pins'' (red blinking boxes), are to be dropped by drag-and-drop onto the ends of the symbol.
    - In the example this means: Pin_pos to ''+'', Pin_neg to ''-''
  - Saving the Macro
    - Save the macro in a folder that makes sense. A project folder or similar is recommended. The suggested subfolder of TINA TI should be avoided.
  - How to use
    - Once the macro has been created, it can be integrated directly as a component.
    - If the component is to be used again, this is possible via Copy&Paste or via menu: ''Insert'' >> ''Macro...''.
    - Please note that the parameters of the macro have to be set by the simulation. \ ''Double click >> SubCkt-Parameters >> ... >>'' Input

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<WRAP right><imgcaption imageNo3 | Macro Wizard: file selection>{{electronic_circuitry:tina_macrowiz01.jpg?500}}</imgcaption></WRAP>
<WRAP right><imgcaption imageNo4 | Macro Wizard: symbol selection>{{electronic_circuitry:tina_macrowiz02.jpg?250}}</imgcaption></WRAP>
<WRAP right><imgcaption imageNo5 | Macro Wizard: pin selection>{{electronic_circuit_technology:tina_macrowiz03.jpg?250}}</imgcaption></WRAP>
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