Unterschiede
Hier werden die Unterschiede zwischen zwei Versionen angezeigt.
Beide Seiten der vorigen Revision Vorhergehende Überarbeitung Nächste Überarbeitung | Vorhergehende Überarbeitung Nächste Überarbeitung Beide Seiten der Revision | ||
electrical_engineering_1:network_analysis [2021/10/12 20:48] slinn |
electrical_engineering_1:network_analysis [2023/11/23 03:02] mexleadmin |
||
---|---|---|---|
Zeile 1: | Zeile 1: | ||
- | ====== 4. Analysis of direct current networks | + | ====== 4 Analysis of Networks |
- | < | + | < |
- | 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. | + | 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 imageNo1 >). The concepts shown below can also be applied to these networks. | + | In addition, networks also occur in other areas, for example, the momentum flux through a truss or the heat flux through individual hardware elements (<imgref imageNo1> |
- | On the {{https://de.wikipedia.org/ | + | On the {{https://en.wikipedia.org/ |
< | < | ||
- | === Goals === | + | === Learning Objectives |
- | After this lesson, you should: | + | By the end of this section, you will be able to: |
- | + | - < | |
- | - <del>be able to determine the number of nodes, number of (tree and connecting) branches, and number of meshes.</ | + | - < |
- | - <del>be able to construct a complete tree from an electrical network.</ | + | - < |
- | - <del>be able to understand the branch current method, | + | - understand |
- | - Understand | + | |
- | + | ||
- | </ | + | |
- | + | ||
- | <callout type=" | + | |
- | + | ||
- | Due to the shortened semester, only the subchapter [[: | + | |
</ | </ | ||
- | ===== 4.1 Preliminary | + | ===== 4.1 Preliminary |
- | ==== Preparation of the circuit | + | ==== Preparation of the Circuit |
- | < | + | < |
Before the network analysis can be tackled, the circuit must be suitably prepared (cf. <imgref imageNo10 >): | Before the network analysis can be tackled, the circuit must be suitably prepared (cf. <imgref imageNo10 >): | ||
Zeile 38: | Zeile 32: | ||
- Clarify what is given and what is sought | - Clarify what is given and what is sought | ||
- Draw a circuit | - Draw a circuit | ||
- | - Add counting | + | - Add voltage and current |
- | - First draw current and voltage arrows at all sources according to the generator arrow system. | + | - 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. | + | - 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. | - 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: | - Select suitable current and voltage designations. If not already given, then: | ||
Zeile 46: | Zeile 40: | ||
- Do not insert any signs in front of the designators in the circuit. | - 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?" | + | In real applications, it is useful to specify the number of variables ("what is wanted?" |
+ | 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> | It often helps to draw the drawing several times (at least in your head) to have enough space for the identifiers (cf. <imgref imageNo10> | ||
Zeile 52: | Zeile 47: | ||
~~PAGEBREAK~~ ~~CLEARFIX~~ | ~~PAGEBREAK~~ ~~CLEARFIX~~ | ||
- | ==== Graph and Trees ==== | + | ==== Graphs |
- | < | + | < |
- | In the chapter [[: | + | In the chapter [[: |
+ | 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, | ||
- | 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 loops are used as before. A **tree** | + | 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 loops are used as before. A **tree** |
+ | Now a tree is characterized precisely by the fact that it contains __no__ | ||
Tree 3 in <imgref imageNo11> | Tree 3 in <imgref imageNo11> | ||
Zeile 64: | Zeile 61: | ||
The branches in complete trees are now distinguished according to their membership: | The branches in complete trees are now distinguished according to their membership: | ||
- | * **tree branches** | + | * **tree branches** belong to the complete tree (solid lines in <imgref imageNo11> |
* **Connecting branches** | * **Connecting branches** | ||
- | Why does the swing to graph theory make sense now? The trick is that by defining the complete tree, all loops have just been removed. Conversely, a new (independent) loop can be created by each connecting branch. So if the number of independent loop equations $m$ is sought, this is just equal to the number of connecting branches. | + | Why does the excursion |
To do this, proceed as follows: | To do this, proceed as follows: | ||
Zeile 77: | Zeile 74: | ||
Thus, the number of independent loop equations $m$ is findable by counting the nodes $k$ and branches $z$ over $m = v = z - k + 1$. | Thus, the number of independent loop 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:// | + | This explanation can also be heard again in [[https:// |
~~PAGEBREAK~~ ~~CLEARFIX~~ | ~~PAGEBREAK~~ ~~CLEARFIX~~ | ||
- | ===== 4.2 Branch Current | + | ===== 4.2 Branch Current |
- | < | + | < |
- | In the branch current method now " | + | The branch current method |
+ | Specifically, | ||
* for all nodes $k$ respectively the equation: $\sum_{k=0}^{N_k}{I_k}=0$ | * for all nodes $k$ respectively the equation: $\sum_{k=0}^{N_k}{I_k}=0$ | ||
Zeile 92: | Zeile 90: | ||
This forms a linear system of equations. This can then be considered as a matrix equation and solved with the rules of (mathematical) art. | 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> | + | <WRAP onlyprint> |
+ | |||
+ | For the example (<imgref imageNo12> | ||
~~PAGEBREAK~~ ~~CLEARFIX~~ | ~~PAGEBREAK~~ ~~CLEARFIX~~ | ||
The matrices still need to be corrected for the voltage and current sources!!! | The matrices still need to be corrected for the voltage and current sources!!! | ||
- | |||
- | <WRAP group>< | ||
=== Example of nodal equations === | === Example of nodal equations === | ||
- | \begin{align*} \sum\limits_{k=0}^{N_k}{I_k}=0 | + | \begin{align*} \sum\limits_{k=0}^{N_k}{I_k}=0 \end{align*} |
- | Setting up the individual equations: \begin{align*} \scriptsize\text{node ' | + | Setting up the individual equations: |
+ | \begin{align*} | ||
+ | \scriptsize\text{node ' | ||
+ | \scriptsize\text{node ' | ||
+ | \scriptsize\text{node ' | ||
+ | \scriptsize\text{node ' | ||
+ | \scriptsize\text{node ' | ||
+ | \scriptsize\text{node ' | ||
+ | \end{align*} | ||
- | Sorting | + | Sorting |
+ | \begin{align*} \begin{smallmatrix} | ||
+ | \text{node ' | ||
+ | \text{node ' | ||
+ | \text{node ' | ||
+ | \text{node ' | ||
+ | \text{node ' | ||
+ | \text{node ' | ||
+ | \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*} | + | 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*} | ||
- | </ | ||
=== Example of loop equations === | === Example of loop equations === | ||
- | \begin{align*} \sum\limits_{m=0}^{N_m}{U_m}=0 | + | \begin{align*} \sum\limits_{m=0}^{N_m}{U_m}=0 |
- | Setting up the individual equations: \begin{align} \scriptsize\text{loop ' | + | Setting up the individual equations: |
+ | \begin{align*} | ||
+ | \scriptsize\text{Masche | ||
+ | \scriptsize\text{Masche | ||
+ | \scriptsize\text{Masche | ||
+ | \scriptsize\text{Masche | ||
+ | \scriptsize\text{Masche | ||
+ | \quad \\ | ||
+ | \end{align*} | ||
- | Sorting stresses into columns: \begin{align*} \begin{mallmatrix} \text{loop ' | ||
- | 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 & | + | Sorting voltages into columns: |
+ | \begin{align*} | ||
+ | \begin{smallmatrix} | ||
+ | \text{Masche ' | ||
+ | \text{Masche ' | ||
+ | \text{Masche ' | ||
+ | \text{Masche ' | ||
+ | \text{Masche ' | ||
+ | \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 | ||
+ | 0 & +R_1 & -R_2 & -R_3 & 0 & 0 | ||
+ | 0 & 0 & +R_2 & 0 & +R_4 & 0 | ||
+ | 0 & 0 & 0 | ||
+ | 0 & 0 & 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*} | ||
+ | |||
+ | These matrices can be solved using, for example, the [[https://en.wikipedia.org/ | ||
+ | |||
+ | ~~PAGEBREAK~~ ~~CLEARFIX~~ | ||
+ | |||
+ | </ | ||
- | These matrices can be solved using, for example, the {{https:// | ||
- | === another examples | + | === Another example |
<WRAP group> <WRAP half column> | <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 loops | + | In video 1 you will learn: |
- | </ | + | |
- | {{youtube> | + | 2. drawing in and designating the nodes |
+ | |||
+ | 3. draw in and label the loops | ||
+ | |||
+ | 4. draw and label the mesh currents | ||
+ | |||
+ | 5. drawing in and designating the mesh voltages | ||
+ | |||
+ | for a simple circuit | ||
+ | |||
+ | </ | ||
+ | |||
+ | {{youtube> | ||
</ | </ | ||
- | Video 2 describes the following steps: \\ \\ 4. draw and label the branch currents \\ 5. drawing in and designating the branch voltages | + | In video 2 you will learn : |
- | </ | + | \\ 1. writing down the given circuit and sizes |
- | {{youtube> | + | 2. drawing in and designating the nodes |
+ | |||
+ | 3. draw in and label the loops | ||
+ | |||
+ | 4. draw and label the branch currents | ||
+ | |||
+ | 5. drawing in and designating the branch voltages | ||
+ | |||
+ | for a more complex circuit | ||
+ | |||
+ | </ | ||
+ | |||
+ | {{youtube> | ||
</ | </ | ||
- | Video 3 describes the following steps: | + | Video 3 describes the following steps: |
- | </ | + | 1. write down the given circuit and sizes |
- | {{youtube> | + | 2. draw in and designate the nodes |
+ | |||
+ | 3. draw in and label the loops | ||
+ | |||
+ | 4. draw and label the branch currents | ||
+ | |||
+ | 5. draw in and designate the branch voltages | ||
+ | |||
+ | 6. set up node equations and loop equations | ||
+ | |||
+ | 7. convert to matrix notation | ||
+ | |||
+ | 8. solve the matrix | ||
+ | |||
+ | </ | ||
+ | |||
+ | {{youtube> | ||
</ | </ | ||
- | In the [[https:// | + | In the [[https:// |
\\ | \\ | ||
- | 8. inserting the numerical values | + | 9. inserting the numerical values |
- | 9. calculating the matrix with a calculator | + | |
+ | 10. calculating the matrix with a calculator | ||
</ | </ | ||
- | <panel type=" | + | <panel type=" |
- | {{youtube> | + | {{youtube> |
</ | </ | ||
- | <panel type=" | + | ===== 4.3 Mesh Current Method ===== |
- | {{youtube> | + | In the [[https://en.wikipedia.org/ |
- | + | ||
- | </WRAP>< | + | |
- | + | ||
- | ===== 4.3 loop flow method ===== | + | |
- | + | ||
- | In the loop flow method, only for all loops $m$ 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 loop currents. | The advantage here is that the number of equations to be solved is reduced to the number of independent loop currents. | ||
- | These can also be considered | + | These can also be considered matrix equations and can be solved with the rules of (mathematical) art. |
<WRAP group> <WRAP half column> | <WRAP group> <WRAP half column> | ||
- | In video 1, the loop flow method is applied | + | In video 1, the mesh current |
- | + | ||
- | <WRAP important> | + | |
- | </ | + | </ |
{{youtube> | {{youtube> | ||
Zeile 191: | Zeile 286: | ||
</ | </ | ||
- | Also in video 2, the loop flow method is applied | + | Also in video 2, the mesh current |
- | </ | + | </ |
{{youtube> | {{youtube> | ||
Zeile 203: | Zeile 298: | ||
</ | </ | ||
- | ===== 4.4 Nodal potential method | + | ===== 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. | + | In the [[https:// |
The advantage here is that the number of equations to be solved is reduced to the number of existing nodes (minus 1). | 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 | + | These can also be considered matrix equations and can be solved with the rules of (mathematical) art. |
<WRAP group> <WRAP half column> | <WRAP group> <WRAP half column> | ||
Zeile 233: | Zeile 328: | ||
</ | </ | ||
- | ===== 4.5 Superposition | + | ===== 4.5 Superposition |
- | The superposition principle shall first be illustrated by some examples | + | The superposition principle shall first be illustrated by some examples: |
- | <callout title=" | + | <callout title=" |
- | **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? | + | **Task**: Three students are to fill a pool. If Alice has 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. </ | + | 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 title=" | + | <callout title=" |
- | < | + | < |
- | **Task**:A mechanical, linear spring is deflected with masses $m_1$ and $m_2$ in the Earth' | + | **Task**:A mechanical, linear spring is displaced due to masses $m_1$ and $m_2$ in the Earth' |
Again, a linear law is used here: \begin{align*} \vec{s}= f(\vec{F}) = - D \cdot \vec{F} \end{align*} | 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*} </ | + | 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 icon=" | + | <callout icon=" |
+ | The total effect is then the sum of the individual effects. </ | ||
- | For electrical engineering this principle was described by {{https://de.wikipedia.org/ | + | For electrical engineering this principle was described by [[https://en.wikipedia.org/ |
- | > The currents in the branches | + | > The currents in the branches |
<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. | <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. | ||
Zeile 263: | Zeile 365: | ||
The " | The " | ||
- | - Choose next source '' | + | - Choose |
- Replace all ideal sources with their respective equivalent resistors: | - Replace all ideal sources with their respective equivalent resistors: | ||
- ideal voltage sources by short circuits | - ideal voltage sources by short circuits | ||
- ideal current sources by an open line | - ideal current sources by an open line | ||
- Calculate the partial currents sought in the branches considered. | - Calculate the partial currents sought in the branches considered. | ||
- | - Go to the next source '' | + | - Go to the next source '' |
- Add up the partial currents in the branches under consideration, | - Add up the partial currents in the branches under consideration, | ||
Zeile 277: | Zeile 379: | ||
{{youtube> | {{youtube> | ||
- | more complex example of the superposition method | + | A more complex example of the superposition method |
{{youtube> | {{youtube> | ||
Zeile 287: | Zeile 389: | ||
=== Example === | === Example === | ||
- | < | + | < |
~~PAGEBREAK~~ ~~CLEARFIX~~ | ~~PAGEBREAK~~ ~~CLEARFIX~~ | ||
+ | === Introduction to Nodal, Mesh and Superposition Method === | ||
+ | {{youtube> | ||
- | <panel type=" | ||
- | <WRAP right> | + | ~~PAGEBREAK~~ ~~CLEARFIX~~ |
- | 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 $U_{sens} \in [-15...15V]$, | + | ===== Exercises ===== |
- | For conditioning, | + | # |
- | - Find the relationship between | + | 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 $U_{\rm sens} \in [-15...15~\rm V]$, and the microcontroller |
- | - 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 | + | |
- | - 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:// | + | |
- | </WRAP></ | + | For conditioning, |
- | {{page>task_4.5.2_with_calculation&nofooter}}{{page>aufgabe_4.5.3&nofooter}}{{page>aufgabe_4.5.4&nofooter}} | + | |
+ | The following simulation shows roughly the situation (the resistor values are not correct). | ||
+ | |||
+ | < | ||
+ | |||
+ | Questions: | ||
+ | |||
+ | 1. Find the relationship between $R_1$, $R_2$, and $R_3$ using superposition. | ||
+ | |||
+ | # | ||
+ | Using superposition, | ||
+ | For these two circuits, we calculate $U_\rm A^{(1)}$ and $U_\rm A^{(2)}$. \\ | ||
+ | To make the calculation simpler, the resistors $R_3$ and $R_{\rm s}$ will be joined to $R_4 =R_3 +R_{\rm s}$. | ||
+ | |||
+ | < | ||
+ | === Circuit 1 : only consider $U_{\rm S}$, ignore $U_{\rm I}$ === | ||
+ | {{drawio> | ||
+ | |||
+ | \begin{align*} | ||
+ | U_{\rm O}^{(1)} | ||
+ | = U_{\rm S} \cdot {{ {{R_2 R_4}\over{R_2 + R_4}} }\over{R_1 + {{R_2 R_4}\over{R_2 + R_4}} }} \\ | ||
+ | & | ||
+ | & | ||
+ | \end{align*} | ||
+ | </callout> | ||
+ | |||
+ | < | ||
+ | === Circuit 2 : only consider $U_{\rm I}$, ignore $U_{\rm S}$ === | ||
+ | {{drawio>electrical_engineering_1: | ||
+ | |||
+ | \begin{align*} | ||
+ | U_{\rm O}^{(2)} | ||
+ | = U_{\rm I} \cdot {{ {{R_1 R_2}\over{R_1 + R_2}} }\over{R_4 + {{R_1 R_2}\over{R_1 + R_2}} }} \\ | ||
+ | & | ||
+ | & | ||
+ | \end{align*} | ||
+ | </ | ||
+ | |||
+ | < | ||
+ | === Superposition: | ||
+ | \\ | ||
+ | These two intermediate voltages for the single sources have to be summed up as $U_{\rm O}= U_{\rm O}^{(1)} + U_{\rm O}^{(2)}$. \\ | ||
+ | When deeper investigated, | ||
+ | We can also simplify further when looking at often-used sub-terms (here: $R_2$) | ||
+ | |||
+ | \begin{align*} | ||
+ | U_{\rm O} | ||
+ | U_{\rm O} \cdot (R_4 R_1 + R_4 R_2+ R_1 R_2 ) & | ||
+ | U_{\rm O} \cdot ({{R_1 R_4}\over{R_2}} + R_4 + R_1 ) & | ||
+ | \end{align*} | ||
+ | |||
+ | The formula $(1)$ is the general formula to calculate the output voltage $U_{\rm O}$ for a changing input voltage $U_{\rm I}$, where the supply voltage $U_{\rm S}" is constant. \\ | ||
+ | </ | ||
+ | |||
+ | Now, we can use the requested boundaries: | ||
+ | - For the minimum input voltage $U_{\rm I}= -15 ~\rm V$, the output voltage shall be $U_{\rm O} = 0 ~\rm V$ | ||
+ | - For the maximum input voltage $U_{\rm I}= +15 ~\rm V$, the output voltage shall be $U_{\rm O} = 3.3 ~\rm V$ | ||
+ | |||
+ | This leads to two situations: | ||
+ | |||
+ | < | ||
+ | === Situation I : $U_{\rm I,min}= -15 ~\rm V$ shall create $U_{\rm O,min} = 0 ~\rm V$ === | ||
+ | \\ | ||
+ | We put $U_{\rm A} = 0 ~\rm V$ in the formula $(1)$ : | ||
+ | \begin{align*} | ||
+ | 0 & | ||
+ | - U_{\rm I,min} \cdot R_1 & | ||
+ | | ||
+ | \end{align*} | ||
+ | |||
+ | So, with formula $(2)$, we already have a relation between $R_1$ and $R_4$. Yeah 😀 \\ | ||
+ | The next step is situation 2 | ||
+ | </ | ||
+ | |||
+ | < | ||
+ | === Situation II : $U_{\rm I,max}= +15 ~\rm V$ shall create $U_{\rm O,max} = 3.3 ~\rm V$ === | ||
+ | \\ | ||
+ | We use formula $(2)$ to substitute $R_1 = k_{14} \cdot R_4 $ in formula $(1)$, and: | ||
+ | \begin{align*} | ||
+ | U_{\rm O,max} \cdot (k_{14}{{ R_4^2}\over{R_2}} + R_4 + k_{14} R_4 ) &= | ||
+ | U_{\rm O,max} \cdot (k_{14}{{ R_4 }\over{R_2}} + 1 + k_{14} | ||
+ | | ||
+ | | ||
+ | {{ R_4 }\over{R_2}} | ||
+ | \end{align*} | ||
+ | |||
+ | So, another relation for $R_4$ and $R_2$. | ||
+ | </ | ||
+ | |||
+ | So, to get values for the relations, we have to put in the values for the input and output voltage conditions. For $k_{14}$ we get by formula $(2)$: | ||
+ | \begin{align*} | ||
+ | k_{14} = {{R_1}\over{R_4}} | ||
+ | \end{align*} | ||
+ | |||
+ | This value $k_{14}$ we can use for formula $(3)$: | ||
+ | \begin{align*} | ||
+ | {{ R_4 }\over{R_2}} &= {{5 ~\rm V + 15 ~\rm V \cdot {{1}\over{3}} }\over{ 3.3 ~\rm V \cdot {{1}\over{3}} }} - {{1 + {{1}\over{3}} }\over{ {{1}\over{3}} }} \\ | ||
+ | & | ||
+ | k_{42} | ||
+ | \end{align*} | ||
+ | |||
+ | We could now - theoretically - arbitrarily choose one of the resistors, e.g., $R_2$, and then calculate the other two. \\ | ||
+ | |||
+ | But we must consider another boundary, a boundary for $R_{\rm S}$. The maximum voltage and the maximum current are given for the sensor. By this, we can calculate $R_{\rm S}$: | ||
+ | \begin{align*} | ||
+ | R_{\rm S} & | ||
+ | &= 15 ~\rm k\Omega | ||
+ | \end{align*} | ||
+ | |||
+ | Therefore, $R_4 = R_{\rm S} + R_3$ must be larger than this. \\ | ||
+ | |||
+ | # | ||
+ | |||
+ | # | ||
- | ---- | + | The sensor resistance is |
+ | \begin{align*} | ||
+ | R_S &= 15 {~\rm k\Omega}\\ | ||
+ | \end{align*} | ||
- | '' | + | We can choose $R_3$ arbitrarily. Here I choose a nice value to get integer values for $R_3$ and $R_1$: |
+ | \begin{align*} | ||
+ | R_3 &= 45 {~\rm k\Omega}\\ | ||
+ | R_1 &= {{1}\over{3}} | ||
+ | R_2 &= {{1}\over{5.09}}(R_3 + 15 {~\rm k\Omega}) = 11.8 {~\rm k\Omega} | ||
+ | \end{align*} | ||
+ | # | ||
- | ---- | ||
- | ---- | + | * Find the relationship between $R_1$, $R_2$, and $R_3$ using the star-delta transformation. |
+ | * What is the input resistance $R_{\rm in}(R_1, R_2, R_3)$ of the circuit (viewed from the sensor)? | ||
+ | * What is the maximum allowed input resistance ($R_{\rm in}(R_1, R_2, R_3)$) 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:// | ||
+ | * | ||
+ | # | ||
- | ---- | ||
- | ---- | ||
- | ---- | + | {{page> |
+ | {{page> | ||
+ | {{page> | ||