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 momentum flux through a truss or the heat flux through individual hardware elements (Abbildung 1, or an example for heat flow through electronics ). The concepts shown below can also be applied to these networks.
On the wiki page for network analysis the different methods are described very well in a compact way
In order to understand the two-terminal theory / one-port theory, we first have to understand what a Terminal and port is.
So, have a look to Abbildung 1:
But, how could this help us in simplifying circuits?
Well: Usually, the voltage over or the current into one component or a group of component has to be found.
Now, it is practical, that
A linear part is here a cirucit consisting of linear components. In general, ohmic resistors, sources, capacitors and inductors are linear - here, we only look onto resistors. (non-linear are most of the semiconductor components, like diodes).
So, what can we do? Once you search for a distinct voltage or current:
Voilà, we have a way to find our desired voltage or current in a complicated circuitry.
When one is only interested in the resistance of a complex circuit, do as follows:
The superposition principle shall first be illustrated by some examples:
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.
Task:A mechanical, linear spring is displaced due to masses $m_1$ and $m_2$ in the Earth's gravitational field (see Abbildung 3). 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*}
For electrical engineering this principle was described by Hermann_von_Helmholtz:
The currents in the branches of a linear network are equal to the sum of the partial currents in the branches concerned caused by the individual sources.
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:
x
x=x+1
1), and to point 2, as long as the partial currents of all sources have not been calculated.This procedure is explained again in more detail using examples in the two videos on the right.
Simple view of the superposition principle
A more complex example of the superposition method
Abb. 4: example circuit with superposition
Here is a simple exercise …
Here is the solution of the Exercise 1
Here is another simple exercise …
Here is the solution of the Exercise 2
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 input can read values in the range $U_{\rm uC} \in [0...3.3~\rm V]$. The sensor can supply a maximum current of $I_{\rm sens, max}=1~\rm mA$. For the internal resistance of the microcontroller, input applies: $R_{\rm uC} \rightarrow \infty$
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 $R_1$ to a supply voltage $U_{\rm s}$.
The following simulation shows roughly the situation. Be aware that the resistor and voltage values are not correct!
Questions:
1. Find the relationship between $R_1$, $R_2$, and $R_3$ using superposition.
\begin{align*} U_{\rm O}^{(1)} &= U_{\rm S} \cdot {{R_2||R_4}\over{R_1 + R_2||R_4}} = 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}} }} \\ &= U_{\rm S} \cdot {{ R_2 R_4 }\over{R_1 (R_2 + R_4)+ R_2 R_4 }} \\ &= U_{\rm S} \cdot {{ R_2 R_4 }\over{R_1 R_2 + R_1 R_4 + R_2 R_4 }} \\ \end{align*}
\begin{align*} U_{\rm O}^{(2)} &= U_{\rm I} \cdot {{R_1||R_2}\over{R_4 + R_1||R_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}} }} \\ &= U_{\rm I} \cdot {{ R_1 R_2 }\over{R_4 (R_1 + R_2)+ R_1 R_2 }} \\ &= U_{\rm I} \cdot {{ R_1 R_2 }\over{R_4 R_1 + R_4 R_2+ R_1 R_2 }} \\ \end{align*}
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, one can see that the denominator for both $U_{\rm O}^{(1)}$ and $U_{\rm O}^{(2)}$ is the same.
We can also simplify further when looking at often-used sub-terms (here: $R_2$)
\begin{align*} U_{\rm O} &= {{ 1 }\over{R_4 R_1 + R_4 R_2+ R_1 R_2 }} \cdot (U_{\rm S} \cdot R_2 R_4 + U_{\rm I} \cdot R_1 R_2 ) \\ U_{\rm O} \cdot (R_4 R_1 + R_4 R_2+ R_1 R_2 ) &= U_{\rm S} \cdot R_2 R_4 + U_{\rm I} \cdot R_1 R_2 \\ \\ U_{\rm O} \cdot ({{R_1 R_4}\over{R_2}} + R_4 + R_1 ) &= U_{\rm S} \cdot R_4 + U_{\rm I} \cdot R_1 \tag 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:
This leads to two situations:
We put $U_{\rm A} = 0 ~\rm V$ in the formula $(1)$ :
\begin{align*}
0 &= U_{\rm S} \cdot R_4 + U_{\rm I,min} \cdot R_1 \\
- U_{\rm I,min} \cdot R_1 &= U_{\rm S} \cdot R_4 \\
{{R_1}\over{R_4}} &=-{{U_{\rm S}}\over {U_{\rm I,min}}} = k_{14} \tag 2 \\
\end{align*}
So, with formula $(2)$, we already have a relation between $R_1$ and $R_4$. Yeah 😀
The next step is situation 2
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 S} \cdot R_4 + U_{\rm I,max} \cdot k_{14} R_4 \\
U_{\rm O,max} \cdot (k_{14}{{ R_4 }\over{R_2}} + 1 + k_{14} ) &= U_{\rm S} + U_{\rm I,max} \cdot k_{14} \\
k_{14}{{ R_4 }\over{R_2}} + 1 + k_{14} &= {{U_{\rm S} + U_{\rm I,max} \cdot k_{14} }\over{ U_{\rm O,max} }} \\
k_{14}{{ R_4 }\over{R_2}} &= {{U_{\rm S} + U_{\rm I,max} \cdot k_{14} }\over{ U_{\rm O,max} }} - (1 + k_{14})\\
{{ R_4 }\over{R_2}} &= {{U_{\rm S} + U_{\rm I,max} \cdot k_{14} }\over{ k_{14} U_{\rm O,max} }} - {{1 + k_{14} }\over{k_{14}}} \tag 3 \\
\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}} =-{{5 ~\rm V}\over {-15 ~\rm V }} = {{1}\over{3}} \\ \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}} }} \\ &= {{10}\over{1.1}} - 4 \\ k_{42} &\approx 5.09 \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} &= {{ U_{\rm OC} }\over{ I_{\rm SC} }} = {{ U_{\rm S,max} }\over{ I_{\rm S,max} }} = {{ 15 ~\rm V }\over{ 1 ~\rm mA }} \\ &= 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_3 + 15 {~\rm k\Omega}) = 20 {~\rm k\Omega} \\ R_2 &= {{1}\over{5.09}}(R_3 + 15 {~\rm k\Omega}) = 11.8 {~\rm k\Omega} \end{align*}
Based on the E24 series, the following values are next to the calculated ones: \begin{align*} R_3^0 &= 43 {~\rm k\Omega}\\ R_1^0 &= {{1}\over{3}} (R_3 + 15 {~\rm k\Omega}) = 20 {~\rm k\Omega} \\ R_2^0 &= {{1}\over{5.09}}(R_3 + 15 {~\rm k\Omega}) = 12 {~\rm k\Omega} \end{align*}
2. Find the relationship between $R_1$, $R_2$, and $R_3$ by investigating Kirchhoff's nodal rule for the node where $R_1$, $R_2$, and $R_3$ are interconnected.
The potential of the node is $U_\rm O$. Therefore the currents are:
This led to the formula based on the Kirchhoff's nodal rule:
\begin{align*} \Sigma I = 0 &= I_1 + I_2 + I_3 \\ 0 &= {{U_{\rm S} - U_{\rm O}}\over{R_1}} + {{U_{\rm I} - U_{\rm O}}\over{R_4}} - {{U_\rm O}\over{R_2}} \end{align*}
The formula can be rearranged, with all terms containing $ U_{\rm O}$ on the left side: \begin{align*} {{U_{\rm O}}\over{R_1}} + {{U_{\rm O}}\over{R_2}} + {{U_{\rm O}}\over{R_4}} &= {{U_{\rm S}}\over{R_1}} + {{U_{\rm I}}\over{R_4}} \\ U_{\rm O}\cdot \left( {{1}\over{R_1}} + {{1}\over{R_2}} + {{1}\over{R_4}} \right) &= {{U_{\rm S}}\over{R_1}} + {{U_{\rm I}}\over{R_4}} \\ \end{align*}
Both sides can be multiplied by $\cdot R_1$, $\cdot R_2$, $\cdot R_4$ - in order to get rid of the fractions : \begin{align*} U_{\rm O}\cdot \left( {{R_1 R_2 R_4 }\over{R_1}} + {{R_1 R_2 R_4 }\over{R_2}} + {{R_1 R_2 R_4 }\over{R_4}} \right) &= R_1 R_2 R_4 \cdot {{U_{\rm S}}\over{R_1}} + R_1 R_2 R_4 \cdot {{U_{\rm I}}\over{R_4}} \\ U_{\rm O}\cdot \left( R_2 R_4 + R_1 R_4 + R_1 R_2 \right) &= R_2 R_4 \cdot U_{\rm S} + R_1 R_2 \cdot U_{\rm I} \\ U_{\rm O} &= {{R_2}\over{R_2 R_4 + R_1 R_4 + R_1 R_2 }} \left( R_4 \cdot U_{\rm S} + R_1 \cdot U_{\rm I} \right)\\ \end{align*}
The last formula was just the result we also got by the superposition but by more thinking.
So, sometimes there is an easier way…
3. What is the input resistance $R_{\rm in}(R_1, R_2, R_3)$ of the circuit (viewed from the sensor)?
\begin{align*} R_{\rm in}(R_1, R_2, R_3) &= R_3 + R_1 || R_2 \\ &= R_3 + {{R_1 R_2}\over{R_1 + R_2}} \end{align*}
4. What is the minimum allowed input resistance ($R_{\rm in, min}(R_1, R_2, R_3)$) for the sensor to still deliver current?
\begin{align*} R_{\rm in, min} &= {{U_{\rm sense}}\over{I_{\rm sense, max}}} \\ &= \rm {{15 V}\over{1 mA}} \\ &= 15 k\Omega \\ \end{align*}
A circuit is given with the following parameters
$R_1=5 ~\Omega$
$U_1=2 ~\rm V$
$I_2=1 ~\rm A$
$R_3=20 ~\Omega$
$U_3=8 ~\rm V$
$R_4=10 ~\Omega$
Determine the open circuit voltage between A and B using the principle of superposition.
The components can be moved in order to understand the circuit s bit better.
For the open circuit, no current is flowing through any resistor. Therefore, the effect is: $U_{AB,1} = U_1$
(current) source $I_2$
Also here, the components can be shifted for a better understanding:
Here, the current source $I_2$ creates a voltage drop $U_{AB_2}$ on the resistor $R_2$ : $U_{\rm AB,2} = - R_1 \cdot I_2$
(Voltage) source $U_3$
Again, rearranging the circuit might help for an understanding:
In this case, between the unloaded outputs $\rm A$ and $\rm B$ there will be an unloaded voltage divider given by $R_3$ and $R_4$.
On $R_1$ there is no voltage drop since there is no current flow out of the unloaded outputs.
Therefore:
\begin{align*} U_{\rm AB,3} = \frac{R_4}{R_3 + R_4} \cdot U_3 \end{align*}
resulting voltage
\begin{align*} U_{\rm AB} &= U_1 - R_1 \cdot I_2 + \frac{R_4}{R_3 + R_4} \cdot U_3 \\ \end{align*}
A circuit is given with the following parameters
$R_1=5 ~\Omega$
$U_1=2 ~{\rm V}$
$I_2=1 ~{\rm A}$
$R_3=20 ~\Omega$
$U_3=8 ~{\rm V}$
$R_4=10 ~\Omega$
Determine the open circuit voltage between A and B using the principle of superposition.
\begin{align*} U_{\rm AB,1} = \frac{R_4}{R_1+R_4} U_1 = \frac{10~\Omega}{5~\Omega+10~\Omega} \cdot 2~{\rm V} = 1.33~{\rm V} \end{align*} Case 2: For this case is $U_1 = 0~{\rm V}$ and $U_3 = 0~{\rm V}$. The voltage is at $R_3$.
\begin{align*} U_{\rm AB,2} = R_3 I_2 = 20~\Omega \cdot 1~{\rm A} = 20~{\rm V} \end{align*} Case 3: For this case is $U_1 = 0~{\rm V}$ and $I_2 = 0~{\rm A}$. The voltage comes from the source $U_3$.
\begin{align*} U_{\rm AB,3} = 8~{\rm V} \end{align*} Superposition means adding the voltages of all three cases. \begin{align*} U_{\rm AB} = U_{\rm AB,1} + U_{\rm AB,2} + U_{\rm AB,3} = 1.33~{\rm V} + 20~{\rm V} + 8~{\rm V} \end{align*}
Introduction to Superposition Method (starting from 34:24, Befor there are other methods not covered here)