Write Down Loop-Impedance Matrix for Current-Controlled Current Sources

In summary: First, the current at node A is ##i_1 = i_0 \frac {v_A} {R_3}##. From this, we can find the voltage at node A by solving for ##v_A##: ##v_A = i_0 \frac {v_A} {R_3} + 2i_1##. Next, the current at node B is ##i_1 = \frac {v_B} {R_3} + 2i_1##. From this, we can find the voltage at node B by solving for ##v_B##: ##v_B = \frac {v_B} {R_3
  • #1
cianfa72
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TL;DR Summary: How to write down the loop-impedance matrix in case of Current-Controlled Current Sources

Hi,
as in the following linear network, I would like to write down the loop-impedance matrix for it where there is a CCCS ##2i_1## source.

Capture.JPG


The starting point is the network elements equation: ##V(s) = E(s) + Z(s)I(s) - Li(0+) + \frac {1} {s}v_c(0+)##

How can we 'fit' the CCCS source in the branch-impedance matrix ##Z(s)## ?

Thank you.
 
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  • #3
No, it is a question related to that book in that specific example.
 
  • #4
It's been about 3-4 decades since I did this stuff, and I hated it at the time. I still would hate it if I wasn't ignoring it, LOL. So, sorry I don't have a useful answer.

However, you may find the answer, at least for special cases by working out the Kirchhoff equations and reforming them into the canonical matrix form. i.e. work the problem backwards until you see a pattern. For this network it's easy to solve for the CCS in terms of the voltages, which should give you an impedance term to put in your matrix.

My guess is that this won't always work out in general. I can easily imagine, singular and non-linear situations. There's a reason network theory likes to use the LPN assumption. Which is why most EEs won't use it. Small active networks are what EEs work with, big passive networks aren't very common. Which is why I've forgotten most all of this stuff, and why experienced EEs aren't likely to comment.

I would bet there's some IEEE paper that discussed this. Maybe email the author of this, or another Network Theory guy and ask them?
 
  • #5
cianfa72 said:
No, it is a question related to that book in that specific example.
It's still a schoolwork-type question, so I need to move it to the Homework Help forums (PF rules). I'll leave a long-expiring redirect here in the EE forum to ensure that it gets good views.
 
  • #6
DaveE said:
For this network it's easy to solve for the CCS in terms of the voltages, which should give you an impedance term to put in your matrix.
Do you mean tranform the CCCS in a Voltage controlled source?
 
  • #7
cianfa72 said:
Do you mean tranform the CCCS in a Voltage controlled source?
Sort of, I meant solve some of the network equations for the voltage across the current source. It looks, for this example, that you can find the current as a linear function of the voltage across it, hence an impedance.

But, I'm not sure how that will help with your basic question; the general case.
 
  • #8
DaveE said:
Sort of, I meant solve some of the network equations for the voltage across the current source. It looks, for this example, that you can find the current as a linear function of the voltage across it, hence an impedance.
Can you help me in writing down such linear function between current and voltage across the CCCS source? Thanks.
 
  • #9
Use KCL at the central node, then KVL on the loop with the voltage source. That will solve for the current i1 and the voltage at the central node. To find the other voltage you can then solve the outer loop.
 
  • #10
DaveE said:
Use KCL at the central node, then KVL on the loop with the voltage source. That will solve for the current i1 and the voltage at the central node. To find the other voltage you can then solve the outer loop.
Call the nodes starting from the left A,B,C.

The KCL at central node B is ##i_1 = \frac {v_B} {R_3} + 2i_1## and the KVL is ##v_1 = v_A = R_2i_1 + v_B##. From these we can solve for ##i_1## and ##v_B##.

Then from the KCL at node C ##\frac {v_1 - v_C} {R_5} + 2i_1 = \frac {v_C} {R_4}## we solve for ##v_C## and then we get ##v_C - v_B## i.e. the voltage across the CCCS source.

The above is the complete solution for the voltage across the CCCS; in order to get the linear non homogeneous relation between current and voltage across it suffice just the first and the last KCL (KCL at node B and at node C).
 
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  • #11
This sort of network can also be solved "graphically" with various source transformations, as below. This is why I dislike and never use the systematic linear algebra approach IRL. It's great if you're a SPICE program, but a lot of unnecessary work if you can just pick apart the network sort of by intuition and equivalents. Of course it only works for small networks.

Also you can use superposition to solve the outer loop quickly since the current source is independent of the outer loop values.

PXL_20230619_194300560.jpg
 
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  • #12
DaveE said:
Also you can use superposition to solve the outer loop quickly since the current source is independent of the outer loop values.
Sorry, do you mean as outer loop the loop made of ##v_1##, ##R_4## and ##R_5## ?
 
  • #13
cianfa72 said:
Sorry, do you mean as outer loop the loop made of ##v_1##, ##R_4## and ##R_5## ?
yes
 
  • #14
DaveE said:
Also you can use superposition to solve the outer loop quickly since the current source is independent of the outer loop values.
Superposition means sum the effect in the network due to the voltage source with the CCCS opened plus that due to the CCCS with the voltage source shorted.

Is the above correct?
 
  • #15
cianfa72 said:
Superposition means sum the effect in the network due to the voltage source with the CCCS opened plus that due to the CCCS with the voltage source shorted.

Is the above correct?
yes, exactly! You can set all of the independent sources except one to 0, then find the effect of that source on the output (or the whole network). Then you can simply add the results together to solve for when multiple sources are non-zero. This is essentially the definition of "linear".
 
  • #16
DaveE said:
yes, exactly! You can set all of the independent sources except one to 0, then find the effect of that source on the output (or the whole network). Then you can simply add the results together to solve for when multiple sources are non-zero. This is essentially the definition of "linear".
Sorry, but in this network there is just one independent source (the voltage source ##v_1##).

Anyway the weird thing for me is that to get the linear non-homogeneous relation between current and voltage across the CCCS we have to solve for KCL and/or KVL. Basically the impedance term for the CCCS we put into the ##Z## matrix is itself the 'result' of solving two KCL for the network !
 
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  • #17
cianfa72 said:
Sorry, but in this network there is just one independent source (the voltage source v1).
I was referring to the general concept. Also, as I said before, when you are solving the outer loop, the current source is effectively independent, since it doesn't depend on parameters in that loop. As shown in the second step of my circuit manipulations above, this network can nearly be split into two parts.

However, you are correct, that method won't work for the inner T section.
 
  • #18
cianfa72 said:
Sorry, but in this network there is just one independent source (the voltage source ##v_1##).

Anyway the weird thing for me is that to get the linear non-homogeneous relation between current and voltage across the CCCS we have to solve for KCL and/or KVL. Basically the impedance term for the CCCS we put into the ##Z## matrix is itself the 'result' of solving two KCL for the network !
Yes, it's not the real answer to your question. A solution algorithm can't depend on already having a solution. My suggestion was to look at a solved example and work backwards with the linear algebra method to see how that works. Unfortunately, this example may be a bit too simple for that.
 
  • #19
DaveE said:
Also, as I said before, when you are solving the outer loop, the current source is effectively independent, since it doesn't depend on parameters in that loop. As shown in the second step of my circuit manipulations above, this network can nearly be split into two parts.
Ah ok, now I see what you meant.
 
  • #20
Just a point to be highlighted: from my knowledge Superposition is feasible only if the network has just one solution (i.e. the coefficient matrix of the solving system is not singular). Otherwise it could exist infinitely many solutions for the network when all the sources are non-zero and no solution when only one source is non-zero.
 
  • #21
cianfa72 said:
Just a point to be highlighted: from my knowledge Superposition is feasible only if the network has just one solution (i.e. the coefficient matrix of the solving system is not singular). Otherwise it could exist infinitely many solutions for the network when all the sources are non-zero and no solution when only one source is non-zero.
Yes, OK. Can you show me an example of a LPN that has multiple solutions for identical ICs? Let's exclude the trivial case of parallel voltage sources and series current sources, which are generally forbidden. Superposition requires linearity and vice-versa.
 
  • #22
I found this trivial example having infinitely many solutions when the loop-impedance matrix becomes singular. In that case to get solutions the voltage ##V_g## has to be zero otherwise the system is incompatible.

20230622_142903.jpg


Another trivial case is a 'cut' of current generators.
 
  • #23
It's not passive. Yes, loops with only voltage sources and capacitors, or a cutset of only current sources and inductors doesn't have a general solution.
 
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Related to Write Down Loop-Impedance Matrix for Current-Controlled Current Sources

What is a loop-impedance matrix in the context of current-controlled current sources?

A loop-impedance matrix, also known as the impedance matrix or Z-matrix, is a representation of the impedance relationships between different loops in an electrical circuit. For current-controlled current sources (CCCS), it helps analyze how the impedance in one loop affects the currents in other loops.

How do you construct a loop-impedance matrix for a circuit with current-controlled current sources?

To construct a loop-impedance matrix for a circuit with CCCS, you need to follow these steps: identify all the loops in the circuit, write down the loop equations using Kirchhoff's Voltage Law (KVL), and express the voltages in terms of the loop currents and impedances. Then, arrange these equations in matrix form where the elements represent the impedances between loops.

What role do current-controlled current sources play in the loop-impedance matrix?

Current-controlled current sources introduce dependencies between the currents in different loops. They affect the off-diagonal elements of the loop-impedance matrix, as the current in one loop can influence the current in another loop through the CCCS. This requires careful consideration when writing the loop equations.

Can you provide an example of a simple loop-impedance matrix for a circuit with a current-controlled current source?

Consider a circuit with two loops where Loop 1 has an impedance Z11 and Loop 2 has an impedance Z22. If there is a current-controlled current source in Loop 2 controlled by the current in Loop 1, the loop-impedance matrix might look like this: \[ \begin{bmatrix} Z11 & -Z12 \\ -Z21 & Z22 \end{bmatrix} \]where Z12 and Z21 represent the mutual impedance introduced by the current-controlled current source.

How do you solve for loop currents using the loop-impedance matrix?

To solve for the loop currents, you need to set up the system of linear equations represented by the loop-impedance matrix and the voltage sources in the loops. This can be written in the form \(\mathbf{Z} \mathbf{I} = \mathbf{V}\), where \(\mathbf{Z}\) is the impedance matrix, \(\mathbf{I}\) is the vector of loop currents, and \(\mathbf{V}\) is the vector of loop voltages. You can then solve this system using matrix algebra techniques, such as Gaussian elimination or matrix inversion, to find the loop currents.

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