At present, I don't have Skilling's book at hand. But I feel that transformation of constant (i. e. independent) sources does rely only on the linear behavior of resistors, and this is secured by Ohm's law. From linearity follows that if a voltage source and a current source are indistinguishable (from the outside) at two different states, they are indistinguishable in every possible state. Thus, it's sufficient to show that a voltage source and a current source would cause the same open circuit voltage and the same short circuit current in order to prove that they are equivalent to each other.

So, if I'm not mistaken, source transformation comes first, superposition theorem later on.

Dear Joerg,

Thank you for your answer. However, I have some more doubt and therefore I am enclosing two files that include the relevant topics that are discussed in Skilling’s book. The first file (skilling 1) explains the equivalence of the current source representation and the voltage source representation when the generator is feeding a passive load. However, when the load is active, that is, for multiple source environments some more arguments are necessary in order to prove the required equivalence. This is explained in the second file (skilling 2). And that explains my question about the inherent circular dependency.

Regards,

Anup

Dear Anup,

thank you for providing Skilling's proofs!  I see your point. Apparently, Skilling didn't; otherwise, he could have easily avoided the circular reasoning:

By introducing the voltage VL at the load, one can transform equations (6-24) and (6-25) into

IL = (V - VL) / ZV   and   IL = (I * ZI - VL) / ZI

As with (6-24) and (6-25), these equations are identical if

ZV = ZI = Z   and   V = I * Z   resp.   I = V / Z.

But ZL has disappeared from the equations, so the properties of the load (i. e. the relationship between voltage at and current through the load) do not affect the source transformation. The load can be a passive or an active linear network or even a non-linear one. As a consequence, the explicit extension of the source transformation to active networks in chapter 11, after proving the superposition theorem, can be omitted.

So, I still think that the general validity of source transformation does not depend on the superposition theorem, and this point of view is supported by the fact that source transformation has a larger scope of validity (external non-linear networks included) than superposition.

Regards,

Joerg

PS. I appreciate Skilling because, as far as I know, he was the first author to point out that Maxwell's divergence laws were experimentally proved only for static fields, in "Fundamentals of electric waves", 1942. In the present case, I guess he made an error which probably many of us have committed sometimes (I know I have): Trying to simplify things for students (an explicit resistor gives a clearer picture than an unspecific load) beyond the point where inconsistencies are introduced.

Dear Joerg,

Thank you for your interest in this thread. In your answer you have modified two equations by assuming VLV  and VLI equal (please see the attached figure) The load network being active we may not be allowed to take such assumption until the two source representations are proved equivalent.

Regards,

Anup

Dear Anup,

Skilling does not differentiate between ILV and ILI. So, he assumes that for given V, ZV and ILV there exist I and ZI in such a way that ILV = ILI, and I think this is a sound assumption, with the only requirement to the external network that it behaves in a deterministic way. Then, Kirchhoff's laws and Ohm's law give us for the voltage source

VLV = V - IL * ZV   and for the current source

VLI = I * ZI - IL * ZI.

With   ZV = ZI = Z   and   V = I * Z   follows   VLV = VLI, regardless of the nature of the external network.

BTW, does Skilling mention the validity of source transformation for the case of external non-linear networks later on?

Regards,

Joerg

Dear Joerg,

Please look at the equations (6-24) and (6-25) again and there related explanation. Skilling has not assumed ILV=ILI. He has shown that under the conditions

ZV= ZI and V=I*ZI.

these two currents are equal. And for this reason he claimed that the two configurations are equivalent. He, however,  overloaded the identifier IL to represent both the currents.

Regards,

Anup

Dear Anup,

my point is: We don't have to look inside the external network in order to determine the relation between voltage and current at the port of the sources. If we replace "active network" in your latest picture by "black box", Kirchhoff's voltage law still applies to each mesh even if one element in the mesh is the port of the black box, and Kirchhoff's current law still applies to each point in the circuit even if current is flowing from this point to a terminal of the black box or in the opposite direction. Equally, Ohm's law applies to ZV and ZI. And taken together, under the known conditions, we get two identical equations describing the relation between voltage and current at the port of the sources.

In other words, under the conditions ZV = ZI and V = I * Z both sources have the same current-voltage characteristic, regardless of the content of the external black box.

Do we agree that source transformation is valid even with external non-linear networks?

Regards,

Joerg

Dear Anup,

assuming ILV = ILI and/or VLV = VLI is, of course, quite unnecessary because the topic is the complete current-voltage characteristic, not a single point of it. So, here comes at last what I should have written at the outset (discussions really help to clear up formulations), more formally, without any overloading, independent of Skilling, and without knowledge of the content of the external network:

For the voltage source we have:

KCL: current everywhere the same = ILV

KVL: V = VZV + VLV

OL: VZV = ILV * ZV

result: VLV = V - ILV * ZV

with ZV = ZI = Z:   VLV = V - ILV * Z

For the current source we have:

KCL: I = IZI + ILI

KVL: voltage everywhere the same = VLI

OL: IZI = VLI / ZI

result: VLI = I * ZI - ILI * ZI

with ZV = ZI = Z:   VLI = I * Z - ILI * Z

and with V = I * Z:   VLI = V - ILI * Z

Obviously, the dependence of the voltage VLV on the current ILV is identical to the dependence of the voltage VLI on the current ILI. I. e., both sources have the same current-voltage characteristic, and are therefore by no means (passive, active, non-linear networks) distinguishable from the outside.

Regards,

Joerg

Please refer to the attached figure. For voltage source representation you have derived

VLV = V - ILV * Z

And for the current source representation you worked out the similar equation as

VLI = V - ILI * Z

In order to prove the necessary equivalence one should show that either VLV = VLI or ILV = ILI.. This is necessary because if the two representations are equivalent then they should produce identical effects on the same active source. This part is absent in your derivation. This is expected because in your entire derivation  only the left portion, that is, the source configurations are  considered.

I postpone the discussion on non linearity until this issue is settled.

If you take an arbitrary active network with known details (on the right side of your figures) and treat the source circuit (on the left of your figures) as a black box, without knowing the I-V characteristic of this black box, then your set of equations is one equation short so you cannot solve it. Introducing the I-V characteristic of the source circuit increments the number of equations in your set by one, which yields (for a linear network) a unique solution. For a non-linear network, the set of equations yields a unique set of solutions. The details of the source circuit are irrelevant. 

In a pictorial way: If you graphically determine the operating point of a certain network and a certain source, you represent the I-V characteristic of the network (at the port connecting it to the source) by a curve (a straight line if linear), and you represent the I-V characteristic of the source by a straight line. The operating point(s) is | are at the intersection(s) of curve and line. If the network remains the same, and the I-V characteristic of the source remains the same, and, as a consequence, the whole graph remains the same, how can any intersection change its location?

In short, the behavior of an ideal one-port network (no dependencies on time, temperature, etc.) is completely characterized by its I-V characteristic, and this I-V characteristic is independent of the properties of any external network.

If you don't agree, please describe in detail a counterexample.

Dear Joerg

Sorry I could not understand the relevance of your answer in connection with the present issue, that is the sequence of arguments  that are necessary to prove the validity of source transformation.

Anup

Dear Anup,

I'm equally sorry. Simultaneously, I'm curious if anyone else here on RG knows an answer you could accept, and if so, what it might be.

Joerg

They say, mathematics is a universal language. So, I'll try a last time with as little prose as possible (sorry for the repetition of answer 8, it's just for the sake of clarity). Prerequisites: Kirchhoff's laws, Ohm's law, elementary algebra:

For the voltage source we have:

KCL: current everywhere the same = ILV

KVL: V = VZV + VLV

OL: VZV = ILV * ZV

=> VLV = V - ILV * ZV

with ZV = ZI = Z:   VLV = V - ILV * Z

For the current source we have:

KCL: I = IZI + ILI

KVL: voltage everywhere the same = VLI

OL: IZI = VLI / ZI

=> VLI = I * ZI - ILI * ZI

with ZV = ZI = Z:   VLI = I * Z - ILI * Z

and with V = I * Z:   VLI = V - IIL * Z

Each source forms a one-port network. The external network is a one-port network, too. Voltage and current at the port of the external network are named VN and IN. If the external network is linear (passive or active), the I-V characteristic of its port can be written in the form:

VN = a + b * IN   (a = 0 if passive)

We'll get a solution if b != -Z.

Case 1: Voltage source connected to external network

KCL: IN = ILV

KVL: VN = VLV

=> a + b * ILV = V - Z * ILV

=> ILV = (V - a) / (b + Z)

Case 2: Current source connected to external network

KCL: IN = ILI

KVL: VN = VLI

=> a + b * ILI = V - Z * ILI

=> ILI = (V - a) / (b + Z)

Ergo: ILV = ILI

=> VLV = VLI

q.e.d.

Dear Joerg,

Your Mathematics is absolutely correct. You have assumed that the linear network can be modeled by the equation

VN = a + b * IN  

which is nothing but Thevenin's theorem.  That is instead of assuming superposition theorem we are assuming Thevenin's theorem. If we are permitted to do so then we can apply it on the current source representation itself to get the required voltage source representation Please see the attached figure..

However I again thank you for your interest in this thread.

With sincere regards,

Anup.

Dear Anup,

I think my assumption was weaker than Thevenin's theorem (the values of a and b remained unknown; the assumption was: A linear network has a linear I-V characteristic.) but it was an assumption anyway. I feel I can do with a still weaker assumption (continuous I-V characteristic (at least piecewise)).

After the first part of answer 13 above, until "Voltage and current at the port of the external network are named VN and IN.", the reasoning continues:

If the behavior of the external network does not depend on time (present or past (hysteresis)), and if the I-V characteristic of its port is a continuous function, then according to the Weierstrass approximation theorem, it can be arbitrarily well approximated by a polynomial equation of finite degree n:

VN = a0 + a1 * IN + a2 * IN2 + ... + an * INn

Case 1: Voltage source connected to external network

KCL: IN = ILV

KVL: VN = VLV

=> V - Z * ILV = a0 + a1 * ILV + a2 * ILV2 + ... + an * ILVn

=> a0 - V + (a1 + Z) * ILV + a2 * ILV2 + ... + an * ILVn = 0   (1)

Case 2: Current source connected to external network

KCL: IN = ILI

KVL: VN = VLI

=> V - Z * ILI = a0 + a1 * ILI + a2 * ILI2 + ... + an * ILIn

=> a0 - V + (a1 + Z) * ILI + a2 * ILI2 + ... + an * ILIn = 0   (2)

Because the constants in (1) and (2) are identical, the sets of solutions are identical for ILV and ILI.

If the I-V characteristic of the port of the external network is only piecewise continuous, the above can be shown for each piece separately.

This argument is more general than that in answer 13 because it is not limited to linear external networks.

With kind regards,

Joerg

Sorry I made a mistake in my last post. Thevenin’s theorem cannot be used to prove the equivalence of the voltage and the current source representation related to source transformation. It is so because all different methods described by you, method using Thevenin’s theorem and the one described in Valkenberg’s book (Network Analysis by Van Valkenberg) rely upon some abstraction (input volt-amp characteristics) of the associated active one port. All these methods finally conclude that the input current and the voltage of this active one port remain invariant under source transformation.   Skilling, on the other hand, considers the circuit details by using Superposition theorem. As a matter of fact, the method explicitly considers contribution of each and every source in the circuit.  Therefore Skilling’s proof ensures its applicability of such transformation in circuit analysis using either KCL or KVL.

To achieve the required abstraction one may use either experimental technique or analytical method. In analytical method a linear active network is modeled by the set of circuit equations which in turn requires source transformation for the circuits containing both voltage and current sources. Therefore, problem of circular dependency cannot be avoided. In experimental method, as you have explained, the circular dependency can be avoided. However, both these abstractions do not allow us to prove the invariance of the entire circuit solution (every loop currents or every node voltages) under source transformation.  

Thank you very much for your interest in this thread. Without your discussion I wouldn’t have realized the matter in such detail.