# generalized Thevenin?

Discussion in 'Electronic Design' started by Jon Slaughter, Jul 19, 2007.

1. ### Jon SlaughterGuest

Is there a generalized Thevenin's theorem for arbitrary "black boxes"?

i.e., Suppose I have something like

---> I
V +---[ ]--- 0

where [] is a black box.

I should be able to write something like

V = Z(t, V, I)*I

which sorta resembles ohms law. V and I generally depend on t.

if [ ] is a resistor then Z(t,V,I) = R and in general Z also depends on a
set of parameters.

But what about more complex black boxes?

If its a resistor and a capacitor then what?

V ---||---/\/\/\/\---- 0

Then Z(t,V,I) = ?

For passive components is Z a linear differential equation?

Any other ways to simplify such expressions?

The reason I ask is I have a circuit that has a lot of these "paths" that
are connected in some way but each path is the same configuration with only
the "constants" of the components that are different.

Thanks,
Jon

2. ### EeyoreGuest

What ??????????

Is Thevenin/Norton too complicated for you ?

3. ### Jon SlaughterGuest

hmm... damn, I thought I put you on the ignore list ;/ you keep changing
your email or something?

4. ### EeyoreGuest

Please feel free to ignore ppl you might learn something from.

I'm sure it's all the rage in today's dumbed down world.

Graham

5. ### EeyoreGuest

Whatever was wrong with (R -jwC) ?

Graham

6. ### Jon SlaughterGuest

My what a big ego you sure do have! I guess it makes up for all the other
little things you have in your life.

7. ### RobertGuest

There's Middlebrook's Theorems but I don't think that's what you're asking
about.

http://en.wikipedia.org/wiki/Extra_element_theorem

Robert

8. ### Jon SlaughterGuest

Actually it looks very similar and it seems close to my problem. Essentially
I have a circuit where each "branch" looks identical(Actually its not but
uses identical topology... its almost fractal like) and I am trying to use
that symmetry to make it easier to solve. I'll have to read up on it to see
what exactly it doing though.

Thanks,
Jon

9. ### Jon SlaughterGuest

hmm... actually it doesn't seem to be what I want. My problem is similar to
his but each branch in the graph has the same "structure".

For example, take any graph and treat an edge as a black box that contains
passive elements. Then each edge is described by a linear differential
equation L_E where E is the edge indicator which really only depends on the
passive elements characteristics. Then is there a way to simplify the
graph/solve the system for the voltages and currents?

for example, you all know about the delta wye transformations... lets take
osmething similar

/\
Z1 Z2
/ \
--Z3--

Now Z1, Z2, and Z3 are a series of passive components. They all are
structurally equivilent. So if Z1 is a resistor than Z2 and Z3 are
resistors(although not necessarily of the same resistance). If Z3 is a
capacitor and inductor then so are Z2 and Z1.

All we know are the "voltage inputs" and I want to be able to calulate the
other variables in the circuit(node voltages and currents).

So essentially between any two nodes we have Vb - Va = Z*I which is just
ohms law in some sense but Z is somewhat arbitrary. The problem is that in
general Z depends on the nodes, current, and time along with all the other
components that exist on that branch.

You can think of it very similar to solving the graphs of resistors that
most people do in basic electronics or physics courses except I want to
replace them with more general components. In the case of a resistor its
very simple. In the case of a capacitor it is not. Since I want to take
into account the transients I get system of integeral equations(because of
the non-constant/non-sinusodial input voltages and unknown node voltages).
Of course these integral equations are equivilent to a system of
differential equations.

Essentially you can think of each branch representing a linear differential
equation. But each linear DE "looks" the same as every other one except for
the constant coefficients and it might depend on different currents and
voltages. It seems though I should be able to recursively simplify the
circuit until I find all the unknowns. well, this should be obvious but the
issue is, is the size. Just hoping for a way to reduce the complexity
because of the "symmetry" that exists.

Anyways, probably no way to do what I want but I can wish...

Thanks,
Jon

10. ### Tony WilliamsGuest

Well..... (R -j/wC) would be better.

11. ### Jens TingleffGuest

That's a brave post for someone who berates others for not listening to him
while going on about things to learn...

Best Regards

Jens

PS: I know, I shouldn't have, but I couldn't resist 12. ### Jens TingleffGuest

In a passive circuit, the complexity of finding the solution grows with the
number of nodes - sorry :-( Think of finding the voltages along a string of
resistors. For two resistors, the voltage at the middle point depends on the
value of two resistors. For three resistors (same structure in each branch...)
both of the two internal voltages depend on the value of three resistors, etc
etc. (However, if you know in advance that all the resistors have the same
value, you can easily calculate the solutions for all the internal nodes.)

Solving the network to find all node voltages / branch currents can certainly be
done the "hard way." This uses admittance matrices, as in i = Y v (with "v" the
vector of internal voltages and "i" the vector of external currents and "Y" the
admittance matrix).

http://users.ecs.soton.ac.uk/mz/CctSim/chap1_2.htm (for instance)

I think there's actually hope for solving your problem if you have a very simple
and regular structure. In that case, you might be able to find the inverse of
the Y matrix (this is what you need in order to solve the circuit equations) in
a way simpler than the general way. An advantage of this approach is that you
don't need to use the actual elements in each branch, you can use a symbol for
each distinct branch admittance and enter the actual branch admittances in the
solution.

Symbolic simplification of matrix inversions is pretty difficult to do, though.

So, the short answer is that going from simple circuits like a resistive divider
to more general circuits is not simple.

[..]

Worse - the voltages at the ends of this branch depend on all other voltages.
Actually, for a circuit with only capacitors, it's not different from a purely
resistive circuit (just a lot more unknown voltages at DC...).
For some very special circuits, yes (e.g. resistor strings, delta or star shaped
circuits, etc). In general, I think not.
It's worth thinking about this. I studied in a group where the professor made a
Big Discovery as a young student when hearing about a deceptively simple result
at a conference (for all passive circuits, the sum of voltages in the solution
is always constant - or something) and asking himself if that could really be
true and then wondering what the consequences were. The answer was the companion
circuit and its use for sensitivity analysis . He thought it was so trivial
that he didn't publish it right away. Tut tut. Two lessons in that anecdote ;-)

Good luck

Best Regards

Jens

 You look it up. Annoyingly, I cannot remember right now.

OOps !

Graham

14. ### John LarkinGuest

For a linear 2-terminal circuit, what you're looking for is the
Laplace transform of the impedance. This can express any impedance as
an algebraic equation using "s"

http://en.wikipedia.org/wiki/Laplace_transform#s-Domain_equivalent_circuits_and_impedances

John

15. ### Jon SlaughterGuest

Well, luckily I can use a CAS to do most of the work. I'm actually not
interested in extremly complex circuits so it won't actually be to much
work. But of course its still not that easy ;/
Yes, at all nodes there will be unknown voltages except at end nodes were
the voltages will be known. But because all paths/branches/edges "look the
same" all the equations will be the same and there is probably someway to
exploit this.

Well, for a capacitor if you want to know the transient voltages it is much
more complicated because your dealing with integrals over unknowns.

No, I do not mean that the topology is simplifiable but that because of the
symmetry of the equations it reduces the overall complexity. Its like
comparing a network of resistors to something else. Because they are all
resistors it makes it "easy" to solve(sure it could still be a bitch
though). In this case we don't have a resistor but its like a resistor in
that each component is similar to all the others. (its not like I have some
transistors on one branch, and inductor on another, etc...

Well, I doubt that will happen here ;/ Chances are I won't be able to make
any progress but its always fun to try Thanks,
Jon

16. ### Jon SlaughterGuest

Yes, I think that might work. If I laplace transform all the equations then
I'll end up with a system of linear equations which can be solved pretty
easily.

i.e., say for a capacitor in series with a resistor then we have

(V2 - V1) - Vc = R*C*dVc/dt

taking the laplace of each one results in a similar equation

v2 - v1 - vc = R*C*(s*vc - vc(0-))

or

v1 + (1 + RCs)vc - v2 = R*C*Vc(0-)

which can treat as a system(assuming one of the v's are unknown besides vc)

of course the issue is going to be actually transforming the functions but
maybe I don't have to worry about that to much.

I'll think about it some more and see what I come up with.

Thanks,
Jon

17. ### Ray-RogersGuest

You could try Tellegen's theorem, unfortunately I can't find an
explanation on the web. I had a link but can't seem to find it now.
I will not try explain it here because it looks ridiculous at first
reading (at least it did to me).

Ray

18. ### Jure NewsgroupsGuest

here :
http://www.ieee.org/web/aboutus/history_center/biography/tellegen.html

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