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generalized Thevenin?

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

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  1. 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.

  2. Eeyore

    Eeyore Guest

    What ??????????

    Is Thevenin/Norton too complicated for you ?
  3. hmm... damn, I thought I put you on the ignore list ;/ you keep changing
    your email or something?
  4. Eeyore

    Eeyore Guest

    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.

  5. Eeyore

    Eeyore Guest

    Whatever was wrong with (R -jwC) ?

  6. 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. Robert

    Robert Guest

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

  8. 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.


  9. 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
    / \

    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...

  10. Well..... (R -j/wC) would be better.
  11. That's a brave post for someone who berates others for not listening to him
    while going on about things to learn...

    Best Regards


    PS: I know, I shouldn't have, but I couldn't resist :)
  12. 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). (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

    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[1]) 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


    [1] You look it up. Annoyingly, I cannot remember right now.
  13. Eeyore

    Eeyore Guest

    OOps !

  14. John Larkin

    John Larkin Guest

    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"

  15. 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 ;)

  16. 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

    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-))


    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.

  17. Ray-Rogers

    Ray-Rogers Guest

    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).

  18. here :
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