help find total impedance with opamp

Discussion in 'Homework Help' started by dacrazyazn, Apr 28, 2009.

  1. dacrazyazn

    Thread Starter Member

    Mar 30, 2009
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    hey guys, i got a question, this may look simple, but i dont know where to start completely. I tried to simplify down the circuit but that doesnt really work just made it look more confusing..

    now im trying to work with the idea of Zin=Vin/Iin but the problem is i dont know what kind of opamp this is used for. it would be nice if i can get a hint.. just a small hint on how i could go about this. thanks!
     
  2. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    Denote the top node voltage as V1. Then denote descending node voltages below as V2, V3, V4 & V5. V5 is the node on Z5 above ground.

    Suppose the two op amps are ideal.

    Each op amp will drive the network to force the following conditions :-

    V5 = V3 = V1

    So the current in Z5, Iz5 = V5/Z5 = V1/Z5

    So V4 = V5 + Z4 x V1/Z5 = V1(1+Z4/Z5)

    So the current in Z3, Iz3 = (V3 - V4)/Z3 = (V1 - V4)/Z4

    and so on ...

    Eventually you'll find the current into node 1 {I1} entirely in terms of V1, and then you'll obtain Z1 = V1/I1

    Tricky question!
     
  3. t_n_k

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    Mar 6, 2009
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    Correction on terminology at the end ....

    Not Z1 = V1/I1

    Rather Zin = V1/I1
     
  4. Skeebopstop

    Active Member

    Jan 9, 2009
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    Another thought process to come to the same answer as t n k is this:

    Assume ideal op-amps.

    That means the output impedance of the op-amp is 0.

    V1/I1 = Z1 = Zin = Vin/Iin.

    Yippeeee kayaaaa Motha $!%#!
     
  5. t_n_k

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    Mar 6, 2009
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    Sorry another correction required ...

    The line

    So the current in Z3, Iz3 = (V3 - V4)/Z3 = (V1 - V4)/Z4

    should read

    So the current in Z3, Iz3 = (V3 - V4)/Z3 = (V1 - V4)/Z3

    Awfully difficult keep track of things I'm afraid!
     
  6. t_n_k

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    Mar 6, 2009
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    Dacrazyazn,

    Just in case you think I'm stating the obvious - as perhaps Skeebopstop is suggesting - you should end up with an answer for Zin which is a relationship involving only the terms Z1, Z2, Z3, Z4 & Z5.

    :rolleyes:
     
  7. The Electrician

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    Oct 9, 2007
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  8. t_n_k

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    Mar 6, 2009
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    Hi Electrician,

    So there was something wrong with my method. Never analyzed this circuit before.

    Oh well - such is life.

    I have

    Zin = (Z1*Z3*Z5)/(Z2*Z4)

    What's the correct answer?
     
  9. The Electrician

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    Oct 9, 2007
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  10. dacrazyazn

    Thread Starter Member

    Mar 30, 2009
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    I worked it out and it was perfect! Thank you!
     
    Last edited: Apr 29, 2009
  11. t_n_k

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    Mar 6, 2009
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    Glad to hear you got it.

    Thanks Electrician - I was in serious self-doubt for a while.
     
  12. The Electrician

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    OK. Now for something completely different, derive an expression for the impedance at the top of the stack, if the opamp gains are not ∞, but are finite gains of A.
     
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