help find total impedance with opamp

Thread Starter


Joined Mar 30, 2009
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!



Joined Mar 6, 2009
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!


Joined Jan 9, 2009
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 $!%#!


Joined Mar 6, 2009
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!


Joined Mar 6, 2009

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.



Joined Mar 6, 2009
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?
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.