# help find total impedance with opamp

#### dacrazyazn

Joined Mar 30, 2009
22
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!

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#### t_n_k

Joined Mar 6, 2009
5,455
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!

#### t_n_k

Joined Mar 6, 2009
5,455
Correction on terminology at the end ....

Not Z1 = V1/I1

Rather Zin = V1/I1

#### Skeebopstop

Joined Jan 9, 2009
358
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 \$!%#!

#### t_n_k

Joined Mar 6, 2009
5,455
Sorry another correction required ...

The line

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

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

Awfully difficult keep track of things I'm afraid!

#### t_n_k

Joined Mar 6, 2009
5,455
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.

#### t_n_k

Joined Mar 6, 2009
5,455
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)

#### dacrazyazn

Joined Mar 30, 2009
22
I worked it out and it was perfect! Thank you!

Last edited:

#### t_n_k

Joined Mar 6, 2009
5,455
I worked it out and it was perfect! Thank you!
Glad to hear you got it.

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

#### The Electrician

Joined Oct 9, 2007
2,834
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.