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Finding the input impedance of CCCS

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Pretty immediate in this case. The output resistance with feedback removed is R1, the loop gain with port (terminals of RL) shorted is A and the loop gain with port open is 0. The result is R1*(1+A).
 
Good day everyone! I need your help on this one. I'd like to derive the equation for the closed-loop input impedance of the circuit attached. The final form of the equation is also shown below.

View attachment 157314
View attachment 157313

Here's what I have tried
View attachment 157316

I was stuck. Can you tell what other route I can take to go about this derivation? Thanks!

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Ratch
 

With a current for its input then its input voltage barely changes. Then its input impedance at low frequencies is extremely low (a virtual ground).
 

With a current for its input then its input voltage barely changes. Then its input impedance at low frequencies is extremely low (a virtual ground).

I don't understand your statement. Assuming an ideal op amp and no reactive components, the voltage is directly proportional to the input current according to my calculations above unless they are in error. The input impedance should be as given in my previous post unless I goofed in my calculation.

Ratch
 

The admittance matrix for the circuit can be written by inspection, then inverted. The (1,1) element of the inverse is Zin:
 

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Hi,

I agree with audioguru.
The voltage will be almost zero.

But yes, even if small it will be proportional to input current.

Klaus
 

Hi,

I agree with audioguru.
The voltage will be almost zero.

But yes, even if small it will be proportional to input current.

Klaus

Yes, the input voltage of the circuit will be small, if the amplification factor of the op amp is high, thereby causing the input impedance of the circuit to be low. This is shown by the formulas shown above.

Ratch
 

Sorry, I don't understand these last posts. It has been already proven that Zin = R2/(1+A*B) that means Zin decreases as A increases.
 

Amplification does not increase. The opamp already has a gain of almost infinity for a modern opamp. But some old opamps do not have much gain. Aren't you talking about modern opamps?
The input signal is a very high impedance current source and it is feeding the extremely low input impedance of the opamp inverting input. Then the input signal voltage is so low that we say it is zero.
 

Sorry, I don't understand these last posts. It has been already proven that Zin = R2/(1+A*B) that means Zin decreases as A increases.

I don't know where the formula Zin = R2/(1+A*B) comes from, or what the definition of A and B are. However, both calculations of The Electrician and myself agree on what Zin is. Yes, Zin approaches zero as the op amp amplification approaches infinity. What don't you understand about my derivation of the nodal equations of the circuit?

Ratch

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Amplification does not increase. The opamp already has a gain of almost infinity for a modern opamp. But some old opamps do not have much gain. Aren't you talking about modern opamps?
The input signal is a very high impedance current source and it is feeding the extremely low input impedance of the opamp inverting input. Then the input signal voltage is so low that we say it is zero.

Amplification changes as op amps age or are replaced. High values of gain of several thousand or million are not near infinity. I am not focused on any particular op amp. Both the plus and minus terminals of an op amp are very high as is any voltage amplifier. The circuit under discussion has a very low input impedance, however.

Ratch
 

I don't know where the formula Zin = R2/(1+A*B) comes from, or what the definition of A and B are. However, both calculations of The Electrician and myself agree on what Zin is. Yes, Zin approaches zero as the op amp amplification approaches infinity. What don't you understand about my derivation of the nodal equations of the circuit?

Ratch

Ratch

Please have a look to #9. The formula was derived there.
You analysis is perfectly clear. Is just another approach to reach the solution.
 

Please have a look to #9. The formula was derived there.
You analysis is perfectly clear. Is just another approach to reach the solution.

I had a hard time following your calculations, but it looks like you came up with the same solution as The Electrician and I did. I think the superposition method is on solid theoretical ground, but it entails multiple times as much work because you have to do multiple calculations for each excitation source.

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Ratch
 

Looks to me the shortest solution was given in #5
 

Looks to me the shortest solution was given in #5

Except that the value of the solution does not agree with what 3 other members have come up with.

Capture.JPG

Ratch
 

Except that the value of the solution does not agree with what 3 other members have come up with.

View attachment 157377

Ratch



Well, it is very easy to verify. Just build a simulation schematic with ideal vcvs for the amplifier, put a large gain, look for the input impedance and check the expressions against simulation. Takes no more than 5 minutes.
 

Well, it is very easy to verify. Just build a simulation schematic with ideal vcvs for the amplifier, put a large gain, look for the input impedance and check the expressions against simulation. Takes no more than 5 minutes.

Read post #35. Ratch made a silly mistake, probably because, like me, he has old eyes. :oops:
 

Yes, I read it. That's not a problem. I'm not saying other derivations are wrong, I said mine was correct and simple. Also, the result is arranged in a useful for design purposes way without need for algebraic manipulations.

Don't we all want to have young eyes? :wink:
 
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Yes, I did have trouble reading that light script and small scrawl. Sorry for the mistake.

Ratch
 

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