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Finding the output impedance of Current Controlled Current Source

  • Thread starter paulmdrdo
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  • #1
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Summary:: I was trying to derive the equation for the output impedance of the ICIS circuit.

1579023923640.png

Here's what I've tried so far.
1579024060734.png


What I have derived does not agree with the one provided in my book.
1579024119464.png

Since it is a current source the negative feedback must increase the output impedance. The one I derived says otherwise. Can you tell me what I am missing? TIA!
 

Answers and Replies

  • #2
berkeman
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I2 cannot be zero. Why not? :smile:
 
  • #3
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I2 cannot be zero. Why not? :smile:
I assumed that the input impedance Rin of the opamp is so large. I2 is very small to have an effect.
 
  • #4
berkeman
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What do you think the relationship is between Iin and I2?
 
  • #5
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What do you think the relationship is between Iin and I2?
I did not include Iin because I learned that when you are finding the output impedance of an amplifier all sources must be set to zero. Short circuit for voltage sources and open circuit for current sources. That is Why I2 is zero. Have I misunderstood something?

Here are the steps I was taught to go about finding Zo
turn off all sources
Apply a test voltage at the output
Find Ix. Then Zo = Vx/Ix
 
  • #6
eq1
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I did not include Iin because I learned that when you are finding the output impedance of an amplifier all sources must be set to zero. Short circuit for voltage sources and open circuit for current sources. That is Why I2 is zero. Have I misunderstood something?

Here are the steps I was taught to go about finding Zo
turn off all sources
Apply a test voltage at the output
Find Ix. Then Zo = Vx/Ix
Opening the input and using a test voltage on the output is often a way to go but it won't work here. To see why: with the input open, assume Vx!=0. then write your equations and you should find a contradiction. (you didn't because I think you have a couple of sign errors but the IL equation is definitely wrong) You can try the alternative, use a text Ix and find the corresponding Vx, but this should lead Zo=0, which is also not correct.

Your teacher should've taught you another way which is also common.
 
  • #7
eq1
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Does anyone know what the ICIS acronym means? I would've called this a TIA with T-feedback. I googled it and ICIS seems like it is a notation from Chegg's text which I never used.
 
  • #8
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Does anyone know what the ICIS acronym means? I would've called this a TIA with T-feedback. I googled it and ICIS seems like it is a notation from Chegg's text which I never used.
ICIS stands for Current(I)Controlled Current(I) Source.

What's wrong with the equation of IL?
 
  • #9
eq1
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Actually, I am wrong. Your technique will work but the math is a bit harder that way which is why I didn't think do to it like that. You didn't get the correct answer because your initial set of equations has a few mistakes. Specifically look at the node that joins R1, R2 and RL (the voltage is not zero at that net) and note that the opamp is wired as an inverting amplifier.
 
  • #10
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Yes. It is not zero. That is why I have the expression of the voltage at that node.
1579029932137.png
 

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  • #11
eq1
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Let's label that node Va. If we agree that Va is non-zero then how can IL=Vx/(RL+R1)?
 
  • #12
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By KVL.
1579030815625.png
 
  • #13
eq1
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Oh right. I forgot we're using I2=0 so that works. I did KCL at that node which gives a slightly different set of equations. I think your answer is right it's just in a different form than the book's.

What does your book say A_vol is?

I solved it in mathematica with this line of code.

Zo=Simplify[Vx/Ix /.
Solve[{Va/R1 == (Vx - Va)/RL,
Ix == (Vx - (-A*Va))/Ro + (Vx - Va)/RL}, {Vx, Ix}]]

The answer was this. Which is the same as yours but rearranged.

{((R1 + RL) Ro)/(R1 + A R1 + RL + Ro)}
 
  • #14
eq1
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I just realized where we both messed up. We both ignored the labels on the initial problem. I am still assuming the circuit is a TIA but it's not. It's a current controlled current source. The output variable is the current through RL, not the output voltage of the opamp like in a TIA. If you connect a test source to the right output then you'll get the correct answer.
 
  • #15
berkeman
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I did not include Iin because I learned that when you are finding the output impedance of an amplifier all sources must be set to zero. Short circuit for voltage sources and open circuit for current sources. That is Why I2 is zero. Have I misunderstood something?
Well for one thing, Iin is not a source affecting the output impedance directly. And when you set it to zero, you lose information about some of the currents and voltages in the output circuit.

Maybe this problem can be worked other ways, but I would solve the output circuit for the quiescent operating point given some load impedance, and use the following equation to calculate the output impedance (your output port is across R1, correct?):

[tex]Z_o = \frac{dV_o}{dI_o}[/tex]
 
  • #16
eq1
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(your output port is across R1, correct?):
I got tricked by the circuit diagram too. I kept thinking it was a TIA.

The ICIS note in the upper corner means it's a current controlled current source and the output label is given as the current through RL. Because it's a current source the question wants the Norton resistance, not Thevenin. The text is doing the derivative technique but it's hidden because the quiescent is zero everywhere. The technique is add Ix in parallel with RL and look at the corresponding change in differential voltage on RL, this is Vx.

Zo=(Vx-0)/(Ix-0)=Vx/Ix and one gets the answer from the book.
 

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