# Amplifier Current Gain Question

#### frozone45

Joined Apr 16, 2016
17
Hi, I had a question about finding current gain. The question is part "ii" in the attached "Problem Picture.pdf" file. The correct answer for the question is: infinity. My work for the problem is shown in the attached "Work for Problem.pdf" file.

I don't understand how the answer of infinity makes sense, because, if io is nonzero, then io/iin should be undefined, and if io is zero, then io/iin should be indeterminate, since iin is always zero.

Would anyone have an idea as to why the correct answer is infinity for the current gain, io/iin?

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#### MrAl

Joined Jun 17, 2014
8,975
Hello there,

Current gain would be defined as:
Ag=Iout/Iin

so write an expression for Iout and for Iin, then see what you get.
A hint would be to call the dependent source a constant voltage source with variable say "Vx" and write the equations as if it was a constant source of value Vx, then later substitute what it really is which is shown on your schematics.

Using a few tricks i came up with this expression:
Ag=10000/(21*(Rf-1000))

and we can see that when Rf=1000 the denominator goes to zero which makes the limit infinite, so i tend to believe the declaration that it is infinite in the problem question and assumed answer because Rf is declared to be equal to 1000 Ohms.
This also tells me that the result of "infinite" probably depends on circuit values being precise just as they are given in the problem statement. In other words, change the value of some component and we probably dont get infinity anymore.

But the main thing to do is write an expression for Iin, and another for Iout, then take the ratio Iout/Iin and see what comes out of it. I think you will see it come out to what you expect it to be.

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• frozone45

#### frozone45

Joined Apr 16, 2016
17
Thank you for the advice, MrAl. I am trying to use your advice to solve the problem again.

#### The Electrician

Joined Oct 9, 2007
2,887
The output current is not really indeterminate unless the input is driven by a current source. Since the input resistance is infinite, if a current source is the providing the input, the voltage at the input will be infinite. If the input is driven by a voltage source, the output current is quite determinate even though the input current iin is zero.

• frozone45

#### MrAl

Joined Jun 17, 2014
8,975
Thank you for the advice, MrAl. I am trying to use your advice to solve the problem again.
Hi,

You're welcome, and another hint is that going into this problem you know the voltage across all resistors almost right away, so you can use Ohm's Law and sum the two currents for Iin, hten do Iout.
If we call the dependent source Vx, then the current through the resistors is:
Ii=Vin/Ri
iRf=(Vin-Vx)/Rf
Iout=Vx/(Ro+RL)

Note that to get the current through Rf we have to find the difference voltage that appears across it.
Maybe that will help a little more.

It's also interesting, if you get time, is to try to find out what resistor values, if changed, will change the current gain such that it is no longer infinite, and what vlaues dont matter as the current gain will stay infinite even after changing those.

• frozone45

#### frozone45

Joined Apr 16, 2016
17

#### The Electrician

Joined Oct 9, 2007
2,887

#### frozone45

Joined Apr 16, 2016
17
What did you get for part 1?
I attached the amplifier I got to this message.

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#### The Electrician

Joined Oct 9, 2007
2,887
I attached the amplifier I got to this message.
That's what I get. Note that your final result is a two-port: https://en.wikipedia.org/wiki/Two-port_network. The parameters you have determined for your problem are the elements of an admittance matrix description of your two-port. Your Rin is y11, your transconductance (the value .06) is y21, your Rout is y22 and y12 is zero.

You seem to do well at sort of problem, so in case you want to exercise you skill a little more try this. Show the three parameters of your final result in symbolic form using Ri, Rf, Rm and Ro from the original circuit.