# Amplifier Current Gain Question

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#### frozone45

##### New Member
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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• Problem Picture.pdf
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• Work for Problem.pdf
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Maybe a better way to think of it is not so much as strict mathematical terms since infinites usually have more nuanced meaning in reality than just "undefined". In this case, it just means that Vs needs to supply no current (or rather, it's current is always forced to zero by something else). So no matter how much current Vs tries to supply, it can't supply it.

The simplest way of thinking about infinite as just a very large number is not very accurate and only works for simple cases. It breaks down when you get to more advanced stuff which is why it's not really making sense here. In reality, infinite isn't a large number as much as the concept of something constantly getting larger without ever being a set value. This means that you can perform math operations on it against another "number" that either increases equal to, faster, or slower than that particular infinite. This nuance is what lets you come up with real-world results that make sense when working with infinite. If we're talking in terms of pure math, stuff like 0/0 or zero x infinite is undefined, but not really. You can kind of manipulate things and play tricks to get an answer out of it that reflects what happens in the real world (not always, but that's where the limits of human understanding are).

In not-so-strict hand-wavey pure math terms thinking about it in reverse: If the current gain is actually infinite but the voltage across Ri is finite, then what must the current supplied by Vs be? Well, it certainly can't be non-zero because you don't have an infinite voltage. It HAS to be zero because in the realm of crazy math zero multipled by the right kind of infnite perfectly counteract each other to become a finite number (BTW zero x infinite does not equal to zero in general in case you were not aware. In general it's undefined because the process of zeroing is up against infinite's process of increasing and without more information we can't decide which one is winning).

But what does this really mean? Well, remember when I said infinite was not a fixed number but a process of continually increasing so you could never catch up to it? There are many ways you can do this such as the rate of increase. In this specific case, this sort of infinite can sort of be thought of as "something intelligent" is constantly adjusting itself to always just barely stay ahead of you. That is infinite in the sense that you can never reach the end, but neither is it running on forever. It's a real finite value that you just can't catch up to because of your frame of reference (an outside observer that the infinite is not staying just ahead of can just go in and measure it just fine and it would be a value). So the gain of infinite is not really a fixed gain that is an infinite number, but a gain that is actually a finite number that is changing and constantly being adjusted to be "just enough" to force the Iin to be zero. In that way, everything stays defined but at the same time Vs can never overcome the obstacle of pushing current ("catching up" to the infinite gain which is actually just a finite gain that's always one step ahead of it).

People write PhD thesis on this stuff so what I told you is about all I know. I don't know if this answered your question or not but maybe it will help you approach strange scenarios like this with a more flexible mindset.

Of course, in reality there are real world limits to this circuit like the power supply voltages so a result of infinite would mean "as much as it possibly can" rather than something as nebulous as "undefined". It's just that you used approximations in your equations rather than entering all the real-world nitty gritty data that would have given you a more specific answer, so you instead you ended up with infinite as an approximation to a very large real-world number.

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• frozone45
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?

I don't believe the answer is infinity. As you can see from the solution of the loop currents, the i3 loop divided by the i1 loop is 0.57143 no matter what the source voltage is. Ratch

• frozone45
I don't believe the answer is infinity. As you can see from the solution of the loop currents, the i3 loop divided by the i1 loop is 0.57143 no matter what the source voltage is.
View attachment 111819

Ratch
What is the flaw then in the OP's work? Because what he did seems to make sense to me (but maybe that's why I always avoid nodal analysis)

EDIT: Oh, wait a second....okay, now I'm confused because Ratchit's solution makes sense just looking at the right loop but OP's solution makes sense looking at the left loop.

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• frozone45
What is the flaw then in the OP's work? Because what he did seems to make sense to me (but maybe that's why I always avoid nodal analysis)

EDIT: Oh, wait a second....okay, now I'm confused because Ratchit's solution makes sense just looking at the right loop but OP's solution makes sense looking at the left loop.

I ignored his work because he avers that Iin equals zero, which is not true. Iin and Io are directly proportional to Vs, as I have shown in my loop analysis of the circuit.

Ratch

• frozone45
I don't believe the answer is infinity. As you can see from the solution of the loop currents, the i3 loop divided by the i1 loop is 0.57143 no matter what the source voltage is.
View attachment 111819

Ratch

Yeah, but that still doesn't explain why his method doesn't work (which is basically what he's asking). You can derive your result from a KVL of loop 3 and ignore everything else. I'm not sure why that's any more valid that what the OP was trying to do by just looking at the one node.

He's making a false assumption somewhere but I can't figure out where it is. For example, if he assumes that the voltage across Rf drops from right to left because if he assumed it drops from left to right he gets another strange answer (Iin = 2*Ii) whereas your results give Ii = 2*Iin

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• frozone45
Yeah, but that still doesn't explain why his method doesn't work (which is basically what he's asking). You can derive your result from a KVL of loop 3 and ignore everything else. I'm not sure why that's any more valid that what the OP was trying to do by just looking at the one node.

He's making a false assumption somewhere but I can't figure out where it is. For example, if he assumes that the voltage across Rf drops from right to left because if he assumed it drops from left to right he gets another strange answer (Iin = 2*Ii) whereas your results give Ii = 2*Iin

In loop analysis, assume all loop currents have the same direction. It's clockwise in my work. Then, I like to do a KVL in the opposite direction with respect to the assumed current direction so I don't run into so many negative voltages. But, a KVL in either direction will give the same results. The problem has 3 unknowns and three equations. A solution is not in doubt, but you can check it by substitution. I don't know what mistakes the OP made, but he avers without proof that the source current is zero. My solution shows that the source current and the output current are dependent on vs. Perhaps you should scrutinize my solution for mistakes, and then compare to what the OP has done. I cannot understand how the OP is getting his results.

Ratch

• frozone45
In loop analysis, assume all loop currents have the same direction. It's clockwise in my work. Then, I like to do a KVL in the opposite direction with respect to the assumed current direction so I don't run into so many negative voltages. But, a KVL in either direction will give the same results. The problem has 3 unknowns and three equations. A solution is not in doubt, but you can check it by substitution. I don't know what mistakes the OP made, but he avers without proof that the source current is zero. My solution shows that the source current and the output current are dependent on vs. Perhaps you should scrutinize my solution for mistakes, and then compare to what the OP has done. I cannot understand how the OP is getting his results.

Ratch
No, I agree with your solution but I am also most familiar with loop analysis. My concern is how to make the nodal analysis work since it's giving strange answers but I can't figure out whats actually going wrong.

You say that he avers that the current is zero with no proof, but he showed his work which is his "proof". So what's wrong with his "proof"? You say you can't understand how the OP got a current of zero but you said yourself that you didn't even look at his work. You just said "nope, that answer is impossible" which isn't helpful to him. Nodal analysis/KCL should be possible.

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• frozone45
No, I agree with your solution but I am also most familiar with loop analysis. My concern is how to make the nodal analysis work since it's giving strange answers but I can't figure out whats actually going wrong.

You say that he avers that the current is zero with no proof, but he showed his work which is his "proof". So what's wrong with his "proof"? You say you can't understand how the OP got a current of zero but you said yourself that you didn't even look at his work. You just said "nope, that answer is impossible" which isn't helpful to him. Nodal analysis/KCL should be possible.

You want nodal analysis? No sweat, you got it. As you can see, the current gain is the same as it was for the loop analysis. In the OP's analysis, how can ii, the current present in R200, possibly be zero when there is a vs voltage across it?

Ratch

• frozone45
Thank you for the advice, Ratchit and dknguyen. I'm trying to use what each of you mentioned above to solve the problem again.

You want nodal analysis? No sweat, you got it.

View attachment 111834
As you can see, the current gain is the same as it was for the loop analysis.

Ratch
Thanks.

In the OP's analysis, how can ii, the current present in R200, possibly be zero when there is a vs voltage across it?

Ratch
I saw that and agree but at the same time could not see anything wrong with the way the OP went about his math.

• frozone45
Thanks.

I saw that and agree but at the same time could not see anything wrong with the way the OP went about his math.

Both my previous node and loop equations are in error. The corrected equations are shown below. Both node and loop calculations show that the input current existing in Vs (Iin) is zero at all values of Vs. That makes the circuit have a current gain of infinity. Sorry for the mistake. Check and see if you agree with the new eq's. Ratch

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• frozone45
Both my previous node and loop equations are in error. The corrected equations are shown below. Both node and loop calculations show that the input current existing in Vs (Iin) is zero at all values of Vs. That makes the circuit have a current gain of infinity. Sorry for the mistake. Check and see if you agree with the new eq's.
View attachment 111837
Ratch
Thanks. Will look over tommorrow.

• frozone45
Maybe a better way to think of it is not so much as strict mathematical terms since infinites usually have more nuanced meaning in reality than just "undefined". In this case, it just means that Vs needs to supply no current (or rather, it's current is always forced to zero by something else). So no matter how much current Vs tries to supply, it can't supply it.

The simplest way of thinking about infinite as just a very large number is not very accurate and only works for simple cases. It breaks down when you get to more advanced stuff which is why it's not really making sense here. In reality, infinite isn't a large number as much as the concept of something constantly getting larger without ever being a set value. This means that you can perform math operations on it against another "number" that either increases equal to, faster, or slower than that particular infinite. This nuance is what lets you come up with real-world results that make sense when working with infinite. If we're talking in terms of pure math, stuff like 0/0 or zero x infinite is undefined, but not really. You can kind of manipulate things and play tricks to get an answer out of it that reflects what happens in the real world (not always, but that's where the limits of human understanding are).

In not-so-strict hand-wavey pure math terms thinking about it in reverse: If the current gain is actually infinite but the voltage across Ri is finite, then what must the current supplied by Vs be? Well, it certainly can't be non-zero because you don't have an infinite voltage. It HAS to be zero because in the realm of crazy math zero multipled by the right kind of infnite perfectly counteract each other to become a finite number (BTW zero x infinite does not equal to zero in general in case you were not aware. In general it's undefined because the process of zeroing is up against infinite's process of increasing and without more information we can't decide which one is winning).

But what does this really mean? Well, remember when I said infinite was not a fixed number but a process of continually increasing so you could never catch up to it? There are many ways you can do this such as the rate of increase. In this specific case, this sort of infinite can sort of be thought of as "something intelligent" is constantly adjusting itself to always just barely stay ahead of you. That is infinite in the sense that you can never reach the end, but neither is it running on forever. It's a real finite value that you just can't catch up to because of your frame of reference (an outside observer that the infinite is not staying just ahead of can just go in and measure it just fine and it would be a value). So the gain of infinite is not really a fixed gain that is an infinite number, but a gain that is actually a finite number that is changing and constantly being adjusted to be "just enough" to force the Iin to be zero. In that way, everything stays defined but at the same time Vs can never overcome the obstacle of pushing current ("catching up" to the infinite gain which is actually just a finite gain that's always one step ahead of it).

People write PhD thesis on this stuff so what I told you is about all I know. I don't know if this answered your question or not but maybe it will help you approach strange scenarios like this with a more flexible mindset.

Of course, in reality there are real world limits to this circuit like the power supply voltages so a result of infinite would mean "as much as it possibly can" rather than something as nebulous as "undefined". It's just that you used approximations in your equations rather than entering all the real-world nitty gritty data that would have given you a more specific answer, so you instead you ended up with infinite as an approximation to a very large real-world number.

Would that very large real-world number that you mentioned at the end of the post above be equal to io/iin, with iin being an extremely small positive number close to zero?

Would that very large real-world number that you mentioned at the end of the post above be equal to io/iin, with iin being an extremely small positive number close to zero?
That is possible too. It just depends on whether the large number is larger than the small number is small (much like the rate of infinite increasing faster or slower). It's a bit like how with op-amps we assume infinite gain or infinite input impedance when that's not actually the case.

But I was talking generally. If there was a false assumption in your work as there seems to be...though I don't know what it is yet that takes precedence. In the end it's just an mathematical approximation of reality where we ignored the limits to something because it's too complex to deal with the them at the same time as everything else, and easier when we can just apply a judgement of the limits after we've done all the work. A bit like how you might use a sine-wave to model a real world power supply...but in the real world a power supply didn't exist for all time into both the past and future like an actual sinewave does, but it's too cumbersome to worry about that so we just pretend it does and use our common sense judgement afterwards

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• frozone45
That is possible too. It just depends on whether the large number is larger than the small number is small (much like the rate of infinite increasing faster or slower). It's a bit like how with op-amps we assume infinite gain or infinite input impedance when that's not actually the case.

But I was talking generally. If there was a false assumption in your work as there seems to be...though I don't know what it is yet that takes precedence. In the end it's just an mathematical approximation of reality where we ignored the limits to something because it's too complex to deal with the them at the same time as everything else, and easier when we can just apply a judgement of the limits after we've done all the work. A bit like how you might use a sine-wave to model a real world power supply...but in the real world a power supply didn't exist for all time into both the past and future like an actual sinewave does, but it's too cumbersome to worry about that so we just pretend it does and use our common sense judgement afterwards

Whoa, this is an circuit analysis question, not a philosophical conundrum. The current feedback from the dependent voltage source causes the Ri resistor to undergo a voltage drop of Vs. Therefore, no current exists into or out of the Vs voltage source because the voltage across Ri exactly counteracts the voltage source Vs. Any finite current output, no matter how small, is infinitely more than zero, because Io divided by zero is infinity. For that reason, the circuit will have a infinite current gain for any nonzero Vs applied to it.

Ratch

• frozone45
Thanks for all the help, Ratchit and dknguyen. I really appreciate it.

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