# Tricky little op amp.

Thread Starter

#### KevinEamon

Joined Apr 9, 2017
281
Hi guys

So I'm just going through these possible questions for the exam. As well as massive circuit diravation, my lecturer likes to throw in the odd tricky lil question. I found this one and thought its just the type of thing he would put in, to trip you up.

Now I've really confused myself here or there's something weird happening that I can't see.

Is this right?

#### danadak

Joined Mar 10, 2018
4,057
Yes, assuming Aol >> (R1 + R2).

Regards, Dana.

Thread Starter

#### KevinEamon

Joined Apr 9, 2017
281
Thanks Dana. I just wasn' sure about it at all.

#### crutschow

Joined Mar 14, 2008
28,172
Yes.
The ideal op amp maintains 0V between the two inputs when in the linear region thus, for that circuit, the output voltage must therefore equal the input voltage.

#### WBahn

Joined Mar 31, 2012
26,398
Yes, assuming Aol >> (R1 + R2).

Regards, Dana.
What does it mean for the dimensionless gain to be much, much larger than the sum of two resistances?

#### WBahn

Joined Mar 31, 2012
26,398
Hi guys

So I'm just going through these possible questions for the exam. As well as massive circuit diravation, my lecturer likes to throw in the odd tricky lil question. I found this one and thought its just the type of thing he would put in, to trip you up.

Now I've really confused myself here or there's something weird happening that I can't see.

Is this right? View attachment 152356
Your conclusion is correct, but I don't follow the reasoning you took to get there.

You second figure appears to be modeling the circuit from the inverting input to the output. But you hard tie the output to ground. Why?

Your very first equation is Vo = 0 V.

Where did that come from? I'm assuming from that second circuit diagram, but what is the justification for doing so.

Isn't it at odds with the later conclusion that Vo = Vi since Vi is not constrained to be 0 V?

Then to get from your next to last equation to your last equation, you have to be claiming that Vo = i·R1. What is the basis for this claim?

So you've made two or three mistakes and somehow magically come up with the right answer?

While I have seen that happen (in fact, one time I had a student make three independent and common mistakes when analyzing a circuit and legitimately get the correct answer to four sig figs) the more common reason is that the person knew what the answer was supposed to be and, consciously or unconsciously, forced their work to produce that result. Did you know what the answer to this problem was supposed to be?

Thread Starter

#### KevinEamon

Joined Apr 9, 2017
281
Well I'd love to be able to say you were wrong but of course your not. I don't have the answer to that one but I did simulate it.

Mind you the answer cane from the equation. I tried other equations but the vi kept cancelling no matter what I tried and obviously the question is phased in such a way to make you express vo in terms of vi.

I think it ended up something like r2/r1 = vo

I also wasn't sure how to do a loop equation and not tie vo to ground. Where else would it go?

#### WBahn

Joined Mar 31, 2012
26,398
Well I'd love to be able to say you were wrong but of course your not. I don't have the answer to that one but I did simulate it.

Mind you the answer cane from the equation.
How? I don't see it.

Again, in order to go from the next to last line to the last line, you have to be claiming that

vo = i·R1

On what basis are you making this claim?

Just looking at either of your circuit drawings shows that

vo = vi - i·R1

Once you have this, then you just need to find the current i.

Since it is an ideal opamp, no current flows into the opamp inputs, which means that whatever current flows in R1 is the same current that flows in R2.

So how much current flows in R2?

What is the voltage across R2?

I tried other equations but the vi kept cancelling no matter what I tried and obviously the question is phased in such a way to make you express vo in terms of vi.

I think it ended up something like r2/r1 = vo
If it ended up like that, what should have immediately SCREAMED out to you? Hint, think UNITS!

I also wasn't sure how to do a loop equation and not tie vo to ground. Where else would it go?
If you tie it to ground, then you are FORCING vo to be ZERO volts. That is a different circuit (and a particularly useless one, at that).

You simply have a voltage between vo and ground. If you like, you can represent it by a voltage source whose output is vo. This isn't changing the circuit -- in fact, it is a faithful model of the circuit because this source is nothing more than the output of the opamp! Remember, the basic model for an opamp is a voltage-controlled voltage source.

#### danadak

Joined Mar 10, 2018
4,057
What does it mean for the dimensionless gain to be much, much larger than the sum of two resistances?
I solved the problem the good old fashion way, assuming inside the OpAmp
was a diff amp with large G, called Aol, and then found the limiting case as
Aol approaches infinity. So yes the inequality dimensions do not match, its
the relative magnitude of their sum that matters.

The shortcut answer is from the approximation that if an OpAmp has - feedback
then V+ = V-, then V+ = Vi = V-, then Ir2 = 0, so Ir1 = 0, so Vi = V- = Vo

Regards, Dana.

Last edited:

#### ebp

Joined Feb 8, 2018
2,332
Think carefully about what crutschow said at #4, and remember that op amp circuits are analyzed in terms of currents in the surrounding components (ignoring input bias currents, offset voltage, bandwidth issues, etc.).

By inspection:
• both ends of R2 are at the same voltage if the amp is in the linear region
• if both ends are at the same voltage, there is no current flow through R2
• the current through R1 must be in the same direction and the same magnitude as the current through R2
• therefore, the current through R1 is zero
• therefore the voltage across R1 is zero
• therefore the amp output voltage is the same as the input voltage
all contingent on the original condition "... in the linear region"
So, is that a valid assumption? What would cause the circuit to be outside of the linear region?

Thread Starter

#### KevinEamon

Joined Apr 9, 2017
281
Om goodness I really must stop using my phone to do forum posts, I think it's time I admitted my eyes are slowly dying day by day. Sorry let me just read over this stuff now guys

Thread Starter

#### KevinEamon

Joined Apr 9, 2017
281

#### WBahn

Joined Mar 31, 2012
26,398
I solved the problem the good old fashion way, assuming inside the OpAmp
was a diff amp with large G, called Aol, and then found the limiting case as
Aol approaches infinity. So yes the inequality dimensions do not match, its
the relative magnitude of their sum that matters.

The shortcut answer is from the approximation that if an OpAmp has - feedback
then V+ = V-, then V+ = Vi = V-, then Ir2 = 0, so Ir1 = 0, so Vi = V- = Vo

Regards, Dana.
No, it's NOT the relative magnitude of their sum that matters.

What if, like many physical quantities, we had many different units for resistance. Just as we have meters, angstroms, and light-years, we could have a unit of resistance based on electrons per second or on Avogadro's number of electrons per second. Would we still just ignore the units and only look at the relative magnitude of their sums?

That the dimensions don't match means the equation is MEANINGLESS and you did something wrong.

Keeping mind that the only non-ideal parameter used in your analysis is finite gain (meaning we are still assuming that there are no voltage or current limitations being placed on the output), let's say that we have Av = 1000. While this might be unreasonably low for modern packaged opamps, that doesn't matter in terms of whether your results are valid. Plus, few of the opamps that I worked with in IC design had gains much more than 10 (yes, ten -- there just wasn't enough room to put circuits that had much more than that).

According to your result, the claim that Vo ~= Vin should be pretty good if R1 = 10 Ω and R2 = 0.01 Ω since R1 + R2 = 10.01 Ω which is much, much less than 1,000. By the same token, your result says that we can't that Vo ~= Vin if R1 = 10 kΩ and R2 = 100 kΩ since now R1 + R2 is 110 kΩ which is much, much LARGER than Av.

So let's see how these hold up. Let's choose Vin = 10 V. What is Vout for both cases.

Assuming I have done the analysis correctly, I get the following

Av = 1000, Vin = 10 V

R1 = 10 Ω, R2 = 0.01 Ω: Vo = 4.98 V

R1 = 10 kΩ, R2 = 100 kΩ: Vo = 9.99 V

It would appear that the results are the reverse of what you claim.

So does that just mean that Av has to be much smaller than R1 + R2?

No, I could easily come up with counter examples that would contradict that claim, too.

If you do the analysis carefully, you will discover that the actual constraint is that

Av >> 1 + R1/R2 or

Av - 1 >> R1/R2

Notice how this is dimensionally consistent. That means it doesn't matter what units we use for resistance (as long as they are the same, of course).

In the first case I used, R1/R2 = 1000. So it's not surprising that vo was that far off (or that the gain turned out to be right at one-half what was expected).

In the second case, R1/R2 = 0.1.

The takeaway, once again, is that ANY equation that is dimensionally inconsistent is fundamentally meaningless.

#### WBahn

Joined Mar 31, 2012
26,398
What does "LUL" mean?

You got to a nonsensical answer and at least recognized it and stopped. By showing the work you used to get there, we can help you spot it.

The mistake is yet another common 8th grade (well, possibly 4th or so) mistake having to do with the distributive property.

Hint, you are claiming that

A - (B - C) = A - B - C

This is a mistake that units tracking can't catch. Unfortunately there are a few common mistakes for which this is true. That's why the companion to tracking units is always asking if the result makes sense.

Once you fix the mistake you will end up with the result that 0 = 0.

While useless, it is also true and hence doesn't necessarily indicate that you did something wrong, just that you went down a path that didn't yield anything useful.

The reason here is that you are only working with part of the circuit and trying to force it to give you the result for the whole thing. It can't. So you ended up with a result that basically says, "anything is possible," because the result depends on things you haven't taken into account yet.

Go to your second line and then write the current in terms of the other half of the circuit, namely Vi and R2.

#### WBahn

Joined Mar 31, 2012
26,398
Think carefully about what crutschow said at #4, and remember that op amp circuits are analyzed in terms of currents in the surrounding components (ignoring input bias currents, offset voltage, bandwidth issues, etc.).

By inspection:
• both ends of R2 are at the same voltage if the amp is in the linear region
• if both ends are at the same voltage, there is no current flow through R2
• the current through R1 must be in the same direction and the same magnitude as the current through R2
• therefore, the current through R1 is zero
• therefore the voltage across R1 is zero
• therefore the amp output voltage is the same as the input voltage
all contingent on the original condition "... in the linear region"
So, is that a valid assumption? What would cause the circuit to be outside of the linear region?
Also contingent on the assumption that the gain is infinite, or at least sufficiently large as to make the resulting error term insignificant.

#### crutschow

Joined Mar 14, 2008
28,172
Also contingent on the assumption that the gain is infinite, or at least sufficiently large as to make the resulting error term insignificant.
The first sentence in his posted homework says it's an "Ideal op-amp".

Thread Starter

#### KevinEamon

Joined Apr 9, 2017
281
Lul is the short version of lol. I'm making it a thing. One day it'll be what all the cool kids are saying, you'll see.

Ahhhmmm kk btw I wasn't away all this time fixing my "4th grade error"...
I dropped like four years there in one shot... times are tough...jezz

#### WBahn

Joined Mar 31, 2012
26,398
The first sentence in his posted homework says it's an "Ideal op-amp".
Agreed. Just trying to keep the entire discussion from becoming seemingly contradictory.

#### WBahn

Joined Mar 31, 2012
26,398
Lul is the short version of lol. I'm making it a thing. One day it'll be what all the cool kids are saying, you'll see.

Ahhhmmm kk btw I wasn't away all this time fixing my "4th grade error"...
I dropped like four years there in one shot... times are tough...jezz

View attachment 152405
How on earth does 0 = 0 result in the claim that Vo = Vin?

What would you have concluded if it had turned out that 1 = 1 or 2 = 2?

Those are all perfectly true results that convey zero information about the problem.

And we ALL make 4th grade math errors, even 1st grade math errors, on an all too frequent basis.

Here just a couple weeks ago I was puzzled why I couldn't get two results to agree and after looking at it for the better part of fifteen minutes I realized it was because 18 + 17 is not, in fact, 45.

#### danadak

Joined Mar 10, 2018
4,057
The limiting equation I got was

Vo = (KR2Vi)/(R1 + R2 + KR2)
= R2Vi/[R2+(R1+R2)/K]

So is dimension-ally correct. And in the limit the solution.

K is gain, dimensionless. So if K >> R1 + R2 is an accurate statement.

Regards, Dana.

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