Tricky little op amp.

KevinEamon

Joined Apr 9, 2017
284

WBahn

Joined Mar 31, 2012
26,398
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.
No, it is not, because what matters is NOT that (R1 + R2)/K be small, but rather that it be small COMPARED to R2.

Again, let's use the previous counter example using YOUR equation above.

Is Vo ~= V1 if K = 1000, R1 = 10 Ω, and R2 = 0.01 Ω?

R1 + R2 = 10.01 Ω

This is certainly very small compared to K = 1000.

But your equation for the output says that Vo = 0.500 Vi, which certainly cannot be said to be close to Vo = Vi.

Whatever you define as the requirement for K being much, much larger than the magnitude of (R1 + R2) and whatever you define as the the gain of Vo/Vi being close enough to 1, I can easily find values of R1 and R2 that are counterexamples. I can do so even if I let you pick a minimum value of K. I may have to resort to unrealizable values of R1 and R2 to do so, but that's beside the point because the claim is based on a circuit model that is ideal except for finite opamp gain.

And, again, what does it even MEAN for the magnitude of the resistance to be much smaller than a dimensionless number?

What is the magnitude of 1000 mΩ. Is it 1000? or is it 1? What if we use a unit of resistance that resolves instead of V per C/s it resolved to V per electron/second? What would the magnitude of that resistance be?

This is perhaps most intuitively seen by looking at a physical analog in which the resistances are distances instead of a resistance (only because we are so used to only having one unit for resistance).

What would it mean to say that we need

K >> d1 + d2

If K is 1,000,000 and d1 = 10 miles and d2 = 20 miles, would that satsify the requirement?

What about if d1 = 10,000,000 mm and d2 = 20,000,000 mm?

What if the boundary between acceptable and unacceptable values of K were equality. What would the gain need to be in these to cases?

Once again, ANY claim based upon a dimensionally inconsistent claim is fundamentally meaningless.

Joined Mar 10, 2018
4,057
Actually I do have an error -

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.
What I should have said is as K > 00 (infinity, very high G). The equation is dimensionally correct,
anytime a constant (K) modifies a dimensioned value the dimensions are maintained and expressed.

In your case you use low values of K, so you are correct in your calculated example.

Regards, Dana.

WBahn

Joined Mar 31, 2012
26,398
Actually I do have an error -

What I should have said is as K > 00 (infinity, very high G). The equation is dimensionally correct,
anytime a constant (K) modifies a dimensioned value the dimensions are maintained and expressed.

In your case you use low values of K, so you are correct in your calculated example.

Regards, Dana.
There is no problem at all with

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

It is perfectly dimensionally consistent, regardless of the value of K.

But the inequality

K >> R1 + R2

is NOT dimensionally consistent and is utterly meaningless.

The only way you can salvage things is to require not that K be large or even that K approach infinity, but by requiring that K must BE infinite. Otherwise, whatever finite value of K you choose, I can find values for R1 and R2 that cause the claim to be false. All I have to do is pick a sufficiently large value of R1/R2 by choosing R1 to be "small" compared to K and then choosing R2 to be so much smaller than that as to drive the ratio through the roof. The reason I can do this is simple. It is NOT the relationship of K to the sum of R1+R2 that matters; it is the relationship of dimensionless K to the dimensionless ratio R1/R2 that matters.

And my using a low value of K has nothing to do with it.

What value of K do you want to pick? 100k? 1 million? 100 billion? Whatever you choose, I can choose values of R1 and R2 that result in your constraint being satisfied but yet Vout not being close to Vin.

Let's pick K = 10 million. So I then pick R1 = 10 kΩ and R2 to be 0.1 mΩ.

The gain is 1000x the "magnitude" of R1+R2.

What is the circuit gain?

Let's run the numbers.

The gain with these values is 0.0909.

Notice that neither of these resistance values is that unreasonable. I've used 0.1 mΩ resistors in opamp circuits. Neither are the resulting currents unreasonable at all. If the input is 10 V, then the output would be 0.91 V and the current through the two resistors would be less than 1 mA.

Now let's go the other way.

I claim that what matters is that K >> 1 + (R1/R2).

So let's pick a low value of K, say 100.

That means that we need R1/R2 to be much smaller than 99. Let's make it 0.1 by choosing R2 = 10 kΩ and R1 = 1 kΩ.

So we have K = 100 compared to R1 + R2 = 11000 Ω in pretty flagrant and gross violation of your requirement. But what is the gain?

Let's run the numbers.

The gain with these values is 0.989, which is very close to 1. And the currents are likewise quite reasonable at only 10 microamps.

So clearly having a gain much larger than most real opamps doesn't guaranteed that the gain is anywhere close to unity while having a gain much lower than nearly any real opamp doesn't prevent the gain from being close to unity.

Another subtle point that is easy to miss is that

K >> 1 + (R1/R2)

is NOT the same thing as

K - 1 >> R1/R2

For instance, if K = 2, the first one shows that it doesn't matter what we do with R1 and R2, we aren't getting the left side to be any better than just twice the right side.

But with K = 2 we can get the left side of the bottom one to be however much larger than the right side by just picking R2 to be that many times larger than R1.

Joined Mar 10, 2018
4,057
I agree with what you have said about K >> R1 + R2 not being properly dimensioned.

Vo = R2Vi/[R2+(R1+R2)/K] that Vo approaches Vi as result, irrespective of R values.
And as you point out when it is infinity Vo == Vi.

Actually K is not dimensionless, in fact its Volts/Volt, but effectively dimensionless
in the scheme of things.

It would be instructive to run the sensitivities of Vo with respect to R12, R2, and K
as well.

Regards, Dana.

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OBW0549

Joined Mar 2, 2015
3,566
Now I've really confused myself here or there's something weird happening that I can't see.
Such a simple circuit, so much confusion. Nothing weird going on here:
• The op amp is ideal: no input offset voltage, no input bias current, and gain is infinite.
• The op amp is in a negative feedback circuit, so start by assuming its output is in the linear region.
• The op amp's (+) input is connected to Vi.
• Since we're operating in the linear region, the (-) input must be at the same voltage as the (+) input.
• If the (-) input is at the same voltage as the (+) input, it is therefore at the same voltage as Vi.
• Thus there is no voltage across R2, and therefore no current through it.
• With no current through R2 and no current into or out of the op amp's (-) input, there can be no current through R1.
• If no current is flowing through R1, there is no voltage drop across it.
• Therefore, the voltage at Vo must equal the voltage at the op amp's (-) input, which is equal to Vi.
• Ergo, Vo = Vi.
There, nothing to it except for some simple, step-by-step logic and application of basic principles.

Arjay Abacan

Joined Jul 17, 2019
1
it goes like this

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MrAl

Joined Jun 17, 2014
8,613
What does it mean for the dimensionless gain to be much, much larger than the sum of two resistances?
Hi,

Maybe this is one of those times when dimensional analysis does not quite work. We could look into this though maybe there is an explanation.

If we set open loop gain A=n*(R1+R2) then maybe we can say that 'n' is a conductance. Then the expression reduces to:
vo=(n*vi*R2)/(n*R2+1)

and here we see we got rid of R1.
Now since 'n' is a conductance, the units are:
v=(1/R*v*R)/(1/R*R+1)=v

So the question is can we explain why 'n' has to be a conductance other than it makes the units work out.

WBahn

Joined Mar 31, 2012
26,398
Dimensional analysis worked just fine and revealed that the claim was fundamentally meaningless.

crutschow

Joined Mar 14, 2008
28,542
Such a simple circuit, so much confusion. Nothing weird going on here:
• The op amp is ideal: no input offset voltage, no input bias current, and gain is infinite.
• The op amp is in a negative feedback circuit, so start by assuming its output is in the linear region.
• The op amp's (+) input is connected to Vi.
• Since we're operating in the linear region, the (-) input must be at the same voltage as the (+) input.
• If the (-) input is at the same voltage as the (+) input, it is therefore at the same voltage as Vi.
• Thus there is no voltage across R2, and therefore no current through it.
• With no current through R2 and no current into or out of the op amp's (-) input, there can be no current through R1.
• If no current is flowing through R1, there is no voltage drop across it.
• Therefore, the voltage at Vo must equal the voltage at the op amp's (-) input, which is equal to Vi.
• Ergo, Vo = Vi.
There, nothing to it except for some simple, step-by-step logic and application of basic principles.
I agree.
The academics seem to be over-complicating the solution.

MrAl

Joined Jun 17, 2014
8,613
Dimensional analysis worked just fine and revealed that the claim was fundamentally meaningless.
Hello again,

The reason i said that was because the claim worked numerically. Why it's not dimensionally accurate i dont know.

WBahn

Joined Mar 31, 2012
26,398
Hello again,

The reason i said that was because the claim worked numerically. Why it's not dimensionally accurate i dont know.
It was dimensionally inaccurate because a mistake was made -- a mistake that the messed up units shouted at full volume and was roundly ignored.

And, it didn't work numerically. I showed counter-examples where it didn't work. I showed that (ignoring the units) the open-loop gain being much greater than the sum of the two resistances was neither necessary nor sufficient for the closed-loop gain to be close to unity. I showed a case where K was much greater than the sum yet the closed-loop gain was no where near unity and I showed a case where K was much smaller than the sum yet the closed-loop gain was very close to unity.

So how does that amount to the claim working numerically?

MrAl

Joined Jun 17, 2014
8,613
Hello,

The claim was (A=Aol):
A>>R1+R2

Now if A has to be greater than the sum, then set:
A=a*(R1+R2)

and now 'a' is a factor that controls how much greater R1+R2 is.
Then the solution reduces to:
vo=(a*vi*R2)/(a*R2+1)

and with 'a' getting higher and higher, the '1' in the denominator is absorbed by the a*R2 there, so we end up with:
vo=a*R2/(a*R2)*vi

which of course reduces to:
vo=vi

However, if we make 'a' too small, that '1' stays significant and we might end up with:
vo=R2/(R2+1)*vi

and now we have to depend on R2 alone to get near to vo=vi.
Before we only had to depend on R1+R2 getting larger.
So i guess if you want to say that A can be small if R2 is very large that might work too.
So we might also include R2>>1 or rather R2>>R1.
This makes sense because 'a' and R2 are mutiplicatively symmetrical.

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The Electrician

Joined Oct 9, 2007
2,866
The gain expression can be manipulated into a form which helps to see what's going on:

The green and red parts can be seen as error terms. Consider the second parenthetical. If R1>>>R2, the red part approaches 1, the second parenthetical approaches zero, and the gain approaches zero. If R2>>>R1, the red part approaches zero, the second parenthetical approaches unity, and the gain is given by the first parenthetical.

The second parenthetical can never be more than 1, so the gain can never be greater than the first parenthetical. Thus the gain can never be unity unless k becomes large (making the green part zero), no matter what R1 and R2 are.

WBahn

Joined Mar 31, 2012
26,398
Hello,

The claim was (A=Aol):
A>>R1+R2

Now if A has to be greater than the sum, then set:
A=a*(R1+R2)
Note that 'a' has to have units of inverse resistance (or conductance).

and now 'a' is a factor that controls how much greater R1+R2 is.
Then the solution reduces to:
vo=(a*vi*R2)/(a*R2+1)

and with a getting higher and higher, the '1' in the denominator is absorbed by the a*R2 there, so we end up with:
vo=a*R2/(a*R2)*vi

which of course reduces to:
vo=vi
Is a = 1,000 S big enough to satisfy A >> (R1 + R2)?

If so, then let's pick A = 100,000 and now R1 + R2 must just be less than 100, right?

Now, is that 100 Ω? 100 mΩ? 100 MΩ?

Let's assume (and that's all it is, since there NO reason to prefer one possible unit of resistance over any other possible unit of resistance) that we mean 100 Ω.

So let's make R1 = 91 Ω and R2 = 0.001 Ω.

Is 100,000 >> (91 Ω + 0.001 Ω) ?

This gives us an 'a' value of 100,000 / (91 Ω + 0.001 Ω) = 1099. Is that high enough?

If so, then wouldn't we expect our output to be very close to our input?

Let's see.

vo=(a*vi*R2)/(a*R2+1)

vo = (1,099 S* 0.001 Ω * vi) / (1,099 S* 0.001 Ω + 1)

vo = (1.099 * vi) / (1.099 + 1) = 0.524 * vi

Doesn't seem very close to unity gain to me.

So, clearly, having A >> (R1 + R2) is not a sufficient condition to have Vo ~= Vi.

The fallacy of the conclusion that A >> (R1 + R2) comes from the fact that what we need is NOT for (R1+R2)/A to become very small, it's that we need it to become very small RELATIVE to the value of R2.

Vo = R2Vi/[R2+(R1+R2)/A]

In order for Vo -> Vi, we need R2 + (R1+R2)/A to be ~= R2.

That that means that we need

(R1+R2)/A << R2

Oh, look, we are comparing a resistance to a resistance -- the units work out. What a shocker!

Thus we need A >> (R1 + R2) / R2 = 1 + (R1/R2)

So let's choose A to be 10 (not 100,000) and let's just swap R1 and R2 (leaving the sum R1 + R2 unchanged).

So you value of 'a' is now A / (R1 + R2) = 10 / (0.001 Ω + 91 Ω) = 0.1099 S

By your criteria, Vo shouldn't be anywhere close to Vi -- and it should certainly be much further away than the gain of 0.524 we had when 'a' was 1099.

Let's see.

vo=(a*vi*R2)/(a*R2+1)

vo = (0.1099 S* 91 Ω * vi) / (0.1099 S* 91 Ω + 1)

vo = (10.001 * vi) / (10.001 + 1) = 0.909 * vi

So, even with a gain of just 10 we got MUCH closer to unity gain that with a gain of 100,000, despite the fact that our gain was much less than the sum of the two resistors whereas before it was much, much, much greater.

So, clearly, having A >> (R1 + R2) is not a necessary condition to have Vo ~= Vi.

Well, if having A >> (R1 + R2) is neither necessary nor sufficient, then what meaning does it have?

The fact that it is dimensionally inconsistent forcing you to compare things that fundamentally cannot be compared (i.e., incommensurable) screams out that it is fundamentally meaningless!!

When you end up with something that is dimensionally inconsistent, you need to take another look at what the real situation is, instead of ignoring the fact that you are working with garbage and trying to salvage it by putting patch after patch on it, such as what you do below.

However, if we make a too small, that '1' stays significant and we might end up with:
vo=R2/(R2+1)*vi

and now we have to depend on R2 alone to get near to vo=vi.
Before we only had to depend on R1+R2 getting larger.
So i guess if you want to say that A can be small if R2 is very large that might work too.
So we might also include R2>>1
What does it mean for a resistance to be much greater than 1?

That's like saying that the length of a ship must be much greater than 1? 1 what? Foot? Mile? Light-year? Angstrom? It's meaningless.

MrAl

Joined Jun 17, 2014
8,613
Hello again,

Why did you leave off the end of that last quote?

Another idea might be A>>R1/R2.
There's your dimensional correctness for ya

But sure, no problem whatever you want to do, you can come up with the conditions then that's fine.

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WBahn

Joined Mar 31, 2012
26,398
The gain expression can be manipulated into a form which helps to see what's going on:

View attachment 181904

The green and red parts can be seen as error terms. Consider the second parenthetical. If R1>>>R2, the red part approaches 1, the second parenthetical approaches zero, and the gain approaches zero. If R2>>>R1, the red part approaches zero, the second parenthetical approaches unity, and the gain is given by the first parenthetical.

The second parenthetical can never be more than 1, so the gain can never be greater than the first parenthetical. Thus the gain can never be unity unless k becomes large (making the green part zero), no matter what R1 and R2 are.
Of course the gain can never be unity unless k become large -- in fact, the gain can never be unity unless k is identically infinite!

But you can get close to unity even with quite moderate values of k. With k=10 you can get a gain of 90+% and with k=100 you can get 99+%, which is good enough for nearly all applications where you need a unity gain amplifier.

The format you present breaks the issues into two pieces, one focusing on k irrespective of the resistance values and one focusing on k relative to the ratio of the resistance values. While that format has a lot to say for it, it obscures the coupling between the open-loop gain and the degree to which the value of R2 exceeds R1.

Another format that is perhaps more practical (because it doesn't break out the issues separately and instead explicitly shows how they are coupled) is

$$\frac{V_o}{V_i} \; = \; \frac{1}{1 \; + \; \frac{\( 1 \; + \; \frac{R_1}{R_2}$$}{k}}\)

Here we can see at a glance that to get the gain to approach unity, we need the second term in the denominator to become small and hence we nee

$$k \; >> \; 1 \; + \; \frac{R_1}{R_2}$$

Once we decide how close we need the gain to be to unity, we can trivially determine the needed open loop gain given the ratio or the needed ratio given the open-loop gain.

WBahn

Joined Mar 31, 2012
26,398
Hello again,

Why did you leave off the end of that last quote?
Because you must have edited after I quoted it -- I started my reply yesterday at about 4:30 pm (about half an hour after you posted) but had to leave to run my daughter around and just now got back to finishing the response. I didn't note that you had changed anything. So here's the last part as you have it now:

So i guess if you want to say that A can be small if R2 is very large that might work too.
Can we say that? Let's see.

A can be small: Let's make it 100
if R2 is very large: What is "very large"? Is 10,000 Ω very large?
Since no constrain has yet been placed on R1, let's make it 10 MΩ.

So now the closed-loop gain is 0.0833

Does't look like this patch helped out that much.

So we might also include R2>>1 or rather R2>>R1.
This makes sense because 'a' and R2 are mutiplicatively symmetrical.
Well, which is it?

Is it R2 >> 1 or is it R2 >> R1

Again, what does it mean for R2 to be much greater than 1. You keep operating under the notion that 1 can ONLY mean 1 OHM. What is the basis for this claim? There is NOTHING magical or fundamental about the ohm. No more so than saying that the length of something has to be greater than 1 somehow necessarily means that we absolutely must be talking about meters and not angstroms or light-years.

Another idea might be A>>R1/R2.
There's your dimensional correctness for ya
Close, but not quite.

Let's choose A = 0.1 and R1 = 1 mΩ and R2 = 1 kΩ.

Is 0.1 >> 0.001 Ω / 1000 Ω = 0.0000001

Sure seems like it to me.

So what's the open-loop gain? It's only 0.091

Once again, the requirement is

A >> 1 + R1/R2

Notice that the 1 that you oh so conveniently choose to ignore places an absolute floor on the open-loop gain irrespective of the ratio of the resistances. You can only compensate for low gain by choosing your resistor ratio wisely up to a point.

But sure, no problem whatever you want to do, you can come up with the conditions then that's fine.
At least the conditions I came up with actually work. If you are happy coming up with conditions that don't work, I just hope that I don't ever have to rely on anything you designed based on them.

MrAl

Joined Jun 17, 2014
8,613
Hi,

I was only guessing that's it.

A >> 1 + R1/R2

is fine with me. But what are the units for that '1' in that?
Is it dimensionless can we assume that
If we can, how do we know we can?

WBahn

Joined Mar 31, 2012
26,398
Hi,

I was only guessing that's it.
There's no need to guess -- just do that math.

A >> 1 + R1/R2

is fine with me. But what are the units for that '1' in that?
Is it dimensionless can we assume that
If we can, how do we know we can?
If it weren't dimensionless, then the units would be there. If it lacks dimensions, then it is dimensionless (seems rather obvious when put that way, doesn't it?).

This, of course, is only if the units are properly tracked in the first place. If someone is routinely sloppy with units, then it's not surprising that they would never have confidence whether a number in an equation is dimensionless or not because they are used to seeing dimensioned quantities presented as pure numbers and then just tacking on whatever units they think the answer should have regardless of the units that their work actually results in.

That's how we slam hundred million dollar space probes into planets or let airliners filled with passengers run out of fuel mid-flight.