Got asked this question in an interview, didn't have a good answer.

[KingNothing]

Can you answer it?

Why do op-amps require internal feedback?

 

[uart]

Why do op-amps require internal feedback?

A. To save us from the bother of having to provide an external compensation network.

More Details. Opamps have very high gain and multiple stages, so without compensations you are bound to get open loop gain greater than unity and 180 degree phase shift at some frequency. This means that the opamp will require some form of compensation network (either internal or external) if any significant amount of feedback is to be used (which of course it is the case for most opamp circuits, unity gain circuits being the worst as they have 100% of the output signal feed back to the input).

 

[turbo]

OP, you ought to Google "op-amp". Wiki has a side-bar that may clear up your question.

 

[berkeman]

Can you answer it?

Why do op-amps require internal feedback?

That's a pretty good interview question. I may borrow it sometime...

First you need to find the feedback element, and then you need to think about what it is used for...

http://www.sentex.net/~mec1995/gadgets/741/741equiv.gif

[PLAIN]http://www.sentex.net/~mec1995/gadgets/741/741equiv.gif

 

[Jiggy-Ninja]

I'm confused. Is this about feedback or compensation?

When I hear feedback, I usually think of the negative feedback you use to control gain, or the positive feedback to make it self-oscillate.

 

[berkeman]

I'm confused. Is this about feedback or compensation?

Why does it have to be either/or?

 

[Jiggy-Ninja]

Why does it have to be either/or?

The question is "Why do op amps require internal feedback?"

By my understanding of feedback, they don't. It would defeat most of the purpose of an op amp, which is that you can design your own feedback to make it do any number of things.

Compensation is a different matter. Internal compensation is important for op amp stability and preventing positive feedback, but that's different from an internal feedback network.

It's either poor, ambiguous wording in the original question, or I'm hopeless confused about something I don't understand. I give the odds about 60-40 between those two.

 

[berkeman]

The question is "Why do op amps require internal feedback?"

By my understanding of feedback, they don't. It would defeat most of the purpose of an op amp, which is that you can design your own feedback to make it do any number of things.

Compensation is a different matter. Internal compensation is important for op amp stability and preventing positive feedback, but that's different from an internal feedback network.

It's either poor, ambiguous wording in the original question, or I'm hopeless confused about something I don't understand. I give the odds about 60-40 between those two.

Probably a little better wording for a more precise question might be "How do many opamps use internal feedback, and why does it improve performance?" Because as uart points out, you can get uncompensated opamps for special applications. But there are advantages for doing at least one kind of feedback inside the opamp IC (where capacitances are low...).

I'll post a spoiler link for anybody who has given up looking at the equivalent circuit that I posted above.

Spoiler

See "Dominant Pole" at wikipedia.org page on Frequency_compensation

BTW, in a way the interview question is worded well, because it will force the person to think a bit about opamps in general, and try to remember all of the reasons that opamps do use internal feedback for something. It's a good "conversation starter".

 

[berkeman]

Quiz Question -- which component in the LM741 equivalent circuit in post #4 is used for the feedback element?

 

[Studiot]

which component in the LM741 equivalent circuit in post #4 is used for the feedback element?

Answer it's the most expensive component on the chip!

Thereby hangs a tale...

 

[berkeman]

Answer it's the most expensive component on the chip!

Great answer. You're hired!

 

[berkeman]

And here is a final spoiler for those who are interested. It goes to Ninja's compensation question/comments:

Spoiler

Look at "Class A Gain Stage" at wikipedia.org's page on Operational_amplifier

It also gives a hint about Studiot's "expensive" comment...

 

[Jiggy-Ninja]

Maybe I'm just getting hung up on the terminology. Everyone is saying "feedback", but it sounds like they're talking about "compensation", and it's throwing me off, like if a bowler started talking about touchdowns.

 

[berkeman]

Maybe I'm just getting hung up on the terminology. Everyone is saying "feedback", but it sounds like they're talking about "compensation", and it's throwing me off, like if a bowler started talking about touchdowns.

Compensation is generally done in the feedback, somewhere. For the opamp, the "dominant pole" compensation is generally done on-chip. Forming the dominant pole with an internal feedback element is the most efficient way to do it, generally.

 

[KingNothing]

Oh, so it's about compensation. I really struggled with the question and couldn't come up with an answer. I had no idea he was referring to the compensation for the dominant pole.

If he had said "internal compensation", I would have answered correctly.

 

[berkeman]

Oh, so it's about compensation. I really struggled with the question and couldn't come up with an answer. I had no idea he was referring to the compensation for the dominant pole.

If he had said "internal compensation", I would have answered correctly.

Bonus Question -- So why did Studiot say this?

Answer it's the most expensive component on the chip!

 

[turbo]

Bonus Question -- So why did Studiot say this?

Because it's not more expensive to fabricate, but because with gain factors of over 10K failure of this component will be responsible for chip rejection more than variance in any other component.

 

[berkeman]

Because it's not more expensive to fabricate, but because with gain factors of over 10K failure of this component will be responsible for chip rejection more than variance in any other component.

Interesting, I'll have to think about that one. I don't think that's what Studiot had in mind, at least, that's not what I am thinking...

 

[Averagesupernova]

Off the top of my head I would say that a HUGE amount of effort went into this part of the device in order to stabilize the part over virtually an infinite number of design configurations. It is expensive because of the effort that went into design, not the cost of production. I believe this borders on what Turbo-1 is saying, maybe it is exactly what he is saying.

 

[DragonPetter]

My guess is that the internal feedback gives the opamp the desired ideal characteristic of infinite open loop gain which is assumed in many basic calculations of opamp applications

The infinite gain is what let's you design your own feedback loops while maintaining stability. This also improves your bandwidth when you limit the gain, since since more bandwidth is a trade off to less gain.

Edit:
Also, the infinite gain helps cancel out the common mode signal. If you solve the the transfer function of an op-amp using KCL/KVL you get an equation with an extra term at the end where it is the common mode voltage difference times the OPEN LOOP gain of the op amp. So internal feedback is giving the opamp a more ideal gain to improve its common mode rejection.

PS I am going all this off the top of my head as if it were an interview, no cheating :P

 

[DragonPetter]

Ok I went back and checked my old notebook, and for the sake of anyone reading my response and using it as fact, I need to clarify.

the common mode rejection ratio CMRR is the ratio of the internal "open-loop" gain of the opamp and its common mode gain.

So the larger your internal gain, the higher the CMRR

 

[dlgoff]

I'm thinking it would be the cost to make R5 (39kΩ) since it needs to be precise in order to provide the current mirror with a good reference current.

 

[berkeman]

I'll post a spoiler solution to all of this tomorrow, unless Studiot wants to post it tonight. Fun thread.

 

[berkeman]

I'm thinking it would be the cost to make R5 (39kΩ) since it needs to be precise in order to provide the current mirror with a good reference current.

Good thought, but no, not R5. R5 does set several bias currents through mirrors, but the accuracy of those bias currents is not real critical.

 

[berkeman]

My guess is that the internal feedback gives the opamp the desired ideal characteristic of infinite open loop gain which is assumed in many basic calculations of opamp applications

The infinite gain is what let's you design your own feedback loops while maintaining stability. This also improves your bandwidth when you limit the gain, since since more bandwidth is a trade off to less gain.

Edit:
Also, the infinite gain helps cancel out the common mode signal. If you solve the the transfer function of an op-amp using KCL/KVL you get an equation with an extra term at the end where it is the common mode voltage difference times the OPEN LOOP gain of the op amp. So internal feedback is giving the opamp a more ideal gain to improve its common mode rejection.

PS I am going all this off the top of my head as if it were an interview, no cheating :P

Ok I went back and checked my old notebook, and for the sake of anyone reading my response and using it as fact, I need to clarify.

the common mode rejection ratio CMRR is the ratio of the internal "open-loop" gain of the opamp and its common mode gain.

So the larger your internal gain, the higher the CMRR

You need to take phase shift into account when talking about "infinite open loop gain". Does that change your response at all?

 

[berkeman]

Because it's not more expensive to fabricate, but because with gain factors of over 10K failure of this component will be responsible for chip rejection more than variance in any other component.

Off the top of my head I would say that a HUGE amount of effort went into this part of the device in order to stabilize the part over virtually an infinite number of design configurations. It is expensive because of the effort that went into design, not the cost of production. I believe this borders on what Turbo-1 is saying, maybe it is exactly what he is saying.

The manufacturing test of the LM741 die will catch any issues. What else makes a die expensive? (Well, in relative terms.)

 

[Jiggy-Ninja]

What I want to know is why some of those transistors have their collectors shorted to their base. I saw that in the schematic for a voltage reference too. What's the point of that?

I'd try and answer the other question everyone is distracted by, but all I see is a mess of parts, and I've never really gone through the 741 schematic to figure out what makes it tick.

 

[DragonPetter]

You need to take phase shift into account when talking about "infinite open loop gain". Does that change your response at all?

Not that I can think of.. a phase shift would imply a pole or zero in the frequency response which would change the phase margin of its internal feedback and shift the maximum frequency of operation which would change the stability characteristics of the opamp (assuming the opamp feedback transfer function looks like a low pass). I don't know what phase shift you're talking about though, unless its just the phase shift of the internal feedback of the opamp. But anyway, a feedback system internal to the opamp is the necessary ingredient for generating the open loop gain, and that's how it ties open-loop gain into the original question.

My only insight was that an ideal opamp has infinite open-loop gain, and this is why there is internal feedback of an opamp, to make it closer to an ideal opamp.

And the reasons an ideal opamp has infinite open-loop gain is to increase stability when it is bandwidth limited and to maximize its CMRR.

 

[uart]

What I want to know is why some of those transistors have their collectors shorted to their base. I saw that in the schematic for a voltage reference too. What's the point of that?

That's one half of a simple current mirror. The BJT's with the shorted CB are acting like diodes. This branch sets the current for the mirror and it's partner follows it.

 

[berkeman]

Thereby hangs a tale...

Seems like a good time for the tale...

 

[vk6kro]

The unusual thing about the 741 was that it had a capacitor in it (across R7 and R8).

Before this chip, opamps had to have an external compensating capacitor and the quality of this capacitor (or lack of quality) affected the stability of many circuits.

It limits the high frequency gain of the chip so that internal phase shifts do not turn negative feedback into positive feedback.

It is still very rare to see capacitors in integrated circuits.

 

[berkeman]

It is still very rare to see capacitors in integrated circuits.

And costly at that size.

We have capacitors in our analog ASICS very often now, but not that large.

 

[Jiggy-Ninja]

That's one half of a simple current mirror. The BJT's with the shorted CB are acting like diodes. This branch sets the current for the mirror and it's partner follows it.

Is there a specific reason it's done that way? Couldn't they just make it a normal diode and save from having to make one P/N region?

EDIT: So the answer is the cap? 30 pF is considered large for an IC?

 

[uart]

Is there a specific reason it's done that way? Couldn't they just make it a normal diode and save from having to make one P/N region?

The idea is that the two transistors are very closely matched, basically side by side on the same die with identical doping profiles.

The fact that it's a transistor, even with the CB junction externally shorted, still ensures that when operating it has an excess minority carrier concentration profile that closely matches that of it's partner.

 

[berkeman]

EDIT: So the answer is the cap? 30 pF is considered large for an IC?

Yes. That's my answer, and I'm sticking to it!