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Message started by summi on Oct 28^th, 2011, 9:22am

Title: Question on FB
Post by summi on Oct 28^th, 2011, 9:22am

Dear Forum,
I have a question regarding the attached ckt. With the shown polarity of both opamp i am sure it is -ve fB. Buyt let us say if reverse the polarity of both opamp, for me it seems it would work, can any please comment on this. It's theory question so hope no one will comment on the bias or some practical issue. My only intension is to learn Feedback polatiry.

BR,
Summi.

Title: Re: Question on FB
Post by raja.cedt on Oct 28^th, 2011, 9:50am

hello,
yes, if you reverse the polarities still +ve feedback lesser gain compared to -ve FB. So it will work. The better way to check this kind of doubts is replace op amps with VCVS and simulate. Do that it.

Thanks,
Raj.

Title: Re: Question on FB
Post by summi on Oct 29^th, 2011, 12:41am

dear raja.cedt,
could you please explain bit clear, because when i change op2 polarity how can you garrantee -ve feedback around op2? i understand entire loop will be -ve feedback.

BR,
Summi.

Title: Re: Question on FB
Post by buddypoor on Oct 29^th, 2011, 1:46am

Hi Summi,

may I try to explain?

For a composite circuit to be stable, both opamps are checked separately.
In case of changed polarities for both opamps:
* opamp 1: It is stable because its feedback loop is closed via an inverting gain (opamp 2).
* opamp 2: It has two feedback loops. One negative (closed via opamp 1 that is connected to the neg. terminal of opamp 2) and one positive via the two resistors. It will be stable because the negatibe FB loop dominates over the positive one. The negative loop gain is much larger (because of the open-loop gain of opamp 1) than the positive one (determined by the resistors). Thus, the net loop gain is negative (as required).
_______________

Regards

Title: Re: Question on FB
Post by rajkumar palwai on Oct 29^th, 2011, 1:52am

Summi,
When op2 polarity is reversed, RC & RD combination provide local +ve feedback to it. But the RA & RB combination go through the OP1 and provide -ve feedback to OP2. As the OP1 has large gain, the total -ve feedback is always greater than the +ve feedback and hence the OP2 is stable.

@raja,
Correct me if my analysis is wrong.

Title: Re: Question on FB
Post by summi on Oct 29^th, 2011, 3:17am

Dear all,
Thanks you very much for your reply, So now i understood that both kind of connections gives same result (from stability point of view). Correct me if am worng.

Br,
Summi.

Title: Re: Question on FB
Post by raja.cedt on Oct 29^th, 2011, 3:34am

@rajkumar: what you are saying is correct.

@summi: For a circuit to function you need _ve Fb.
From dc -ve feedback point of view you are correct both are same, but i will go with the the schematic what you have posted rather than changing polarity because if you analyze you get more benefits from stability point of view, i guess you get More UGB.

Thanks,
raj.

Title: Re: Question on FB
Post by buddypoor on Oct 29^th, 2011, 7:34am

Summi,

I think Raj is correct in his doubts - as far as RF stability is concerned.

Explanation: You always have to discriminate between two different stability issues:

* DC stability concerns the bias point only - and that can be evaluated by simple visual inspection (as we did up to now).

*However, RF stability (against oscillations) is quite another story.
And in this respect, both alternatives are very critical. This kind of stability (dynamic stability) strongly depends on the open-loop gain function vs. frequency of both opamps as well as on the chosen resistor ratios. I think, the dynamic stability can be evaluated by simulation only.
However, in case of changed opamp polarities (referred to the drawing presented) you must not trust the BODE diagram as one part of the circuit (opamp 2) is unstable by itself. Instead, you have to check the Nyquist diagram.

Title: Re: Question on FB
Post by raja.cedt on Oct 29^th, 2011, 7:42am

hello buddypoor,
Now i got a question, for the both cases i am getting different transfer function (i took simple integrator model for every opamp, and op1 UGB lesser than op2 for better Phase margin).

I guess it's not so surprise, because those two ckts are identical only from DC feedback point of view only not other aspects.

Thanks,
Raj.

Title: Re: Question on FB
Post by raja.cedt on Oct 29^th, 2011, 8:04am

sorry forgot to write some thing....in the inverted case i am getting RHZ, so tried with nyquist and found no encirclements...Buddypoor could you please verify this?

Thanks,
Raj.

Title: Re: Question on FB
Post by summi on Oct 31^st, 2011, 2:44am

dear forum,
thanks a lot for helping me, through this discussion i had learned a lot about feedback. I want to find the relation between opamp bandwidths for stable operation, please suggest me how to find.

BR,
Summi.

Title: Re: Question on FB
Post by raja.cedt on Oct 31^st, 2011, 2:49am

roughly you can say inner loop should settle faster than main loop, so op2 BW has to be higher than op1, make sure op2bw=8*op1bw, it works

Thanks,
raj.

Title: Re: Question on FB
Post by raja.cedt on Oct 31^st, 2011, 1:23pm

hello all,
i donno why, the polarity reverse case seems to be unstable, any comments on this?

Thanks,
Raj.

Title: Re: Question on FB
Post by summi on Nov 5^th, 2011, 1:03pm

Drea raja.cedt,
On what basis you are telling ratio 8?

Br,
Summi.

Title: Re: Question on FB
Post by raja.cedt on Nov 5^th, 2011, 1:13pm

hello Summi,
Sorry 8 is bit difficult to understand without clear explanation.
think about internal loop with op2, assume op2 UGB is wu2, so closed loop 3db BW is wu2/2, now in the overal loop this appers like non-dominate pole so at least wu1(op1 UGB) to be equal to wu2/2 to have 45deg phase margin, so ratio 2 is okay, but if you have wu1=wu2/8 then good PM.

Please go through 2nd order system basics.

Thanks,
Raj.

Title: Re: Question on FB
Post by loose-electron on Nov 6^th, 2011, 5:25am

Simplify the thinking a little here:

OP2 and its attached circuitry do not matter.

Why? Its a fixed positive gain.

Thats like saying the first op-amp has a bit more gain in it, and noting else beyond that.

Analyze it from there.

Title: Re: Question on FB
Post by raja.cedt on Nov 6^th, 2011, 6:00am

hello loose electron,
you are corect, but it adds adds Phase also.

Thanks,
raj.

Title: Re: Question on FB
Post by Alexandar on Nov 7^th, 2011, 12:21am

The second opamp has local feedback. In general that means less gain and a high frequency pole. The latter means that it most likely won't be dominant. So you don't really need to consider the 'phase'.

Title: Re: Question on FB
Post by raja.cedt on Nov 7^th, 2011, 12:25am

hello alex,

I think you have to consider phase also because you said high okay but how high compared to wp1..otherwise could you please tell me the relations between both UGB's

Thanks,
Raj.

Title: Re: Question on FB
Post by buddypoor on Nov 7^th, 2011, 12:38am

Of course, Raja.cedt is right in saying that the second opamp matters.
Contrary to loose-electron's statement ("fixed gain") the second opamp strongly influences the stability of the two-opamp combination (called "composite amplifier") due to its phase shift.
In this regard, the GBW of the second opamp as well as both resistor ratios are of importance. A simple ac simulations can reveal the problem areas.

Title: Re: Question on FB
Post by Alexandar on Nov 7^th, 2011, 12:59am

well raja. By broad banding your second opamp its pole moves up. That's the only thing we can tell. How much that is the question, and as we have no details, it's difficult to judge whether it is going to be dominant or not. I agree, and although I suspect it is not of importance, it can go either way...

Title: Re: Question on FB
Post by buddypoor on Nov 7^th, 2011, 2:19am

Lex wrote on Nov 7^th, 2011, 12:59am:

I don't think that it is a question of "dominance"! The effect is simply that the frequency response of the second stage (with feedback) adds to the response of the open-loop gain of the 1st opamp.
Thus, this results in a composite opamp combination that has (at least) two poles within the active region. For a good phase margin the overall feedback should be sufficiently small - resulting in a relatively high overall gain. As mentioned already - details depend on resistor ratios and opamp properties (GBW in particular).

Title: Re: Question on FB
Post by Alexandar on Nov 7^th, 2011, 2:34am

buddypoor wrote on Nov 7^th, 2011, 2:19am:

....

I don't think that it is a question of "dominance"! The effect is simply that the frequency response of the second stage (with feedback) adds to the response of the open-loop gain of the 1st opamp.
Thus, this results in a composite opamp combination that has (at least) two poles within the active region.

...

I underlined some words in your reply. The first sentence I agree. The second sentence can be true, but doesn't have to be. So the word 'thus' is out of place, in my opinion. If you replace 'thus' by 'for example' it'd be fine =)

Title: Re: Question on FB
Post by buddypoor on Nov 7^th, 2011, 3:37am

Hi Alexander,

I suppose you are somewhat more familiar with the secrets of the english language than I am. Thus, I agree with you.
However, for a reasonable design (that makes sense) and for a second opamp with a GBW that is not much larger than that of the 1st opamp there will be a 2nd pole in the active region of the resulting opamp combination.
Don`t forget, the composite opamp should have better properties (more gain for higher frequencies) if compared with a single opamp. Otherwise the whole combination makes no sense. Of course, I can give the 2nd opamp with feedback a gain of 0 or perhaps 6 dB. But, where is the improvement?
Therefore, the 2nd opamp should provide an additional gain of - let's say - at least 12...20 dB. And if its own GBW is not larger at least by a factor of 10 (compared to opamp 1) there will be a second pole within the active region of the combination.
Don't you agree?

Title: Re: Question on FB
Post by buddypoor on Nov 7^th, 2011, 4:10am

In this context (properties of a composite two-opamp combination) it is perhaps worth mentioning that many, many papers have been published in the past dealing with methods (active, passive) to stabilize such circuits - for amplifier as well as for integrator applications.
These methods are summarized under the name "active/passive phase compensation".
By the way: A very detailed treatement of 4 different methods to combine two opamps with the aim to extend the usable frequency range was published already in 1987 by Mikhael and Michael:
(Composite operational amplifiers: Generation and finite-gain applications, IEEE Transactions Circuits & Systems, CAS34, No.5, may 1987)

Title: Re: Question on FB
Post by Alexandar on Nov 7^th, 2011, 6:54am

buddypoor wrote on Nov 7^th, 2011, 3:37am:

Well first of all, the second opamp might have other features than gain. Say for example a good driving capability or so. You could mix a low noise opamp with a low distortion opamp in this way.

But assuming that the second opamp would purely be used for gain, why placing the local feedback at all? To me it would seem better to get rid of that local feedback (find a way of preserving the negative sign), and use some decent frequency compensation, e.g. placing a capacitor in parallel with the resistor in the fb network (i.e. placing a zero in the fb). Much more efficient, imho.

Thx for the info. I will take a look at it some time. =)

Title: Re: Question on FB
Post by raja.cedt on Nov 7^th, 2011, 8:11am

hello buddypoor,
thanks for such a nice pap.

Thanks,
raj.

Title: Re: Question on FB
Post by buddypoor on Nov 7^th, 2011, 10:12am

Hi Alexander,

Qote: Well first of all, the second opamp might have other features than gain. Say for example a good driving capability or so. You could mix a low noise opamp with a low distortion opamp in this way.

Yes, I agree with you. More gain - and thus - hopefully more bandwidth is only one of several possible reasons. For example, often the second opamp is replaced by a current feedback amplifier because of its good slew rate properties.

Quote: But assuming that the second opamp would purely be used for gain, why placing the local feedback at all? To me it would seem better to get rid of that local feedback (find a way of preserving the negative sign), and use some decent frequency compensation, e.g. placing a capacitor in parallel with the resistor in the fb network (i.e. placing a zero in the fb). Much more efficient, imho.

I doubt if this is really a good suggestion. If this would be true, you could compensate each opamp with 2 poles in the active region using such a parallel capacitor. Or do you speak about two completely different opamps with different GBW's?
But I know several other composite configurations that have similar (or even better) properties. Some are treated/investigated in the mentioned article (and eqipped with stabilizing features).

Title: Re: Question on FB
Post by loose-electron on Nov 7^th, 2011, 12:58pm

Short of special characteristics the second opamp is not needed.

If its chip level, that second set of "special" properties can be put into the first opamp.

Second order effects like additional noise, additional phase, (and a bunch of other things) are not even being considered here.

Just the basic architecture.

If you want to get into he relative pole placement of things, additive phase, additive noise figure, contributions of offset and systematic mismatches etc etc etc etc you are going to need a lot more details to define it.

Whet was described in the original question can be done with a single opamp.

Title: Re: Question on FB
Post by loose-electron on Nov 7^th, 2011, 1:00pm

Also, if you want a special output stage, close the feedback path with that device in the feedback loop, that way the high gain of the opamp gets used to maintain linearity of the output device.

Title: Re: Question on FB
Post by buddypoor on Nov 8^th, 2011, 12:50am

Hello loose-electron,

from the practical point of view (that means: application oriented) I agree with you. However, there are some other aspects.
For example: Understanding the principles connected with terms like "Bode-Diagram, Stability, influence of poles and zeros,...".
And, according to my understanding, this was the reason behind the original question. And in this context, I think it is interesting to investigate the pros and cons of the presented two-opamp configuration. I Like to remind you that a lot of papers were published in the past dealing with exactly this (and some similar) configuration(s).

Beside these arguments, I doubt if you really can find the best combinationof all relevant parameters within one single chip: high gain, low offset, low noise, high slew rate, high output current capability, high input and low output resistance, ....
Regards

Title: Re: Question on FB
Post by loose-electron on Nov 8^th, 2011, 4:11am

BuddyP -

Not a problem, we get many questions here that can be answered easily opening a book on the basics, or similar.

In this particular question it was pretty globabal in nature, and the structure did not make a lot of sense unless there was a special need in the application.