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summi

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Question on FB
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.

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raja.cedt Senior Fellow Offline Posts: 1516 Germany	Re: Question on FB Reply #1 - 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.
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summi Community Member Offline Posts: 38	Re: Question on FB Reply #2 - 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.
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buddypoor

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Re: Question on FB
Reply #3 - 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

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LvW (buddypoor: In memory of the great late Buddy Rich)

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rajkumar palwai Community Member Offline Posts: 46 Sunnyvale, California	Re: Question on FB Reply #4 - 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.
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summi Community Member Offline Posts: 38	Re: Question on FB Reply #5 - 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.
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raja.cedt

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Re: Question on FB
Reply #6 - 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.

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buddypoor

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Re: Question on FB
Reply #7 - 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.

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LvW (buddypoor: In memory of the great late Buddy Rich)

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raja.cedt

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Re: Question on FB
Reply #8 - 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.

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raja.cedt Senior Fellow Offline Posts: 1516 Germany	Re: Question on FB Reply #9 - 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.
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summi Community Member Offline Posts: 38	Re: Question on FB Reply #10 - 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.
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raja.cedt Senior Fellow Offline Posts: 1516 Germany	Re: Question on FB Reply #11 - 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.
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raja.cedt Senior Fellow Offline Posts: 1516 Germany	Re: Question on FB Reply #12 - 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.
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summi Community Member Offline Posts: 38	Re: Question on FB Reply #13 - Nov 5^th, 2011, 1:03pm Drea raja.cedt, On what basis you are telling ratio 8? Br, Summi.
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raja.cedt

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Re: Question on FB
Reply #14 - 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.

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