The Designer's Guide Community
Forum
Welcome, Guest. Please Login or Register. Please follow the Forum guidelines.
Mar 24th, 2019, 5:53am
Pages: 1 2 
Send Topic Print
stability for ckts having two feedback loops (Read 24947 times)
justdoit
New Member
*
Offline



Posts: 6
IIT-Kharagpur,India
stability for ckts having two feedback loops
Dec 19th, 2007, 10:18pm
 
Hi all ,
Can any one give me the references for doing the stability analysis if there are multiple feedback loops (i have two feedback loops in my circuit ) in circuit ...  Thanks in advance
Back to top
 
 

Haribabu,
M.tech.
IIT-Kharagpur,
India.
View Profile justdoit   IP Logged
buddypoor
Community Fellow
*****
Offline



Posts: 529
Bremen, Germany
Re: stability for ckts having two feedback loops
Reply #1 - Dec 20th, 2007, 9:24am
 
Hi Haribabu !

You have picked up a very interesting point in control theory since this question is not answered in most text books. I face this problem since several years and I came to the conclusion that there is no other way than to do  an open loop analysis for each of the possible loops.
In your case, if you have two feedback loops, there are three possible options:
1.)Loop 1 open, loop 2 closed
2.)Loop 2 open, loop 1 closed
3.) Both loops open.

Than, the most critical stability margin determines the system margin. With other words, the loop with the least margin is dominant.

Interestingly, to reduce the system to a single loop system prior to the open-loop-analysis gives exactly the same results. The reason is that - in your example - there are exactly three different ways to do this (with three different loop gains and, hence, three different margins).
I hope this could help a little.
Good luck and greetings from Germany.
Lutz



Back to top
 
 

LvW (buddypoor: In memory of the great late Buddy Rich)
View Profile   IP Logged
HdrChopper
Community Fellow
*****
Offline



Posts: 493

Re: stability for ckts having two feedback loops
Reply #2 - Dec 22nd, 2007, 8:31pm
 
Hi Haribabu,

This topic is a very tough one. However the analysis of multi-loop feedback systems can be simplified to a single loop case if any of this two conditions is met:

1) All the feedback loops have a break point in common. In this case, by breaking the loops at that point the stability analysis is performed as if it were a single loop case.

2) A multi-loop feedback which comprises one general feedback network and several inner (local) feedback loops can be analyzed by just breaking the general feedback network (single loop) if each of its inner feedback networks is stable by itself (which must be left intact during the analisys).

The paper "Determination of stability using return ratios in balanced fully differential feedback circuits" by Paul Hurst, IEEE TRANSACTIONS ON CIRCUITS AND SYSTEMS-11: ANALOG AND DIGITAL SIGNAL PROCESSING, VOL. 42, NO. 12, DECEMBER 1995 , has examples of the cases mentioned above

Hope this helps
tosei
Back to top
 
 

Keep it simple
View Profile   IP Logged
buddypoor
Community Fellow
*****
Offline



Posts: 529
Bremen, Germany
Re: stability for ckts having two feedback loops
Reply #3 - Dec 26th, 2007, 12:32am
 
The above reply from TOSEI is, of course, completely correct as it takes two special cases into consideration.
The first example refers to the common case that an amplifier has one negative and one positive feedback circuit. The second example can be found in control circuits with one or more "local" loops and one overall feedback loop, which than is the most "slowest" of all.
But, what is the procedure if none of these cases applies ? There are a lot of cases where, for example, two loops exist which cannot discriminated by "inner loop" resp. "outer loop". In this case, I think, the procedure has to be as described in my first reply to HARIBABU on Dec. 20th.
Best wishes and a happy new year to all community members
Lutz (Germany)
Back to top
 
 

LvW (buddypoor: In memory of the great late Buddy Rich)
View Profile   IP Logged
HdrChopper
Community Fellow
*****
Offline



Posts: 493

Re: stability for ckts having two feedback loops
Reply #4 - Jan 14th, 2008, 10:56am
 
Hi Lutz,

From your previous reply, you suggested that

buddypoor wrote on Dec 20th, 2007, 9:24am:
Hi Haribabu !


In your case, if you have two feedback loops, there are three possible options:
1.)Loop 1 open, loop 2 closed
2.)Loop 2 open, loop 1 closed
3.) Both loops open.

Than, the most critical stability margin determines the system margin. With other words, the loop with the least margin is dominant.




My question is, how do you account for the influence of the two loops closed at the same time in your analysis?. I understand that by analyzing them in a separate way, the one with the least margin will dominate. However, things might be even worse when considering all the cases you enumerated at the same time, because of the relative margin between the different options. Could you please clarify?

Thanks
Tosei
Back to top
 
 

Keep it simple
View Profile   IP Logged
Eugene
Senior Member
****
Offline



Posts: 262

Re: stability for ckts having two feedback loops
Reply #5 - Jan 14th, 2008, 10:41pm
 
I thought there was already a discussion of this somewhere in the Forum but I could not find it. Oh well, perhaps it's just a case of deja vu. Anyway, the most direct method of assessing stability of multiloop systems is to compute the Eigen values of the closed loop system matrix. This is not that practical in electric circuits because the poles are often widely separated and the order of the system can also be quite high. Spectre and SPICE have tools for computing closed loop poles but they sometimes list RHP (i.e. unstable) poles when none exist. Such poles are usually in close proxmity to closed loop zeros such that they nearly cancel.

If you insist on using more classical frequency domain methods, there is a little known procedure called sequential loop closures, or sequential return differences. The procedure is as follows:
1. Open all loops such that the resulting system is stable.
2. Assess the loop gain of one loop with all other loops open. Keep track of the number of clockwise encirclements of the Nyquist point.
3. Close that loop and assess the loop gain of the next loop. Keep track of the net number of encirlcements of the Nyquist point.
4. Close that loop and do the same for the next loop. And so on.
5. If the net number of encirclements (clockwise - counter clockwise) equals zero, the system is stable. If the net is greater than zero, the system is unstable.

The only problem with this method is that it is hard to identify a single phase margin or gain margin. You could select the minimum phase margin and minimum gain margin as you assessed each loop but you may get different numbers if you select a different sequence. Despite this shortcoming, the procedure is mathematically rigorous.

This method is often mis-applied, most commonly when common mode feedback loops are involved. Suppose we have two interacting loops called loop one and loop two. The mistake is to start with loop one closed while you assess loop two. You then look at loop one with loop two closed.  The problem is that the logic is circular because you do not know how many RHP poles you are starting with. If loop one has one RHP pole, loop two MUST encircle the Nyquist point exactly once counter clockwise to make the closed loop system stable. In short, the mis-applied method is the same as saying loop one is stable because loop two is stable and loop two is stable because loop one is stable, therefore the system is stable. Imagine two brothers going to court saying "I'm telling the truth because my brother never lies and he says I'm telling the truth". The other brother says the same thing. Does that prove they are both truthful? To get the net encirclements at the end of the procedure, you must start with all loops open; you must start knowing for sure that one brother never lies.
Back to top
 
 
View Profile   IP Logged
Frank Wiedmann
Community Fellow
*****
Offline



Posts: 642
Munich, Germany
Re: stability for ckts having two feedback loops
Reply #6 - Jan 14th, 2008, 11:31pm
 
Eugene wrote on Jan 14th, 2008, 10:41pm:
I thought there was already a discussion of this somewhere in the Forum but I could not find it.

You may have been thinking of http://www.designers-guide.org/Forum/YaBB.pl?num=1163532257/1#1 (see the last paragraph of reply #1). By the way, for Bode's method, you must open the loops not by cutting a wire but by setting the controlled sources to zero (so that the impedances you are seeing are not changed). This is very difficult with traditional circuit simulators because the controlled sources are usually inside transistor models and you do not have direct access to them.
Back to top
 
 
View Profile WWW   IP Logged
buddypoor
Community Fellow
*****
Offline



Posts: 529
Bremen, Germany
Re: stability for ckts having two feedback loops
Reply #7 - Jan 15th, 2008, 2:45am
 
Hello Haribabu!

Q:My question is, how do you account for the influence of the two loops closed at the same time in your analysis?

A: Stability margins (e.g. phase margin) can be defined only for an open loop. It is defined as the amount of additional (parasitic) phase shift which must be introduced into this loop in order to make the system instable if the loop is closed again. Therefore, when both loops of a two-loop system are closed you cannot perform any additional stability analysis (hopefully I did understand your question correctly).  

Eugene wrote on Jan 14th, 2008, 10:41pm:
I thought there was already a discussion of this somewhere in the Forum but I could not find it. ................................................................................
...............................
If you insist on using more classical frequency domain methods, there is a little known procedure called sequential loop closures, or sequential return differences. The procedure is as follows:
1. Open all loops such that the resulting system is stable.
2. Assess the loop gain of one loop with all other loops open. Keep track of the number of clockwise encirclements of the Nyquist point.
3. Close that loop and assess the loop gain of the next loop. Keep track of the net number of encirlcements of the Nyquist point.
4. Close that loop and do the same for the next loop. And so on.
5. If the net number of encirclements (clockwise - counter clockwise) equals zero, the system is stable. If the net is greater than zero, the system is unstable.

The only problem with this method is that it is hard to identify a single phase margin or gain margin. You could select the minimum phase margin and minimum gain margin as you assessed each loop but you may get different numbers if you select a different sequence. Despite this shortcoming, the procedure is mathematically rigorous........................................................................
................................................................................
...........


Of course, I completely agree with the procedure as quoted above.
One short comment to the "problem" as mentioned above: If one system has three different feedback loops it certainly will exhibit three different stability margins. Therefore. to ask for a single system margin is - for my opinion - a more or less philosophical question, because in reality additional parasitic phase shifts will occur not only in one of these loops.
I think the whole subject seems to be a very interesting one and I appreciate the discussion about it.
Lutz

[color=#006600][/color]
Back to top
 
 

LvW (buddypoor: In memory of the great late Buddy Rich)
View Profile   IP Logged
buddypoor
Community Fellow
*****
Offline



Posts: 529
Bremen, Germany
Re: stability for ckts having two feedback loops
Reply #8 - Jan 15th, 2008, 2:49am
 
Hello Tosei,

I apologize for the error I have made in context with your name (Haribabu was the originator of this topic).
Lutz
Back to top
 
 

LvW (buddypoor: In memory of the great late Buddy Rich)
View Profile   IP Logged
Eugene
Senior Member
****
Offline



Posts: 262

Re: stability for ckts having two feedback loops
Reply #9 - Jan 15th, 2008, 8:28am
 
Frank,
Thanks for the reference. However, I was thinking of another Forum thread. Perhaps I was originally thinking of some e-mail exchanges.

You raises an excellent point regarding the application of classical frequency domain stability analysis to transistor circuits. It can indeed be hard to find convenient break points within each loop. But that is what Spectre's stability probe is for is it not? I think the problem arises in multiloop systems because Spectre's stability probe does not actually break the loop; I don't think you can use the probe to break the loops you have not yet assessed.

Back to top
 
 
View Profile   IP Logged
rajdeep
Senior Member
****
Offline



Posts: 220
UK
Re: stability for ckts having two feedback loops
Reply #10 - Jan 15th, 2008, 9:34am
 
Hi all,

I found this discussion fascinating!! Well, I am not a circuit designer but has grown tremendous interest in it for last couple of months or so. I mainly deal with verification of mixed signal circuits.

Can anyone suggest me a good book where the issue of control theory and its application in circuit design have been discussed in detail??

I understand handling multi-loop systems has not been covered that much, but even the single-loop analysis would be okay for me to start with. I have the book by Allen and Hollberg, but somehow it does not provide me with enough FEEL of the thing (must be due to my shortcomings). I have started to look into Razavi and like it very much. But for the control part I would prefer to have a book that looks at the problem in a more generic way first and then shows its application in ckt design. Is there any book or good lecture report anything available?

Too much asking... and sorry for any digression! Let the good talk continue..

Rajdeep
IIT Kharagpur!!
Back to top
 
 

Design is fun, verification is a requirement.
View Profile   IP Logged
HdrChopper
Community Fellow
*****
Offline



Posts: 493

Re: stability for ckts having two feedback loops
Reply #11 - Jan 15th, 2008, 11:29am
 
buddypoor wrote on Jan 15th, 2008, 2:45am:
Of course, I completely agree with the procedure as quoted above.
One short comment to the "problem" as mentioned above: If one system has three different feedback loops it certainly will exhibit three different stability margins. Therefore. to ask for a single system margin is - for my opinion - a more or less philosophical question, because in reality additional parasitic phase shifts will occur not only in one of these loops.

[color=#006600][/color]


Lutz:

I'm not completely sure about your statement concerning the validity of determining a single system margin. Actually my question was more or less pointing towards that very concept: I was not completely sure - from the method you suggested - you were accounting for the "cross-effects" between loops as for overall stability (or in other words for a single system's margin). Although you might have several loops in your system, I do not see why a single margin for the system could not be determined. From Eugene's answer it looks superposition is the answer (see below)

Eugene:

Thanks for pointing out the sequential loop closures. From your description I understand you are simply applying superposition as for the stability criteria; i.e.  counting the net (cumulative) number of encirclements around the -1 point in the Nyquist plot resulting from each individual loop assessment looks to me like superposition. Am I interpreting it correctly? If so, then my first answer as for how to account the relative influence of each loop into the other ones is simply: superposition. This would not be crazy since, at the end of the day, we are analyzing linearly-modeled systems.

Thanks
Tosei
Back to top
 
 

Keep it simple
View Profile   IP Logged
Frank Wiedmann
Community Fellow
*****
Offline



Posts: 642
Munich, Germany
Re: stability for ckts having two feedback loops
Reply #12 - Jan 15th, 2008, 1:04pm
 
Eugene wrote on Jan 15th, 2008, 8:28am:
Frank,
Thanks for the reference. However, I was thinking of another Forum thread. Perhaps I was originally thinking of some e-mail exchanges.

You raises an excellent point regarding the application of classical frequency domain stability analysis to transistor circuits. It can indeed be hard to find convenient break points within each loop. But that is what Spectre's stability probe is for is it not? I think the problem arises in multiloop systems because Spectre's stability probe does not actually break the loop; I don't think you can use the probe to break the loops you have not yet assessed.

Spectre's stb analysis works differently but it can only handle single loops. The probe is simply a dummy element like a voltage source with 0V that you specify when you set up the stb analysis, so of course it cannot be used to break the loop.

If anyone wants to dig really deep into this loop-gain stuff (single loop only), you can find on my webpage http://www.geocities.com/frank_wiedmann/loopgain.html some additional thoughts about the method used by the stb analysis (I call it "Tian's method" there) and a somewhat different method for analyzing loop gain called the "General Feedback Theorem" (GFT), which was developed by R. David Middlebrook. The associated discussion starting at http://groups.yahoo.com/group/Design-Oriented_Analysis_D-OA/message/40 might also be worth reading, even if some of the points are probably mainly of academic interest.
Back to top
 
 
View Profile WWW   IP Logged
Eugene
Senior Member
****
Offline



Posts: 262

Re: stability for ckts having two feedback loops
Reply #13 - Jan 15th, 2008, 10:15pm
 
I don't think I would refer to sequential loop closures as a superposition method. I say that because I would apply superpostion to a very special kind of multiloop system. If all feedback loops have one node in common such that all loops could be broken at that single node, then you can indeed define a single loop gain and that loop gain is the SUPERPOSITION (i.e. sum) of all the individual loop gains. But in general, the sum of all loop gains is only part of the story. If you apply Mason's rule to find the overall loop gain, the second term in the expansion gives you the  "superposition" case I described above. The third term is the sum of the products of all non-touching loops; the fourth term is the sum of the products of all three-non-touching loops; and so on. My point is that in general the loop gain is not a simple superposition of all the individual loop gains.

Now that I think of it, I suppose you could use Mason's rule to define a single loop gain for a multiloop system and that loop gain could be used to define a unique phase margin and gain margin. It's just that Mason's rule can be tricky to apply and I don't know of any simulator that has a Mason's rule calculator.
Back to top
 
 
View Profile   IP Logged
Frank Wiedmann
Community Fellow
*****
Offline



Posts: 642
Munich, Germany
Re: stability for ckts having two feedback loops
Reply #14 - Jan 15th, 2008, 11:59pm
 
The difficulty with applying Mason's rule might be that usually not all paths in an electronic circuit are unidirectional (consider e.g. resistive feedback). The General Feedback Theorem can be used to transform a circuit into an equivalent diagram that contains only unidirectional paths (which may or may not correspond to actual components of the circuit), but it only applies for single loops.

In my opinion, when looking at phase margin or gain margin, you should always examine how sensitive they are to variations of your circuit elements in order to see how far you are away from instability. The well-known relationship between the phase margin and the step response of a circuit was derived for a rather special single-loop configuration (see the links I gave above for details), I certainly would not expect it to be valid for complicated multi-loop configurations.
Back to top
 
 
View Profile WWW   IP Logged
Pages: 1 2 
Send Topic Print
Copyright 2002-2019 Designer’s Guide Consulting, Inc. Designer’s Guide® is a registered trademark of Designer’s Guide Consulting, Inc. All rights reserved. Send comments or questions to editor@designers-guide.org. Consider submitting a paper or model.