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# Frequency compensation and dominant poles

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#### Jone

dominant closed-loop poles definition

As is well known, only the dominant loop poles can be used for compensating a negative feedback amplifier. To determine the dominant poles one then compares the sum of the loop poles with the sum of the system poles. The set of loop poles which sum is larger (less negative) than the sum of the system poles (more negative) are the dominant ones dominant. Why is this true?

compensation des poles dominants

Since nobody has answered, I'll give some background information.

The loop gain (return ratio) of an amplifier can be written as

Aβ(s) = Aβ(0)N(s)/D(s), where Aβ(0) is the DC loop gain.

Now the closed-loop gain of an amplifier is given by

A_cl = A_ideal • -Aβ(s)/(1-Aβ(s)), where A_ideal = 1/β.
Substituting the above in A_cl gives

A_cl = A_ideal • Aβ(0)N(s)/[D(s) - Aβ(0)N(s)]

So the poles of the closed loop gain (system poles) is given by the characteristic equation CP(s) = D(s) - Aβ(0)N(s) = 0.
To simplify loop gain calculations, we'll neglect C_µ (C_gd for FETs) in the hybrid pi-model. This is equivalent to a case with n poles and no finite zeros, that is, N(s) = 1 and
D(s) = (s-p1)(s-p2)...(s-pn).

Substituting this for the characteristic equation gives (after some manipulations)

s^n - s^(n-1)Σpi + ... + [1-Aβ(0)](s-p1)...(s-pn) = 0, where Σpi = p1+p2 + ... pn. Compensating an amplifier means that we want to move the system poles in a certain position, that is, the characteristic equation should be

CP2(s) = (s-ps1)(s-ps2)...(s-psn) = s^n - s^(n-1)s^(n-1)Σpsi - ... = 0, where psi are the wanted system poles.

We can now compare CP(s) with CP2(s) and to compensate the amplifier we want Σpi = Σpsi. According to my text, it is not always possible to move all poles such that Σpi = Σpsi. It only works if Σpi ≥ Σpsi, that is if the sum of the loop poles is larger or equal than the sum of the wanted system poles. So, why is this true? I guess one should be able to understand that from the above? I should also mention that I have little knowledge of analog electronics, I just started learning frequency compensation.

I must confess, that
1.) I am interested in system theory (which is touched by your question),
and unfortunately
2.) rather lazy. That means that for me it is to difficult and it cost too much of my time to read the equation and symbols and "hieroglyphes" of your posting.
______________
If you could prepare a nice pdf document using a formula editor it would be much easier to read your contribution. Sorry.

LvW

dominant frequency

To make things more readable I've prepared a pdf of the above problem statement.

1.) I don´t understand the equation below (4); is it correct ?
2.) I don´t understand the meaning of equation in the second line of the last paragraph ( Σ=Σ). Of course, it is true that X1=X1 - but what does it help ?

Sorry for some mistakes in the above pdf. I've corrected the errors and put in some more details.

The material I've presented is from some old lecture notes I've taken. I remember we were only provided the results. Now I want to understand the details behind all this, and I know there is almost no litterature on this. I find this material extremely unique and powerful, especially as nobody else seem to know about it. It is part of the the so called Structured Analog Design, a quite new direction in the field.

Hi, Jone !

Question: Why do you write in Equ. (4) D(s)=0 ?

More than that: Your statement/assumption "loop poles are complex-conjugate" is not correct. Poles of the loop gain can be simple and real; and they turn into complex poles by feedback.

Hi LvW. You're right, poles don't necessary occur in conjugate pairs. From the little experience I have, I can only tell that loop poles are typically real (neglecting C_mu, Cgd). If I make that assumption I get eq (6). If the poles are single complex I can't cancel the (-1)^n. It is not written in my text what assumptions that have been made, so I'm guessing a little bit. The pdf is updated with corrections. Anyway, the main question is why only dominant poles can be moved?

And what about eqn. (4) ? D(s)=0 ? Is this correct ? I don´t think so.

Yes, I corrected (4) as well.

Maybe there are still some misunderstandings. Solving eq. (3) gives you the system poles. (4) is no equation but a factorization of loop poles. I've just denoted it in two ways, in case someone is not familiar with the product notation (pi). This factorization is then substituted into (3). How do they contradict?

(3) says: D(s)-Aβ=0
(4) and (5) say: D(s)=0

Thus, Aβ=0 ! Correct, or not ?

No, (4) says just D(s) = (1-s/p_l1)...(1-s/p_ln), i.e. it is no equation but a factorization of poles. Please download the new pdf in the above post.

OK, I did not know that you have corrected it in your old posting.

Hi Jone,

I believe you have made another error in equ. (4).
If you develop D(s) in linear terms it yields: (s-p1)(s-p2).....=Π(s-pi).

As a result the right part of equ. (5) is only Aβ (without the Π).

Perhaps this clarifies your problem ?

Yes, I've tried that before. However, it seems harder to obtain equation (6) (which I assume is correct) by factorizing like (s-p1)...(s-pn). Also in the text I have they factorize it like I did.

Although the derivations bother me somewhat, I'm more interested in getting answers for questions 1 and 2 in the pdf.

Or maybe I'll have to contact my old professor...

Jone said:
Yes, I've tried that before. However, it seems harder to obtain equation (6) (which I assume is correct) by factorizing like (s-p1)...(s-pn). Also in the text I have they factorize it like I did.
Although the derivations bother me somewhat, I'm more interested in getting answers for questions 1 and 2 in the pdf.
Or maybe I'll have to contact my old professor...

I cannot follow you. Are you trying to find an easy method for calculations - even if it is mathematically not correct ? (I have tried it ...... it seems harder....).
This seems to be a strange approach.

As the answers to both questions depend on your formulas resp. results one cannot give any answer until it is clear if results are OK or not. Sorry for that.
Regards

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