Phase margin

Sep 11, 2008 3 Replies
S

If i consider the open loop gain of an opamp with negative feedback composed of 2 poles (with negative real part) and one zero (with positive real part) and if i want a phase margin of 45=B0, from its definition



360=B0-tan^-1[-A(jwo)F(jwo)]=3D45=B0 i have
+180=B0-tan^-1[A(jwo)F(jwo)]=3D+180=B0-[-tan^-1(w/p1)-tan^-1(w/p2) -tan^-1(= w/ z)]=3D45=B0 where +180 is obtained from 360=B0-tan^-1[negative constant]=3D360=B0- tan^-1[-1]=3D360=B0-180=B0 Instead i've found as solution
+-180=B0-tan^-1(w/p1)-tan^-1(w/p2) -tan^-1(w/z) =3D45=B0 what's wrong? thanks



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E

Are you trying to do one of those cute pole-zero compensations with the extra pole inside the op-amp itself.

Fun isn't it ? I just hack it with simulations. I know why the lecturers at Uni said they didn't fancy getting into it !

Graham

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J
360°-tan^-1[-A(jwo)F(jwo)]=45° i have +180°-tan^-1[A(jwo)F(jwo)]=+180°-[-tan^-1(w/p1)-tan^-1(w/p2) -tan^-1(w/ z)]=45° where +180 is obtained from 360°-tan^-1[negative constant]=360°- tan^-1[-1]=360°-180° Instead i've found as solution +-180°-tan^-1(w/p1)-tan^-1(w/p2) -tan^-1(w/z) =45° what's wrong? thanks =================

wtf is all that?

maybe rewrite them using better notation tan^-1 is also known as arctan, A(jwo) looks like a function such as f(x) and jwo looks like a single object.

Also define all your symbols so it will be easier to decipher.

in anycase maybe

formatting link
will help

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A

This appears to say:

360 - arctan(-k) = 360 - arctan(-1) = 360 - 180 = 180

Surely arctan(-1) = -45 and arctan(-k) arctan(-1) unless k=1

No?

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