# Phase margin

Discussion in 'Electronic Basics' started by [email protected], Sep 11, 2008.

1. ### Guest

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°, from its
definition
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°
+-180°-tan^-1(w/p1)-tan^-1(w/p2) -tan^-1(w/z) =45°
what's wrong?
thanks

2. ### EeyoreGuest

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

3. ### Jon SlaughterGuest

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°, from its
definition
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°
+-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 http://www.intersil.com/data/an/an9415.pdf will help

4. ### Andrew HolmeGuest

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?