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A question about multiplication of gain_bandwidth in closed loop opamp

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akbarza

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Hi
in section 12.6.1 of gray and meyer book (Analysis and Design of Analog Integrated Circuits) the below text has been written:
Therefore, the poles p1d and p2d of the DM half-circuit are given by (9.32)
and (9.33). Assume that the op amp is operating in a feedback loop, the feedback factor fdm
for the DM feedback loop is frequency-independent, and the right-half-plane zero has been
eliminated. Then to achieve 45◦ phase margin, the magnitude of the DM loop gain should be
unity at the frequency |p2d|. Since |gain|×frequency is constant from |p1d| to |p2d| due to the
one-pole roll-off there, we can write
|adm0.fdm.p1d| = 1 · |p2d| (12.73)

anyone can explain to me:
1) why between p1d and p2d poles, the |gain|*frequency is constant?i know in open loop opamp in this range, gain is roll of as 20db/decade
2) the relation 12.73 how is obtained ? note that this relation is written for closed loop opamp.
thanks
 

Hi,

1) why between p1d and p2d poles, the |gain|*frequency is constant?i know in open loop opamp in this range, gain is roll of as 20db/decade
You almost answered it on your own.
you say: gain drops by 20dB/decade.
Now a decade is a factor of 10
and 20dB is a factor of 10, too

While frequency rises ... the gain drops. And because both with the same rate (but in opposite direction) you get a constant gain_frequency_product.

Klaus
 

hi KLAUST
I tried to calculate to obtain this value(-20db/dec) in paper, but i couldn't. i started with open loop gain with A_op= A_0/{(1+s/wp1)(1+s/wp2)}.
suppose we accept the value -20db/dec in open_loop gain.the question is about in closed loop_gain. also i tried to obtain the relation between gain and poles in closed loop gain, but unfortunately i was unsuccessful.
thanks for your answer.
 

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