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[Design of Analog CMOS Integrated Circuits by Behzad Razavi] enter image description here

The author explains that if the feedback (beta) is made weaker, the plot shifts down and hence, we hit the GX earlier now, thereby improving phase margin.

My question is, by decreasing the feedback factor (hence decreasing loop gain), the position of the closed loop pole should push out to a higher frequency and give us the 'bandwidth extension' property of closed-loop feedback systems. Why doesn't that happen here?

AlfroJang80
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    Is this positive feedback or negative feedback? I don't see the context here. – jonk Oct 22 '22 at 20:11
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    _"My question is, by decreasing the feedback factor"_ - What is this 'feedback factor' you speak of, and where is it decreased? – Bruce Abbott Oct 22 '22 at 20:17
  • I don't know. This is the start of the section on multi-pole systems. I don't have any more information. – AlfroJang80 Oct 22 '22 at 20:35

1 Answers1

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To understand this, consider the open loop gain plot and the 1/beta plots as separate plots on the same frequency/magnitude plot.

I expect you are familiar with the typical shape of the open loop gain plot and then add to this the horizontal plot of the 1/beta plot.

Now if we consider the 1/beta plot to be approximately equal to a closed loop gain plot bearing in mind that the closed loop gain plot will gradually reduce as frequency increases until it reaches the open loop gain plot where it will be -3 dB down and it will then follow the open loop gain plot downwards.

Now if we reduce beta (to improve stability) we are actually moving the 1/beta plot upwards on the graph and as a result the closed loop gain is also increasing.

This has the effect that the closed loop gain plot will meet the open loop gain plot at a lower frequency.

By reducing beta we have increased the closed loop gain, moving the closed loop pole down in frequency reducing bandwidth in the process.