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Closed loop amplifer stability question

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ljy4468

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Hi all
I've a question about closed loop amp' stability
In case of Open loop Amp, Phase margin is determined at frequency at gain=0db
And I know why.

But in closed loop system,
The Stability of the system is defined at the frequency where the open loop gain of the amplifier intercepts the closed loop gain response
I can't understand why....
why not unity gain frequency but intercept point frequency??

please tell me why
 

Phase margin is determined in open loop response at unity gain frequancy.
If your PM is good than closed loop system is stable.
I cant open pdf.
 

I cannot see the attachement.

I think it is because when the open loop gain is high enough, the close loop gain will be 0dB. And when the two loops are crossing. that means the open loop gain is lower than 0dB. So, at the intersection, we can check the phase margin.

can you post the attachement again?
 

attachment:(page 5, an underlined part)
https://ww1.microchip.com/downloads...mplifier graph gain bandwidth crossing point"


But, If AB is high enough(A=openloop gain, B=feedback factor), closedloop gain is about 1/B
For example, A=100dB, B=0.5,
Then closed loop gain is about (20log(2)]=6dB.

There are difference between unity gain frequency and 6dB gain frequency(crossing point).
So, when determine closed loop amplifer' stability, WHAT frequency could I use??(0dB gain or 6dB gain)
And why??
[In the attatchment, use 6dB gain]

Thanks in advance
 

For stability it is important point where loop gain |βA|=1 (0dB of loop gain βA). Phase distance from this point to 180 degrees is phase margin.
You can also analyze stability if you plot independantly A and 1/β characteristics.
This is possible when you can clearly see difference between A and beta circuit. Some advantage is that you can see 1/β characteristics which is ideal closed loop characteristics for infinite A.
A and 1/β characteristics are cut at |βA|=1, which is same point where closed loop is 0dBm.
 

hello ,
this is just my thoughts nothing for sure,
CLosed loop=A/(1+beta*A)
loop gain=1+beta*A
what u are talking about is (A in dB)-(Closed loop in dB) "note subtraction in dB is division in ratio"
which is the same as loop gain 1+beta *A
i.e. both are the same consition
 

I think this figure will help you.
22_1159895661.JPG
 

94_1159897329.jpg


Thanks all your kindly answers
But I can't understand clearly, sorry

In above picture, I modeled b=1/2(green line),b=1/100(blue line)
open loop(red line) with matlab.

Then, If I made closed loop amp with b=1/100(blue line),
1)crossing point is A (@phasemargin ~ 125)
2)unity gain point is B (@phase margin ~ 90) as you see.

THen, which is PHASE margin of This closed loop amp(b=1/100)?
125 degree or 90 degree??
In many articles, 125 degree is right. Am I right??
And I want to know WHY.

Thanks in advance
 

I'll try to explain.

Your open loop gain is A, your loop gain is βA and your close loop gain is about 1/β - if βA>>1.

Now we have,

(open loop gain)÷(close loop gain)=(loop gain), that's A÷1/β=βA.

If we express above in dB:

20log(A÷1/β)=A(dB)-1/β(dB)=βA(dB) = loop gain(dB)

Now we know the loop gain is the difference between open loop gain and close loop gain on Bode plot. Then "the frequency where the open loop gain of the amplifier intercepts the closed loop gain" is where your loop gain drops to 0dB.


Let's talk about phase margin. It is defined as 180-(βA phase shift) when |βA|=1(0dB). So if we assume β is a real number, then your βA phase shift on Bode plot is same as your A(open loop gain) phase shift on Bode plot. So at the interception point we just derived above, read out your open loop phase shift, use 180 minus the phase shift, you get your answer!!!

In your case, if β=1/100, phase margin should be 90 degree.
 

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