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Why we check Phase margin at the 0db

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vovan76

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phase margin

Hi
I have strange question
According to Theory I check phase margin at the freq when Gain cross 0db
Why 0db
and What about next Bode plot is the sircuit stable or not and why
 

why 0db

Hi

You have to look at the closed loop transfer function: H(s)=\[\frac{A(s)}{1+A(s)\beta(s)}\]

where A(s) is your forward gain and \[\beta(s)\] is your feedback factor.

The interesting quantity is the open-loop factor \[A(s)\beta(s)\] as this is in the denominator. Making this quantity equal 1(0dB) and negative(phase>180 degrees) makes the transfer function go to infinity ie. unstable.

This is why you always look at the open-loop transfer function and check the phase change at 0dB and hope it is not more than 180 degrees.

In your plot the program tells you it is stable, you can see that when you cross the 0dB the phase change is only 116.3 degrees.

Regards
 
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textbook on gain and phase margins

Thank you for replay
I think that you not completely right
When gain =0db and phase >180 G(S)*H(s) ≠8800; -1 and 1+ G(S)*H(s)≠8800;0
:arrow:not a pole of the close loop Transfer function
right
 

negative feedback concept phase margin

Hi

I can not really read this??
When gain =0db and phase >180 G(S)*H(s) ≠8800; -1 and 1+ G(S)*H(s)≠8800;0
:arrow:not a pole of the close loop Transfer function
 

why phase margin ?

Thank you for replay
I think that you not completely right

When |LT(s)| =0db and phase >180 LT(s) not equal to -1
and 1+ LT(s) not equal to 0
therefore not a pole of the close loop Transfer function
 

change phase margin

Hi

I do not understand what you mean by
When |LT(s)| =0db and phase >180 LT(s) not equal to -1
?
 

phase margin diagram

Hi
At the your replay yo say

"The interesting quantity is the open-loop factor as this is in the denominator. Making this quantity equal 1(0dB) and negative(phase>180 degrees) makes the transfer function go to infinity ie. unstable."

Accor ding to this I understand ,when open-loop factor=1 and phase >180 it make transfer function not stable ,but you don't explain why ?
Why phase >180 make loop unstable exactly at the 0db gain?
 

phase margin when not at 0db

Hi

This is because when A(s)\[\beta\](s)=-1, H(s)=\[\frac{A(s)}{1+A(s)\beta(s)}\]-->infinity

The |A(s)\[\beta\](s)|=1 (0dB) limit is used because it is the boundary between the stable and unstable state, IF the phase shift is equal or more than 180 degrees the circuit is unstable otherwise it is stable. If |A(s)\[\beta\](s)| is larger than 1 and the phase shift is still 180 degrees or more the circuit is unstable.

This can not be explained from the Bode plots, but need to evaluated from the Nyquist diagram.

I think it is much more informative if you take a look at this chapter
**broken link removed**

Hope it helps
 
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phase and amplitude margin

Hi
According to you
"If |A(s)(s)| is larger than 1 and the phase shift is still 180 degrees or more the circuit is unstable"

If you look on attached picture ,there are the area of the Bode plot where Gain is more than 0db and phase more than 180 ,but is still stable according to Matlab
how you can explain it
Thanks
 

how to check phase margin

Hi

Sure you are right, but it seems that the Bode plot method sometimes is not enough to evaluate a system. Especially for the kind you have, where the phase go beyond the 180 degrees and amplitude response is also larger than 1.

Check this post, I believe he is having the same issue as you:


Hope this helps
 

phase margin over 180

tyassin said:
Hi
Sure you are right, but it seems that the Bode plot method sometimes is not enough to evaluate a system. Especially for the kind you have, where the phase go beyond the 180 degrees and amplitude response is also larger than 1.

Yes tyassin, you are right. "Sometimes" the Bode plot is not enough and the complete Nyquist criterion has to be applied.
In fact, the BODE stability check is based on this Nyquist criterion in its simplified version. And this version applies under the condition that the 180deg-line is crossed only once.
In your example this is not the case - and therefore the Bode diagramm gives no indication on stability. For example the Gain margin in the diagram would be negative - indicating instability (if one does not know about the mentioned restrictions).
However, as you have mentioned: The system under discussion is stable - and this can be checked with the help of the Nyquist diagram .
Regards
LvW
 

Thank to all

If Bode stability analysis not always can give answer ,
how I can check stability in real life ,i.e. if i should check stability to real system (BG,LDO ) and i don't know all poles and zeros ( placement of the poles and zeros can change due to corners, temperatures ,load changes ...)
Till now i open loop with iprobe from Candence library and phase margin
but if i have not regular Bode plot(for example phase cross 180 not one time)
what i should do?

I don't think that Nyquist is appropriated for real system
 

vovan76 said:
If Bode stability analysis not always can give answer ,
how I can check stability in real life ,i.e. if i should check stability to real system (BG,LDO ) and i don't know all poles and zeros ( placement of the poles and zeros can change due to corners, temperatures ,load changes ...)
Till now i open loop with iprobe from Candence library and phase margin
but if i have not regular Bode plot(for example phase cross 180 not one time)
what i should do?

I don't think that Nyquist is appropriated for real system

1.) What do you mean with "real system" ? Hardware ?
2.) If you are not informed about all pole and zero location, of course, you cannot perform any stability check (by simulation or calculation) as you don´t know the frequency response of your system.
3.) The concept of phase resp. amplitude margin is applicable only for a "continuous" loop gain frequency response with only one crossing of the 180 deg resp. the 0 dB line.
4.) It is possible to transfer the Nyquist theorem to the BODE plot. In this case the number of "positive" and "negative" crossings is to be evaluated.
5.) The Nyquist stability check is applicable to all systems - even if the loop gain is unstable (but the closed loop is stable). But, of course, it is a "theoretical" tool.
Real hardware needs other methods.
Regards
 

Vovan76, the reason you check stability at 0db gain is that, as was said before, the denominator of the gain becomes 0 and the closed loop gain is infinite. That means, even with no input applied, your system will produce some output (because of noise, disturbances etc.). This is of course valid when you have an inherent inversion in the loop and you get additional -180 deg phase shift for high frequencies, so that the total shift around the loop is 360. And BTW, usually we call Aβ not the open-loop gain but just the loop gain.
Apart from this can you do a quick simulation and tell us the result. Set up a transient simulation and inject a disturbance in your loop. Either a current source injecting and then removing a short pulse of current or a voltage source in series with the loop also injecting a short pulse. Make sure the rise and fall times of the pulse are rather short and the amplitude of the pulse is small enough that the circuit can be considered working in the small-signal regime. Instead of pulse you can also do a sharp step. Keep the input to the circuit unchanged at the value of your input bias (i.e. an AC ground). I think your system will oscillate.
 

sutapanaki said:
......................
And BTW, usually we call Aβ not the open-loop gain but just the loop gain.
.................................
.

Thank you, sutapanaki, for this point. Very often I have seen and read some confusion about this parameter (even in textbooks and papers from professionals).

Let me summarize:
- open-loop gain is the gain of one stage (e.g. opamp) with no feedback (open loop);
for opamps this is called sometimes Ao.

- loop gain is simply the gain of the complete feedback loop (βAo for opamps with feedback)..
 

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