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[SOLVED] non-dominant pole in second order systems

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fateme m

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Why is it that if the non-dominant pole is 3 times larger than the gain crossover frequency, the phase margin will be about 70 degrees?
is there a book that this is discussed?
 

Where have you seen that ?
You can find an equivalent transfer function with only the dominants poles. Please show the example you are talking about.
 

Thanks for replying,
I am supposed to prove through mathematical relations that if the open loop transfer function of an op-amp has 2 poles in LHP, and one is 3 times larger than the gain bandwidth product, then PM=70 degrees.
 

The exact value of PM is ≈71.6° for this case. This is discussed in Willy Sansen "Analog Design Essential".
You could also make a simple calculations for it - or simply plot arcus tangens function rescaled to angle degrees.
 
Very easy to visualize in a simulation first and then figure out how to calculate second.

You'll see that a single pole filter (RC) has a phase shift of ~18.4 degrees at 1/3rd its cutoff.

So assuming the dominant pole is contributing 90 degrees, this additional pole adds another 18.4 for a total of 108.4 degrees. 180-108.4 = ~70 degrees margin.
 

Dominik Przyborowski said:
The exact value of PM is ≈71.6° for this case. This is discussed in Willy Sansen "Analog Design Essential".
You could also make a simple calculations for it - or simply plot arcus tangens function rescaled to angle degrees.
Click to expand...

Thanks for replying,
I found it thanks
 
Last edited:

Yes, there is in a page 161.
Plot a function of phase for two pole system:
\[\phi(\nu) = 180^o - 180^o \left [ arctg(\frac{\nu}{|p_d|}) + arctg(\frac{\nu}{p_{|nd|}}) \right ] / \pi \],
and look for the frequency of p_nd/3.
 

I have solved your problem by thinking the other way around. I will update the whole process shortly.
Until then, some conclusions I have found:
  • pole 2 = 3*Gain*BW makes phase margin ≈ 72º
  • By making pole 2 = 2.6 *Gain*BW you will have phase margin ≈ 70º

Here is an experimental proof.



The whole process in order to get to the value 2.6 times the G*BW will be uploaded shortly.
 
Thanks for the time you spent on this! Actually in the book it isn't calculated enough, more the results are shown.
I'd be really thankful if you updated sooner
 

Here you have it. If you have any question about the calculation process, feel free to ask.
 

Attachments

  • PhaseMargin problem.pdf
    482.4 KB · Views: 113
Last edited:
I can't thank you enough! You have indeed helped me ! I understood fully:thumbsup:
 

CataM said:
I have solved your problem by thinking the other way around. I will update the whole process shortly.
Until then, some conclusions I have found:
  • pole 2 = 3*Gain*BW makes phase margin ≈ 72º
  • By making pole 2 = 2.6 *Gain*BW you will have phase margin ≈ 70º

Here is an experimental proof.



The whole process in order to get to the value 2.6 times the G*BW will be uploaded shortly.
Click to expand...

May I ask how you added those other lines like"closed loop stable? " or "PM=70.1"?? because mine doesn't show such information.
Thank you in advance.

- - - Updated - - -

ohoh i should right click and choose the charactristics tht I want to be shown
thanks
 

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