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?
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.
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.
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.
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