I have found a definition of phase margin of amplifier system from Texas Instruments application report. This definition looks like this: $$\phi = tan^{-1} (A \beta)$$ where \$ A\$ is amplifiers open-loop gain (aka direct gain) and \$ \beta \$ is feedback return signal ratio - or \$ A \beta \$ known as loop gain. Now, \$ A\beta \$ would typically be a value ranging \$ 1000000 \$ to \$ 10000 \$ (in opamp amplifier systems, where open-loop gain is usually around \$ 120 dB \$).

Such values of \$ A\beta \$ inserted into upper definition of phase margin always equals (approximately) \$ \phi = 90° \$. So, using that equation for definition of phase margin must be definitely wrong, because it is not possible, for amplifier's phase margin to be \$ 90° \$ in all scenarios possible. Unless we would be discussing an example with \$ A\beta < 100 \$, which is very unlikely to happen.

Also, it would seem more logical if phase margin definition equation would be described as a function, dependent on poles of amplifier or \$s\$, damping factor or \$\zeta\$, frequency or \$\omega\$, etc.

I know how to find phase margin (and gain margin) from already drawn Bode plot, but I cannot solve it, using mathematical ways, not graphical.

Can anyone tell me, if this is the actual formula for calculation of phase margin? Or are there more data needed to solve such case? Would "fully defined" transfer function provide enough data for proper calculation of phase margin?

  • 1
    \$\begingroup\$ I do not know which Texas application report you are referring to. However, the "definition" as given by you cannot be true. You say that you are able to find the phase margin of a system with feedback. But in this case, it should be clear to you that the mentioned definition is not correct. Are you aware that the lopp gain will be unity at a certain frequency? \$\endgroup\$
    – LvW
    Commented Aug 27, 2018 at 14:32
  • \$\begingroup\$ For a second-order system, there is a fixed relation between phase margin and damping factor. \$\endgroup\$
    – LvW
    Commented Aug 27, 2018 at 14:34
  • \$\begingroup\$ @LvW Here: ti.com/lit/an/sloa021a/sloa021a.pdf Pages 6-7 are explaining about stability of current feedback amplifier, including definition of phase margin. \$\endgroup\$
    – lucenzo97
    Commented Aug 27, 2018 at 15:02
  • \$\begingroup\$ Yes - of course. But did you realize that the definition of the phase margin is not in accordance with the first two lines of your question? I am afraid you misunderstood something. \$\endgroup\$
    – LvW
    Commented Aug 27, 2018 at 15:12
  • \$\begingroup\$ Try to watch this youtube.com/watch?v=kC8FYL8gr3E and Lec44 and 45 \$\endgroup\$
    – G36
    Commented Aug 29, 2018 at 17:44

1 Answer 1


Now, Aβ would typically be a value ranging 1000000 to 10000 (in opamp amplifier systems, where open-loop gain is usually around 120dB).

That's at DC where nobody really worries about phase margin because it's never going to be an issue. Look at a typical open-loop response of an op-amp: -

enter image description here

Picture source and other relevant information that could be useful.

At 1 kHz the open loop response might have dropped to 60 dB (G = 1000). At 1 MHz, the o/l gain is only 10 and this is an area where quite a few op-amp circuits have problems. Arctan of 10 (assuming a unity gain situation) is 84 degrees and consistent with the graph above. At 10 MHz the gain is unity and arctan of 1 is 45 degrees i.e. not a million miles off.

If you know the T.F. of the forward gain device and you know the feedback T.F. then certainly you can calculate phase margin by considering the loop broken with a signal being injected at the input and the output being taken from where the break is. But you have to respect impedances and loading when you open the loop and sometimes it can be difficult to realize.

  • \$\begingroup\$ How to mathematically calculate phase margin of an amplifier? I have never seen a direct formula to calculate it somehow... \$\endgroup\$
    – lucenzo97
    Commented Aug 28, 2018 at 18:30
  • \$\begingroup\$ @keno it is rather dependent on the specific op-amp. Looking at the graph in my answer, the phase rapidly approaches a constant 90 degrees at a low frequency and stays like this until at some seemingly arbitrary high frequency a 2nd pole emerges and starts advancing the phase towards -180. The first pole (at LF) is well-documented to be found by using the GBW product specified in most op-amp data sheets but the HF pole is not very well covered by most data sheets. \$\endgroup\$
    – Andy aka
    Commented Aug 29, 2018 at 7:13

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