Take the 2-minute tour ×
Electrical Engineering Stack Exchange is a question and answer site for electronics and electrical engineering professionals, students, and enthusiasts. It's 100% free, no registration required.

I was reading datasheet of MCP6072. I saw a parameter namely "phase margin" (table 1-2, page 4). As far as I know, "phase margin" is a control engineering term, and implies the phase difference between input and output when the gain is unity. I don't understand the meaning of this term in opamp terminology. The typical phase margin of this opamp is given to be 57o. That what does it mean?

share|improve this question

3 Answers 3

The phase margin PM is a measure for the stability of a system with feedback. And, thus, it also applies to operational amplifiers. The PM is defined for the LOOP GAIN of the system - that means: open the loop at a suitable node and measure/simulate the gain and the phase around the complete loop. Then ,the PM is the DIFFERENCE between the measured phase and -360 deg (that means: The "distance" to the oscillation condition, positive feedback) at the frequency which gives unity loop gain. Without taking the phase inversion at the inverting input into account, the PM is the "distance" to -180 deg.

Now, for an opamp the most critical situation arises for 100% feedback (unity gain operation). In this case, the feedback factor is unity and the Loop gain is identical to the open-loop gain Ao of the opamp. Normally, only this condition is used to specify the PM in the opamp´s data sheet.

Summary: The PM as given for an opamp is the DIFFERENCE between the opamp´s phase shift and -180 deg at the unity-gain frequency.

share|improve this answer

An op-amp is a "control system" and the phase margin is defined as the difference from -180 degrees of the phase of the open-loop transfer function when the magnitude is unity. This allows you to predict the stability and response of the system when you close the loop with a given amount of feedback.

An op-amp is typically compensated with a dominant pole, so the transfer function is approximately Ao/(1+s/wo) where wo is the dominant pole frequency and Ao is the DC gain- This implies a phase margin of 90 degrees. In practice, the Ft of the transistors causes additional phase shift and the dominant pole is set so that the 0dB crossover occurs with reasonable phase margin.

share|improve this answer
I disagree, the phase margin is calculated with negative feedback and the gain of the feedback network is unity. –  Vladimir Cravero Apr 20 at 8:16
I also disagree. The phase margin is the DIFFERENCE between the opamps open-loop phase and -180deg.at the unity-gain frequency. –  LvW Apr 20 at 8:26
@LvW Yep, good catch, I meant to say the difference between the phase and -180, which is really -360 degrees. Will edit to fix. –  John D Apr 20 at 16:19
@VladimirCravero LvW has it right, it's not the feedback network, it's the entire open loop transfer function. On a datasheet they can't know what you're going to wrap around an op-amp, so they specify phase margin of the open loop transfer function of the amplifier itself. Make a unity gain buffer out of it and you will know the phase margin. Pure resistive feedback to give more gain will be more stable than that, i.e. the datasheet number is the worst case in "normal" operation. (Though there are plenty of situations like capacitive loads that can cause trouble.) –  John D Apr 20 at 16:34
The OL transfer function of an opamp characterized by a dominant pole has a phase margin that's practically 90°. You've got your pole at a few Hz, cross the 0db at some 100kHz so the phase is 90° for all practical purposes. If you close the loop instead, and \$\beta=1\$, then the pole shifts very near the unity gain frequency and the phase margin decreases. –  Vladimir Cravero Apr 20 at 16:43

Phase Margin is the amount of phase shift margin at unity gain which could cause instability or Oscillation.

90 deg is theoretical ideal, 0 is NG , 45 deg will have some overshoot, 60 deg is practical solution. Phase margin shows tradeoff between rise time and overshoot.

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.