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Everything I read on the web, concerning this question, says that in order to measure or set the phase margin of an amplifier the loop must be opened and the phase lag around the loop measured when the gain around the loop equals -1 (unity loop gain).

This seems to be almost universal advice with its resulting practical problems such as DC offset across the break and creating the correct load for the output of the opened loop.

My question - Can the phase margin be measured in a closed loop setup, why is an open loop always recommended for this?

Classical amplifier theory says that for an amplifier to be stable the loop phase must be less than 180 degrees lag by the time the loop gain gets down to 1 (unity). To give sufficient phase margin the value of the compensation capacitor is often set to give a loop phase of, say, 135 degrees at unity loop gain (phase margin = 45 degrees).

Consider the following amplifier:-

schematic

simulate this circuit – Schematic created using CircuitLab

Loop Gain = BAol

where:-

Aol = Vout/Vdiff

and

B = Vfeedback/Vout

Therefore loop gain = Vfeedback/Vdiff (using pk to pk values).

Loop phase is the phase of Vfeedback to Vdiff.

It appears that I can measure both loop gain and loop phase in a closed loop configuration.

Why can't I just increase the input signal frequency until Vdiff = Vfeedback (unity loop gain) and then measure the loop phase with a 2 channel oscilloscope (phase of Vfeedback to Vdiff) to determine or set the phase margin? (How short of 180 degrees the loop phase is).

Is there anything with this apparently very practical closed loop method of determining (or setting) phase margin. Why is such a much more practically difficult open loop technique so commonly recommended?

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  • \$\begingroup\$ Measure the loop frequency response, \$\frac{output}{error}\$, with the loop closed. \$\endgroup\$
    – Chu
    May 25, 2019 at 18:08

4 Answers 4

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Vd isn't possible to be measured easily, because it's not against GND. Differential mode oscilloscope is a must. Of course you in theory can calculate Vd=V1-Vf, if you store simultaneously measured V1 and Vf. Waveform math shouldn't be a big deal if you have that capable oscilloscope.

But that math has a problem: The accuracy. Vf and V1 can both be acceptably accurate alone, but their calculated difference can have so high error percentage that it's useless. Think, if V1 and Vf both have 1% error, but to different directions and Vd is only about 2% of V1.

Another problem: An oscilloscope in the -input of the opamp is a capacitive load. One must be well aware how to be sure that it doesn't cause substantial phase shift.

Conclusion: Not theoretically impossible, but needs some work and good enough equipment. I guess a creative enough person can find a way around the accuracy limitations of ordinary tools. He for example swaps channels and compensates errors with more math and introduces a calibration method.

Why this isn't recommended? I guess measuring A and B separately is easier. At least at high frequencies where the absolute value of A isn't especially high. At DC it can be several decades higher.

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Maybe. What you have outlined is a Return Ratio and not Loop Gain. Usually close and mistakenly used interchangeably along with other approximations such as the 'open loop' method and variations using replica loading. Much of the confusion lies in the lack of proper definition of terms. Aol and Beta are terms with origins in ideal block modelling carried over onto the real world where they are not quite complete. Aol becomes a function of loading and of the feedforward element, here R2, as well as its internal impedance, ro. Beta is also dependant on loading as well as input impedance of the opamp... Another complication arises if the opamp is not strictly unilateral and an internal feedback path is active.

If you were to set V1=0 and in a simulation, drive Vfeedback (Vfbk) with a unity ac swept source and capture Vout, form Hfwd=Vout/Vfbk|v1=0, then drive the system backward with the source at vout and capture Vfbk, forming Hbkwd=Vfbk/Vout|v1=0, then LG==Hfwd*Hbkwd. See Ochoa et. al, 2019 Mid West Symposium on Ckts and Systems. This works in simulation since ac analysis in SPICE is linearized and the 'virtual short' at the amp input is not an issue.

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For a second-order system there is a known relationship between (a) the phase margin PM and (b) closed-loop gain peaking (GP) and overshoot (OS) in the time domain (step response). These relationships can also be used as a good approximation for higher-order systems as long as there is a dominating pole pair which allows to treat the system as if it would be of 2nd-order.

Examples:

GP=0.3 dB (2.4 dB) for PM=60 deg (45 deg).

OS=8.8% (23%) for PM=60 deg (45 deg).

Comment: The above applies to the case when real measurements are wanted. When you want to simulate the PM under closed-loop conditions, there are other methods. One novel method is to introduce an additional phase shift into the loop and to evaluate the closed-loop behaviour. This method does not disturb the DC operating point and does not introduce any loading errors.

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First approximation ... EE&O The inverting stage does it simply.
One can see also the "double" pole.

enter image description here

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  • \$\begingroup\$ Vo/eps is the open loop response. To get the loop response (to determine phase margin) it would be necessary to multiply that open loop response by beta which is equal to R2/(R1+R2). \$\endgroup\$
    – user173271
    Aug 22, 2022 at 6:35
  • \$\begingroup\$ Right. Just drawing this "line" into the graph? \$\endgroup\$
    – Antonio51
    Aug 22, 2022 at 6:45
  • \$\begingroup\$ Sorry. Confusing "open loop" and "loop" gain. Phase not affected. \$\endgroup\$
    – Antonio51
    Aug 22, 2022 at 7:52
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    \$\begingroup\$ That's quite all right! Since asking the question in May 2019, I have come up with a practical technique for measuring phase margin with the loop closed for a non-inverting amp but it requires having a facility on your oscilloscope to display CH1 - CH2. So, I attach CH1 to the non-inv input and CH2 to the inv input (both probe grounds at 0V). Then I increase the input frequency until CH2 level is equal to CH1 - CH2 (loop gain = 1). I can then look at the phase lag between CH1 - CH2 and CH2 (loop phase) and see how far short of -180 degrees it is. That is the phase margin. \$\endgroup\$
    – user173271
    Aug 22, 2022 at 9:45
  • \$\begingroup\$ Note that if the "error" voltage is too low, I wonder if it is not possible to use an attenuator just before the -input of the op-amp (usable if the open-loop gain of the op-amp is really too large). \$\endgroup\$
    – Antonio51
    Aug 22, 2022 at 11:53

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