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I am not able to understand what significance phase margin would hold in case of open loop. Shouldn't it be just phase? If it was loop gain I'd understand specifying phase margin to denote stability. Why then, do the datasheets of some op amps like the TLV365 specify it as below:

enter image description here

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    \$\begingroup\$ Please explain what makes sense to you i.e. which type of circumstances make more sense to specify phase margin if not opem-loop. \$\endgroup\$
    – Andy aka
    Commented Jun 17 at 14:33
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    \$\begingroup\$ To add to the previous comment, if you had a closed loop gain of some amount less than the open loop gain, how would you show that on the plot? \$\endgroup\$ Commented Jun 17 at 14:35
  • \$\begingroup\$ @Andyaka edited my post to elaborate \$\endgroup\$ Commented Jun 17 at 14:42
  • \$\begingroup\$ I usually determine the Aol poles from the phase margin. For this op-amp it is about 150Hz. \$\endgroup\$
    – slimcolt
    Commented Jun 17 at 14:57
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    \$\begingroup\$ @needbrainscratched - Hi, Where did the image come from? To comply with the site rule on referencing, details of the original source of copied / adapted material must be provided by you, next to each item. If the original source is online (webpage, PDF, video etc.) please edit the question & add its name & link (URL) (e.g. website name + PDF title + its URL). If the original source was a book or other offline material, please edit the question & add the best reference you can e.g. title, author(s), page, edition etc. TY \$\endgroup\$
    – SamGibson
    Commented Jun 17 at 15:57

3 Answers 3

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Yes - it is a mistake. The wording just below the diagram ("Gain and phase vs. frequency") is correct. The phase margin (PM) is just one point of the phase response (it is app PM=60deg).

However, for your understanding it is important to realize that the phase margin is found in the Bode diagram for the loop gain only - and only in case of 100% feedback the loop gain is identical with the opamps open-loop gain Aol (this is the most critical case).

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  • \$\begingroup\$ another thing I don't get is, why does it start from 90? Shouldn't it start from 0? \$\endgroup\$ Commented Jun 18 at 14:13
  • \$\begingroup\$ No - the phase response starts at 180deg (most right axis). The most left axis is for the magnitude only. \$\endgroup\$
    – LvW
    Commented Jun 18 at 14:17
  • \$\begingroup\$ oh right...my bad. But still 180 degree is also strange. Does that mean it offers complete inversion in open loop? \$\endgroup\$ Commented Jun 18 at 14:31
  • \$\begingroup\$ Yes - I understand your question. But there is a good answer: The phase response is most important for finding the stability margin of the opamp when it has negative feedback. In this case, it is the phase response of the loop gain (Aol of the opamp multplied with the feedback factor) which matters. And for negative feedback the loop gain starts at small frequencies with app. -180deg. And the "critical point" for stability analyses is the frequency where the phase crosses the 0 deg line. \$\endgroup\$
    – LvW
    Commented Jun 18 at 15:02
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Shouldn't it be just phase?

I agree.

It makes no sense to draw a graph like this and call it phase margin because, phase margin usually only occurs in one place i.e. where the open-loop gain is unity.

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The only difference I see is that they've subtracted 180° from the open-loop phase shift to get 'phase margin' on some op-amp datasheets.

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