I designed a multiple feedback bandpass filter using the Analog filter wizard. The design is purely education therefore I used ideal op-amp. Fc = 50KHz; DC gain = 6dB; Q = 5;

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I followed the instructions included in the AD tutorial “Stability of Op-Amp Circuits” and simulated the circuit in the same fashion. LTspice: Stability of Op Amp Circuits | Analog Devices

My recollection from control systems is that the negative feedback circuit is stable as long as we do not invert the signal by 180° with a magnitude equal to or greater than 0dB to prevent oscillation. If correct my 0dB point is at 10Mhz and I have 93° of phase margin. Which is sufficient. Additional circuit never reaches the -180° phase shift. However, at 90KHz the phase shift equals 164° with 22.7 dB of gain. I would imagine that it is a very low phase margin of only 26° with 22.7dB of gain. I vaguely remember the “rule of thumb” for phase margin as 45° and gain margin as 20dB.

If the above statements are correct are they applicable for filters?

If yes how can I improve the phase margin? Would the introduction of a delay in order to shift the phase response be a good idea?

  • \$\begingroup\$ The gain is not zero at that point. As for the improvement in the phase response (in case it's needed), why not follow the advice in the video you linked? If you're worried about a "wobbly" initial transient, that's the inherent impulse response. \$\endgroup\$ Commented Apr 4, 2022 at 13:56
  • \$\begingroup\$ @aconcernedcitizen Thank you for your reply. "The gain is not zero at that point." are you referring to the point where the phase shift is equal to 164°? Isn't that worse as we get nearly 180° of phase shift with loads of gain? Regarding the video, I was not sure if this solution is specific to the problem presented which is driving a large capacitance at the output of the op-amp. I will give it a go and experiment. \$\endgroup\$
    – Wintermute
    Commented Apr 4, 2022 at 14:07
  • 1
    \$\begingroup\$ Since you already have enough phase margin, there's nothing to correct for. That the phase goes as low as it does, that's a direct consequence of the response of the transfer function -- it's a sharp bandpass. Making the phase smoother will result in a wider transition width for the filter. Think of it this way: if that would have been an inverse filter (inverse Pascal, inverse Chebyshev, Cauer/elliptic), you would have had zeroes, and the phase would have jumped 180 degrees. Would you have considered that to be unstable? \$\endgroup\$ Commented Apr 4, 2022 at 14:15
  • \$\begingroup\$ As to your last sentence - the phase margin of app 90 deg (see my detailed answer) is caused by the opamp only (opamp phase shift at very large frequencies) and not by the frequency-dependent filter circuitry. Shifting the phase response in negative direction would be the worst solution. \$\endgroup\$
    – LvW
    Commented Apr 4, 2022 at 15:35

1 Answer 1


The simulation results as shown in your problem description look weird (rising phase characteristics). I suppose, it is your intention to produce a Bode diagram for the loop gain, correct? Your ac source (input signal at the inv. node) has an inverting polarity (causing the rising phase) - nevertheless, you can perform a stability check.

However - your interpretation of the results is wrong. The stability check requires to verify the phase response relativ to 360 deg (0 deg) and NOT to 180 deg. Note that the correct definition for the loop gain contains the signal inversion at the "-" opamp input. And the loop gain as simulated by you does, of course, include this signal inversion.

Remember: The oscillation condition (stability limit) is defined for zero phase shift of the loop gain function.

(In some documentation - in particular in control theory papers - the loop gain is used without the phase inversion at the summing junction, which gives rise to some confusion, as we can see here)

Therefore - the magnitude crosses the 0dB axis at app. 10 MHz with a phase shift of app. 90deg. Hence, the phase margin is 90 deg.

  • \$\begingroup\$ Thank you for your reply. Yes, this is correct I intended to produce a Bode plot of open-gain to use this circuit with different real-world op-amps. I have changed the orientation of V1 and the resulting Bode plot has not changed. \$\endgroup\$
    – Wintermute
    Commented Apr 4, 2022 at 14:51

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