I am trying to determine if a Sallen-Key filter that I designed will be stable. In order to do this, I am trying to simulate the filter in LTSpice and measure its open-loop phase margin.

The filter is designed to have a cutoff frequency of 150kHz, but the operational amplifier that I would like to use, the LMV358, has a GBW of 1MHz. This means that the open loop gain at high frequencies will be very low.

I have set up the simulation below to try and estimate the open-loop gain (V(FB) / V(INB)):

Simulation Schematic

Simulation Results

As you can see the magnitude of the open-loop gain seems about right. But the phase seems wrong to me.

  • Why is the phase increasing?
  • Why is it starting out at -180°?

Analytically, I can estimate the frequency of the Op-Amp dominant pole response and that introduced by the output capacitive load, which in this simulation seems tobe fairly small.

The poles introduced by the filter however are estimated under the assumption of infinite open-loop gain, which is not valid anymore here since the loop gain drops to low values around the supposed filter cutoff frequency. How can I estimate the effect of C2 and C3 in this case?

Also, one final question:

  • Should I use another operational amplifier with a higher GBW?
  • \$\begingroup\$ I don't think the GBW is much of an issue; but many other things could be. What kind of signal amplitude will be fed in to the circuit (and needs to come out of it as you want gain of 1)? \$\endgroup\$
    – Justme
    Commented Jul 19, 2023 at 20:33
  • \$\begingroup\$ @Justme are you thinking about saturation of the opamp? Or slew rate limitation? It should not saturate in its bias point, and the small-signal input should be rather small. Either way, the signal should be a 5V DC value max. The filter is only there to clean it up a bit. \$\endgroup\$ Commented Jul 19, 2023 at 20:43
  • \$\begingroup\$ The output amplitude starts to drop at 50 kHz or so already. At 200 kHz amplitude can be only below 2Vpp. And if it is powered with 5V, the input supports voltages only up to 4V. \$\endgroup\$
    – Justme
    Commented Jul 19, 2023 at 20:51

1 Answer 1


That is a classical representation of a typical open loop gain and phase shift. No need to worry; this is as you would expect with a "twist" at high-frequencies.

Why is the phase increasing?

The phase begins at -180° and by 1Hz has risen to -90&deg. This is because an open-loop op-amp exhibits the behaviour of a very high-gain system fed into a low-frequency low-pass filter (think integrator). Pretty normal and nothing strange happening at all.

Above 10 kHz other things take over and, feed-through of your input signal will be one of these because the op-amp output impedance is no-longer less than a few ohms as seen at low frequencies and you will get your input signal passing through C2 to the output or, in your test circuit, V3 will couple to the no-inverting input.

Why is it starting out at -180°?

Because your circuit is inverting.

Should I use another operational amplifier with a higher GBW?

Use your simulator to measure how close to the ideal response is and, if not close enough then choose a faster op-amp <-- you have all the tools to hand.

One final thing; don't forget to bias your proper input signal to a DC level of 2.5 volts.

  • \$\begingroup\$ Thank you! I did not fully realize what I was doing when I put the small signal on the inverting input. Of course it's inverting. Just one more question, how exactly does the increasing output impedance affect phase? It would seem that a phase lag of 180° is introduced - why? \$\endgroup\$ Commented Jul 19, 2023 at 20:41
  • \$\begingroup\$ I'm not sure I understand your question about phase lag. \$\endgroup\$
    – Andy aka
    Commented Jul 19, 2023 at 23:32
  • \$\begingroup\$ what I don't understand is why the phase increases at the end of the plot above. This results in a phase margin of -47° and I would like to try and understand the cause of that in order to correct it. \$\endgroup\$ Commented Jul 20, 2023 at 5:15
  • 1
    \$\begingroup\$ @RiccardoFagiolo An op-amp will naturally degrade the phase as frequency gets higher but, in the case of a sallen key filter it can get worse more rapidly due to the input (the real input) coupling into the less effective output node due to the output node's impedance becoming less-effective due to loss of gain. So, two effects in play: normal op-amp behaviour of a second pole in the response at higher frequencies and the signal input coupling to the output node via the feedback capacitor. \$\endgroup\$
    – Andy aka
    Commented Jul 20, 2023 at 8:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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