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From "Learning the Art of Electronics" by T. Hayes and P. Horowitz, Cambridge University Press (p. 360):enter image description here

As per above figure, the LF411 op-amp imposes a -90 degree phase shift on output voltage. It is stated that to prevent parasitic oscillations, we must avoid including an RC circuit within the (negative) feedback loop, which induces an additional -90 degree phase shift at large frequencies, bringing the total phase shift to -180 degrees at some large frequency.

However, for a purely sinusoidal signal if the voltage fed back to the inverting input has any phase shift (except 0 degrees), it will violate the op-amp rule that for negative feedback, difference in voltages at the inverting and non-inverting inputs should be zero.

Why the emphasis on avoiding precisely -180 degree phase shift, vis-a-vis parasitic oscillations ?

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2 Answers 2

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180° of phase shift is true positive feedback and you get an oscillator. A bit less than 180° means that you see a lot of oscillatory ringing when you try and amplify step edges (such as a square wave) and, if you plotted the closed-loop frequency response you would see a big undesirable peak in the spectrum.

Around 120° of phase shift is about the point where most amplifiers would be acceptable in a lot of circuits but, it's still not ideal. 90° or less is pretty good and acceptable.

Op-amps are far from ideal. Here's a simulation of an op-amp that has a 1 MHz unity gain bandwidth and an open loop gain of 1,000,000. X1 is the diff input followed by an open loop gain of 1 million followed by an RC network that limits the BW to 1 MHz. A unity gain buffer follows it: -

enter image description here

Switch SW1 is open circuit hence, there is no feedback. When SW1 closes (later on), it becomes a 10 MΩ resistor. Notice that the phase shift is getting close to -90° by 100 Hz but, if I close the loop to make ×11 amplifier we see this significantly different result: -

enter image description here

Do you see any problems with this closed-loop amplifier given that the resulting output phase shift remains reasonable close to 0° well past 10 kHz? Here's what the input and output look like when driving a 100 Hz 1 volt p-p square wave: -

enter image description here

Do you see any ringing or overshoot on the output? Let's try a 100 kHz square wave: -

enter image description here

No overshoot problems here but, what if I added an extra pole (R3, C2): -

enter image description here enter image description here

Above 100 kHz, the phase shift is marching towards -180° yet, the overshoot is only somewhat undesirable and, probably liveable-with in many applications. And, as a bonus waveform, here's the closed-loop spectral response for the above: -

enter image description here

And clearly, if you see a peak in the response then, there is bound to be some overshoot but, of course, the open loop response gave us that information.

So, in summary, op-amps with significant open-loop phase shift show very little phase change (input to output) when the loop is closed and, even if we really push things by adding an extra pole around 200 kHz, the overshoot is liveable-with for many applications.

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  • \$\begingroup\$ Thank you. If one has <90 degree phase shift, wouldnt it still violate the opamp rule ? In which case, we should get parasitic oscillations even if the phase shift is small. Or do we necessarily need "true" positive feedback for parasitic oscillations ? \$\endgroup\$
    – Frost
    Commented May 7, 2023 at 11:28
  • \$\begingroup\$ As I explained in my answer, oscillatory ringing (is that what you call parasitic oscillations?) is not a problem up to around 120 degrees of phase shift but, gets gradually worse as you move towards 180 degrees whereupon the circuit will oscillate. All op-amps will naturally produce a phase shift of about 90 deg at under a few hundred Hz so, liken this to an op-amp integrator and, ask yourself the question if an integrator will produce undesirable overshoot (the answer is no). \$\endgroup\$
    – Andy aka
    Commented May 7, 2023 at 11:39
  • \$\begingroup\$ Do you mean \$\delta V_{in}=0\$ as in \$\delta V_{in}=0\$? If so, then real op-amps violate this all the time by having non infinite open-loop gain and offset voltages and input bias currents? \$\endgroup\$
    – Andy aka
    Commented May 7, 2023 at 11:43
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    \$\begingroup\$ True positive feedback causes oscillation. I've add pictures to help you understand. \$\endgroup\$
    – Andy aka
    Commented May 7, 2023 at 11:53
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    \$\begingroup\$ Thank you @Andy aka for the detailed answer, which is accepted. I think the point is well made, assuming that opamp input voltage difference is zero is fraught with issues. \$\endgroup\$
    – Frost
    Commented May 10, 2023 at 17:39
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An ideal op-amp with infinite gain has infinitesimal input voltage. That does not mean zero input voltage and that’s a crucial distinction. Infinitesimal periodic signals have finite phase (!). It’s the amplitude that’s infinitesimal, and that’s neither zero nor finite by the way.

So there’s no problem in analysis of ideal circuits as long as such analysis uses appropriate mathematical tools. The “op amp input voltage is zero ideally” in closed feedback is a simplification that has too many misleading implications. Your question is a case in point.

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  • \$\begingroup\$ A befitting answer. Strange things might happen "at large frequencies"... \$\endgroup\$
    – V.V.T
    Commented May 8, 2023 at 3:31

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