I'm trying to get a better understanding of frequency compensation for op-amps. I'll start off with a very specific questions and then get into the context. For a certain op-amp, open loop gain is 90dB and GBWP is 10MHz, so the first pole in open-loop mode is at about 300Hz. With a nominal closed-loop gain of 1, the closed loop gain won't start to roll off until 10MHz. Does this mean that the first pole of the system has been moved to 10MHz, or is it still down at 300Hz? The important consideration is the phase - is there a pole in the amplifier that will cause 90 degree phase shift by shortly after 300Hz, or will the phase be about 0 out into the MHz?

This plays into dominant-pole compensation. If the pole is at 10MHz, I could have a low-pass filter at about 100Hz to cause around 100dB of attenuation, or less than 1 loop gain, by the time the 10MHz pole brings the total phase shift to 180 degrees. But if the first pole of the system is still at 300Hz, the phase shift from that plus the new "dominant" pole will cause 180 degrees phase shift when the loop gain is still far too high and cause instability.

I assume the pole does move, or else dominant-pole compensation would be pretty useless for anything outside of DC. But I was talking to someone who insists that the pole is still down at 300Hz causing significant phase shift before we even get into the kHz.

  • 1
    \$\begingroup\$ Yes, the pole has been moved. e.g. OLTF: \$\frac{K}{1+s}\$, with a pole at \$\small \omega =1\$, gives a unity feedback CLTF: \$\frac{K}{(1+K)+s}\$, with the pole moved to \$\small \omega = 1+K\$, but with the gain reduced to \$\frac{K}{1+K}\$. Note, this is the unity feedback case; there is, generally, a gain/bandwidth trade-off. \$\endgroup\$
    – Chu
    May 16, 2019 at 6:46

4 Answers 4


But I was talking to someone who insists that the pole is still down at 300Hz causing significant phase shift before we even get into the kHz.

Of course physically the circuit still has a pole at 300 Hz. However, through the use of feedback we can make the pole of the circuit (as we see it from the outside) move to a different frequency at the cost of gain.

About the phase shift, what this person forgets is that phase and frequency are related. A 90 degrees phase shift at 1 MHz is 250 ns, while at 300 Hz that 250 ns is only 0.027 degrees.

Also the phase shift is compensated for by the feedback loop. So: ...significant phase shift before we even get into the kHz is true but that is irrelevant as the loop is compensating for that phase shift.

If the phase shift from the 300 Hz pole was an issue then it would be impossible to apply the feedback and still get a stable amplifier.

You might want to ask this person who insists about this phase shift to explain why amplifiers with feedback can still be stable. If there is so much phase shift at a few kHz then how can the system be stable? (it cannot, that person doesn't understand how feedback works)

  • \$\begingroup\$ What's the mechanism that causes the feedback network to be unable to compensate for high group delay at 10+ MHz? \$\endgroup\$
    – ttshaw1
    May 16, 2019 at 15:15
  • \$\begingroup\$ At higher frequencies the loop gain isn't high enough to compensate. I suggest that you study linear systems with feedback. Many books have been written on the subject so get studying. \$\endgroup\$ May 16, 2019 at 15:51

In the feedback equation for the opamp. with the possibility of huge phaseshifts (a few opamps have 200+ dB gain at DC, with many poles of rolloff right above DC), the system is still stable because

large_G_at_1000 degrees / [ 1 + large_G_at_1000 degrees ]

has a denominator far away from Nyquist point, for low frequencies.


Many of your questions can be answered with use of a circuit simulator, such as LTSpice.

You would set up a basic AC simulation as shown in the figure below: enter image description here

Now, take a look at the frequency response of the op-amp operating closed loop as a unity-gain non-inverting amplifier:

enter image description here

Due to the negative feedback being applied to the op-amp, the internal 300 Hz pole appears to have been relocated to 10 MHz (-3 dB frequency).

In reality, the 300 Hz pole is still there, and you can see this if you look at the differential potential being developed across the op-amp's input terminals:

enter image description here

After the 300 Hz pole, the open-loop gain of the op-amp is declining at 20 dB/dec. As result, with diminishing loop-gain, the opamp-op has limited in its ability to drive the output in response to an input signal. For example, at 1 MHz the differential input voltage is 1/10 of the output voltage, this is due to the absence of high open-loop gain (only 20 dB at 1 MHz).


The open loop frequency response is fixed and does not change. The dominant pole is at a fixed low frequency say 10Hz where the open loop phase shift is 45 degrees. The open loop phase shift increases to 90 degrees by about 10 times the dominant pole frequency (90 degrees open loop phase shift by about 100Hz) and remains at 90 degrees until the second open loop pole rears its ugly head at say 900kHz adding another 90 degrees of open loop lag (45 degrees extra at the second pole giving 135 degrees in total at the second pole). Therefore open loop phase lag is 90 degrees over most of the op amp's bandwidth.

So that is the open loop response which is the relationship between the output and the difference signal between the op amp's actual input terminals.

In addition to that is the closed loop frequency response. The closed loop response creates a closed loop pole at the frequency where the closed loop gain is 3dB down. At this point the closed loop gain is virtually equal to the open loop gain. (closed loop gain curve meets the open loop gain curve). At this point the closed loop frequency response has 45 degree phase lag.

So that is the closed loop response which is the relationship between the output and the input to the circuit.

Edit: The above frequency values are for an op amp with a GBW of 1MHz.


Stability of an amplifier depends on the loop gain.

loop gain = Beta * Aol

Where Beta = the feedback fraction = Ri/(Rf+Ri) = 1/(noise gain)

and Aol = open loop gain

The idea of dominant pole compensation is to get the loop gain down below unity before the loop phase lag exceeds 180 degrees. This can be achieved if the dominant pole is placed at a low enough frequency, from where the open loop gain rolls off at 6dB/octave (20dB/decade).

With an amplifier configured for unity noise gain (Beta = 1)and a purely resistive feedback network, stability depends solely on the gain and phase of the open loop response.

If an amplifier is to be configured for a noise gain (same as signal gain for a non-inverting amp) greater than unity then this actually reduces the loop gain and increases stability (increases phase margin). In this case the compensation capacitor can be reduced in value, increasing the dominant pole frequency and increasing the open loop gain at all frequencies. The overall result is that the loop gain still gets down below unity before the loop phase lag gets above 180 degrees.

(Beta has been reduced, open loop gain has been increased keeping the loop gain constant and keeping the phase margin constant).

The advantage of doing this is that the bandwidth of the amplifier has been increased. This technique can be taken advantage of in decompensated op amps and externally compensated op amps.


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