# Why doesn't compensation capacitance add additional phase for OpAmp?

The following image depicts the transfer function of an compensated against an uncompensated OpAmp.

If the internal Miller-compensation capacitor results in an additional pole at low frequency I would expect it to yield an ADDITIONAL phase shift to the existing phase shift of the uncompensated circuit. So why is the resulting phase shift smaller as compared to the uncompensated circuit for f>20kHz?

This seems to be true, because a simulation with LT Spice gives the same result. Obviously, the system transfer function is not a simple linear cascading of low-pass elements, because in this case I would get additional phase change. Can it be understood qualitatively why the compensation "masks out" the intrinsic frequency response?

• Look up "pole splitting" Jun 17, 2022 at 11:19
• This is exactly I was looking for. Thx. Jun 17, 2022 at 13:06
• this would be the accepted answer ;-) I learned so much and I understood the first time the principle behind. Jun 18, 2022 at 19:56
• Glad I could help! Jun 18, 2022 at 20:44

If the internal Miller-compensation capacitor results in an additional pole at low frequency I would expect it to yield an ADDITIONAL phase shift to the existing phase shift of the uncompensated circuit. So why is the resulting phase shift smaller as compared to the uncompensated circuit for f>20kHz?

The compensation capacitor doesn't create an additional pole, it modifies an existing one.

You are thinking of a circuit like this:-

simulate this circuit – Schematic created using CircuitLab

When it is actually like this:-

simulate this circuit

There is no additional pole, just one pole being forced to a much lower frequency than the others, causing gain to drop below 0 while the total phase shift is still less than 180°.

• Excellent answer. If I may add another way to see why the miller capacitor doesn't create another pole: The miller cap will be connected to an input capacitor to ground, and another output capacitor to ground as well. This forms a 3-capacitor loop. Now, only 2 of them can have an independent initial condition, and the one left can only adopt those. Therefore, 2 independent initial conditions = 2 poles in the system. Sep 19, 2022 at 11:06

Why doesn't compensation capacitance add additional phase for OpAmp?

And

So why is the resulting phase shift smaller as compared to the uncompensated circuit for f>20kHz?

At low frequencies, the additional pole turns the compensator into an integrator and, its phase shift is now lagging by 90° compared to the uncompensated circuit. Above 20 kHz the compensated circuit maintains the 90° phase shift all the way up to about 1 MHz whereas the uncompensated circuit's phase angle has dropped to 180° (at 1 MHz) and this will cause instability at this frequency.

Because the compensated circuit's closed-loop gain is so much lower than the uncompensated circuit gain (at the higher frequencies), it can maintain its intended phase performance (as an integrator) up to much higher frequencies compared to the uncompensated circuit and, importantly, will remain stable when used.

In short: The additional low-frequency pole causes the magnitude to be below the critical value of 0dB before the other poles (with their associated phase shifts) come into the play.

And you are right that the overall phase shift (large frequencies) is larger than without this extra pole - but this fact does not create any problems as far as stability is concerned. (This assumes that there is no zero-pole compensation which - in addition - is introduced into some devices)

This is more of a speculative answer, not sure about this but thought I'd give an answer a shot anyhow!

The dominant pole compensation basically acts like a low pass filter having a transfer function of 1/(1+TS).

A transfer function has a zero where the transfer function goes to a value of zero and for the low pass filter this occurs when S=infinity.

Therefore one could say, when considering the amplifier's transfer function with dominant pole compensation and ignoring the higher frequency poles, that the open loop response levels off when its magnitude gets quite low because of this zero. A low pass filter can't go on attenuating for ever!

A zero in the open loop gain response, when it gets quite close to 0dB (ignoring higher frequency poles) would effectively reduce the open loop gain to a slope of 0dB/decade and reduce the open loop lag to zero degrees.

Therefore I am suggesting that, closer to the 0dB frequency, the reducing phase lag from the compensation capacitor is cancelled out by the increasing phase lag from the next higher frequency pole keeping the open loop phase lag at about -90 degrees until the 2nd high frequency pole occurs which causes the open loop phase lag to increase beyond -90 degrees.

I wondered what you fellow "StackExchangers" with keener brains and more focused knowledge than my own would make of this answer.