1

Can anybody briefly explain about Miller compensation?

  • What is the need of pole splitting?
  • How does it improve both the unity gain bandwidth and phase margin?
JRE
  • 67,678
  • 8
  • 104
  • 179
Camila's voice
  • 87
  • 9
  • Do you mean [this](https://www.allaboutcircuits.com/technical-articles/miller-frequency-compensation/)? – Andy aka Feb 15 '21 at 11:02
  • Yes,but in the website it's little bit complex for me...can anyone simplify it? – Camila's voice Feb 15 '21 at 11:04
  • How about [this](https://en.wikipedia.org/wiki/Operational_amplifier#Frequency_compensation)? – Finbarr Feb 15 '21 at 11:05
  • You need to focus on what specifically you don't understand. This is a Q and A site and you need to be coherent with your question else you'll get opinions and then your question will be closed. – Andy aka Feb 15 '21 at 11:05
  • Yeah... But I don't understand how pole splitting increase unity gain bandwidth. It decrease the original full power bandwidth in some extent through increasing PM but does it improve the unity gain BW? – Camila's voice Feb 15 '21 at 11:08
  • 1
    It doesn't increase the unity gain bandwidth; it decreases it. It ensures that the phase change as frequency rises cannot reach 180 degrees before the open-loop gain drops below unity. – Andy aka Feb 15 '21 at 11:16
  • Hence, it gives more stability in compensation – Camila's voice Feb 15 '21 at 11:19
  • 1
    It (the miller compensation) makes an op-amp stable when you apply negative feedback. – Andy aka Feb 15 '21 at 11:20
  • Okay,thank you for assisting – Camila's voice Feb 15 '21 at 11:22

1 Answers1

3

How does it improves the unity gain bandwidth and phase margin both?

It doesn't improve the unity gain bandwidth; it reduces the unity gain bandwidth. It reduces it to avoid the prospect of the op-amp turning into an oscillator when negative feedback is applied.

However it does improve the phase margin by ensuring that the overall open-loop phase response does not slip towards 180° before the open-loop gain has dropped below unity.

Thus, when negative feedback is applied to the op-amp, the circuit does not turn into an oscillator: -

enter image description here

The compensation (blue) adds a pole at a low frequency and ensures that the uncompensated response (red) falls below unity gain before the phase angle has a chance to reach 180 °.

Andy aka
  • 434,556
  • 28
  • 351
  • 777
  • Can you explain me how have you associated the phase angles in the picture (90deg, 180deg) with the magnitude curve? Is it simply linked to the transfer function formula, or can it be found by the magnitude plot? – Kinka-Byo Feb 16 '21 at 07:12
  • 1
    @Kinka-Byo for each pole, there will be a tendency to shift the phase response by 90 degrees. For example, take the red uncompensated line; at low frequencies the phase response will be 0 degrees and then, at the first pole (vertical thin black line), the phase will have changed to 45 degrees and during the first slope downwards it will become 90 degrees. Then when the 2nd pole arrives, the phase angle will have become 90 deg + 45 deg = 135 deg and, as frequency rises more it will tend to become 180 degrees. It is linked to the TF and can be assumed from the bode plot. – Andy aka Feb 16 '21 at 08:47
  • Perfect, thank you very much – Kinka-Byo Feb 16 '21 at 08:51