Pole Splitting - Why does output pole not also decrease to lower frequency?

Question

This is a follow-on question.

I have a question regarding pole Splitting. If there was an existing capacitor on node E and on node A, I often see that books say that by adding a Miller cap, the pole at node E will decrease and the pole at node A will increase.

I understand why E decreases, but why does A increase? Node A will also see a higher capacitance than before.

@VoltageSpike yes. Frequency pole. Like the inverse of the R and C on any node — AlfroJang80, Oct 27 '22 at 23:43
@AlfroJang80 Have you read [this web page](https://www.allaboutcircuits.com/technical-articles/pole-splitting-and-miller-frequency-compensation/) on the topic, yet? — jonk, Oct 28 '22 at 00:03

score 0 · Answer 1 · answered Oct 28 '22 at 04:01

The output pole is (originally; without feedback) basically 1/(2.π.ROUT*COUT). Because of the feedback given by the output capacitor, the effective output impedance of the amplifier is reduced by the feedback factor (from the feedback capacitor and the input capacitance). This lowering in output impedance effectively raises the pole formed by the output capacitor.

A very good explanation is at https://web.stanford.edu/class/archive/ee/ee214/ee214.1042/Handouts/HO9POLESPLIt.pdf

Pole Splitting - Why does output pole not also decrease to lower frequency?

1 Answers1