whyn't calculate the pole of output in two step amp use mill

why not calculate the pole of output in two step amp use miller effect.

They get it for the dominant pole from miller effect.But for the output pole
,they just set it short and get the result. Why can't using miller . It seems
cannot get the same results

Re: whyn't calculate the pole of output in two step amp use

It works at the input because the gain is high and flat and the phase is zero.
The miller multiplication in principle is true everywhere., but when calculating the non-dominant pole you need to understand that the gain has already fallen enough and the phase of the gain has already reached close to 90deg.
Being a shunt-shunt feedback, the miller capacitor can only reduce impedances at both the ports. While it does the job by increasing the capacitance at the input (Pole moves in)., it reduces the resistance at the output (pole moves out) (remember 90deg phase makes the cap look like a res). Here reducing the impedance drives the poles differently mainly because the phase of the gain is different.

I have assumed that the input pole is dominant to start with., It is easy to see if it is not the case as well

Re: whyn't calculate the pole of output in two step amp use

the output pole expression is easily asserted if you calculate the small-signal model transfert function.
but intuitively, with the capacitor between the gate and drain of the mos, for high frequencies it behaves like a diode.
considering that a pole is conductance/capacitante, you can see that the conductance seen at the output node is roughly gm of the transistor multiplied by the capacitive divider between Cc and CE.
so, the output pole will be approximately gm*(CE/(Cc+CE)) / (Cc +CL)