How to decrease the Miller Capacitor

[lallaby]

miller capacitor

Hello Everyone,

How to decrease the Miller Capacitor in a two-stage opamp which has a large gm at the input differential pair?

I think the reason for large Cc is as follows:
Since UGB is decided by gm/Cc, a very large Cc which is around 60 pF is used to move the UGB to a value which is lower than the second pole.

Is that right? And how to decrease the Cc while keep the large gm at the input pair.

Thanks!

Regards,
Lallaby

 

[yxo]

In order to keep the same PM while decreasing Cc, you should increase gm of the output stage. I.e. you should increase current of the second stage. Two stage opamp we consider as a two-pole system. So, increaseing gm of the output stage you keep the same distance between dominant and non-dominant poles.
Another words, GBW= gm1/2piCc and fnd=gm2/2piCload

 

[lallaby]

In order to keep the same PM while decreasing Cc, you should increase gm of the output stage. I.e. you should increase current of the second stage. Two stage opamp we consider as a two-pole system. So, increaseing gm of the output stage you keep the same distance between dominant and non-dominant poles.
Another words, GBW= gm1/2piCc and fnd=gm2/2piCload

Thank you for your answer.

Another question, increasing the current of the output stage will decrease its gain, right?

 

[yxo]

sure

Added after 5 minutes:

although, gain of an inverter is gm*Req. So, gm increase Req decrease... It shouldn´t be a big diference

 

[sutapanaki]

Actually, gain will drop as square root of current, assuming square law MOS devices. In other words, this is the gain-bandwidth trade-off in voltage amplifiers - you increase the bandwidth and you pay with lower gain.

 

[liusupeng]

it depend on your load capacitance also. if ur load capacitance is very small, the miller capacitance required will be small