hi,
its difficult to explain like this, but still i will try.
gain=effective gm*Rout. Effective gm=gm1.
Rout=gm2*r02*r01 but due to gain boosting gm2 will be magnified by Av, so Rout=gm2*R02*Av*R01.
Hence total gain=gm1*gm2*r01*r02*Av

Why don't you try with small signal analysis..

Thanks.

small signal analysis seems the way to go (unless by inspection?)

I was able to find component values, but not sure what KVL or KCL should we use to get it solved, any help?

you can simply understand this amplifier as 3 stages of amplification.

the first is the M1 transistor which sees at its drain a small signal voltage (e.g. V1) equal to the Vi multiplied by the M1's gain (A1) and is equal to gm1xro1.

This signal V1 is amplified again by Av  and applied to the gate of M2, so the gate voltage of M2 is simply V1xAv

The third stage is M2 which has a Gate Source voltage equal to V2-V1, however V2 is much larger than V1 so that it can be ignored for simplification. Then the drain voltage of M2 in this case is V2 multiplied with M2's gain (A2) which is nothing else than gm2xro2.

As a result the Vo is then  VixA1xAvxA2, and hence the A=Vo/Vi=A1xAvxA2 or gm1xro1xAvxgm2xro2

best regards,

aamar

Hi,

a very nice tool for this is Symbolic Spice, SSPICE. It is an old DOS program and gives analythic results like this on any circuit.



Bye Stephan

hi,
please refer basic micro-electronics books by razavi. Here FB works like if source of nmos M2 decreases then op-amp will decrease it's gate, BTW feedback polarity independent of nmos or pmos.

Thanks.

raja.cedt wrote on Jul 9^th, 2011, 9:16pm:


Hey, so is the polarity of the feedback for the gain boost diagram posted by author Positive?