I've been going through Design of Analog CMOS integrated circuits 2nd Edition by Razavi. I'm currently at the Cascode amplifier section. The author does an example for calculating the gain for the circuit below, but I don't understand how he got his end result.
He first starts by obtaining Id2, noting the the current generated from M1 is split between Rp and the impedance seen looking into the source of M2, which is
$$1/(gm2+gmb2)$$
$$Id2 = gm1Vin\frac{(gm2+gmb2)Rp}{1+(gm2+gmb2)Rp}$$
I don't understand how he comes to that conclusion for the current of Id2.
Next, he just states the voltage gain is given by:
$$Av = -gm1\frac{(gm2+gmb2)RdRp}{1+(gm2+gmb2)Rp}$$
I attempted to use the Lemma technique to solve it (Av = -GmRout) but I can't seem to get anywhere. I think I'm missing an important step he's doing to obtain that gain.
Could anyone provide some insight into the process he is doing to obtain those results (Id2, and Av)? Thank you.
Best Answer
Since gmb2 is usually ignored when the substrate connection is properly grounded, I'll ignore it and make thing slightly simpler. So gmb2 = 0
Then looking "up" into the source of M2 I see:
$$1/(gm_2)$$
Looking right from the drain of M1 I see:
$$Rp$$
The current coming out of the drain of M1 is:
$$Id_1 = gm_1Vin$$
Now this current splits between that 1/gm2 and Rp:
$$Id_2 = Id_1\frac{gm_2Rp}{1+gm_2Rp}$$
(Note that this uses the "current dividing into two resistances" formula. It is similar to the voltage divider formula but then for current.)
Then filling in Id1 that makes:
$$Id_2 = gm_1Vin\frac{gm_2Rp}{1+gm_2Rp}$$
You can now replace gm2 with (gm2 + gmb2) if you like, the end result remains the same.
Drain current Id2 does not split anymore after M2, all the current goes into the drain load resistor Rd. So the voltage gain Av is simply Id2 as derived above multiplied by the value of the load resistor Rd. With a negative sign as a rising Vin results in a falling Vout.