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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.

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  • \$\begingroup\$ What is 'gmb2' in the small-signal model? \$\endgroup\$ – analogsystemsrf Aug 22 '17 at 4:13
  • \$\begingroup\$ @analogsystemsrf gmb2 is the gm resulting from the "bulk gate". The bulk/substrate of the MOSFET can act as a (secondary) gate. If the bulk/substrate is grounded then one can usually discard this effect so then gmb2 = 0 \$\endgroup\$ – Bimpelrekkie Aug 22 '17 at 7:55
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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.

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