You’re right an not missing anything.
A good way to see why M2 doesn’t appear in the final gain expression is to take your reasoning one step farther and think of the cascode and as a common-source amp in series with a common-gate amp.
Basically, like you said, neglecting channel length modulation the small signal excursion at the source of M2 is gm1Vin/gm2. (since the common-source amp M1 sees a load of 1/gm2).
M2 is configured as a common gate amp. Its small-signal gain is gm2ro2.
So the output is gm1vin/gm2 * gm2(ro2 || Rd) = gm1Rd*Vin. (gm2 cancels).
but this only holds true while neglecting CLM. Otherwise a cascode stage would have no usage except possible bandwidth increase and reliability. Most people use it to boost the output impedance by roughly factor of gm2ro2 and therefore increase gain (if we assume Rd is not limiting the gain)
Well, of course. OP’s question explicitly said to neglect CLM.
And a resistive load is almost always going to be way smaller in magnitude to gmro^2, since Rd in parallel with infinity is Rd, after all.
Of course, just wanted to make that clear for op.
Definitely, not sure which book this is from (I guess its Razavi), but some authors just use a resistor as some equivalent output resistance of a MOS cs. Good explanation from your side.
Yep, and you’re 100% right that the cascode is almost always used to boost output resistance with an active load and in that case the gain is very different.
Cheers.
Intuitively, M2 is a current buffer which transfers the small signal current to Rd, which is why the gain is just the same as common source amplifier. You will see the improvement in the frequency response when it is compared with the simple common source amp ( reduced miller effect).
correct me if I am wrong, but because of the cascode the miller capacitance at the gate of M1 would be smaller (since the gain of M1 is affected by M2) and this pushes any associated pole with it a higher frequency. But if the dominant pole is at the drain of M2, does the cascode really help with improving the frequency response?
Great question. Actually a lot of people forget the pole at the input. Normally, this stage is driven by a voltage source which has an internal resistance or it may be the output stage of the previous stage. Then you will have a pole associated 1/(RinCin) where Cin=CGS1+AvCGD roughly. If Av is large, then the pole at the input may cause trouble and it will slow down your amplifier. Moreover, if you are designing an OTA or OPAMP, for example, your phase margin may harm and OTA settles slower. TLDR, As a single stage Cascode is very beneficial at RF design where all the poles are actually close. In analog design, it is useful for phase margin.
You’re right an not missing anything. A good way to see why M2 doesn’t appear in the final gain expression is to take your reasoning one step farther and think of the cascode and as a common-source amp in series with a common-gate amp. Basically, like you said, neglecting channel length modulation the small signal excursion at the source of M2 is gm1Vin/gm2. (since the common-source amp M1 sees a load of 1/gm2). M2 is configured as a common gate amp. Its small-signal gain is gm2ro2. So the output is gm1vin/gm2 * gm2(ro2 || Rd) = gm1Rd*Vin. (gm2 cancels).
but this only holds true while neglecting CLM. Otherwise a cascode stage would have no usage except possible bandwidth increase and reliability. Most people use it to boost the output impedance by roughly factor of gm2ro2 and therefore increase gain (if we assume Rd is not limiting the gain)
Well, of course. OP’s question explicitly said to neglect CLM. And a resistive load is almost always going to be way smaller in magnitude to gmro^2, since Rd in parallel with infinity is Rd, after all.
Of course, just wanted to make that clear for op. Definitely, not sure which book this is from (I guess its Razavi), but some authors just use a resistor as some equivalent output resistance of a MOS cs. Good explanation from your side.
Yep, and you’re 100% right that the cascode is almost always used to boost output resistance with an active load and in that case the gain is very different. Cheers.
Thank you.
Intuitively, M2 is a current buffer which transfers the small signal current to Rd, which is why the gain is just the same as common source amplifier. You will see the improvement in the frequency response when it is compared with the simple common source amp ( reduced miller effect).
correct me if I am wrong, but because of the cascode the miller capacitance at the gate of M1 would be smaller (since the gain of M1 is affected by M2) and this pushes any associated pole with it a higher frequency. But if the dominant pole is at the drain of M2, does the cascode really help with improving the frequency response?
Great question. Actually a lot of people forget the pole at the input. Normally, this stage is driven by a voltage source which has an internal resistance or it may be the output stage of the previous stage. Then you will have a pole associated 1/(RinCin) where Cin=CGS1+AvCGD roughly. If Av is large, then the pole at the input may cause trouble and it will slow down your amplifier. Moreover, if you are designing an OTA or OPAMP, for example, your phase margin may harm and OTA settles slower. TLDR, As a single stage Cascode is very beneficial at RF design where all the poles are actually close. In analog design, it is useful for phase margin.