So I came across this interesting MOSFET question and thought I'd give it a try. I seem to be stuck on a couple of parts however, hopefully someone can see something I'm not seeing. Here is the circuit:

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Here's my attempt at a solution for it and where I've been getting stuck:

(a) In the low frequency regime, capacitor CL becomes an open circuit. Performing a small signal analysis on M1 results in the following gain equation: Av=(Vo/Vsig)=gm1*Req, where Req=(r01||r02) since there is no load resistor RL at the output. gm1 is calculable since ID1 can be determined from the current mirror formed by Mb and M2 and the provided value of ID3. But the question specifies that r01=r02=infinity, however. Would this mean that the low frequency gain is indeed infinity?

(b) We can consider the high-frequency regime to determine the 3dB freq and therefore must consider the internal capacitances of M1. Since there is no signal resistance (Rsig=0), you can use the amplifier output to determine the 3dB freq. In particular, fH=1/(2pi(CL+Cgd1)*Req), where CL includes the relevant capacitances from the other transistors. But the result is fH=0 since Req=(ro1||r02)=infinity.

To me, these answers don't seem correct, but I cannot figure out where I am going wrong. The fact that this circuit doesn't contain a load resistance and neglects channel length modulation is puzzling.

  • \$\begingroup\$ What are r01 and r02? The resistances of M1 and M2? Something else? What are ID1 and ID3, the drain currents of M1 and M3? Those are my guesses, but please verify. \$\endgroup\$ Commented Nov 24, 2020 at 3:05
  • \$\begingroup\$ It's been a long time since I've looked at a circuit like this... But I think M3 is the key. What does it do? This may answer a lot of your questions \$\endgroup\$
    – Matt
    Commented Nov 24, 2020 at 4:35
  • \$\begingroup\$ @JTaft121 You wrote: "But the question specifies that r01=r02=infinity, however. Would this mean that the low frequency gain is indeed infinity?" Do you know what channel length modulation is? If not, you should research it, and it's relationship to the output resistance of FET current mirrors \$\endgroup\$ Commented Nov 24, 2020 at 4:41
  • \$\begingroup\$ @MathKeepsMeBusy r01 and r02 are the output resistances of M1 and M2, respectively. Yes ID1 and ID3 are drain currents of M1 and M3. Am familiar with CLM, I assumed it was negligible here since Va=infinity. Will look into its relation to current mirror outputs, thanks \$\endgroup\$
    – JTaft121
    Commented Nov 24, 2020 at 5:25
  • \$\begingroup\$ @Matt From what I understand, the output of the current mirror and the drain current of M3 sum to equal the drain current of M1. \$\endgroup\$
    – JTaft121
    Commented Nov 24, 2020 at 5:30

1 Answer 1


I think the key is that M3 provides a finite load resistance. It's diode-connected, which acts a resistance of 1/gm. The low-frequency gain is therefore the ratio of gm of M1/M3.


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