tm123
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Posts: 67
Chicago, IL
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To find the output impedance of this circuit, replace the load capacitor with a 1A AC current source pointing in to the circuit. After an AC simulation plot the the voltage across this current source. Since Z=V/I and I=1A, plotting the real and imaginary part of the voltage across the current source is the same as plotting the impedance. Realize that this is the SERIES EQUIVALENT output impedance, in other words, R+j*X. You can convert this to parallel equivalent impedance which is what I normally do since I find it easier to work with. Also, always make sure you inject only 1 AC source at a time in your circuit. You would be surprised how many times I have seen this mistake.
Now, the use of the large inductor can help in debugging this problem. For example, place a large inductor (maybe 100H, large enough so the impedance of the inductor is extremely high at your frequency of interest) in series with the drain of M2. Use the same procedure to find the output impedance as described above, and you should see that the impedance is now the output impedance of your cascode device only. You can compare this to your calculation, then put the inductor in series with the drain of the cascode and see the output impedance of your PMOS current mirror only.
It seems you have solved the issue with the capacitor on the gate of the current mirror. Reviewing your calculations, this makes sense since I see you add capacitances Cgd and Cdb of M2 which is incorrect unless the current mirror gate is at AC ground. I will let Kevin elaborate since it was his suggestion :)
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