I was watching one of Razavi's videos https://youtu.be/oC2otEX3i0I?t=553 Minute 9:13

He has this circuit

enter image description here

He says that at higher frequencies, the effect of the capacitor becoems more pronounced and that the capacitor will be providing some of the drain current and hence the resistor Rd will have less current through it. The resistor has less current through it, thus a smaller voltage drop across it. Then, he says that because of that the Vout will drop. That makes no sense to me.

My reasoning: $$V_{out} = V_{DD} - R_DI_R$$

Now surely, if I_R decreases as Razavi said it would, then of course the voltage drop across the resistor would decrease and hence we would be doing Vdd minus a smaller number, so Vout should increase and not decrease

So which is it, would Vout increase or decrease at higher frequencies where capacitor has an effect?

  • \$\begingroup\$ CL is in parallel with Rd, which you aren't including in your reasoning. CL impedance decreases with frequency which means your output is shorted to ground. \$\endgroup\$ Nov 14, 2018 at 17:08

2 Answers 2


If you ignore non-linear effects, i.e., small signal, Vout neither falls nor rises, it stays the same.

The problem lies in that we are speaking about different things. The DC value of Vout (which is the equation you wrote) does not change.

However the small-signal (AC) magnitude of Vout goes down as you increase the frequency. More and more of the AC current goes through the capacitor and less and less of the AC current goes through the resistor.

But this AC signal is riding on the DC operating point. And that does not change.

Nonetheless, if you consider non-linear effects, there will be a change on the DC operating point due to distortion of the AC signal. Consider the components introduced by a quadratic distortion of a sinusoid. This will produce both a second harmonic and a DC component that changes with frequency and input magnitude.


View that collector node like this


simulate this circuit – Schematic created using CircuitLab

and we see the node is a Low Pass Filter.


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