Consider the following common-gate stage designed in CMOS:

enter image description here

Calculating input impedance from small-signal model

If we replace the circuit with it's small-signal equivalent and then apply a voltage source Vx and measure current Ix with body effect included, we obtain the following: enter image description here

Calculating input impedance just be inpsection (intuitively) in the original circuit

I know it isn't usually done like this and that input impedance is almost always defined in small-signal.

However, just by looking at the circuit and thinking if I applied an increase voltage step at the source of M1, and considering that if the source increases, then VBS (body-source voltage assuming body is 0V) decreases by becoming more negative and VGS (gate-source voltage) decreases, hence you would think that the current decreases as well just by the square-law equations.

enter image description here

So, an increase voltage at source of M1 has led to a decrease in current, this would lead me to assume that the circuit has a negative input impedance looking into the source.

Why is this not consistent with the first approach?


1 Answer 1


The current indeed gets smaller absolutely, when you raise the input voltage. But as it is already negative (coming into the voltage source), that means that it gets less negative, so mathematically it increases. So the input impedance is positive.


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