# Small signal equivalent circuit - MOSFET

Let's consider the following amplifier circuit:

Now, if we would analyze small signal operation, we could represent the circuit with small signal equivalent:

The part that bothers me is the PMOS representation in my workbook. Shouldn't the voltage controlled current source of the PMOS transistor (index 2 in the drawing) be rotated so that the current goes from its source to drain?

• Is gm2 positive or negative? Sep 4, 2018 at 15:21
• @ElliotAlderson Transconductance is a positive value.
– A6EE
Sep 4, 2018 at 17:39

For this circuit:

simulate this circuit – Schematic created using CircuitLab

The voltage gain is:

$\large \frac{V_{OUT}}{r_x} + g_{m1}*V_{IN} - g_{m2}*(-V_{OUT}) = 0$

Where $r_x = r_{o1}||r_{o2}$

$$\frac{V_{OUT}}{V_{IN}} = - \frac{g_{m1}*r_x}{1 + g_{m2}*r_x} = - g_{m1}*\left(r_{o1}||r_{o2}||\frac{1}{g_{m2}}\right)$$

And now let us analysis this circuit:

simulate this circuit

As you can see I used the N-MOS small-signal equivalent circuit for the P-MOS this time.

$\large \frac{V_{OUT}}{r_x} + g_{m1}*V_{IN} + g_{m2}*V_{OUT} = 0$

And the voltage gain is exactly the same as before.

$\frac{V_{OUT}}{V_{IN}} = - \frac{g_{m1}*r_x}{1 + g_{m2}*r_x} = - g_{m1}*\left(r_{o1}||r_{o2}||\frac{1}{g_{m2}}\right)$

So to conclude it may sound strange at first glance but P-MOS circuit small signal model is identical to N-MOS.

We have the same situation with the BJT's

Why are the current directions in the hybrid-$\pi$ model for BJT the same for both NPN and PNP?

Applying hybrid-pi model of an npn-BJT to a pnp BJT in small signal analysis

Part annoying me in your text book is that they didn’t substitute the diode connected PMOS with a resistor (1/gm2)//ro2

For your question, the small signal model remains same in both nome and pmos because it is actually a current source whose direction is from source to drain, it’s value should be (gmvsg) which is same as (gm-vgs) which is same as (gm*vgs) in opposite direction (from drain to source)