Let's consider the following amplifier circuit:

enter image description here

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

enter image description here

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?

  • \$\begingroup\$ Is gm2 positive or negative? \$\endgroup\$ – Elliot Alderson Sep 4 '18 at 15:21
  • \$\begingroup\$ @ElliotAlderson Transconductance is a positive value. \$\endgroup\$ – A6EE Sep 4 '18 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


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