Electronic – Transconductance of a MOSFET

Looking into the drain, the small-signal resistance is $$r_{id} = r_o = \frac{\lambda^{-1}+V_{DS}}{I_D}$$ if the source is at AC common (common-source configuration).

If the AC resistance from source to common is \$R_{ts} \ne 0\$, the small-signal resistance looking into the drain is

$$r_{id} = r_o \left(1 + \frac{R_{ts}}{r_s} \right) + R_{ts}$$

where

$$r_s = \frac{1}{g_m}$$

Looking into the source, the small-signal resistance is

$$r_{is} = r_s$$

The above assumes the body is connected to the source.

I understand why r02 is in parallel with Rf, but what is the 1/gm2 resistor doing there?

The lower right circuit is drawn oddly and further, seems to mix AC and DC sources which is an error.

If I were teaching this circuit, I would draw the AC circuit, with Q1 and Q2 replaced by their small-signal T-models, as follows

^{simulate this circuit – Schematic created using CircuitLab}

Now, is it clear why \$r_{s2} = \frac{1}{g_{m2}}\$ is there?

Edit: on second thought, I don't understand also why the first r0 is at the drain of the first transistor, and the second r0 is at the source of the second transistor.

\$r_o\$ connects to the drain and source.

Since, for Q1, the source is grounded, \$r_{o1}\$ connects from D1 to ground.

Since, for Q2, the drain is AC grounded, \$r_{o2}\$ connects from S2 to AC ground.

Quite often it is the gate-source capacitance of a MOSFET that is the dominant factor on how quickly a MOSFET can be turned on or off. Quite high currents have to be injected into the gate capacitance to quickly change the gate voltage so, if the transconductance is (say) twice as much on MOSFET A than MOSFET B then, to adequately switch a particular load current, you only need to change the gate voltage by half the amount compared to MOSFET B.

This usually results in an increase of switching speed for a given current injected into the gate.