MOSFET resistance

Question

I'm trying to understand how MOSFET resistances work, but I'm seeing a lot of things that don't always fit together (probably due to my lack of understanding). Specifically the amplifier configurations (CS, CG, CD). Is this correct:

1. Looking into the gate, resistance is infinity.
2. Looking into the source, resistance is r0
3. Looking into the drain, resistance is r_ds = 1/g_m

Is this view flawed somehow? Thanks!

Here is an example from Sedra Smith (which is a bit more involved but really, I have no idea what's going on here).

For the purpose of determining the close-loop gain of this amplifier with feedback, the A circuit is shown:

I understand why r02 is in parallel with Rf, but what is the 1/gm2 resistor doing there? Why is the first transistor a current source and a resistor, and the second a pair of resistors?

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.

Answer 1

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.

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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.