Degenerated common source output resistance by inspection

Question

In Razavi's Design of Analog CMOS Integrated Circuits (2nd edition), on page 65 he describes a way to derive the output resistance of a source-degenerated CS stage by inspection:

I understand how to derive this result from the small signal model but I don't understand the reasoning/intuition behind the inspection analysis.

He starts off by saying that if you apply a \$\Delta V\$ at the drain and measure the resulting \$ \Delta I\$, then all the \$ \Delta I \$ must pass through the degeneration resistor \$R_S\$. Sounds good, since the current has nowhere else to go.

He then derives a voltage divider equivalent circuit as shown in the linked image. However, he uses the result that the resistance looking into the source of a MOSFET is \$ \frac{1}{g_m+g_{mb}} \$ to insert a resistor in parallel with \$ R_S \$, which I don't understand. My confusion is that the resistance looking into the source of a MOSFET is derived when you have an ideal independent source applied at that terminal and calculate the voltage/current ratio (typical Thevenin resistance). How is that scenario applicable here? This analysis started off by saying that all of the \$ \Delta I\$ must pass through the resistor \$ R_S\$ but the model derived in figure 3.30(c) seems to imply that some of it is going through this other impedance.

(A side-doubt that I have is that here \$ \lambda \$ is not assumed to be zero since \$r_o\$ is finite but in then in this case resistance looking into the source shouldn't be exactly \$ \frac{1}{g_m+g_{mb}} \$, not sure if that makes a difference here.)

Answer 1

My question is answered. I wrote up some notes fully explaining what's going on with this inspection analysis and clearing up some other details that were confusing me. Hopefully others will find them helpful.

The short answer is that when you use visual shortcuts like drawing the \$r_o\$ in parallel with the MOSFET symbol, you're separating out the MOSFET into two parts: one part which holds the dependence on \$v_{ds}\$ only (\$r_o\$) and one part that holds the dependence on \$v_{gs}\$ only (the traditional MOSFET symbol). At the source, the resistor \$ R_S \$ senses \$ \Delta V_{RS} \$ and carries away a current proportional to it (and no other voltages). The MOSFET symbol (since it "holds" only the VCCSs \$g_m v_{gs} \$ and \$ g_m v_{bs}\$) will also sense the voltage \$ \Delta V_{RS}\$ and carry away from the source a current proportional to it (and no other voltages). This means the MOSFET symbol has the exact same function as the resistor \$R_S\$, so you can model the MOSFET symbol as a resistor and put it in parallel as shown.