Small-signal of CS stage with source degeneration

Question

Can someone explain how Fig 3.30 (c) was created? In particular, the \$\large\frac{1}{g_m + g_{mb}}\$ resistor to GND.

I know you see \$\large\frac{1}{g_m + g_{mb}}\$ when looking into the source, but why is it connected to GND in small-signal?

Answer 1

Figs. 3.30(a-c) are graphs representing a single transistor amplifier as it behaves w.r.t. small AC constituents of the total circuit voltages and currents, i.e. DC-bias plus (small) AC-signal components. Constant voltage sources, which provide bias voltages, are omitted, because these are just shorts for (small) AC signals.

While this fact, discussed earlier in comments, is a direct answer to you question, notice that the formula for the source voltage \$V_S\$ variation given the drain voltage \$V_D\$ variation (formula 3.69) can be derived in a one-step, straightforward way, and truly by inspection, without bothering about "lookin into/looking from" considerations.

First, you replace the finite-output-resistance transistor with its pi model, exactly as Razavi did it, only instead of a "part that holds the dependence on vgs only (the traditional MOSFET symbol)", as the poster of Degenerated common source output resistance by inspection puts it, we use (explicitly) a voltage controlled current source \$-g·V_S\$ in accordance with the MOSFET pi model.

^{simulate this circuit – Schematic created using CircuitLab}

The Thévenin equivalent of \$VCCS||r_o\$ is a voltage controlled voltage source \$g·r_o·V_S\$ in series with \$r_o\$

^{simulate this circuit}

With the current through VCVS equal to the current through the \$R_S\$ component, the voltage across VCVS (\$g·r_o·V_S\$) is equal to the voltage drop across the resistor \$g·r_o·R_S\$. We arrive at the equivalent circuit

^{simulate this circuit}

Now, with this voltage divider circuit, we can calculate the voltage variation \$ΔV_S\$ at the transistor source caused by the voltage variation \$ΔV\$ at the transistor drain, truly by inspection: $$ ΔV_S = ΔV{R_S \over (r_o+R_S+g·R_S·r_o)} $$ Substituting explicit expressions for \$R_S||(1/g)\$ terms in Razavi's formula 3.69, you can make sure for yourself that this formula is exactly Razavi's formula, only written in a much simpler way.

This derivation does not mean to deprecate the value of Razavi's analysis: further in his text, his narrative may need to analyze the other current branches for additional analysis, and the reader would appreciate the inclusion of \$R_S||(1/g)\$ terms. Still, the alternative derivations are useful because they help develop a circuit analysis intuition.