Small-signal output resistance of MOS common-source stage with source degeneration

Question

This is a bit of a long question so bear with me. In chapter 3 of Razavi's Design of Analog CMOS Integrated Circuits, he introduces the CS stage with source degeneration. He draws the small-signal model and says that the output resistance is as follows (Eqn 3.65):

That makes complete sense to me and I am able to derive that via the small-signal model. Next, Razavi provides another technique to derive the same small-signal output resistance without drawing the small-signal model but rather incorporating it with the large-signal drawing as a means to quickly inspect circuits and gain some intuition. To do this, he applies a ΔV and measures ΔI as shown below,

I understand how he has transitioned from Fig 3.30(a) to (b) and to (c). My confusion is regarding the final step where he calculates the output resistance itself,

I can understand what he is doing but what I don't understand is that as soon as he simplifies the circuit to Fig. 3.30(c), we have a circuit that comprises only of resistors.

Why can you not simply at this stage (Fig 3.30(c)) just write that, $$ R_{out} = r_o + Rs||\frac{1}{g_m + g_{mb}}$$

Why does this not give the same result as Eqn 3.65?

Answer 1

You are just measuring the total output resistance, in the absence of any stimulus, but the book says (and shows) that you are using \$\Delta V_{RS}\$ as the output of the \$\Delta V\$ input. Which means what you've written is just the denominator. You now have a resistive divider formed by the equivalent \$R_{in}=r_o\$ and \$R_{out}=R_S||\dfrac{1}{g_m+g_{mb}}\$, the latter being the one with \$\Delta V_{RS}\$ across: \$\Delta V_{RS}=\Delta V\dfrac{R_{out}}{R_{in}+R_{out}}=...\$ (I'll let you fill in the blanks; hint -- look at eq. 3.69).

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$$\begin{align} R_{eq}&=R_S||\dfrac{1}{g_m+g_{mb}} \\ &=\dfrac{R_S}{(g_m+g_{mb})R_S+1}\tag{1} \\ \dfrac{\Delta V_{RS}}{\Delta V}&=\dfrac{R_{eq}}{R_{eq}+r_o} \\ &=\dfrac{R_S}{(1+(g_m+g_{mb})R_S)r_o+R_S}\tag{2} \\ \Delta V_{RS}&=\Delta V\dfrac{R_S}{(1+(g_m+g_{mb})R_S)r_o+R_S}\tag{3} \end{align}$$

This leads to 3.70 and onwards.