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

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?

• To show that it's similar to 3.65? Mar 13, 2021 at 22:30
• Sorry, maybe I wasn't clear. What I meant is that - how come I can't just take my Rout = ro + (Rs Parallel with gm and gmb)? Why does that not give the same answer? Mar 13, 2021 at 22:37

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).
\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}
• Don't forget this part: "Since the current through $R_S$ must change with $\Delta I$ [...]" -- this is key. A variation of V means variation of I, the source potential changes, thus the operating point changes, This tells you that the applied voltage, $\Delta V$ influences the current and, thus, the whole resistance. In (c) you don't just have resistors, you have resistors plus $\Delta V$. This is why the current is calculated in 3.70, and why 3.72 uses this current to divide the voltage (calculated before). Mar 13, 2021 at 23:33