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):

enter image description here

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,

enter image description here

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,

enter image description here

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?

  • \$\begingroup\$ To show that it's similar to 3.65? \$\endgroup\$ Commented Mar 13, 2021 at 22:30
  • \$\begingroup\$ 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? \$\endgroup\$ Commented Mar 13, 2021 at 22:37

1 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).

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

This leads to 3.70 and onwards.

  • \$\begingroup\$ Hmm. But the last highlighted equation in the book (Eqn 3.72) is defining ΔV/ΔI which should be the total output resistance, right? They are applying a ΔV at the output and measuring ΔI. Yet the answer for total output resistance (= ΔV/ΔI = Eqn 3.72) is different to what I got based on Fig. 3.30(c) \$\endgroup\$ Commented Mar 13, 2021 at 22:58
  • \$\begingroup\$ @AlfroJang80 Can you take it from here? They are calculating the resistance, so it makes sense to divide V/I, but they are using the previously calculated V. \$\endgroup\$ Commented Mar 13, 2021 at 23:20
  • \$\begingroup\$ 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). \$\endgroup\$ Commented Mar 13, 2021 at 23:33
  • \$\begingroup\$ That's quite interesting. The applied ΔV is influencing the total output resistance since the source voltage is changing. I would have thought that the transition of M1 into a 1/gm +1/gmb resistance to ground would have handled that phenomenon. I need to think about this some more - it's not sitting well in my brain. Thank you for your help. \$\endgroup\$ Commented Mar 14, 2021 at 3:06
  • \$\begingroup\$ I think what seems to be confusing me is that in the large signal view of Fig 3.30a, all of the deltaI current flows through Rs. However, after breaking it into the small-signal, that is no longer true. \$\endgroup\$ Commented Mar 14, 2021 at 3:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.