Not a homework problem, I'm refreshing before semester starts. Problem is from chapter 7 of Razavi Fundamentals. Given are \$V_{th} = 0.4\text{ V}\$, \$\mu_nC_{ox} = 200\text{ µA/V}^2\$, \$\mu_pC_{ox} = 100\text{ µA/V}^2\$.


Part (a) I figured out, \$A_v = g_{m1}\left(r_{o1} \parallel r_{o2}\right)\$ and \$r_{o1}\$ and \$r_{o2}\$ are \$1/(\lambda I_d)\$. That gets me \$g_{m1}\$, and I set $$g_{m1} = \sqrt{2 \mu_nC_{ox} \times (W/L)_1 \times I_d}$$ and solve for \$(W/L)_1\$. I got 7.8125.

I'm now struggling on part (b). The equation for bias \$I_d\$ of each transistor is $$\frac{1}{2} \mu C_{ox} \frac{W}{L} \times (V_{GS}-V_{TH})^2 \times (1 + \lambda V_{DS})$$ both equal to each other (obviously using the PMOS version for the PFET).

Where I'm having problems is how I go about finding \$V_{GS}\$ for M1 so that I can find \$V_{DS}\$ for M1[attachimg=1], and where \$V_{\text{out}}\$ is biased at (i.e. what is \$V_{DS}\$ or \$V_{SD}\$?). The drain-source voltage for M1 and M2 add up to \$V_{DD}\$ but that's all I know. I was going to assume \$V_{\text{out}}\$ sits at \$V_{DD}/2\$ but that seems wrong. Any pointers?

  • 1
    \$\begingroup\$ You do not need to know the Vgs1. In part B you need to solve for Vb at Id = 0.5mA. \$\endgroup\$
    – G36
    Sep 10, 2021 at 16:48

1 Answer 1


from Razavi (1st ed. eq 2.28). \$ g_m \$ (with \$\lambda\$ included) is equal to $$\sqrt{\frac{(2\times \mu_n\times C_{ox} \times \frac {W}{L}\times I_d)}{(1+\lambda\times V_{DS})}}$$ Since you know \$g_m,\mu_n ,C_{ox},\lambda,\frac{W}{L},\$ and \$I_d\$ (from your comments and problem statement(s) parts a and b) you can get \$V_{DS1}\$.

If you know \$V_{DS1}\$, then you can get \$V_{DS2}\$ (given fixed supply). If you know \$V_{DS2}\$, then you have enough information to get \$V_{GS2}\$ from the \$I_d\$ equation (including \$V_{DS}\$ and \$\lambda\$ effects).

You were on the right track.

  • \$\begingroup\$ Thank you!! I think I was just doubting myself, sometimes the answer is right there but seems like it shouldn't be haha. \$\endgroup\$
    – Shredder
    Sep 12, 2021 at 4:56

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