The circuit produces noise at the output in a bandwidth given by $R_D$ and $C_L$. From the noise model shown, the reader can derive a Thevenin equivalent for the circuit in the dashed box, obtaining the output noise spectrum as

$$
\begin{aligned}
\overline{V_{n, \text { out }}^2} & =\left(\overline{V_{n, R D}^2}+R_D^2 \overline{I_{n, M 1}^2}\right) \frac{1}{R_D^2 C_L^2 \omega^2+1} \\
& =\left(4 k T R_D+4 k T \gamma g_m R_D^2\right) \frac{1}{R_D^2 C_L^2 \omega^2+1}
\end{aligned}
$$

Since we know that the integral of $4 k T R_D /\left(R_D^2 C_L^2 \omega^2+1\right)$ from 0 to $\infty$ yields a value of $k T / C_L$, we manipulate the transistor noise contribution as follows:

$$
\overline{V_{n, \text { out }}^2}=\frac{4 k T R_D}{R_D^2 C_L^2 \omega^2+1}+\gamma g_m R_D \frac{4 k T R_D}{R_D^2 C_L^2 \omega^2+1}
$$


Integration from 0 to $\infty$ thus gives

$$
\begin{aligned}
\overline{V_{n, \text { out }, \text { tot }}^2} & =\frac{k T}{C_L}+\gamma g_m R_D \frac{k T}{C_L} \\
& =\left(1+\gamma g_m R_D\right) \frac{k T}{C_L}
\end{aligned}
$$


This noise must be divided by $g_m^2 R_D^2$ and equated to $\left(100 \mu \mathrm{~V}_{\mathrm{rms}}\right)^2$. We also note that $1 /\left(2 \pi R_D C_L\right)=1 \mathrm{GHz}$ and $k T=4.14 \times 10^{-21} \mathrm{~J}$ at the room temperature, arriving at

$$
\frac{1+\gamma g_m R_D}{g_m^2 R_D} \cdot \frac{2 \pi k T}{2 \pi R_D C_L}=\left(100 \mu \mathrm{~V}_{\mathrm{rms}}\right)^2
$$

and hence

$$
\frac{1}{g_m}\left(\frac{1}{g_m R_D}+\gamma\right)=384 \Omega
$$


We have some flexibility in the choice of $g_m$ and $R_D$ here. For example, if $g_m R_D=3$ and $\gamma=1$, then $1 / g_m=288 \Omega$ and $R_D=864 \Omega$. With a drain-current budget of $1 \mathrm{~mW} / V_{D D}=1 \mathrm{~mA}$, we can choose $W / L$ so as to obtain this amount of transconductance.

The above choice of the voltage gain and the resulting values of $R_D$ and $g_m$ must be checked against the bias conditions. Since $R_D I_D=864 \mathrm{mV}, V_{D S, \min }=136 \mathrm{mV}$, leaving little headroom for voltage swings. The reader is encouraged to try $g_m R_D=2$ or 4 to see how the voltage headroom depends on the choice of the gain.