Question 7.19: Design a resistively-loaded common-source stage with a total...
Design a resistively-loaded common-source stage with a total input-referred noise voltage of 100 μV_{rms}, a power budget of 1 mW, a bandwidth of 1 GHz, and a supply voltage of 1 V. Neglect channel-length modulation and flicker noise and assume that the bandwidth is limited by the load capacitance.
Illustrated in Fig. 7.45(a), the circuit produces noise at the output in a bandwidth given by R_D and C_L . From the noise model shown in Fig. 7.45(b), 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} & (7.86)\\ &=\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} & (7.87) \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 ∞ yields a value of kT/C_L , we manipulate the transistor noise contribution as follows:
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} (7.88)
Integration from 0 to ∞ 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} &(7.89) \\ &=\left(1+\gamma g_m R_D\right) \frac{k T}{C_L} &(7.90) \end{aligned}This noise must be divided by g_m^2 R_D^2 and equated to \left(100 \mu V _{ rms }\right)^2 . We also note that 1 /\left(2 \pi R_D C_L\right)=1 GHz and kT = 4.14 × 10^{−21} 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 V _{ rms }\right)^2 (7.91)
and hence
\frac{1}{g_m}\left(\frac{1}{g_m R_D}+\gamma\right)=384 \Omega (7.92)
We have some flexibility in the choice of g_m and R_D here. For example, if g_m R_D = 3 and γ = 1, then 1/g_m = 288 Ω and R_D = 864 Ω. With a drain-current budget of 1 mW/V_{DD} = 1 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 mV , V_{D S, \text { min }}=136 mV , leaving little headroom for voltage swings. The reader is encouraged to try g_m R_D=2 \text { or } 4 to see how the voltage headroom depends on the choice of the gain.
