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## Q. 7.11

For the circuit of Fig. 7.29, calculate the input-referred noise voltage.

## Verified Solution

We have

\begin{aligned} \overline{V_{n, i n}^2} &=\frac{\overline{V_{n, o u t}^2}}{A_v^2} &(7.45) \\ &=\left(4 k T \gamma g_m+\frac{K}{C_{o x} W L} \cdot \frac{1}{f} \cdot g_m^2+\frac{4 k T}{R_D}\right) R_D^2 \frac{1}{g_m^2 R_D^2} &(7.46) \\ &=4 k T \frac{\gamma}{g_m}+\frac{K}{C_{o x} W L} \cdot \frac{1}{f}+\frac{4 k T}{g_m^2 R_D}&(7.47) \end{aligned}

Note that the first term in (7.47) can be viewed as the thermal noise of a resistor equal to $γ/(g_m)$ placed in series with the gate. Similarly, the third term corresponds to the noise of a resistor equal to $\left(g_m^2 R_D\right)^{-1}$. We sometimes say the “equivalent thermal noise resistance” of a circuit is equal to $R_T$ , meaning that the total input-referred thermal noise of the circuit in unit bandwidth is equal to $4kT R_T$ .

Why does $\overline{V_{n, i n}^2}$ decrease as $R_D$ increases? This is because the noise voltage due to $R_D$ at the output is proportional to $\sqrt{R_D}$ while the voltage gain of the circuit is proportional to $R_D$.