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## Q. 6.20

Repeat the above example while including both $C_F$ and a capacitor, $C_B,$ from node B to ground.

## Verified Solution

Since we have already obtained the transfer function with $C_F$ present, we must seek the $Z_{out,0}$ and $Z_{in,0}$ corresponding to $C_B$. The arrangement depicted in Fig. 6.48(a) suggests that $Z_{out,0}$=0 because the drain voltage must be zero while $V_1$ is not, requiring an infinite current to flow through $V_1$.

For $Z_{in},0$, we note from Fig. 6.48(b) that $V_{GS} = V_1R_SC_F s/(R_SC_F s + 1)$ and the current flowing through $C_F$ is equal to $V_1/[(C_F s)−1 + R_S]$. A KCL at the drain node gives

$\frac{V_1}{R_D}+\frac{V_1C_F s}{R_SC_F s + 1} + g_m V_1 \frac{R_SC_F s}{R_SC_F s + 1} = I_1$                                           (6.112)

Thus,
$Z_{in},0 = \frac{R_D(R_SC_F s + 1)}{[R_S(1 + g_m R_D) + R_D]C_F s + 1}$                                               (6.113)

Using Eq. (6.111), we write the new transfer function as

$G(s) = −g_m(R_D||r_O )\frac{1 − \frac{1}{g_m}C_F s}{1 + [(1 + g_m R_D)R_S + R_D]C_F s}$                                               (6.111)

$G(s) = −g_m(R_D||r_O )\frac{1 − \frac{C_F}{g_m} s}{1 + [(1 + g_m R_D)R_S + R_D]C_F s} \frac{1}{1+\frac{R_D(R_SC_F s + 1)C_Bs}{[R_S(1 + g_m R_D) + R_D]C_F s + 1}}$

$= −g_m(R_D||r_O )\frac{1 − \frac{C_F}{g_m}s}{[R_S(1 + g_m R_D) + R_D]C_F s + R_D(R_SC_F s + 1)C_Bs + 1}$                                                (6.114) 