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Chapter 6

Q. 6.20

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

Step-by-Step

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)

6.48