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## Q. 6.19

Using the EET, find the transfer function of the circuit in Fig. 6.47(a)

## Verified Solution

We first consider the circuit without $C_F$ and write $H(s) = −g_m(R_D||r_O )$. Next, we find Zout,0 using the setup shown in Fig. 6.47(b), exploiting the condition that $V_{out}$ is zero and so is the current through $R_D$. Since $V_{out} = 0$, we have $V_{GS} = V_1$ and $I_1 = −g_m V_{GS} = −g_m V_1$. That is, $Z_{out},0 = −1/g_m$. Note that we resisted the temptation to

write equations involving $V_{in}$. Also, the negative sign of $Z_{out},0$ does not imply a negative impedance between A and B because $V_{in} ≠ 0$.

For $Z_{in},0$, we have from Fig. 6.47(c), $V_A = I_1R_S = V_{GS}$. A KCL at node B gives the current through $R_D$ as $g_m I_1R_S + I_1,$ and a KVL across $R_D, V_1, and R_S$ leads to $I_1R_D(1 + g_m R_S) − V_1 + I_1R_S = 0.$ It follows that $Z_{in},0 = (1 + g_m R_S)R_D + R_S = (1 + g_m R_D)R_S + R_D$ and

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

We see that the EET beautifully predicts the zero and the pole produced by $C_F$ .