## Chapter 6

## Q. 6.19

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

## Step-by-Step

## 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 .