What's wrong with my FET transfer function?

Question

The following is from the book Design of Analog CMOS Integrated Circuit, page 207.

Find the transfer function of the circuit in Fig. 6.47(a):

Suppose the voltage at node A is \$V_x\$.

with the use of KCL at node B,

$$V_x g_{m1} + \frac{V_{out}} {r_o} = (V_x - V_{out}) s C_F - \frac{V_{out}} {R_D}$$

For \$V_x\$, with voltage divider (\$R_S\$ and \$C_F\$).

$$V_x = \frac{V_{in} - V_{out}} {R_S + 1/(s C_F)} \frac{1} {s C_F} + V_{out} $$

$$\Rightarrow \frac{V_{out}} {V_{in}} = \frac{(s C_F - g_m)} {s C_F R_S (g_m + 1/r_o + 1/R_D) + (s C_F + 1/r_o + 1/R_D)}$$

However, the arthor gave this.

$$G(S) = -g_m (R_D \parallel r_o) \frac{1- \frac{1} {g_m} C_F s} {1 + \biggr[(1 + g_m R_D) R_S + R_D \biggr] C_F s}$$

The zero is the same but the pole is different.

Answer 1

If you check the book's derivation, you can see that the author ignores \$r_o\$ in their derivation of the driving point impedance. The exact \$Z_{in,0}\$ should be \$(1 + g_m \cdot R_s) \cdot (R_D \,||\, r_o) + R_s\$.

Also, whether you can ignore \$r_o\$ or not depends on the relative values of \$R_D\$ and \$r_o\$. As you can see, there is the term \$R_D \,||\, r_o\$. If \$R_D\$ is much smaller than \$r_o\$, you can ignore \$r_o\$. In some cases, for example, if the load \$R_D\$ is a current source which is comparable in magnitude to \$r_o\$, then you can't ignore \$r_o\$.
Hopefully, from this example, you can see the usefulness of the Extra Element Theorem (EET), which can provide more insight than the brute force method you used.