Question 6.15: For the common-gate stage shown in Fig. 6.32(a), calculate t...

Using the equivalent circuit shown in Fig. 6.32(b), we can write the current through R_S as −V_{out}C_L s + V_1C_{in}s. Noting that the voltage across R_S plus V_{in} must equal −V_1, we have

(−V_{out}C_L s + V_1C_{in}s)R_S + V_{in} = −V_1                                                                 (6.70)

That is

V_1 = −\frac{−V_{out}C_L s R_S + V_{in}}{1 + C_{in} R_Ss}                             (6.71)

We also observe that the voltage across r_O minus V_1 equals V_{out}:

r_O (−V_{out}C_L s − g_m V_1) − V_1 = V_{out}                              (6.72)

Substituting for V_1 from (6.71), we obtain the transfer function:

\frac{V_{out}}{V_{in}}(s) =\frac{1 + g_mr_O}{r_OC_LC_{in} R_Ss^2 + [r_OC_L + C_{in} R_S + (1 + g_mr_O )C_L R_S]s + 1}                                             (6.73)

The reader can prove that body effect can be included by simply replacing gm with g_m + g_{mb}. As expected, the gain at very low frequencies is equal to 1 + g_mr_O . For Z_{in}, we can use (6.69) by replacing Z_L  with  1/(C_L s), obtaining

Z_{in} ≈ \frac{Z_L}{(g_m + g_{mb})r_O}+\frac{1}{g_m + g_{mb}}                                                         (6.69)

Z_{in} = \frac{1}{ g_m + g_{mb}} +\frac{1}{C_L s} ·\frac{ 1}{(g_m + g_{mb})r_O}                                (6.74)

We note that as C_L or s increases, Z_{in} approaches 1/(g_m + g_{mb}), and hence the input pole can be defined as

ω_{p,in}=\frac{ 1}{\left(R_S\parallel \frac{1}{g_m + g_{mb}}\right) c_{in}}                                   (6.75)

Why does Z_{in} become independent of C_L at high frequencies? This is because C_L lowers the voltage gain of the circuit, thereby suppressing the effect of the negative resistance introduced by the Miller effect through r_O (Fig. 6.7). In the limit, C_L shorts the output node to ground, and r_O affects the input impedance negligibly.