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## Q. 6.15

For the common-gate stage shown in Fig. 6.32(a), calculate the transfer function and the input impedance,$Z_{in}$. Explain why $Z_{in}$ becomes independent of $C_L$ as this capacitance increases.

## Verified Solution

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.