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## Q. 6.5

Neglecting channel-length modulation, compute the transfer function of the common-gate stage shown in Fig. 6.12(a).

## Verified Solution

In this circuit, the capacitances contributed by $M_1$ are connected from the input and output nodes to ground [Fig. 6.12(b)]. At node $X, C_S = C_{GS} + C_{SB}$, giving a pole frequency

$ω_{in} = \left[(C_{GS} + C_{SB})\left(R_S \parallel \frac{1}{g_m + g_{mb}}\right) \right] ^{-1}$                                (6.18)

Similarly, at node $Y , C_D = C_{DG} + C_{DB}$, yielding a pole frequency

$ω_{out} = [(C_{DG} + C_{DB})R_D] ^{−1}$                                          (6.19)

The overall transfer function is thus given by

$\frac{V_{out}}{V_{in}}(s) = \frac{(g_m + g_{mb})R_D}{1 + (g_m + g_{mb})R_S}. \frac{1}{\left(1+\frac{s}{w_{in}}\right) \left(1+\frac{s}{w_{out}}\right) }$                                   6.20)
where the first fraction represents the low-frequency gain of the circuit. Note that if we do not neglect $r_{O1}$, the input and output nodes interact, making it difficult to calculate the poles.