## Chapter 6

## Q. 6.5

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

## Step-by-Step

## 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.