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## Q. 6.13

Neglecting channel-length modulation and body effect, calculate the transfer function of the circuit shown in Fig. 6.27(a).

## Verified Solution

Let us first identify all of the capacitances in the circuit. At node $X,C_{G D1} and C_{DB2}$ are connected to ground and $C_{GS1} and C_{G D2}$ to Y . At node $Y ,C_{SB1},C_{GS2},$ and $C_L$ are connected to ground. Similar to the source follower of Fig. 6.22(b), this circuit has three capacitances in a loop and hence a second-order transfer function. Using the equivalent circuit shown in Fig. 6.27(b), where $C_X = C_{G D1} +C_{DB2}, C_{XY} = C_{GS1} +C_{G D2}, and C_Y = C_{SB1} +C_{GS2} +C_L$ , we have $V_1C_{XY} s + g_{m1}V_1 = V_{out}C_Y s,$ and hence $V_1 = V_{out}C_Y s/(C_{XY} s + g_{m1}).$ Also, since $V_2 = V_{out}$, the summation of currents at node X gives

$(V_1 + V_{out})C_X s + g_{m2}V_{out} + V_1C_{XY} s = \frac{V_{in} − V_1 − V_{out}}{R_S}$                                   (6.61)

Substituting for $V_1$ and simplifying the result, we obtain

$\frac{V_{out}}{V_{in}}(s)=\frac{g_{m1} + C_{XY} s}{R_Sξ s^2 + [C_Y + g_{m1}R_SC_X + (1 + g_{m2}R_S)C_{XY} ]s + g_{m1}(1 + g_{m2}R_S)}$                                                               (6.62)

where $ξ = C_XC_Y + C_XC_{XY} + C_YC_{XY}$ . As expected, (6.62) reduces to a form similar to (6.51) for $g_{m2} = 0$.