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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 $g_{m2} = 0$.