## Chapter 6

## Q. 6.13

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

## Step-by-Step

## 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.