Repeat the analysis of problem 15 for the network of Fig. 9.83 with the addition of a source resistance and signal source as shown in Fig. 9.87. Plot the gain A_vs = V_oV_s and comment on the change in low-frequency cut-off as compared to problem 18.

Problem 18

Repeat Problem 15 for the common-base configuration of Fig. 9.83. Keep in mind that the common-base configuration is a noninverting network when you consider the Miller effect.

Transcribed Image Text:

C = 18 pF Che = 24 pF Ca = 12 pF Cw, = 8 pF = 10 pF o4V 9-16 V Che = 18 pF Ce = 24 pF Cee = 12 pF %3D %3D Cw. %3D 1.2 ΚΩ 2 3.3 k2 10 µF HE Ve 10µF B = 80 4.7 k2 FIG. 9.83