(a) Shown in Fig. P6.38 is the ac equivalent of a MOS current mirror operating as a highfrequency current amplifi er. Since the channel width of M2 is fi ve times that of M1, the mirror provides a...

(a) Shown in Fig. P6.38 is the ac equivalent of a MOS current mirror operating as a highfrequency current amplifi er. Since the channel width of M2 is fi ve times that of M1, the mirror provides a nominal gain of 5 A/A. Draw the high-frequency small-signal equivalent; then, assuming the diode-connected transistor M1 is biased at 0.2 mA, fi nd the low-frequency gain a0 5 ioyii and use OCTC analysis to estimate f23 dB, given the following parameter values: k2 5 5k1 5 12.5 mA/V2 , 2 5 1 5 0.05 V21 , Cgs2 5 5Cgs1 5 1 pF, Cgd2 5 5Cgd1 5 200 fF, Cdb2 5 5Cdb1 5 50 fF. Explain what makes this circuit a high-frequency type. (b) Repeat, if M2’s drain is terminated on a load RL5 2 kV. Compare with part (a), and comment.