Shown in Fig. P6.37 is the ac equivalent of a bipolar current mirror operating as a high-frequency current amplifi er. Since the emitter area of Q2 is four times that of Q1, the mirror provides a...

Shown in Fig. P6.37 is the ac equivalent of a bipolar current mirror operating as a high-frequency current amplifi er. Since the emitter area of Q2 is four times that of Q1, the mirror provides a nominal gain of 4 A/A. Draw the high-frequency small-signal equivalent; then, assuming the diode-connected transistor Q1 is biased at 0.25 mA, fi nd the low-frequency gain a0 5 ioyii and use OCTC analysis to estimate f23 dB, given the following parameter values: 01 5 02 5 250, F1 5 F2 5 0.25 ns, Cje2 5 4Cje1 5 4 pF, C2 5 4C1 5 1 pF, and Cs2 5 4Cs1 5 6 pF. Explain what makes this circuit a high-frequency type.