HW4solutions

# HW4solutions - 4.53(a VCB2 < 200 mV IC2 RC < 200 mV IC2 <...

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4.53 (a) V CB 2 < 200 mV I C 2 R C < 200 mV I C 2 < 400 μ A V EB 2 = V E 2 = V T ln( I C 2 /I S 2 ) < 741 mV β 2 1 + β 2 I E 2 R C < 200 mV β 2 1 + β 2 1 + β 1 β 1 I C 1 R C < 200 mV I C 1 < 396 μ A V BE 1 = V T ln( I C 1 /I S 1 ) < 712 mV V in = V BE 1 + V EB 2 < 1 . 453 V (b) I C 1 = 396 μ A I C 2 = 400 μ A g m 1 = 15 . 2 mS r π 1 = 6 . 56 kΩ r o 1 = g m 2 = 15 . 4 mS r π 2 = 3 . 25 kΩ r o 2 = The small-signal model is shown below. v in + B 1 r π 1 + v π 1 C 1 g m 1 v π 1 E 1 /E 2 r π 2 v π 2 + B 2 g m 2 v π 2 C 2 v out R C

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5.3 (a) Looking into the base of Q 1 we see an equivalent resistance of r π 1 , so we can draw the following equivalent circuit for finding R in : R in R 1 R 2 r π 1 R in = R 1 + R 2 bardbl r π 1 (b) Looking into the emitter of Q 1 we see an equivalent resistance of 1 g m 1 bardbl r π 1 , so we can draw the following equivalent circuit for finding R in : R in R 1 1 g m 1 bardbl r π 1 R in = R 1 bardbl 1 g m 1 bardbl r π 1 (c) Looking down from the emitter of Q 1 we see an equivalent resistance of 1 g m 2 bardbl r π 2 , so we can draw

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• Spring '08
• SILVA
• Following, Black-and-white films, IC codes, rin, VT ln

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