# Fundamentals of Microelectronics

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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 the following equivalent circuit for finding R in : R in Q 1 V CC 1 g m 2 bardbl r π 2 R in = r π 1 + (1 + β 1 ) parenleftbigg 1 g m 2 bardbl r π 2 parenrightbigg (d) Looking into the base of Q 2 we see an equivalent resistance of r π 2 , so we can draw the following equivalent circuit for finding R in :

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R in Q 1 V CC r π 2 R in = r π 1 + (1 + β 1 ) r π 2
5.4 (a) Looking into the collector of Q 1 we see an equivalent resistance of r o 1 , so we can draw the following equivalent circuit for finding R out : R out R 1 r o 1 R out = r o 1 bardbl R 1 (b) Let’s draw the small-signal model and apply a test source at the output. R B r π 1 + v π 1 g m 1 v π 1 r o 1 v t + i t i t = g m 1 v π 1 + v t r o 1 v π 1 = 0 i t = v t r o 1 R out = v t i t = r o 1 (c) Looking down from the emitter of Q 1 we see an equivalent resistance of 1 g m 2 bardbl r π 2 bardbl r o 2 , so we can draw the following equivalent circuit for finding R out : Q 1 R out 1 g m 2 bardbl r π 2 bardbl r o 2 R out = r o 1 + (1 + g m 1 r o 1 ) parenleftbigg r π 1 bardbl 1 g m 2 bardbl r π 2 bardbl r o 2 parenrightbigg

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(d) Looking into the base of Q 2 we see an equivalent resistance of r π 2 , so we can draw the following equivalent circuit for finding R out : Q 1 R out r π 2 R out = r o 1 + (1 + g m 1 r o 1 ) ( r π 1 bardbl r π 2 )
5.5 (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) Let’s draw the small-signal model and apply a test source at the input. r π 1 + v π 1 v t + i t g m 1 v π 1 R 1 i t = v π 1 r π 1 g m 1 v π 1 v π 1 = v t i t = v t r π 1 + g m 1 v t i t = v t parenleftbigg g m 1 + 1 r π 1 parenrightbigg R in = v t i t = 1 g m 1 bardbl r π 1 (c) From our analysis in part (b), we know that looking into the emitter we see a resistance of 1 g m 2 bardbl r π 2 . Thus, we can draw the following equivalent circuit for finding R in :

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R in Q 1 V CC 1 g m 2 bardbl r π 2 R in = r π 1 + (1 + β 1 ) parenleftbigg 1 g m 2 bardbl r π 2 parenrightbigg (d) Looking up from the emitter of Q 1 we see an equivalent resistance of 1 g m 2 bardbl r π 2 , so we can draw the following equivalent circuit for finding R in : R in Q 1 1 g m 2 bardbl r π 2 V CC R in = r π 1 + (1 + β 1 ) parenleftbigg 1 g m 2 bardbl r π 2 parenrightbigg (e) We know that looking into the base of Q 2 we see R in = r π 2 if the emitter is grounded. Thus, transistor Q 1 does not affect the input impedance of this circuit.
5.6 (a) Looking into the collector of Q 1

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• Spring '08
• MonaHella
• Input impedance, Electrical resistance, Output impedance

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