Fundamentals of Microelectronics

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Unformatted text preview: 5.3 (a) Looking into the base of Q 1 we see an equivalent resistance of r 1 , so we can draw the following equivalent circuit for finding R in : R in R 1 R 2 r 1 R in = R 1 + R 2 bardbl r 1 (b) Looking into the emitter of Q 1 we see an equivalent resistance of 1 g m 1 bardbl r 1 , so we can draw the following equivalent circuit for finding R in : R in R 1 1 g m 1 bardbl r 1 R in = R 1 bardbl 1 g m 1 bardbl r 1 (c) Looking down from the emitter of Q 1 we see an equivalent resistance of 1 g m 2 bardbl r 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 2 R in = r 1 + (1 + 1 ) parenleftbigg 1 g m 2 bardbl r 2 parenrightbigg (d) Looking into the base of Q 2 we see an equivalent resistance of r 2 , so we can draw the following equivalent circuit for finding R in : R in Q 1 V CC r 2 R in = r 1 + (1 + 1 ) r 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) Lets draw the small-signal model and apply a test source at the output. R B r 1 + v 1 g m 1 v 1 r o 1 v t + i t i t = g m 1 v 1 + v t r o 1 v 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 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 2 bardbl r o 2 R out = r o 1 + (1 + g m 1 r o 1 ) parenleftbigg r 1 bardbl 1 g m 2 bardbl r 2 bardbl r o 2 parenrightbigg (d) Looking into the base of Q 2 we see an equivalent resistance of r 2 , so we can draw the following equivalent circuit for finding R out : Q 1 R out r 2 R out = r o 1 + (1 + g m 1 r o 1 ) ( r 1 bardbl r 2 ) 5.5 (a) Looking into the base of Q 1 we see an equivalent resistance of r 1 , so we can draw the following equivalent circuit for finding R in : R in R 1 R 2 r 1 R in = R 1 + R 2 bardbl r 1 (b) Lets draw the small-signal model and apply a test source at the input. r 1 + v 1 v t + i t g m 1 v 1 R 1 i t = v 1 r 1 g m 1 v 1 v 1 = v t i t = v t r 1 + g m 1 v t i t = v t parenleftbigg g m 1 + 1 r 1 parenrightbigg R in = v t i t = 1 g m 1 bardbl r 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 2 . Thus, we can draw the following equivalent circuit for finding R in : R in Q 1 V CC 1 g m 2 bardbl r 2 R in = r 1 + (1 + 1 ) parenleftbigg 1 g m 2 bardbl r 2 parenrightbigg (d) Looking up from the emitter of Q 1 we see an equivalent resistance of 1 g m 2 bardbl r 2 , so we can draw the following equivalent circuit for finding R in : R in Q 1 1 g m 2 bardbl r 2 V CC R in = r 1 + (1 + 1 ) parenleftbigg 1 g m 2 bardbl r 2 parenrightbigg (e) We know that looking into the base of...
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This note was uploaded on 05/03/2009 for the course ECSE 2050 taught by Professor Monahella during the Spring '08 term at Rensselaer Polytechnic Institute.

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ch05sol - 5.3 (a) Looking into the base of Q 1 we see an...

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