This preview shows pages 1–9. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
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 smallsignal 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 smallsignal 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...
View
Full
Document
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.
 Spring '08
 MonaHella

Click to edit the document details