Fundamentals of Microelectronics

This preview shows page 1 - 10 out of 112 pages.

Image of page 1

Subscribe to view the full document.

Image of page 2
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 :
Image of page 3

Subscribe to view the full document.

R in Q 1 V CC r π 2 R in = r π 1 + (1 + β 1 ) r π 2
Image of page 4
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
Image of page 5

Subscribe to view the full document.

(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 )
Image of page 6
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 :
Image of page 7

Subscribe to view the full document.

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.
Image of page 8
5.6 (a) Looking into the collector of Q 1
Image of page 9

Subscribe to view the full document.

Image of page 10
  • Spring '08
  • MonaHella
  • Input impedance, Electrical resistance, Output impedance

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern