# hw6_soln - The equivalent resistance looking into the...

This preview shows pages 1–2. Sign up to view the full content.

UNIVERSITY OF CALIFORNIA AT BERKELEY College of Engineering Department of Electrical Engineering and Computer Sciences Homework #6 Solutions Due Tuesday, October 9, 2007 EE 105 Fall 2007 Prof. Liu Problem 1. Emitter Follower Design R in = r π + (1 + β ) R E = βV T I C + (1 + β ) R E 20 kΩ (1) A v = g m R E 1 + g m R E = I C R E /V T 1 + I C R E /V T 0 . 9 (2) If we pick R in = 20 kΩ and A v = 0 . 9, then we change the greater-than-or-equal-to signs into equal signs and solve the equations simultanously. This gives: I C = 1 . 312 mA R E = 178 . 4 Ω We can solve this more generally by symbolically solving these inequalities as follows: R E I C (20 kΩ) βV T (1 + β ) I C (3) R E ( I C 0 . 9 I C ) 0 . 9 V T (4) R E 9 V T I C (5) (4) comes from re-arranging (1) and (6) comes from re-arranging (2) (with (5) being an inter- mediate step from (2) to (6)). If we pick any reasonble I C , we can find an R E that fits the specifications for our amplifier using the inequalities (4) and (6) ( R E must be chosen to satisfy both inequalities). Problem 2. Emitter Follower Performance Parameters The equivalent resistance looking into the collector/base of

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: The equivalent resistance looking into the collector/base of Q 2 is 1 g m 2 b r π 2 ≈ 1 g m 2 . Thus, this is just like a common emitter ampli±er with an emitter resistor of size 1 g m 2 . A v = g m 1 /g m 2 1 + g m 1 /g m 2 R in = r π 1 + (1 + β ) 1 g m 2 R out = 1 g m 1 b 1 g m 2 Problem 3. Emitter Follower Analysis The equivalent resistance at the emitter (assuming C 2 is a short in small signal analysis) is 1 kΩ b 100 Ω. The 10 kΩ resistor at the base doesn’t a²ect the voltage gain, so we have: A v = g m (1000 b 100) 1 + g m (1000 b 100) 1 We need to fnd I C so that we can calculate g m . This is a biasing problem similar to those From homework 4. V BE = V CC − I B (10 kΩ) − I E (1 kΩ) = V CC − I C β (10 kΩ) − 1 + β β I C (1 kΩ) = V T ln p I C I S + 1 P I C = 1 . 586 mA g m = 61 mS A v = . 847 2...
View Full Document

{[ snackBarMessage ]}

### What students are saying

• 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.

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

• 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.

Dana University of Pennsylvania ‘17, Course Hero Intern

• 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.

Jill Tulane University ‘16, Course Hero Intern