hw6_soln

hw6_soln - The equivalent resistance looking into the...

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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 ±nd an R E that ±ts the speci±cations for our ampli±er using the inequalities (4) and (6) ( R E must be chosen to satisfy both inequalities). Problem 2. Emitter Follower Performance Parameters
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Unformatted text preview: The equivalent resistance looking into the collector/base of Q 2 is 1 g m 2 b r 2 1 g m 2 . Thus, this is just like a common emitter amplier 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 1 + (1 + ) 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 k b 100 . The 10 k resistor at the base doesnt aect 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 k) I E (1 k) = V CC I C (10 k) 1 + I C (1 k) = V T ln p I C I S + 1 P I C = 1 . 586 mA g m = 61 mS A v = . 847 2...
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hw6_soln - The equivalent resistance looking into the...

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