# CEAMP - Common-Emitter Amplifier Example x.y x y Function...

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Common-Emitter Amplifier Example R P xy , () . Function for calculating parallel resistors. R 1 100000 R 2 120000 R C 4300 R E 5600 R S 5000 R L 10000 V p 15 V m 15 V BE 0.65 V T 0.025 β 99 α 0.99 r x 20 r 0 50000 R 3 100 v s 1 With v s = 1, the voltage gain is equal to v o . 1

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DC Bias Solution V BB V p R 2 . V m R 1 . R 1 R 2 V BB 1.3636 = R BB R P R 1 R 2 , R BB 5.4545 10 4 = I E V BB V BE V m R BB 1 β R E I E 2.557 10 3 = V C V p α I E . R C . V C 4.1151 = V B V BE I E R E . V m V B 0.0311 = V C V B 4.1461 = Thus active mode. r e V T I E r e 9.7773 = 2
Exact AC Solution v tb v s R P R 1 R 2 , R S R P R 1 R 2 , . v tb 0.916 = R tb R P R S R P R 1 R 2 , , R tb 4.5802 10 3 = R te R P R E R 3 , R te 98.2456 = 3

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r' e R tb r x 1 β r e r' e 55.7788 = R tc R P R C R L , R tc 3.007 10 3 = r ic r 0 R P r' e R te , 1 α R te . r' e R te r ic 1.3577 10 5 = i csc v tb r' e R P R te r 0 , α R te R te r 0 . i csc 5.8835 10 3 = v o i csc R P R C R P r ic R L , , . v o 17.3084 = A v v o A v 17.3084 = This is the voltage gain. 4
Circuit for r out . r out R P R C r ic , r out 4.168 10 3 = Circuit for r in . r ib r x 1 β () r e R P R te r 0 R tc , .

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## This note was uploaded on 05/26/2011 for the course ECE 3050 taught by Professor Hollis during the Summer '08 term at Georgia Tech.

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CEAMP - Common-Emitter Amplifier Example x.y x y Function...

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