ceamp - Common-Emitter Amplifier Example - Summer 2000 x .y...

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Common-Emitter Amplifier Example - Summer 2000 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 = 0, the voltage gain is equal to v o . 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 = 1
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V C V p α I E . R C . V C 4.1151 = r e V T I E r e 9.7773 = 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 = r ie R tb r x 1 β r e r ie 55.7788 = R tc R P R C R L , R tc 3.007 10 3 = r ic r 0 R P r ie R te , 1 α R te . r ie R te r ic 1.3577 10 5 = i csc v tb r ie 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 = This is the voltage gain. r out R P R C r ic , r out 4.168 10 3 = 2
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r ib r x 1 β () r e R P R te r 0 R tc , . β R te . R tc . R tc r 0 R te r ib 1.0253 10 4 = r in R P r ib R P R 1 R 2 , , r in 8.6309 10 3 = The following solution is based on the
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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 Institute of Technology.

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ceamp - Common-Emitter Amplifier Example - Summer 2000 x .y...

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