# ccamp - Common-Collector Amplifier Example - Summer 2000 x...

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Common-Collector Amplifier Example - Summer 2000 R P xy , () . Function for calculating parallel resistors. R 1 100000 R 2 120000 R C 0 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 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 1.364 = R BB R P R 1 R 2 , R BB 5.455 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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r e V T I E r e 9.777 = AC Solutions This first solution uses the equations involving R tc , even though R tc = 0. It is based on the Thevenin emitter circuit which have v eoc in series with r ieo . 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 4.58 10 3 = R te R P R E R L , R te 3.59 10 3 = r ie R tb r x 1 β r e r ie 55.779 = R tc R C R tc 0 = v eoc v tb r 0 R tc 1 β r ie r 0 R tc 1 β . v eoc 0.915 = r ieo r ie r 0 R tc r ie r 0 R tc 1 β . r ieo 55.717 = v o v eoc R P R E R L , r ieo R P R E R L , . v o 0.901 = This is the voltage gain. r out R P R E r ieo , r out 55.168 = 2
r out R P r ieo R E , r out 55.168 = r ib r x 1 β () r e R P R te R tc r 0 , . β R tc . R te . R tc r 0 R te r ib 3.359 10 5 = r in R P r ib R P R 1 R 2 , , r in 4.693 10 4 = The following solution is based on the emitter equivalent circuit. It is the preferred solution when R tc = 0. Note that this is an exact solution, where
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## ccamp - Common-Collector Amplifier Example - Summer 2000 x...

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