su10q02

# su10q02 - the equation for the emitter voltage v e as a...

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ECE3050Ana logE lectron icsQu iz2 June 2, 2010 Professor Leach Last Name: First Name: Instructions. Print and sign your name in the spaces above. Place a box around answers when appropriate. Honor Code Statement: I have neither given nor received help on this quiz. Initials 1o f2 . G iven : R C =12k , R B =82k , R E =1 . 5k , V + =15V , β =99 , V BE =0 . 65V ,and I C = βI B = αI E (a) Write the equations for V BB and R BB .Whyi s R B not a part of the equation for V BB ? (b) Write the equations for V CC and R CC . (c) Draw the bias equivalent circuit and write the loop equation for I C . (d) Use the equation found above to solve numerically for I C . V BB = V + I C R C R BB = R B + R C R BB isnotparto f V BB because you set I B =0 to solve for V BB . V CC = V + I B R C R CC = R C V BB V EE = I B R BB + V BE + I E R EE = V + I C R C = I C β ( R B + R C )+ V BE + I C α R E I C = V + V BE R C + R B

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2o f2 . (a)Thehybr
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Unformatted text preview: the equation for the emitter voltage v e as a function of v tb , i b , and any appropriate resistors. (b) If i is neglected and i c = g m v π = βi b = αi e , solve the equation obtained above for v e as a function of v tb , i e , and any appropriate resistors. (c) Use the equation obtained above to obtain and draw the Thévenin equivalent circuit seen looking into the emitter. v e = v tb − i b ( R tb + r π ) = v tb − i e 1 + β ( R tb + r π ) = v tb − ( i e − i ) R tb + r π 1 + β ' v tb − i e R tb + r π 1 + β The circuit is a voltage source v tb in series with a resistor ( R tb + r π ) / (1 + β ) . 2...
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su10q02 - the equation for the emitter voltage v e as a...

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