# q03 - For the transistor assume β = 100 V A = 75 and V T =...

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ECE3050Ana logE lectron icsQu iz3 September 9, 2009 Professor Leach Last Name: First Name: Instructions. Print your name in the spaces above. Place a box around any answer. Honor Code Statement: I have neither given nor received help on this quiz. Initials 1o f2 . For V + =+24V , R 1 =1M , R 2 =1 . 5M , K =4 × 10 4 S ,and V TO =1 . 5V ,so lvefor R S for I D =2 . 5mA . Reference equation: i D = K ( v GS V TO ) 2 . V p 24 R 1 1000000 R 2 1500000 K4 1 0 4 . V TO 1.5 I D 0.0025 V GG V p R 2 R 1 R 2 . R GG R p2 R 1 R 2 , V GG 14.4 = R GG 6 10 5 = V GS I D K V TO V GS 4 = V GG V GS I D R S . R S V GG V GS I D R S 4.16 10 3 = 2 of 2. The circuit on the left is the signal circuit for a CE ampli f er. The two circuits on the right show the hybrid- π model for the circuit. For R B =2 . 2k , R C =10k , I C =1 . 5mA ,and V CE =10V , solve for the voltage

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Unformatted text preview: . For the transistor, assume β = 100 , V A = 75 , and V T = 25mV . Reference equations: i c = g m v π = βi b = αi e , r π = V T /I B , g m = I C /V T , I C = βI B , r = ( V A + V CE ) /I C . 1 R B 2200 R C 10000 I C 0.0015 V CE 10 β 100 V T 0.025 I B I C β I B 1.5 10 5 = g m I C V T g m 0.06 = V A 75 r π V T I B r π 1.667 10 3 = r V A V CE I C r 5.667 10 4 = v s 1 i' c v s r π R B r π . g m . i' c 0.026 = v o1 i' c R p2 R C r , . v o1 219.828 = i b i' c β i b 2.586 10 4 = v o2 β i b . R p2 R C r , . v o2 219.828 = 2...
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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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q03 - For the transistor assume β = 100 V A = 75 and V T =...

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