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HW 7 Solutions - Due Wed EE323 Homework 7 Solution Au/2004...

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Due 11/24/2004, Wed 1 EE323 Homework 7--- Solution, Au/2004 5.79 (25 points) (a) (5 points) 2 1 1 0, 2 5 1.6 2 1.8 0, 0.7 0.7 B E C B BE E B BE I I mA I V V V V V V V V V V β α = ∞ = = = = = × = = = = = = − (b) (5 points) 4 3 0.7 ( 5) 4.3 0.7 1.9545 2.2 2.2 5 1.6 5 1.6 1.9545 1.8727 E B BE E E V V V V I mA I V I V − − = = − = = = = = = × = (c) (5 points) Similar to part (b): 6 5 0 , 0.7 E V V V V V = = = − and: 7 0.7 ( 5) 5 1.6 5 1.6 1.8727 2.2 E V I V − − = = × = (d) (5 points) For this pnp transistor, 0.7 EB E B V V V V = = . 8 8 9 0 1.2 1.2 0.7 1.9 5 5 1.9 3.1 0.9394 3.3 3.3 3.3 5 5.1 0.2091 B B E E C C I V V V V V V I mA I V I V = = = + = = = = = = = = − + = − (e) (5 points) For this pnp transistor, 0.7 EB V V = : ( ) 10 11 10 11 12 150 5 5 ( 5) 5 6.224 1.224 150 91 0.7 1.924 5 0.9321 3.3 5 5.1 0.2463 E C C V V V V V V I I mA V I V = − + − − = − + = + = + = = = = = − + = − 5.124 (30 points) (1) (5 points) 1 , 0 E C B I I I β α = ∞ = = = : 100 15 7.5 7.5 0.7 6.8 100 100 B E B BE V V V V V V V = × = = = = + Therefore 6.8 1 6.8 E C E E V V I I mA R K = = = =
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