# Ch17s - Chapter 17 Problem Solutions 17.1 a vI =-1.5 Vi Q1...

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Chapter 17 Problem Solutions 17.1 a. 1 2 1.5 V off, on I i v Q Q = − ( )( ) 1 01 2 02 2 0.7 ( 3.5) 0.56 mA 5 0 3.5 V 3.5 3.5 0.56 2 E E C C E E C i i i v i i v i R − − = = = = = = = or 02 2.38 V v = b. 1 2 1.0 V on, off I i v Q Q = ( ) ( ) 2 02 1 0.7 3.5 0.76 mA 5 0 3.5 V E E C i i i v − − = = = = c. logic 0 at 02 v (low level) 2.38 V = Then 01 1 2.38 3.5 (0.76) C v R = = or 1 1.47 k C R = Ω 17.2 (a) 2 2 2 3 0 0.5 6 K C Q C C i I R R = = = = (b) 1 1 1 3 1 0.5 4 K C Q C C i I R R = = = = (c) 1 1 1 2 2 1 1 2 exp exp exp 1 1 exp BE S T C Q BE BE S T T BE BE T I BE BE V I V i I V V I V V V V V v V V = + = + = So 1 1 1 exp 0.1 1 0.2 0.5 1 exp C Q I T I T i I v V v V = + = = +

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( ) ( ) 1 exp 1 4 0.2 0.026 ln (4) 0.0360 V I T I I v V v v = = = = − 17.3 (a) 0.5 V, I v = 1 Q on, 2 Q off 02 3 V v = = 01 3 (1)(0.5) 2.5 V v = = (b) 0.5 V I v = − 1 Q off, 2 Q on 01 3 V v = 02 3 (1)(0.5) 2.5 V v = = 17.4 (a) 2 on, 1.2 0.7 1.9 E Q v V = − = − ( ) ( )( ) 2 2 2 2 2 2 1.9 5.2 1.32 2.5 1 1.32 0.758 E C C C C C i i mA v V i R R R k − − = = = = − = − = − = Ω (b) 1 on, 0.7 0.7 1.40 E Q v V = − = − ( ) ( )( ) 1 1 1 1 1 1 1.4 5.2 1.52 2.5 1 1.52 0.658 E C C C C C i i mA v V i R R R k − − = = = = − = − = − = Ω (c) 1 2 For 0.7 , on, off in v V Q Q = − 1 0.70 O v V = − 2 2 1 0.7 1.7 O O v v V = − − = − For 1 2 1.7 , off, on in v V Q Q = − 2 0.7 O v V = − 1 1 1 0.7 1.7 O O v v V = − − = − (d) (i) For 0.7 , 1.52 in E v V i mA = − = ( ) ( ) ( )( ) ( )( ) 4 3 4 3 1.7 5.2 1.17 3 0.7 5.2 1.5 3 5.2 1.52 1.17 1.5 5.2 C C E C C i mA i mA P i i i − − = = − − = = = + + = + + or 21.8 P mW = (ii) For 1.7 , 1.32 in E v V i mA = − = ( ) ( ) ( )( ) 4 3 0.7 5.2 1.5 3 1.7 5.2 1.17 3 1.32 1.5 1.17 5.2 C C i mA i mA P − − = = − − = = = + + or 20.7 P mW =