Ch02s - Chapter 2 Problem Solutions 2.1 I 0 R 1K I 10 0 10...

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Chapter 2 Problem Solutions 2.1 R 1 K I 0 10 0 10 I 0.6 V, 20 f V r = = Ω γ ( ) ( ) 0 0 For 10 V, 10 0.6 1 9.4 1 0.02 9.22 I f R v v R r v = = + = + = 0 9.22 0.6 10 2.2 R D I 0 D 0 0 0 0 ln and ln I D D D T D S I T S v v v v i v V i I R v v v V I R = = = = 2.3 (a) 80sin 13.33sin 6 s t v t ω ω = = Peak diode current 13.33 (max) d i R =
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(b) (max) 13.3 V s PIV v = = (c) ( ) [ ] ( ) ( ) 1 1 ( ) 13.33sin 2 13.33 13.33 13.33 cos 1 1 2 2 4.24 o T o o o o o o o v avg v t dt dt T x v avg V π π π π π π = = × = = − − − = = (d) 50% 2.4 ( ) ( ) 0 0 2 max max 2 S S v v V v v V γ γ = = + a. For ( ) ( ) ( ) 0 max 25 V max 25 2 0.7 = 26.4 V S v v = = + 1 1 2 2 160 6.06 26.4 N N N N = = b. For ( ) ( ) 0 max 100 V max 101.4 V S v v = = 1 1 2 2 160 1.58 101.4 N N N N = = From part (a) ( ) ( ) 2 max 2 26.4 0.7 S PIV v V γ = = or 52.1 PIV V = or, from part (b) ( ) 2 101.4 0.7 PIV = or 202.1 PIV V = 2.5 (a) ( ) (max) 12 2(0.7) 13.4 V 13.4 rms (rms) 9.48 V 2 s s s v v v = + = = = (b) ( )( )( ) 2 2 12 2222 F 2 60 0.3 150 M M r C r V V V C f R f V R C C μ = = = = (c) ( ) 2 , peak 1 2 12 12 1 150 0.3 , peak 2.33 A M M d r d V V i R V i π π = + = + = 2.6 (a) ( ) ( ) ( ) ( ) max 12 0.7 12.7 max rms rms 8.98 2 S S S S v V v v v V =