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 ² 10 0 10 ± I 0.6 V, 20 f Vr == Ω γ () 0 0 For 10 V, 10 0.6 1 9.4 10 . 0 2 9.22 I f R vv Rr v ⎛⎞ ⎜⎟ + ⎝⎠ = + = ± 0 9.22 0.6 10 2.2 R D ± I ± 0 ³ ± D ² 0 0 0 0 ln and ln ID D DT D S IT S vvv v i vV i I R v vvV IR =− 2.3 (a) 80sin 13.33sin 6 s t vt ω Peak diode current 13.33 (max) d i R =
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(b) (max) 13.3 V s PIV v == (c) () [] 11 13 .33s in 2 13.33 13.33 13.33 cos 1 1 22 4.24 o T oo o o o v avg v t dt dt T x va v g V π ππ × =− = − = ⎡⎤ ⎣⎦ = ∫∫ (d) 50% 2.4 ()() 00 2m a x m a x 2 SS vv V v v V γγ =+ a. For ( ) 0 max 25 V max 25 2 0.7 = 26.4 V S vv = 160 6.06 26.4 NN = = b. For 0 max 100 V max 101.4 V S = = 160 1.58 101.4 = = From part (a) a x 2 2 6 . 40 . 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) ( m a x )1 ( 0 . 7 3 . 4 V 13.4 rms (rms) 9.48 V 2 s ss v = = = (b) ( ) 2 12 2222 F 260 0 .3 150 MM r Cr VV VC fR fVR CC μ = = = = (c) 2 , peak 1 212 12 1 150 0.3 , peak 2.33 A d r d i RV i ⎢⎥ = 2.6 (a) max 12 0.7 12.7 max rms rms 8.98 2 S S vV v V =+ = = = (b) 12 60 150 0.3 r r fRC fRV = or 4444 CF =
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(c) For the half-wave rectifier () , max 12 12 14 21 5 0 2 0 . 3 MM D r VV i RV ππ ⎛⎞ =+ = + ⎜⎟ ⎝⎠ or 4.58 D iA = 2.8 (a) ( ) peak 15 2 0.7 16.4 V 16.4 rms 11.6 V 2 s s v v = == (b) ( ) 15 2857 F 2 2 60 125 0.35 M r V C fRV μ = 2.9 ± S ± O ± ± O ² 26 0.6 ² 0.6 ² 26 25.4 ² 25.4 0 0 2.10 V B ³ 12 V ² ± i D x 1 x 2 ´ t R i D ± ² ± S ³ ± S 24sin S vt t ω = Now 0 1 T DD ia v g itd t T = We have for 12 x tx ≤≤ 24sin 12.7 D x i R = To find x 1 and x 2 , 1 24sin 12.7 x = 1 2 0.558 0.558 2.584 xr a d x rad π = =− = Then
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() 2 1 2 2 1 1 1 24sin 12.7 2 2 1 24 1 12.7 6.482 4.095 cos or 2 1.19 22 x D x x x x x x ia v g d x R xx R RR R R ⎡⎤ == ⎢⎥ ⎣⎦ ⎛⎞ ⎛ ⎞ =− = ⎜⎟ ⎜ ⎟ ⎝⎠ ⎝ ⎠ π ππ 21 2.584 0.558 Fraction of time diode is conducting 100% 100% = × or Fraction = 32.2% Power rating () ( ) ( ) ( )() 2 1 2 1 2 2 2 1 1 1 2 0 2 2 2 24sin 12.7 2 1 24 sin 2 12.7 24 sin 12.7 2 1s i n 2 24 2 12.7 24 cos 12.7 4 x T avg rms D x x x x x x x x x x PR i i d t d x TR d x R x x R =⋅ = = + + ∫∫ For 1.19 , R then 17.9 avg PW = 2.11 (a) () () 15 150 0.1 max max 15 0.7 or max 15.7 So S R vv V v V γ ==Ω =+ = + = Then 15.7 11.1 2 S vr m s V Now 11 120 10.8 11.1 NN = = (b) ( ) 15 2 2 2 60 150 0.4 MM r r VV VC fRC fRV = or 2083 CF μ = (c) 2 max 2 15.7 0.7 S PIV v V = or 30.7 PIV V = 2.12 R 1 R 2 R L D 2 ± 0 ± ² ± i R 1 R 2 R L ± 0 ± ² ± i For 0 i v > 0 V = Voltage across 1 L i RRv += Voltage Divider 0 1 1 2 L ii L R v +
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± 0 20 2.13 For () 0, 0 i vV γ >= ± ² R 1 R L R 2 ± i ± 0 a. 2 0 21 2 0 || || || 2.2 || 6.8 1.66 k 1.66 0.43 1.66 2.2 L i L L ii RR vv RR R v ⎛⎞ = ⎜⎟ + ⎝⎠ == + 4.3 ± 0 b. 0 00 max rms rms 3.04 2 v V = = 2.14 ( ) 2 3.9 0.975 mA 4 20 3.9 1.3417 mA 12 1.3417 0.975 0.367 mA 0.367 3.9 1.43 mW L R ZZ TZ Z T II I PIV P = = =− = =⋅= = 2.15 (a) ( ) 40 12 0.233 A 120 0.233 12 2.8 W Z I P (b) I R = 0.233 A, I L = (0.9)(0.233) = 0.21 A So 12 0.21 57.1 L L R R = (c) ( ) 0.1 0.233 12 0.28 W PP = = 2.16 ± R L I I I Z I L R i V Z V I V 0 ² 6.3 V, 12 , 4.8 Ii Z VR V Ω =
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a.
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This homework help was uploaded on 02/10/2008 for the course ECE 138 taught by Professor N/a during the Spring '08 term at CSU Fresno.

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

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