Chapter 2 Problem Solutions 2.1 R v = Peak diode current i d (max) = R 1 K 0.6 I 0 I 0

Transcription

1 Chapter Problem Solutons. K γ.6, r f Ω For v, v.6 r + f ( 9.4) +. v v v v v v T ln and S v T ln S v v.3 8snωt (a) vs 3.33snωt Peak dode current d (max)

2 (b) P v s (max) 3.3 (c) T o π vo( avg ) vo() t dt 3.33sn dt T o π o o 3.33 π [ cos x] ( ) o π π π v avg 4.4 o (d) 5%.4 v vs γ vs ( max) v( max) + γ a. For v ( max ) 5 v S ( max) 5 + (.7 ) 6.4 N 6 N 6.6 N 6.4 N b. For v ( max) v S ( max).4 N 6 N.58 N.4 N From part (a) P vs ( max) γ ( 6.4).7 or P 5. P.4.7 or P. d or, from part (b).5 (a) vs (max) + (.7) vs( rms ) vs(rms) 9.48 (b) M M r C f C f r C C μf 6 (.3)( 5) (c) M M d, peak + π r + π 5.3, peak.33 A.6 (a) vs ( max) vs ( max) vs ( rms) vs ( rms) 8.98 M M (b) r C fc f or C 4444 μf r

3 (c) For the half-wave rectfer, max 4.58 A, max M M + 4π + 4π r 5 (.3) or.8 (a) v peak s v s (b) ( rms) M 5 C 857 μf f r.9 S O 5.4 O 5.4. S B S x x t v () t 4snωt S T Now () avg t dt T 4sn x.7 We have for x ωt x To fnd x and x, 4sn x.7 x.558 rad x π rad Then

4 x 4sn x.7 ( avg) dx π x 4 x.7 x ( cos x) xx or.9 x Ω π π x x Fracton of tme dode s conductng % % or Fracton 3.% π π Power ratng T x 4sn x.7 avg rms T π x x P dt dx ( 4) sn x (.7)( 4) sn x+ (.7) dx π x x x snx x x ( 4) (.7)( 4)( cos x) + (.7) x x x π 4 x For.9 Ω, then P 7.9 W avg. (a) 5 5Ω. v max v max or v max 5.7 S o γ S 5.7 Then vs ( rms). N N Now.8 N. N (b) M M 5 C or C 83 μf r fc fr ( 6)( 5)(.4) (c) ( max) ( 5.7).7 P P v γ or 3.7 S. For v > γ oltage across + v oltage vder v v v +

5 .3 v For >, ( γ ) a. v v kω.66 v v.43 v b. ( max) v v rms v( rms) ma ma ma P P.43 mw T T.5 (a) 4.33 A P (.33).8 W (b).33 A, (.9)(.33). A So. 57.Ω P..33 P.8 W (c).6 6.3, Ω, 4.8

6 a ma 5 5 ma 4 9Ω b. P ( 4.8) P 48 mw P 4.8 P 576 mw.7 a. 45. ma 6.3 ma ma b. 4 P ( max) 4 mw ( max) 4 ma mn max 45 4 ( mn) 5 ma kω (c) For 75Ω 57. ma 6.3 ma 3.8 ma max 4 ma mn ma 585Ω 7..8 a. From Eq. (-3) 5[ ] 5[ 5 ] ( max) 5 (.9) (.) ( max).875 A ( mn).875 A From Eq. (-9(b)) 8.8Ω b. P (.875) P.9 W P max.5 P 5 W.9 (a) As approxmaton, assume ( max) and ( mn) ( max) ( nom) + ( max) r + (.875)().375 ( mn) ( nom) + ( mn) r + (.875)().375 are the same as n problem.8.

7 b % eg % % eg.4%. % eg % ( max) ( mn) ( nom) + ( max) ( + ( mn) ) ( nom) ( max) ( mn) ( 3) nom rz nom rz.5 6 max mn. A So 6 6 Now ( max). A, ( mn).6 A 5 PS ( mn) Now mn + max 5 6 or 8 ( mn). A mn +. Then max.+.. A and PS max 6 or 8 PS ( max) Usng Fgure. a. ± 5% 5 5 PS For PS ( mn ): mn + max ma PS mn 5 Ω 5 PS b. For ( max) PS For r + + mn max 75 ma PS ( max) ( max) + ( mn) 5 max max 75 ma max mn Δ.35 c. Δ % eg % % eg 3.5% ( nom). From Equaton (.9(a)) PS ( mn) 4 6 mn + max Also r M M C fc f + rz 8. +.Ω Then r or 8.Ω

8 4 C C 99 μf r ( nom) (.)(.5) 7.95 S ( max) ( nom) A 3 For. A.33 A For A.333 A ( max) + ( max) r (.33)(.5) ( mn) + ( mn) r (.333)(.5) 8.65 Δ.4 Δ.4 % eg % eg 5.% ( nom) 8 M M r C fc fr + r Ω z Then C C.357 F (a) For v, both dodes are conductng v O For v 3, ener not n breakdown, so, v O v 3 For v > 3 ma v 3 vo v.5 At v, v o 3.5,.35 ma O () () (b) For v <, both dodes forward based v. At v, ma v 3 For v > 3,. At v,.35 ma

9 (ma).35 3 ().5 (a) K K K for v 5.7, v v 3 For v > 5.7 v , v +.7 v 5 (.7 v v ) v.7 + v v + + v(.5) v v(.5) v v v 5.7 v 5.7 v 5 v 9.4 () (b) for v 5.7 Then for v > 5.7 v v v v O.5 or 5 ().6v 3.4 For v 5, 5.58 ma

10 (ma) ().6 (a) For off, vo () Then for v vo 3.33 For v > 4.3, vo vl.7; For v, v 9.3 o O () (b) For v 4.3, 4.3 v vo o For v > 4.3, + 3 Whch yelds v.65 For v,.895 ma (ma).895 () 4.3 ().7 O For v 3,.75 A + v () +.7.5

11 b. O O 6.8 K γ.6 v 5snωt O O a. γ 3 γ.6.4 b. γ 5 γ

12 One possble example s shown. gn 4 AO AO wll tend to block the transent sgnals z wll lmt the voltage to +4 and.7. Power ratngs depends on number of pulses per second and duraton of pulse..3 O () 4 (a) O () 35 (b) 5.33 C O x a. For γ x.7 b. For γ.7 x..34 C O.35

13 O B O B B 3 7 O For Fgure P.3(a) B B 3 O.37 a ma ( 9.5) 8.93 b ma ( 9.5) 4.8 c. Same as (a) d. ( ) (.5 ) ( 9.5 ).964 ma ( 9.5) ma.38 a. b.

14 ma c. ( 9.5) (.5) ma ( 9.5) 5.8 d. ( 9.5) (.5).964 ma.48 ma a.,, 3, on ma ma ma 3 3 b. 5 and on, 3 off ( 9.5) (.5) ma.6 ma c. 5, off,, 3 on ma ma ma 3 3 d. 5, off,, 3 on ma ma ma (a) on, off, 3 on So.6 (.6) Now.6,.5 ma + + 6

15 ma ma (b) on, on, 3 off So , 6 or.833 ma 4.4 ( 5) ma ma (c) All dodes are on 4.4, ma kω 4.4 (.6) ma 5 kω.6 ( 5) 3.5 ma 3.93 kω For v small, both dodes off v O v.99v When v vo.6, turns on. So we have v.99v.6 v.66, vo.6 v.6 vo v vo vo v.6 For on + whch yelds vo v.6 When v O.6, turns on. Then.6 v 3.9 v.6 vo v vo vo vo.6 Now for v > v Whch yelds vo ; For v vo.5.4 K 3 4 K K

16 For v >. when and 4 turn off ma v kω v v for 4.65 v a. 5 k Ω, kω and on ma b. k Ω, 5 k Ω, off, on f both dodes on (a) A.7, O.4 (.7).7 ma.4 ( 5).7 ma ma (b) off, on.7 ( 5).6 ma 5+ O 5 (.6) 5 O. A off,.45

17 (a) on, off.7.93 ma 5 O (b) on, off.7.86 ma 5 5 O.46 5 ( +.7) ma K a b K K K a. 5, ode off a 7.5, b 5.5, 5 ode on b. b b a a + + a b b + + b b b ma.6.48 v, off, on.5.5 ma 5 vo (.5)( 5) vo 7.5 for v 7.5 For v > 7.5, Both and on v vo vo.5 vo + or v vo( 5.5) When v o, turns off v ( 5.5) For v >.5, v o

18 .49 a. b. 4.4, 3.8 c. 4.4, 3.8 ogc level degrades as t goes through addtonal logc gates..5 a. 5 b..6,. c..6,. ogc sgnal degrades as t goes through addtonal logc gates..5 AN O AN ma Ω. 68.7Ω PS K PS PS,.8 ma ηeφa Ph () A 3.75 cm A