Microelectronics: Circuit Analysis and Design, 4 th edition Chapter 2. Chapter (a) For. Then ( ) 06. rad 10. Also ( ) π. Now 2.

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1 . (a) >. 6 <. 6,, (.6) (b) (ii) [ sin( ωt).6] Now Chapter.6 sin ω t. ( ωt) π rad ωt rad Then ( ) 6 lso ( ) π T T ( avg) () t dt [ sin x.6]dx π π.989π.989π cos x.6x π.9π.9π ( )( ).6(.989π.9π ) π. ( avg) 89 [ ] π (iii) ( peak) sin ; id ( max ) m (iv) P. v v v D i D v vd T ln and id S v v v T ln S

2 .3 (a) S (peak) peak 6. ( ) 7 i D 6.7 (c) 6.97 sin ωt. 7 (b) ( peak) 8. 4 (d) m.7 sin( ω t).45 ( ωt) ω t % % 48.7% π π ( avg) [ 6.97 sin x.7]dx.33π ( ).9869π.9869π ( 6.97) cos x.7x π.33π.33π [( 6.97)( ).7(.9738π )] π avg 5. ( ) 6 ( avg) 5.6 (e) id ( avg).53 m.4 (a) () t 5sin ωt.7 9 5sin ωt t rad 5 ωt rad ( ω ) sin π ( ) π.776π π ( avg) [ 5sin x 9.7]dx.38π.776π π x.38π avg. ( ) π π ( 5) cos x 9.7 [( 5)( ) 9.7(.553π )].38π.968 i D ( avg).8. Ω (b) % % 7.6% 36

3 .5 (a) i( peak ) ( peak ) Ω. 5sin ωt 9. (b) () t 7 ( ω t ) 38π ; ( ωt ) 776π..776π π. ( avg) [ 5sin x 9.7]dx.38π r from Problem.4, ( avg) (.968). 956 ( avg).956 i ( avg).436 D 4.47 (c) % % 7.6% 36.6 (a) ( peak) (b) S N N Ω. C f 6 r ( )( 6)(.5) (c) ( max) (.7) γ 6667 μ F P S.7 γ S ( max) ( max) γ ( ) ( ) + ( ) v v v v + S v max 5 v max a. S N 6 N 6.6 N 6.4 N ( ) ( ) v max v max.4 b. S N 6 N.58 N.4 N ( ) ( ) P vs max 6.4. From part (a) γ 7 or P 5. P (.4).7 or, from part (b) or P..8 (a) v (max) + (.7) 3.4 v s s 3.4 rms vs(rms) 9.48 ( )

4 (b) r C f f C C C μf d ( )( )( ) (c) M M id, peak + π r ( ) + π 5.3 i, peak.33 r.9 (a) vs ( max) vs ( max) vs ( rms) vs ( rms) 8.98 r C fc f r ( 6)( 5)(.3) C 4444 μf (b) or M M id, max + 4π + 4π r 5 (.3) (c) the half-wave rectifier i D, max 4.58 or. (a) ( peak) (b) C 6 μ F fr ( 6)( 5)(.5) (c) P (a).3. 3 (b) So.3 r. 586 fc r 6 ( 6)( )( 35 ).3 6 ( 6)( )( 35 ) r.49.3 (c) C 53 μ F fr ( 6)( )(.4)

5 . (a) ( peak) 8.5( ). (b) S C max..7.3 M F f r ( 6)( )(.5) (c) P ( peak ) (.) S γ.3 (a) v s ( ) ( ) peak vs ( rms).6 5 C 857 μf f r ( 6)( 5)(.35) (b).4.5 (a) S. 8 N 3.3 N.8 (b) 4 Ω.5 r 3 % r (.3)( ). 36 C.6 F fr ( 6)( 4)(.36) (c) ( ) ( ) M + M i + D peak π π r peak 3. i D ( ) 3

6 (d) i ( avg) D (.36) ( ) r M π + M π + π M r π i D ( avg). 539 (e) P (a) 9 + (.8). 6 (b) S N N Ω. 9 C 467 μ F f 6 9. (c) i ( peak ) D (d) i ( avg) D r ( )( )( ) () 9 9 M + π M + π r (.) 9 () 9 r M π + M π + π M r π 9 9. i D ( avg). 67 (e) P ( max ) S γ.7 v > i γ oltage across + vi

7 v vi vi + oltage Divider.8 i, ( ) v > γ a. v vi kω.66 v vi.43 vi ( ) v max v( rms) v( rms) 3.4 b (a). 975 m m m Z Z ZZ P (.367)( 3.9). 43 mw 3.9 (b). 39 m m Z (.95)( 3.9) 3. 7 mw P Z

8 . (a) 4 Z.33 P (.33)( ).8 W (b).33, (.9)(.33).. 57.Ω So P (.)(.33)( ) P.8 W (c). (a) P Z ZZ 4 z ( 5.4) Z ( max ) m So 5 z m (b) m.5 So ( max ) m ( min ) m 5.4 Then ( min) Ω ( max) 4 Ω.3759 So Ω. a. b. 45. m 6.3 m m Z Z 4 PZ max 4 mw max 4 m min max 45 4 ( ) Z ( ) ( ) ( ) ( min) 5 m Z kω (c) i 75Ω 57. m 6.3 m Z 3.8 m Z ( max) 4 m ( min) m 585Ω 7.

9 .3 a. From Eq. (.3) 5[ ] 5[ 5 ] Z ( max) 5.9. ( )( ) ( )( ) Z ( max).875 Z ( min).875 i i 8.8Ω From Eq. (.8(b)) b. PZ (.875)( ) PZ.9 W P ( max) (.5)( ) P 5 W.4 (a) 5.6 Z 83. m Z (.83)( 3) PZ ZZ (.83)( 5.85). 486 W 5.6 (b) ( ) So Then m m.5 nd Z m P Z (.558)( 5.769). 3 W (c) 5.6 Z.8 m Z (.8)( 3) P Z (.8)( 5.96). 7 W 5.6 (d) ( ) So Then 9. 4 m;. 4 m m Z (.93)( 5.88). 547 W P Z

10 .5 (a) Set (b) 7.5 Z m; 7. 5 m m i 57 Ω i + (.)( ) (.)( ) Z Z ( ) (.9)( ) ( ) Then, Source eg % 4.4% 3..8 (c) k Ω, , Z 7. 7 m ( )( ) Then, oad eg %.5% So ( max) ( min) ( nom) ( ) + ( max) ( ( ) + ( min) z) ( nom) ( max) ( min) ( 3) % eg % Now Z nom Z rz nom Z r Z Z.5 6 max min. ( ) ( ) Z 6 6 ( max)., ( min).6 5 PS ( min) Z ( min) + ( max) i Z Now Z ( min). Z ( min) +. or

11 ( ) PS max Z i Z ( max ).+.. Z ( max) + ( min) Then and PS ( max) 6 8 PS ( max) 4.3 or Using Figure.9 a. PS ± 5% 5 PS 5 PS ( min ) : Z ( min) + ( max) m PS ( min) Z 5 i i Ω 5 5 ( max) ( max) 75 m ( max) b. PS i ( min) Z ( max) 75 m Z Z ZrZ (.5)( 5) ( max) (.75)( 5).5 ( min) (.5)( 5) 9.9 Δ.35 Δ % eg % % eg 3.5% ( nom) c..8 From Equation (.8(a)) PS ( min) Z 4 6 i Z ( min) + ( max) i 8.Ω or lso r C fc fr i + rz 8. +.Ω Then 4 C C 99 μf 6 ( )( )(.).9 Z Z + ZrZ Z ( nom) 8 8 Z + (.)(.5) Z 7.95 S ( max) Z ( nom) 8 i.333 i 3. Z.33 Z.333

12 ( max) + ( max) ( )( ) ( min) + ( min) ( )( ) r Z Z Z Z Z rz Δ.4 Δ.4 % eg % eg 5.% ( nom) 8 r C fc fr i + rz Ω C C.357 F ( 6)( 3.5)(.8) Then , 3 > 3, and (.5).5 3 (.5) < 6. 3, and (.5) (.5) (a) v, both diodes are conducting v v 3, i, Zener not in breakdown, so v v 3 v > 3 i m v 3 vo ( ) v.5 t v, v o 3.5, i.35 m

13 (b) v <, both diodes forward biased v i. t v, i m v 3 v > 3, i. t v, i.35 m.3 (a) 5 5 for v 5.7, v v 3 v > 5.7 v ( +.7) 5 +, 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

14 (b) i for.7 D v 5 Then for v > 5.7 v v v v.5 id i or D.6v 3.4 v 5, i D 5.58 m.33 (a) (i) B.8.,.,. (ii) (b) (i) (ii) B.8. 5,. 5,. 5 B.8. 5,. 5. 5,.8 B.,..,.34 v 3.7 3, i.75 + v i() +.7.5

15 b..35 (a) (i) B , 5. 7 (ii) (b) (i) (ii) 5. 7, B , , B , 4. 3, 4. 3 B , 5. 7, a. γ γ.6 b. γ

16 γ ne possible example is shown. will tend to block the transient signals D z will limit the voltage to +4 and.7. Power ratings depends on number of pulses per second and duration of pulse..39 (a) Square wave between +4 and. (b) Square wave between + 35 and 5. (c) Square wave between + 5 and a. γ.7 x b. γ.7. x

17 .4 Circuit similar to Figure.3(a) with B..4 n steady-state, ( sin ωt + 5).43 (i) B 5, n steady-state, sin ωt 5 (ii) B ( ) 5, n steady-state, ( sin ω 5) t.44 a. b..6 D D.94 m D ( ) D 5.6 D D.44 m D ( ) D c. Same as (a) d. ( ) (.5 ) ( 9.5 ).964 m ( ) D D D D.48 m.45 a. b. c. D D ( ) ( ) ( ) m ( ) ( ) ( ) D D m D D

18 d. ( 9.5) (.5).964 m D D D D.48 m ( 9.5) a. D, D, D3, on m D D D D 7.6 m.5 D3 D + D ( 7.6).589 D3 4.6 m b. 5 D and D on, D3 off ( 9.5) (.5) m D D D D.6 m D3 ( 9.5) (.45)( 9.5) c. 5, D off, D, D 3 on m D D 7.6 m.5 D D3 D D3 7. m 4.4 d. 5, D off, D, D 3 on m D D 3.6 m.5 D D3 D D3 3. m (a) 4.4, D. 5 k Ω m 4.4 (.6).6,.5 k Ω

19 m.6 ( 5) k Ω. (b) ssume all diodes conducting , D. 5 m 4.4 (.6).6,.5 m 4 Then m Then (c) Diode D ( 5) D 3 m.75 m D cutoff D.6, Then.6 + (.6) 9 D.6 (.)( 3) 6. 7 (d) Diode ( 5) D 3.76 m.65 m D cutoff D , D. 833 m ( 5) m 3 D (.44)( 6) 5. 7.m Then.m.48 (a) D D.5 m ( ).7.7 D.5 (b)., D 5 D D.8 k Ω + D. D + D 5 D D k Ω (c), D 5 D 5 D + D 5 D + D 5 D 4.67 m.833m D k Ω

20 .49 (a) and D on D ( 5) Then D. m.3.7 ( 5) D 6. m. (b) D cutoff, D.7 + ()( 5.5) 5 8. D 5 m, 5.7 (c), D. 5 m Then m D ( 5).7 D k Ω.5 (a) (i) 5, and D on ( +.6) D (ii) 5 ( ) (b) (i) 5, 4. 4 (ii) 5,. 6.5

21 v >. when D and D 4 turn off m v kω 4.65 ( ) v v for 4.65 v (a) ll diodes on ( 5).7 ( ) ( ) Then D. m.7.7 ( 5) D.5 m ( ) D3.5 m 4 (b) D cutoff, D 5.7 ( 5).7 ( ) ( ) ( 5) Then D. m ( ) D3.6 m 5. (c) D and D cutoff, D D 5.7 ( ) 4.3 D m ( 3.5)( 6.5) 5

22 .53 a. 5 k Ω, kω D and D on.7 ( ) D D.86 m b. k Ω, 5 k Ω, D off, D on D.7 ( ) f both diodes on (a).7,.4 (.7).7 m.4 ( 5).7 m 5 + D D.7.7 D.65 m (b) D off, D on.7 ( 5).6 m 5+ 5 (.6)( ) D off, D.55 (a) D on, D off.7 D.93 m 5

23 (b) D on, D off.7 D.86 m ( +.7) D D.349 m.57 (a) Diode is cutoff,, B 3 (b) Diode is conducting, D D D.7 5 B B B.7 B and. 4 ( ) + B B 5 B B + D 5.. So D. 8 m (c) Diode is cutoff, D.5.5 () 5. 5, ( 4). D (d) Diode is conducting, B D.7 8 B B B.7 B and. 5 ( ) 85 + B B 8 B B Then D [ 8 (.85) ]. 3 m

24 .58 v, D off, D on.5.5m 5 v.5 5 v 7.5 for 7.5 o ( )( ) v > 7.5, Both D and D on v vo vo.5 vo v vo( 5.5) or When v o, D turns off v ( )( ) o v v >.5, vo.59 (a). 5,,. 5 D D D3 D D 3 (b). 5, D on; m D ( )( ) 3 (c) 3, and D conducting, D D ( ) Then D. 7 m D.65 m 6 (d) 5, all diodes conducting So. 9 ( ).9.7 Then D. 75 m D. m D3.5 m 4.6 (a) D for < 4. 5 D m for > 4. 5 (b) D for < 9 D m for > 9

25 .6 a. b. 4.4, 3.8 c. 4.4, 3.8 ogic level degrades as it goes through additional logic gates..6 a. 5 b..6,. c..6,. ogic signal degrades as it goes through additional logic gates..63 ( ND ) ( 3 ND 4) m Ω. 68.7Ω (.75) r f. k Ω 5 r f Ω Ω

26 .67,.8 m PS PS ( )( ) ηeφ Ph 7 ()( )( ) cm