Electronic Devices and Circuit Theory

Instructor s Resource Manual to accompany Electronic Devices an Circuit Theory Tenth Eition Robert L. Boylesta Louis Nashelsky Upper Sale River, New Jersey Columbus, Ohio

Copyright 2009 by Pearson Eucation, Inc., Upper Sale River, New Jersey 07458. Pearson Prentice Hall. All rights reserve. Printe in the Unite States of America. This publication is protecte by Copyright an permission shoul be obtaine from the publisher prior to any prohibite reprouction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recoring, or likewise. For information regaring permission(s), write to: Rights an Permissions Department. Pearson Prentice Hall is a traemark of Pearson Eucation, Inc. Pearson is a registere traemark of Pearson plc Prentice Hall is a registere traemark of Pearson Eucation, Inc. Instructors of classes using Boylesta/Nashelsky, Electronic Devices an Circuit Theory, 10 th eition, may reprouce material from the instructor s text solutions manual for classroom use. 10 9 8 7 6 5 4 3 2 1 ISBN-13: 978-0-13-503865-9 ISBN-10: 0-13-503865-0

Contents Solutions to Problems in Text 1 Solutions for Laboratory Manual 185 iii

Chapter 1 1. Copper has 20 orbiting electrons with only one electron in the outermost shell. The fact that the outermost shell with its 29 th electron is incomplete (subshell can contain 2 electrons) an istant from the nucleus reveals that this electron is loosely boun to its parent atom. The application of an external electric fiel of the correct polarity can easily raw this loosely boun electron from its atomic structure for conuction. Both intrinsic silicon an germanium have complete outer shells ue to the sharing (covalent boning) of electrons between atoms. Electrons that are part of a complete shell structure require increase levels of applie attractive forces to be remove from their parent atom. 2. Intrinsic material: an intrinsic semiconuctor is one that has been refine to be as pure as physically possible. That is, one with the fewest possible number of impurities. 3. Negative temperature coefficient: materials with negative temperature coefficients have ecreasing resistance levels as the temperature increases. Covalent boning: covalent boning is the sharing of electrons between neighboring atoms to form complete outermost shells an a more stable lattice structure. 4. W Q (6 C)(3 ) 18 J 5. 48 e 48(1.6 10 19 J) 76.8 10 19 J Q W 19 76.8 10 J 6.40 10 19 C 12 6.4 10 19 C is the charge associate with 4 electrons. 6. GaP Gallium Phosphie E g 2.24 e ZnS Zinc Sulfie E g 3.67 e 7. An n-type semiconuctor material has an excess of electrons for conuction establishe by oping an intrinsic material with onor atoms having more valence electrons than neee to establish the covalent boning. The majority carrier is the electron while the minority carrier is the hole. A p-type semiconuctor material is forme by oping an intrinsic material with acceptor atoms having an insufficient number of electrons in the valence shell to complete the covalent boning thereby creating a hole in the covalent structure. The majority carrier is the hole while the minority carrier is the electron. 8. A onor atom has five electrons in its outermost valence shell while an acceptor atom has only 3 electrons in the valence shell. 9. Majority carriers are those carriers of a material that far excee the number of any other carriers in the material. Minority carriers are those carriers of a material that are less in number than any other carrier of the material. 1

10. Same basic appearance as Fig. 1.7 since arsenic also has 5 valence electrons (pentavalent). 11. Same basic appearance as Fig. 1.9 since boron also has 3 valence electrons (trivalent). 12. 13. 14. For forwar bias, the positive potential is applie to the p-type material an the negative potential to the n-type material. 15. T K 20 + 273 293 k 11,600/n 11,600/2 (low value of D ) 5800 (5800)(0.6) kd TK I D I s e 1 50 10 9 293 e 1 50 10 9 (e 11.877 1) 7.197 ma 16. k 11,600/n 11,600/2 5800 (n 2 for D 0.6 ) T K T C + 273 100 + 273 373 (5800)(0.6 ) k / T K 373 9.33 e e e 11.27 10 3 I I e 5 μa(11.27 10 3 1) 56.35 ma / ( k T K s 1) 17. (a) T K 20 + 273 293 k 11,600/n 11,600/2 5800 kd (5800)( 10 ) T K 293 I D I s e 1 0.1μA e 1 0.1 10 6 (e 197.95 1) 0.1 10 6 (1.07 10 86 1) 0.1 10 6 0.1μA I D I s 0.1 μa (b) The result is expecte since the ioe current uner reverse-bias conitions shoul equal the saturation value. 18. (a) x y e x 0 1 1 2.7182 2 7.389 3 20.086 4 54.6 5 148.4 (b) y e 0 1 (c) For 0, e 0 1 an I I s (1 1) 0 ma 2

19. T 20 C: I s 0.1 μa T 30 C: I s 2(0.1 μa) 0.2 μa (Doubles every 10 C rise in temperature) T 40 C: I s 2(0.2 μa) 0.4 μa T 50 C: I s 2(0.4 μa) 0.8 μa T 60 C: I s 2(0.8 μa) 1.6 μa 1.6 μa: 0.1 μa 16:1 increase ue to rise in temperature of 40 C. 20. For most applications the silicon ioe is the evice of choice ue to its higher temperature capability. Ge typically has a working limit of about 85 egrees centigrae while Si can be use at temperatures approaching 200 egrees centigrae. Silicon ioes also have a higher current hanling capability. Germanium ioes are the better evice for some RF small signal applications, where the smaller threshol voltage may prove avantageous. 21. From 1.19: F @ 10 ma I s 75 C 25 C 125 C 1.1 0.85 0.6 0.01 pa 1 pa 1.05 μa F ecrease with increase in temperature 1.1 : 0.6 1.83:1 I s increase with increase in temperature 1.05 μa: 0.01 pa 105 10 3 :1 22. An ieal evice or system is one that has the characteristics we woul prefer to have when using a evice or system in a practical application. Usually, however, technology only permits a close replica of the esire characteristics. The ieal characteristics provie an excellent basis for comparison with the actual evice characteristics permitting an estimate of how well the evice or system will perform. On occasion, the ieal evice or system can be assume to obtain a goo estimate of the overall response of the esign. When assuming an ieal evice or system there is no regar for component or manufacturing tolerances or any variation from evice to evice of a particular lot. 23. In the forwar-bias region the 0 rop across the ioe at any level of current results in a resistance level of zero ohms the on state conuction is establishe. In the reverse-bias region the zero current level at any reverse-bias voltage assures a very high resistance level the open circuit or off state conuction is interrupte. 24. The most important ifference between the characteristics of a ioe an a simple switch is that the switch, being mechanical, is capable of conucting current in either irection while the ioe only allows charge to flow through the element in one irection (specifically the irection efine by the arrow of the symbol using conventional current flow). 25. D 0.66, I D 2 ma D 0.65 R DC 325 Ω I 2 ma D 3

26. At I D 15 ma, D 0.82 D 0.82 R DC 54.67 Ω I D 15 ma As the forwar ioe current increases, the static resistance ecreases. 27. D 10, I D I s 0.1 μa D 10 R DC I D 0.1 μa 100 MΩ D 30, I D I s 0.1 μa D 30 R DC 300 MΩ I 0.1μA D As the reverse voltage increases, the reverse resistance increases irectly (since the ioe leakage current remains constant). 28. (a) r (b) r Δ ΔI 0.79 0.76 0.03 15 ma 5 ma 10 ma 26 m 26 m 2.6 Ω 10 ma I D 3 Ω (c) quite close 29. I D 10 ma, D 0.76 D 0.76 R DC 76 Ω I 10 ma r Δ ΔI D R DC >> r 0.79 0.76 0.03 15 ma 5 ma 10 ma 3 Ω 30. I D 1 ma, r I D 15 ma, r Δ ΔI Δ ΔI 0.72 0.61 55 Ω 2 ma 0 ma 0.8 0.78 2 Ω 20 ma 10 ma 26 m 31. I D 1 ma, r 2 2(26 Ω) 52 Ω vs 55 Ω (#30) I D 26 m 26 m I D 15 ma, r 1.73 Ω vs 2 Ω (#30) 15 ma I D 32. r av Δ ΔI 0.9 0.6 13.5 ma 1.2 ma 24.4 Ω 4

33. r Δ ΔI 0.8 0.7 0.09 22.5 Ω 7 ma 3 ma 4 ma (relatively close to average value of 24.4 Ω (#32)) 34. r av Δ ΔI 0.9 0.7 0.2 14 ma 0 ma 14 ma 14.29 Ω 35. Using the best approximation to the curve beyon D 0.7 : Δ 0.8 0.7 0.1 r av 4 Ω ΔI 25 ma 0 ma 25 ma 36. (a) R 25 : C T 0.75 pf R 10 : C T 1.25 pf ΔC Δ T R 1.25 pf 0.75 pf 0.5 pf 10 25 15 0.033 pf/ (b) R 10 : C T 1.25 pf R 1 : C T 3 pf ΔC Δ T R 1.25 pf 3 pf 1.75 pf 10 1 9 0.194 pf/ (c) 0.194 pf/: 0.033 pf/ 5.88:1 6:1 Increase sensitivity near D 0 37. From Fig. 1.33 D 0, C D 3.3 pf D 0.25, C D 9 pf 38. The transition capacitance is ue to the epletion region acting like a ielectric in the reversebias region, while the iffusion capacitance is etermine by the rate of charge injection into the region just outsie the epletion bounaries of a forwar-biase evice. Both capacitances are present in both the reverse- an forwar-bias irections, but the transition capacitance is the ominant effect for reverse-biase ioes an the iffusion capacitance is the ominant effect for forwar-biase conitions. 5

39. D 0.2, C D 7.3 pf 1 1 X C 3.64 kω 2π fc 2 π(6 MHz)(7.3 pf) D 20, C T 0.9 pf 1 1 X C 29.47 kω 2π fc 2 π(6 MHz)(0.9 pf) 40. I f 10 10 kω 1 ma t s + t t t rr 9 ns t s + 2t s 9 ns t s 3 ns t t 2t s 6 ns 41. 42. As the magnitue of the reverse-bias potential increases, the capacitance rops rapily from a level of about 5 pf with no bias. For reverse-bias potentials in excess of 10 the capacitance levels off at about 1.5 pf. 43. At D 25, I D 0.2 na an at D 100, I D 0.45 na. Although the change in I R is more than 100%, the level of I R an the resulting change is relatively small for most applications. 44. Log scale: T A 25 C, I R 0.5 na T A 100 C, I R 60 na The change is significant. 60 na: 0.5 na 120:1 Yes, at 95 C I R woul increase to 64 na starting with 0.5 na (at 25 C) (an ouble the level every 10 C). 6

45. I F 0.1 ma: r 700 Ω I F 1.5 ma: r 70 Ω I F 20 ma: r 6 Ω The results support the fact that the ynamic or ac resistance ecreases rapily with increasing current levels. 46. T 25 C: P max 500 mw T 100 C: P max 260 mw P max F I F Pmax 500 mw I F 714.29 ma F 0.7 Pmax 260 mw I F 371.43 ma 0.7 F 714.29 ma: 371.43 ma 1.92:1 2:1 47. Using the bottom right graph of Fig. 1.37: I F 500 ma @ T 25 C At I F 250 ma, T 104 C 48. ΔZ 49. T C +0.072% 100% Z ( T1 T0) 0.75 0.072 100 10 ( T1 25) 7.5 0.072 T 25 1 7.5 T 1 25 0.072 104.17 T 1 104.17 + 25 129.17 50. T C ΔZ 100% ( T T ) Z 1 0 (5 4.8 ) 100% 0.053%/ C 5 (100 25 ) 7

51. (20 6.8 ) 100% 77% (24 6.8 ) The 20 Zener is therefore 77% of the istance between 6.8 an 24 measure from the 6.8 characteristic. At I Z 0.1 ma, T C 0.06%/ C (5 3.6 ) 100% 44% (6.8 3.6 ) The 5 Zener is therefore 44% of the istance between 3.6 an 6.8 measure from the 3.6 characteristic. At I Z 0.1 ma, T C 0.025%/ C 52. 53. 24 Zener: 0.2 ma: 400 Ω 1 ma: 95 Ω 10 ma: 13 Ω The steeper the curve (higher I/) the less the ynamic resistance. 54. T 2.0, which is consierably higher than germanium ( 0.3 ) or silicon ( 0.7 ). For germanium it is a 6.7:1 ratio, an for silicon a 2.86:1 ratio. 55. Fig. 1.53 (f) I F 13 ma Fig. 1.53 (e) F 2.3 56. (a) Relative efficiency @ 5 ma 0.82 @ 10 ma 1.02 1.02 0.82 100% 24.4% increase 0.82 ratio: 1.02 0.82 1.24 (b) Relative efficiency @ 30 ma 1.38 @ 35 ma 1.42 1.42 1.38 100% 2.9% increase 1.38 ratio: 1.42 1.38 1.03 (c) For currents greater than about 30 ma the percent increase is significantly less than for increasing currents of lesser magnitue. 8

57. (a) 0.75 3.0 0.25 From Fig. 1.53 (i) 75 (b) 0.5 40 58. For the high-efficiency re unit of Fig. 1.53: 0.2 ma 20 ma C x 20 ma x 100 C 0.2 ma/ C 9

Chapter 2 1. The loa line will intersect at I D E 8 R 330 Ω 24.24 ma an D 8. (a) (b) (c) 0.92 I 21.5 ma R E 8 0.92 7.08 0.7 I 22.2 ma R E 8 0.7 7.3 0 I 24.24 ma R E 8 0 8 For (a) an (b), levels of an I are quite close. Levels of part (c) are reasonably close but as expecte ue to level of applie voltage E. E 5 2. (a) I D R 2.27 ma 2.2 kω The loa line extens from I D 2.27 ma to D 5. 0.7, I 2 ma E 5 (b) I D R 10.64 ma 0.47 kω The loa line extens from I D 10.64 ma to D 5. 0.8, I 9 ma E 5 (c) I D R 27.78 ma 0.18 kω The loa line extens from I D 27.78 ma to D 5. 0.93, I 22.5 ma The resulting values of 3. Loa line through 11.25 ma. I D 11.25 ma with R are quite close, while I extens from 2 ma to 22.5 ma. I 10 ma of characteristics an D 7 will intersect I D axis as E 7 R R 7 11.25 ma 0.62 kω 10

E 30 0.7 4. (a) I D I R D 13.32 ma R 2.2 kω D 0.7, R E D 30 0.7 29.3 E 30 0 (b) I D D R 2.2 kω D 0, R 30 13.64 ma Yes, since E T the levels of I D an R are quite close. 5. (a) I 0 ma; ioe reverse-biase. (b) 20Ω 20 0.7 19.3 (Kirchhoff s voltage law) I 19.3 20 Ω 0.965 A (c) I 10 1 A; center branch open 10 Ω 6. (a) Dioe forwar-biase, Kirchhoff s voltage law (CW): 5 + 0.7 o 0 o 4.3 4.3 I R I D o R 1.955 ma 2.2 kω (b) Dioe forwar-biase, 8 0.7 I D 1.24 ma 1.2 kω+ 4.7 kω o 4.7 kω + D (1.24 ma)(4.7 kω) + 0.7 6.53 7. (a) o 2 k Ω(20 0.7 0.3) 2 kω+ 2 kω 1 2 (20 1 ) 1 (19 ) 9.5 2 10 + 2 0.7 ) 11.3 (b) I 1.915 ma 1.2 kω+ 4.7 kω 5.9 kω IR (1.915 ma)(4.7 kω) 9 o 2 9 2 7 11

8. (a) Determine the Thevenin equivalent circuit for the 10 ma source an 2.2 kω resistor. E Th IR (10 ma)(2.2 kω) 22 R Th 2. 2kΩ (b) Dioe forwar-biase I D 20 + 5 0.7 2.65 ma 6.8 kω Kirchhoff s voltage law (CW): + o 0.7 + 5 0 o 4.3 Dioe forwar-biase 22 0.7 I D 6.26 ma 2.2 kω + 1.2 kω o I D (1.2 kω) (6.26 ma)(1.2 kω) 7.51 9. (a) (b) I o 1 12 0.7 11.3 o 2 0.3 o 1 10 + 0.3 + 0.7 9 10 0.7 0.3 9 1.2 kω+ 3.3 kω 4.5 kω 2 ma, o (2 ma)(3.3 kω) 6.6 2 10. (a) Both ioes forwar-biase I R 20 0.7 4.106 ma 4.7 kω Assuming ientical ioes: I 4.106 ma I D R 2.05 ma 2 2 o 20 0.7 19.3 (b) Right ioe forwar-biase: I D 15 + 5 0.7 8.77 ma 2.2 kω o 15 0.7 14.3 11. (a) Ge ioe on preventing Si ioe from turning on : 10 0.3 9.7 I 9.7 ma 1 kω 1 kω 16 0.7 0.7 12 2.6 (b) I 0.553 ma 4.7 kω 4.7 kω o 12 + (0.553 ma)(4.7 kω) 14.6 12

12. Both ioes forwar-biase: 0.7, 0.3 o 1 o 2 20 0.7 I 1 kω 19.3 19.3 ma 1 kω 1 kω I 0.47 kω 0.7 0.3 0.851 ma 0.47 kω I(Si ioe) I 1 kω I 0.47 kω 19.3 ma 0.851 ma 18.45 ma 13. For the parallel Si 2 kω branches a Thevenin equivalent will result (for on ioes) in a single series branch of 0.7 an 1 kω resistor as shown below: I 2 kω 6.2 3.1 ma 2 kω I2 k 3.1 ma I D Ω 1.55 ma 2 2 14. Both ioes off. The threshol voltage of 0.7 is unavailable for either ioe. o 0 15. Both ioes on, o 10 0.7 9.3 16. Both ioes on. o 0.7 17. Both ioes off, o 10 18. The Si ioe with 5 at the cathoe is on while the other is off. The result is o 5 + 0.7 4.3 19. 0 at one terminal is more positive than 5 at the other input terminal. Therefore assume lower ioe on an upper ioe off. The result: o 0 0.7 0.7 The result supports the above assumptions. 20. Since all the system terminals are at 10 the require ifference of 0.7 across either ioe cannot be establishe. Therefore, both ioes are off an o +10 as establishe by 10 supply connecte to 1 kω resistor. 13

21. The Si ioe requires more terminal voltage than the Ge ioe to turn on. Therefore, with 5 at both input terminals, assume Si ioe off an Ge ioe on. The result: o 5 0.3 4.7 The result supports the above assumptions. c 2 22. c 0.318 m m 6.28 0.318 0.318 I m m 6.28 R 2.2 kω 2.85 ma 23. Using c 0.318( m T ) 2 0.318( m 0.7 ) Solving: m 6.98 10:1 for m : T c 2 24. m 6.28 0.318 0.318 I L max 6.28 6.8 kω 0.924 ma 14

I max (2.2 kω) 6.28 2.855 ma 2.2 kω I I + I max (2.2 kω) 0.924 ma + 2.855 ma 3.78 ma Dmax Lmax 25. m 2 (110 ) 155.56 c 0.318 m 0.318(155.56 ) 49.47 26. Dioe will conuct when v o 0.7 ; that is, 10 k Ω( v ) v o 0.7 i 10 kω+ 1 kω Solving: v i 0.77 For v i 0.77 Si ioe is on an v o 0.7. For v i < 0.77 Si ioe is open an level of v o is etermine by voltage ivier rule: 10 k Ω( v ) v o i 10 kω+ 1 kω 0.909 v i For v i 10 : v o 0.909( 10 ) 9.09 When v o 0.7, vr v max i 0.7 max 10 0.7 9.3 9.3 I R 9.3 ma max 1 kω 10 I max (reverse) 0.909 ma 1 kω+ 10 kω 15

27. (a) P max 14 mw (0.7 )I D 14 mw I D 20 ma 0.7 (b) 4.7 kω 56 kω 4.34 kω R 160 0.7 159.3 I max 159.3 36.71 ma 4.34 kω Imax 36.71 ma (c) I ioe 18.36 ma 2 2 () Yes, I D 20 ma > 18.36 ma (e) I ioe 36.71 ma I max 20 ma 28. (a) m 2 (120 ) 169.7 L m i m 2 D 169.7 2(0.7 ) 169.7 1.4 168.3 c 0.636(168.3 ) 107.04 (b) PI m (loa) + D 168.3 + 0.7 169 (c) I D (max) L m 168.3 R 1 kω L 168.3 ma () P max D I D (0.7 )I max (0.7 )(168.3 ma) 117.81 mw 29. 16

30. Positive half-cycle of v i : oltage-ivier rule: 2.2 k Ω( i ) max o max 2.2 kω+ 2.2 kω 1 ( imax ) 2 1 (100 ) 2 50 31. Positive pulse of v i : Top left ioe off, bottom left ioe on 2.2 kω 2.2 kω 1.1 kω 1.1 k Ω(170 ) o peak 1.1 kω + 2.2 kω 56.67 Negative pulse of v i : Top left ioe on, bottom left ioe off 1.1 k Ω(170 ) o peak 1.1 kω + 2.2 kω 56.67 c 0.636(56.67 ) 36.04 c 0.636 m 0.636 (50 ) 31.8 32. (a) Si ioe open for positive pulse of v i an v o 0 For 20 < v i 0.7 ioe on an v o v i + 0.7. For v i 20, v o 20 + 0.7 19.3 For v i 0.7, v o 0.7 + 0.7 0 Polarity of v o across the 2.2 kω resistor acting as a loa is the same. oltage-ivier rule: 2.2 k Ω( i ) max o max 2.2 kω+ 2.2 kω 1 ( imax ) 2 1 (100 ) 2 50 17

(b) For v i 5 the 5 battery will ensure the ioe is forwar-biase an v o v i 5. At v i 5 v o 5 5 0 At v i 20 v o 20 5 25 For v i > 5 the ioe is reverse-biase an v o 0. 33. (a) Positive pulse of v i : 1.2 k Ω(10 0.7 ) o 1.2 kω+ 2.2 kω Negative pulse of v i : ioe open, v o 0 3.28 (b) Positive pulse of v i : o 10 0.7 + 5 14.3 Negative pulse of v i : ioe open, v o 0 34. (a) For v i 20 the ioe is reverse-biase an v o 0. For v i 5, v i overpowers the 2 battery an the ioe is on. Applying Kirchhoff s voltage law in the clockwise irection: 5 + 2 v o 0 v o 3 (b) For v i 20 the 20 level overpowers the 5 supply an the ioe is on. Using the short-circuit equivalent for the ioe we fin v o v i 20. For v i 5, both v i an the 5 supply reverse-bias the ioe an separate v i from v o. However, v o is connecte irectly through the 2.2 kω resistor to the 5 supply an v o 5. 18

35. (a) Dioe on for v i 4.7 For v i > 4.7, o 4 + 0.7 4.7 For v i < 4.7, ioe off an v o v i (b) Again, ioe on for v i 4.7 but v o now efine as the voltage across the ioe For v i 4.7, v o 0.7 For v i < 4.7, ioe off, I D I R 0 ma an 2.2 kω IR (0 ma)r 0 Therefore, v o v i 4 At v i 0, v o 4 v i 8, v o 8 4 12 36. For the positive region of v i : The right Si ioe is reverse-biase. The left Si ioe is on for levels of v i greater than 5.3 + 0.7 6. In fact, v o 6 for v i 6. For v i < 6 both ioes are reverse-biase an v o v i. For the negative region of v i : The left Si ioe is reverse-biase. The right Si ioe is on for levels of v i more negative than 7.3 + 0.7 8. In fact, v o 8 for v i 8. For v i > 8 both ioes are reverse-biase an v o v i. i R : For 8 < v i < 6 there is no conuction through the 10 kω resistor ue to the lack of a complete circuit. Therefore, i R 0 ma. For v i 6 v R v i v o v i 6 For v i 10, v R 10 6 4 an i R 4 10 kω 0.4 ma For v i 8 v R v i v o v i + 8 19

For v i 10 v R 10 + 8 2 2 an i R 0.2 ma 10 kω 37. (a) Starting with v i 20, the ioe is in the on state an the capacitor quickly charges to 20 +. During this interval of time v o is across the on ioe (short-current equivalent) an v o 0. When v i switches to the +20 level the ioe enters the off state (open-circuit equivalent) an v o v i + v C 20 + 20 +40 (b) Starting with v i 20, the ioe is in the on state an the capacitor quickly charges up to 15 +. Note that v i +20 an the 5 supply are aitive across the capacitor. During this time interval v o is across on ioe an 5 supply an v o 5. When v i switches to the +20 level the ioe enters the off state an v o v i + v C 20 + 15 35. 20

38. (a) For negative half cycle capacitor charges to peak value of 120 0.7 119.3 with polarity. The output v o is irectly across the on ioe resulting in v o 0.7 as a negative peak value. For next positive half cycle v o v i + 119.3 with peak value of v o 120 + 119.3 239.3. (b) For positive half cycle capacitor charges to peak value of 120 20 0.7 99.3 with polarity. The output v o 20 + 0.7 20.7 For next negative half cycle v o v i 99.3 with negative peak value of v o 120 99.3 219.3. Using the ieal ioe approximation the vertical shift of part (a) woul be 120 rather than 119.3 an 100 rather than 99.3 for part (b). Using the ieal ioe approximation woul certainly be appropriate in this case. 39. (a) τ RC (56 kω)(0.1 μf) 5.6 ms 5τ 28 ms (b) 5τ 28 ms 2 T 1 ms 2 0.5 ms, 56:1 (c) Positive pulse of v i : Dioe on an v o 2 + 0.7 1.3 Capacitor charges to 10 + 2 0.7 11.3 Negative pulse of v i : Dioe off an v o 10 11.3 21.3 21

40. Solution is network of Fig. 2.176(b) using a 10 supply in place of the 5 source. 41. Network of Fig. 2.178 with 2 battery reverse. 42. (a) In the absence of the Zener ioe 180 Ω(20 ) L 180 Ω+ 220 Ω 9 L 9 < Z 10 an ioe non-conucting 20 Therefore, I L I R 220 Ω+ 180 Ω with I Z 0 ma an L 9 50 ma (b) In the absence of the Zener ioe 470 Ω(20 ) L 470 Ω+ 220 Ω 13.62 L 13.62 > Z 10 an Zener ioe on Therefore, L 10 an R s 10 I / R 10 /220 Ω 45.45 ma Rs Rs s I L L /R L 10 /470 Ω 21.28 ma an I Z I I L 45.45 ma 21.28 ma 24.17 ma R s (c) P Z max 400 mw Z I Z (10 )(I Z ) 400 mw I Z 40 ma 10 I L min IR I s Z 45.45 ma 40 ma 5.45 ma max L 10 R L 1,834.86 Ω I 5.45 ma Lmin Large R L reuces I L an forces more of I R s to pass through Zener ioe. () In the absence of the Zener ioe RL (20 ) L 10 R L + 220 Ω 10R L + 2200 20R L 10R L 2200 R L 220 Ω 22