Online Instructor’s Manual for Electronic Devices and Circuit Theory Eleventh Edition PDF
Document Details
2013
Robert L. Boylestad Louis Nashelsky
Tags
Summary
This document is an instructor's manual for an electronic devices and circuit theory textbook. The eleventh edition by Robert L. Boylestad and Louis Nashelsky provides solutions to problems and a laboratory manual. It covers fundamental concepts and principles in electronics and circuits.
Full Transcript
Online Instructor’s Manual for Electronic Devices and Circuit Theory Eleventh Edition Robert L. Boylestad Louis Nashelsky Boston Columbus Indianapolis New York San Fra...
Online Instructor’s Manual for Electronic Devices and Circuit Theory Eleventh Edition Robert L. Boylestad Louis Nashelsky Boston Columbus Indianapolis New York San Francisco Upper Saddle River Amsterdam Cape Town Dubai London Madrid Milan Munich Paris Montreal Toronto Delhi Mexico City Sao Paulo Sydney Hong Kong Seoul Singapore Taipei Tokyo Copyright 2013 Pearson Education, Inc., publishing as Prentice Hall, 1 Lake Street, Upper Saddle River, New Jersey, 07458. All rights reserved. Manufactured in the United States of America. This publication is protected by Copyright, and permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. To obtain permission(s) to use material from this work, please submit a written request to Pearson Education, Inc., Permissions Department, 1 Lake Street, Upper Saddle River, New Jersey 07458. Many of the designations by manufacturers and seller to distinguish their products are claimed as trademarks. Where those designations appear in this book, and the publisher was aware of a trademark claim, the designations have been printed in initial caps or all caps. 10 9 8 7 6 5 4 3 2 1 ISBN10: 0-13-278373-8 ISBN13: 978-0-13-278373-6 Contents Solutions to Problems in Text 1 Solutions for Laboratory Manual 209 iii iv Chapter 1 1. Copper has 20 orbiting electrons with only one electron in the outermost shell. The fact that the outermost shell with its 29th electron is incomplete (subshell can contain 2 electrons) and distant from the nucleus reveals that this electron is loosely bound to its parent atom. The application of an external electric field of the correct polarity can easily draw this loosely bound electron from its atomic structure for conduction. Both intrinsic silicon and germanium have complete outer shells due to the sharing (covalent bonding) of electrons between atoms. Electrons that are part of a complete shell structure require increased levels of applied attractive forces to be removed from their parent atom. 2. Intrinsic material: an intrinsic semiconductor is one that has been refined to be as pure as physically possible. That is, one with the fewest possible number of impurities. Negative temperature coefficient: materials with negative temperature coefficients have decreasing resistance levels as the temperature increases. Covalent bonding: covalent bonding is the sharing of electrons between neighboring atoms to form complete outermost shells and a more stable lattice structure. 3. 4. a. W = QV = (12 µC)(6 V) = 72 μJ 1 eV b. 72 × 106 J = 19 = 2.625 × 1014 eV 1.6 10 J 5. 48 eV = 48(1.6 1019 J) = 76.8 1019 J W 76.8 1019 J Q= = = 2.40 1018 C V 3.2 V 6.4 1019 C is the charge associated with 4 electrons. 6. GaP Gallium Phosphide Eg = 2.24 eV ZnS Zinc Sulfide Eg = 3.67 eV 7. An n-type semiconductor material has an excess of electrons for conduction established by doping an intrinsic material with donor atoms having more valence electrons than needed to establish the covalent bonding. The majority carrier is the electron while the minority carrier is the hole. A p-type semiconductor material is formed by doping an intrinsic material with acceptor atoms having an insufficient number of electrons in the valence shell to complete the covalent bonding thereby creating a hole in the covalent structure. The majority carrier is the hole while the minority carrier is the electron. 8. A donor atom has five electrons in its outermost valence shell while an acceptor atom has only 3 electrons in the valence shell. 1 9. Majority carriers are those carriers of a material that far exceed 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. 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 forward bias, the positive potential is applied to the p-type material and the negative potential to the n-type material. kTK (1.38 1023 J/K)(20C 273C) 15. a. VT q 1.6 1019 C 25.27 mV b. I D I s (eVD / nVT 1) 40 nA(e(0.5 V) / (2)(25.27mV) 1) 40 nA(e9.89 1) 0.789 mA k (TK ) (1.38 1023 J/K)(100C 273C) 16. a. VT q 1.6 1019 32.17 mV b. I D I s (eVD / nVT 1) 40 nA(e(0.5 V) / (2)(32.17 mV) 1) 40 nA(e7.77 1) 11.84 mA 17. a. TK = 20 + 273 = 293 kT (1.38 1023 J/K)(293) VT K q 1.6 1019 C 25.27 mV b. I D I s (eVD / nVT 1) 0.1 A e 10/(2)(25.27 mV) 1 = 0.1 A(e197.86 1) 0.1 A 2 kTK (1.38 1023 J/K)(25C 273C) 18. VT q 1.6 1019 C =25.70 mV ID = I s (eVD / nVT 1) 8mA = I s (e(0.5V) / (1)(25.70 mV) 1) I s (28 108 ) 8 mA Is = 28.57 pA 2.8 108 19. I D I s (eVD / nVT 1) 6 mA 1 nA(eVD /(1)(26 mV) 1) 6 106 eVD / 26 mV 1 eVD / 26 mV 6 106 1 6 106 log e eVD / 26 mV log e 6 106 VD = 15.61 26 mV VD = 15.61(26 mV) 0.41 V 20. (a) x y = ex 0 1 1 2.7182 2 7.389 3 20.086 4 54.6 5 148.4 (b) y = e0 = 1 (c) For x = 0, e0 = 1 and I = Is(1 1) = 0 mA 21. T = 20C: Is = 0.1 A T = 30C: Is = 2(0.1 A) = 0.2 A (Doubles every 10C rise in temperature) T = 40C: Is = 2(0.2 A) = 0.4 A T = 50C: Is = 2(0.4 A) = 0.8 A T = 60C: Is = 2(0.8 A) = 1.6 A 1.6 A: 0.1 A 16:1 increase due to rise in temperature of 40C. 22. For most applications the silicon diode is the device of choice due to its higher temperature capability. Ge typically has a working limit of about 85 degrees centigrade while Si can be used at temperatures approaching 200 degrees centigrade. Silicon diodes also have a higher current handling capability. Germanium diodes are the better device for some RF small signal applications, where the smaller threshold voltage may prove advantageous. 3 23. From 1.19: 75C 25C 100C 200C VF 1.1 V 0.85 V 1.0 V 0.6 V @ 10 mA Is 0.01 pA 1 pA 1 A 1.05 A VF decreased with increase in temperature 1.7 V: 0.65 V 2.6:1 Is increased with increase in temperature 2 A: 0.1 A = 20:1 24. An “ideal” device or system is one that has the characteristics we would prefer to have when using a device or system in a practical application. Usually, however, technology only permits a close replica of the desired characteristics. The “ideal” characteristics provide an excellent basis for comparison with the actual device characteristics permitting an estimate of how well the device or system will perform. On occasion, the “ideal” device or system can be assumed to obtain a good estimate of the overall response of the design. When assuming an “ideal” device or system there is no regard for component or manufacturing tolerances or any variation from device to device of a particular lot. 25. In the forward-bias region the 0 V drop across the diode at any level of current results in a resistance level of zero ohms – the “on” state – conduction is established. 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 conduction is interrupted. 26. The most important difference between the characteristics of a diode and a simple switch is that the switch, being mechanical, is capable of conducting current in either direction while the diode only allows charge to flow through the element in one direction (specifically the direction defined by the arrow of the symbol using conventional current flow). 27. VD 0.7 V, ID = 4 mA V 0.7 V RDC = D = 175 I D 4 mA 28. At ID = 15 mA, VD = 0.82 V V 0.82 V RDC = D = 54.67 I D 15 mA As the forward diode current increases, the static resistance decreases. 4 29. VD = 10 V, ID = Is = 0.1 A V 10 V RDC = D = 100 M I D 0.1 A VD = 30 V, ID = Is= 0.1 A V 30 V RDC = D = 300 M I D 0.1 A As the reverse voltage increases, the reverse resistance increases directly (since the diode leakage current remains constant). 30. ID = 10 mA, VD = 0.76 V V 0.76 V RDC = D = 76 I D 10 mA Vd 0.79 V 0.76 V 0.03 V rd = =3 I d 15 mA 5 mA 10 mA RDC >> rd Vd 0.79 V 0.76 V 0.03 V 31. (a) rd = =3 I d 15 mA 5 mA 10 mA 26 mV 26 mV (b) rd = = 2.6 ID 10 mA (c) quite close Vd 0.72 V 0.61 V 32. ID = 1 mA, rd = = 55 I d 2 mA 0 mA Vd 0.8 V 0.78 V ID = 15 mA, rd = =2 I d 20 mA 10 mA 26 mV 33. ID = 1 mA, rd = 2 = 2(26 ) = 52 vs 55 (#30) I D 26 mV 26 mV ID = 15 mA, rd = = 1.73 vs 2 (#30) ID 15 mA Vd 0.9 V 0.6 V 34. rav = = 24.4 I d 13.5 mA 1.2 mA Vd 0.8 V 0.7 V 0.09 V 35. rd = = 22.5 I d 7 mA 3 mA 4 mA (relatively close to average value of 24.4 (#32)) 5 Vd 0.9 V 0.7 V 0.2 V 36. rav = = 14.29 I d 14 mA 0 mA 14 mA 37. Using the best approximation to the curve beyond VD = 0.7 V: Vd 0.8 V 0.7 V 0.1 V rav = =4 I d 25 mA 0 mA 25 mA 38. Germanium: 0.42 V 0.3 V rav 4 30 mA 0 mA GaAa: 1.32 V 1.2 V rav 4 30 mA 0 mA 39. (a) VR = 25 V: CT 0.75 pF VR = 10 V: CT 1.25 pF CT 1.25 pF 0.75 pF 0.5 pF = 0.033 pF/V VR 10 V 25 V 15 V (b) VR = 10 V: CT 1.25 pF VR = 1 V: CT 3 pF CT 1.25 pF 3 pF 1.75 pF = 0.194 pF/V VR 10 V 1 V 9V (c) 0.194 pF/V: 0.033 pF/V = 5.88:1 6:1 Increased sensitivity near VD = 0 V 40. From Fig. 1.33 VD = 0 V, CD = 3.3 pF VD = 0.25 V, CD = 9 pF 6 41. The transition capacitance is due to the depletion region acting like a dielectric in the reverse- bias region, while the diffusion capacitance is determined by the rate of charge injection into the region just outside the depletion boundaries of a forward-biased device. Both capacitances are present in both the reverse- and forward-bias directions, but the transition capacitance is the dominant effect for reverse-biased diodes and the diffusion capacitance is the dominant effect for forward-biased conditions. 42. VD = 0.2 V, CD = 7.3 pF 1 1 XC = = 3.64 k 2 fC 2 (6 MHz)(7.3 pF) VD = 20 V, CT = 0.9 pF 1 1 XC = = 29.47 k 2 fC 2 (6 MHz)(0.9 pF) C (0) 8 pF 43. CT 1 VR / VK 1 5 V / 0.7 V n 1/2 8pF 8 pF 8 pF 1/2 (1+7.14) 8.14 2.85 2.81 pF C (0) 44. CT 1 V /V R k 10 pF 4 pF = 1 VR /0.7 V 1/3 (1 VR /0.7 V)1/3 2.5 1 VR /0.7 V (2.5)3 15.63 VR /0.7 V 15.63 1 14.63 VR (0.7)(14.63) 10.24 V 10 V 45. If = = 1 mA 10 k ts + tt = trr = 9 ns ts + 2ts = 9 ns ts = 3 ns tt = 2ts = 6 ns 7 46. 47. a. As the magnitude of the reverse-bias potential increases, the capacitance drops rapidly from a level of about 5 pF with no bias. For reverse-bias potentials in excess of 10 V the capacitance levels off at about 1.5 pF. b. 6 pF c. At VR 4 V, CT 2 pF C (0) CT 1 VR / Vk n 6 pF 2 pF 1 4V/0.7 V n 1 4 V 0.7 V 3 n (6.71) n 3 n log10 6.71 log10 3 n(0.827) 0.477 0.477 n 0.58 0.827 48. At VD = 25 V, ID = 0.2 nA and at VD = 100 V, ID 0.45 nA. Although the change in IR is more than 100%, the level of IR and the resulting change is relatively small for most applications. 49. Log scale: TA = 25C, IR = 0.5 nA TA = 100C, IR = 60 nA The change is significant. 60 nA: 0.5 nA = 120:1 Yes, at 95C IR would increase to 64 nA starting with 0.5 nA (at 25C) (and double the level every 10C). 50. IF = 0.1 mA: rd 700 IF = 1.5 mA: rd 70 IF = 20 mA: rd 6 The results support the fact that the dynamic or ac resistance decreases rapidly with increasing current levels. 8 51. T = 25C: Pmax = 500 mW T = 100C: Pmax = 260 mW Pmax = VFIF P 500 mW IF = max = 714.29 mA VF 0.7 V Pmax 260 mW IF = = 371.43 mA VF 0.7 V 714.29 mA: 371.43 mA = 1.92:1 2:1 52. Using the bottom right graph of Fig. 1.37: IF = 500 mA @ T = 25C At IF = 250 mA, T 104C 53. VZ 54. TC = +0.072% = 100% VZ (T1 T0 ) 0.75 V 0.072 = 100 10 V(T1 25) 7.5 0.072 = T1 25 7.5 T1 25 = = 104.17 0.072 T1 = 104.17 + 25 = 129.17 VZ 55. TC = 100% VZ (T1 T0 ) (5 V 4.8 V) = 100% = 0.053%/C 5 V(100 25) (20 V 6.8 V) 56. 100% = 77% (24 V 6.8 V) The 20 V Zener is therefore 77% of the distance between 6.8 V and 24 V measured from the 6.8 V characteristic. 9 At IZ = 0.1 mA, TC 0.06%/C (5 V 3.6 V) 100% = 44% (6.8 V 3.6 V) The 5 V Zener is therefore 44% of the distance between 3.6 V and 6.8 V measured from the 3.6 V characteristic. At IZ = 0.1 mA, TC 0.025%/C 57. 58. 24 V Zener: 0.2 mA: 400 1 mA: 95 10 mA: 13 The steeper the curve (higher dI/dV) the less the dynamic resistance. 59. VK 2.0 V, which is considerably higher than germanium ( 0.3 V) or silicon ( 0.7 V). For germanium it is a 6.7:1 ratio, and for silicon a 2.86:1 ratio. 1.6 1019 J 19 60. 0.67 eV 1.072 10 J 1 eV hc hc (6.626 1034 Js)(3 108 ) m/s Eg Eg 1.072 1019 J 1850 nm Very low energy level. 61. Fig. 1.53 (f) IF 13 mA Fig. 1.53 (e) VF 2.3 V 62. (a) Relative efficiency @ 5 mA 0.82 @ 10 mA 1.02 1.02 0.82 100% = 24.4% increase 0.82 1.02 ratio: = 1.24 0.82 (b) Relative efficiency @ 30 mA 1.38 @ 35 mA 1.42 1.42 1.38 100% = 2.9% increase 1.38 1.42 ratio: = 1.03 1.38 10 (c) For currents greater than about 30 mA the percent increase is significantly less than for increasing currents of lesser magnitude. 0.75 63. (a) = 0.25 3.0 From Fig. 1.53 (i) 75 (b) 0.5 = 40 64. For the high-efficiency red unit of Fig. 1.53: 0.2 mA 20 mA C x 20 mA x= = 100C 0.2 mA/ C 11 Chapter 2 E 12 V 1. The load line will intersect at ID = = 16 mA and VD = 12 V. R 750 (a) VDQ 0.85 V I DQ 15 mA VR = E VDQ = 12 V 0.85 V = 11.15 V (b) VDQ 0.7 V I DQ 15 mA VR = E VDQ = 12 V 0.7 V = 11.3 V (c) VDQ 0 V I DQ 16 mA VR = E VDQ = 12 V 0 V = 12 V For (a) and (b), levels of VDQ and I DQ are quite close. Levels of part (c) are reasonably close but as expected due to level of applied voltage E. E 6V 2. (a) ID = = 30 mA R 0.2 k The load line extends from ID = 30 mA to VD = 6 V. VDQ 0.95 V, I DQ 25.3 mA E 6V (b) ID = = 12.77 mA R 0.47 k The load line extends from ID = 12.77 mA to VD = 6 V. VDQ 0.8 V, I DQ 11 mA E 6V (c) ID = = 8.82 mA R 0.68 k The load line extends from ID = 8.82 mA to VD = 6 V. VDQ 0.78 V, I DQ 78 mA The resulting values of VDQ are quite close, while I DQ extends from 7.8 mA to 25.3 mA. 3. Load line through I DQ = 10 mA of characteristics and VD = 7 V will intersect ID axis as 11.3 mA. E 7V ID = 11.3 mA = R R 7V with R = = 619.47 k 0.62 kΩ standard resistor 11.3 mA 12 E VD 30 V 0.7 V 4. (a) ID = IR = = 19.53 mA R 1.5 k VD = 0.7 V, VR = E VD = 30 V 0.7 V = 29.3 V E VD 30 V 0 V (b) ID = = 20 mA R 1.5 k VD = 0 V, VR = 30 V Yes, since E VT the levels of ID and VR are quite close. 5. (a) I = 0 mA; diode reverse-biased. (b) V20 = 20 V 0.7 V = 19.3 V (Kirchhoff’s voltage law) I(20 Ω) = 19.3 V = 0.965 A 20 V(10 Ω) = 20 V 0.7 V = 19.3 V I(10 Ω) = 19.3 V = 1.93 A 10 I = I(10 Ω) + I(20 Ω) = 2.895 A 10 V (c) I = = 1 A; center branch open 10 6. (a) Diode forward-biased, Kirchhoff’s voltage law (CW): 5 V + 0.7 V Vo = 0 Vo = 4.3 V Vo 4.3 V IR = ID = = 1.955 mA R 2.2 k (b) Diode forward-biased, 8 V + 6 V 0.7 V ID = = 2.25 mA 1.2 k 4.7 k Vo = 8 V (2.25 mA)(1.2 kΩ) = 5.3 V 10 k(12 V 0.7 V 0.3 V) 7. (a) Vo = = 9.17 V 2 k 10 k (b) Vo = 10 V 13 8. (a) Determine the Thevenin equivalent circuit for the 10 mA source and 2.2 k resistor. ETh = IR = (10 mA)(2.2 k) = 22 V RTh = 2. 2k Diode forward-biased 22 V 0.7 V ID = = 4.84 mA 2.2 k 2.2 k Vo = ID(1.2 k) = (4.84 mA)(1.2 k) = 5.81 V (b) Diode forward-biased 20 V + 20 V 0.7 V ID = = 5.78 mA 6.8 k Kirchhoff’s voltage law (CW): +Vo 0.7 V + 20 V = 0 Vo = 19.3 V 9. (a) Vo1 = 12 V – 0.7 V = 11.3 V Vo2 = 1.2 V (b) Vo1 = 0 V Vo2 = 0 V 10. (a) Both diodes forward-biased Si diode turns on first and locks in 0.7 V drop. 12 V 0.7 V IR = 2.4 mA 4.7 k ID = IR = 2.4 mA Vo = 12 V 0.7 V = 11.3 V (b) Right diode forward-biased: 20 V + 4 V 0.7 V ID = = 10.59 mA 2.2 k Vo = 20 V 0.7 V = 19.3 V 11. (a) Si diode “on” preventing GaAs diode from turning “on”: 1 V 0.7 V 0.3 V I= = 0.3 mA 1 k 1 k Vo = 1 V 0.7 V = 0.3 V 16 V 0.7 V 0.7 V + 4 V 18.6 V (b) I = = 3.96 mA 4.7 k 4.7 k Vo = 16 V 0.7 V 0.7 V = 14.6 V 14 12. Both diodes forward-biased: Vo1 = 0.7 V, Vo2 = 0.7 V 20 V 0.7 V 19.3 V I1 k = = = 19.3 mA 1 k 1 k I0.47 k = 0 mA I = I1 kΩ I0.47 kΩ = 19.3 mA 0 mA = 19.3 mA 13. 1 k(9.3 V) Superposition: Vo1 (9.3 V) 3.1 V 1 k 2 k 16 k(8.8 V) Vo2 (8.8 V) 2.93 V 1 k 2 k Vo = Vo1 Vo2 = 6.03 V 9.3 V 6.03 V ID = = 1.635 mA 2 k 14. Both diodes “off”. The threshold voltage of 0.7 V is unavailable for either diode. Vo = 0 V 15. Both diodes “on”, Vo = 10 V 0.7 V = 9.3 V 16. Both diodes “on”. Vo = 0.7 V 17. Both diodes “off”, Vo = 10 V 18. The Si diode with 5 V at the cathode is “on” while the other is “off”. The result is Vo = 5 V + 0.7 V = 4.3 V 19. 0 V at one terminal is “more positive” than 5 V at the other input terminal. Therefore assume lower diode “on” and upper diode “off”. The result: Vo = 0 V 0.7 V = 0.7 V The result supports the above assumptions. 20. Since all the system terminals are at 10 V the required difference of 0.7 V across either diode cannot be established. Therefore, both diodes are “off” and Vo = +10 V as established by 10 V supply connected to 1 k resistor. 15 21. The Si diode requires more terminal voltage than the Ge diode to turn “on”. Therefore, with 5 V at both input terminals, assume Si diode “off” and Ge diode “on”. The result: Vo = 5 V 0.3 V = 4.7 V The result supports the above assumptions. Vdc 2V 22. Vdc = 0.318 Vm Vm = = 6.28 V 0.318 0.318 Vm 6.28 V Im = = 3.14 mA R 2 k 23. Using Vdc 0.318(Vm VT) 2 V = 0.318(Vm 0.7 V) Solving: Vm = 6.98 V 10:1 for Vm:VT Vdc 2V 24. Vm = = 6.28 V 0.318 0.318 6.28 V I Lmax = = 0.628 mA 10 k 16 6.28 V Imax(2 k) = = 3.14 mA 2 k I Dmax I Lmax + Imax(2 k) = 0.678 mA + 3.14 mA = 3.77 mA 25. Vm = 2 (120 V) = 169.68 V Vdc = 0.318Vm = 0.318(169.68 V) = 53.96 V 26. Diode will conduct when vo = 0.7 V; that is, 1 k(vi ) vo = 0.7 V = 1 k 1 k Solving: vi = 1.4 V For vi 1.4 V Si diode is “on” and vo = 0.7 V. For vi < 1.4 V Si diode is open and level of vo is determined by voltage divider rule: 1 k(vi ) vo = = 0.5 vi 1 k 1 k For vi = 10 V: vo = 0.5(10 V) = 5 V When vo = 0.7 V, vRmax vimax 0.7 V = 10 V 0.7 V = 9.3 V 9.3 V I Rmax = 9.3 mA 1 k 10 V Imax(reverse) = = 0.5 mA 1 k 1 k 17 27. (a) Pmax = 14 mW = (0.7 V)ID 14 mW ID = = 20 mA 0.7 V (b) Imax = 2 × 20 mA = 40 mA (c) 4.7 k 68 k = 4.4 kΩ VR = 160 V 0.7 V = 159.3 V 159.3 V Imax = = 36.2 mA 4.4 k I Id = max = 18.1 mA 2 (d) Total damage, 36.2 mA > 20 mA 28. (a) Vm = 2 (120 V) = 169.7 V VLm = Vim 2VD = 169.7 V 2(0.7 V) = 169.7 V 1.4 V = 168.3 V Vdc = 0.636(168.3 V) = 107.04 V (b) PIV = Vm(load) + VD = 168.3 V + 0.7 V = 169 V VLm 168.3 V (c) ID(max) = = 168.3 mA RL 1 k (d) Pmax = VDID = (0.7 V)Imax = (0.7 V)(168.3 mA) = 117.81 mW 29. 100 V Imax = = 45.45 mA 2.2 k 18 30. Positive half-cycle of vi: Voltage-divider rule: 2.2 k(Vimax ) Vomax = 2.2 k 2.2 k 1 = (Vimax ) 2 1 = (100 V) 2 = 50 V Polarity of vo across the 2.2 k resistor acting as a load is the same. Voltage-divider rule: 2.2 k(Vimax ) Vomax = 2.2 k 2.2 k 1 = (Vimax ) 2 1 = (100 V) 2 = 50 V Vdc = 0.636Vm = 0.636 (50 V) = 31.8 V 31. Positive pulse of vi: Top left diode “off”, bottom left diode “on” 2.2 k 2.2 k = 1.1 k 1.1 k(170 V) Vopeak = = 56.67 V 1.1 k 2.2 k Negative pulse of vi: Top left diode “on”, bottom left diode “off” 1.1 k(170 V) Vopeak = = 56.67 V 1.1 k 2.2 k Vdc = 0.636(56.67 V) = 36.04 V 32. (a) Si diode open for positive pulse of vi and vo = 0 V For 20 V < vi 0.7 V diode “on” and vo = vi + 0.7 V. For vi = 20 V, vo = 20 V + 0.7 V = 19.3 V For vi = 0.7 V, vo = 0.7 V + 0.7 V = 0 V 19 (b) For vi 8 V the 8 V battery will ensure the diode is forward-biased and vo = vi 8 V. At vi = 8 V vo = 8 V 8 V = 0 V At vi = 20 V vo = 20 V 8 V = 28 V For vi > 8 V the diode is reverse-biased and vo = 0 V. 33. (a) Positive pulse of vi: 1.8 k(12 V 0.7 V) Vo = = 5.09 V 1.8 k 2.2 k Negative pulse of vi: diode “open”, vo = 0 V (b) Positive pulse of vi: Vo = 12 V 0.7 V + 4 V = 15.3 V Negative pulse of vi: diode “open”, vo = 0 V 34. (a) For vi = 20 V the diode is reverse-biased and vo = 0 V. For vi = 5 V, vi overpowers the 4 V battery and the diode is “on”. Applying Kirchhoff’s voltage law in the clockwise direction: 5 V + 4 V vo = 0 vo = 1 V (b) For vi = 20 V the 20 V level overpowers the 5 V supply and the diode is “on”. Using the short-circuit equivalent for the diode we find vo = vi = 20 V. For vi = 5 V, both vi and the 5 V supply reverse-bias the diode and separate vi from vo. However, vo is connected directly through the 2.2 k resistor to the 5 V supply and vo = 5 V. 20 35. (a) Diode “on” for vi 4.7 V For vi > 4.7 V, Vo = 4 V + 0.7 V = 4.7 V For vi < 4.7 V, diode “off” and vo = vi (b) Again, diode “on” for vi 3.7 V but vo now defined as the voltage across the diode For vi 3.7 V, vo = 0.7 V For vi < 3.7 V, diode “off”, ID = IR = 0 mA and V2.2 k = IR = (0 mA)R = 0 V Therefore, vo = vi 3 V At vi = 0 V, vo = 3 V vi = 8 V, vo = 8 V 3 V = 11 V 36. For the positive region of vi: The right Si diode is reverse-biased. The left Si diode is “on” for levels of vi greater than 5.3 V + 0.7 V = 6 V. In fact, vo = 6 V for vi 6 V. For vi < 6 V both diodes are reverse-biased and vo = vi. For the negative region of vi: The left Si diode is reverse-biased. The right Si diode is “on” for levels of vi more negative than 7.3 V + 0.7 V = 8 V. In fact, vo = 8 V for vi 8 V. For vi > 8 V both diodes are reverse-biased and vo = vi. iR: For 8 V < vi < 6 V there is no conduction through the 10 k resistor due to the lack of a complete circuit. Therefore, iR = 0 mA. For vi 6 V vR = vi vo = vi 6 V For vi = 10 V, vR = 10 V 6 V = 4 V 4V and iR = = 0.4 mA 10 k For vi 8 V vR = vi vo = vi + 8 V 21 For vi = 10 V vR = 10 V + 8 V = 2 V 2 V and iR = = 0.2 mA 10 k 37. (a) Starting with vi = 20 V, the diode is in the “on” state and the capacitor quickly charges to 20 V+. During this interval of time vo is across the “on” diode (short-current equivalent) and vo = 0 V. When vi switches to the +20 V level the diode enters the “off” state (open-circuit equivalent) and vo = vi + vC = 20 V + 20 V = +40 V (b) Starting with vi = 20 V, the diode is in the “on” state and the capacitor quickly charges up to 15 V+. Note that vi = +20 V and the 5 V supply are additive across the capacitor. During this time interval vo is across “on” diode and 5 V supply and vo = 5 V. When vi switches to the +20 V level the diode enters the “off” state and vo = vi + vC = 20 V + 15 V = 35 V. 22 38. (a) For negative half cycle capacitor charges to peak value of 120 V = 120 V with polarity. The output vo is directly across the “on” diode resulting in vo = 0 V as a negative peak value. For next positive half cycle vo = vi + 120 V with peak value of vo = 120 V + 120 V = 240 V. (b) For positive half cycle capacitor charges to peak value of 120 V 20 V = 100 V with polarity. The output vo = 20 V = 20 V For next negative half cycle vo = vi 100 V with negative peak value of vo = 120 V 100 V = 220 V. 39. (a) = RC = (56 k)(0.1 F) = 5.6 ms 5 = 28 ms T 1 ms (b) 5 = 28 ms = = 0.5 ms, 56:1 2 2 (c) Positive pulse of vi: Diode “on” and vo = 2 V + 0.7 V = 1.3 V Capacitor charges to 12 V + 2 V 0.7 V = 13.3 V Negative pulse of vi: Diode “off” and vo = 12 V 13.3 V = 25.3 V 40. Solution is network of Fig. 2.181(b) using a 10 V supply in place of the 5 V source. 23 41. Network of Fig. 2.178 with 2 V battery reversed. 42. (a) In the absence of the Zener diode 180 (20 V) VL = =9V 180 220 VL = 9 V < VZ = 10 V and diode non-conducting 20 V Therefore, IL = IR = = 50 mA 220 180 with IZ = 0 mA and VL = 9 V (b) In the absence of the Zener diode 470 (20 V) VL = = 13.62 V 470 220 VL = 13.62 V > VZ = 10 V and Zener diode “on” Therefore, VL = 10 V and VRs = 10 V I Rs VRs / Rs 10 V/220 = 45.45 mA IL = VL/RL = 10 V/470 = 21.28 mA and IZ = I Rs IL = 45.45 mA 21.28 mA = 24.17 mA (c) PZ max = 400 mW = VZIZ = (10 V)(IZ) 400 mW IZ = = 40 mA 10 V I Lmin = I Rs I Z max = 45.45 mA 40 mA = 5.45 mA VL 10 V RL = = 1,834.86 I Lmin 5.45 mA Large RL reduces IL and forces more of I Rs to pass through Zener diode. (d) In the absence of the Zener diode RL (20 V) VL = 10 V = RL 220 10RL + 2200 = 20RL 10RL = 2200 RL = 220 24 VL 12 V 43. (a) VZ = 12 V, RL = = 60 I L 200 mA RLVi 60 (16 V) VL = VZ = 12 V = RL RS 60 Rs 720 + 12Rs = 960 12Rs = 240 Rs = 20 (b) PZ max = VZ I Z max = (12 V)(200 mA) = 2.4 W VL VZ 44. Since IL = is fixed in magnitude the maximum value of I Rs will occur when IZ is a RL RL maximum. The maximum level of I Rs will in turn determine the maximum permissible level of Vi. PZ max 400 mW I Zmax = 50 mA VZ 8V VL VZ 8V IL = = 36.36 mA RL RL 220 I Rs = IZ + IL = 50 mA + 36.36 mA = 86.36 mA Vi VZ I Rs Rs or Vi = I Rs Rs + VZ = (86.36 mA)(91 ) + 8 V = 7.86 V + 8 V = 15.86 V Any value of vi that exceeds 15.86 V will result in a current IZ that will exceed the maximum value. 45. At 30 V we have to be sure Zener diode is “on”. RLVi 1 k(30 V) VL = 20 V = RL Rs 1 k Rs Solving, Rs = 0.5 k 50 V 20 V 20 V At 50 V, I RS = 60 mA, IL = = 20 mA 0.5 k 1 k IZM = I RS IL = 60 mA 20 mA = 40 mA 46. For vi = +50 V: Z1 forward-biased at 0.7 V Z2 reverse-biased at the Zener potential and VZ2 = 10 V. Therefore, Vo = VZ1 VZ2 = 0.7 V + 10 V = 10.7 V 25 For vi = 50 V: Z1 reverse-biased at the Zener potential and VZ1 = 10 V. Z2 forward-biased at 0.7 V. Therefore, Vo = VZ1 VZ2 = 10.7 V For a 5 V square wave neither Zener diode will reach its Zener potential. In fact, for either polarity of vi one Zener diode will be in an open-circuit state resulting in vo = vi. 47. Vm = 1.414(120 V) = 169.68 V 2Vm = 2(169.68 V) = 339.36 V 48. The PIV for each diode is 2Vm PIV = 2(1.414)(Vrms) 26 Chapter 3 1. 2. A bipolar transistor utilizes holes and electrons in the injection or charge flow process, while unipolar devices utilize either electrons or holes, but not both, in the charge flow process. 3. Forward- and reverse-biased. 4. The leakage current ICO is the minority carrier current in the collector. 5. 6. 7. 8. IE the largest IB the smallest IC IE 1 9. IB = I C IC = 100IB 100 IE = IC + IB = 100IB + IB = 101IB I 8 mA IB = E = 79.21 A 101 101 IC = 100IB = 100(79.21 A) = 7.921 mA 10. 11. IE = 5 mA, VCB = 1 V: VBE = 800 mV VCB = 10 V: VBE = 770 mV VCB = 20 V: VBE = 750 mV The change in VCB is 20 V:1 V = 20:1 The resulting change in VBE is 800 mV:750 mV = 1.07:1 (very slight) V 0.9 V 0.7 V 12. (a) rav = = 25 I 8 mA 0 (b) Yes, since 25 is often negligible compared to the other resistance levels of the network. 13. (a) IC IE = 3.5 mA (b) IC IE = 3.5 mA (c) negligible: change cannot be detected on this set of characteristics. (d) IC IE 27 14. (a) Using Fig. 3.7 first, IE 2 mA Then Fig. 3.8 results in IC 2 mA (b) Using Fig. 3.8 first, IE 5 mA Then Fig. 3.7 results in VBE 0.77 V (c) Using Fig. 3.10(b) IE = 5 mA results in VBE 0.81 V (d) Using Fig. 3.10(c) IE = 5 mA results in VBE = 0.7 V (e) Yes, the difference in levels of VBE can be ignored for most applications if voltages of several volts are present in the network. 15. (a) IC = IE = (0.998)(4 mA) = 3.992 mA (b) IE = IC + IB IC = IE IB = 2.8 mA 0.02 mA = 2.78 mA I 2.78 mA dc = C = 0.993 IE 2.8 mA 0.98 (c) IC = IB = IB (40 A) = 1.96 mA 1 1 0.98 I 1.96 mA IE = C = 2 mA 0.993 16. 17. 18. (a) Fig. 3.13(b): IB 35A Fig. 3.13(a): IC 3.6 mA (b) Fig. 3.14(a): VCE 12 V Fig. 3.14(b): VBE 0.75 V IC 2 mA 19. (a) dc = = 111.11 I B 18 A 111.11 (b) = = 0.991 1 111.11 1 (c) ICEO = 0.3 mA (d) ICBO = (1 )ICEO = (1 0.991)(0.3 mA) = 2.7 A 20. (a) Fig. 3.14(a): ICEO 0.3 mA (b) Fig. 3.14(a): IC 1.35 mA I 1.35 mA dc = C = 135 IB 10 A 28 135 (c) = = 0.9926 1 136 ICBO (1 )ICEO = (1 0.9926)(0.3 mA) = 2.2 A I C 5.25 mA 21. (a) dc = = 87.5 IB 60 A I C 3.25 mA (b) dc = = 108.3 IB 30 A I C 1.35 mA (c) dc = = 135 IB 10 A (d) Yes, highest at lowest levels of I C. I C 5.9 mA 4.6 mA 22. (a) ac = = = 65 I B 70 A 50 A I C 4.1 mA 2.3 mA (b) ac = = = 90 I B 40 A 20 A I C 2.4 mA 0.3 mA (c) ac = = = 105 I B 20 A 0 A (d) Yes, highest at lowest levels of IC. (e) IC dc ac VCE dc/ac a. 5.25 mA 87.5 65 4V 1.35 b. 3.25 mA 108.3 90 7V 1.20 c. 1.35 mA 135 105 10 V 1.29 dc > ac although ratio consistent over scope of characteristics. I C 2.9 mA 23. dc = = 116 IB 25 A 116 = = 0.991 1 116 1 IE = IC/ = 2.9 mA/0.991 = 2.93 mA 0.980 24. (a) = = 49 1 1 0.980 120 (b) = = 0.992 1 120 1 29 IC 2 mA (c) IB = = 16.66 A 120 IE = IC + IB = 2 mA + 16.66 A 2.017 mA 25. 26. Ve = Vi Vbe = 2 V 0.1 V = 1.9 V V 1.9 V Av = o = 0.95 1 Vi 2V VE 1.9 V Ie = = 1.9 mA (rms) RE 1 k 27. Output characteristics: Curves are essentially the same with new scales as shown. Input characteristics: Common-emitter input characteristics may be used directly for common-collector calculations. 28. PCmax = 35 mW = VCEIC PCmax 35 mW IC = I Cmax , VCE = = = 5.83 V I Cmax 6 mA PCmax 35 mW VCE = VCEmax , IC = = = 2.33 mA VCEmax 20 V PCmax 35 mW VCE = 10 V, IC = = 3.5 mA VCE 10 V PCmax 35 mW IC = 5 mA, VCE = = 7.00 V IC 5 mA PCmax 35 mW VCE = 15 V, IC = = 2.91 mA VCE 12 V 30 PCmax 42 mW 29. IC = I Cmax , VCE = =6V I Cmax 7 mA PCmax 42 mW VCB = VCBmax , IC = = 2.1 mA VCBmax 20 V PCmax 42 mW IC = 4 mA, VCB = = 10.5 V IC 4 mA PCmax 42 mW VCB = 10 V, IC = = 2.8 mA VCB 15 V 30. The operating temperature range is 55C TJ 150C 9 F = C + 32 5 9 = (55C) + 32 = 67F 5 9 F = (150C) + 32 = 302F 5 67F TJ 302F 31. I Cmax = 200 mA, VCEmax = 30 V, PDmax = 625 mW PDmax 625 mW IC = I Cmax , VCE = = = 3.125 V I Cmax 200 mA PDmax 625 mW VCE = VCEmax , IC = = 20.83 mA VCEmax 30 V 31 PDmax 625 mW IC = 100 mA, VCE = = 6.25 V IC 100 mA PDmax 625 mW VCE = 20 V, IC = = 31.25 mA VCE 20 V 32. From Fig. 3.23 (a) ICBO = 50 nA max max avg = min 2 50 150 200 = = 2 2 = 100 ICEO ICBO = (100)(50 nA) = 5 A 33. hFE (dc) with VCE = 1 V, T = 25C IC = 0.1 mA, hFE 0.43(100) = 43 IC = 10 mA, hFE 0.98(100) = 98 hfe(ac) with VCE = 10 V, T = 25C IC = 0.1 mA, hfe 72 IC = 10 mA, hfe 160 For both hFE and hfe the same increase in collector current resulted in a similar increase (relatively speaking) in the gain parameter. The levels are higher for hfe but note that VCE is higher also. 34. As the reverse-bias potential increases in magnitude the input capacitance Cibo decreases (Fig. 3.23(b)). Increasing reverse-bias potentials causes the width of the depletion region to A increase, thereby reducing the capacitance C . d 32 35. (a) At IC = 1 mA, hfe 120 At IC = 10 mA, hfe 160 (b) The results confirm the conclusions of problems 21 and 22 that beta tends to increase with increasing collector current. 36. Tj = +125°C: 1.45 norm Tj = +25°C: 0.95 norm Tj = 55°C: 0.5 norm Yes, 34% drop between +125°C and 25°C. 65.5% drop between +125°C and 55°C. I C 16 mA 12.2 mA 3.8 mA 37. (a) ac = = = 190 I B VCE 3 V 80 A 60 A 20 A IC 12 mA (b) dc = = 201.7 I B 59.5 A 4 mA 2 mA 2 mA (c) ac = = 200 18 A 8 A 10 A I C 3 mA (d) dc = = 230.77 I B 13 A (e) In both cases dc is slightly higher than ac ( 10%) (f)(g) In general dc and ac increase with increasing IC for fixed VCE and both decrease for decreasing levels of VCE for a fixed IE. However, if IC increases while VCE decreases when moving between two points on the characteristics, chances are the level of dc or ac may not change significantly. In other words, the expected increase due to an increase in collector current may be offset by a decrease in VCE. The above data reveals that this is a strong possibility since the levels of are relatively close. 33 Chapter 4 VCC VBE 16 V 0.7 V 15.3 V 1. (a) I BQ = 30 A RB 510 k 510 k (b) I CQ I BQ = (120)(30 A) = 3.6 mA (c) VCEQ VCC I CQ RC = 16 V (3.6 mA)(1.8 k) = 9.52 V (d) VC = VCEQ = 6.48 V (e) VB = VBE = 0.7 V (f) VE = 0 V 2. (a) IC = IB = 80(40 A) = 3.2 mA VRC VCC VC 12 V 6 V 6V (b) RC = = 1.875 k IC IC 3.2 mA 3.2 mA VRB 12 V 0.7 V 11.3 V (c) RB = = 282.5 k IB 40 A 40 A (d) VCE = VC = 6 V 3. (a) IC = IE IB = 4 mA 20 A = 3.98 mA 4 mA (b) VCC = VCE + ICRC = 7.2 V + (3.98 mA)(2.2 k) = 15.96 V 16 V I C 3.98 mA (c) = = 199 200 IB 20 A VRB VCC VBE 15.96 V 0.7 V (d) RB = = 763 k IB IB 20 A VCC 16 V 4. I Csat = = 5.93 mA RC 2.7 k 21 V 5. (a) Load line intersects vertical axis at IC = = 7 mA 3 k and horizontal axis at VCE = 21 V. VCC VBE 21 V 0.7 V (b) IB = 25 A: RB = = 812 k IB 25 A (c) I CQ 3.4 mA, VCEQ 10.75 V 34 I C 3.4 mA (d) = = 136 IB 25 A 136 136 (e) = = 0.992 1 136 1 137 VCC 21 V (f) I Csat = = 7 mA RC 3 k (g) (h) PD = VCEQ I CQ = (10.75 V)(3.4 mA) = 36.55 mW (i) Ps = VCC(IC + IB) = 21 V(3.4 mA + 25 A) = 71.92 mW (j) PR = Ps PD = 71.92 mW 36.55 mW = 35.37 mW VCC VBE 16 V 0.7 V 6. (b) I BQ = 30 µA RB 510 k VCEQ = 8.5 V, I CQ = 4.1 mA I C 4.1 mA (c) = = 136.67 IB 30 A 16 V 0.7 V 7. I BQ = 16.81 µA 910 k I CQ I BQ = (120)(16.81 µA) = 2.017 mA (from characteristics) VEQ = 11.5 V, I CQ = 2.4 mA VCC VBE 20 V 0.7 V 19.3 V 8. (a) I BQ = RB ( 1) RE 270 k (126)2.2 k 547.2 k = 35.27 A (b) I CQ I BQ = (125)(35.27 A) = 4.41 mA (c) VCEQ = VCC IC(RC + RE) = 20 V (4.41 mA)(470 k + 2.2 k) = 20 V 11.77 V = 8.23 V (d) VC = VCC ICRC = 20 V (4.41 mA)(470 k) = 20 V 2.07 V = 17.93 V (e) VB = VCC IBRB = 20 V (35.27 A)(270 k) = 20 V 9.52 V = 10.48 V (f) VE = VC VCE = 17.93 V 8.23 V = 9.7 V 35 9. (a) I BQ = 35.27 µA using from problem 8. 20 V I sat = = 7.49 mA 2.2 k 470 (b) I CQ = 4.7 mA, VCEQ = 7.5 V IC 4.7 mA (c) = 133.25 I B 35.27 A (d) reasonably close (e) different and Q pt. VCC VC 12 V 7.6 V 4.4 V 10. (a) RC = = 2.2 k IC 2 mA 2 mA VE 2.4 V (b) IE IC: RE = = 1.2 k I E 2 mA VRB VCC VBE VE 12 V 0.7 V 2.4 V 8.9 V (c) RB = = 356 k IB IB 2 mA/80 25 A (d) VCE = VC VE = 7.6 V 2.4 V = 5.2 V (e) VB = VBE + VE = 0.7 V + 2.4 V = 3.1 V VE 2.1 V 11. (a) IC IE = = 3.09 mA RE 0.68 k I C 3.09 mA = = 154.5 IB 20 A (b) VCC = VRC + VCE + VE = (3.09 mA)(2.7 k) + 7.3 V + 2.1 V = 8.34 V + 7.3 V + 2.1 V = 17.74 V VRB VCC VBE VE 17.74 V 0.7 V 2.1 V (c) RB = IB IB 20 A 14.94 V = = 747 k 20 A VCC 20 V 20 V 12. I Csot = 7.49 mA RC RE 470 2.2 k 2.67 k 36 VCC 24 V 13. (a) I Csat = 6.8 mA = RC RE RC 1.2 k 24 V RC + 1.2 k = = 3.529 k 6.8 mA RC = 2.33 k IC 4 mA (b) = = 133.33 I B 30 A VRB VCC VBE VE 24 V 0.7 V (4 mA)(1.2 k) (c) RB = IB IB 30 A 18.5 V = = 616.67 k 30 A (d) PD = VCEQ I CQ = (10 V)(4 mA) = 40 mW (e) P = I C2 RC = (4 mA)2(2.33 k) = 37.28 mW 14. (a) Problem 1: I CQ = 3.6 mA, VCEQ = 6.48 V (b) I BQ = 30 A (the same) I CQ I BQ = (180)(30 A) = 5.4 mA VCEQ VCC I CQ RC = 16 V (5.4 mA)(1.8 k) = 6.28 V 5.4 mA 3.6 mA (c) %IC = 100% = 50% 3.6 mA 9.52 V 6.28 V %VCE = 100% = 51.6% 6.28 V About 50% change for each. (d) Problem 8: I CQ = 4.41 mA, VCEQ = 8.23 V ( I BQ = 35.27 A) VCC VBE 20 V 0.7 V (e) I BQ = 28.19 A RB ( 1) RE 270 k (187.5 1)(2.2 k) I CQ I BQ = (187.5)(28.19 A) = 5.29 mA = VCC IC(RC + RE) = 20 V (5.29 mA)(470 k + 2.2 k) = 5.88 V 37 5.29 mA 4.41 mA (f) %IC = 100% = 20% 4.41 mA 5.88 V 8.23 V %VCE = 100% = 28.6% 8.23 V (g) For both IC and VCE the % change is less for the emitter-stabilized. ? 15. RE 10R2 (80)(0.68 k) 10(9.1 k) 54.4 k 91 k (No!) (a) Use exact approach: RTh = R1 R2 = 62 k 9.1 k = 7.94 k RV (9.1 k)(16 V) ETh = 2 CC = 2.05 V R2 R1 9.1 k 62 k ETh VBE 2.05 V 0.7 V I BQ RTh ( 1) RE 7.94 k (81)(0.68 k) = 21.42 A (b) I CQ I BQ = (80)(21.42 A) = 1.71 mA (c) VCEQ = VCC I CQ (RC + RE) = 16 V (1.71 mA)(3.9 k + 0.68 k) = 8.17 V (d) VC = VCC ICRC = 16 V (1.71 mA)(3.9 k) = 9.33 V (e) VE = IERE ICRE = (1.71 mA)(0.68 k) = 1.16 V (f) VB = VE + VBE = 1.16 V + 0.7 V = 1.86 V 16. (a) RTh 62 k 9.1 k = 7.94 k 9.1 k(16 V) ETh = 2.05 V 9.1 k + 62 k ETh VBE 2.05 V 0.7 V I BQ RTh ( 1) RE 7.94 k (140 1)(0.68 k) = 13 μA (vs. 21.42 μA) I CQ I BQ = (140)(13 A) = 1.82 mA (vs. 1.71 mA) VCEQ = VCC I CQ (RC + RE) = 16 V (1.82 mA)(3.9 k + 0.68 k) = 7.66 V (vs. 8.17 V) 38 VC = VCC ICRC = 16 V (1.82 mA)(3.9 k) = 8.9 V (vs. 9.33 V) VE = IERE ICRE = (1.82 mA)(0.68 k) = 1.24 V (vs. 1.16 V) VB = VE + VBE = 1.24 V + 0.7 V = 1.94 V (vs. 1.86 V) (b) I BQ affected the most VCC VC 18 V 12 V 17. (a) IC = = 1.28 mA RC 4.7 k (b) VE = IERE ICRE = (1.28 mA)(1.2 k) = 1.54 V (c) VB = VBE + VE = 0.7 V + 1.54 V = 2.24 V VR1 (d) R1 = : VR1 = VCC VB = 18 V 2.24 V = 15.76 V I R1 VB 2.24 V I R1 I R2 = 0.4 mA R2 5.6 k VR1 15.76 V R1 = = = 39.4 k I R1 0.4 mA 18. (a) IC = IB = (100)(20 A) = 2 mA (b) IE = IC + IB = 2 mA + 20 A = 2.02 mA VE = IERE = (2.02 mA)(1.2 k) = 2.42 V (c) VCC = VC + ICRC = 10.6 V + (2 mA)(2.7 k) = 10.6 V + 5.4 V = 16 V (d) VCE = VC VE = 10.6 V 2.42 V = 8.18 V (e) VB = VE + VBE = 2.42 V + 0.7 V = 3.12 V (f) I R1 I R2 I B 3.12 V = + 20 A = 380.5 A + 20 A = 400.5 A 8.2 k V VB 16 V 3.12 V R1 = CC = 32.16 k I R1 400.5 A 39 VCC 16 V 16 V 19. I Csat = = = 3.49 mA RC RE 3.9 k 0.68 k 4.58 k 20. (a) Testing RE 10 Rz (140)(0.68 k ) 10(9.1 k ) 95.2 k 91 k satisfied (barely) Applying the approximation approved: RV (9.1 k)(16 V) VB = 2 CC = 2.05 V (vs. 1.95 V) R1 R2 62 k 9.1 k VE = VB VBE = 2.05 V 0.7 V = 1.35 V (vs. 1.24 V) V 1.35 V IC IE = E = 1.99 mA (vs. 1.82 mA) RE 0.68 k VC = VCC ICRC = 16 V (1.99 mA)(3.9 k) = 8.24 V (vs 8.9 V) VCE = VC VE = 8.24 V 1.35 V = 6.89 V (vs 7.66 V) 21. (a) RE 10R2 (120)(1 k) 10(8.2 k) 120 k 82 k (checks) RV (8.2 k)(18 V) VB = 2 CC = 3.13 V R1 R2 39 k 8.2 k VE = VB VBE = 3.13 V 0.7 V = 2.43 V V 2.43 V IC IE = E = 2.43 mA RE 1 k (b) VCE = VCC IC(RC + RE) = 18 V (2.43 mA)(3.3 k + 1 k) = 7.55 V IC 2.43 mA (c) IB = = 20.25 A 120 (d) VE = IERE ICRE = (2.43 mA)(1 k) = 2.43 V (e) VB = 3.13 V 22. (a) RTh = R1 R2 = 39 k 8.2 k = 6.78 k R V 8.2 k(18 V) ETh = C CC = 3.13 V R1 R2 39 k 8.2 k ETh VBE 3.13 V 0.7 V IB = RTh ( 1) RE 6.78 k (121)(1 k) 2.43 V = = 19.02 A 127.78 k IC = IB = (120)(19.02 A) = 2.28 mA (vs. 2.43 mA #16) 40 (b) VCE = VCC IC(RC + RE) = 18 V (2.28 mA)(3.3 k + 1 k) = 18 V 9.8 V = 8.2 V (vs. 7.55 V #16) (c) 19.02 A (vs. 20.25 A #16) (d) VE = IERE ICRE = (2.28 mA)(1 k) = 2.28 V (vs. 2.43 V #16) (e) VB = VBE + VE = 0.7 V + 2.28 V = 2.98 V (vs. 3.13 V #16) The results suggest that the approximate approach is valid if Eq. 4.33 is satisfied. R2 9.1 k(16 V) 23. (a) VB = VCC = 2.05 V R1 R2 62 k 9.1 k VE = VB VBE = 2.05 V 0.7 V = 1.35 V V 1.35 V IE = E = 1.99 mA RE 0.68 k I CQ IE = 1.99 mA VCEQ = VCC IC (RC + RE) = 16 V (1.99 mA)(3.9 k + 0.68 k) = 16 V 9.11 V = 6.89 V I CQ 1.99 mA I BQ = = 24.88 A 80 (b) From Problem 12: I CQ = 1.71 mA, VCEQ = 8.17 V, I BQ = 21.42 A (c) The differences of about 14% suggest that the exact approach should be employed when appropriate. VCC 24 V 24 V 24. (a) I Csat 7.5 mA RC RE 3RE RE 4RE 24 V 24 V RE = = 0.8 k 4(7.5 mA) 30 mA RC = 3RE = 3(0.8 k) = 2.4 k (b) VE = IERE ICRE = (5 mA)(0.8 k) = 4 V (c) VB = VE + VBE = 4 V + 0.7 V = 4.7 V RV R2 (24 V) (d) VB = 2 CC , 4.7 V = R2 R1 R2 24 k R2 = 5.84 k IC 5 mA (e) dc = = 129.8 I B 38.5 A 41 (f) RE 10R2 (129.8)(0.8 k) 10(5.84 k) 103.84 k 58.4 k (checks) 25. (a) From problem 12b, IC = 1.71 mA From problem 12c, VCE = 8.17 V (b) changed to 120: From problem 12a, ETh = 2.05 V, RTh = 7.94 k ETh VBE 2.05 V 0.7 V IB = RTh ( 1) RE 7.94 k + (121)(0.68 k) = 14.96 A IC = IB = (120)(14.96 A) = 1.8 mA VCE = VCC IC(RC + RE) = 16 V (1.8 mA)(3.9 k + 0.68 k) = 7.76 V 1.8 mA 1.71 mA (c) % I C 100% = 5.26% 1.71 mA 7.76 V 8.17 V % VCE 100% = 5.02% 8.17 V (d) 11c 11f 20c %IC 49.83% 34.59% 5.26% %VCE 48.70% 46.76% 5.02% Fixed-bias Emitter Voltage- feedback divider (e) Quite obviously, the voltage-divider configuration is the least sensitive to changes in . 26. (a) Problem 21: Approximation approach: I CQ = 2.43 mA, VCEQ = 7.55 V Problem 22: Exact analysis: I CQ = 2.28 mA, VCEQ = 8.2 V The exact solution will be employed to demonstrate the effect of the change of . Using the approximate approach would result in %IC = 0% and %VCE = 0%. Problem 22: ETh = 3.13 V, RTh = 6.78 k ETH VBE 3.13 V 0.7 V 2.43 V IB = RTh ( 1) RE 6.78 k (180 1)1 k 187.78 k = 12.94 A IC = IB = (180)(12.94 A) = 2.33 mA VCE = VCC IC(RC + RE) = 18 V (2.33 mA)(3.3 k + 1 k) = 7.98 V 42 2.33 mA 2.28 mA %IC = 100% = 2.19% 2.28 mA 7.98 V 8.2 V %VCE = 100% = 2.68% 8.2 V For situations where RE > 10R2 the change in IC and/or VCE due to significant change in will be relatively small. %IC = 2.19% vs. 49.83% for problem 14. %VCE = 2.68% vs. 48.70% for problem 14. Voltage-divider configuration considerably less sensitive. (b) The resulting %IC and %VCE will be quite small. VCC VBE 16 V 0.7 V 27. (a) IB = RB ( RC RE ) 270 k + (120)(3.6 k 1.2 k) = 18.09 A (b) IC = IB = (120)(18.09 A) = 2.17 mA (c) VC = VCC ICRC = 16 V (2.17 mA)(3.6 k) = 8.19 V VCC VBE 16 V 0.7 V 28. (a) I CQ = 3.19 mA RC RE 3.6 k 1.2 k (b) I CQ = 3.19 mA vs 2.17 mA (not too close) (c) R RC RE 4.8 k , RF / 270 k /120 2.25 k R 2( RF / ) (VCC VBE ) VCC VBE (d) Yes, I CQ I BQ RF ( RC RE ) RF / RC RE VCC VBE For RC RE RF / , I CQ RC RE 16 V 0.7 V (e) I CQ = 2.58 mA vs. 3.19 mA (much closer) 270 k 3.6 k 1.2 k 240 43 VCC VBE 30 V 0.7 V 29. (a) IB = = 12.47 A RB ( RC RE ) 550 k 180(8.2 k 1.8 k) IC = IB = (180)(12.47 A) = 2.24 mA (b) VC = VCC ICRC = 30 V (2.24 mA)(8.2 k) = 30 V 18.37 V = 11.63 V (c) VE = IERE ICRE = (2.24 mA)(1.8 k) = 4.03 V (d) VCE = VCC IC(RC + RE) = 30 V (2.24 mA)(8.2 k + 1.8 k) = 7.6 V 30. (a) R RC RE 8.2 k 1.8 k 10 k RF / 550 k/180 3.06 k R 3.27( RF / ) should be close VCC VBE 30 V 0.7 V (b) I CQ = 2.93 mA RC RE 8.2 k 1.8 k I CQ = 2.93 mA relatively close to 2.24 mA VCC VBE 22 V 0.7 V 31. (a) IB = RB ( RC RE ) 470 k (90)(9.1 k 9.1 k) = 10.09 A IC = IB = (90)(10.09 A) = 0.91 mA VCE = VCC IC(RC + RE) = 22 V (0.91 mA)(9.1 k + 9.1 k) = 5.44 V VCC VBE 22 V 0.7 V (b) = 135, IB = RB ( RC RE ) 470 k (135)(9.1 k 9.1 k) = 7.28 A IC = IB = (135)(7.28 A) = 0.983 mA VCE = VCC IC(RC + RE) = 22 V (0.983 mA)(9.1 k + 9.1 k) = 4.11 V 0.983 mA 0.91 mA (c) % I C 100% = 8.02% 0.91 mA 4.11 V 5.44 V % VCE 100% = 24.45% 5.44 V (d) The results for the collector feedback configuration are closer to the voltage-divider configuration than to the other two. However, the voltage-divider configuration continues to have the least sensitivities to change in . 44 32. 1 M = 0 , RB = 150 k VCC VBE 12 V 0.7 V IB = RB ( RC RE ) 150 k (180)(4.7 k 3.3 k) = 7.11 A IC = IB = (180)(7.11 A) = 1.28 mA VC = VCC ICRC = 12 V (1.28 mA)(4.7 k) = 5.98 V Full 1 M: RB = 1,000 k + 150 k = 1,150 k = 1.15 M VCC VBE 12 V 0.7 V IB = RB ( RC RE ) 1.15 M (180)(4.7 k 3.3 k) = 4.36 A IC = IB = (180)(4.36 A) = 0.785 mA VC = VCC ICRC = 12 V (0.785 mA)(4.7 k) = 8.31 V VC ranges from 5.98 V to 8.31 V 33. (a) VE = VB VBE = 4 V 0.7 V = 3.3 V VE 3.3 V (b) IC IE = = 2.75 mA RE 1.2 k (c) VC = VCC ICRC = 18 V (2.75 mA)(2.2 k) = 11.95 V (d) VCE = VC VE = 11.95 V 3.3 V = 8.65 V VRB VC VB 11.95 V 4 V (e) IB = = 24.09 A RB RB 330 k IC 2.75 mA (f) = = 114.16 I B 24.09 A VCC VEE VBE 6 V + 6 V 0.7 V 34. (a) IB = RB ( 1) RE 330 k (121)(1.2 k) = 23.78 A IE = ( + 1)IB = (121)(23.78 A) = 2.88 mA VEE + IERE VE = 0 VE = VEE + IERE = 6 V + (2.88 mA)(1.2 k) = 2.54 V 45 82 k(12 V) 35. (a) VB = 9.46 V 82 k 22 k VE VB VBE = 9.46 V 0.7 V = 8.76 V V 8.76 V IE = E = 7.3 mA RE 1.2 k IE 7.3 mA IB = = 65.77 μA 1 111 I C I B (110)(65.77 A) = 7.23 mA (b) VB 9.46 V, VC = 12 V, VE = 8.76 V (c) VBC = VB VC = 9.46 V 12 V = 2.54 V VCE = VC VE = 12 V 8.76 V = 3.24 V VEE VBE 12 V 0.7 V 36. (a) IB = RB ( 1) RE 9.1 k (80 1)15 k = 9.26 A (b) IC = IB = (80)(9.26 A) = 0.741 mA (c) VCE = VCC + VEE IC(RC + RE) = 16 V + 12 V (0.741 mA)(27 k) =8V (d) VC = VCC ICRC = 16 V (0.741 mA)(12 k) = 7.11 V 8 V 0.7 V 7.3 V 37. (a) IE = = 3.32 mA 2.2 k 2.2 k (b) VC = 10 V (3.32 mA)(1.8 k) = 10 V 5.976 = 4.02 V (c) VCE = 10 V + 8 V (3.32 mA)(2.2 k + 1.8 k) = 18 V 13.28 V = 4.72 V 38. (a) VE = 4 V 0.7 V = 3.3 V V V 3.3 V IE = E E = 3 mA RE RE 1.1 k IC IE = 3 mA VC = 8 V = VCC ICRC = 14 V (3 mA)RC RC = 2 k IE 3 mA (b) IE = 3 mA, IB = = 32.97 μA 1 91 46 (c) VBC = VB VC = 4 V 8 V = 4 V VCE = VC VE = 8 V 3.3 V = 4.7 V 39. (a) RE > 10R2 not satisfied Use exact approach: Network redrawn to determine the Thevenin equivalent: 510 k RTh = = 255 k 2 18 V + 18 V I= = 35.29 A 510 k 510 k ETh = 18 V + (35.29 A)(510 k) =0V 18 V 0.7 V IB = 255 k (130 1)(7.5 k) = 13.95 A (b) IC = IB = (130)(13.95 A) = 1.81 mA (c) VE = 18 V + (1.81 mA)(7.5 k) = 18 V + 13.58 V = 4.42 V (d) VCE = 18 V + 18 V (1.81 mA)(9.1 k + 7.5 k) = 36 V 30.05 V = 5.95 V VRB VC VBE 8 V 0.7 V 40. (a) IB = = 13.04 A RB RB 560 k VCC VC 18 V 8 V 10 V (b) IC = = = 2.56 mA RC 3.9 k 3.9 k IC 2.56 mA (c) = = 196.32 I B 13.04 A (d) VCE = VC = 8 V 47 IC 2.5 mA 41. IB = = 31.25 A 80 VRB VCC VBE 12 V 0.7 V RB = = 361.6 k IB IB 31.25 A VRC VCC VC VCC VCEQ 12 V 6 V 6V RC = IC IC I CQ 2.5 mA 2.5 mA = 2.4 k Standard values: RB = 360 k RC = 2.4 k VCC 42. I Csat = = 10 mA RC RE 20 V 20 V 20 V = 10 mA = 10 mA 5RE = = 2 k 4RE RE 5 RE 10 mA 2 k RE = = 400 5 RC = 4RE = 1.6 k I 5 mA IB = C = 41.67 A 120 20 V 0.7 V 5 mA(0.4 k) 19.3 2 V RB = VRB/IB = 41.67 A 41.67 A = 415.17 k Standard values: RE = 390 , RC = 1.6 k, RB = 430 k VE VE 3V 43. RE = = 0.75 k I E I C 4 mA VRC VCC VC VCC (VCEQ VE ) RC = IC IC IC 24 V (8 V + 3 V) 24 V 11 V 13 V = = 3.25 k 4 mA 4 mA 4 mA VB = VE + VBE = 3 V + 0.7 V = 3.7 V RV R (24 V) VB = 2 CC 3.7 V = 2 2 unknowns! R2 R1 R2 R1 use RE 10R2 for increased stability (110)(0.75 k) = 10R2 R2 = 8.25 k Choose R2 = 7.5 k 48 Substituting in the above equation: 7.5 k(24 V) 3.7 V = 7.5 k R1 R1 = 41.15 k Standard values: RE = 0.75 k, RC = 3.3 k, R2 = 7.5 k, R1 = 43 k 1 1 44. VE = VCC = (28 V) = 5.6 V 5 5 V 5.6 V RE = E = 1.12 k (use 1.1 k) I E 5 mA VCC 28 V VC = VE + 5.6 V = 14 V + 5.6 V = 19.6 V 2 2 VRC = VCC VC = 28 V 19.6 V = 8.4 V VRC8.4 V RC = = 1.68 k (use 1.6 k) IC 5 mA VB = VBE + VE = 0.7 V + 5.6 V = 6.3 V RV R (28 V) VB = 2 CC 6.3 V = 2 (2 unknowns) R2 R1 R2 R1 IC 5 mA = = 135.14 I B 37 A RE = 10R2 (135.14)(1.12 k) = 10(R2) R2 = 15.14 k (use 15 k) (15.14 k)(28 V) Substituting: 6.3 V = 15.14 k R1 Solving, R1 = 52.15 k (use 51 k) Standard values: RE = 1.1 k RC = 1.6 k R1 = 51 k R2 = 15 k 4.7 k(20 V) 45. (a) VB1 = 4.14 V 4.7 k 18 k VE1 = 4.14 V 0.7 V = 3.44 V 3.44 V I C1 I E1 = 3.44 mA 1 k VC1 20 V (3.44 mA)(2.2 k) = 12.43 V 3.3 k(20 V) VB2 = 2.61 V 3.3 k 22 k VE2 = 2.61 V 0.7 V = 1.91 V 49 1.91 V I E2 I C2 = 1.59 mA 1.2 k VC2 20 V (1.59 mA)(2.2 k) = 16.5 V I C1 3.44 mA (b) I B1 = 21.5 μA, I C1 I E1 = 3.44 mA 160 I C2 1.59 mA I B2 = 17.67 μA, I C2 I E2 = 1.59 mA 90 46. (a) D 12 = (50)(75) = 3750 VCC VBE1 VBE2 18 V 0.7 V 0.7 V (b) I B1 RB ( D 1) RE 2.2 M (3750+1)470 = 4.19 μA I B2 (1 1) I B1 = (50 + 1)(4.19 µA) = 213.69 μA (c) I C1 1 I B1 = (50)(4.19 µA) = 0.21 mA I C2 2 I B2 = (75)(213.69 µA) = 16.03 mA (d) VC1 = 18 V, VC2 = 18 V VE2 I E RE I C2 RE (16.03 mA)(470 ) = 7.53 V VE1 VE2 + 0.7 V = 7.53 V + 0.7 V = 8.23 V 3.3 k(22 V) 47. (a) VB1 = 4.48 V 3.3 k 4.7 k 8.2 k VE1 VB1 0.7 V = 3.78 V VE1 3.78 V I E1 I E2 I C2 I C1 = 3.44 mA RE 1.1 k I C1 I B1 = 57.33 μA, I C1 = 3.44 mA 60 I C3 I B2 = 28.67 μA, I C2 = 3.44 mA 120 (b) VB1 = 4.48 V (4.7 k 3.3 k)(22 V) VB2 = 10.86 V 3.3 k 4.7 k 8.2 k VE1 = 3.78 V, VC1 VB2 0.7 V = 10.16 V VE@ = 10.16 V, VC2 = 22 V (3.44 mA)(2 2 k) = 14.43 V 50 VCC VEB1 12 V 0.7 V 48. (a) I B1 = 2.45 μA RB 1 2 RC 1.8 M (80)(160)(220 ) I C1 1 I B1 (80)(2.45 A) = 196 μA I B2 I C1 = 196 μA I C2 2 I B2 (160)(196 A) = 31.36 mA (b) VB1 I B1 RB (2.45 A)(1.8 M) = 4.41 V VE1 12 V I C (220 ) 12 V (31.36 mA)(220 ) = 5.1 V VB2 VBE2 = 0.7 V, VC2 VE1 = 5.1 V, VC1 VB2 = 0.7 V, VE2 = 0 V 18 V 0.7 V 49. I2 k = = 8.65 mA I 2 k 50. For current mirror: I(3 k) = I(2.4 k) = I = 2 mA 51. 6 V I B RB VBE I E RE 0 6 V I B 100 k 0.7 V ( 1) I B 1.2 k 0 6 V 0.7 V I B (100 k (120 1)(1.2 k)) 0 6 V 0.7 V IB = 21.61 µA 100 k 145 k I = I C I B (120)(21.61 A) = 2.59 mA 4.3 k 52. VB (18 V) = 9 V 4.3 k 4.3 k VE = 9 V 0.7 V = 9.7 V 18 V (9.7 V) IE = = 4.6 mA = I 1.8 k VZ VBE 5.1 V 0.7 V 53. IE = = 3.67 mA RE 1.2 k VCC VBE 12 V 0.7 V 11.3 V 54. IB = = 22.16 A RB 510 k 510 k IC = IB = (100)(22.16 A) = 2.216 mA VC = VCC + ICRC = 12 V + (2.216 mA)(3.3 k) = 4.69 V VCE = VC = 4.69 V 51 55. RE 10R2 (220)(0.75 k) 10(16 k) 165 k 160 k (checks) Use approximate approach: 16 k(22 V) VB = 3.59 V 16 k + 82 k VE = VB + 0.7 V = 3.59 V + 0.7 V = 2.89 V IC IE = VE/RE = 2.89/0.75 k = 3.85 mA I 3.85 mA IB = C = 17.5 A 220 VC = VCC + ICRC = 22 V + (3.85 mA)(2.2 k) = 13.53 V V VBE 8 V 0.7 V 7.3 V 56. IE = = 2.212 mA RE 3.3 k 3.3 k VC = VCC + ICRC = 12 V + (2.212 mA)(3.9 k) = 3.37 V VCC 10 V 57. I Csat = 4.167 mA RC 2.4 k From characteristics I Bmax 31 A Vi VBE 10 V 0.7 V IB = = 51.67 A RB 180 k 51.67 A 31 A, well saturated Vo = 10 V (0.1 mA)(2.4 k) = 10 V 0.24 V = 9.76 V 52 5V 58. I Csat = 8 mA = RC 5V RC = = 0.625 k 8 mA IC 8 mA I Bmax = sat = 80 A 100 Use 1.2 (80 A) = 96 A 5 V 0.7 V RB = = 44.79 k 96 A Standard values: RB = 43 k RC = 0.62 k 59. (a) From Fig. 3.23c: IC = 2 mA: tf = 38 ns, tr = 48 ns, td = 120 ns, ts = 110 ns ton = tr + td = 48 ns + 120 ns = 168 ns toff = ts + tf = 110 ns + 38 ns = 148 ns (b) IC = 10 mA: tf = 12 ns, tr = 15 ns, td = 22 ns, ts = 120 ns ton = tr + td = 15 ns + 22 ns = 37 ns toff = ts + tf = 120 ns + 12 ns = 132 ns The turn-on time has dropped dramatically 168 ns:37 ns = 4.54:1 while the turn-off time is only slightly smaller 148 ns:132 ns = 1.12:1 60. (a) Open-circuit in the base circuit Bad connection of emitter terminal Damaged transistor (b) Shorted base-emitter junction Open at collector terminal (c) Open-circuit in base circuit Open transistor 53 61. (a) The base voltage of 9.4 V reveals that the 18 k resistor is not making contact with the base terminal of the transistor. If operating properly: 18 k(16 V) VB = 2.64 V vs. 9.4 V 18 k 91 k As an emitter feedback bias circuit: VCC VBE 16 V 0.7 V IB = R1 ( 1) RE 91 k (100 1)1.2 k = 72.1 A VB = VCC IB(R1) = 16 V (72.1 A)(91 k) = 9.4 V (b) Since VE > VB the transistor should be “off” 18 k(16 V) With IB = 0 A, VB = = 2.64 V 18 k 91 k Assume base circuit “open” The 4 V at the emitter is the voltage that would exist if the transistor were shorted collector to emitter. 1.2 k(16 V) VE = =4V 1.2 k 3.6 k 62. (a) RB, IB, IC, VC (b) , IC (c) Unchanged, I Csat not a function of (d) VCC, IB, IC (e) , IC, VRC , VRE , VCE ETh VBE E VBE 63. (a) IB = Th RTh ( 1) RE RTh RE E VBE ETh VBE IC = IB = Th RTh RE RTh R E RTh As , , IC, VRC VC = VCC VRC and VC (b) R2 = open, IB, IC VCE = VCC IC(RC + RE) and VCE 54 (c) VCC, VB, VE, IE, IC (d) IB = 0 A, IC = ICEO and IC(RC + RE) negligible with VCE VCC = 20 V (e) Base-emitter junction = short IB but transistor action lost and IC = 0 mA with VCE = VCC = 20 V 64. (a) RB open, IB = 0 A, IC = ICEO 0 mA and VC VCC = 18 V (b) , IC, VRC , VRE , VCE (c) RC, IB, IC, VE (d) Drop to a relatively low voltage 0.06 V (e) Open in the base circuit 65. (a) S ( I CO ) = 120 (b) S (VBE ) /RB 120 / 510 k = 235 × 10-6 S (c) S ( ) I C1 /1 = 3.6 mA/120 = 30 × 10-6 A (d) I C S ( I CO )I CO S (VBE )VBE S ( ) (120)(10 A 0.2 A) ( 2.35 104 S )(0.5 V 0.7 V) (30 106 A)(150 120) 2.12 mA (1 RB /RE ) 125(1 270 k/2.2 k) 66. (a) S ( I CO ) RB /RE 125 270 k/2.2 k = 62.44 / RE 125 / 2.2 k (b) S (VBE ) RB / RE 125 270 k / 2.2 k = 229.3 × 106 S I C1 (1 RB / RE ) 4.41 mA(1 122.73) (c) S ( ) 1 (2 RB / RE ) 125(156.25 122.73) = 15.65 × 10-6 A 55 (d) I C S ( I CO )I CO S (VBE )VBE S ( ) (62.44)(10 A 0.2 A) ( 229.36 106 S)(0.5 V 0.7 V) + 15.65 106 A(156.25 125) = 1.03 mA (1 RTh / RE ) (1 7.94 k / 0.68 k) 67. (a) S(ICO) = (80) RTh / RE 80 7.94 k / 0.68 k = 11.06 / RE 80 / 0.68 k (b) S(VBE) = RTh / RE 80 7.94 k / 0.68 k = 1280 106S I C1 (1 RTh / RE ) 1.71 mA(1 + 7.94 k / 0.68 k) (c) S() = 1 (2 RTh / RE ) 80(100 7.94 k / 0.68 k) = 2.43 106 A (d) IC = S(ICO)ICO + S(VBE) VBE + S() = (11.06)(10 A 0.2 A) + (1.28 103S)(0.5 V 0.7 V) + (2.43 106)(100 80) = 0.313 mA (1 RB / RC ) 196.32(1 560 k / 3.9 k) 68. (a) S(ICO) = ( + 1) RB / RC 196.32 560 k / 3.9 k = 83.69 / RC 196.32 / 3.9 k (b) S(VBE) = ?
