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Electronic Devices And Circuit Theory 11th Edition Solutions.rar [UPD]


2. Copyright 2009 by Pearson Education, Inc., Upper SaddleRiver, New Jersey 07458. Pearson Prentice Hall. All rightsreserved. Printed in the United States of America. This publicationis protected by Copyright and permission should be obtained fromthe publisher prior to any prohibited reproduction, storage in aretrieval system, or transmission in any form or by any means,electronic, mechanical, photocopying, recording, or likewise. Forinformation regarding permission(s), write to: Rights andPermissions Department. Pearson Prentice Hall is a trademark ofPearson Education, Inc. Pearson is a registered trademark ofPearson plc Prentice Hall is a registered trademark of PearsonEducation, Inc. Instructors of classes using Boylestad/Nashelsky,Electronic Devices and Circuit Theory, 10th edition, may reproducematerial from the instructors text solutions manual for classroomuse. 10 9 8 7 6 5 4 3 2 1 ISBN-13: 978-0-13-503865-9 ISBN-10:0-13-503865-0 3. iii Contents Solutions to Problems in Text 1Solutions for Laboratory Manual 185 4. 1 Chapter 1 1. Copper has 20orbiting electrons with only one electron in the outermost shell.The fact that the outermost shell with its 29th electron isincomplete (subshell can contain 2 electrons) and distant from thenucleus reveals that this electron is loosely bound to its parentatom. The application of an external electric field of the correctpolarity can easily draw this loosely bound electron from itsatomic structure for conduction. Both intrinsic silicon andgermanium have complete outer shells due to the sharing (covalentbonding) of electrons between atoms. Electrons that are part of acomplete shell structure require increased levels of appliedattractive forces to be removed from their parent atom. 2.Intrinsic material: an intrinsic semiconductor is one that has beenrefined to be as pure as physically possible. That is, one with thefewest possible number of impurities. Negative temperaturecoefficient: materials with negative temperature coefficients havedecreasing resistance levels as the temperature increases. Covalentbonding: covalent bonding is the sharing of electrons betweenneighboring atoms to form complete outermost shells and a morestable lattice structure. 3. 4. W = QV = (6 C)(3 V) = 18 J 5. 48 eV= 48(1.6 1019 J) = 76.8 1019 J Q = W V = 19 76.8 10 J 12 V = 6.401019 C 6.4 1019 C is the charge associated with 4 electrons. 6. GaPGallium Phosphide Eg = 2.24 eV ZnS Zinc Sulfide Eg = 3.67 eV 7. Ann-type semiconductor material has an excess of electrons forconduction established by doping an intrinsic material with donoratoms having more valence electrons than needed to establish thecovalent bonding. The majority carrier is the electron while theminority carrier is the hole. A p-type semiconductor material isformed by doping an intrinsic material with acceptor atoms havingan insufficient number of electrons in the valence shell tocomplete the covalent bonding thereby creating a hole in thecovalent structure. The majority carrier is the hole while theminority carrier is the electron. 8. A donor atom has fiveelectrons in its outermost valence shell while an acceptor atom hasonly 3 electrons in the valence shell. 9. Majority carriers arethose carriers of a material that far exceed the number of anyother carriers in the material. Minority carriers are thosecarriers of a material that are less in number than any othercarrier of the material. 5. 2 10. Same basic appearance as Fig. 1.7since arsenic also has 5 valence electrons (pentavalent). 11. Samebasic appearance as Fig. 1.9 since boron also has 3 valenceelectrons (trivalent). 12. 13. 14. For forward bias, the positivepotential is applied to the p-type material and the negativepotential to the n-type material. 15. TK = 20 + 273 = 293 k =11,600/n = 11,600/2 (low value of VD) = 5800 ID = Is 1 D K kV T e =50 109 (5800)(0.6) 293 1e = 50 109 (e11.877 1) = 7.197 mA 16. k =11,600/n = 11,600/2 = 5800 (n = 2 for VD = 0.6 V) TK = TC + 273 =100 + 273 = 373 (5800)(0.6 V) / 9.33373KkV T e e e= = = 11.27 103 I= / ( 1)KkV T sI e = 5 A(11.27 103 1) = 56.35 mA 17. (a) TK = 20 +273 = 293 k = 11,600/n = 11,600/2 = 5800 ID = Is 1 D K kV T e =0.1A (5800)( 10 V) 293 1e = 0.1 106 (e197.95 1) = 0.1 106 (1.071086 1) 0.1 106 0.1A ID = Is = 0.1 A (b) The result is expectedsince the diode current under reverse-bias conditions should equalthe saturation value. 18. (a) x y = ex 0 1 1 2.7182 2 7.389 320.086 4 54.6 5 148.4 (b) y = e0 = 1 (c) For V = 0 V, e0 = 1 and I= Is(1 1) = 0 mA 6. 3 19. T = 20C: Is = 0.1 A T = 30C: Is = 2(0.1A) = 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(0.8A) = 1.6 A 1.6 A: 0.1 A 16:1 increase due to rise in temperature of40C. 20. For most applications the silicon diode is the device ofchoice due to its higher temperature capability. Ge typically has aworking limit of about 85 degrees centigrade while Si can be usedat temperatures approaching 200 degrees centigrade. Silicon diodesalso have a higher current handling capability. Germanium diodesare the better device for some RF small signal applications, wherethe smaller threshold voltage may prove advantageous. 21. From1.19: 75C 25C 125C VF @ 10 mA Is 1.1 V 0.01 pA 0.85 V 1 pA 0.6 V1.05 A VF decreased with increase in temperature 1.1 V: 0.6 V1.83:1 Is increased with increase in temperature 1.05 A: 0.01 pA =105 103 :1 22. An ideal device or system is one that has thecharacteristics we would prefer to have when using a device orsystem in a practical application. Usually, however, technologyonly permits a close replica of the desired characteristics. Theideal characteristics provide an excellent basis for comparisonwith the actual device characteristics permitting an estimate ofhow well the device or system will perform. On occasion, the idealdevice or system can be assumed to obtain a good estimate of theoverall response of the design. When assuming an ideal device orsystem there is no regard for component or manufacturing tolerancesor any variation from device to device of a particular lot. 23. Inthe forward-bias region the 0 V drop across the diode at any levelof current results in a resistance level of zero ohms the on stateconduction is established. In the reverse-bias region the zerocurrent level at any reverse-bias voltage assures a very highresistance level the open circuit or off state conduction isinterrupted. 24. The most important difference between thecharacteristics of a diode and a simple switch is that the switch,being mechanical, is capable of conducting current in eitherdirection while the diode only allows charge to flow through theelement in one direction (specifically the direction defined by thearrow of the symbol using conventional current flow). 25. VD 0.66V, ID = 2 mA RDC = 0.65 V 2 mA D D V I = = 325 7. 4 26. At ID = 15mA, VD = 0.82 V RDC = 0.82 V 15 mA D D V I = = 54.67 As the forwarddiode current increases, the static resistance decreases. 27. VD =10 V, ID = Is = 0.1 A RDC = 10 V 0.1 A D D V I = = 100 M VD = 30 V,ID = Is= 0.1 A RDC = 30 V 0.1 A D D V I = = 300 M As the reversevoltage increases, the reverse resistance increases directly (sincethe diode leakage current remains constant). 28. (a) rd = 0.79 V0.76 V 0.03 V 15 mA 5 mA 10 mA d d V I = = = 3 (b) rd = 26 mV 26 mV10 mADI = = 2.6 (c) quite close 29. ID = 10 mA, VD = 0.76 V RDC =0.76 V 10 mA D D V I = = 76 rd = 0.79 V 0.76 V 0.03 V 15 mA 5 mA 10mA d d V I = = 3 RDC >> rd 30. ID = 1 mA, rd = 0.72 V 0.61 V2 mA 0 mA d d V I = = 55 ID = 15 mA, rd = 0.8 V 0.78 V 20 mA 10 mAd d V I = = 2 31. ID = 1 mA, rd = 26 mV 2 DI = 2(26 ) = 52 vs 55(#30) ID = 15 mA, rd = 26 mV 26 mV 15 mADI = = 1.73 vs 2 (#30) 32.rav = 0.9 V 0.6 V 13.5 mA 1.2 mA d d V I = = 24.4 8. 5 33. rd = 0.8V 0.7 V 0.09 V 7 mA 3 mA 4 mA d d V I = = 22.5 (relatively close toaverage value of 24.4 (#32)) 34. rav = 0.9 V 0.7 V 0.2 V 14 mA 0 mA14 mA d d V I = = = 14.29 35. Using the best approximation to thecurve beyond VD = 0.7 V: rav = 0.8 V 0.7 V 0.1 V 25 mA 0 mA 25 mA dd V I = = 4 36. (a) VR = 25 V: CT 0.75 pF VR = 10 V: CT 1.25 pF1.25 pF 0.75 pF 0.5 pF 10 V 25 V 15 V T R C V = = = 0.033 pF/V (b)VR = 10 V: CT 1.25 pF VR = 1 V: CT 3 pF 1.25 pF 3 pF 1.75 pF 10 V 1V 9 V T R C V = = = 0.194 pF/V (c) 0.194 pF/V: 0.033 pF/V = 5.88:16:1 Increased sensitivity near VD = 0 V 37. From Fig. 1.33 VD = 0V, CD = 3.3 pF VD = 0.25 V, CD = 9 pF 38. The transitioncapacitance is due to the depletion region acting like a dielectricin the reverse- bias region, while the diffusion capacitance isdetermined by the rate of charge injection into the region justoutside the depletion boundaries of a forward-biased device. Bothcapacitances are present in both the reverse- and forward-biasdirections, but the transition capacitance is the dominant effectfor reverse-biased diodes and the diffusion capacitance is thedominant effect for forward-biased conditions. 9. 6 39. VD = 0.2 V,CD = 7.3 pF XC = 1 1 2 2 (6 MHz)(7.3 pF)fC = = 3.64 k VD = 20 V, CT= 0.9 pF XC = 1 1 2 2 (6 MHz)(0.9 pF)fC = = 29.47 k 40. If = 10 V10 k = 1 mA ts + tt = trr = 9 ns ts + 2ts = 9 ns ts= 3 ns tt = 2ts= 6 ns 41. 42. As the magnitude of the reverse-bias potentialincreases, the capacitance drops rapidly from a level of about 5 pFwith no bias. For reverse-bias potentials in excess of 10 V thecapacitance levels off at about 1.5 pF. 43. At VD = 25 V, ID = 0.2nA and at VD = 100 V, ID 0.45 nA. Although the change in IR is morethan 100%, the level of IR and the resulting change is relativelysmall for most applications. 44. Log scale: TA = 25C, IR = 0.5 nATA = 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). 10. 7 45. IF = 0.1 mA:rd 700 IF = 1.5 mA: rd 70 IF = 20 mA: rd 6 The results support thefact that the dynamic or ac resistance decreases rapidly withincreasing current levels. 46. T = 25C: Pmax = 500 mW T = 100C:Pmax = 260 mW Pmax = VFIF IF = max 500 mW 0.7 VF P V = = 714.29 mAIF = max 260 mW 0.7 VF P V = = 371.43 mA 714.29 mA: 371.43 mA =1.92:1 2:1 47. Using the bottom right graph of Fig. 1.37: IF = 500mA @ T = 25C At IF = 250 mA, T 104C 48. 49. TC = +0.072% = 1 0 100%( ) Z Z V V T T 0.072 = 1 0.75 V 100 10 V( 25)T 0.072 = 1 7.5 25TT1 25 = 7.5 0.072 = 104.17 T1 = 104.17 + 25 = 129.17 50. TC = 1 0() Z Z V V T T 100% = (5 V 4.8 V) 5 V(100 25 ) 100% = 0.053%/C 11. 851. (20 V 6.8 V) (24 V 6.8 V) 100% = 77% The 20 V Zener istherefore 77% of the distance between 6.8 V and 24 V measured fromthe 6.8 V characteristic. At IZ = 0.1 mA, TC 0.06%/C (5 V 3.6 V)(6.8 V 3.6 V) 100% = 44% The 5 V Zener is therefore 44% of thedistance between 3.6 V and 6.8 V measured from the 3.6 Vcharacteristic. At IZ = 0.1 mA, TC 0.025%/C 52. 53. 24 V Zener: 0.2mA: 400 1 mA: 95 10 mA: 13 The steeper the curve (higher dI/dV) theless the dynamic resistance. 54. VT 2.0 V, which is considerablyhigher than germanium ( 0.3 V) or silicon ( 0.7 V). For germaniumit is a 6.7:1 ratio, and for silicon a 2.86:1 ratio. 55. Fig. 1.53(f) IF 13 mA Fig. 1.53 (e) VF 2.3 V 56. (a) Relative efficiency @ 5mA 0.82 @ 10 mA 1.02 1.02 0.82 0.82 100% = 24.4% increase ratio:1.02 0.82 = 1.24 (b) Relative efficiency @ 30 mA 1.38 @ 35 mA 1.421.42 1.38 1.38 100% = 2.9% increase ratio: 1.42 1.38 = 1.03 (c) Forcurrents greater than about 30 mA the percent increase issignificantly less than for increasing currents of lessermagnitude. 12. 9 57. (a) 0.75 3.0 = 0.25 From Fig. 1.53 (i) 75 (b)0.5 = 40 58. For the high-efficiency red unit of Fig. 1.53: 0.2 mA20 mA C x = x = 20 mA 0.2 mA/ C = 100C 13. 10 Chapter 2 1. The loadline will intersect at ID = 8 V 330 E R = = 24.24 mA and VD = 8 V.(a) QDV 0.92 V QDI 21.5 mA VR = E QDV = 8 V 0.92 V = 7.08 V (b) QDV0.7 V QDI 22.2 mA VR = E QDV = 8 V 0.7 V = 7.3 V (c) QDV 0 V QDI24.24 mA VR = E QDV = 8 V 0 V = 8 V For (a) and (b), levels of QDVand QDI are quite close. Levels of part (c) are reasonably closebut as expected due to level of applied voltage E. 2. (a) ID = 5 V2.2 k E R = = 2.27 mA The load line extends from ID = 2.27 mA to VD= 5 V. QDV 0.7 V, QDI 2 mA (b) ID = 5 V 0.47 k E R = = 10.64 mA Theload line extends from ID = 10.64 mA to VD = 5 V. QDV 0.8 V, QDI 9mA (c) ID = 5 V 0.18 k E R = = 27.78 mA The load line extends fromID = 27.78 mA to VD = 5 V. QDV 0.93 V, QDI 22.5 mA The resultingvalues of QDV are quite close, while QDI extends from 2 mA to 22.5mA. 3. Load line through QDI = 10 mA of characteristics and VD = 7V will intersect ID axis as 11.25 mA. ID = 11.25 mA = 7 VE R R =with R = 7 V 11.25 mA = 0.62 k 14. 11 4. (a) ID = IR = 30 V 0.7 V2.2 k DE V R = = 13.32 mA VD = 0.7 V, VR = E VD = 30 V 0.7 V = 29.3V (b) ID = 30 V 0 V 2.2 k DE V R = = 13.64 mA VD = 0 V, VR = 30 VYes, 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(Kirchhoffs voltage law) I = 19.3 V 20 = 0.965 A (c) I = 10 V 10 =1 A; center branch open 6. (a) Diode forward-biased, Kirchhoffsvoltage law (CW): 5 V + 0.7 V Vo = 0 Vo = 4.3 V IR = ID = 4.3 V 2.2k oV R = = 1.955 mA (b) Diode forward-biased, ID = 8 V 0.7 V 1.2 k4.7 k + = 1.24 mA Vo = V4.7 k + VD = (1.24 mA)(4.7 k) + 0.7 V =6.53 V 7. (a) Vo = 2 k (20 V 0.7V 0.3V) 2 k 2 k + = 1 2 (20 V 1 V)= 1 2 (19 V) = 9.5 V (b) I = 10 V 2V 0.7 V) 11.3 V 1.2 k 4.7 k 5.9k + = + = 1.915 mA V = IR = (1.915 mA)(4.7 k) = 9 V Vo = V 2 V = 9V 2 V = 7 V 15. 12 8. (a) Determine the Thevenin equivalent circuitfor the 10 mA source and 2.2 k resistor. ETh = IR = (10 mA)(2.2 k)= 22 V RTh = 2. 2k Diode forward-biased ID = 22 V 0.7 V 2.2 k 1.2 k+ = 6.26 mA Vo = ID(1.2 k) = (6.26 mA)(1.2 k) = 7.51 V (b) Diodeforward-biased ID = 20 V + 5 V 0.7 V 6.8 k = 2.65 mA Kirchhoffsvoltage law (CW): +Vo 0.7 V + 5 V = 0 Vo = 4.3 V 9. (a) 1oV = 12 V0.7 V = 11.3 V 2oV = 0.3 V (b) 1oV = 10 V + 0.3 V + 0.7 V = 9 V I =10 V 0.7 V 0.3 V 9 V 1.2 k 3.3 k 4.5 k = + = 2 mA, 2oV = (2 mA)(3.3k) = 6.6 V 10. (a) Both diodes forward-biased IR = 20 V 0.7 V 4.7 k= 4.106 mA Assuming identical diodes: ID = 4.106 mA 2 2 RI = = 2.05mA Vo = 20 V 0.7 V = 19.3 V (b) Right diode forward-biased: ID = 15V + 5 V 0.7 V 2.2 k = 8.77 mA Vo = 15 V 0.7 V = 14.3 V 11. (a) Gediode on preventing Si diode from turning on: I = 10 V 0.3 V 9.7 V1 k 1 k = = 9.7 mA (b) I = 16 V 0.7 V 0.7 V 12 V 2.6 V 4.7 k 4.7 k= = 0.553 mA Vo = 12 V + (0.553 mA)(4.7 k) = 14.6 V 16. 13 12. Bothdiodes forward-biased: 1oV = 0.7 V, 2oV = 0.3 V I1 k = 20 V 0.7 V 1k = 19.3 V 1 k = 19.3 mA I0.47 k = 0.7 V 0.3 V 0.47 k = 0.851 mAI(Si diode) = I1 k I0.47 k = 19.3 mA 0.851 mA = 18.45 mA 13. Forthe parallel Si 2 k branches a Thevenin equivalent will result (foron diodes) in a single series branch of 0.7 V and 1 k resistor asshown below: I2 k = 6.2 V 2 k = 3.1 mA ID = 2 k 3.1 mA 2 2 I = =1.55 mA 14. Both diodes off. The threshold voltage of 0.7 V isunavailable for either diode. Vo = 0 V 15. Both diodes on, Vo = 10V 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 theother is off. The result is Vo = 5 V + 0.7 V = 4.3 V 19. 0 V at oneterminal 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 differenceof 0.7 V across either diode cannot be established. Therefore, bothdiodes are off and Vo = +10 V as established by 10 V supplyconnected to 1 k resistor. 17. 14 21. The Si diode requires moreterminal voltage than the Ge diode to turn on. Therefore, with 5 Vat both input terminals, assume Si diode off and Ge diode on. Theresult: Vo = 5 V 0.3 V = 4.7 V The result supports the aboveassumptions. 22. Vdc = 0.318 Vm Vm = dc 2 V 0.318 0.318 V = = 6.28V Im = 6.28 V 2.2 k mV R = = 2.85 mA 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 24. Vm = dc 2V 0.318 0.318 V = = 6.28 V maxLI = 6.28 V 6.8 k = 0.924 mA 18. 15Imax(2.2 k) = 6.28 V 2.2 k = 2.855 mA max maxD LI I= + Imax(2.2 k)= 0.924 mA + 2.855 mA = 3.78 mA 25. Vm = 2 (110 V) = 155.56 V Vdc =0.318Vm = 0.318(155.56 V) = 49.47 V 26. Diode will conduct when vo= 0.7 V; that is, vo = 0.7 V = 10 k ( ) 10 k 1 k iv + Solving: vi =0.77 V For vi 0.77 V Si diode is on and vo = 0.7 V. For vi 18.36 mA (e) Idiode =36.71 mA Imax = 20 mA 28. (a) Vm = 2 (120 V) = 169.7 V mLV = miV2VD = 169.7 V 2(0.7 V) = 169.7 V 1.4 V = 168.3 V Vdc = 0.636(168.3V) = 107.04 V (b) PIV = Vm(load) + VD = 168.3 V + 0.7 V = 169 V (c)ID(max) = 168.3 V 1 k mL L V R = = 168.3 mA (d) Pmax = VDID = (0.7V)Imax = (0.7 V)(168.3 mA) = 117.81 mW 29. 20. 17 30. Positivehalf-cycle of vi: Voltage-divider rule: maxoV = max 2.2 k ( ) 2.2 k2.2 k iV + = max 1 ( ) 2 iV = 1 (100 V) 2 = 50 V Polarity of voacross the 2.2 k resistor acting as a load is the same.Voltage-divider rule: maxoV = max 2.2 k ( ) 2.2 k 2.2 k iV + = max1 ( ) 2 iV = 1 (100 V) 2 = 50 V Vdc = 0.636Vm = 0.636 (50 V) = 31.8V 31. Positive pulse of vi: Top left diode off, bottom left diodeon 2.2 k 2.2 k = 1.1 k peakoV = 1.1 k (170 V) 1.1 k 2.2 k + =56.67 V Negative pulse of vi: Top left diode on, bottom left diodeoff peakoV = 1.1 k (170 V) 1.1 k 2.2 k + = 56.67 V Vdc =0.636(56.67 V) = 36.04 V 32. (a) Si diode open for positive pulseof vi and vo = 0 V For 20 V 5 Vthe diode is reverse-biased and vo = 0 V. 33. (a) Positive pulse ofvi: Vo = 1.2 k (10 V 0.7 V) 1.2 k 2.2 k + = 3.28 V Negative pulseof vi: diode open, vo = 0 V (b) Positive pulse of vi: Vo = 10 V 0.7V + 5 V = 14.3 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 = 5V, vi overpowers the 2 V battery and the diode is on. ApplyingKirchhoffs voltage law in the clockwise direction: 5 V + 2 V vo = 0vo = 3 V (b) For vi = 20 V the 20 V level overpowers the 5 V supplyand the diode is on. Using the short-circuit equivalent for thediode we find vo = vi = 20 V. For vi = 5 V, both vi and the 5 Vsupply reverse-bias the diode and separate vi from vo. However, vois connected directly through the 2.2 k resistor to the 5 V supplyand vo = 5 V. 22. 19 35. (a) Diode on for vi 4.7 V For vi > 4.7V, Vo = 4 V + 0.7 V = 4.7 V For vi 8 V both diodes arereverse-biased and vo = vi. iR: For 8 V < vi < 6 V there isno conduction through the 10 k resistor due to the lack of acomplete circuit. Therefore, iR = 0 mA. For vi 6 V vR = vi vo = vi6 V For vi = 10 V, vR = 10 V 6 V = 4 V and iR


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