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

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

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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 < 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 21. 18 (b) For vi 5 V the 5 V battery willensure the diode is forward-biased and vo = vi 5 V. At vi = 5 V vo= 5 V 5 V = 0 V At vi = 20 V vo = 20 V 5 V = 25 V For vi > 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 < 4.7 V, diode off and vo =vi (b) Again, diode on for vi 4.7 V but vo now defined as thevoltage across the diode For vi 4.7 V, vo = 0.7 V For vi < 4.7V, diode off, ID = IR = 0 mA and V2.2 k = IR = (0 mA)R = 0 VTherefore, vo = vi 4 V At vi = 0 V, vo = 4 V vi = 8 V, vo = 8 V 4 V= 12 V 36. For the positive region of vi: The right Si diode isreverse-biased. The left Si diode is on for levels of vi greaterthan 5.3 V + 0.7 V = 6 V. In fact, vo = 6 V for vi 6 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 = 4 V 10 k = 0.4 mAFor vi 8 V vR = vi vo = vi + 8 V 23. 20 For vi = 10 V vR = 10 V + 8V = 2 V and iR = 2 V 10 k = 0.2 mA 37. (a) Starting with vi = 20 V,the diode is in the on state and the capacitor quickly charges to20 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-circuitequivalent) and vo = vi + vC = 20 V + 20 V = +40 V (b) Startingwith vi = 20 V, the diode is in the on state and the capacitorquickly charges up to 15 V+. Note that vi = +20 V and the 5 Vsupply are additive across the capacitor. During this time intervalvo is across on diode and 5 V supply and vo = 5 V. When vi switchesto the +20 V level the diode enters the off state and vo = vi + vC= 20 V + 15 V = 35 V. 24. 21 38. (a) For negative half cyclecapacitor charges to peak value of 120 V 0.7 V = 119.3 V withpolarity . The output vo is directly across the on diode resultingin vo = 0.7 V as a negative peak value. For next positive halfcycle vo = vi + 119.3 V with peak value of vo = 120 V + 119.3 V =239.3 V. (b) For positive half cycle capacitor charges to peakvalue of 120 V 20 V 0.7 V = 99.3 V with polarity . The output vo =20 V + 0.7 V = 20.7 V For next negative half cycle vo = vi 99.3 Vwith negative peak value of vo = 120 V 99.3 V = 219.3 V. Using theideal diode approximation the vertical shift of part (a) would be120 V rather than 119.3 V and 100 V rather than 99.3 V for part(b). Using the ideal diode approximation would certainly beappropriate in this case. 39. (a) = RC = (56 k)(0.1 F) = 5.6 ms 5 =28 ms (b) 5 = 28 ms 2 T = 1 ms 2 = 0.5 ms, 56:1 (c) Positive pulseof vi: Diode on and vo = 2 V + 0.7 V = 1.3 V Capacitor charges to10 V + 2 V 0.7 V = 11.3 V Negative pulse of vi: Diode off and vo =10 V 11.3 V = 21.3 V 25. 22 40. Solution is network of Fig.2.176(b) using a 10 V supply in place of the 5 V source. 41.Network of Fig. 2.178 with 2 V battery reversed. 42. (a) In theabsence of the Zener diode VL = 180 (20 V) 180 220 + = 9 V VL = 9 V< VZ = 10 V and diode non-conducting Therefore, IL = IR = 20 V220 180 + = 50 mA with IZ = 0 mA and VL = 9 V (b) In the absence ofthe Zener diode VL = 470 (20 V) 470 220 + = 13.62 V VL = 13.62 V> VZ = 10 V and Zener diode on Therefore, VL = 10 V and sRV = 10V / 10 V/220s sR R sI V R= = = 45.45 mA IL = VL/RL = 10 V/470 =21.28 mA and IZ = sRI IL = 45.45 mA 21.28 mA = 24.17 mA (c) maxZP =400 mW = VZIZ = (10 V)(IZ) IZ = 400 mW 10 V = 40 mA minLI = maxsRZI I = 45.45 mA 40 mA = 5.45 mA RL = min 10 V 5.45 mA L L V I = =1,834.86 Large RL reduces IL and forces more of sRI to pass throughZener diode. (d) In the absence of the Zener diode VL = 10 V = (20V) 220 L L R R + 10RL + 2200 = 20RL 10RL = 2200 RL = 220 26. 23 43.(a) VZ = 12 V, RL = 12 V 200 mA L L V I = = 60 VL = VZ = 12 V = 60(16 V) 60 L i L s s R V R R R = + + 720 + 12Rs = 960 12Rs = 240 Rs= 20 (b) maxZP = VZ maxZI = (12 V)(200 mA) = 2.4 W 44. Since IL = LZ L L V V R R = is fixed in magnitude the maximum value of sRI willoccur when IZ is a maximum. The maximum level of sRI will in turndetermine the maximum permissible level of Vi. max max 400 mW 8 V ZZ Z P I V = = = 50 mA IL = 8 V 220 L Z L L V V R R = = = 36.36 mAsRI = IZ + IL = 50 mA + 36.36 mA = 86.36 mA s i Z R s V V I R = orVi = sR sI R + VZ = (86.36 mA)(91 ) + 8 V = 7.86 V + 8 V = 15.86 VAny value of vi that exceeds 15.86 V will result in a current IZthat will exceed the maximum value. 45. At 30 V we have to be sureZener diode is on. VL = 20 V = 1 k (30 V) 1 k L i L s s R V R R R =+ + Solving, Rs = 0.5 k At 50 V, 50 V 20 V 0.5 kSRI = = 60 mA, IL =20 V 1 k = 20 mA IZM = SRI IL = 60 mA 20 mA = 40 mA 46. For vi =+50 V: Z1 forward-biased at 0.7 V Z2 reverse-biased at the Zenerpotential and 2ZV = 10 V. Therefore, Vo = 1 2Z ZV V+ = 0.7 V + 10 V= 10.7 V 27. 24 For vi = 50 V: Z1 reverse-biased at the Zenerpotential and 1ZV = 10 V. Z2 forward-biased at 0.7 V. Therefore, Vo= 1 2Z ZV V+ = 10.7 V For a 5 V square wave neither Zener diodewill reach its Zener potential. In fact, for either polarity of vione 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 V48. The PIV for each diode is 2Vm PIV = 2(1.414)(Vrms) 28. 25Chapter 3 1. 2. A bipolar transistor utilizes holes and electronsin the injection or charge flow process, while unipolar devicesutilize either electrons or holes, but not both, in the charge flowprocess. 3. Forward- and reverse-biased. 4. The leakage current ICOis the minority carrier current in the collector. 5. 6. 7. 8. IEthe largest IB the smallest IC IE 9. IB = 1 100 CI IC = 100IB IE =IC + IB = 100IB + IB = 101IB IB = 8 mA 101 101 EI = = 79.21 A 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 changein VCB is 20 V:1 V = 20:1 The resulting change in VBE is 800 mV:750mV = 1.07:1 (very slight) 12. (a) rav = 0.9 V 0.7 V 8 mA 0 V I = =25 (b) Yes, since 25 is often negligible compared to the otherresistance levels of the network. 13. (a) IC IE = 4.5 mA (b) IC IE= 4.5 mA (c) negligible: change cannot be detected on this set ofcharacteristics. (d) IC IE 29. 26 14. (a) Using Fig. 3.7 first, IE7 mA Then Fig. 3.8 results in IC 7 mA (b) Using Fig. 3.8 first, IE5 mA Then Fig. 3.7 results in VBE 0.78 V (c) Using Fig. 3.10(b) IE= 5 mA results in VBE 0.81 V (d) Using Fig. 3.10(c) IE = 5 mAresults in VBE = 0.7 V (e) Yes, the difference in levels of VBE canbe ignored for most applications if voltages of several volts arepresent 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 dc = 2.78 mA2.8 mA C E I I = = 0.993 (c) IC = IB = 0.98 1 1 0.98 BI = (40 A) =1.96 mA IE = 1.96 mA 0.993 CI = = 2 mA 16. 17. Ii = Vi/Ri = 500mV/20 = 25 mA IL Ii = 25 mA VL = ILRL = (25 mA)(1 k) = 25 V Av = 25V 0.5 V o i V V = = 50 18. Ii = 200 mV 200 mV 20 100 120 i i s V RR = = + + = 1.67 mA IL = Ii = 1.67 mA VL = ILR = (1.67 mA)(5 k) =8.35 V Av = 8.35 V 0.2 V o i V V = = 41.75 19. 20. (a) Fig.3.14(b): IB 35A Fig. 3.14(a): IC 3.6 mA (b) Fig. 3.14(a): VCE 2.5 VFig. 3.14(b): VBE 0.72 V 30. 27 21. (a) = 2 mA 17 A C B I I = =117.65 (b) = 117.65 1 117.65 1 = + + = 0.992 (c) ICEO = 0.3 mA (d)ICBO = (1 )ICEO = (1 0.992)(0.3 mA) = 2.4 A 22. (a) Fig. 3.14(a):ICEO 0.3 mA (b) Fig. 3.14(a): IC 1.35 mA dc = 1.35 mA 10 A C B I I= = 135 (c) = 135 1 136 = + = 0.9926 ICBO (1 )ICEO = (1 0.9926)(0.3mA) = 2.2 A 23. (a) dc = 6.7 mA 80 A C B I I = = 83.75 (b) dc =0.85 mA 5 A C B I I = = 170 (c) dc = 3.4 mA 30 A C B I I = = 113.33(d) dc does change from pt. to pt. on the characteristics. Low IB,high VCE higher betas High IB, low VCE lower betas 24. (a) ac = 5 VC CEB I VI = = 7.3 mA 6 mA 1.3 mA 90 A 70 A 20 A = = 65 (b) ac = 15V C CEB I VI = = 1.4 mA 0.3 mA 1.1mA 10 A 0 A 10 A = = 110 (c) ac =10 V C CEB I VI = = 4.25 mA 2.35 mA 1.9 mA 40 A 20 A 20 A = = 95(d) ac does change from point to point on the characteristics. Thehighest value was obtained at a higher level of VCE and lower levelof IC. The separation between IB curves is the greatest in thisregion. 31. 28 (e) VCE IB dc ac IC dc/ac 5 V 80 A 83.75 65 6.7 mA1.29 10 V 30 A 113.33 95 3.4 mA 1.19 15 V 5 A 170 110 0.85 mA 1.55As IC decreased, the level of dc and ac increased. Note that thelevel of dc and ac in the center of the active region is close tothe average value of the levels obtained. In each case dc is largerthan ac, with the least difference occurring in the center of theactive region. 25. dc = 2.9 mA 25 A C B I I = = 116 = 116 1 116 1 =+ + = 0.991 IE = IC/ = 2.9 mA/0.991 = 2.93 mA 26. (a) = 0.987 0.9871 1 0.987 0.013 = = = 75.92 (b) = 120 120 1 120 1 121 = = + + =0.992 (c) IB = 2 mA 180 CI = = 11.11 A IE = IC + IB = 2 mA + 11.11A = 2.011 mA 27. 28. Ve = Vi Vbe = 2 V 0.1 V = 1.9 V Av = 1.9 V 2 Vo i V V = = 0.95 1 Ie = 1.9 V 1 k E E V R = = 1.9 mA (rms) 29.Output characteristics: Curves are essentially the same with newscales as shown. Input characteristics: Common-emitter inputcharacteristics may be used directly for common-collectorcalculations. 32. 29 30. maxCP = 30 mW = VCEIC IC = maxCI , VCE =max max C C P I = 30 mW 7 mA = 4.29 V VCE = maxCEV , IC = max max CCE P V = 30 mW 20 V = 1.5 mA VCE = 10 V, IC = max 30 mW 10 V C CE PV = = 3 mA IC = 4 mA, VCE = maxC C P I = 30 mW 4 mA = 7.5 V VCE =15 V, IC = max 30 mW 15 V C CE P V = = 2 mA 31. IC = maxCI , VCE =max max 30 mW 6 mA C C P I = = 5 V VCB = maxCBV , IC = max max 30mW 15 V C CB P V = = 2 mA IC = 4 mA, VCB = max 30 mW 4 mA C C P I == 7.5 V VCB = 10 V, IC = max 30 mW 10 V C CB P V = = 3 mA 33. 3032. The operating temperature range is 55C TJ 150C F = 9 C 5 + 32 =9 ( 55 C) 5 + 32 = 67F F = 9 (150 C) 5 + 32 = 302F 67F TJ 302F 33.maxCI = 200 mA, maxCEV = 30 V, maxDP = 625 mW IC = maxCI , CEV =max max D C P I = 625 mW 200 mA = 3.125 V VCE = maxCEV , IC = maxmax 625 mW 30 V D CE P V = = 20.83 mA IC = 100 mA, VCE = max 625 mW100 mA D C P I = = 6.25 V VCE = 20 V, IC = max 625 mW 20 V D CE P V= = 31.25 mA 34. From Fig. 3.23 (a) ICBO = 50 nA max avg = min max2 + = 50 150 2 + = 200 2 = 100 ICEO ICBO = (100)(50 nA) = 5 A 34.31 35. 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 hfethe same increase in collector current resulted in a similarincrease (relatively speaking) in the gain parameter. The levelsare higher for hfe but note that VCE is higher also. 36. As thereverse-bias potential increases in magnitude the input capacitanceCibo decreases (Fig. 3.23(b)). Increasing reverse-bias potentialscauses the width of the depletion region to increase, therebyreducing the capacitance A C d = . 37. (a) At IC = 1 mA, hfe 120 AtIC = 10 mA, hfe 160 (b) The results confirm the conclusions ofproblems 23 and 24 that beta tends to increase with increasingcollector current. 39. (a) ac = 3 V C CEB I VI = = 16 mA 12.2 mA3.8 mA 80 A 60 A 20 A = = 190 (b) dc = 12 mA 59.5 A C B I I = =201.7 (c) ac = 4 mA 2 mA 2 mA 18 A 8 A 10 A = = 200 (d) dc = 3 mA13 A C B I I = = 230.77 (e) In both cases dc is slightly higherthan ac ( 10%) (f)(g) In general dc and ac increase with increasingIC for fixed VCE and both decrease for decreasing levels of VCE fora fixed IE. However, if IC increases while VCE decreases whenmoving between two points on the characteristics, chances are thelevel of dc or ac may not change significantly. In other words, theexpected increase due to an increase in collector current may beoffset by a decrease in VCE. The above data reveals that this is astrong possibility since the levels of are relatively close. 35. 32Chapter 4 1. (a) 16 V 0.7 V 15.3 V 470 k 470 kQ CC BE B B V V I R == = = 32.55 A (b) Q QC BI I= = (90)(32.55 A) = 2.93 mA (c) Q QCE CCC CV V I R= = 16 V (2.93 mA)(2.7 k) = 8.09 V (d) VC = QCEV = 8.09 V(e) VB = VBE = 0.7 V (f) VE = 0 V 2. (a) IC = IB = 80(40 A) = 3.2mA (b) RC = 12 V 6 V 6 V 3.2 mA 3.2 mA CR CC C C C V V V I I = = == 1.875 k (c) RB = 12 V 0.7 V 11.3 V 40 A 40 A BR B V I = = = 282.5k (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 (c)= 3.98 mA 20 A C B I I = = 199 200 (d) RB = 15.96 V 0.7 V 20 A BRCC BE B B V V V I I = = = 763 k 4. satCI = 16 V 2.7 k CC C V R = =5.93 mA 5. (a) Load line intersects vertical axis at IC = 21 V 3 k= 7 mA and horizontal axis at VCE = 21 V. (b) IB = 25 A: RB = 21 V0.7 V 25 A CC BE B V V I = = 812 k (c) QCI 3.4 mA, QCEV 10.75 V 36.33 (d) = 3.4 mA 25 A C B I I = = 136 (e) = 136 136 1 136 1 137 = =+ + = 0.992 (f) satCI = 21 V 3 k CC C V R = = 7 mA (g) (h) PD = QQCE CV I = (10.75 V)(3.4 mA) = 36.55 mW (i) Ps = VCC(IC + IB) = 21V(3.4 mA + 25 A) = 71.92 mW (j) PR = Ps PD = 71.92 mW 36.55 mW =35.37 mW 6. (a) 20 V 0.7 V ( 1) 510 k (101)1.5 kQ CC BE B B E V V IR R = = + + + = 19.3 V 661.5 k = 29.18 A (b) Q QC BI I= =(100)(29.18 A) = 2.92 mA (c) QCEV = VCC IC(RC + RE) = 20 V (2.92mA)(2.4 k + 1.5 k) = 20 V 11.388 V = 8.61 V (d) VC = VCC ICRC = 20V (2.92 mA)(2.4 k) = 20 V 7.008 V = 13 V (e) VB = VCC IBRB = 20 V(29.18 A)(510 k) = 20 V 14.882 V = 5.12 V (f) VE = VC VCE = 13 V8.61 V = 4.39 V 7. (a) RC = 12 V 7.6 V 4.4 V 2 mA 2 mA CC C C V V I= = = 2.2 k (b) IE IC: RE = 2.4 V 2 mA E E V I = = 1.2 k (c) RB =12 V 0.7 V 2.4 V 8.9 V 2 mA/80 25 A BR CC BE E B B V V V V I I = == = 356 k (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 37. 34 8. (a) IC IE = 2.1 V 0.68 k E E V R == 3.09 mA = 3.09 mA 20 A C B I I = = 154.5 (b) VCC = CRV + VCE + VE= (3.09 mA)(2.7 k) + 7.3 V + 2.1 V = 8.34 V + 7.3 V + 2.1 V = 17.74V (c) RB = 17.74 V 0.7 V 2.1 V 20 A BR CC BE E B B V V V V I I = == 14.94 V 20 A = 747 k 9. sot 20 V 20 V 2.4 k 1.5 k 3.9 k CC C C EV I R R = = = + + = 5.13 mA 10. (a) satCI = 6.8 mA = 24 V 1.2 k CCC E C V R R R = + + RC + 1.2 k = 24 V 6.8 mA = 3.529 k RC = 2.33 k(b) = 4 mA 30 A C B I I = = 133.33 (c) RB = 24 V 0.7 V (4 mA)(1.2 k) 30 A BR CC BE E B B V V V V I I = = = 18.5 V 30 A = 616.67 k (d)PD = Q QCE CV I = (10 V)(4 mA) = 40 mW (e) P = 2 C CI R = (4 mA)2(2.33 k) = 37.28 mW 11. (a) Problem 1: QCI = 2.93 mA, QCEV = 8.09 V(b) QBI = 32.55 A (the same) Q QC BI I= = (135)(32.55 A) = 4.39 mAQ QCE CC C CV V I R= = 16 V (4.39 mA)(2.7 k) = 4.15 V 38. 35 (c)%IC = 4.39 mA 2.93 mA 2.93 mA 100% = 49.83% %VCE = 4.15 V 8.09 V8.09 V 100% = 48.70% Less than 50% due to level of accuracy carriedthrough calculations. (d) Problem 6: QCI = 2.92 mA, QCEV = 8.61 V (QBI = 29.18 A) (e) 20 V 0.7 V ( 1) 510 k (150 1)(1.5 k )Q CC BE B BE V V I R R = = + + + + = 26.21 A Q QC BI I= = (150)(26.21 A) =3.93 mA QCEV = VCC IC(RC + RE) = 20 V (3.93 mA)(2.4 k + 1.5 k) =4.67 V (f) %IC = 3.93 mA 2.92 mA 2.92 mA 100% = 34.59% %VCE = 4.67V 8.61 V 8.61 V 100% = 46.76% (g) For both IC and VCE the % changeis less for the emitter-stabilized. 12. 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 ETh = 2 2 1 (9.1 k )(16 V) 9.1 k 62 k CCRV R R = + + = 2.05 V 2.05 V 0.7 V ( 1) 7.94 k (81)(0.68 k )Q Th BEB Th E E V I R R = = + + + = 21.42 A (b) Q QC BI I= = (80)(21.42 A)= 1.71 mA (c) QCEV = VCC QCI (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 39. 36 13. (a) IC = 18 V 12 V 4.7 k CC C CV V R = = 1.28 mA (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 (d) R1 = 1 1 : R R V I1RV = VCC VB = 18 V 2.24 V = 15.76 V 1RI 2 2 2.24 V 5.6 k B R V I R= = = 0.4 mA R1 = 1 1 R R V I = 15.76 V 0.4 mA = 39.4 k 14. (a) IC= IB = (100)(20 A) = 2 mA (b) IE = IC + IB = 2 mA + 20 A = 2.02 mAVE = 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 V2.42 V = 8.18 V (e) VB = VE + VBE = 2.42 V + 0.7 V = 3.12 V (f) 12R R BI I I= + = 3.12 V 8.2 k + 20 A = 380.5 A + 20 A = 400.5 A R1= 1 16 V 3.12 V 400.5 A CC B R V V I = = 32.16 k 15. satCI = 16 V3.9 k 0.68 k CC C E V R R = + + = 16 V 4.58 k = 3.49 mA 40. 37 16.(a) RE 10R2 (120)(1 k) 10(8.2 k) 120 k 82 k (checks) VB = 2 1 2(8.2 k )(18 V) 39 k 8.2 k CCR V R R = + + = 3.13 V VE = VB VBE =3.13 V 0.7 V = 2.43 V IC IE = 2.43 V 1 k E E V R = = 2.43 mA (b)VCE = VCC IC(RC + RE) = 18 V (2.43 mA)(3.3 k + 1 k) = 7.55 V (c) IB= 2.43 mA 120 CI = = 20.25 A (d) VE = IERE ICRE = (2.43 mA)(1 k) =2.43 V (e) VB = 3.13 V 17. (a) RTh = R1 || R2 = 39 k || 8.2 k =6.78 k ETh = 1 2 8.2 k (18 V) 39 k 8.2 k C CCR V R R = + + = 3.13 VIB = 3.13 V 0.7 V ( 1) 6.78 k (121)(1k ) Th BE Th E E V R R = + + += 2.43 V 127.78 k = 19.02 A IC = IB = (120)(19.02 A) = 2.28 mA (vs.2.43 mA #16) (b) VCE = VCC IC(RC + RE) = 18 V (2.28 mA)(3.3 k + 1k) = 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) Theresults suggest that the approximate approach is valid if Eq. 4.33is satisfied. 18. (a) VB = 2 1 2 9.1 k (16 V) 62 k 9.1 k CC R V R R= + + = 2.05 V VE = VB VBE = 2.05 V 0.7 V = 1.35 V IE = 1.35 V 0.68k E E V R = = 1.99 mA QCI IE = 1.99 mA 41. 38 QCEV = VCC IC (RC +RE) = 16 V (1.99 mA)(3.9 k + 0.68 k) = 16 V 9.11 V = 6.89 V QBI =1.99 mA 80 QCI = = 24.88 A (b) From Problem 12: QCI = 1.71 mA, QCEV= 8.17 V, QBI = 21.42 A (c) The differences of about 14% suggestthat the exact approach should be employed when appropriate. 19.(a) sat 24 V 24 V 7.5 mA 3 4 CC C C E E E E V I R R R R R = = = = ++ RE = 24 V 24 V 4(7.5 mA) 30 mA = = 0.8 k 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 (d) VB = 2 2 1 CCR V R R+ , 4.7 V = 2 2 (24 V)24 k R R + R2 = 5.84 k (e) dc = 5 mA 38.5 A C B I I = = 129.8 (f)RE 10R2 (129.8)(0.8 k) 10(5.84 k) 103.84 k 58.4 k (checks) 20. (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 IB =2.05 V 0.7 V ( 1) 7.94 k + (121)(0.68 k ) Th BE Th E E V R R = + += 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 42. 39 (c) 1.8 mA 1.71 mA %1.71 mA CI = 100% = 5.26% 7.76 V 8.17 V % 8.17 V CEV = 100% = 5.02%(d) 11c 11f 20c %IC 49.83% 34.59% 5.26% %VCE 48.70% 46.76% 5.02%Fixed-bias Emitter feedback Voltage- divider (e) Quite obviously,the voltage-divider configuration is the least sensitive to changesin . 21. I.(a) Problem 16: Approximation approach: QCI = 2.43 mA,QCEV = 7.55 V Problem 17: Exact analysis: QCI = 2.28 mA, QCEV = 8.2V The exact solution will be employed to demonstrate the effect ofthe change of . Using the approximate approach would result in %IC= 0% and %VCE = 0%. (b) Problem 17: ETh = 3.13 V, RTh = 6.78 k IB =3.13 V 0.7 V 2.43 V ( 1) 6.78 k (180 1)1 k 187.78 k TH BE Th E E VR R = = + + + + = 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 (c) %IC =2.33 mA 2.28 mA 2.28 mA 100% = 2.19% %VCE = 7.98 V 8.2 V 8.2 V 100%= 2.68% For situations where RE > 10R2 the change in IC and/orVCE due to significant change in will be relatively small. (d) %IC= 2.19% vs. 49.83% for problem 11. %VCE = 2.68% vs. 48.70% forproblem 11. (e) Voltage-divider configuration considerably lesssensitive. II. The resulting %IC and %VCE will be quite small. 43.40 22. (a) IB = 16 V 0.7 V ( ) 470 k + (120)(3.6 k 0.51 k ) CC BE BC E V V R R R = + + + = 15.88 A (b) IC = IB = (120)(15.88 A) = 1.91mA (c) VC = VCC ICRC = 16 V (1.91 mA)(3.6 k) = 9.12 V 23. (a) IB =30 V 0.7 V ( ) 6.90 k 100(6.2 k 1.5 k ) CC BE B C E V V R R R = + ++ + = 20.07 A IC = IB = (100)(20.07 A) = 2.01 mA (b) VC = VCC ICRC= 30 V (2.01 mA)(6.2 k) = 30 V 12.462 V = 17.54 V (c) VE = IEREICRE = (2.01 mA)(1.5 k) = 3.02 V (d) VCE = VCC IC(RC + RE) = 30 V(2.01 mA)(6.2 k + 1.5 k) = 14.52 V 24. (a) IB = 22 V 0.7 V ( ) 470k (90)(9.1 k 9.1 k ) CC BE B C E V V R R R = + + + + = 10.09 A IC =IB = (90)(10.09 A) = 0.91 mA VCE = VCC IC(RC + RE) = 22 V (0.91mA)(9.1 k + 9.1 k) = 5.44 V (b) = 135, IB = 22 V 0.7 V ( ) 470 k(135)(9.1 k 9.1 k ) CC BE B C E V V R R R = + + + + = 7.28 A IC =IB = (135)(7.28 A) = 0.983 mA VCE = VCC IC(RC + RE) = 22 V (0.983mA)(9.1 k + 9.1 k) = 4.11 V (c) 0.983 mA 0.91 mA % 0.91 mA CI =100% = 8.02% 4.11 V 5.44 V % 5.44 V CEV = 100% = 24.45% (d) Theresults for the collector feedback configuration are closer to thevoltage-divider configuration than to the other two. However, thevoltage-divider configuration continues to have the leastsensitivities to change in . 44. 41 25. 1 M = 0 , RB = 150 k IB =12 V 0.7 V ( ) 150 k (180)(4.7 k 3.3 k ) CC BE B C E V V R R R = ++ + + = 7.11 A IC = IB = (180)(7.11 A) = 1.28 mA VC = VCC ICRC = 12V (1.28 mA)(4.7 k) = 5.98 V Full 1 M: RB = 1,000 k + 150 k = 1,150k = 1.15 M IB = 12 V 0.7 V ( ) 1.15 M (180)(4.7 k 3.3 k ) CC BE B CE V V R R R = + + + + = 4.36 A IC = IB = (180)(4.36 A) = 0.785 mAVC = VCC ICRC = 12 V (0.785 mA)(4.7 k) = 8.31 V VC ranges from 5.98V to 8.31 V 26. (a) VE = VB VBE = 4 V 0.7 V = 3.3 V (b) IC IE = 3.3V 1.2 k E E V R = = 2.75 mA (c) VC = VCC ICRC = 18 V (2.75 mA)(2.2k) = 11.95 V (d) VCE = VC VE = 11.95 V 3.3 V = 8.65 V (e) IB =11.95 V 4 V 330 k BR C B B B V V V R R = = = 24.09 A (f) = 2.75 mA24.09 A C B I I = = 114.16 27. (a) IB = 6 V + 6 V 0.7 V ( 1) 330 k(121)(1.2 k ) CC EE BE B E V V V R R + = + + + = 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 28. (a) IB = 12 V 0.7 V ( 1) 9.1 k(120 1)15 k EE BE B E V V R R = + + + + = 6.2 A (b) IC = IB =(120)(6.2 A) = 0.744 mA (c) VCE = VCC + VEE IC(RC + RE) = 16 V + 12V (0.744 mA)(27 k) = 7.91 V (d) VC = VCC ICRC = 16 V (0.744 mA)(12k) = 7.07 V 45. 42 29. (a) IE = 8 V 0.7 V 7.3 V 2.2 k 2.2 k = =3.32 mA (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 V30. (a) RE > 10R2 not satisfied Use exact approach: Networkredrawn to determine the Thevenin equivalent: RTh = 510 k 2 = 255 kI = 18 V + 18 V 510 k 510 k + = 35.29 A ETh = 18 V + (35.29 A)(510k) = 0 V IB = 18 V 0.7 V 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 + 7.5k) = 36 V 30.05 V = 5.95 V 31. (a) IB = 8 V 0.7 V 560 k BR C BE B BV V V R R = = = 13.04 A (b) IC = CC C C V V R = 18 V 8 V 10 V 3.9 k3.9 k = = 2.56 mA (c) = 2.56 mA 13.04 A C B I I = = 196.32 (d) VCE= VC = 8 V 46. 43 32. IB = 2.5 mA 80 CI = = 31.25 A RB = 12 V 0.7 V31.25 A BR CC BE B B V V V I I = = = 361.6 k RC = 12 V 6 V 6 V 2.5mA 2.5 mA QC Q CC CER CC C C C C V VV V V I I I = = = = = 2.4 kStandard values: RB = 360 k RC = 2.4 k 33. satCI = CC C E V R R+ =10 mA 20 V 4 E ER R+ = 10 mA 20 V 5 ER = 10 mA 5RE = 20 V 10 mA = 2k RE = 2 k 5 = 400 RC = 4RE = 1.6 k IB = 5 mA 120 CI = = 41.67 A RB= VRB/IB = 20 V 0.7 V 5 mA(0.4 k ) 19.3 2 V 41.67 A 41.67 A = =415.17 k Standard values: RE = 390 , RC = 1.6 k, RB = 430 k 34. RE= 3 V 4 mA E E E C V V I I = = 0.75 k RC = ( )QC CC CE ER CC C C CC V V VV V V I I I + = = = 24 V (8 V + 3 V) 24 V 11 V 13 V 4 mA 4mA 4 mA = = = 3.25 k VB = VE + VBE = 3 V + 0.7 V = 3.7 V VB = 2 2 21 2 1 (24 V) 3.7 V =CCR V R R R R R + + 2 unknowns! use RE 10R2 forincreased stability (110)(0.75 k) = 10R2 R2 = 8.25 k Choose R2 =7.5 k 47. 44 Substituting in the above equation: 3.7 V = 1 7.5 k(24 V) 7.5 k R + R1 = 41.15 k Standard values: RE = 0.75 k, RC =3.3 k, R2 = 7.5 k, R1 = 43 k 35. VE = 1 5 CCV = 1 (28 V) 5 = 5.6 VRE = 5.6 V 5 mA E E V I = = 1.12 k (use 1.1 k) VC = 28 V 2 2 CC E VV+ = + 5.6 V = 14 V + 5.6 V = 19.6 V CRV = VCC VC = 28 V 19.6 V =8.4 V RC = 8.4 V 5 mA CR C V I = = 1.68 k (use 1.6 k) VB = VBE + VE= 0.7 V + 5.6 V = 6.3 V VB = 2 2 1 CCR V R R+ 6.3 V = 2 2 1 (28 V)RR R+ (2 unknowns) = 5 mA 37 A C B I I = = 135.14 RE = 10R2(135.14)(1.12 k) = 10(R2) R2 = 15.14 k (use 15 k) Substituting: 6.3V = 1 (15.14 k )(28 V) 15.14 k R + Solving, R1 = 52.15 k (use 51 k)Standard values: RE = 1.1 k RC = 1.6 k R1 = 51 k R2 = 15 k 36. I2 k= 18 V 0.7 V 2 k = 8.65 mA I 37. For current mirror: I(3 k) = I(2.4k) = I = 2 mA 38. QD DSSI I= = 6 mA 48. 45 39. VB 4.3 k ( 18 V) 4.3k 4.3 k + = 9 V VE = 9 V 0.7 V = 9.7 V IE = 18 V ( 9.7 V) 1.8 k =4.6 mA = I 40. IE = 5.1 V 0.7 V 1.2 k Z BE E V V R = = 3.67 mA 41.sat 10 V 2.4 k CC C C V I R = = = 4.167 mA From characteristicsmaxBI 31 A IB = 10 V 0.7 V 180 k i BE B V V R = = 51.67 A 51.67 A31 A, well saturated Vo = 10 V (0.1 mA)(2.4 k) = 10 V 0.24 V = 9.76V 42. satCI = 8 mA = 5 V CR RC = 5 V 8 mA = 0.625 k maxBI = sat 8mA 100 CI = = 80 A Use 1.2 (80 A) = 96 A RB = 5 V 0.7 V 96 A =44.79 k Standard values: RB = 43 k RC = 0.62 k 49. 46 43. (a) FromFig. 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 = 110ns + 38 ns = 148 ns (b) IC = 10 mA: tf = 12 ns, tr = 15 ns, td = 22ns, 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 droppeddramatically 168 ns:37 ns = 4.54:1 while the turn-off time is onlyslightly smaller 148 ns:132 ns = 1.12:1 44. (a) Open-circuit in thebase 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 45. (a) The basevoltage of 9.4 V reveals that the 18 k resistor is not makingcontact with the base terminal of the transistor. If operatingproperly: VB 18 k (16 V) 18 k 91 k + = 2.64 V vs. 9.4 V As anemitter feedback bias circuit: IB = 1 16 V 0.7 V ( 1) 91 k (1001)1.2 k CC BE E V V R R = + + + + = 72.1 A VB = VCC IB(R1) = 16 V(72.1 A)(91 k) = 9.4 V 50. 47 (b) Since VE > VB the transistorshould be off With IB = 0 A, VB = 18 k (16 V) 18 k 91 k + = 2.64 VAssume base circuit open The 4 V at the emitter is the voltage thatwould exist if the transistor were shorted collector to emitter. VE= 1.2 k (16 V) 1.2 k 3.6 k + = 4 V 46. (a) RB, IB, IC, VC (b) , IC(c) Unchanged, satCI not a function of (d) VCC, IB, IC (e) , IC,CRV , ERV , VCE 47. (a) IB = ( 1) Th BE Th BE Th E Th E E V E V R RR R + + + IC = IB = Th BE Th BE ThTh E E E V E V RR R R = + + As ,ThR , IC, CRV VC = VCC CRV and VC (b) R2 = open, IB, IC VCE = VCCIC(RC + RE) and VCE (c) VCC, VB, VE, IE, IC (d) IB = 0 A, IC = ICEOand IC(RC + RE) negligible with VCE VCC = 20 V (e) Base-emitterjunction = short IB but transistor action lost and IC = 0 mA withVCE = VCC = 20 V 48. (a) RB open, IB = 0 A, IC = ICEO 0 mA and VCVCC = 18 V (b) , IC, CRV , ERV , VCE (c) RC, IB, IC, VE (d) Drop toa relatively low voltage 0.06 V (e) Open in the base circuit 51. 4849. IB = 12 V 0.7 V 11.3 V 510 k 510 k CC BE B V V R = = = 22.16 AIC = IB = (100)(22.16 A) = 2.216 mA VC = VCC + ICRC = 12 V + (2.216mA)(3.3 k) = 4.69 V VCE = VC = 4.69 V 50. RE 10R2 (220)(0.75 k)10(16 k) 165 k 160 k (checks) Use approximate approach: VB 16 k (22 V) 16 k + 82 k = 3.59 V VE = VB + 0.7 V = 3.59 V + 0.7 V = 2.89V IC IE = VE/RE = 2.89/0.75 k = 3.85 mA IB = 3.85 mA 220 CI = =17.5 A VC = VCC + ICRC = 22 V + (3.85 mA)(2.2 k) = 13.53 V 51. IE =8 V 0.7 V 7.3 V 3.3 k 3.3 k BE E V V R = = = 2.212 mA VC = VCC +ICRC = 12 V + (2.212 mA)(3.9 k) = 3.37 V 52. (a) S(ICO) = + 1 = 91(b) S(VBE) = 90 470 kBR = = 1.92 104 S (c) S() = 1 1 2.93 mA 90 CI= = 32.56 106 A (d) IC = S(ICO)ICO + S(VBE)VBE + S() = (91)(10 A0.2 A) + (1.92 104 S)(0.5 V 0.7 V) + (32.56 106 A)(112.5 90) =(91)(9.8 A) + (1.92 104 S)(0.2 V) + (32.56 106 A)(22.5) = 8.92 104A + 0.384 104 A + 7.326 104 A = 16.63 104 A 1.66 mA 52. 49 53. Forthe emitter-bias: (a) S(ICO) = ( + 1) (1 / ) (1 510 k /1.5 k ) (1001) ( 1) / (100 1) 510 k /1.5 k B E B E R R R R + + = + + + + + =78.1 (b) S(VBE) = 100 ( 1) 510 k (100 1)1.5 kB ER R = + + + + =1.512 104 S (c) S() = 1 1 2 (1 / ) 2.92 mA(1 + 340) (1 / ) 100(1125 340) C B E B E I R R R R + = + + + + = 21.37 106 A (d) IC =S(ICO)ICO + S(VBE)VBE + S() = (78.1)(9.8 A) + (1.512 1014 S)(0.2 V)+ (21.37 106 A)(25) = 0.7654 mA + 0.0302 mA + 0.5343 mA = 1.33 mA54. (a) RTh = 62 k || 9.1 k = 7.94 k S(ICO) = ( + 1) 1 / (1 7.94 k/0.68 k ) (80 1) ( 1) / (80 1) 7.94 k /0.68 k Th E Th E R R R R + += + + + + + = (81)(1 11.68) 81 11.68 + + = 11.08 (b) S(VBE) = 80 (1) 7.94 k (81)(0.68 k )Th ER R = + + + = 80 7.94 k 55.08 k + = 1.27103 S (c) S() = 1 1 2 (1 / ) 1.71 mA(1 + 7.94 k /0.68 k ) (1 / )80(1 100 7.94 k /0.68 k ) C Th E Th E I R R R R + = + + + + = 1.71mA(12.68) 80(112.68) = 2.41 106 A (d) IC = S(ICO)ICO + S(VBE) VBE +S() = (11.08)(10 A 0.2 A) + (1.27 103 S)(0.5 V 0.7 V) + (2.41 106A)(100 80) = (11.08)(9.8 A) + (1.27 103 S)(0.2 V) + (2.41 106A)(20) = 1.09 104 A + 2.54 104 A + 0.482 104 A = 4.11 104 A = 0.411mA 53. 50 55. For collector-feedback bias: (a) S(ICO) = ( + 1) (1 /) (1 560 k /3.9 k ) (196.32 1) ( 1) / (196.32 1) 560 k /3.9 k B C BC R R R R + + = + + + + + = (197.32) 1 143.59 (197.32 143.59) + + =83.69 (b) S(VBE) = 196.32 ( 1) 560 k (196.32 1)3.9 kB CR R = + + ++ = 1.477 104 S (c) S() = 1 1 2 ( ) 2.56 mA(560 k 3.9 k ) ( ( 1))196.32(560 k 3.9 k (245.4 1)) C B C B C I R R R R + + = + + + + =4.83 106 A (d) IC = S(ICO)ICO + S(VBE) VBE + S() = (83.69)(9.8 A) +(1.477 104 S)(0.2 V) + (4.83 106 A)(49.1) = 8.20 104 A + 0.295 104A + 2.372 104 A = 10.867 104 A = 1.087 mA 56. Type S(ICO) S(VBE)S() Collector feedback 83.69 1.477 104 S 4.83 106 A Emitter-bias78.1 1.512 104 S 21.37 106 A Voltage-divider 11.08 12.7 104 S 2.41106 A Fixed-bias 91 1.92 104 S 32.56 106 A S(ICO): Considerablyless for the voltage-divider configuration compared to the otherthree. S(VBE): The voltage-divider configuration is more sensitivethan the other three (which have similar levels of sensitivity).S(): The voltage-divider configuration is the least sensitive withthe fixed-bias configuration very sensitive. In general, thevoltage-divider configuration is the least sensitive with thefixed-bias the most sensitive. 57. (a) Fixed-bias: S(ICO) = 91, IC= 0.892 mA S(VBE) = 1.92 104 S, IC = 0.0384 mA S() = 32.56 106 A,IC = 0.7326 mA (b) Voltage-divider bias: S(ICO) = 11.08, IC =0.1090 mA S(VBE) = 1.27 103 S, IC = 0.2540 mA S() = 2.41 106 A, IC= 0.0482 mA 54. 51 (c) For the fixed-bias configuration there is astrong sensitivity to changes in ICO and and less to changes inVBE. For the voltage-divider configuration the opposite occurs witha high sensitivity to changes in VBE and less to changes in ICO and. In total the voltage-divider configuration is considerably morestable than the fixed-bias configuration. 55. 52 Chapter 5 1. (a)If the dc power supply is set to zero volts, the amplification willbe zero. (b) Too low a dc level will result in a clipped outputwaveform. (c) Po = I2 R = (5 mA)2 2.2 k = 55 mW Pi = VCCI = (18V)(3.8 mA) = 68.4 mW (ac) 55 mW (dc) 68.4 mW o i P P = = = 0.80480.4% 2. 3. xC = 1 1 2 2 (1 kHz)(10 F)fC = = 15.92 f = 100 kHz: xC= 0.159 Yes, better at 100 kHz 4. 5. (a) Zi = 10 mV 0.5 mA i i V I= = 20 (=re) (b) Vo = IcRL = IcRL = (0.98)(0.5 mA)(1.2 k) = 0.588 V(c) Av = 0.588 V 10 mV o i V V = = 58.8 (d) Zo = (e) Ai = o i I I =e e I I = = 0.98 (f) Ib = Ie Ic = 0.5 mA 0.49 mA = 10 A 56. 53 6.(a) re = 48 mV 3.2 mA i i V I = = 15 (b) Zi = re = 15 (c) IC = Ie =(0.99)(3.2 mA) = 3.168 mA (d) Vo = ICRL = (3.168 mA)(2.2 k) = 6.97V (e) Av = 6.97 V 48 mV o i V V = = 145.21 (f) Ib = (1 )Ie = (10.99)Ie = (0.01)(3.2 mA) = 32 A 7. (a) re = 26 mV 26 mV (dc) 2 mAEI= = 13 Zi = re = (80)(13 ) = 1.04 k (b) Ib = 1 1 C e e eI I I I = == + + = 2 mA 81 = 24.69 A (c) Ai = o L i b I I I I = IL = ( )o b oL r I r R + Ai = o b o L o b o L r I r R r I r R + = + = 40 k (80)40 k 1.2 k + = 77.67 (d) Av = 1.2 k 40 k 13 L o e R r r = = 1.165 k13 = 89.6 57. 54 8. (a) Zi = re = (140)re = 1200 re = 1200 140 =8.571 (b) Ib = 30 mV 1.2 k i i V Z = = 25 A (c) Ic = Ib = (140)(25A) = 3.5 mA (d) IL = (50 k )(3.5 mA) 50 k 2.7 k o c o L r I r R = ++ = 3.321 mA Ai = 3.321 mA 25 A L i I I = = 132.84 (e) Av = o i L ii V A R V Z = = (2.7 k ) (132.84) 1.2 k = 298.89 9. (a) re: IB = 12V 0.7 V 220 k CC BE B V V R = = 51.36 A IE = ( + 1)IB = (60 +1)(51.36 A) = 3.13 mA re = 26 mV 26 mV 3.13 mAEI = = 8.31 Zi = RB|| re = 220 k || (60)(8.31 ) = 220 k || 498.6 = 497.47 ro 10RC Zo =RC = 2.2 k (b) Av = 2.2 k 8.31 C e R r = = 264.74 (c) Zi = 497.47(the same) Zo = ro || RC = 20 k || 2.2 k = 1.98 k (d) Av = 1.98 k8.31 C o e R r r = = 238.27 Ai = AvZi/RC = (238.27)(497.47 )/2.2 k= 53.88 58. 55 10. Av = C e R r re = 4.7 k ( 200) C v R A = = 23.5re = 26 mV EI IE = 26 mV 26 mV 23.5er = = 1.106 mA IB = 1.106 mA 191 EI = + = 12.15 A IB = CC BE B V V R VCC = IBRB + VBE = (12.15A)(1 M) + 0.7 V = 12.15 V + 0.7 V = 12.85 V 11. (a) IB = 10 V 0.7 V390 k CC BE B V V R = = 23.85 A IE = ( + 1)IB = (101)(23.85 A) =2.41 mA re = 26 mV 26 mV 2.41 mAEI = = 10.79 IC = IB = (100)(23.85A) = 2.38 mA (b) Zi = RB || re = 390 k || (100)(10.79 ) = 390 k ||1.08 k = 1.08 k ro 10RC Zo = RC = 4.3 k (c) Av = 4.3 k 10.79 C e Rr = = 398.52 (d) Av = (4.3 k ) (30 k ) 3.76 k 10.79 10.79 C o e R rr = = = 348.47 12. (a) Test RE 10R2 (100)(1.2 k) ? 10(4.7 k) 120 k> 47 k (satisfied) Use approximate approach: VB = 2 1 2 4.7 k(16 V) 39 k 4.7 k CCR V R R = + + = 1.721 V VE = VB VBE = 1.721 V0.7 V = 1.021 V IE = 1.021 V 1.2 k E E V R = = 0.8507 mA re = 26 mV26 mV 0.8507 mAEI = = 30.56 59. 56 (b) Zi = R1 || R2 || re = 4.7 k|| 39 k || (100)(30.56 ) = 1.768 k ro 10RC Zo RC = 3.9 k (c) Av =3.9 k 30.56 C e R r = = 127.6 (d) ro = 25 k (b) Zi(unchanged) =1.768 k Zo = RC || ro = 3.9 k || 25 k = 3.37 k (c) Av = ( ) (3.9 k) (25 k ) 3.37 k 30.56 30.56 C o e R r r = = = 110.28 (vs. 127.6)13. RE ? 10R2 (100)(1 k) 10(5.6 k) 100 k > 56 k (checks!) &ro 10RC Use approximate approach: Av = 3.3 k 160 C C e e v R R r rA = = = 20.625 re = 26 mV 26 mV 26 mV 20.625 E E e I I r = = =1.261 mA IE = E E V R VE = IERE = (1.261 mA)(1 k) = 1.261 V VB =VBE + VE = 0.7 V + 1.261 V = 1.961 V VB = 5.6 k 5.6 k 82 k CCV + =1.961 V 5.6 k VCC = (1.961 V)(87.6 k) VCC = 30.68 V 14. Test RE10R2 (180)(2.2 k) ? 10(56 k) 396 k < 560 k (not satisfied) Useexact analysis: (a) RTh = 56 k || 220 k = 44.64 k ETh = 56 k (20 V)220 k 56 k + = 4.058 V IB = 4.058 V 0.7 V ( 1) 44.64 k (181)(2.2 k) Th BE Th E E V R R = + + + 60. 57 = 7.58 A IE = ( + 1)IB =(181)(7.58 A) = 1.372 mA re = 26 mV 26 mV 1.372 mAEI = = 18.95 (b)VE = IERE = (1.372 mA)(2.2 k) = 3.02 V VB = VE + VBE = 3.02 V + 0.7V = 3.72 V VC = VCC ICRC = 20 V IBRC = 20 V (180)(7.58 A)(6.8 k) =10.72 V (c) Zi = R1 || R2 || re = 56 k || 220 k || (180)(18.95 k) =44.64 k || 3.41 k = 3.17 k ro < 10RC Av = C o e R r r = (6.8 k )(50 k ) 18.95 = 315.88 15. (a) IB = 20 V 0.7 V ( 1) 390 k (141)(1.2k ) CC BE B E V V R R = + + + = 19.3 V 559.2 k = 34.51 A IE = ( +1)IB = (140 + 1)(34.51 A) = 4.866 mA re = 26 mV 26 mV 4.866 mAEI == 5.34 (b) Zb = re + ( + 1)RE = (140)(5.34 k) + (140 + 1)(1.2 k) =747.6 + 169.9 k = 169.95 k Zi = RB || Zb = 390 k || 169.95 k =118.37 k Zo = RC = 2.2 k (c) Av = C b R Z = (140)(2.2 k ) 169.95 k= 1.81 (d) Zb = re + ( 1) / 1 ( )/ C o E C E o R r R R R r + + + += 747.6 (141) 2.2 k / 20 k 1 (3.4 k )/ 20 k + + 1.2 k 61. 58 =747.6 + 144.72 k = 145.47 k Zi = RB || Zb = 390 k || 145.47 k =105.95 k Zo = RC = 2.2 k (any level of ro) Av = 1 1 C e C b o oo Cio R r R Z r rV RV r + + = + = (140)(2.2 k ) 5.34 2.2 k 1 145.47 k20 k 20 k 2.2 k 1 20 k + + + = 2.117 0.11 1.11 + = 1.81 16. Eventhough the condition ro 10RC is not met it is sufficiently close topermit the use of the approximate approach. Av = C C C b E E R R RZ R R = = = 10 RE = 8.2 k 10 10 CR = = 0.82 k IE = 26 mV 26 mV3.8er = = 6.842 mA VE = IERE = (6.842 mA)(0.82 k) = 5.61 V VB = VE+ VBE = 5.61 V + 0.7 V = 6.31 V IB = 6.842 mA ( 1) 121 EI = + =56.55 A and RB = 20 V 6.31 V 56.55 A BR CC B B B V V V I I = = =242.09 k 17. (a) dc analysis the same re = 5.34 (as in #15) (b) Zi= RB || Zb = RB || re = 390 k || (140)(5.34 ) = 746.17 vs. 118.37 kin #15 Zo = RC = 2.2 k (as in #15) (c) Av = 2.2 k 5.34 C e R r = =411.99 vs 1.81 in #15 (d) Zi = 746.17 vs. 105.95 k for #15 Zo = RC|| ro = 2.2 k || 20 k = 1.98 k vs. 2.2 k in #15 62. 59 Av = 1.98 k5.34 C o e R r r = = 370.79 vs. 1.81 in #15 Significant differencein the results for Av. 18. (a) IB = ( 1) CC BE B E V V R R + + = 22V 0.7 V 21.3 V 330 k (81)(1.2 k 0.47 k ) 465.27 k = + + = 45.78 AIE = ( + 1)IB = (81)(45.78 A) = 3.71 mA re = 26 mV 26 mV 3.71 mAEI= = 7 (b) ro < 10(RC + RE) Zb = re + ( 1) / 1 ( )/ C o E C E o Rr R R R r + + + + = (80)(7 ) + (81) 5.6 k / 40 k 1 6.8 k / 40 k + +1.2 k = 560 + 81 0.14 1 0.17 + + 1.2 k (note that ( + 1) = 81 RC/ro= 0.14) = 560 + [81.14 /1.17]1.2 k = 560 + 83.22 k = 83.78 k Zi =RB || Zb = 330 k || 83.78 k = 66.82 k Av = 1 1 C e C b o o C o R rR Z r r R r + + + = (80)(5.6 k ) 7 5.6 k 1 83.78 k 40 k 40 k 1 5.6k /40 k + + + = (5.35) 0.14 1 0.14 + + = 4.57 19. (a) IB = 16 V 0.7V 15.3 V ( 1) 270 k (111)(2.7 k ) 569.7 k CC BE B E V V R R = = + ++ 63. 60 = 26.86 A IE = ( + 1)IB = (110 + 1)(26.86 A) = 2.98 mA re= 26 mV 26 mV 2.98 mAEI = = 8.72 re = (110)(8.72 ) = 959.2 (b) Zb =re + ( + 1)RE = 959.2 + (111)(2.7 k) = 300.66 k Zi = RB || Zb = 270k || 300.66 k = 142.25 k Zo = RE || re = 2.7 k || 8.72 = 8.69 (c)Av = 2.7 k 2.7 k 8.69 E E e R R r = + + 0.997 20. (a) IB = 8 V 0.7V ( 1) 390 k (121)5.6 k CE BE B E V V R R = + + + = 6.84 A IE = ( +1)IB = (121)(6.84 A) = 0.828 mA re = 26 mV 26 mV 0.828 mAEI = =31.4 ro < 10RE: Zb = re + ( 1) 1 / E E o R R r + + = (120)(31.4) + (121)(5.6 k ) 1 5.6 k /40 k + = 3.77 k + 594.39 k = 598.16 k Zi= RB || Zb = 390 k || 598.16 k = 236.1 k Zo RE || re = 5.6 k ||31.4 = 31.2 (b) Av = ( 1) / 1 / E b E o R Z R r + + = (121)(5.6 k)/598.16 k 1 5.6 k / 40 k + = 0.994 (c) Av = 0 i V V = 0.994 Vo =AvVi = (0.994)(1 mV) = 0.994 mV 64. 61 21. (a) Test RE ? 10R2(200)(2 k) 10(8.2 k) 400 k 82 k (checks)! Use approximate approach:VB = 8.2 k (20 V) 8.2 k 56 k + = 2.5545 V VE = VB VBE = 2.5545 V0.7 V 1.855 V IE = 1.855 V 2 k E E V R = = 0.927 mA IB = 0.927 mA (1) (200 + 1) EI = + = 4.61 A IC = IB = (200)(4.61 A) = 0.922 mA (b)re = 26 mV 26 mV 0.927 mAEI = = 28.05 (c) Zb = re + ( + 1)RE =(200)(28.05 ) + (200 + 1)2 k = 5.61 k + 402 k = 407.61 k Zi = 56 k|| 8.2 k || 407.61 k = 7.15 k || 407.61 k = 7.03 k Zo = RE || re =2 k || 28.05 = 27.66 (d) Av = 2 k 2 k 28.05 E E e R R r = + + =0.986 22. (a) IE = 6 V 0.7 V 6.8 k EE BE E V V R = = 0.779 mA re =26 mV 26 mV 0.779 mAEI = = 33.38 (b) Zi = RE || re = 6.8 k || 33.38= 33.22 Zo = RC = 4.7 k (c) Av = (0.998)(4.7 k ) 33.38 C e R r = =140.52 23. = 75 1 76 = + = 0.9868 65. 62 IE = 5 V 0.7 V 4.3 V 3.9 k3.9 k EE BE E V V R = = = 1.1 mA re = 26 mV 26 mV 1.1 mAEI = =23.58 Av = (0.9868)(3.9 k ) 23.58 C e R r = = 163.2 24. (a) IB = 12V 0.7 V 220 k 120(3.9 k ) CC BE F C V V R R = + + = 16.42 A IE = (+ 1)IB = (120 + 1)(16.42 A) = 1.987 mA re = 26 mV 26 mV 1.987 mAEI= = 13.08 (b) Zi = re || F v R A Need Av! Av = 3.9 k 13.08 C e R r= = 298 Zi = (120)(13.08 ) || 220 k 298 = 1.5696 k || 738 = 501.98Zo = RC || RF = 3.9 k || 220 k = 3.83 k (c) From above, Av = 29825. Av = C e R r = 160 RC = 160(re) = 160(10 ) = 1.6 k Ai = F F C RR R + = 19 19 = 200 200(1.6 k ) F F R R + 19RF + 3800RC = 200RF RF= 3800 181 CR = 3800(1.6 k ) 181 = 33.59 k IB = CC BE F C V V R R +IB(RF + RC) = VCC VBE 66. 63 and VCC = VBE + IB(RF + RC) with IE =26 mV 26 mV 10er = = 2.6 mA IB = 2.6 mA 1 200 1 EI = + + = 12.94 AVCC = VBE + IB(RF + RC) = 0.7 V + (12.94 A)(33.59 k + (200)(1.6 k))= 5.28 V 26. (a) Av: Vi = Ibre + ( + 1)IbRE Io + I = IC = Ib but Ii= I + Ib and I = Ii Ib Substituting, Io + (Ii Ib) = Ib and Io = ( +1)Ib Ii Assuming ( + 1)Ib Ii Io ( + 1)Ib and Vo = IoRC = ( + 1)IbRCTherefore, ( 1) ( 1) o b C i b e b E V I R V I r I R + = + + b C be b E I R I r I R + and Av = o C C i e E E V R R V r R R + (b) ViIb(re + RE) For re RE Vi IbRE Now Ii = I + Ib = i o b F V V I R +67. 64 Since Vo Vi Ii = o b F V I R + or Ib = Ii + o F V R and Vi =IbRE Vi = REIi + o E F V R R but Vo = AvVi and Vi = REIi + v i E FAV R R or Vi v E i F A R V R = REIi Vi [ ]1 v E E i F A R R I R =so Zi = ( )1 i E E F v Ei F v E F V R R R A RI R A R R = = + Zi = ii V I = x || y where x = RE and y = RF/|Av| with Zi = ( )( / ) / EF v E F v R R Ax y x y R R A = + + Zi E F E v F R R R A R+ Zo: SetVi = 0 Vi = Ibre + ( + 1)IbRE Vi Ib(re + RE) = 0 since , re + RE 0Ib = 0 and Ib = 0 68. 65 Io = 1 1o o o C F C F V V V R R R R + = +and Zo = 1 1 1 o C F o C F C F V R R I R R R R = = ++ = RC || RF(c) Av 2.2 k 1.2 k C E R R = = 1.83 Zi (90)(1.2 k )(120 k )(90)(1.2 k )(1.83) 120 k E F E v F R R R A R = + + = 40.8 k Zo RC|| RF = 2.2 k || 120 k = 2.16 k 27. (a) IB = 9 V 0.7 V (39 k 22 k )(80)(1.8 k ) CC BE F C V V R R = + + + = 8.3 V 8.3 V 61 k 144 k 205k = + = 40.49 A IE = ( + 1)IB = (80 + 1)(40.49 A) = 3.28 mA re = 26mV 26 mV 3.28 mAEI = = 7.93 Zi = 1FR || er = 39 k || (80)(7.93 ) =39 k || 634.4 = 0.62 k Zo = RC || 2FR = 1.8 k || 22 k = 1.66 k (b)Av = 2 1.8 k 22 k 7.93 C F e e R RR r r = = = 1.664 k 7.93 = 209.8228. Ai = 60 29. Ai = 100 30. Ai = AvZi/RC = (127.6)(1.768 k)/3.9 k= 57.85 31. (c) Ai = (140)(390 k ) 390 k 0.746 k B B b R R Z = + += 139.73 (d) Ai = i v C Z A R = (370.79)(746.17 )/2.2 k = 125.7669. 66 32. Ai = AvZi/RE = (0.986)(7.03 k)/2 k = 3.47 33. Ai = o e ie I I I I = = = 0.9868 1 34. Ai = AvZi/RC = (298)(501.98 )/3.9 k =38.37 35. Ai = ( 209.82)(0.62 k ) 1.8 k i v C Z A R = = 72.27 36.(a) IB = 18 V 0.7 V 680 k CC BE B V V R = = 25.44 A IE = ( + 1)IB =(100 + 1)(25.44 A) = 2.57 mA re = 26 mV 2.57 mA = 10.116 3.3 k10.116NL C v e R A r = = = 326.22 Zi = RB || re = 680 k ||(100)(10.116 ) = 680 k || 1,011.6 = 1.01 k Zo = RC = 3.3 k (b) (c)LvA = 4.7 k 4.7 k 3.3 kNL L v L o R A R R = + + (326.22) = 191.65(d) LiA = (1.01 k ) ( 191.65) 4.7 kL i v L Z A R = = 41.18 (e) LvA= ( ) 100(1.939 k ) ( ) 100(10.116 ) b C Lo i b e I R RV V I r = == 191.98 Zi = RB || re = 1.01 k IL = ( )C b C L R I R R + = 41.25IbIb = B i B e R I R r+ = 0.9985Ii LiA = o bL L i i b i I II I I I II = = = (41.25)(0.9985) = 41.19 Zo = RC = 3.3 k 70. 67 37. (a) NLvA= 326.22 LvA = NL L v L o R A R R+ RL = 4.7 k: LvA = 4.7 k (326.22) 4.7 k 3.3 k + = 191.65 RL = 2.2 k: LvA = 2.2 k ( 326.22)2.2 k 3.3 k + = 130.49 RL = 0.5 k: LvA = 0.5 k ( 326.22) 0.5 k 2.3k + = 42.92 As RL, LvA (b) No change for Zi, Zo, and NLvA ! 38. (a)IB = 12 V 0.7 V 1 M CC BE B V V R = = 11.3 A IE = ( + 1)IB =(181)(11.3 A) = 2.045 mA re = 26 mV 26 mV 2.045 mAEI = = 12.71 NLVA= 3 k 12.71 C e R r = = 236 Zi = RB || re = 1 M || (180)(12.71 ) =1 M || 2.288 k = 2.283 k Zo = RC = 3 k (b) (c) No-load: Av = NLvA =236 (d) 2.283 k ( 236) 2.283 k 0.6 ks NL i v v i s Z A A Z R = = ++ = 186.9 (e) Vo = IoRC = IbRC Vi = Ibre Av = 3 k 12.71 o b C C i be e V I R R V I r r = = = = 236 s o o i v s i s V V V A V V V = =Vi = (1 M ) 2.288 k ( ) (1 M ) 2.288 k 0.6 k e s s e s r V V r R =+ + = 0.792 Vs svA = (236)(0.792) = 186.9 (same results) 71. 68 (f)No change! (g) 2.283 k ( 236) ( ) 2.283 k 1 ks NL i v v i s Z A A ZR = = + + = 164.1 Rs, svA (h) No change! 39. (a) IB = 24 V 0.7 V500 k CC BE B V V R = = 41.61 A IE = ( + 1)IB = (80 + 1)(41.61 A) =3.37 mA re = 26 mV 26 mV 3.37 mAEI = = 7.715 NLvA = 4.3 k 7.715 L eR r = = 557.36 Zi = RB || re = 560 k || (80)(7.715 ) = 560 k ||617.2 = 616.52 Zo = RC = 4.3 k (b) (c) LvA = 2.7 k ( 557.36) 2.7 k4.3 kNL o L v i L o V R A V R R = = + + = 214.98 s o o i v s i s VV V A V V V = = Vi = 616.52 616.52 1 k i s s i s Z V V Z R = + + =0.381 Vs ( 214.98)(0.381)svA = = 81.91 (d) 1 k 616.52 ( 81.91) 2.7ks s s i i v L R Z A A R + + = = = 49.04 (e) LvA = 5.6 k ( 557.36)5.6 k 4.3 kNL o L v i L o V R A V R R = = + + = 315.27 i s V V thesame = 0.381 s o i v i s V V A V V = = (315.27)(0.381) = 120.12 AsRL, svA 72. 69 (f) LvA the same = 214.98 616.52 616.52 0.5 k i i si s V Z V Z R = = + + = 0.552 s o i v i s V V A V V = =(214.98)(0.552) = 118.67 As Rs, svA (g) No change! 40. (a) Exactanalysis: ETh = 2 1 2 16 k (16 V) 68 k 16 k CC R V R R = + + =3.048 V RTh = R1 || R2 = 68 k || 16 k = 12.95 k IB = 3.048 V 0.7 V( 1) 12.95 k (101)(0.75 k ) Th BE Th E E V R R = + + + = 26.47 A IE= ( + 1)IB = (101)(26.47 A) = 2.673 mA re = 26 mV 26 mV 2.673 mAEI= = 9.726 2.2 k 9.726NL C v e R A r = = = 226.2 Zi = 68 k || 16 k|| re = 12.95 k || (100)(9.726 ) = 12.95 k || 972.6 = 904.66 Zo =RC = 2.2 k (b) (c) LvA = 5.6 k ( 226.2) ( ) 5.6 k 2.2 kNL L v L o RA R Z = + + = 162.4 (d) LiA = L i v L Z A R = (162.4) (904.66 ) 5.6k = 26.24 73. 70 (e) LvA = 2.2 k 5.6 k 9.726 C e e R R r = = 162.4Zi = 68 k || 16 k || 972.6 re = 904.66 LiA = L i v L Z A R = (162.4)(904.66 ) 5.6 k = 26.24 Zo = RC = 2.2 k Same results! 41. (a)LvA = NL L v L o R A R Z+ RL = 4.7 k: LvA = 4.7 k ( 226.4) 4.7 k2.2 k + = 154.2 RL = 2.2 k: LvA = 2.2 k ( 226.4) 2.2 k 2.2 k + =113.2 RL = 0.5 k: LvA = 0.5 k ( 226.4) 0.5 k 2.2 k + = 41.93 RL,LvA (b) Unaffected! 42. (a) IB = 18 V 0.7 V ( 1) 680 k (111)(0.82 k) CC BE B E V V R R = + + + = 22.44 A IE = ( + 1)IB = (110 +1)(22.44 A) = 2.49 mA re = 26 mV 26 mV 2.49 mAEI = = 10.44 3 k10.44 0.82 kNL C v e E R A r R = = + + = 3.61 Zi RB || Zb = 680 k|| (re + ( + 1)RE) = 680 k || (610)(10.44 ) + (110 + 1)(0.82 k) =680 k || 92.17 k = 81.17 k Zo RC = 3 k (b) 74. 71 (c) LvA = 4.7 k (3.61) 4.7 k 3 kNL o L v i L o V R A V R R = = + + = 2.2 s o o i v si s V V V A V V V = = Vi = 81.17 k ( ) 81.17 k 0.6 k i s s i s Z VV Z R = + + = 0.992 Vs svA = (2.2)(0.992) = 2.18 (d) None! (e) LvAnone! 81.17 k 81.17 k 1 k i i s i s V Z V Z R = = + + = 0.988 svA =(2.2)(0.988) = 2.17 Rs, svA , (but only slightly for moderatechanges in Rs since Zi is typically much larger than Rs) 43. Usingthe exact approach: IB = ( 1) Th BE Th E E V R R + + ETh = 2 1 2 CCR V R R+ = 2.33 V 0.7 V 10.6 k (121)(1.2 k ) + = 12 k (20 V) 91 k12 k + = 2.33 V = 10.46 A RTh = R1 || R2 = 91 k || 12 k = 10.6 k IE= ( + 1)IB = (121)(10.46 A) = 1.266 mA re = 26 mV 26 mV 1.266 mAEI= = 20.54 (a) 1.2 k 20.54 1.2 kNL E v e E R A r R = + + = 0.983 Zi= R1 || R2 || (re + ( +1)RE) = 91 k || 12 k || ((120)(20.54 ) +(120 + 1)(1.2 k)) = 10.6 k || (2.46 k + 145.2 k) = 10.6 k || 147.66k = 9.89 k Zo = RE || re = 1.2 k || 20.54 = 20.19 75. 72 (b) (c)LvA = 2.7 k (0.983) 2.7 k 20.19NL L v L o R A R Z = + + = 0.9769.89 k (0.976) 9.89 k 0.6 ks L i v v i s Z A A Z R = = + + = 0.92(d) LvA = 0.976 (unaffected by change in Rs) 9.89 k (0.976) 9.89 k1 ks L i v v i s Z A A Z R = = + + = 0.886 (vs. 0.92 with Rs = 0.6k) As Rs, svA (e) Changing Rs will have no effect on NLvA , Zi, orZo. (f) LvA = ( ) 5.6 k (0.983) 5.6 k 20.19NL L v L o R A R Z = + += 0.979 (vs. 0.976 with RL = 2.7 k) 9.89 k (0.979) ( ) 9.89 k 0.6ks L i v v i s Z A A Z R = = + + = 0.923 (vs. 0.92 with RL = 2.7 k)As RL, LvA , svA 44. (a) IE = 6 V 0.7 V 2.2 k EE BE E V V R = =2.41 mA re = 26 mV 26 mV 2.41 mAEI = = 10.79 NLvA = 4.7 k 10.79 C eR r = = 435.59 Zi = RE || re = 2.2 k || 10.79 = 10.74 Zo = RC = 4.7k (b) (c) LvA = 5.6 k (435.59) 5.6 k 4.7 kNL L v L o R A R R = + += 236.83 Vi = 10.74 ( ) 10.74 100 i s s i s Z V V Z R = + + = 0.097Vs s o i v i s V V A V V = = (236.83)(0.097) = 22.97 76. 73 (d) Vi= Ie re Vo = IoRL Io = 4.7 k ( ) 4.7 k 5.6 k eI + = 0.4563Ie LvA =o i V V = (0.4563 ) 0.4563(5.6 k ) 10.79 e L e e I R I r + = =236.82 (vs. 236.83 for part c) :svA 2.2 k || 10.79 = 10.74 Vi =10.74 ( ) 10.74 100 i s s i s Z V V Z R = + + = 0.097 Vs s o i v is V V A V V = = (236.82)(0.097) = 22.97 (same results) (e) LvA =2.2 k (435.59) 2.2 k 4.7 kNL L v L o R A R R = + + = 138.88 0 10.74, 10.74 500s i i i v i s s i s V V V Z A V V V Z R = = = + + =0.021 svA = (138.88)(0.021) = 2.92 svA very sensitive to increasein Rs due to relatively small Zi; Rs, svA LvA sensitive to RL; RL,LvA (f) Zo = RC = 4.7 k unaffected by value of Rs! (g) Zi = RE ||re = 10.74 unaffected by value of RL! 45. (a) 1 1 k ( 420) 1 k 3.3k NLL v v L o R A A R R = = + + = 97.67 2 2.7 k ( 420) 2.7 k 3.3 kNLL v v L o R A A R R = = + + = 189 (b) LvA = 1 2v vA A =(97.67)(189) = 18.46 103 1 1 2 1 s o io o v s i i s V VV V A V V VV = = = 2 1 i v v s V A A V Vi = 1 k ( ) 1 k 0.6 k i s s i s Z V VZ R = + + = 0.625 svA = (189)(97.67)(0.625) = 11.54 103 77. 74 (c)1 ( 97.67)(1 k ) 1 k v i i L A Z A R = = = 97.67 2 ( 189)(1 k ) 2.7k v i i L A Z A R = = = 70 (d) 1 2Li i iA A A= = (97.67)(70) = 6.84103 (e) No effect! (f) No effect! (g) In phase 46. (a) 2 1 1 2 11.2 k (1) 1.2 k 20NL i v v i o Z A A Z Z = = + + = 0.984 2vA = 2 22.2 k ( 640) 2.2 k 4.6 kNL L v L o R A R Z = + + = 207.06 (b) 1 2Lvv vA A A= = (0.984)(207.06) = 203.74 s L i v v i s Z A A Z R = + =50 k ( 203.74) 50 k 1 k + = 199.75 (c) 1 1 1 2 i i v i Z A A Z = =(50 k ) (0.984) 1.2 k = 41 2 2 2 i i v L Z A A R = = (1.2 k ) (207.06) 2.2 k = 112.94 (d) 1 L L i i v L Z A A R = = (203.74) (50 k) 2.2 k = 4.63 103 78. 75 (e) A load on an emitter-followerconfiguration will contribute to the emitter resistance (in fact,lower the value) and therefore affect Zi (reduce its magnitude).(f) The fact that the second stage is a CE amplifier will isolateZo from the first stage and Rs. (g) The emitter-follower has zerophase shift while the common-emitter amplifier has a 180 phaseshift. The system, therefore, has a total phase shift of 180 asnoted by the negative sign in front of the gain for TvA in part b.47. For each stage: VB = 6.2 k 24 k 6.2 k + (15 V) = 3.08 V VE = VB0.7 V = 3.08 V 0.7 V = 2.38 V IE IC = 2.38 V 1.5 k E E V R = = 1.59mA VC = VCC ICRC = 15 V (1.59 mA)(5.1 k) = 6.89 V 48. re = 26 mV 26mV 1.59 mAEI = = 16.35 2iR = R1 || R2 || re = 6.2 k || 24 k ||(150)(16.35 ) = 1.64 k 2 1 5.1 k 1.64 k 16.35 C i v e R R A r = = =75.8 2 5.1 k 16.35 C v e R A r = = = 311.9 Av = 1 2v vA A =(75.8)(311.9) = 23,642 49. 1 3.9 k (20 V) 3.9 k 6.2 k 7.5 k BV = ++ = 4.4 V 2 6.2 k 3.9 k (20 V) 3.9 k 6.2 k 7.5 k BV + = + + = 11.48V 1 1E BV V= 0.7 V = 4.4 V 0.7 V = 3.7 V 1 1 1 3.7 V 1 k E C E E VI I R = = = 3.7 mA 2 2E CI I 2CV = VCC ICRC = 20 V (3.7 mA)(1.5 k)= 14.45 V 79. 76 50. re = 26 mV 26 mV 3.7 mAEI = = 7 1 e v e r A r= = 1 2 1.5 k 7 E v e R A r = = 214 1 2Tv v vA A A= = (1)(214) =214 Vo = Tv iA V = (214)(10 mV) = 2.14 V 51. Ro = RD = 1.5 k (Vo(from problem 50) = 2.14 V) Vo(load) = 10 k ( ) ( 2.14 V) 10 k 1.5k L o o L R V R R = + + = 1.86 V 52. IB = (16 V 1.6 V) (6000)(510 )2.4 M CC BE D E B V V R R = + + = 14.4 V 5.46 M = 2.64 A IC IE =DIB = 6000(2.64 A) = 15.8 mA VE = IERE = (15.8 mA)(510 ) = 8.06 V53. From problem 69, IE = 15.8 mA re = 26 26 V 15.8 mAEI = = 1.65Av = 510 1.65 510 E e E R r R = + + = 0.997 1 54. dc: 16 V 1.6 V2.4 M (6000)(510 ) CC BE B B D E V V I R R = + + = 2.64 A(6000)(2.64 A)C D BI I = = = 15.84 mA 2 2 26 mV 26 mV 15.84 mA e Er I = = = 1.64 ac: 2 (6000)(1.64 ) 9.84 ki D eZ r = = 1 9.84 k i bV I = 1 ( )( ) (6000) (200 ) 9.84 k = 121.95 i o D b C V V I R Vi == and o v i V A V = 121.95 80. 77 55. IB = 1 1 2 16 V 0.7 V 1.5 M(160)(200)(100 ) CC EB B E V V R R = + + = 3.255 A IC 12IB =(160)(200)(3.255A) 104.2 mA 2CV = VCC ICRC = 16 V (104.2 mA)(100 )= 5.58 V 1BV = IBRB = (3.255 A)(1.5 M) = 4.48 V 56. From problem55: 1EI = 0.521 mA 1 26 mV 26 mV (mA) 0.521 mA e E r I = = = 49.9 1ii eR r= = 160(49.9 ) = 7.98 k Av = 1 1 2 1 2 (160)(200)(100 )(160)(200)(100 ) 7.98 k C C i R R R = + + = 0.9925 Vo = AvVi =0.9975 (120 mV) = 119.7 mV 57. re = (dc) 26 mV 26 mV 1.2 mAEI = =21.67 re = (120)(21.67 ) = 2.6 k 58. 59. 60. 61. 62. (a) Av = o i VV = 160 Vo = 160 Vi (b) Ib = (1 )i re o i re v i i re v ie ie ie Vh V V h AV V h A h h h = = = ( )4 1 (2 10 )(160) 1 k iV Ib = 9.68104 Vi (c) Ib = 1 k iV = 1 103 Vi 81. 78 (d) % Difference = 3 4 3 110 9.68 10 1 10 i i i V V V 100% = 3.2 % (e) Valid firstapproximation 63. % difference in total load = 1/L L oe L R R h R100% = 2.2 k (2.2 k 50 k ) 2.2 k 100% = 2.2 k 2.1073 k 2.2 k 100% =4.2 % In this case the effect of 1/hoe can be ignored. 64. (a) Vo =180Vi (hie = 4 k, hre = 4.05 104 ) (b) Ib = 4 (4.05 10 )(180 ) 4 ki iV V = 2.32 104 Vi (c) Ib = 4 k i i ie V V h = = 2.5 104 Vi (d) %Difference = 4 4 4 2.5 10 2.32 10 2.5 10 i i i V V V 100% = 7.2%(e) Yes, less than 10% 65. From Fig. 5.18 min max hoe: 1 S 30 S Avg= (1 30) S 2 + = 15.5 S 66. (a) hfe = = 120 hie re = (120)(4.5 ) =540 hoe = 1 1 40 kor = = 25 S (b) re 1 k 90 ieh = = 11.11 = hfe =90 ro = 1 1 20 Soeh = = 50 k 82. 79 67. (a) re = 8.31 (from problem9) (b) feh = = 60 ieh = re = (60)(8.31 ) = 498.6 (c) Zi = RB || hie= 220 k || 498.6 = 497.47 Zo = RC = 2.2 k (d) Av = (60)(2.2 k )498.6 fe C ie h R h = = 264.74 Ai hfe = 60 (e) Zi = 497.47 (thesame) Zo = ro || RC, ro = 1 25 S = 40 k = 40 k || 2.2 k = 2.09 k(f) Av = ( ) (60)(2.085 k ) 498.6 fe o C ie h r R h = = 250.90 Ai =AvZi/RC = (250.90)(497.47 )/2.2 k = 56.73 68. (a) 68 k || 12 k =10.2 k Zi = 10.2 k || hie = 10.2 k || 2.75 k = 2.166 k Zo = RC ||ro = 2.2 k || 40 k = 2.085 k (b) Av = fe C ie h R h CR = RC || ro =2.085 k = (180)(2.085 k ) 2.75 k = 136.5 Ai = o o i i i i I I I I II = = 10.2 k 1 10.2 k 2.68 k ef oe L h h R + + = 180 (0.792) 1 (25S)(2.2 k ) + = 135.13 83. 80 69. (a) Zi = RE || hib = 1.2 k || 9.45= 9.38 Zo = RC || 1 obh = 2.7 k || 6 1 A 1 10 V = 2.7 k || 1 M 2.7k (b) Av = ( 1/ ) ( 0.992)( 2.7 k ) 9.45 fb C ob ib h R h h = =284.43 Ai 1 (c) = hfb = (0.992) = 0.992 = 0.992 1 1 0.992 = = 124re = hib = 9.45 ro = 1 1 1 A/Vobh = = 1 M 70. (a) 4 (180)(2 10)(2.2 k ) 2.75 k 1 (1 25 S)(2.2 k ) fe re L i ie oe L h h R Z h h R= = + + 2.68 k= Zi = 10.2 k || iZ = 2.12 k 4 1 1 ( / ) 25 S (180)(210 )/ 2.75 k 83.75 k o oe fe re ie Z h h h h = = = Zo = 2.2 k ||83.75 k = 2.14 k (b) ( )4 (180)(2.2 k ) ( ) 2.75 k (2.75 k )(25 S)(180)(2 10 ) 2.2 k fe L v ie ie oe fe re L h R A h h h h h R = = ++ = 140.3 (c) (180) 170.62 1 1 (25 S)(2.2 k ) fe i oe L h A h R = == + + Ai = 10.2 k (170.62) 10.2 k 2.68 k o o i i ii I I I I II = =+ = 135.13 84. 81 71. (a) Zi = hie = 1 fe re L oe L h h R h R + =0.86 k 4 (140)(1.5 10 )(2.2 k ) 1 (25 S)(2.2 k ) + = 0.86 k 43.79 =816.21 iZ = RB || Zi = 470 k || 816.21 = 814.8 (b) Av = ( ) fe L ieie oe fe re L h R h h h h h R + = 4 (140)(2.2 k ) 0.86 k ((0.86 k)(25 S) (140)(1.5 10 ))2.2 k + = 357.68 (c) Ai = 140 1 1 (25 S)(2.2k ) feo i oe L hI I h R = = + + = 132.70 o o i i ii i I I I A II I= = Ii = 470 k 470 k 0.816 k iI + = (132.70)(0.998) i i I I = 0.998= 132.43 (d) Zo = ( ) 1 /( )oe fe re ie sh h h h R + = ( )4 1 25 S(140)(1.5 10 )/(0.86 k 1 k ) + = 1 13.71 S 72.9 k 72. (a) 4 (0.997)(1 10 )(2.2 k ) 9.45 1 1 (0.5 A/V)(2.2 k ) fb rb L i ib ob Lh h R Z h h R = = + + 9.67= Zi = 1.2 k || iZ = 1.2 k || 9.67 = 9.59(b) 0.997 0.996 1 1 (0.5 A/V)(2.2 k ) fb i ob L h A h R = = = + +1.2 k ( 0.996) 1.2 k 9.67 k o i i ii I I A II = = + 0.988 85. 82(c) ( )4 ( ) ( 0.997)(2.2 k ) 9.45 (9.45 )(0.5 A/V) ( 0.997)(1 10 )(2.2 k ) fb L v ib ib ob fb rb L h R A h h h h h R = + = + = 226.61(d) 4 1 / 1 0.5 A/V ( 0.997)(1 10 )/9.45 90.5 k o ob fb rb ib Z h hh h = = = Zo = 2.2 k || oZ = 2.15 k 73. 74. (a) hfe (0.2 mA) 0.6(normalized) hfe (1 mA) = 1.0 % change = (0.2 mA) (1mA) (0.2 mA) fefe fe h h h 100% = 0.6 1 0.6 100% = 66.7% (b) hfe(1 mA) = 1.0 hfe(5mA) 1.5 % change = (1mA) (5 mA) (1 mA) fe fe fe h h h 100% = 1 1.51 100% = 50% 75. Log-log scale! (a) Ic = 0.2 mA, hie = 4(normalized) Ic = 1 mA, hie = 1(normalized) % change = 4 1 4 100% =75% (b) Ie = 5 mA, hie = 0.3 (normalized) % change = 1 0.3 1 100% =70% 86. 83 76. (a) hoe = 20 S @ 1 mA Ic = 0.2 mA, hoe = 0.2(hoe @ 1mA) = 0.2(20 S) = 4 S (b) ro = 1 1 4 Soeh = = 250 k 6.8 k Ignore1/hoe 77. (a) Ic = 10 mA, hoe = 10(20 S) = 200 S (b) ro = 1 1 200Soeh = = 5 k vs. 8.6 k Not a good approximation 78. (a) hre(0.1 mA)= 4(hre(1 mA)) = 4(2 104 ) = 8 104 (b) hreVce = hreAv Vi = (8 104)(210)Vi = 0.168 Vi In this case hreVce is too large a factor to beignored. 79. (a) hfe (b) hoe (c) hoe 30 (normalized) to hoe 0.1(normalized) at low levels of Ic (d) mid-region 80. (a) hie is themost temperature-sensitive parameter of Fig. 5.33. (b) hoeexhibited the smallest change. (c) Normalized: hfe(max) = 1.5,hfe(min) = 0.5 For hfe = 100 the range would extend from 50 to150certainly significant. (d) On a normalized basis re increasedfrom 0.3 at 65C to 3 at 200Ca significant change. (e) Theparameters show the least change in the region 0 100C. 87. 84 81.(a) Test: RE 10R2 70(1.5 k) 10(39 k) 105 k ? 390 k No! RTh = 39 k|| 150 k = 30.95 k ETh = 39 k (14 V) 39 k 150 k + = 2.89 V IB =2.89 V 0.7 V ( 1) 30.95 k (71)(1.5 k ) Th BE Th E E V R R = + + + =15.93 A VB = ETh IBRTh = 2.89 V (15.93 A)(30.95 k) = 2.397 V VE =2.397 V 0.7 V = 1.697 V and IE = 1.697 V 1.5 k E E V R = = 1.13 mAVCE = VCC IC(RC + RE) = 14 V 1.13 mA(2.2 k + 1.5 k) = 9.819 VBiasing OK (b) R2 not connected at base: IB = 0 14 V 0.7 V ( 1) 150k + (71)(1.5 k ) CC B E V R R = + + = 51.85 A VB = VCC IBRB = 14 V(51.85 A)(150 k) = 6.22 V as noted in Fig. 5.187. 88. 85 Eq. (6.1)is valid! Chapter 6 1. 2. From Fig. 6.11: VGS = 0 V, ID = 8 mA VGS= 1 V, ID = 4.5 mA VGS = 1.5 V, ID = 3.25 mA VGS = 1.8 V, ID = 2.5mA VGS = 4 V, ID = 0 mA VGS = 6 V, ID = 0 mA 3. (a) VDS 1.4 V (b)rd = 1.4 V 6 mA V I = = 233.33 (c) VDS 1.6 V (d) rd = 1.6 V 3 mA VI = = 533.33 (e) VDS 1.4 V (f) rd = 1.4 V 1.5 mA V I = = 933.33 (g)ro = 233.33 rd = [ ] [ ] 2 2 233.33 233.33 0.56251 1 ( 1 V) ( 4 V)o GS P r V V = = = 414.81 (h) rd = [ ] 2 233.33 233.33 0.251 ( 2 V)( 4 V) = = 933.2 (i) 533.33 vs. 414.81 933.33 vs 933.2 4. (a) VGS =0 V, ID = 8 mA (for VDS > VP) VGS = 1 V, ID = 4.5 mA ID = 3.5 mA(b) VGS = 1 V, ID = 4.5 mA VGS = 2 V, ID = 2 mA ID = 2.5 mA 89. 86(c) VGS = 2 V, ID = 2 mA VGS = 3 V, ID = 0.5 mA ID = 1.5 mA (d) VGS= 3 V, ID = 0.5 mA VGS = 4 V, ID = 0 mA ID = 0.5 mA (e) As VGSbecomes more negative, the change in ID gets progressively smallerfor the same change in VGS. (f) Non-linear. Even though the changein VGS is fixed at 1 V, the change in ID drops from a maximum of3.5 mA to a minimum of 0.5 mAa 7:1 change in ID. 5. The collectorcharacteristics of a BJT transistor are a plot of output currentversus the output voltage for different levels of input current.The drain characteristics of a JFET transistor are a plot of theoutput current versus input voltage. For the BJT transistorincreasing levels of input current result in increasing levels ofoutput current. For JFETs, increasing magnitudes of input voltageresult in lower levels of output current. The spacing betweencurves for a BJT are sufficiently similar to permit the use of asingle beta (on an approximate basis) to represent the device forthe dc and ac analysis. For JFETs, however, the spacing between thecurves changes quite dramatically with increasing levels of inputvoltage requiring the use of Shockleys equation to define therelationship between ID and VGS. satCV and VP define the region ofnonlinearity for each device. 6. (a) The input current IG for aJFET is effectively zero since the JFET gate-source junction isreverse-biased for linear operation, and a reverse-biased junctionhas a very high resistance. (b) The input impedance of the JFET ishigh due to the reverse-biased junction between the gate andsource. (c) The terminology is appropriate since it is the electricfield established by the applied gate to source voltage thatcontrols the level of drain current. The term field is appropriatedue to the absence of a conductive path between gate and source (ordrain). 7. VGS = 0 V, ID = IDSS = 12 mA VGS = VP = 6 V, ID = 0 mAShockleys equation: VGS = 1 V, ID = 8.33 mA; VGS = 2 V, ID = 5.33mA; VGS = 3 V, ID = 3 mA; VGS = 4 V, ID = 1.33 mA; VGS = 5 V, ID =0.333 mA. 8. For a p-channel JFET, all the voltage polarities inthe network are reversed as compared to an n-channel device. Inaddition, the drain current has reversed direction. 90. 87 9. (b)IDSS = 10 mA, VP = 6 V 10. VGS = 0 V, ID = IDSS = 12 mA VGS = VP =4 V, ID = 0 mA VGS = 2 PV = 2 V, ID = 4 DSSI = 3 mA VGS = 0.3VP =1.2 V, ID = 6 mA VGS = 3 V, ID = 0.75 mA (Shockleys equation) 11.(a) ID = IDSS = 9 mA (b) ID = IDSS(1 VGS/VP)2 = 9 mA(1 (2 V)/(3.5V))2 = 1.653 mA (c) VGS = VP = 3.5 V, ID = 0 mA (d) VGS < VP =3.5 V, ID = 0 mA 12. VGS = 0 V, ID = 16 mA VGS = 0.3VP = 0.3(5 V) =1.5 V, ID = IDSS/2 = 8 mA VGS = 0.5VP = 0.5(5 V) = 2.5 V, ID =IDSS/4 = 4 mA VGS = VP = 5 V, ID = 0 mA 13. VGS = 0 V, ID = IDSS =7.5 mA VGS = 0.3VP = (0.3)(4 V) = 1.2 V, ID = IDSS/2 = 7.5 mA/2 =3.75 mA VGS = 0.5VP = (0.5)(4 V) = 2 V, ID = IDSS/4 = 7.5 mA/4 =1.875 mA VGS = VP = 4 V, ID = 0 mA 14. (a) ID = IDSS(1 VGS/VP)2 = 6mA(1 (2 V)/(4.5 V))2 = 1.852 mA ID = IDSS(1 VGS/VP)2 = 6 mA(1 (3.6V)/(4.5 V))2 = 0.24 mA (b) VGS = 3 mA 1 ( 4.5 V) 1 6 mA D P DSS I VI = = 1.318 V VGS = 5.5 mA 1 ( 4.5 V) 1 6 mA D P DSS I V I = =0.192 V 15. ID = IDSS(1 VGS/VP)2 3 mA = IDSS(1 (3 V)/(6 V))2 3 mA =IDSS(0.25) IDSS = 12 mA 91. 88 16. From Fig. 6.22: 0.5 V < VP< 6 V 1 mA < IDSS < 5 mA For IDSS = 5 mA and VP = 6 V: VGS= 0 V, ID = 5 mA VGS = 0.3VP = 1.8 V, ID = 2.5 mA VGS = VP/2 = 3 V,ID = 1.25 mA VGS = VP = 6 V, ID = 0 mA For IDSS = 1 mA and VP = 0.5V: VGS = 0 V, ID = 1 mA VGS = 0.3VP = 0.15 V, ID = 0.5 mA VGS =VP/2 = 0.25 V, ID = 0.25 mA VGS = VP = 0.5 V, ID = 0 mA 17. VDS =maxDSV = 25 V, ID = max max 120 mW 25 V D DS P V = = 4.8 mA ID =IDSS = 10 mA, VDS = max 120 mW 10 mA D DSS P I = = 12 V ID = 7 mA,VDS = max 120 mW 7 mA D D P I = = 17.14 V 92. 89 2.5 mA 18. VGS =0.5 V, ID = 6.5 mA VGS = 1 V, ID = 4 mA Determine ID above 4 mAline: 2.5 mA 0.5 V 0.3 V x = x = 1.5 mA ID = 4 mA + 1.5 mA = 5.5 mAcorresponding with values determined from a purely graphicalapproach. 19. Yes, all knees of VGS curves at or below |VP| = 3 V.20. From Fig 6.25, IDSS 9 mA At VGS = 1 V, ID = 4 mA ID = IDSS(1VGS/VP)2 D DSS I I = 1 VGS/VP GS P V V = 1 D DSS I I VP = 1 V 4 mA11 9 mA GS D DSS V I I = = 3 V (an exact match) 21. ID = IDSS(1VGS/VP)2 = 9 mA(1 (1 V)/(3 V))2 = 4 mA, which compares very wellwith the level obtained using Fig. 6.25. 22. (a) VDS 0.7 V @ ID = 4mA (for VGS = 0 V) r = 0.7 V 0 V 4 mA 0 mA DS D V I = = 175 (b) ForVGS = 0.5 V, @ ID = 3 mA, VDS = 0.7 V r = 0.7 V 3 mA = 233 (c) rd =2 2 175 (1 / ) (1 ( 0.5 V)/( 3 V) o GS P r V V = = 252 vs. 233 frompart (b) 23. 24. The construction of a depletion-type MOSFET and anenchancement-type MOSFET are identical except for the doping in thechannel region. In the depletion MOSFET the channel is establishedby the doping process and exists with no gate-to-source voltageapplied. As the gate-to-source voltage increases in magnitude thechannel decreases in size until pinch-off occurs. The enhancementMOSFET does not have a channel established by the doping sequencebut relies on the gate-to-source voltage to create a channel. Thelarger the magnitude of the applied gate-to-source voltage, thelarger the available channel. 93. 90 DI = 3.34 mA DI = 4.67 mA DI =6 mA 25. 26. At VGS = 0 V, ID = 6 mA At VGS = 1 V, ID = 6 mA(1 (1V)/(3 V))2 = 2.66 mA At VGS = +1 V, ID = 6 mA(1 (+1 V)/(3 V))2 = 6mA(1.333)2 = 10.667 mA At VGS = +2 V, ID = 6 mA(1 (+2 V)/(3 V))2 =6 mA(1.667)2 = 16.67 mA From 1 V to 0 V, ID = 3.34 mA while from +1V to +2 V, ID = 6 mA almost a 2:1 margin. In fact, as VGS becomesmore and more positive, ID will increase at a faster and fasterrate due to the squared term in Shockleys equation. 27. VGS = 0 V,ID = IDSS = 12 mA; VGS = 8 V, ID = 0 mA; VGS = 2 PV = 4 V, ID = 3mA; VGS = 0.3VP = 2.4 V, ID = 6 mA; VGS = 6 V, ID = 0.75 mA 28.From problem 20: VP = 1 V 1 V 1 V 1 1.2139514 mA 1 1.473 11 9.5 mAGS D DSS V I I + + + = = = = 1 0.21395 4.67 V 29. ID = IDSS(1VGS/VP)2 IDSS = ( ) 2 2 4 mA (1 ( 2 V) ( 5 V))1 D GS P I V V = =11.11 mA 30. From problem 14(b): VGS = 20 mA 1 ( 5 V) 1 2.9 mA D PDSS I V I = = (5 V)(1 2.626) = (5 V)(1.626) = 8.13 V 31. From Fig.6.34, maxDP = 200 mW, ID = 8 mA P = VDSID and VDS = max 200 mW 8mAD P I = = 25 V VGS ID 1 V 2.66 mA 0 6.0 mA +1 V 10.67 mA +2 V16.67 mA 94. 91 32. (a) In a depletion-type MOSFET the channelexists in the device and the applied voltage VGS controls the sizeof the channel. In an enhancement-type MOSFET the channel is notestablished by the construction pattern but induced by the appliedcontrol voltage VGS. (b) (c) Briefly, an applied gate-to-sourcevoltage greater than VT will establish a channel between drain andsource for the flow of charge in the output circuit. 33. (a) ID =k(VGS VT)2 = 0.4 103 (VGS 3.5)2 VGS ID 3.5 V 0 4 V 0.1 mA 5 V 0.9mA 6 V 2.5 mA 7 V 4.9 mA 8 V 8.1 mA (b) ID = 0.8 103 (VGS 3.5)2 VGSID 3.5 V 0 4 V 0.2 mA 5 V 1.8 mA 6 V 5.0 mA 7 V 9.8 mA 8 V 16.2 mAFor same levels of VGS, ID attains twice the current level as part(a). Transfer curve has steeper slope. For both curves, ID = 0 mAfor VGS < 3.5 V. 34. (a) k = ( ) (on) 2 2 (on) 4 mA (6 V 4 V) DGS T I V V = = 1 mA/V2 ID = k(VGS VT)2 = 1 103 (VGS 4 V)2 (b) VGSID For VGS < VT = 4 V, ID = 0 mA 4 V 0 mA 5 V 1 mA 6 V 4 mA 7 V9 mA 8 V 16 mA (c) VGS ID (VGS < VT) 2 V 0 mA 5 V 1 mA 10 V 36mA 95. 92 35. From Fig. 6.58, VT = 2.0 V At ID = 6.5 mA, VGS = 5.5V: ID = k(VGS VT)2 6.5 mA = k(5.5 V 2 V)2 k = 5.31 104 ID = 5.31104 (VGS 2)2 36. ID = ( ) 2 (on)GS Tk V V and ( ) 2 (on) D GS T I VV k = (on) D GS T I V V k = VT = (on)GSV DI k = 4 V 3 3 mA 0.4 10 =4 V 7.5 V = 4 V 2.739 V = 1.261 V 37. ID = k(VGS VT)2 DI k = (VGSVT)2 DI k = VGS VT VGS = VT + DI k = 5 V + 3 30 mA 0.06 10 = 27.36V 38. Enhancement-type MOSFET: ID = 2 ( )GS Tk V V 2 ( ) ( )D GS TGS T GS GS dI d k V V V V dV dV = D GS dI dV = 2k(VGS VT) 1 96. 93Depletion-type MOSFET: ID = IDSS(1 VGS/VP)2 D GS dI dV = 2 1 GS DSSGS P Vd I dV V 2 1 0GS GS DSS P GS P V d V I V dV V = 1 PV = 1 2 1GS DSS P P V I V V = 2 1DSS GS P P I V V V = 2 1DSS GSP P P P I VVV V V D GS dI dV = 2 2 ( )DSS GS P P I V V V For both devices D GSdI dV = k1(VGS K2) revealing that the drain current of each willincrease at about the same rate. 39. ID = k(VGS VT)2 = 0.45 103(VGS (5 V))2 = 0.45 103 (VGS + 5 V)2 VGS = 5 V, ID = 0 mA; VGS = 6V, ID = 0.45 mA; VGS = 7 V, ID = 1.8 mA; VGS = 8 V, ID = 4.05 mA;VGS = 9 V, ID = 7.2 mA; VGS = 10 V, ID = 11.25 mA 41. 42. (a) (b)For the on transistor: R = 0.1 V 4 mA V I = = 25 ohms For the offtransistor: R = 4.9 V 0.5 A V I = = 9.8 M Absolutely, the highresistance of the off resistance will ensure Vo is very close to 5V. 43. 97. 94 Chapter 7 1. (a) VGS = 0 V, ID = IDSS = 12 mA VGS =VP = 4 V, ID = 0 mA VGS = VP/2 = 2 V, ID = IDSS/4 = 3 mA VGS =0.3VP = 1.2 V, ID = IDSS/2 = 6 mA (b) (c) QDI 4.7 mA QDSV = VDD QDDI R = 12 V (4.7 mA)(1.2 k) = 6.36 V (d) QDI = IDSS(1 VGS/VP)2 = 12mA(1 (1.5 V)/(4 V))2 = 4.69 mA QDSV = VDD QD DI R = 12 V (4.69mA)(1.2 k) = 6.37 V excellent comparison 2. (a) QDI = IDSS 2 (1 /)GS PV V = 10 mA( ) 2 1 ( 3 V) ( 4.5 V) = 10 mA(0.333)2 QDI = 1.11mA (b) QGSV = 3 V (c) VDS = VDD ID(RD + RS) = 16 V (1.11 mA)(2.2 k)= 16 V 2.444 V = 13.56 V VD = VDS = 13.56 V VG = QGSV = 3 V VS = 0V 98. 95 3. (a) 14 V 9 V 1.6 kQ DD D D D V V I R = = = 3.125 mA (b)VDS = VD VS = 9 V 0 V = 9 V (c) ID = IDSS(1 VGS/VP)2 VGS = 1 D PDSS I V I VGS = (4 V) 3.125 mA 1 8 mA = 1.5 V VGG = 1.5 V 4. QGSV =0 V, ID = IDSS = 5 mA VD = VDD IDRD = 20 V (5 mA)(2.2 k) = 20 V 11V = 9 V 5. VGS = VP = 4 V QDI = 0 mA and VD = VDD QD DI R = 18 V(0)(2 k) = 18 V 6. (a)(b)VGS = 0 V, ID = 10 mA VGS = VP = 4 V, ID =0 mA VGS = 2 PV = 2 V, ID = 2.5 mA VGS = 0.3VP = 1.2 V, ID = 5 mAVGS = IDRS ID = 5 mA: VGS = (5 mA)(0.75 k) = 3.75 V (c) QDI 2.7 mAQGSV 1.9 V (d) VDS = VDD ID(RD + RS) = 18 V (2.7 mA)(1.5 k + 0.75k) = 11.93 V VD = VDD IDRD = 18 V (2. 2b1af7f3a8