Solution Manual For Electronic Devices And Circuit Theory, 11th Edition

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1Chapter 11.Copper has 20 orbiting electrons with only one electron in the outermost shell. The fact thatthe outermost shell with its 29thelectron is incomplete (subshell can contain 2 electrons) anddistant from the nucleus reveals that this electron is loosely bound to its parent atom. Theapplication of an external electric field of the correct polarity can easily draw this looselybound electron from its atomic structure for conduction.Both intrinsic silicon and germanium have complete outer shells due to the sharing (covalentbonding) of electrons between atoms. Electrons that are part of a complete shell structurerequire increased levels of applied attractive forces to be removed from their parent atom.2.Intrinsic material: an intrinsic semiconductor is one that has been refined to be as pure asphysically possible. That is, one with the fewest possible number of impurities.Negative temperature coefficient: materials with negative temperature coefficients havedecreasing resistance levels as the temperature increases.Covalent bonding: covalent bonding is the sharing of electrons between neighboring atoms toform complete outermost shells and a more stable lattice structure.3.4.a.W=QV= (12μC)(6 V) =72μJb.72 × 106J =191 eV1.610J=2.625 × 1014eV5.48 eV = 48(1.61019J) =76.81019JQ=WV=1976.810J3.2 V=2.401018C6.41019C is the charge associated with 4 electrons.6.GaPGallium PhosphideEg=2.24 eVZnSZinc SulfideEg=3.67 eV7.Ann-type semiconductor material has an excess of electrons for conduction established bydoping an intrinsic material with donor atoms having more valence electrons than needed toestablish the covalent bonding. The majority carrier is the electron while the minority carrieris the hole.Ap-type semiconductor material is formed by doping an intrinsic material with acceptoratoms having an insufficient number of electrons in the valence shell to complete the covalentbonding thereby creating a hole in the covalent structure. The majority carrier is the holewhile the minority carrier is the electron.8.A donor atom has five electrons in its outermost valence shell while an acceptor atom hasonly 3 electrons in the valence shell.

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29.Majority carriers are those carriers of a material that far exceed the number of any othercarriers in the material.Minority carriers are those carriers of a material that are less in number than any other carrierof the material.10.Same basic appearance as Fig. 1.7 since arsenic also has 5 valence electrons (pentavalent).11.Same basic appearance as Fig. 1.9 since boron also has 3 valence electrons (trivalent).12.13.14.For forward bias, the positive potential is applied to thep-type material and the negativepotential to then-type material.15.a.2319(1.3810J/K)(20 C273 C)1.610CKTkTVq25.27 mVb./(0.5 V) / (2)(25.27mV)9.89(1)40 nA(1)40 nA(1)DTVnVDsIIeee0.789 mA16.a.2319()(1.3810J/K)(100 C273 C)1.610KTk TVq32.17 mVb./(0.5 V) / (2)(32.17 mV)7.77(1)40 nA(1)40 nA(1)DTVnVDsIIeee11.84 mA17.a.TK= 20 + 273 = 2932319(1.3810J/K)(293 )1.610CKTkTVq25.27 mVb./10/(2)(25.27 mV)197.86(1)0.1A1= 0.1A(1)DTVnVDsIIeee0.1A

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318.2319(1.3810J/K)(25 C273 C)1.610C=25.70 mVKTkTVqID=/(1)DTVnVsIe8mA =(0.5V) / (1)(25.70 mV)8(1)(2810 )ssIeI88 mA2.810sI=28.57 pA19./(1)DTVnVDsIIe/(1)(26 mV)6 mA1 nA(1)DVe/ 26 mV66101DVe/ 26 mV666101610DVe/ 26 mV6eelogelog610DV26 mVDV= 15.61VD= 15.61(26 mV)0.41 V20.(a)xy=ex0112.718227.389320.086454.65148.4(b)y=e0= 1(c)Forx= 0,e0= 1 andI=Is(11) =0 mA21.T= 20C:Is= 0.1AT= 30C:Is= 2(0.1A) = 0.2A (Doubles every 10C rise in temperature)T= 40C:Is= 2(0.2A) = 0.4AT= 50C:Is= 2(0.4A) = 0.8AT= 60C:Is= 2(0.8A) =1.6A1.6A: 0.1A16:1 increase due to rise in temperature of 40C.22.For most applications the silicon diode is the device of choice due to its higher temperaturecapability. Ge typically has a working limit of about 85 degrees centigrade while Si can beused at temperatures approaching 200 degrees centigrade. Silicon diodes also have a highercurrent handling capability. Germanium diodes are the better device for some RF small signalapplications, where the smaller threshold voltage may prove advantageous.

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423.From 1.19:75C25C100C200CVF@ 10 mAIs1.1 V0.01 pA0.85 V1 pA1.0 V1A0.6 V1.05AVFdecreased with increase in temperature1.7 V: 0.65 V2.6:1Isincreased with increase in temperature2A: 0.1A =20:124.An “ideal” device or system is one that has the characteristics we would prefer to have whenusing a device or system in a practical application. Usually, however, technology onlypermits a close replica of the desired characteristics. The “ideal” characteristics provide anexcellent basis for comparison with the actual device characteristics permitting an estimate ofhow well the device or system will perform. On occasion, the “ideal” device or system can beassumed to obtain a good estimate of the overall response of the design. When assuming an“ideal” device or system there is no regard for component or manufacturing tolerances or anyvariation from device to device of a particular lot.25.In the forward-bias region the 0 V drop across the diode at any level of current results in aresistance level of zero ohms – the “on” state – conduction is established. In the reverse-biasregion the zero current level at any reverse-bias voltage assures a very high resistance levelthe open circuit or “off” stateconduction is interrupted.26.The most important difference between the characteristics of a diode and a simple switch isthat the switch, being mechanical, is capable of conducting current in either direction whilethe diode only allows charge to flow through the element in one direction (specifically thedirection defined by the arrow of the symbol using conventional current flow).27.VD0.7 V,ID= 4 mARDC=0.7 V4 mADDVI=17528.AtID= 15 mA,VD= 0.82 VRDC=0.82 V15 mADDVI=54.67As the forward diode current increases, the static resistance decreases.

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529.VD=10 V,ID=Is=0.1ARDC=10 V0.1ADDVI=100 MVD=30 V,ID=Is=0.1ARDC=30 V0.1ADDVI=300 MAs the reverse voltage increases, the reverse resistance increases directly (since the diodeleakage current remains constant).30.ID= 10 mA,VD= 0.76 VRDC=0.76 V10 mADDVI=76rd=0.79 V0.76 V0.03 V15 mA5 mA10 mAddVI=3RDC>>rd31.(a)rd=0.03 V0.79 V0.76 V15 mA5 mA10 mAddVI=3(b)rd=26 mV26 mV10 mADI=2.6(c)quite close32.ID= 1 mA,rd=0.72 V0.61 V2 mA0 mAddVI=55ID= 15 mA,rd=0.8 V0.78 V20 mA10 mAddVI=233.ID= 1 mA,rd=26 mV2DI= 2(26) =52vs 55(#30)ID= 15 mA,rd=26 mV26 mV15 mADI=1.73vs 2(#30)34.rav=0.9 V0.6 V13.5 mA1.2 mAddVI=24.435.rd=0.8 V0.7 V0.09 V7 mA3 mA4 mAddVI=22.5(relatively close to average value of 24.4(#32))

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636.rav=0.9 V0.7 V0.2 V14 mA0 mA14 mAddVI=14.2937.Using the best approximation to the curve beyondVD= 0.7 V:rav=0.8 V0.7 V0.1 V25 mA0 mA25 mAddVI=438.Germanium:0.42 V0.3 V430 mA0 mAavrGaAa:1.32 V1.2 V430 mA0 mAavr39.(a)VR=25 V:CT0.75 pFVR=10 V:CT1.25 pF1.25 pF0.75 pF0.5 pF10 V25 V15 VTRCV=0.033 pF/V(b)VR=10 V:CT1.25 pFVR=1 V:CT3 pF1.25 pF3 pF1.75 pF10 V1 V9 VTRCV=0.194 pF/V(c)0.194 pF/V: 0.033 pF/V = 5.88:16:1Increased sensitivity nearVD= 0 V40.From Fig. 1.33VD= 0 V,CD=3.3 pFVD= 0.25 V,CD=9 pF

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741.The transition capacitance is due to the depletion region acting like a dielectric in the reverse-bias region, while the diffusion capacitance is determined by the rate of charge injection intothe region just outside the depletion boundaries of a forward-biased device. Bothcapacitances are present in both the reverse- and forward-bias directions, but the transitioncapacitance is the dominant effect for reverse-biased diodes and the diffusion capacitance isthe dominant effect for forward-biased conditions.42.VD= 0.2 V,CD= 7.3 pFXC=1122(6 MHz)(7.3 pF)fC=3.64 kVD=20 V,CT= 0.9 pFXC=1122(6 MHz)(0.9 pF)fC=29.47 k43.1/21/2(0)8 pF15 V / 0.7 V1/8pF8 pF8 pF2.85(1+7.14)8.14TnRKCCVV2.81 pF44.1/31/3(0)1/10 pF4 pF =1/0.7 V(1/0.7 V)2.5TRkRRCCVVVV31/0.7 V(2.5)15.63/0.715.63114.63(0.7)(14.63)RRRVVVV10.24 V45.If=10 V10 k= 1 mAts+tt=trr= 9 nsts+ 2ts= 9 nsts=3 nstt= 2ts=6 ns

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846.47.a.As the magnitude of the reverse-bias potential increases, the capacitance drops rapidlyfrom a level of about 5 pF with no bias. For reverse-bias potentials in excess of 10 V thecapacitance levels off at about 1.5 pF.b.6 pFc.At4 V,2 pFRTVC (0)1/TnRkCCVV6 pF2 pF14V/0.7 Vn101014 V0.7 V3(6.71)3log6.71log3(0.827)0.477nnnn0.4770.827n0.5848.AtVD=25 V,ID=0.2 nA and atVD=100 V,ID0.45 nA. Although the change inIRis morethan 100%, the level ofIRand the resulting change is relatively small for most applications.49.Log scale:TA= 25C,IR=0.5 nATA= 100C,IR=60 nAThe change is significant.60 nA: 0.5 nA =120:1Yes, at 95CIRwould increase to 64 nA starting with 0.5 nA (at 25C)(and double the level every 10C).50.IF= 0.1 mA:rd700IF= 1.5 mA:rd70IF= 20 mA:rd6The results support the fact that the dynamic or ac resistance decreases rapidly withincreasing current levels.

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951.T= 25C:Pmax= 500 mWT =100C:Pmax= 260 mWPmax=VFIFIF=max500 mW0.7 VFPV=714.29 mAIF=max260 mW0.7 VFPV=371.43 mA714.29 mA: 371.43 mA = 1.92:12:152.Using the bottom right graph of Fig. 1.37:IF= 500 mA @T= 25CAtIF= 250 mA,T104C53.54.TC= +0.072% =10100%()ZZVVTT0.072 =10.75 V10010 V(25)T0.072 =17.525TT125=7.50.072= 104.17T1= 104.17+ 25=129.1755.TC=10()ZZVVTT100%=(5 V4.8 V)5 V(10025 ) 100% =0.053%/C56.(20 V6.8 V)(24 V6.8 V)100% = 77%The 20 V Zener is therefore77% of the distance between 6.8 V and 24 V measured fromthe 6.8 V characteristic.

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10AtIZ= 0.1 mA,TC0.06%/C(5 V3.6 V)(6.8 V3.6 V)100% = 44%The 5 V Zener is therefore44% of the distance between 3.6 V and 6.8 V measured from the3.6 V characteristic.AtIZ= 0.1 mA,TC0.025%/C57.58.24 V Zener:0.2 mA:4001 mA:9510 mA:13The steeper the curve (higherdI/dV) the less the dynamic resistance.59.KV2.0 V, which is considerably higher than germanium (0.3 V) or silicon (0.7 V). Forgermanium it is a 6.7:1 ratio, and for silicon a 2.86:1 ratio.60.0.67 eV191.610J1 eV191.07210J 34819(6.62610Js)(310 ) m/s1.07210JgghchcEE1850 nmVery low energy level.61.Fig. 1.53 (f)IF13 mAFig. 1.53 (e)VF2.3 V62.(a)Relative efficiency @ 5 mA0.82@ 10 mA1.021.020.820.82100% =24.4%increaseratio:1.020.82= 1.24(b)Relative efficiency @ 30 mA1.38@ 35 mA1.421.421.381.38100% =2.9%increaseratio:1.421.38=1.03

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11(c)For currents greater than about 30 mA the percent increase is significantly less than forincreasing currents of lesser magnitude.63.(a)0.753.0= 0.25From Fig. 1.53 (i)75(b)0.5=4064.For the high-efficiency red unit of Fig. 1.53:0.2 mA20 mACxx=20 mA0.2 mA/ C= 100C

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12Chapter 21.The load line will intersect atID=12 V750ER= 16 mA andVD= 12 V.(a)QDV0.85 VQDI15 mAVR=EQDV= 12 V0.85 V =11.15 V(b)QDV0.7 VQDI15 mAVR=EQDV= 12 V0.7 V =11.3 V(c)QDV0 VQDI16 mAVR=EQDV= 12 V0 V =12 VFor (a) and (b), levels ofQDVandQDIare quite close. Levels of part (c) are reasonably closebut as expected due to level of applied voltageE.2.(a)ID=6 V0.2 kER= 30 mAThe load line extends fromID= 30 mA toVD= 6 V.QDV0.95 V,QDI25.3 mA(b)ID=6 V0.47 kER= 12.77 mAThe load line extends fromID= 12.77 mA toVD= 6 V.QDV0.8 V,QDI11 mA(c)ID=6 V0.68 kER= 8.82 mAThe load line extends fromID= 8.82 mA toVD= 6 V.QDV0.78 V,QDI78 mAThe resulting values ofQDVare quite close, whileQDIextends from 7.8 mA to 25.3 mA.3.Load line throughQDI= 10 mA of characteristics andVD= 7 V will intersectIDaxis as11.3 mA.ID= 11.3 mA =7 VERRwithR=7 V11.3 mA= 619.47 k0.62 kΩstandard resistor

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134.(a)ID=IR=30 V0.7 V1.5 kDEVR=19.53 mAVD=0.7 V,VR=EVD= 30 V0.7 V =29.3 V(b)ID=30 V0 V1.5 kDEVR=20 mAVD=0 V,VR=30 VYes, sinceEVTthe levels ofIDandVRare quite close.5.(a)I=0 mA; diode reverse-biased.(b)V20= 20 V0.7 V = 19.3 V (Kirchhoff’s voltage law)I(20Ω) =19.3 V20= 0.965 AV(10Ω) = 20 V0.7 V = 19.3 VI(10Ω) =19.3 V10= 1.93 AI=I(10Ω) +I(20Ω)=2.895 A(c)I=10 V10=1 A; center branch open6.(a)Diode forward-biased,Kirchhoff’s voltage law (CW):5 V + 0.7 VVo= 0Vo=4.3 VIR=ID=4.3 V2.2 koVR=1.955 mA(b)Diode forward-biased,ID=8 V + 6 V0.7 V1.2 k4.7 k=2.25 mAVo= 8 V(2.25 mA)(1.2 kΩ) =5.3 V7.(a)Vo=10 k(12 V0.7 V0.3 V)2 k10 k=9.17 V(b)Vo=10 V

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148.(a)Determine the Thevenin equivalent circuit for the 10 mA source and 2.2 kresistor.ETh=IR= (10 mA)(2.2 k) = 22 VRTh= 2. 2kDiode forward-biasedID=22 V0.7 V2.2 k2.2 k=4.84 mAVo=ID(1.2 k)= (4.84 mA)(1.2 k)=5.81 V(b)Diode forward-biasedID=20 V + 20 V0.7 V6.8 k=5.78 mAKirchhoff’s voltage law (CW):+Vo0.7 V + 20 V = 0Vo=19.3 V9.(a)1oV= 12 V – 0.7 V =11.3 V2oV=1.2 V(b)1oV=0 V2oV=0 V10.(a)Both diodes forward-biasedSi diode turns on first and locks in 0.7 V drop.12 V0.7 V4.7 kRI= 2.4 mAID=IR=2.4 mAVo= 12 V0.7 V =11.3 V(b)Right diode forward-biased:ID=20 V + 4 V0.7 V2.2 k=10.59 mAVo= 20 V0.7 V =19.3 V11.(a)Si diode “on” preventing GaAs diode from turning “on”:I=1 V0.7 V0.3 V1 k1 k=0.3 mAVo= 1 V0.7 V =0.3 V(b)I=16 V0.7 V0.7 V + 4 V18.6 V4.7 k4.7 k=3.96 mAVo= 16 V0.7 V0.7 V =14.6 V

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1512.Both diodes forward-biased:1oV=0.7 V,2oV=0.7 VI1 k=20 V0.7 V1 k=19.3 V1 k= 19.3 mAI0.47 k= 0 mAI=I1 kΩI0.47 kΩ= 19.3 mA0 mA=19.3 mA13.Superposition:11 k(9.3 V)(9.3 V)3.1 V1 k2 koV216 k(8.8 V)(8.8 V)2.93 V1 k2 koVVo=12ooVV=6.03 VID=9.3 V6.03 V2 k=1.635 mA14.Both diodes “off”. The threshold voltage of 0.7 V is unavailable for either diode.Vo=0 V15.Both diodes “on”,Vo= 10 V0.7 V =9.3 V16.Both diodes “on”.Vo=0.7 V17.Both diodes “off”,Vo=10 V18.The Si diode with5 V at the cathode is “on” while the other is “off”. The result isVo=5 V + 0.7 V =4.3 V19.0 V at one terminal is “more positive” than5 V at the other input terminal. Thereforeassume lower diode “on” and upper diode “off”.The result:Vo= 0 V0.7 V =0.7 VThe result supports the above assumptions.20.Since all the system terminals are at 10 V the required difference of 0.7 V across either diodecannot be established. Therefore, both diodes are “off” andVo=+10 Vas established by 10 V supply connected to 1 kresistor.

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