半导体物理与器件第三版(尼曼)13章答案

1、 Chapter13 # ProblemSolutions13.1Sketch13.2Sketch133p-channelJFET一Silicon(a)応N。(1.6xl0_w)(0.5xl0_*)，(3xl0u)一2e-2(11.7)(8.85x10-)or%=5.7Nowr=(0.0259)111orVbl=0884Vso(5xl0”)(3xl)(1.5x10)=-=5.79-0.884orr;.=49ir(b)_，_乙一N+N)ah=a(i)ForN匚NThena-h=0.5x1O42(11.7)(8.85x10)(0.884+1-乙J(1.6xl0)(3:d0)ora-h=0.5X10

2、-4-(4.31xl(T)(L884-乙.)ora_h=0.215y)nForN=1/,%=-2.5Va_h=0.0653p)n(in)For乙=1匚乙严一a_h=-0.045urnwhichimpliesnoundepletedregion13.4p-channelJFET一GaAs(a)_?(0.5力旷)(3x10”)一2e-2(13.1)(8.85x10)orNow(5x1018)(3x10k)3(0.0259)1“(1.8x109or乙=1.35/SO,=-=5.18-1.35orr；.=3.837(b)r/xqin_,_広亿一N+N)an=aeNL.aJ(0For=17,.=0The

3、na-/?=0.5x1O42(13.1)(8.85x10)(1.35+(1.6x10-w)(3xl0u)ora-h=0.5X10-4-(4.83xl0-10)(2.35-.)pwhichyieldsa-力=(J.163帥(h)ForN十心=-2-5Va_h=0.016ju?2(hi)For卞十心=-5Ta_h=-0096(Amwhichimpliesnoundepletedregion13.5/、eaN,2e(1.6xl0)(0.5xl巧(8x10”)2(11/0(8.85x10)or乙严15(b)0.2x10*=0.5x1O4_2(13.1)(885xl(T“)(匕_巧)(L6xl0)(8x

4、l0)whichcanbewrittenas%1严=1.8llxlO1亿一匕Jor人-=49WNow #r=(0.0259)ln(3xlOw)(8xlOx,)(1.8X10*)1 #GSso0.2x10*=0.5x1O42(11.7)(&85丹0-“)低-诊)(1.6x1严)(8xlO”)oror=i.36rThen%=-4.97=1.36-497or #9x10“=1.618x10*(人-)whichyields乙一乙=5.56口Now13.7(Q%orr=(0.0259)ln(3xlOw)(8xlO1/=7.66/r=(0.0259)ln13.6ForGaAs:(a)2G(1.6x10_w

5、)(0.5x10_*)，(8x10h)2(13.1)(885x10H)or(b)1/2(5x1018)(3x10h)(1.8x10丁or人=1352VThen,=-=1863-1.352orr;.=0.511r(b)(i)広亿+N)eNaora-/2=(0.3xW*)l(13.1)(8.85xl(T)(L35汀(1.6x10_w)(3xlOl) whichyieldsa_h=4.45x104cm(11)a=(03xl()7)2(13.1)(8.85xl0_1*)(1351+l)(1.6x10_w)(3xl0u)whichyieldsa_h=-3.7x10cmwhichimpliesnounde

6、pletedregion(in)For=-ir,=2Z,a_h=-0.0051Umwhichimpliesnoundepletedregion13.9%=(00259)ln(5x1018)(4x10u)13.8(a)n-chamielJFET一Siliconea2Na(1.6x1019)(0.35x10_4)2(4x101=0-892-3.79orF；=-2.90F(b)2血+N-N)1/2or=0.35x10*(13.1)(885x1(T)(1.359+F；q-兀(1.6x10_19)(4x101s=0,wefind%=-1.125r(b)ForVns=1r.wefind%=-0.1257

7、#Wehavea-A=0.35x10*2(11.7)(8.85xl(T】4)(0.892+-%):(1.6x10_19)(4x1016)ora-h=Q.35xl-(3.24x1严)(0892(0ForN=0匕十，a-力=0.102y)n(ii)For=-ir,.=1V,a_h=0.044pin13.10(a)r心(帆,)唏(1000)(1.6xl0)(10u)2 # or2(11.7)(885x10)乙厂193/Also人=(0.0259)ln(厅)2)(1.5x101*)2(ii)For乙、=-0.265I)1(sat)=0.140mA(in)ForVGs=-0.537=IiA(sat)=0

8、.061mA(lv)ForVGs=-0.795V=Inl(sat)=0.0145mA13.11or=0874VNow抵(血)=乙厂亿-乙)=1.93-0.874+rvor乙、(sw)=1.06+%WehavewhereorPO193in=-=0874-1.93or乙二一1.067Then(1).=0=乙3)=LOWG(A=1.60mS1(li)rv=-f;,=-o.265r=4乙、(sm)=0.795/1(m)乙=一乙=0.53/=人2N(sH)=053/3(lv)冬、=一乙=一0.795/=4乙、(sm)=d265/(c)人3)(1)For%=0=g(sm)=0.258mA(乙-N)/唤&(

9、加s)004530.523-0.26505900.371-0.5307270.236-0.79509450.112-1.06100Then13.12n-chamielJFET一GaAs(a)euNWa(1.6xl0*w)(8000)(2xl01，)(30 xl0)(035x10)lOxlO-4orG讥=269xl(rS抵(血)=啜-亿-N)Wehaver=味2e(1.6.rl0_19)(0.35xl0_4)1(2xl0n)2(13.1)(8.85xl01*)or匚严L697 13.13Wefind #orr=(0.0259)ln(5x1018)(2x10h)(1.8x10丁Wehave乙=13

10、4VThen=1-03mA,VM=1.93V0.874VThemaximumtransconductanceoccurswhen gM5(max)=0.524mS400X10-4cmgMV(max)=1.48mS乙二匕一曝34-169orVp=0.35VWetlienobtain诊(如)=1.69-(1.34-Q；J=0.35+NForVGs=0=r,ls()=03571Forr=-r=-o.i75r=乙、(sm)=0.175/(c)(8000)(1.6x10-19)(2x10u)J一6(13.1)(885x1O1*)(30 x104)(0.35x10_4)3(lOxlO-4)orZrl=1.

11、51mAThen/,3)=15卜彳罟|(必)Forr(v=0=Zz4OaZ)=0.0504mAandforrv=-0.175T=Z/n()=0.0123=0Thengmax)or孤=0.524加SForW=4004，WehaveorgMS=13.1mSIcm=1.31mS/mm13.14Themaximumtransconductanceoccursfor%=0sowehave(a)WefoundGol=2.69mS,=1.34V,f=1.69VThen&(max)=(2.69)1orgv(max)=0.295mSThisisforachannellengthofZ=10側.Ifthechan

12、nellengthisreducedtoL=2中n,thengM5(max)=(0.29513.15n-channelMESFET-GaAs(a)ea2N,V=2e(16x1O_19)(O.5x1O4)J(1.5x1Ou)2(131)(8.85x1Cf)or诊=2.5Now%=叽-九(Nz、f47xlO17A0=71n-(00259)1111/UJI1.5x10whereor仪=0.0892Fsothatr=0.90-0.0892=0.8117Then乙=乙一乙严08112.59orVr=1.78V(b)Ifr.0forann-chamieldevice,thedeviceisadepleti

13、onmodeMESFET.BytrialanderrorN=8.1xl01?cmz(b)AtT=400K,/、V=(0.0259M=0.034531300JThenF；=0.89-(0.03453)In7.24x1八8.1x1015)-(845x1017)(8.Lx1015)whichbecomesF；.=+0.051r13.17Wehaveinwhere #13.16n-channelMESFET-GaAs(a)WewantF；=+0.10TThen%=叽-九Now/、f4.7x1017)0=(00259)ln=0058口I5x1中Then #(00259)111or(0.0259)ln4.

14、7x1中+(8.45x1中=0.79卩产兀一二叽一叽一5so/N(r=0.10=0.89-nd丿whichcanbewrittenas4.7x1017M丿(1.6xl0_19)(0.35xl04)2iVt+=0.89-0.102(13.1)(8.85xW11)r=080-0.058=0.7427Fora-/2=(08xl0_l)2(13.1/8.85%101*)(0.742-0.5)(1.6x10_19)(5x101-VWewant匕=0=化十=叽Device1:Nd=3x1cmThen（、f4.7x1017A0（）=（00259）ln=0.0713|0.5=0.85（0.0259）111M丿

15、-（4.31x10”）见Bytrialanderror,wefindN=5.45x1USemiconductorPhysicsandDevicesBasicPrinciples,3rdeditionSolutionsManualChapter13ProblemSolutions # #SemiconductorPhysicsandDevicesBasicPrinciples,3rdeditionSolutionsManualChapter13ProblemSolutions #九=（00259）ln4.7x1中3x1017丿=0.0116口l（13.1）（8.85xl严）（j.8187）T（1

16、6x1严）10）ora=0.199HtnDevice2:N.=3x1017cnT、aThensotliatF；v=089-0.0116=087847Nowr-iina=枫_2（13.1）（8.85%10_1+）（0.8784）（1.6x10ls）（3xlO17）ora=0.0651xm13.19V-V-V=d-6-VWewantVt=0.5Tso13.20n-chamielMESFET一silicon（a）Foragoldcontact,0“=082V.Wefind/、2.8x1019I0=（0.0259）ln=0.206T10”丿and%=聯一忙=0.82-0.206=0.614VWithV

17、lys=0=0.35/Wefind=0.075x10=aeN,usothat6t=O.O75xlO-4P2（11.7）（8.85x10_h）（0.614一0.35）+（1.6x1严）（10”）ora=0.26pinNowea2N,乙二乙一N4-十2Gor”c.(1血1严)(0如1旷)(10“)V=0614z2(11.7)(8.85xl0-14)WeobtainVT=0.092V氐(血)=乙-(人-匕JNow瞪$伽)=035-0.092or乙(sH)=0.258/13.21(a)nchannelMESFETsiliconandz、2.8x10190=(0.0259)ln=0.188口”2x1中J

18、soVht=080-0.188=r=0612/Nowea2NtV=.2g(1.6x10w)(0.4xl0_*)，(2xl0u)2(11.7)(8.85x101*)orWefind-247or7=-1.867and抵(如)=乙厂亿-乙)=2.47-(0612-(-1)orF；JY(saO=0.8587For=4.5r,additionaldonoratomsmustber%(sat)=(0.5)(012)Or乙(sB)=0.06/andfor冬.3)=(or乙.(sH)=0.127(C)厶(sm)乂仇-叮For%=15乙=Ijsat)=(2.51)(006)2OrIl)1(sat)=9.Q4nA

19、andfor%=2匕.=g(saf)=(2.51)(0.12)orI1)=36ApA13.23Wehave比=乂化-乙)sothat1.75X10-3=2(050-0.25)whichgivesiperrk=3.5x10AIV=2aLWeobtain(3.5x10_j)(2)(0.35x10_4)(10*)IP=(8000)(13.1)(885xlO-1*)or护=26.4pin(b)儿3)=叽-凉ForVGS=04Vyn(saf)=(3.5x1)(0.4-0.25)or易3)=7&2For=065/,n(saf)=(3.5x1)(065-0.25)orIiA(sat)=0.56mA13.24

20、Computerplot13.25ComputerplotL1AZ=0.90=1L2LWehaveWefindeaNtV=切2e(16x1严)(04xl(r)(3xl0)2(11.7)(885x10)or4=3.7irr=(00259)ln(1019)(3.xl0u)andor乙=0902Vsothat乙3)=3.71-0.902=2.81/Then_2(11.7)(8.85xl0_1*)(5-2.81)(1.6x10_19)(3x10u)orAZ=0.307UmNowL1AZ=0.90=1L2LorAZ=1-0.9=0.10LsoAZ叮心刘旷L=2(0.10)2(0.10)orL=1.54中

21、nSemiconductorPhysicsandDevicesBasicPrinciples,3rdeditionSolutionsManualChapter13ProblemSolutions # #13.26Wehave2/=厶一丄2Or #SemiconductorPhysicsandDevicesBasicPrinciples,3rdeditionSolutionsManualChapter13ProblemSolutions13.27We斶ethmt鶯“(品疋Assumingthatweareinthesaturationregion,then=Isat)and=IA(sat).We

22、canwnte13.28(a)SaturationoccurswhenE=1x10*VIcmAsafirstapproximation,letLIsat)=I!A(sat)1-9AThenF；JV=E-L=(h-10l)(2.xl0-4)orIfL,thenWehavethat1+2L船3)=Ain2w(N-N(sm)抵(sm)=X%whichcanbewrittenasinIfwewritez：】(sm)=伽)(1+码thenbycomparingequations,wehave1/21_乙(佃)%Theparameterisnotindependentof%.Define12Z2gx=an

23、dconsidertliefunction乙(saf)whichisdirectlyproportionalto.vfM1.50.2221.750.2452.00.2502.250.2472.500.2402.750.231300.222A.WefindthatSothatAisnearlyaconstant(b)Wehavethat皿+n%)Tand(5x1018)(4x10u(1.5x10”)orr=(0.0259)lnr=0892VForVGS=0,weobtainP2(11.7)(885x10)(0.892+2)L(1.6x1严)(4xl0”)or力=0.306pjn(C)Wetlie

24、nfindIl)x(sat)=eNJva-hJW=(1.6x10-w)(4.v10u)(107)(050-0.306)x(10)(30 xl0-4)orInl(sat)=3.72mA(d)SemiconductorPhysicsandDevicesBasicPrinciples,3rdeditionSolutionsManualChapter13ProblemSolutions #(1000)(1.6x10_19)(4x10b)J6(11.7)(885x10u)(3Oxl(T*)(o.5xl(ry(2X10-4)orZrl=2AmAAlso(L6xl0)(0.5xlf(4x10)一2(11.7

25、)(8.85x1严)or=7-73VThenorIiA(sat)=9.08mAIlA(sat)=9.08=ID1(sat)=1&2mA1/(b)Ifvelocitysaturationoccurs,tlientherelationIlA(sat)x(1/厶)doesnotapply.13.30(a)v=“E=(8000)(5xl0)=4xl0?cm!sThenL2x10C=v4x10or(b)Assumev=107cmIsThenL2X10-4h=30psSemiconductorPhysicsandDevicesBasicPrinciples,3rdeditionSolutionsManua

26、lChapter13ProblemSolutions # #SemiconductorPhysicsandDevicesBasicPrinciples,3rdeditionSolutionsManualChapter13ProblemSolutions # #13.29(a)IfZ=1R”，tliensaturationwilloccurwhenr;s=E-L=(io*)(ixio-4)=irWefind13.31(a)v=pE=(1000)(10*)=107cw/L_2xl(fv107hOps2w(人+N-Z)1/2(b)Forv.r=10?cmls.SemiconductorPhysics

27、andDevicesBasicPrinciples,3rdeditionSolutionsManualChapter13ProblemSolutions # #L2X10-4Wehave=0.892Vandfor匕、.=0,weobtainSemiconductorPhysicsandDevicesBasicPrinciples,3rdeditionSolutionsManualChapter13ProblemSolutions # #%=(0.0259)ln(5xlOls)(3xlOu)2(11.7)(885x10*)(0892+1)(1.6x1严)(4xl0”)or仁=0.247pmThe

28、nS3)=eN九(a-忙)W=(1.6x1019)(4x101IMi=042pA(c)For7DS=57=IJX；=0.50pA1333(a)Theidealtraiisconductancefor匕、=0is2g(+-F；JineNLaJ2(11.7)(8.85xl0*)(0.884+-(-1.21)1/2whereSemiconductorPhysicsandDevicesBasicPrinciples,3rdeditionSolutionsManualChapter13ProblemSolutions # #Vol=orG讥=504mSWefind(1.6x10)(3xl0H)or%=(

29、4.31H严)(209+(a)For乙=0=,0=0.30pmFor乙、.=1/=,x“=0.365pm(C)“ForVi)s=57=0.553pmThedepletionregionvolumeis(护)+(xJ(%)(0)(1.6x1019)(4500)(7x101Vol=162x1011cm(b)For=17=Vol=1.74x1011cd(C)For%=57=Vol=2.08x10-11cinThegenerationcurrentisor诊=4.35/Wehave/、f4.7x1017J0=(00259)ln=00497I7x10”)sothat人=%忙=0.89-0.049=0.8

30、41T环82加SorIMi=(2A).Vol(a)ForN=0=IMi=0.39pA(b)Withasourceresistance=0.80=久1+(282)匚SemiconductorPhysicsandDevicesBasicPrinciples,3rdeditionSolutionsManualChapter13ProblemSolutions #fi=2nel!whichyieldsr=88.70s(C)pLLLAoA(欲/)(0.3xl()Y)(5xl0*)soL=(887)(1.6x10)(4500)(7x10)x(03xl0_*)(5xl04)orL=0.67pin13.35(

31、a)Foraconstantmobility(1.6x1O)(55oo)(i7)(o“I巧】2兀(13.1)(8.85xl广)(10*orft=755GHzSaturationvelocitymodelSemiconductorPhysicsandDevicesBasicPrinciples,3rdeditionSolutionsManualChapter13ProblemSolutions # #SemiconductorPhysicsandDevicesBasicPrinciples,3rdeditionSolutionsManualChapter13ProblemSolutions #

32、#13.34f,=2nC(wheregWLsuf2itL(8.85x10)(5x104)(1.5x10)Assumingvv=107/,wefind107or力=15.9GHzSemiconductorPhysicsandDevicesBasicPrinciples,3rdeditionSolutionsManualChapter13ProblemSolutions # #SemiconductorPhysicsandDevicesBasicPrinciples,3rdeditionSolutionsManualChapter13ProblemSolutions # #13.360.3x1O4SemiconductorPhysicsandDevicesBasicPrinciples,3rdeditionSolutionsManualChapter13ProblemSolutions # #(Q%(