固态电子器件---课后答案

东南大学电子科学与工程学院习题答案SolidStateElectronicDevices王磊1/92012-06-06电子器件习题答案Chapter66.11816021021010ln0.026ln0.81(1.510)adiNNkTVVqn2191642141.61010(110)7.662211.88.8510dpqNaVV07.660.816.85PPVVVV36.853.85DGGDGPVVVVVV6.6PGPGPGmPGDDDPGPGGDPDPDPGGDPGGDPGGDPGPGDGDmPGDPGPDPDVVVVVVgsaturationgVVGVIgVVVGVVVVVVVGIVVVVVVVVVVVVVVVVVmVVxxforxmmmxxVVVVVVVVGI2/12/12/102/102/302/32/32/32/32/32/322/32/301010(1123231123;);1(!2)1(1132326.7SincedEdx,dVEdx,itisreadilyseenfromthefiguresthat东南大学电子科学与工程学院习题答案SolidStateElectronicDevices王磊2/92012-06-06(0)dqNWE21(0)22ddsqNWqNWVW2saWqN6.9计算方法见Example6-16.11Beawareitisap-channelMOSFET.Noticethesign.2idTmsFiiQQVCCFromFig.6-17,0.1msV101019922102101.6103.210/iQqCcm14721023.98.85106.910/501010iiCFcmd181010ln0.0259ln0.471.510dFiNkTVqn,20.94sFV1461918211.88.85100.943.5101.610120ssmdWcmqN东南大学电子科学与工程学院习题答案SolidStateElectronicDevices王磊3/92012-06-0619186721.610103.49105.610/ddmQqWCcmN97773.2105.581020.10.941.866.9106.910idTmsFiiQQVVCCForpchanneldevice,VGshouldbenegativetoformthechannel.Since0TV,thedeviceneedtobeenhancedtobeformthechannel.Soit'saenhancement-modedevice.iBTiiQqFVCC7122196.9101.857.98101.610iTBCVFcmq6.12ImagineaMOSFETwiththeoxidethicknessofx'andФms=0,QoxcanberegardedasQ'iofthisMOSFET.Soitisreadilyshownthat/ioxoxFBiiiQQQxVCxCdForarbitrarydistribution,theintegrationofthesheetchargesyields000()'1'()xdddiFBxiiidQxdxxxVxdxCCdCd6.18同学们自行计算，下节习题课提问。6.192idTmsFiiQQVCC1.5msV101019925105101.610810/iQqCcmxx'0Qox东南大学电子科学与工程学院习题答案SolidStateElectronicDevices王磊4/92012-06-0614721023.98.85103.4510/1001010iiCFcmd181010ln0.0259ln0.471.510dFiNkTVqn,20.94sFVSincethereisasubstratebiasof2.5BVV,1461918211.88.8510(0.942.5)6.7101.261010ssBmdVWcmqN-19186621.610106.7101.0710/amdqQCcWmN96773.2101.071021.50.942.533.45103.4510idTmsFiiQQVVCC182010/pCcm,2210182200/1.510/102.2510/innpCcmdeepinthesubstrate.UsingthedefinitionofVT,182010/itnpCcmattheinterfaceRegardingthedepletionregionasaquasiP-Njunction,andusingtheP-Ntheoryinanalogy,2.524020.02602.25103.9210/BqVkTitpneeCcmattheinterface.6.21SinceGDTVVV,6226501020011510.10.1197522102niiDGTDDiiZCCIVVVVCLSinceGDTVVV622465010200113111022102niiDGTiiZCCIVVCLxEFEFnEFpEiFFBVE东南大学电子科学与工程学院习题答案SolidStateElectronicDevices王磊5/92012-06-066.22Inlinearregion:kN*0.25(4-VT)=0.4*10-3kN*0.25(5-VT)=0.8*10-3VT=3V,andkN=1.6×10-3(A/V2)Insaturationregion:VG=5V,ID~1.5mAVG=4V,ID~0.7mA6.250iTiQVC,soionspeciesshouldbeNtype(likephosphorusion)toaccommodatepositiveiQ14821023.98.85108.6310/4001010iiCFcmd8122198.631021.08101.610iTPCVFcmq191261.6101.08102001.7281020BqFAQIAttimplantEnergyis50keVasgiven.6.292016021021010ln0.026ln0.93(1.510)adiNNkTVVqn)(2;21)(2regionsaturationVVkIVVkIregionlinearVVVkITGNDTGNDDTGNDSDxnsxpsxpdxndL东南大学电子科学与工程学院习题答案SolidStateElectronicDevices王磊6/92012-06-06Beforepunchthrough,VSDmainlyfallsacrossthedepletionregionofthedrainjunction,sothevoltageacrossthesourcejunctionisnearlyzero.SincedaNN,xnsandxndcanbenegelected145019162211.88.85103.48101.61010psaVxcmqNPunchthroughrequires4551103.48106.5210pspdxLxcmWith02()DpsaVVxqN,21916520141.610106.52100.932.332211.811.8510apsDqNxVVVChapter77.5Wehaveknown,pn/nn=10,np/pn=1/2,Wb/Lpn=1/10821988.0)]1.0sinh(1.0*5.0*2)1.0[cosh(]sinh[coshtanh1111pbADnppnpnnppbnpbADnppnpnnpLWNNLLLWBLWNNLL7.6(a)SinceitisaBJT(assumeap-n-pone),itisreasonabletobelievethatBbpWL,221sec1sec12bbppbbppWWhLLBWBWhLLAlso,lineardistributionisvalidinthebase,hence东南大学电子科学与工程学院习题答案SolidStateElectronicDevices王磊7/92012-06-06112105pEbpEbqADpWqADpWSo15,15CCII.(b)Assumeap-n-pBJT.SinceemitterandbasewidthismuchshorterthanL,linearapproximationisvalid.//10.1100101EBEBpqVkTnppppbeEPqVkTnENnbnnnpeqADpeDpDDWWIqADIDWnDDneW,11110EPnENEPpIDIIDor111111110110010bnnnneppppWnWpsec1bpWBhLwhen0bpWL.(c)SincewidthsaremuchlargerthanL,linearapproximationisinvalid.111111tanh1100ppnnnbnnpnpppnppLLnWnLpLLpsec0bpWBhLwhen1bpWL.ItisnotaBJTnow.7.8(1)∆𝑝(𝑥𝑛)=𝐶1𝑒𝑥𝑛𝐿𝑝⁄+𝐶2𝑒−𝑥𝑛𝐿𝑝⁄𝑄𝑝=A∫𝑞∆p(𝑥𝑛)𝑑𝑥𝑛𝑊𝑏0=A𝑞∫(𝐶1𝑒𝑥𝑛𝐿𝑝⁄+𝐶2𝑒−𝑥𝑛𝐿𝑝⁄)𝑑𝑥𝑛𝑊𝑏0=A𝑞𝐶1𝐿𝑝𝑒𝑥𝑛𝐿𝑝⁄|0𝑊𝑏−A𝑞𝐶2𝐿𝑝𝑒−𝑥𝑛𝐿𝑝⁄|0𝑊𝑏=A𝑞𝐿𝑝[𝐶1(𝑒𝑊𝑏𝐿𝑝⁄−1)+𝐶2(1−𝑒−𝑊𝑏𝐿𝑝⁄)]=A𝑞𝐿𝑝[∆𝑝𝐶−∆𝑝𝐸𝑒−𝑊𝑏𝐿𝑝⁄𝑒𝑊𝑏𝐿𝑝⁄−𝑒−𝑊𝑏𝐿𝑝⁄(𝑒𝑊𝑏𝐿𝑝⁄−1)+∆𝑝𝐸𝑒𝑊𝑏𝐿𝑝⁄−∆𝑝𝐶𝑒𝑊𝑏𝐿𝑝⁄−𝑒−𝑊𝑏𝐿𝑝⁄(1−𝑒−𝑊𝑏𝐿𝑝⁄)]=⋯=A𝑞𝐿𝑝(∆𝑝𝐸+∆𝑝𝐶)tanh𝑊𝑏