半导体物理与器件第三版(尼曼)06章答案

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1、rdSemiconductorPhysicsandDevices:BasicPrinciples,3editionChapter6SolutionsManualProblemSolutionsChapter6ProblemSolutions42.2510x93−−1G=⇒Gx=112510.cms6.1−62010xn-typesemiconductor,low-injectionsothat(c)13δpx51093−−1R′==RG==112510.xcms−6τ10pOor6.419−−31Rxc′=510msb662510.−34gb3108ghcxx(a)

2、Eh===ν−10λ630010x6.2orn−19(a)R=OExJ=31510.Thisistheenergyof1nOτphoton.nOandNow21821011WJ=⇒/s31710.xphotons/snb10gi43−n===10cm+3O16Volume=101()()..=01cmp10OThenThen18431710.x1010−−31g=⇒R=⇒Rxc=510msnO−7nO01.210x193(b)gxe=−−31710./hpairscms12δn10(b)18−−31Rn==−7orRxcn=510msδδτnpg===bg31710

3、.x19bg1010x−6τ210xnOsoor181014−3∆RRRx=−=510−510x⇒δδnp==31710.xcmnnnO18−−31∆Rxc≈510msn6.5Wehave6.3∂pp(a)Recombinationratesareequal+=−∇•Fg+−ppnp∂tτpOO=ττandnOpOJepe=−µΕDp∇16−3pppnN==10cmOdTheholeparticlecurrentdensityis22b1510.x10gnJi43−+pp===2.2510xcmF==−µpDpΕ∇O16pppn10()+eOSoNow164+()1

4、02.2510x∇•Fp=µ∇•Ε−D∇•∇pppp=−6Wecanwriteτ2010xnOor∇•()ppΕΕ=•∇+∇•pΕ+6andτ=88910.xsnO2∇•∇=∇pp(b)GenerationRate=RecombinationRateSoso65rdSemiconductorPhysicsandDevices:BasicPrinciples,3editionChapter6SolutionsManualProblemSolutionsBychargeneutrality+()2δδδn=≡⇒pnnp∇=()()δ∇δ∇•Fp=µΕΕ•∇+∇•p−D∇

5、ppppThen22∂()()δδn∂pand∇=()δδnp∇()and=∂p2p∂t∂t=−µ()ΕΕ•∇+∇•pp+Dp∇+−gpppAlso∂tτppnWecanthenwriteggg=≡,=≡Rnp2ττDp∇−µ()ΕΕ•∇+∇pp•pnppThenwecanwritepp∂2+−=gp(1)Dn∇−•()δµΕΕ∇+()δnp∇•ppτ∂tp∂()δn+−=gR6.6∂tFromEquation[6.18]and2∂pp(2)Dn∇+•()δµΕΕ∇+()δnn∇•+nn=−∇•Fg+−pp()∂tτ∂δnp+−=gR∂p∂tForsteady-st

6、ate,=0MultiplyEquation(1)byµnandEquation(2)∂tnThenbyµp,andthenaddthetwoequations.p+0=−∇•FgR+−Wefindppp2andforaone-dimensionalcase,bgµµnD+∇pD()δnnppn+dF+−µµ()()pnΕ•∇δnp2019np=−=−gR10210x⇒ppdx∂()δn+++bgbgµµnpnp()g−=+RnpµµnpdFp19−−31∂t=810xcmsdxDividebybgµµnp+,thennpFµµnD+pDInppn26.7∇()δn

7、FromEquation[6.18],HGµµnp+KJnp+dFp19Lµµ()pn−O00=−+−2x10np()+MPΕ•∇δndxµµnp+NnpQor+∂()δndFp19−−31+−=()gR=−210xcms∂tdxDefineµµnD+pDDDnp()+nppnnp6.8D′==µµnp+DnDp+Wehavethecontinuityequationsnpnp2()(1)Dp∇−•()δµΕΕ∇+()δpp∇•µµnppn−ppandµ′=pp∂()δµµnpnp++g−=pThenwehaveτ∂tp∂()δnand2()()()Dn′∇+δ