1、Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 8By D. A. Neamen Problem Solutions_1Chapter 88.1In forward biaskTeVISfxpThen212121expeVkTkVISfor2121lnfIeT(a)For , then021fI1ln59.1VormV mV62(b)For , then102fIln59.1VormV mV322_8.2cm415202 108.8.aipoNn3cm515202dino 3tanoVpxetap
2、op(a) V,45.0aV0259.4exp125nxcm9.33.84porcm108.9pxn3(b) V,50aV259.exp2.5ncm14830pxcm13069(c) V5.a259.exp25nxp00.18.4p_8.3cm 516220.804.aipoNn3cm41622.3.dino 3(a) V,9.aV0259.exp24nxpcm10.3.85pcm10.3(b) V10.a259.exp243nxpcm1.930.85pcm1426.3(c) 0nxp_Semiconductor Physics and Devices: Basic Principles, 4
3、th edition Chapter 8By D. A. Neamen Problem Solutions_28.4(a) 162025.aipoNncm3.415202dinocm4.3(i) tanonVpxeor notaxVl4150.l0259.V(ii) n-region - lower doped side(b) 152027.aipoNncm4.3316202dinocm35.7(i) poatanNVl41502.3l059.V61(ii) p-region - lower doped side_8.5(a) tanpopnVLeDxJxtanoaiN281629105058
4、 9.expA/cm849.12A84.103pnxAJIor mA5.(b) tapnonpVLeDxJxtapdiNe0281629.98 05.1expA/cm52.42A.413npxAJIor mA.(c) mA7.65.8_8.6For an silicon diodepnnOaiSDNAeI1261621094 550orA58.1SI(a) For V,0aVtSDIexp0259.18.5orA736.4DI(b) For V,.aV10259.exp185IorA1.SDI_Semiconductor Physics and Devices: Basic Principle
5、s, 4th edition Chapter 8By D. A. Neamen Problem Solutions_38.7021pdnoais DNeJ213194.06. 617615 04824A/cm48.sJ2(a) tasVAIexp0259.exp1568.044A2or mA.I(b) 44168.ssJA81056._8.8 (a) 02pdnoais DNeJ21195.06.81571 085A/cm14.sJ214.0AIA129.(b) tasVIexp(i) 0259.41029.4A763(ii) .exp.14IA502(iii) 0259.69.14IA68_
6、8.9We have1exptSVIor we can write this astSIso that1lnStIVIn reverse bias, is negative, so at, we have90.SI.1ln25VormV6._8.10Case 1: tasVIexp0259.6105.3sA mA1.6sI 1234JmA/cm8053.2Case 2: tasVIexp0259.712or mA093.I31AJsmA/cm922Case 3: tasVIexpSo staAJIln74108.l0259.V6aSemiconductor Physics and Device
7、s: Basic Principles, 4th edition Chapter 8By D. A. Neamen Problem Solutions_4ThenmA174010ssAJICase 4: 259.exp.etaVImA1204.sI8sJAcm57.2_8.11(a) pnonpopnLeDJdipoainoainoN222daponD190.190.daponN.nopda1.05127or 08.daN3.2adN(b) From part (a),120.nopdaD4517or 82.daN35.0ad_8.12The cross-sectional area iscm
8、43105210JIA2We have259.6expexpStDSJVwhich yieldsA/cm1052.SJ2We can writepOdnOaiSNe2We want10.11pOdnOanaDNor7771051052daaN= 10.42.333dawhich yields2.14daNNow21019105.6.5. SJ 7723.4ddNWe findcm1409.7dN3andcm6.a_8.13Plot_Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 8By D. A.
9、 Neamen Problem Solutions_58.14(a)pnonpopnLeDJdipoainoainoN222daponD1We haveand 4.2np1.0posodapn NJ1.042ordapnJ.(b) Using Einsteins relation, we can writedipaininpn NLeJ22apndneeWe haveand dnNapNAlso90.412popDLThen.pnpnJ_8.15(a) p-side; iaFi nNkTEl105.l0259.oreV3.FiEAlso on the n-side;idinNkTl1075.l
10、029.oreV47.FiE(b) We can findcm /s.32059.12nDcm /s83pNowpOdnOaiS DNeJ12210195.06.71619.843orA/cm042.SJ2Then146.1AIorA5026.SWe findtDSVIexp0259.1426.5orA A07.I7.(c) The hole current istDpodip VNenIex12Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 8By D. A. Neamen Problem So
11、lutions_6174210195.6.tDVexp9.8or(A)tDpIe10278.36Then0741.46.15SpJI_8.16(a) dipopnosp NnDeALeI 216208419 5.10506. A32.spI(b) ainonposn NDeL216207419 5.102506. A5.snI(c) 21066.l9biVV74682.0V5974biataditanonNnpxexpe20259.615.062cm43(d) tasnnpn VIxIIep0259.7461025.4A98(e) taspnpVIxIe0259.746xp1342.4A97p
12、nTotalII45137.18.4A402Now pnpnp LxILxI 2e10397.14A586ThenpnpTotalpn LxILxI 212540896.108. A579_8.17(a) The excess hole concentration is given bynonpptaLxVe1exWe findcm4162025.5.dinoNp 3and6.8pOpDLcm m4102.2Then159.0ex5.4n482.pxorcm41410.ex08.3pn 3Semiconductor Physics and Devices: Basic Principles,
13、4th edition Chapter 8By D. A. Neamen Problem Solutions_7(b) We havedxpeDJnp4411082.e082.3xAt cm,4x.3p16.13419pJorA/cm5.0p2(c) We havetanponoVLeDJxWe can determine thatcm and m3105.4po72.10nLThen43972.5.6.noJ029.61exporA/cm2615.0noJ2We can also findA/cm74.pThen at m,3xponJJ596.02.16.0orA/cm39n_8.18(a
14、) Problem 8.7tapoVnexor aitpota Nn21.0ll21.0lniat213504.l59.V20(b) Problem 8.8tanoVpexor notalditN21.0l21.0lidt2105.8ln59.V6230_8.19The excess electron concentration is given bypopnntaLxVep1exThe total number of excess electrons isdnANpp0We may note thatnnnn LxLx00eeThen1ptaonpVANWe find thatcm /s a
15、nd m25nD0.5nLAlsocm415208.8.aipoN3Then44310p1exptaVor1exp1406.tapNThen, we find the total number of excess electrons in the p-region to be:Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 8By D. A. Neamen Problem Solutions_8(a) V, 3.0aV4105.pN(b) V, 47(c) V, 5.a.pSimilarly, t
16、he total number of excess holes in the n-region is found to be1exptanopnVALPWe find thatcm /s and m0.1pD20.pLAlsocm4162015.5.dinoN3Thenexp02.tanVPSo(a) V, 3.aV314.n(b) V, 050(c) V, 5.a 6.nP_8.20kTeVkEVnI agtai xpeexp2ThenkTIgasokTEeVIgaga2121xporI ga212121eWe then have0259.3.xp10263 gEor0259.exp103g
17、EThen321lngoreV769.0E_8.21(a) We havepOdnOaiS DNAeI 12which can be written in the form2iSCIkTEgOcexp30orkTIgSexp3(b) Taking the ratio12312expkTEIggS21312gFor K, , 0T059.k6.38kTFor K, , 424.29.2(i) Germanium: eV6gE6.81.3.0exp312 SIor 8(ii) Silicon: eV2.1gE96.281.3exp30412 SIor 527._Semiconductor Phys
18、ics and Devices: Basic Principles, 4th edition Chapter 8By D. A. Neamen Problem Solutions_98.22Plot_8.23First case:tasfVIexpor V0549.12ln.l4sftINow 309.054.TK8TSecond case:podnoais DNAeI 12294606.510. in 7174or 229.8inNowkTENgciexp23191927 04.08.159.8 T25.exp TT09.31301837.2By trial and error,K5TThe
19、 reverse-bias current is limiting factor._8.24cm3710popDLor m; 10pnLW(a) tanopnpVexJx(i) tanodnVpNxpex1.0tadi2or 21.0lnitaV2105.l59.V160a(ii) tadinppNWAeDIx21542019307619.6expA3156.4pItanponVLeDtaainoNAx217207193 5.06.109.6expA612.nIpI36105.40. A37or mA.I(b) (i) tapoapVnNxnex.taai2or 21.0lniataV2105
20、.l59.V160aSemiconductor Physics and Devices: Basic Principles, 4th edition Chapter 8By D. A. Neamen Problem Solutions_10(ii) tadinppVNWAeDIx21742019307.5.619.6expA516.4pItaainonVNDee215207193 .056.109.6expA4129.nIpI54106.057. A6or mA21.I_8.25(a) We can write for the n-region022pnLdxThe general solut
21、ion is of the formppn LxBAeeThe boundary condition at givesn1extanonVppnpnLxBLAeand the boundary condition at Wgives0nnWxppnpLxBLAeeFrom this equation, we havepnxB2Then, from the first boundary condition, we obtain1exptanoVpnpnLxBLWBe2pnpnx1eWe then obtainpnpntaoLWLxVB2e1ewhich can be written aspnpn
22、pntanoLWxVBeex1We can also findpnpnpntanoLxVAeex1The solution can now be written aspntanoLWVpsih21 pnpn xxeexor finallypnpntano LWxVpsihi1exSemiconductor Physics and Devices: Basic Principles, 4th edition Chapter 8By D. A. Neamen Problem Solutions_11(b)nxppdeDJ= pntaopLWVsih1nxpnpxco1Then1exthtapnpn
23、op VLeDJ_8.26tDiDVnIexp2For the temperature range K,320Tneglect the change in and .cNThenkekTEIDgDxpeVgTaking the ratio of currents, but maintaining a constant, we haveDI21exp1kTeVEDggWe then have21ggWe haveK , V and30T60.1DeV, V59.1k59.ekTK ,eV, V0267.T0267.K ,3eV, V02763.3kT02763.3ekTFor K ,10267.
24、59.0DVwhich yieldsV82DFor K ,3T02763.159.01Dwhich yieldsV3D_8.27(a) We can writekTenCIaiDxp2where C is a constant, independent of temperature.As a first approximation, neglect the variation of and with temperature cNover the range of interest. We can then writetagDVkTEIexpe1Cg1where is another const
25、ant, independent of temperature. We findDagIkTeVE1lnorgaICe1l_8.28(a) 0021pdnais DNAeI 211945.6.10 716716040A532.sISemiconductor Physics and Devices: Basic Principles, 4th edition Chapter 8By D. A. Neamen Problem Solutions_12(b) 02WAenIigWe find 210665.4ln9.biVV760and 2/12daRbisNeW19406.57.857.12/16
26、14cm5019.6WThen 75101429 genIA3.7(c) 41506.0.sgenI_8.29(a) Set ,genSI0002 21WAenDNAipdnai 7167164254i7509.so 131325.9528incm47Then 2920.i319104.8T25.exp306947.1By trial and error,K67TWe have 02WAenIigs7514194 09.63.6.10 Then AgensI60.2or A314(b) From Problem 8.28A5.sIA107genSo tagentas VIVI2xpxpta10
27、32.5taV2exp103.715032.exptatV48.ta1053ln2taV6.0_8.305029.nnekTDcm /s5.142cm /s7.0p 2(a)(i) 0021pdnais DNAeI 261948.6.1 81686 07.570257A.sISemiconductor Physics and Devices: Basic Principles, 4th edition Chapter 8By D. A. Neamen Problem Solutions_13(ii) tasDVIexp0259.6150.2A76(iii) .8exp.2DIA91083(iv
28、) 25.05.2IA67(b) 02WenIig261108.7l59.biVV631194.3.22/161607cm502.4(i)Then8561912.4.1genIA409.6(ii) taroecVI2xp059.6e1064A93.(iii) 2.8xp14recIA7058.(iv) 59.e614recIA53._8.31Using results from Problem 8.30, we findV, A,4.0aV1604.7dIA, A135recI 1035.TIV, A6.a 12.dA, A9rec 9.6V, A8.0aV90.dIA, A713recI 7
29、1.3TIV, A.a 6.dA, A54rec 5.2V. A2.aV09.dIA, A416recI 216.TI_8.32Plot_8.33Plot_8.34We have thatpnROpOi2Let and inWe can writekTEFiniexandpFpiiWe also haveaFpiFineVEEso thatinapieThenkTnFiixEeVinai pSemiconductor Physics and Devices: Basic Principles, 4th edition Chapter 8By D. A. Neamen Problem Solutions_14Defineand kTeVakTEFinThen the recombination rate can be written asiiiiOneneRa2orai21To find the
