2022年半导体物理与器件第四版课后习题答案 .pdf

Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 8 By D. A. Neamen Problem Solutions _ 1 Chapter 8 8.1 In forward bias kTeVIISfexpThen 212121expexpexpVVkTekTeVIkTeVIIISSffor 2121lnffIIekTVV(a) For 1021ffII, then 10ln0259.021VVor 6.5921VVmV60mV (b) For 10021ffII, then 100ln0259.021VVor 3.11921VVmV120mV _ 8.2 4152102108125.2108105 .1aipoNnncm3515210210125.1102105.1dinoNnpcm3tanonnVVpxpexptapoppVVnxnexp(a)45.0aVV, 0259.045.0exp10125.15nnxp121095.3cm30259.045.0exp108125. 24ppxnor 111088.9ppxncm3(b)55.0aVV, 0259.055.0exp10125.15nnxp141088.1cm30259.055.0exp108125. 24ppxn131069.4cm3(c)55.0aVV 0259.055.0exp10125.15nnxp00259.055.0exp108125. 24ppxn0_ 8.3 516262101. 8104108.1aipoNnncm34162621024. 310108 .1dinoNnpcm3(a)90.0aVV, 0259.090. 0exp1024. 34nnxp11100. 4cm30259.090.0exp101.85ppxn名师资料总结 - - -精品资料欢迎下载 - - - - - - - - - - - - - - - - - - 名师精心整理 - - - - - - - 第 1 页，共 21 页 - - - - - - - - - Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 8 By D. A. Neamen Problem Solutions _ 2 10100.10cm3(b)10. 1aVV 0259.010.1exp1024.34nnxp141003.9cm30259.010. 1exp101 .85ppxn141026.2cm3(c)0nnxp0ppxn_ 8.4 (a)162102105105 .1aipoNnn3105.4cm3152102105105.1dinoNnp4105.4cm3(i)tanonnVVpxpexpor nonntapxpVVln415105. 41051.0ln0259.0599.0V (ii) n-region - lower doped side (b)152102107105 .1aipoNnn410214.3cm3162102103105.1dinoNnp3105.7cm3(i) poatanNVV1. 0ln41510214. 31071. 0ln0259. 06165.0V (ii) p-region - lower doped side _ 8.5 (a)tanponpnVVLneDxJexptanonaiVVDNenexp28162619105205105108. 1106. 10259.010.1exp849.1A/cm2849.1103pnnxAJIA or 85.1nImA (b)tapnopnpVVLpeDxJexptappdiVVDNenexp0281626191080.910108. 1106. 10259.010.1exp521.4A/cm2521.4103nppxAJIA or 52.4pImA (c)37.652.485.1pnIIImA _ 8.6 For an pnsilicon diode 名师资料总结 - - -精品资料欢迎下载 - - - - - - - - - - - - - - - - - - 名师精心整理 - - - - - - - 第 2 页，共 21 页 - - - - - - - - - Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 8 By D. A. Neamen Problem Solutions _ 3 nOnaiSDNAenI12616210194102510105.1106 .110or 15108.1SIA (a) For 5 .0aVV, taSDVVIIexp0259.05.0exp108.115or 71036.4DIA (b) For 5. 0aVV, 10259.05 .0exp108.115DIor 15108.1SDIIA _ 8.7 0211ppdnonaisDNDNenJ21319104. 2106 .1617615102481021102901041410568.1sJA/cm2(a)tasVVAJIexp0259.025. 0exp10568.1104441044.2A or 244. 0ImA (b)4410568. 110ssAJII810568. 1A _ 8.8 (a)0211ppdnonaisDNDNenJ21019105.1106 .1815717108101081102510511110145.5sJA/cm211410145.5102ssAJI1410029.1A (b)tasVVIIexp(i)0259.045. 0exp10029.114I71061.3A (ii)0259. 055.0exp10029. 114I51072. 1A (iii)0259.065.0exp10029.114I41016. 8A _ 8.9 We have 1exptSVVIIor we can write this as tSVVIIexp1so that 1lnStIIVVIn reverse bias, Iis negative, so at 名师资料总结 - - -精品资料欢迎下载 - - - - - - - - - - - - - - - - - - 名师精心整理 - - - - - - - 第 3 页，共 21 页 - - - - - - - - - Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 8 By D. A. Neamen Problem Solutions _ 4 90.0SII, we have 90.01ln0259.0Vor 6.59VmV _ 8.10 Case 1: tasVVIIexp0259.065.0exp1050.03sI1510305. 6sIA1210305.6mA 41210210305.6AIJss810153.3mA/cm2Case 2: tasVVIIexp0259.070. 0exp10212or 093.1ImA 312101102AIJss9102mA/cm2Case 3: tasVVAJIexpSo staAJIVVln74101080.0ln0259.06502.0aVV Then 1174101010ssAJImA Case 4: 0259.072. 0exp20.1exptasVVII1210014. 1sImA 81210210014.1ssJIA51007.5cm2_ 8.11 (a)pnopnponnponpnnLpeDLneDLneDJJJdipopainonainonNnDNnDNnD222daponnopNNDD1190.0190.01daponnopNNDD190.01noppondaDDNN1111.01051010257707857.0daNNor 73.12adNN(b)From part (a), 120.01noppondaDDNN410510102577828.2daNNor 354.0adNN_ 名师资料总结 - - -精品资料欢迎下载 - - - - - - - - - - - - - - - - - - 名师精心整理 - - - - - - - 第 4 页，共 21 页 - - - - - - - - - Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 8 By D. A. Neamen Problem Solutions _ 5 8.12 The cross-sectional area is 43105201010JIAcm2We have 0259.065. 0exp20expStDSJVVJJwhich yields 1010522.2SJA/cm2We can write pOpdnOnaiSDNDNenJ112We want 10.0111pOpdnOnanOnaDNDNDNor 777105101105251105251daaNNN=10. 010472.410071.710071.7333daNNwhich yields 23.14daNNNow 2101910105. 1106. 110522.2SJ771051011052523.141ddNNWe find 141009.7dNcm3and 161001.1aNcm3_ 8.13 Plot _ 8.14 (a) pnopnponnponpnnLpeDLneDLneDJJJdipopainonainonNnDNnDNnD222daponnopNNDD11We have 4 .21npnpDDand 1. 01ponoso dapnnNNJJJ1. 014 .2111or dapnnNNJJJ04.211(b) Using Einsteins relation, we can write dippainnainnpnnNnLeNnLeNnLeJJJ222appndndnNeLLNeNeWe have 名师资料总结 - - -精品资料欢迎下载 - - - - - - - - - - - - - - - - - - 名师精心整理 - - - - - - - 第 5 页，共 21 页 - - - - - - - - - Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 8 By D. A. Neamen Problem Solutions _ 6 dnnNeand appNeAlso 90.41.04.2popnonpnDDLLThen 90.4pnpnpnnJJJ_ 8.15 (a) p-side; iaFFinNkTEEln1015105.1105ln0259.0or 329.0FFiEEeV Also on the n-side; idFiFnNkTEEln1017105.110ln0259.0or 407.0FiFEEeV (b) We can find 4.320259.01250nDcm2/s 29.80259.0320pDcm2/s Now pOpdnOnaiSDNDNenJ11221019105 .1106.17176151029.8101104 .321051or 1110426.4SJA/cm2Then 11410426.410SSAJIor 1510426.4SIA We find tDSVVIIexp0259.05.0exp10426.415or 61007.1IA07. 1A (c) The hole current is tDpopdipVVDNAenIexp121742101910110105 .1106.1tDVVexp1029. 87or tDpVVIexp10278.316(A) Then 0741.010426.410278.31516SppJJII_ 8.16 (a)dipoppnopspNnDeALpeDAI2162108419105. 1105.110810105106.11410342.1spIA (b)ainonnponsnNnDeALneDAI2162107419105105.110225105106.11510025. 4snIA (c)2101616105. 1105. 1105ln0259.0biV746826.0V 名师资料总结 - - -精品资料欢迎下载 - - - - - - - - - - - - - - - - - - 名师精心整理 - - - - - - - 第 6 页，共 21 页 - - - - - - - - - Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 8 By D. A. Neamen Problem Solutions _ 7 5 9 7 4 6.07 4 6 8 2 6.08.08.0biaVVV taditanonnVVNnVVpxpexpexp20259. 059746.0exp105 .1105. 116210141056.1cm3(d)tasnnnpnVVIxIxIexp0259.059746.0exp10025.4155101981. 4A (e)taspnpVVIxIexp0259.059746.0exp10342.1144103997.1A pnTotalIII45103997.1101981.4410820.1A Now ppnppnpLLxILxI21ex p2121exp103997.145104896.8A Then pnpTotalpnnLxIILxI212154104896. 810820. 1510710.9A _ 8.17 (a) The excess hole concentration is given by nonnpppptanoLxVVpexp1expWe find 41621021025.210105. 1dinoNnpcm3and 61001.08pOppDL410828.2cm 828.2m Then 10259.0610.0exp1025.24np410828.2expxor 41410828. 2exp1081.3xpncm3(b) We have dxpdeDJnpp441410828.2exp10828.210808.3xeDpAt 4103xcm, 828. 23exp10828. 210808.38106. 1341419pJor 5966.03pJA/cm2(c) We have tanponnoVVLneDJexpWe can determine that 名师资料总结 - - -精品资料欢迎下载 - - - - - - - - - - - - - - - - - - 名师精心整理 - - - - - - - 第 7 页，共 21 页 - - - - - - - - - Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 8 By D. A. Neamen Problem Solutions _ 8 3105 .4poncm3and 72.10nLm Then 43191072.10105. 423106. 1noJ0259.0610.0expor 2615.0noJA/cm2We can also find 724. 1poJA/cm2Then at 3xm, 33ppononJJJJ5966.0724.12615.0or 39.13nJA/cm2_ 8.18 (a) Problem 8.7 tapopVVnnexpor aiatpoptaNnNVnnVV21 .0lnln221. 0lniatnNV213215104.21041.0ln0259.0205.0V (b) Problem 8.8 tanonVVppexpor nontappVVlndidtNnNV21.0ln221 .0lnidtnNV210215105.11081. 0ln0259.0623.0V _ 8.19 The excess electron concentration is given by poppnnnntapoLxVVnexp1expThe total number of excess electrons is dxnANpp0We may note that nnnnLLxLdxLx00expexpThen 1exptaponpVVnALNWe find that 25nDcm2/s and 0.50nLm Also 41521021081. 2108105. 1aipoNnncm3Then 443108125.2100.5010pN1exptaVVor 1exp1406.0tapVVNThen, we find the total number of excess electrons in the p-region to be: (a)3.0aVV, 41051.1pN(b)4.0aVV, 51017. 7pN名师资料总结 - - -精品资料欢迎下载 - - - - - - - - - - - - - - - - - - 名师精心整理 - - - - - - - 第 8 页，共 21 页 - - - - - - - - - Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 8 By D. A. Neamen Problem Solutions _ 9 (c)5. 0aVV, 71040.3pNSimilarly, the total number of excess holes in the n-region is found to be 1exptanopnVVpALPWe find that 0.10pDcm2/s and 0.10pLm Also 41621021025. 210105.1dinoNnpcm3Then 1exp1025.22tanVVPSo (a)3. 0aVV, 31041. 2nP(b)4. 0aVV, 51015.1nP(c)5. 0aVV, 61045.5nP_ 8.20 kTeVkTEVVnIagtaiexpexpexp2Then kTEeVIgaexpso kTEeVkTEeVIIgaga221121expexpor kTEEeVeVIIggaa212121expWe then have 0259.0525.032.0255. 0exp10101010263gEor 0259.059.0exp1023gEThen 3210ln0259.059.0gEor 769.02gEeV _ 8.21 (a) We have pOpdnOnaiSDNDNAenI112which can be written in the form 2iSnCIkTETNNCgOcOexp3003or kTECTIgSexp3(b) Taking the ratio 1231212expexpkTEkTETTIIggSS2131211expkTkTETTgFor 3001TK, 0259.01kT, 61.3811kTFor 4002TK, 03453. 02kT, 96.2812kT(i) Germanium: 66. 0gEeV 96.2861.3866.0exp300400312SSII名师资料总结 - - -精品资料欢迎下载 - - - - - - - - - - - - - - - - - - 名师精心整理 - - - - - - - 第 9 页，共 21 页 - - - - - - - - - Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 8 By D. A. Neamen Problem Solutions _ 10 or 138312SSII(ii) Silicon: 12. 1gEeV 96.2861.3812. 1exp300400312SSIIor 5121017. 1SSII_ 8.22 Plot _ 8.23 First case: tasfVVIIexpor 05049.0102ln50. 0ln4sfatIIVVV Now 3000259.005049.0T8 .584TK Second case: popdnonaisDNDNAenI11221946106. 1105102 .1in71771510101021105251041or 272102519.8inNow kTENNngciexp231919273001004. 1108 .2102519.8T3000259.012.1expTTT0259.030012.1exp300108337.2311By trial and error, 502TK The reverse-bias current is limiting factor. _ 8.24 37101010poppDLcm or 10pLm; pnLW(a)tannopnpVVWpeDxJexp(i)tanodnnVVpNxpexp1.0tadiVVNnexp2or 221.0lnidtanNVV210215105.11021. 0ln0259.05516.0aVV (ii)tadinppVVNnWAeDIexp2154210193102107.0105.110106. 1100259.05516.0exp310565.4pIA tanponnVVLnAeDIexp名师资料总结 - - -精品资料欢迎下载 - - - - - - - - - - - - - - - - - - 名师精心整理 - - - - - - - 第 10 页，共 21 页 - - - - - - - - - Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 8 By D. A. Neamen Problem Solutions _ 11 taainonVVNnDAeexp2172107193102105.110525106.1100259.05516.0exp61026.2nIA pnIII3610565. 41026.2310567.4A or 567. 4ImA (b) (i)tapoappVVnNxnexp1.0taaiVVNnexp2or 221. 0lniatanNVV210215105.11021. 0ln0259. 05516. 0aVV (ii)tadinppVVNnWAeDIexp2174210193102107.0105. 110106.1100259.05516.0exp510565.4pIA taainonnVVNnDAeIexp2152107193102105.110525106.1100259.05516.0exp4102597.2nIA pnIII5410565.4102597.2410716.2A or 2716.0ImA _ 8.25 (a) We can write for the n-region 0222pnnLpdxpdThe general solution is of the form ppnLxBLxApexpexpThe boundary condition at nxxgives 1exptanonnVVpxppnpnLxBLxAexpexpand the boundary condition at nnWxxgives 0nnnWxppnnpnnLWxBLWxAexpexpFrom this equation, we have pnnLWxBA2expThen, from the first boundary condition, we obtain 1exptanoVVp名师资料总结 - - -精品资料欢迎下载 - - - - - - - - - - - - - - - - - - 名师精心整理 - - - - - - - 第 11 页，共 21 页 - - - - - - - - - Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 8 By D. A. Neamen Problem Solutions _ 12 pnpnnLxBLWxBexp2exppnpnLWLxB2exp1expWe then obtain pnpntanoLWLxVVpB2exp1exp1expwhich can be written as pnpnpnntanoLWLWLWxVVpBexpexpexp1expWe can also find pnpnpnntanoLWLWLWxVVpAexpexpexp1expThe solution can now be written as pntanonLWVVppsinh21exppnnpnnLxWxLxWxexpexpor finally pnpnntanonLWLxWxVVppsinhsinh1exp(b) nxxnppdxpdeDJ=pntanopLWVVpeDsinh1expnxxpnnpLxWxLcosh1Then 1expcothtapnpnoppVVLWLpeDJ_ 8.26 tDiDVVnIexp2For the temperature range 320300TK, neglect the change in cNand N. Then kTeVkTEIDgDexpexpkTeVEDgexpTaking the ratio of currents, but maintaining DIa constant, we have 2211expexp1kTeVEkTeVEDgDgWe then have 2211kTeVEkTeVEDgDgWe have 300TK , 60. 01DVV and 0259. 01kTeV, 0259. 01ekTV 名师资料总结 - - -精品资料欢迎下载 - - - - - - - - - - - - - - - - - - 名师精心整理 - - - - - - - 第 12 页，共 21 页 - - - - - - - - - Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 8 By D. A. Neamen Problem Solutions _ 13 310TK , 02676.02kTeV, 02676.02ekTV 320TK , 02763.03kTeV, 02763.03ekTV For 310TK , 02676.012. 10259.060.012.12DVwhich yields 5827.02DVV For 320TK , 02763.012. 10259.060.012.13DVwhich yields 5653. 03DVV _ 8.27 (a)We can write kTeVnCIaiDexp2where C is a constant, independent of temperature. As a first approximation, neglect the variation of cNand Nwith temperature over the range of interest. We can then write tagDVVkTECIexpexp1kTeVECagexp1where 1Cis another constant, independent of temperature. We find DagICkTeVE1lnor DgaICekTEV1ln_ 8.28 (a)00211ppdnnaisDNDNAenI210194105.1106. 11071671610101041102510411510323. 2sIA (b)02WAenIigenWe find 2101616105.1104104ln0259.0biV7665.0V and 2/12dadaRbisNNNNeVVW1914106.157665. 01085.87 .1122/116161616104104104104510109.6Wcm Then 751019410210109.6105. 1106 .110genI1110331. 7A (c)415111016.310323. 210331.7sgenII_ 8.29 (a)Set genSII, 0002211WAenDNDNAenippdnnai名师资料总结 - - -精品资料欢迎下载 - - - - - - - - - - - - - - - - - - 名师精心整理 - - - - - - - 第 13 页，共 21 页 - - - - - - - - - Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 8 By D. A. Neamen Problem Solutions _ 14 7167161010104110251041in75010210109.62Wso 131321050.2109528.3100545.3in1410734. 4cm3Then 292102407. 2in319193001004.1108.2TTT0 2 5 9.03 0 012.1exp300106947.7310By trial and error, 567TK We have 02WA e nIIig e ns751419410210109.610734. 4106 .110Then gensII610314.2A or 314. 2gensIIA (b)From Problem 8.28 1510323.2sIA 1110331.7genIA So tagentasVVIVVII2expexptaVVe x p10323.215taVV2exp10331. 711151110323. 210331.72expexptataVVVV4101558.32exptaVV4101558.3ln2taVV5366. 0V _ 8.30 55000259. 0nnekTD5.142cm2/s 70. 52200259. 0pDcm2/s (a) (i)00211ppdnnaisDNDNAenI26194108. 1106.110281681610270.510711025.1421071221050.1sIA (ii)tasDVVIIexp0259.06. 0exp1050.1221210726.1A (iii)0259.08 .0exp1050.122DI910896.3A (iv)0259.00 .1exp1050.122DI610795.8A 名师资料总结 - - -精品资料欢迎下载 - - - - - - - - - - - - - - - - - - 名师精心整理 - - - - - - - 第 14 页，共 21 页 - - - - - - - - - Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 8 By D. A. Neamen Problem Solutions _ 15 (b)02WAenIigen261616108.1107107ln0259. 0biV263.1V 1914106.13263.11085. 81.132W2/116161616107107107107510201. 4cm (i)Then 856194102210201.4108. 1106. 1102genI1410049.6A (ii)tarorecVVII2exp0259.026.0exp10614910436.6A (iii)0259. 028 .0exp10614recI710058.3A (iv)0259. 020.1exp10614recI510453.1A _ 8.31 Using results from Problem 8.30, we find 4 .0aVV, 161064.7dIA, 101035.1recIA, 101035.1TIA 6 .0aVV, 121073.1dIA 91044. 6recIA, 91044.6TIA 8 .0aVV, 91090. 3dIA 71006. 3recIA, 71010.3TIA 0. 1aVV, 61080.8dIA 51045.1recIA, 51033.2TIA 2. 1aVV. 21099.1dIA 41090. 6recIA, 21006.2TIA _ 8.32 Plot _ 8.33 Plot _ 8.34 We have that ppnnnnpRnOpOi2Let OnOpOand inpnWe can write kTEEnnFiFniexpand kTEEnpFpFiiexpWe also have aFpFiFiFneVEEEEso that FiFnaFpFiEEeVEEThen kTEEeVnpFiFnaiexpkTEEkTeVnFiFnaiexpexp名师资料总结 - - -精品资料欢迎下载 - - - - - - - - - - - - - - - - - - 名师精心整理 - - - - - - - 第 15 页，共 21 页 - - - - - - - - - Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 8 By D. A. Neamen Problem Solutions _ 16 Define kTeVaaand kTEEFiFnThen the recombination rate can be written as iiiiOiiineennenneenenRaa2or eeeenRaaOi21To find the maximum recombination rate, set 0ddR121eeeddenaaOior 22110eeeenaaOieeeawhich simplifies to 2210eeeeeeenaaaOiThe denominator is not zero, so we have eeea0or aee22aThen the maximum recombination rate becomes 22max21aaaaeeeenROi2221aaaeeenOior 1212maxaaeenROiwhich can be written as 12exp21expmaxkTeVkTeVnRaOaiIf ekTVa, then we can neglec