半导体物理与器件第三版(尼曼)05章答案

Semiconductor Physics and Devices Basic Principles 3rd edition Chapter 5 Solutions Manual Problem Solutions 53 Chapter 5 Problem Solutions 5 1 a ncm O 10 163 and p n n x O i O 2 6 2 16 18 10 10 bg pxcm O 324 10 43 b Jen nO For GaAs doped at Ncm d 10 163 ncmVs 7500 2 Then Jx 16 107500 1010 1916 bgb g or JA cm 120 2 b i pcm O 10 163 nxcm O 324 10 43 ii For GaAs doped at Ncm a 10 163 p cmVs 310 2 Jep pO 16 10310 1010 1916 xbgb g JA cm 4 96 2 5 2 a VIRR 1001 R 100 b R L A L RA 10 100 10 3 3 b g 001 1 cm c eN nd or 00116 101350 19 xNdbg or Nxcm d 4 63 10 133 d ep pO 00116 10480 19 xpObg or pxcmNNN Oada 130 1010 14315 or Nxcm a 113 10 153 Note For the doping concentrations obtained the assumed mobility values are valid 5 3 a R L A L A and eN nd For Nxcm d 5 10 163 ncmVs 1100 2 Then R xx 01 16 101100 5 10100 10 19164 2 bgbgb g or Rx 1136 10 4 Then I V Rx 5 1136 10 4 ImA 044 b In this case Rx 1136 10 3 Then I V Rx 5 1136 10 3 ImA 4 4 c V L For a 5 010 50 Vcm And vd n 1100 50 or vxcm s d 55 10 4 For b V L Vcm 5 001 500 And vd 1100 500vxcm s d 55 10 5 Semiconductor Physics and Devices Basic Principles 3rd edition Chapter 5 Solutions Manual Problem Solutions 54 5 4 a GaAs R L A V I k L A 10 20 05 Now eN pa For Ncm a 10 173 p cmVs 210 2 Then 16 10210 10336 1917 1 xcmbgb g So LR Ax 500 336 85 10 8 bg or Lm 14 3 b Silicon For Ncm a 10 173 p cmVs 310 2 Then 16 10310 104 96 1917 1 xcmbgb g So LR Ax 500 4 96 85 10 8 bg or Lm 211 5 5 a V L Vcm 3 1 3 v v dnn d 10 3 4 or ncmVs 3333 2 b vd n 800 3 or vxcm s d 2 4 10 3 5 6 a Silicon For 1 kVcm vxcm s d 12 10 6 Then t d vx t d 10 12 10 4 6 txs t 833 10 11 For GaAs vxcm s d 7 5 10 6 Then t d vx t d 10 7 5 10 4 6 txs t 133 10 11 b Silicon For 50 kVcm vxcm s d 9 5 10 6 Then t d vx t d 10 9 5 10 4 6 txs t 105 10 11 GaAs vxcm s d 7 10 6 Then t d vx t d 10 7 10 4 6 txs t 143 10 11 5 7 For an intrinsic semiconductor iinp en bg a For NNcm da 10 143 np cmVscmVs 1350480 22 Then i xx 16 1015 101350480 1910 bgbg or i xcm 4 39 10 6 1 b For NNcm da 10 183 np cmVscmVs 300130 22 Then i xx 16 1015 10300130 1910 bgbg or i xcm 103 10 6 1 5 8 a GaAs epxp pOpO 516 10 19 bg From Figure 5 3 and using trial and error we find pxcmcmVs Op 13 10240 1732 Then Semiconductor Physics and Devices Basic Principles 3rd edition Chapter 5 Solutions Manual Problem Solutions 55 n n p x x O i O 2 6 2 17 18 10 13 10 bg or nxcm O 2 49 10 53 b Silicon 1 en nO or n ex O n 11 8 16 101350 19 bg or nxcm O 579 10 143 and p n n x x O i O 2 10 2 14 15 10 579 10 bg pxcm O 389 10 53 Note For the doping concentrations obtained in part b the assumed mobility values are valid 5 9 iinp en bg Then 1016 101000600 619 xnibg or nKxcm i 300391 10 93 Now nN N E kT iCV g2 F H G I K J exp or EkT N N n x g CV i F H G I K J L N M M O Q P P ln ln 2 19 2 9 2 00259 10 391 10 b g bg or EeV g 1122 Now nK i 219 2 50010 1122 00259 500 300 L N M O Q P b g af exp 515 10 26 x or nKxcm i 5002 27 10 133 Then i xx 16 102 27 101000600 1913 bgbg so i Kxcm500581 10 3 1 5 10 a i Silicon iinp en bg i xx 16 1015 101350480 1910 bgbg or i xcm 4 39 10 6 1 ii Ge i xx 16 102 4 1039001900 1913 bgbg or i xcm 2 23 10 2 1 iii GaAs i xx 16 1018 108500400 196 bgbg or i xcm 2 56 10 9 1 b R L A i R x xx 200 10 4 39 1085 10 4 68 bgbg Rx 536 10 9 ii R x xx 200 10 2 23 1085 10 4 28 bgbg Rx 106 10 6 iii R x xx 200 10 2 56 1085 10 4 98 bgbg Rx 9 19 10 12 5 11 a 5 1 eN nd Assume ncmVs 1350 2 Then N x d 1 16 101350 5 19 bg Nxcm d 9 26 10 143 b TKTC 20075 TKTC 400125 From Figure 5 2 TC Ncm d 7510 153 Semiconductor Physics and Devices Basic Principles 3rd edition Chapter 5 Solutions Manual Problem Solutions 56 ncmVs 2500 2 TC Ncm d 12510 153 ncmVs 700 2 Assuming nNxcm Od 9 26 10 143 over the temperature range For TK 200 1 16 102500 9 26 10 1914 xxbgbg 2 7 cm For TK 400 1 16 10700 9 26 10 1914 xxbgbg 9 64 cm 5 12 Computer plot 5 13 a 10Vcmvd n vd 1350 10 vxcm s d 135 10 4 so Tm vxx nd 1 2 1 2 108 9 11 10135 10 2312 2 bgbg or TxJxeV 897 1056 10 278 b 1 kVcm vxcm s d 1350 1000135 10 6 Then Txx 1 2 108 9 11 10135 10 314 2 bgbg or TxJxeV 897 1056 10 234 5 14 a nN N E kT iCV g2 F H G I K J exp 2 101 10 110 00259 1919 xxbgbgexp F H I K 7 18 10847 10 1993 xnxcm i For Ncmnncm diO 1010 143143 Then Jen nO 16 101000 10100 1914 xbgb g or JA cm 160 2 b A 5 increase is due to a 5 increase in electron concentration So nx NN n O dd i F H I K 105 10 22 14 2 2 We can write 105 105 105 10 1413 2 13 2 2 xxxni bg bg so nx i 226 525 10 F H I K F H G I K J 2 101 10 300 1919 3 xx TE kT g bgbgexp which yields 2 625 10 300 110 12 3 x T kT F H I K F H I K exp By trial and error we find TK 456 5 15 a enep nOpO and n n p O i O 2 Then en p ep ni O pO 2 To find the minimum conductivity d dp en p e O ni O p 0 1 2 2 which yields pn Oi n p F H G I K J 1 2 Answer to part b Substituting into the conductivity expression min en n en ni inp pinp 2 1 2 1 2 bg bg which simplifies to min 2eni np The intrinsic conductivity is defined as Semiconductor Physics and Devices Basic Principles 3rd edition Chapter 5 Solutions Manual Problem Solutions 57 iinpi i np enen bg The minimum conductivity can then be written as min 2 inp np 5 16 e ni 1 Now 1 1 1 50 1 5 5 50 010 2 2 1 2 1 2 F H G I K J F H G I K J exp exp E kT E kT g g or 010 1 2 1 2 12 exp F H G I K J L N M O Q P E kTkT g kT100259 kT200259 330 300 002849 F H I K 1 2 19 305 1 2 17 550 12 kTkT Then Eg19 30517 55010 ln or EeV g 1312 5 17 1111 123 1 2000 1 1500 1 500 000050000066700020 or 316 2 cmVs 5 18 n T T F H I K F H I K 1300 300 1300 300 3 23 2 a At TK 200 n 1300 1837 ncmVs 2388 2 b At TK 400 n 1300 065 ncmVs 844 2 5 19 1111 250 1 500 0006 12 Then 167 2 cmVs 5 20 Computer plot 5 21 Computer plot 5 22 JeD dn dx eD xn nnn F H G I K J 5 100 0010 14 01916 1025 5 100 0010 19 14 F H G I K J x xn bg Then 0190010 16 1025 5 100 19 14 x xn bg which yields nxcm0025 10 143 5 23 JeD dn dx eD n x nn F H G I K J 16 1025 1010 0010 19 1615 xbg or JA cm 036 2 For Acm 005 2 IAJ 005 036 ImA 18 Semiconductor Physics and Devices Basic Principles 3rd edition Chapter 5 Solutions Manual Problem Solutions 58 5 24 JeD dn dx eD n x nnn so F H G I K J 40016 10 106 10 04 10 19 1716 4 xD x x nbg or 40016Dn Then Dcms n 25 2 5 25 JeD dp dx p F H I K L NM O QP F H G I K J eD d dx x L eD L pp 101 10 16 16 16 1010 10 10 10 1916 4 x x bg b g or JA cm 16 2 constant at all three points 5 26 JxeD dp dx ppx 0 0 eD L x x p p 10 16 1010 10 5 10 15 1915 4 b g bg b g or JxA cm p 032 2 Now JxeD dn dx nnx 0 0 F H G I K J eD x L xx n n 5 10 16 1025 5 10 10 14 1914 3 bg bg or JxA cm n 02 2 Then JJxJx pn 00322 or JA cm 52 2 5 27 JeD dp dx eD d dp x ppp F H I K L NM O QP 10 22 5 15 exp Distance x is in m so 22 522 5 10 4 xcm Then JeD x x pp F H I K F H I K 10 1 22 5 1022 5 15 4 b g exp F H I K 16 1048 10 22 5 1022 5 1915 4 exp x x xbg b g or J x A cm p F H I K 341 22 5 2 exp 5 28 JeneD dn dx nnn or F H I K L NM O QP 4016 1096010 18 1916 expx x bg F H I K F H I K 16 1025 10 1 18 1018 1916 4 expx x x bg b g Then F H I K L NM O QP F H I K 401536 18 22 2 18 expexp xx Then F H I K F H I K 22 2 18 40 1536 18 exp exp x x F H I K 14 526 18 exp x 5 29 JJJ Tn drfp dif a JeD dp dx p difp and p x x L F H I K 10 15 exp where Lm 12 so JeD L x L p difp exp F H I K F H I K 10 1 15 b g or Semiconductor Physics and Devices Basic Principles 3rd edition Chapter 5 Solutions Manual Problem Solutions 59 J x x x p dif exp F H I K 16 1012 10 12 1012 1915 4 bg b g or J x L A cm p dif exp F H I K 16 2 b JJJ n drfTp dif or J x L n drf exp F H I K 4 816 c Jen n drfnO Then 16 101000 10 1916 x bgb g F H I K 4 816 exp x L which yields F H I K L NM O QP 31exp x L Vcm 5 30 a Jen xeD dn x dx nn Now ncmVs 8000 2 so that Dcms n 00259 8000207 2 Then 10016 108000 12 19 xn xbg 16 10207 19 x dn x dx bg which yields 100154 10331 10 1417 xn xx dn x dx Solution is of the form n xAB x d F H I K exp so that dn x dx B d x d F H I K exp Substituting into the differential equation we have 100154 10 14 F H I K L NM O QP expxAB x d bg F H I K 331 10 17 exp x d B x d bg This equation is valid for all x so 100154 10 14 xA or Ax 65 10 15 Also 154 10 14 expxB x d F H I K F H I K 331 10 0 17 exp x d B x d bg which yields dxcm 2 15 10 3 At x 0 en n 050 so that 5016 108000 12 19 xABbg which yields Bx 324 10 15 Then n xxx x d cm F H I K 65 10324 10 15153 exp b At xnxx 0 065 10324 10 1515 Or nxcm0326 10 153 At xm 50 nxx5065 10324 10 50 215 1515 F H I K exp or nxcm50618 10 153 c At xm 50 Jen drtn 50 16 108000 618 1012 1915 xx bgbg or JxA cm drf 5094 9 2 Then Jx dif 5010094 9 JxA cm dif 5051 2 Semiconductor Physics and Devices Basic Principles 3rd edition Chapter 5 Solutions Manual Problem Solutions 60 5 31 nn EE kT i FFi F H I K exp a EEaxb FFi b 04 0151004 3 ab g so that ax 2 5 10 2 Then EExx FFi 042 5 10 2 So nn xx kT i F H G I K J exp 042 5 10 2 b JeD dn dx nn F H G I K J F H G I K J eD n x kT xx kT ni 2 5 10042 5 10 22 exp Assume TK kTeV 30000259 and nxcm i 15 10 103 Then J xxx n 16 1025 15 102 5 10 00259 19102 bg bgbg F H G I K J exp 042 5 10 00259 2 xx or Jx xx n F H G I K J 579 10 042 5 10 00259 4 2 exp i At x 0 JxA cm n 2 95 10 32 ii At xm 5 JA cm n 237 2 5 32 a JeneD dn dx nnn F H I K 8016 101000 101 1916 x x L bgb g F H G I K J 16 10259 10 19 16 x L bg where Lxcm 10 1010 43 We find F H I K 801616 10 4144 3 x or 801614144 F H I K x L Solving for the electric field we find F H I K 3856 1 x L b For JA cm n 20 2 201614144 F H I K x L Then F H I K 2144 1 x L 5 33 a JeneD dn dx nn Let nNNxJ ddo exp 0 Then 0 ndondo NxD Nxexpexp or 0 Dn n Since DkT e n n So F H I K kT e b Vdx z 0 1 F H I Kz kT e dx 0 1 F H I K L NM O QP F H I K kT e 1 so that V kT e F H I K 5 34 From Example 5 5 x xx 00259 10 1010 00259 10 110 19 1619 3 3 b g bg b g bg Vdx dx x x zz 0 10 0 10 4 3 3 4 00259 10 110 b g bg Semiconductor Physics and Devices Basic Principles 3rd edition Chapter 5 Solutions Manual Problem Solutions 61 F H I K 00259 10 1 10 110 3 3 3 4 0 10 lnb gx 002591011 ln ln or VmV 2 73 5 35 From Equation 5 40 x d d kT eNx dNx dx F H I K F H G I K J 1 Now 100000259 1 F H G I K J Nx dNx dx d d or dNx dx xNx d d 386 100 4 Solution is of the form NxAx d exp and dNx dx Ax d exp Substituting into the differential equation AxxAx exp exp386 100 4 which yields 386 10 41 xcm At x 0 the actual value of Nd0 is arbitrary 5 36 a JJJ ndrfdif 0 JeD dn dx eD dNx dx difnn d F H I K eD L N x L n doexp We have D kT e cms nn F H I K 6000 002591554 2 Then J xx x x L dif F H I K 16 101554 5 10 01 10 1916 4 exp bgbg bg or Jx x L A cm dif F H I K 124 10 52 exp b 0 JJ drfdif Now Jen drfn F H I K L NM O QP 16 106000 5 10 1916 expxx x L bgbg F H I K 48 exp x L We have JJ drfdif so 48124 10 5 exp exp F H I K F H I K x L x x L which yields 2 58 10 3 xVcm 5 37 Computer Plot 5 38 a D kT e F H I K 925 00259 so Dcms 2396 2 b For Dcms 283 2 283 00259 1093 2 cmVs 5 39 We have Lcmm 1010 13 Wcmm 1010 24 dcmm