半导体物理与器件 第三版 (尼曼 著) 电子工业出版社 Chapter 5 课后答案.pdf

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