半导体物理与器件第三版(尼曼)05章答案

1、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, ncmVs7500 2 / Then Jx= ()(

2、 )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 cmVs310 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

3、 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, ncmVs1100 2

4、 / 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 5

5、00 . / 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 cmVs210 2 / Then = ()()16 10210 10336 1917 1 .xcmbgb g So LR Ax

6、=()() 500 336 85 10 8 .bg or Lm= 14.3 (b) Silicon For Ncm a = 10 173 , p cmVs310 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

7、: 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

8、 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 cmVscmVs300130 22 /,/ Then i xx=+ ()16 1015 103

9、00130 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

10、 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 a

11、re 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

12、 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.

13、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 n

14、cmVs=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 ncmVs2500 2 / TC Ncm d = 12510 153 , ncmVs7

15、00 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

16、 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 .

17、 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

18、 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

19、 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 S

20、olutions 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= . kT20

21、0259 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

22、) 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 . . ()() () (

23、 ) = 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

24、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 5.2

25、6 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

26、 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

27、 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, =

28、 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

29、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

30、 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

31、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

32、 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 =16 108

33、000 618 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

34、 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

35、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

36、= 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

37、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

38、 .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

39、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 +=()()AxxAxexp.exp386 100 4 which yields = 386 10 41 .xcm At x = 0, the actual value of Nd0( ) is arbitrar

40、y. 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

41、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= 101