半导体物理与器件第三版(尼曼)12章答案

1、Semiconductor Physics and Devices: Basic Principles, 3rd edition Chapter 12 Solutions Manual Problem Solutions 185 Chapter 12 Problem Solutions 12.1 (a) I V V D GS t = () L N M O Q P 10 2.1 15 exp For VV GS = 05 ., ID= ()() L N M O Q P 10 05 2.1 00259 15 exp . . IxA D = 9.83 10 12 For VV GS = 07 .,

2、IxA D = 388 10 10 . For VV GS = 09 ., IxA D = 154 10 8 . Then the total current is: II TotalD =10 6 b g For VV GS = 05 ., IA Total = 9.83 For VV GS = 07 ., ImA Total = 0388. For VV GS = 09 ., ImA Total = 154 . (b) Power: PIV TotalDD = Then For VV GS = 05 ., PW= 49.2 For VV GS = 07 ., PmW= 194. For V

3、V GS = 09 ., PmW= 77 12.2 We have L eN VsatV a fpDSDS = +() 2 +() fpDS Vsat where fpt a i V N nx = F H G I K J () F H G I K J ln.ln . 00259 10 15 10 16 10 or fp V= 0347. We find 2 2 117 885 10 16 1010 14 1916 1 2 = ()L N M O Q P eN x x a . . / bg bgb g = 0360 1 2 ./ / m V We have VsatVV DSGST ()= (a

4、) For VVVsatV GSDS =()54.25 Then L =+03600347503474.25. or Lm= 00606. If L is 10%of L, then Lm= 0606. (b) For VVVVVsatV DSGSDS =()52125,. Then L =+0360034750347125. or Lm= 0377. Now if L is 10% of L, then Lm= 377. 12.3 L eN VsatV a fpDSDS = +() 2 +() fpDS Vsat where fpt a i V N n x x = F H G I K J (

5、) F H G I K J ln.ln . 00259 4 10 15 10 16 10 or fp V= 0383. and x eN dT fp a = L N M O Q P 4 1 2 / = ()()L N M O Q P 4 117 885 100383 16 104 10 14 1916 1 2 . . / x xx bg bgbg or xm dT = 0157. Then =()QeN x SDadT max Semiconductor Physics and Devices: Basic Principles, 3rd edition Chapter 12 Solution

6、s Manual Problem Solutions 186 = 16 104 100157 10 19164 .xxxbgbgbg or =() QC cm SD max/10 72 Now VQQ t TSDSS ox ox msfp = +() F H G I K J maxaf2 so that V xxx x T = () 1016 103 10400 10 39 885 10 719108 14 . . bgbgbg bg + +()02 0383. or VV T = 187. Now VsatVVV DSGST ()=5187313. We find 2 2 117 885 1

7、0 16 104 10 14 1916 1 2 = ()L N M O Q P eN x xx a . . / bg bgbg = 180 10 5 .x Now Lx VDS=+ 180 100383313 5 . +0383313. or LxVDS=+ 180 1035133513 5 . We obtain VDS Lm() 0 1 2 3 4 5 0 0.0451 0.0853 0.122 0.156 0.188 12.4 Computer plot 12.5 Plot 12.6 Plot 12.7 (a) Assume VsatV DS( )= 1, We have sat DS

8、Vsat L = () We find Lm() satVcm/() 3 1 05 . 025. 013. 333 10 3 .x 10 4 2 10 4 x 4 10 4 x 7.69 10 4 x (b) Assume ncmVs=500 2 /, we have v nsat = Then For Lm= 3, vxcm s= 167 10 6 ./ For Lm= 1, vxcm s= 5 10 6 / For Lm 05 ., vcm s 10 7 / 12.8 We have =() IL LLI DD 1 We may write g I V L LLI L V O D DS D

9、 DS = = () () ()F H G I K J 1 2 = () ()L LL I L V D DS 2 We have L eN VVsat a fpDSfpDS = +() 2 We find = + ()L VeNV DSa fpDS 21 2 (a) For VVVV GSDS =21, , and VsatVVV DSGST ()=20812. Also VVsatVV DSDSDS =+=+=()1212.2. and fp x x V=() F H G I K J 00259 3 10 15 10 0376 16 10 .ln . . Now Semiconductor

10、Physics and Devices: Basic Principles, 3rd edition Chapter 12 Solutions Manual Problem Solutions 187 2 2 117 885 10 16 103 10 14 1916 1 2 = ()L N M O Q P eN x xx a . . / bg bgbg = 02077 1 2 ./ / m V We find L =+0207703762.2037612. = 00726.m Then = + ()L VDS 02077 2 1 03762.2 . . = 00647./m V From th

11、e previous problem, ImALm D =0482., Then gx O = () () 2 200726 048 1000647 2 3 . .bg or gxS O = 167 10 5 . so that r g k O O = 1 59.8 (b) If Lm= 1, then from the previous problem, we would have ImA D = 096., so that gx O = () () 1 100726 096 1000647 2 3 . .bg or gxS O = 7.22 10 5 so that r g k O O =

12、 1 138 . 12.9 (a) Isat WC L VV D nox GST ()= 2 2 af = F H I K( ) 10 2 500 69 101 8 2 . xVGSb ga f or IsatVmA DGS ()()=01731 2 .af and IsatVmA DGS ()()=01731 1 2 . / af (b) Let effO eff C = F H G I K J 1 3/ Where OcmVs=1000 2 / and CxVcm= 2.5 10 4 /. Let eff GS ox V t = We find C t t C x x ox ox ox o

13、x ox ox = = = () 39 885 10 69 10 14 8 . . bg or tA ox = 500 Then VGS eff eff Isat D( ) 1 2 3 4 5 - 4E5 6E5 8E5 10E5 - 397 347 315 292 0 0.370 0.692 0.989 1.27 (c) The slope of the variable mobility curve is not constant, but is continually decreasing. 12.10 Plot 12.11 VV Q C TFB SD ox fp =+ + ()max

14、2 We find fpt a i V N n x x = F H G I K J () F H G I K J ln.ln . 00259 5 10 15 10 16 10 or fp V= 0389. and x eN dT fp a = L N M O Q P 4 1 2 / = ()()L N M O Q P 4 117 885 100389 16 105 10 14 1916 1 2 . . / x xx bg bgbg or xm dT = 0142. Now Semiconductor Physics and Devices: Basic Principles, 3rd edit

15、ion Chapter 12 Solutions Manual Problem Solutions 188 =()QeN x SDadT max = 16 105 100142 10 19164 .xxxbgbgbg or =() QxC cm SD max./114 10 72 Also C t x x x ox ox ox Fcm= = () 39 885 10 400 10 863 10 14 8 82 . ./ bg Then V x x T = + ()112 114 10 863 10 2 0389 7 8 . . . . or VV T = +090. (a) I WC L VV

16、 VV D nox GSTDSDS = 2 2 2 af and VsatVV DSGST ()= We have Ix D =F H I KFH I K( ) 20 2 1 2 400 863 10 8 .bg 2 2 VV VV GSTDSDS af or IVV VVmA DGSTDSDS =()0173 2 2 .af For VVsatVVV DSDSGST =()1, IsatmA D( )= 0173. For VVsatVVV DSDSGST =()2, IsatmA D( )= 0692. (b) For VVcmVs DSn =125400 2 .,/. The curve

17、 for VVV GST = 1 is unchanged. For VVV GST = 2 and 0125VV DS ., the curve in unchanged. For VV DS 125., the current is constant at ImA D =( )() ()0173 2 2 1251250595 2 . When velocity saturation occurs, VsatV DS( )= 125. for the case of VVV GST = 2. 12.12 Plot 12.13 (a) Non-saturation region IC W L

18、VV VV DnoxGSTDSDS = F H I K 1 2 2 2 af We have C t C k ox ox ox ox = and WkWLkL, also VkVVkV GSGSDSDS , So I C k kW kL kVVkVkV Dn ox GSTDSDS = F H I KFH I K 1 2 2 2 afa f Then IkI DD In the saturation region I C k kW kL kVV Dn ox GST = F H I KFH I K 1 2 2 Then IkI DD (b) PI VkIkVk P DDDDDD =a faf 2

19、12.14 IsatWCVVv DoxGSTsat ()=af ()F H I K kW C k kVVv ox GSTsat af or IsatkIsat DD ()() 12.15 (a) (i) IK VV DnGST =()()af 22 01 508. or ImA D = 1764. (ii) ID= F H I K ()( ) 01 06 06 508 2 . . . or ImA D = 0807. (b) (i) P =()( )1764 5. PmW= 882. (ii) P =()()( )080706 5. PmW= 2.42 Semiconductor Physic

20、s and Devices: Basic Principles, 3rd edition Chapter 12 Solutions Manual Problem Solutions 189 (c) Current: Ratio = 0807 1764 0457 . . . Power: Ratio = 2.42 882 0274 . . 12.16 V eN x C r L x r T adT ox j dT j = + L N M O Q P R S T U V W 1 2 1 Now fpt a i V N nx = F H G I K J () F H G I K J ln.ln . 0

21、0259 10 15 10 16 10 or fp V= 0347. and x eN dT fp a = L N M O Q P 4 1 2 / = ()()L N M O Q P 4 117 885 100347 16 1010 14 1916 1 2 . . / x x bg bgb g or xm dT = 030. Also C t x x ox ox ox = = () 39 885 10 450 10 14 8 .bg = 7.67 10 82 xFcm/ Then V xx x T = 16 101003 10 7.67 10 19164 8 .bgb gbg + ()L N

22、M O Q P R S T U V W 03 1 1 2 03 03 1 . . or VV T = 0137. 12.17 V eN x C r L x r T adT ox j dT j = + L N M O Q P R S T U V W 1 2 1 Now fp x x V=() F H G I K J 00259 3 10 15 10 0376 16 10 .ln . . and x eN dT fp a = L N M O Q P 4 1 2 / = ()()L N M O Q P 4 117 885 100376 16 103 10 14 1916 1 2 . . / x xx

23、 bg bgbg or xm dT = 0180. Also C t x x ox ox ox = = () 39 885 10 800 10 14 8 .bg or CxFcm ox = 4.31 10 82 / Then V xxx x T = = 020 16 103 10018 10 4.31 10 19164 8 . .bgbgbg + ()L N M O Q P R S T U V W 06 1 2 018 06 1 . .L or = = 020 0319 . . L which yields Lm= 159. 12.18 We have = +()LLab and from t

24、he geometry (1) arxrx jdTjdS +=+bgbg 2 2 2 and (2) brxrx jdTjdD +=+bgbg 2 2 2 From (1), arrxx jjdSdT +=+bg bg 22 2 so that arxxr jdSdTj =+bg 2 2 which can be written as ar x r x r j dS j dT j =+ F H G I K J F H G I K J L N M M O Q P P 11 22 or Semiconductor Physics and Devices: Basic Principles, 3rd

25、 edition Chapter 12 Solutions Manual Problem Solutions 190 ar x r x r x r j dS j dS j dT j =+ F H G I K J F H G I K J L N M M O Q P P 1 2 1 22 Define 2 22 2 = xx r dSdT j We can then write ar x r j dS j =+ L N M O Q P 1 2 1 2 Similarly from (2), we will have br x r j dD j =+ L N M O Q P 1 2 1 2 wher

26、e 2 22 2 = xx r dDdT j The average bulk charge in the trapezoid (per unit area) is = +F H I K QLeN x LL BadT 2 or = +F H I K QeN x LL L BadT 2 We can write LL L L LL Lab + =+ =+() 2 1 22 1 2 1 2 which is = +() 1 2 ab L Then = +()L NM O QP QeN x ab L BadT 1 2 Now QB replaces ()QSDmax in the threshold

27、 equation. Then V Q C Q C T B ox SD ox = ()max = + ()L NM O QP eN x C ab L eN x C adT ox adT ox 1 2 or V eN x C ab L T adT ox = +() 2 Then substituting, we obtain V eN x C r L x r T adT ox j dS j = + L N M O Q P R S T 2 1 2 1 2 + L N M O Q PUV W 1 2 1 2 x r dD j Note that if xxx dSdDdT =, then = 0 a

28、nd the expression for VT reduces to that given in the text. 12.19 We have = L0 , so Equation (12.25) becomes LL L L L r L x r j dT j + =+ L N M O Q P R S T U V W 22 1 2 11 2 1 or r L x r j dT j 1 2 1 1 2 += L N M O Q P Then Equation (12.26) is = F H I K QeN x BadT 1 2 The change in the threshold vol

29、tage is V Q C Q C T B ox SD ox = ()max or V eN x C eN x C T adT ox adT ox = 1 2a faf af or V eN x C T adT ox = F H I K 1 2 af 12.20 Computer plot 12.21 Computer plot 12.22 V eN x C r L x r T adT ox j dT j = + L N M O Q P R S T U V W 1 2 1 Semiconductor Physics and Devices: Basic Principles, 3rd edit

30、ion Chapter 12 Solutions Manual Problem Solutions 191 + F H I K F H I K L N M O Q P R S T U V W e N k kx C k kr kL kx kr a dT ox j dT j a f 1 2 1 or Vk V TT = 12.23 V eN x C x W T adT ox dT = F H I K We find fp x V=() F H G I K J 00259 10 15 10 0347 16 10 .ln . . and x eN dT fp a = L N M O Q P 4 1 2

31、 / = ()()L N M O Q P 4 117 885 100347 16 1010 14 1916 1 2 . . / x x bg bgb g or xm dT = 030. Also C t x x ox ox ox = = () 39 885 10 450 10 14 8 .bg or CxFcm ox = 7.67 10 82 / Then V xx x T = 16 101003 10 7.67 10 19164 8 .bgb gbg L N M O Q P 203 10 2.5 10 4 4 a fbg. x x or VV T = +0118. 12.24 Additio

32、nal bulk charge due to the ends: QeN LxeN Lxx BadTadTdT = F H I K 1 2 2 2 a f where = 1. Then V eN x C W T adT ox = 2 We find fp x x V=() F H G I K J 00259 3 10 15 10 0376 16 10 .ln . . and x eN dT fp a = L N M O Q P 4 1 2 / = ()()L N M O Q P 4 117 885 100376 16 103 10 14 1916 1 2 . . / x xx bg bgbg

33、 or xm dT = 0180. Also C t x x ox ox ox = = () 39 885 10 800 10 14 8 .bg or CxFcm ox = 4.31 10 82 / Now, we can write W eN x CV adT oxT = 2 a f = () 16 103 10018 10 4.31 10025 19164 2 8 . . xxx x bgbgbg bg or Wm= 144. 12.25 Computer plot 12.26 V eN x C x W T adT ox dT = F H I K Assume that is a cons

34、tant F H I K F H I K F H I K e N k kx C k kx kW a dT ox dT a f or Vk V TT = 12.27 (a) Vxtxx BDox = 6 106 10250 10 668 bgbgbg or Semiconductor Physics and Devices: Basic Principles, 3rd edition Chapter 12 Solutions Manual Problem Solutions 192 VV BD = 15 (b) With a safety factor of 3, VBD= 1 3 15 VV

35、BD = 5 12.28 We want VV G = 20. With a safety factor of 3, then VV BD = 60, so that 606 10 6 =xtoxbg tA ox = 1000 12.29 Snapback breakdown means M = 1, where =() F H I K 018 3 10 10 9 .log I x D and M V V CE BD m = F H G I K J 1 1 Let VVm BD =153,. Now when M VCE = F H I K 1 1 15 3 we can write this

36、 as 1 15 151 3 3 = F H I K V V CE CE Now ID VCE E-8 E-7 E-6 E-5 E-4 E-3 0.0941 0.274 0.454 0.634 0.814 0.994 14.5 13.5 12.3 10.7 8.6 2.7 12.30 One Debye length is L kT e eN D a = L N M O Q P af 1 2/ = ()()L N M O Q P 117 885 1000259 16 1010 14 1916 1 2 . . / x x bg bgb g or Lxcm D = 4.09 10 6 Six De

37、bye lengths: 6 4.09 100246 6 xm =bg. From Example 12.4, we have xm dO = 0336., which is the zero-biased source-substrate junction width. At near punch-through, we will have xLxL dODd +=6 where xd is the reverse-biased drain-substrate junction width. Now 03360246120618.+=xxm dd at near punch-through.

38、 We have x VV eN d biDS a = +L N M O Q P 2 1 2 af / or VV x eN biDS da += 2 2 = () 0618 1016 1010 2 117 885 10 4 2 1916 14 . . xx x bgbgb g bg which yields VVV biDS += 2.95 From Example 12.4, we have VV bi = 0874., so that VV DS = 2.08 which is the near punch-through voltage. The ideal punch-through

39、 voltage was VV DS = 4.9 12.31 V x x V bi =() L N M M O Q P P 00259 103 10 15 10 0902 1916 10 2 .ln . . b gbg bg The zero-biased source-substrate junction width: x V eN dO bi a = L N M O Q P 2 1 2/ = ()()L N M O Q P 2 117 885 100902 16 103 10 14 1916 1 2 . . / x xx bg bgbg or xm dO = 0197. The Debye

40、 length is Semiconductor Physics and Devices: Basic Principles, 3rd edition Chapter 12 Solutions Manual Problem Solutions 193 L kT e eN D a = L N M O Q P af 1 2/ = ()()L N M O Q P 117 885 1000259 16 103 10 14 1916 1 2 . . / x xx bg bgbg or Lxcm D = 2.36 10 6 so that 66 2.36 100142 6 Lxm D = bg. Now

41、xLxL dODd +=6 We have for VV DS = 5, x VV eN d biDS a = +L N M O Q P 2 1 2 af / = +()()L N M O Q P 2 117 885 1009025 16 103 10 14 1916 1 2 . . / x xx bg bgbg or xm d = 0505. Then L =+019701420505. or Lm= 0844. 12.32 With a source-to-substrate voltage of 2 volts, x VV eN dO biSB a = +L N M O Q P 2 1

42、2 af / = +()()L N M O Q P 2 117 885 1009022 16 103 10 14 1916 1 2 . . / x xx bg bgbg or xm dO = 0354. We have 60142Lm D = . from the previous problem. Now x VVV eN d biDSSB a = +L N M O Q P 2 1 2 af / = +()()L N M O Q P 2 117 885 10090252 16 103 10 14 1916 1 2 . . / x xx bg bgbg or xm d = 0584. Then

43、 LxLx dODd =+6 =+035401420584. or Lm= 108. 12.33 (a) fp x x V=() F H G I K J 00259 2 10 15 10 0306 15 10 .ln . . and ms g fp E e = += + F H G I K J F H I K 2 112 2 0306 . . or msV= 0866. Also x eN dT fp a = L N M O Q P 4 1 2 / = ()()L N M O Q P 4 117 885 100306 16 102 10 14 1915 1 2 . . / x xx bg bg

44、bg or xm dT = 0629. Now =()QeN x SDadT max = 16 102 100629 10 19154 .xxxbgbgbg or =() QxC cm SD max/2.01 10 82 We have = = QxxxC cm SS 2 1016 1032 10 111982 bgbg./ Then VQQ t TSDSS ox ox msfp = +() F H G I K J maxaf2 = () 2.01 1032 10650 10 39 885 10 888 14 xxx x . . bgbg bg +()08662 0306. which yie

45、lds VV T = 0478. (b) We need a shift in threshold voltage in the positive direction, which means we must add acceptor atoms. We need VV T = + =()0800478128. Then Semiconductor Physics and Devices: Basic Principles, 3rd edition Chapter 12 Solutions Manual Problem Solutions 194 D VC e x xx I Tox = ()(

46、) a fbg bgbg 128 39 885 10 16 10650 10 14 198 . . or Dxcm I = 4.25 10 112 12.34 (a) fn x V=() F H G I K J 00259 10 15 10 0347 16 10 .ln . . and msms g fn E e = + F H G I K J 2 =+()323250560347. or msV= 0263. Also x eN dT fn d = L N M O Q P 4 1 2 / = ()()L N M O Q P 4 117 885 100347 16 1010 14 1916 1

47、 2 . . / x x bg bgb g or xm dT = 030. Now =()QeN x SDddT max = 16 1010030 10 19164 .xxbgb gbg or =() QxC cm SD max/4.8 10 82 We have = = QxxxC cm SS 5 1016 108 10 111982 bgbg./ Now VQQ t TSDSS ox ox msfn = + +() F H G I K J maxaf2 = + () 4.8 108 10750 10 39 885 10 888 14 xxx x bgbg bg. ()02632 0347.

48、 which becomes VV T = 374. (b) We want VV T = 050. Need to shift VT in the positive direction which means we need to add acceptor atoms. So VV T = =()050374324. Now D VC e x xx I Tox = ()() a fbg bgbg 32439 885 10 16 10750 10 14 198 . . or Dxcm I = 9.32 10 112 12.35 (a) fp x V=() F H G I K J 00259 10 15 10 0288 15 10 .ln . . and x eN dT fp a = L N M O Q P 4 1 2 / = 4 117 885 100288 16 1010 14 1915 1 2 . . / ()()L N M O Q P x x bg bgb g or xm dT = 0863. Now