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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 () F
5、 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 Solutions M
6、anual 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 10 16 10
7、4 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 Vsat L = (
8、) 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 DS = = ()
9、 () ()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 Physics and
10、 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 the previous pro
11、blem, 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 = 1 138 . 12.9 (a
12、) 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 ox ox ox = = = ()
13、 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 2 We find fpt a i
14、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 edition 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 8
