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semisolex03 UCSD ECE 103
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  • Title: semisolex03
  • Type: Notes
  • School: UCSD
  • Course: ECE 103
  • Term: Fall

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Physics Semiconductor and Devices: Basic Principles, 3rd edition Solutions Manual Chapter 3 Exercise Solutions Chapter 3 Exercise Solutions E3.1 1 = 10 sin a + cos a gT = 4 2(1.08) 9.11x10 34 a By trial and error, a = 5.305 rad Now 2 mE 2 so E2 = or E 2 = 6.86 x10 Also 2 mE1 so E1 = or E1 = 2.41x10 Then . E = E 2 E1 = 4.29 150 or E = 2.79 eV 19 19 g F 2I b6.625x10 g H 3K (0.0259)b1.6 x10 g 31 3/ 2 3 19 25 3 19 b 3/ 2 which yields g T = 2.12 x10 m = 2.12 x10 cm E3.3 We have 3 h 2 a = 5.305 (5.305) h 2 2 2 ma 2 = (5.305) 1.054 x10 2 2 9.11x10 b b 34 31 gb5x10 10 g g 2 gT = 4 2 m p * bg h 3 * 3 3/ 2 Ev 2 E v kT za Ev E f 1/ 2 dE J = 4.29 eV h 2 a = ( )2 (h )2 2 ma 2 = 2 9.11x10 b ( )2 1.054 x10 34 31 b gb5x10 g 10 g 2 2 J = 150 eV . b g F 2 I a E E f g= H 3K h or 4 b2 m g F 2 I g= H 3 K 0 (kT ) h 4 b 2 m g F 2 I = H 3 K (kT ) h Then 4 2(0.56)b9.11x10 g F 2 I g= b6.625x10 g H 3K (0.0259)b1.6 x10 g 4 2 m p 3/ 2 3/ 2 which yields T Ev v E v kT * 3/ 2 p 3/ 2 T 3 * 3/ 2 p 3/ 2 3 31 3/ 2 T 34 3 19 3/ 2 or g T = 7.92 x10 m = 7.92 x10 cm 24 18 3 3 E3.2 gc ( E ) Then = gT 4 2 mn h * 3 bg * 3/ 2 3/ 2 E Ec E3.4 (a) fF = 1 + exp or Ec + kT Ec b g z a E E f dE = h 4 b 2 m g F 2 I = H 3K aE E f h 4 2 mn 3 E c + kT 1/ 2 C Ec * 3/ 2 n 3/ 2 3 c FE E I H kT K F 1 = 1 + exp FE E I H kT K c F 6 1 fF = 1 + exp or gT = FI H 0.0259 K 0.30 1 f F = 9.32 x10 4 2 mn 3 b g F 2 I (kT ) H 3K h * 3/ 2 3/ 2 which yields 21 Semiconductor Physics and Devices: Basic Principles, 3rd edition Solutions Manual (b) fF = 1 + exp Chapter 3 Exercise Solutions F 0.30 + 0.0259 I H 0.0259 K 6 1 E3.6 kT = ( 0.0259 ) (a) fF = 1 + exp F 400 I = 0.03453 H 300 K 0.30 f F = 1.69 x10 4 f F = 3.43 x10 E3.5 (a) 1 fF = 1 1 + exp exp F FI H 0.03453K 1 1 FE E I H kT K = 1 = F E E I 1 + expF E E I 1 + exp H kT K H kT K F F FE E I H kT K F 1 (b) fF = 1 + exp F 0.30 + 0.03453I H 0.03453 K 5 f F = 6.20 x10 E3.7 Then 1 fF = so 1 f F = 1.35 x10 (b) 1 fF = or 1 f F = 4.98 x10 7 6 kT = 0.03453 eV F 0.35 I 1 + exp H 0.0259 K 1 1 (a) 1 fF = F 0.35 I 1 + exp H 0.03453K 5 1 1 f F = 3.96 x10 (b) 1 fF = F 0.35 + 0.0259 I 1 + exp H 0.0259 K F 0.35 + 0.03453I 1 + exp H 0.03453 K 5 1 1 f F = 1.46 x10 22

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semisolex04
Path: UCSD >> ECE >> 103 Fall, 2008

Description: Semiconductor Physics and Devices: Basic Principles, 3rd edition Solutions Manual Chapter 4 Exercise Solutions Chapter 4 Exercise Solutions E4.1 no = 2.8 x10 exp 19 Now ni = 2.8 x10 2 F 0.22 I H 0.0259 K 3 b 19 gb 1.04 x10 19 112 . gFH 400 ...
semisolex05
Path: UCSD >> ECE >> 103 Fall, 2008
Description: Semiconductor Physics and Devices: Basic Principles, 3rd edition Solutions Manual Chapter 5 Exercise Solutions Chapter 5 Exercise Solutions E5.1 no = 10 10 = 9 x10 cm 15 14 14 3 (b) = 1.6 x10 b e n N d N a 19 a g(1000)b3x10 g 16 1 f =...
semisolex06
Path: UCSD >> ECE >> 103 Fall, 2008
Description: Semiconductor Physics and Devices: Basic Principles, 3rd edition Solutions Manual Chapter 6 Exercise Solutions Chapter 6 Exercise Solutions E6.1 n(t ) = 5 x10 14 n(t ) = n(0) exp so n(t ) = 10 15 FG t IJ H K F t IJ expG H 1 s K no 15 3 14 3 1...
semisolex11
Path: UCSD >> ECE >> 103 Fall, 2008
Description: Semiconductor Physics and Devices: Basic Principles, 3rd edition Solutions Manual Chapter 11 Exercise Solutions Chapter 11 Exercise Solutions E11.1 E11.5 fp = 0.376 V 16 or F 3x10 IJ = 0.376 V (a) = (0.0259 ) lnG H 15x10 K . R 4(11.7)b8.85x10 g...
semisolex12
Path: UCSD >> ECE >> 103 Fall, 2008
Description: Semiconductor Physics and Devices: Basic Principles, 3rd edition Solutions Manual Chapter 12 Exercise Solutions Chapter 12 Exercise Solutions E12.1 I D1 I D2 or VGS 1 Now FG V IJ H V K = expFG V = F V IJ H expG HV K F I IJ V = V lnG HI K exp GS 1...
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Path: USC >> PSYCH >> 360 Fall, 2008
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