# 11 - Physics Semiconductor and Devices Basic Principles 3rd...

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Physics Semiconductor and Devices: Basic Principles, 3rd edition Solutions Manual Chapter 11 Problem Solutions Chapter 11 Problem Solutions 11.1 (a) p-type, inversion (b) p-type, depletion (c) p-type, accumulation (d) n-type, inversion 11.2 (a) For T = 300 K Silicon: Then x dT or Then QSD ( max ) = 3.76 x10 C / cm -8 2 L 4(16)b8.85x10 g(0.156) OP =M N b1.6x10 gb10 g Q -14 -19 16 1/ 2 x dT = 0.235 m p = Vt ln F 10 IJ = 0.347 V = (0.0259 ) lnG H 15x10 K . Now L 4 OP x =M N eN Q L 4(11.7)b8,85x10 g(0.347) OP =M N b1.6x10 gb10 g Q 16 10 1/ 2 p dT a -14 1/ 2 -19 16 FG N IJ Hn K a i (b) For T = 200 K , Vt = (0.0259 ) F 200 I = 0.01727 V H 300 K 4 - 3 -8 2 -3 Silicon: ni = 7.68 x10 cm We obtain p = 0.442 V and x dT = 0.388 m , QSD ( max ) = 5.4 x10 C / cm GaAs: ni = 1.38 cm We obtain p = 0.631 V and x dT = 0.428 m , QSD ( max ) = 6.85x10 C / cm -8 2 or Germanium: ni = 2.16 x10 cm 10 -3 x dT = 0.30 m Also QSD ( max) = eN a x dT We obtain p = 0.225 V and x dT = 0.282 m , QSD ( max) = 4.5 x10 C / cm -8 -19 2 = 1.6 x10 or b gb10 gb0.30x10 g 16 -4 -8 2 11.3 (a) We want QSD ( max) = 7.5x10 C / cm We have QSD ( max) = eN d xdT where -9 2 QSD ( max) = 4.8 x10 C / cm GaAs: p = ( 0.0259) ln and x dT or Then FG 10 IJ = 0.581 V H 18x10 K . . L 4(131)b8.85x10 g(0.581) OP =M N b1.6x10 gb10 g Q 16 6 -14 1/ 2 -19 16 x dT For n-type silicon, L 4 OP =M N eN Q fn d 1/ 2 and f n = Vt ln FG N IJ Hn K d i 1/ 2 1/ 2 fn d x dT = 0.410 m QSD ( max ) = 6.56 x10 C / cm -8 2 g(11.7)b8.85x10 g N or b7.5x10 g = b6.63x10 g N = 4 1.6 x10 -9 2 -19 -14 -31 d fn QSD ( max) = 7.5x10 -9 b = 4 eN d f n which yields Germanium: p = ( 0.0259) ln FG 10 IJ = 0.156 V H 2.4 x10 K 16 13 N d f n = 8.48 x10 and 13 167 Semiconductor Physics and Devices: Basic Principles, 3rd edition Solutions Manual Chapter 11 Problem Solutions g f n = (0.0259) ln By trial and error F N I H 15x10 K . d 10 ms = - or N d = 3.27 x10 cm 14 -3 (b) s = -2 f n where fn FG E - IJ -0.35 = -b0.56 - g H 2e K F N I = 0.21 = (0.0259 ) ln H 1.5x10 K fn fn d 10 which yields fn = (0.0259) ln Then FG 3.27 x10 IJ = 0.259 V H 15x10 K . 14 10 N d = 4.98 x10 cm 13 -3 (b) p polysilicon gate: + s = -0.518 V or 11.4 p-type silicon (a) Aluminum gate ms = FG E + IJ -0.35 = b0.56 + g H 2e K g fn fn + fn = -0.91 V This is impossible, cannot use a p polysilicon gate. (c) Aluminum gate: Eg - fn ms = - + m 2e or LM FG E + IJ OP N H 2e K Q We have F N IJ = (0.0259) lnFG 6x10 IJ = V lnG H 15x10 K Hn K . ms = - + m g fp 15 a fp t 10 i FG H IJ K Or f p = 0.334 V Then ms = 3.20 - (3.25 + 0.56 + 0.334) or ms = -0.944 V (b) n polysilicon gate: Eg + f p = -(0.56 + 0.334 ) ms = - 2e or ms = -0.894 V + -0.35 = 3.20 - 3.25 + 0.56 - fn which yields b g fn = 0.26 = ( 0.0259) ln or finally, 14 F N I H 15x10 K . d 10 -3 N d = 3.43 x10 cm 11.6 VFB = ms - (a) For t ox = 500 A Then Cox = + FG H IJ K QSS Cox and Cox = ox t ox (c) p polysilicon gate: + ms or F - IJ = (0.56 - 0.334) =G H 2e K Eg fp (3.9) 8.85x10 -14 500 x10 -8 b g = 6.9 x10 -8 F / cm 2 ms = +0.226 V 11.5 We want, for n-type

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## This note was uploaded on 11/26/2009 for the course ECC 101 taught by Professor Horton during the Spring '09 term at Punjab Engineering College.

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11 - Physics Semiconductor and Devices Basic Principles 3rd...

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