HW4Solution - EE105 HOMEWORK2 SOLUTION Spring 2006 1...

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EE105 HOMEWORK2 SOLUTION Spring 2006 1. SOLUTIONS: a) ± n = V th ln N d n i = 0.026 ² ln 3.3 ² 10 17 1 ² 10 10 = 0.45 V ± p + = ³ 0.55 V ´ V FB = ³ ± p + ³ ± n ( ) = 1 V b) C ox = ± ox T ox = 3.9 ² 8.85 ² 10 ³ 14 34.5 ² 10 ³ 7 = 1 ² 10 ³ 7 F / cm 2 V Tn = V FB ³ 2 ´ n ³ 2 q ± si N d 2 ´ n ( ) C ox = 1 ³ 2 ² 0.45 ³ 2 ² 1.6 ² 10 ³ 19 ² 12 ² 8.85 ² 10 ³ 14 ² 3.3 ² 10 17 ² 2 ² 0.45 1 ² 10 ³ 7 = ³ 3.08 V c) V GB =2V: Accumulation C = C ox = 1 ± 10 ² 7 F / cm 2 V GB =0V: Depletion ± qN d X d = C ox V GB ± V FB + qN d 2 ² si X d 2 ³ ´ µ · ¸ ¹ ± X d = C ox V GB ± V FB ( ) qN d + C ox 2 ² si X d 2 ¹ 1 C b = X d ² si = ± 1 + 1 ± 2 C ox 2 V GB ± V FB ( ) q ² si N d C ox ¹ 1 C dep = 1 C b + 1 C ox = 1 C ox 2 ± 2 V GB ± V FB ( ) q ² si N d = 1 1 º 10 ± 7 ( ) 2 ± 2 0 ± 1 ( ) 1.6 º 10 ± 19 º 12 º 8.85 º 10 ± 14 º 3.3 º 10 17
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= 11.65 ± 10 6 cm 2 / F ² C dep = 8.58 ± 10 ³ 8 F / cm 2 V GB =-2V: Depletion 1 C dep = 1 C ox 2 ± 2 V GB ± V FB ( ) q ² si N d = 1 1 ³ 10 ± 7 ( ) 2 ± 2 ± 2 ± 1 ( ) 1.6 ³ 10 ± 19 ³ 12 ³ 8.85 ³ 10 ± 14 ³ 3.3 ³ 10 17 = 14.4 ³ 10 6 cm 2 / F ´ C dep = 6.95 ³ 10 ± 8 F / cm 2 V GB =-4V: Inversion C = C ox = 1 ± 10 ² 7 F / cm 2 2. SOLUTIONS: a) ± p = V th ln N a n i = 0.026 ² ln 7 ² 10 15 1 ² 10 10 = 0.35 V ± n + = 0.55 V ´ V FB = ³ ± n + ³ ± p ( ) = ³ 0.9 V b) C ox = ± ox T ox = 3.9 ² 8.85 ² 10 ³ 14 34.5 ² 10 ³ 7 = 1 ² 10 ³ 7 F / cm 2 V Tn = V FB ³ 2 ´ p + 2 q ± si N a ³ 2 ´ p ( ) C ox = ³ 0.9 + 2 ² 0.35 + 2 ² 1.6 ² 10 ³ 19 ² 12 ² 8.85 ² 10 ³ 14 ² 7 ² 10 15 ² 2 ² 0.35 1 ² 10 ³ 7 = 0.21 V c) V GB =1V: Inversion C = C ox = 1 ± 10 ² 7 F / cm 2 V GB =0V: Depletion
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qN a X d = C ox V GB ± V FB ± qN d 2 ² si X d 2 ³ ´ µ · ¸ ¹ X d = C ox V GB ± V FB ( ) qN a ± C ox 2 ² si X d 2 ¹ 1 C b = X d ² si
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  • Spring '06
  • MingWu

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