5 m l p 124 8 10 8 996 m 15 1010 2 np0

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Unformatted text preview: - 4 cm2. Calculate (a) the ideal reverse- saturation cross- sectional area is 5 x 10 current due to holes, (b) the ideal reverse- saturation current dut to electrons, (c) The hole concentration a x = xn for Va = 0,8Vbi, (d) the electron current at x = xn for Va = 0.8Vbi, and (e) the electron current at x = xn + Lp/2 for Va = 0.8 Vbi. First, we use Table 5.2 to give Dn = 35 cm2/s and Dp = 12.4 cm2/s. Then, we compute a number of items that will be necessary: Ln = 35 ⋅ 2 × 10 −7 = 26.5 µ m L p = 12.4 ⋅ 8 × 10 −8 = 9.96 µ m (1.5 × 1010 )2 np0 = = 4.5 × 10 3 cm −3 16 5 × 10 (1.5 × 1010 )2 pn 0 = = 1.5 × 10 4 cm −3 1.5 × 1016 ⎛ 5 × 1016 ⋅ 1.5 × 1016 ⎞ Vbi = 0.0259 ln ⎜ = 0.747V ⎝ (1.5 × 1010 )2 ⎟ ⎠ 0.8Vbi = 0.598V (a) (b) (d) eD p pn 0 1.6 × 10 −19 ⋅ 12.4 ⋅ 1.5 × 10 4 A=− 5 × 10 −4 = 15 fA...
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This document was uploaded on 02/19/2014.

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