6 10 19 829 225 10 3 4 i sp a 10 328 10 16 a 7 lp 829

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Unformatted text preview: ln ⎜ = 0.330V ⎝ 1.5 × 1010 ⎟ ⎠ ⎝ ni ⎠ Then, (b) We calculate the two contributions to the current as I sn = eDn n p 0 1.6 × 10 −19 ⋅ 32.4 ⋅ 4.5 × 10 4 −4 A= 10 = 4.1 × 10 −15 A −6 Ln 32.4 ⋅ 10 Then we find that the reverse current is eD p pn 0 1.6 × 10 −19 ⋅ 8.29 ⋅ 2.25 × 10 3 −4 I sp = A= 10 = 3.28 × 10 −16 A −7 Lp 8.29 ⋅ 10 I s = −( I sn + I sp ) = 4.43 × 10 −15 A and the forward current is (c) ⎡ ⎛ 0.5 ⎞ ⎤ −6 I = − I s ⎢ exp ⎜ ⎟ − 1⎥ = 1.07 × 10 A . ⎝ 0.0259 ⎠ ⎦ ⎣ ratio = I sp 0.33 = = 0.07 . I sp + I sn 4.43 8.16 Consider an ideal silicon pn junction diode with the geometry shown in Fig. P8.16. The doping concentrations are Na = 5 x 1016 cm- 3 and Nd = 1.5 x 1016 cm- 3, and the minority carrier lifetimes are τ = 2.x 10- 7 s and τ = 8 x 10- 8 s. The n0 p0...
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This document was uploaded on 02/19/2014.

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