This preview shows page 1. Sign up to view the full content.
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...
View
Full
Document
This document was uploaded on 02/19/2014.
 Fall '14

Click to edit the document details