Unformatted text preview: p0
2 × 10 (d) Finally, we have ⎛ p0 ⎞
⎛ 2 × 1016 ⎞
EFi − EF = kBT ln ⎜ ⎟ = 0.0259 ln ⎜
= 0.365 eV ⎝ 1.5 × 1010 ⎟
⎠
⎝ n0 ⎠ 4.22 The Fermi energy level in silicon at T = 300 K is as close to the top of the valence band as to the midgap energy. (a) Is the material n type or p type? (b) Calculate the values of n0 and p0. (a) The material is p type, because the Fermi energy is closer to the valence band than to the conduction band. (b) ⎛E⎞
1.12 ⎞
⎛
p0 = N v exp ⎜ − G ⎟ = 1.04 × 1019 exp ⎜ −
= 2.1 × 1014 cm −3
⎝ 4 ⋅ 0.0259 ⎟
⎠
⎝ 4 kBT ⎠ . ni2 (1.5 × 1010 )2
6
−3
n0 =
=
= 1.07 × 10 cm
p0
2.1 ×...
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This document was uploaded on 02/19/2014.
 Fall '14

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