152 ev 296 1016 which is several times the

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Unformatted text preview: .8 × 10 6 )2 p= = = 1.62 × 10 −3 cm −3 n 2 × 1015 16 p = 10 − 7 × 1015 = 3 × 1015 cm −3 ni2 (1.8 × 10 6 )2 n= = = 1.08 × 10 −3 cm −3 p 3 × 1015 (c) It means that the minority carriers are very rare, having a population of only about 1000 per cubic meter. A p- type silicon material is to have a conductivity of σ = 1.8 mho/cm. If the mobility values are µn = 1250 cm2/Vs and µp = 380 cm2/Vs, what must be the acceptor impurity concentration in the material? First, we must find the number of holes in the material as ni2 eµn σ = neµn + peµ p = + peµ p p For this level, we find the Fermi energy to be 2 ⎛σ⎞ σ 2 µn 16 −3 p= ±⎜ ⎟ − ni µ = 2.96 × 10 cm 2 eµ p ⎝ 2 eµ p...
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