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Unformatted text preview: From the given data, we have that Ix
0.75 × 10 −3
Jx =
=
= 75 A / cm 2
−2
−3
Wd 10 ⋅ 10
V
15 Ex = x =
= 150V / cm
L 10 −1
V
5.8 × 10 − 3
Ey = H =
= 0.58V / cm
W
10 − 2 Then, the Hall constant is given by RH = Ey
J x Bz = 58V / m
5 2 7.5 × 10 A / m ⋅ 0.1 = +7.73 × 10 − 4 m 3 / coul (a) Since the Hall constant is positive, the conductivity is p type. (b) The majority carrier concentration is p= 1
RH e = 1
7.73...
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This document was uploaded on 01/28/2014.
 Winter '14

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