8 10 3 ey h 058v cm w 10 2 then the hall constant

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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.

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