As per Question,
for electron' mobility μe=0.13
for holes
mobility μh=0.05
Concentration of Phosphorus atom i.e.
Donor concentration ND=n=1.5×1010/m3
Concentration of Aluminum atom i.e.
Acceptor impurities concentration NA=p=2.5×1010/m3
Conductivity
σ=q(nμe+pμp)
=1.6×10−19(1.5×1010×0.13+2.5×1010×0.05)
=1.6×10−9(0.195+0.125)
=1.6×10−9×0.320
=0.512×10−9
=5.12×10−8 Ω−1
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