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Electronic Devices and Circuits - Engineering

Q1:

The carrier mobility in a semiconductor is 0.4 m2/Vs. Its diffusion constant at 300k will be (in m2/s).

A 0.43

B 0.16

C 0.04

D 0.01

ANS:B - 0.16

Yes,i.e .0104 answer.

Dn/Ue(mobelity)= t/11600,
Where as t =300 then 300/11600=26*10^-3,
and Ue is .4,
Then Dn(diffusion constant)=.4*26*10^-3=.0104.