Electronics and Communication Engineering - Electronic Devices and Circuits - Discussion
Discussion Forum : Electronic Devices and Circuits - Section 5 (Q.No. 13)
13.
The carrier mobility in a semiconductor is 0.4 m2/Vs. Its diffusion constant at 300k will be (in m2/s).
Discussion:
10 comments Page 1 of 1.
Mohamed Abdel Fatah said:
4 years ago
The right answer is D.
Mohan mayank Chauhan said:
5 years ago
Option D is the correct answer.
LEKHRAJ RANA said:
7 years ago
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.
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.
(2)
Komal said:
7 years ago
0.0104 is the answer.
TUP-Taguig said:
8 years ago
Yeah, it's D.
Manoj said:
8 years ago
Answer is D.
Vijay said:
9 years ago
Right answer is option D.
Abc said:
10 years ago
What is the right answer?
Ankit said:
10 years ago
You are right answer is D.
Please explain if anybody have answer B?
Please explain if anybody have answer B?
(1)
Chestah said:
1 decade ago
Isn't the answer letter D?
D = ukT/q, where kT/q is 26 mV.
26 mV *0.4 m^2/Vs = 0.01 m^2/s.
D = ukT/q, where kT/q is 26 mV.
26 mV *0.4 m^2/Vs = 0.01 m^2/s.
(3)
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