Electronics and Communication Engineering - Electronic Devices and Circuits - Discussion

13. 

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

Answer: Option B

Explanation:

No answer description available for this question.

Chestah said: (Oct 10, 2015)  
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.

Ankit said: (Nov 22, 2015)  
You are right answer is D.

Please explain if anybody have answer B?

Abc said: (Mar 31, 2016)  
What is the right answer?

Vijay said: (May 19, 2016)  
Right answer is option D.

Manoj said: (Oct 14, 2017)  
Answer is D.

Tup-Taguig said: (Mar 1, 2018)  
Yeah, it's D.

Komal said: (May 16, 2018)  
0.0104 is the answer.

Lekhraj Rana said: (Jun 2, 2018)  
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.

Mohan Mayank Chauhan said: (Apr 7, 2021)  
Option D is the correct answer.

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