Electronics and Communication Engineering - Exam Questions Papers - Discussion

7. 

In a semiconductor material. The hole concentration is found to be 2 x 2.5 x 1015 cm-3. If mobility of carriers is 0.13 m2/ v-s. Then find the current density if electric field intensity is 3.62 x 10-19

[A]. 7.6237 x 10-4 A/cm2
[B]. 7.6237 x 10-5 A/cm2
[C]. 7.6237 x 10-3 A/cm2
[D]. none of these

Answer: Option A

Explanation:

Current density J = σE

Where σ = conductivity

Given -> μ = 0.13 m2/v-s = 0.13 x 104 cm2/V sec

P = 2.25 x 1015/cm3

We have, ni = 1.5 x 1010

Also n.p. =

n = /p

= (1.6 x 10-19 x 0.13 x 104 x 2.25 x 1015) x

= (0.468) (4.5 x 1015)

σ = 2.106 x 1015 μ/cm

J = σE

Current density = 2.106 x 1015 x 3.620 x 10-19

= 7.6237 x 10-4 A/m2.


Vasimhasina said: (Jan 25, 2018)  
How to find the value of ni?

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