Electronics and Communication Engineering - Exam Questions Papers - Discussion
Discussion Forum : Exam Questions Papers - Exam Paper 1 (Q.No. 7)
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
Answer: Option
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.
Discussion:
1 comments Page 1 of 1.
Vasimhasina said:
8 years ago
How to find the value of ni?
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