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Exam Questions Papers - Engineering

Q1:

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

ANS:A - 7.6237 x 10-4 A/cm2

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.