Q: A semiconductor has an electron concentration of 8 × 1013 per cm3 an a hole concentration of 5 × 1012 per cm3. The electron mobility is 25,000 cm2 V-1 sec-1 and the hole mobility is 100 cm2 V-1 sec-1
(i) the semiconductor is n-type
(ii) the semiconductor is p-type
(iii) the conductivity is 320 mho cm-1
(iv) the conductivity is 80 mho cm-1
(a) (i), (iii)
(b) (ii), (iii)
(c) (i), (iv)
(d) (ii), (iv)
Sol: (a) ne = 8 × 10^13/cm3 , nh = 5 × 1012/cm3
μe = 25000 cm2 V-1 sec-1, μg = 100 cm2 V-1
σ = ne μe e + nh μh e
= (8 × 1013 × 2500 + 5 × 1012 × 100) × 1.6 × 10-19
= 320 × 10-3 mho/cm