Question

Suppose a pure Si crystal has 5 × 10²⁸ atoms per m³. It is doped with 5 × 10²² atoms per m³ of Arsenic. Calculate majority and minority carrier concentration in the doped silicon.

(Given: n_i = 1.5 × 10¹⁶ m⁻³)

Answer 1

Arsenic is a pentavalent impurity that produces an n-type semiconductor.

Majority carriers (electrons): n ≈ N_D

Donor concentration (N_D) = 5 × 10²² m⁻³

Since N_D ≫ n_i:

n ≈ N_D = 5 × 10²² m⁻³

Mass action law:

n_p = `n_i^2`

p = `n_i^2/n`

= `((1.5 xx 10^16)^2)/(5 xx 10^22)`

= `(2.25 xx 10^32)/(5 xx 10^22)`

= 0.45 × 10¹⁰

= 4.5 × 10⁹ m⁻³

∴ The majority carrier concentration electrons (n) is 5 × 10²² m⁻³, and the minority carrier concentration holes (p) is 4.5 × 10⁹ m⁻³.