What is n_i, using semiconductor w/ electron and hole carrier mobility?

Question

If a semiconductor has 500(Ω-m)^-1, with electron carrier mobility of 1.6 x 10^-1 m²/V-s and hole carrier mobility of 7.5 x 10^-2 m²/V-s, what is n_i?

Answer 1

Hello Maisie;

The formula for intrinsic carrier concentration is fiven as

n_i² = (electron mobility * hole mobility * Boltzmann constant * temperature) / (electron charge * semiconductor intrinsic conductivity)

We have all these properties:

Electron carrier mobility (μ_e) = 1.6 x 10^-1 m²/V-s

Hole carrier mobility (μ_h) = 7.5 x 10^-2 m²/V-s

Boltzmann constant (k) ≈ 1.38 x 10^-23 J/K

Temperature (T) = room temperature ≈ 300 K

Semiconductor intrinsic conductivity (σ_i) = 500 (Ω-m)^-1

So plugging them in;

n_i² = 1.6 x 7.5 x 1.38 x 300 x 500 x 10^{(-1 - 2 -23)}

n_i² = 5.775 x 10^-21

n_i ≈ 7.6 x 10^-11 intrinsic carriers/m³