A sample of semiconductor has a cross sectional area of 1cm^2 and a thickness of 0.1cm.
(a) Determine the number of electron hole pairs that are generated per unit volume per unit time by the uniform absorpotion of 1 Watt of light at a wavelength of 6300 A. Assume each photon creates one electron hole pair.
(b) If the excess minority carrier lifetime is 10μs, what is the steady state excess carrier concentration?

a. 3.17*10¹⁹ b. 3.17*10¹⁴

Explanation:

a. Area A = 1cm²

Thickness h = 0.1cm

Energy of photon E = hf=hc/λ

Where f = frequency; c=speed of light; λ=wavelength = 6300Α=6300*10⁻¹⁰;

Planck's constant h = 6.626*10⁻³⁴ joule-seconds; speed of light c = 3*10⁸

Therefore E = (6.626*10⁻³⁴)*3*10⁸/6300*10⁻¹⁰=3.155*10¹⁹J

1Watt of light releases 3.17*10¹⁸ photons per second.

Volume of sample = Area * Thickness = 1*0.1=0.1cm³

Therefore, number of electron hole pairs that are generated per unit volume per unit time = 3.17*10¹⁸/0.1 = 3.17*10¹⁹photon/cm³-s

b. Steady state excess carrier concentration = 3.17*10¹⁹*10μs

=3.17*10¹⁹*10*10⁻⁶

=3.17*10¹⁴/cm³

diy

explanation: