Question

A p-n photodiode is fabricated from a semiconductor with band gap of 2.8 eV. Can it detect a wavelength of 6000 nm?

Answer 1

Energy band gap of the given photodiode, E_g = 2.8 eV

Wavelength, λ = 6000 nm = 6000 × 10⁻⁹ m

The energy of a signal is given by the relation:

E =

Where,

h = Planck’s constant

= 6.626 × 10⁻³⁴ Js

c = Speed of light

= 3 × 10⁸ m/s

E

= 3.313 × 10⁻²⁰ J

But 1.6 × 10⁻¹⁹ J = 1 eV

∴E = 3.313 × 10⁻²⁰ J

The energy of a signal of wavelength 6000 nm is 0.207 eV, which is less than 2.8 eV − the energy band gap of a photodiode. Hence, the photodiode cannot detect the signal.