Question

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

Answer 1

It is given that a p-n photodiode is fabricated from a semiconductor with band gap of

E g =2⋅8 eV and wavelength λ=6000 nm.

The formula of energy is,

E= hc λ

Substitute the values in above expression.

E= 6⋅626× 10 −34 ×3× 10 8 6000× 10 −9 =3⋅313× 10 −20  J = 3⋅313× 10 −20 1⋅6× 10 −19 E=0⋅207 eV

Hence, the energy of the signal of wavelength 6000 nm is 0⋅207 eV and it is less than the given energy band gap that is 2⋅8 eV. So, the photodiode cannot detect the signal.