Q. 114.6( 10 Votes )

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

Answer :

Given that the Energy band gap of photodiode, Eg = 2.8 eV


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


The energy of a signal is given by the relation:


…. (I)


Where,


h = Planck’s constant = 6.626 × 10–34J-s


c = Speed of light = 3 × 108m/s


So, E = 3.3 × 10–20 J [From equation (I)]


We know that, 1.6 × 10–19J = 1 eV


Therefore,


E = 0.21 eV


The energy of any particular signal of wavelength 6000 nm is 0.21 eV, which is less than given 2.8 eV − the energy band gap of the photodiode. So, the photodiode cannot detect the signal.


(Actually, this is example of application of photodiode)


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