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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?
Solution
Energy band gap of the given photodiode, Eg = 2.8 eV
Wavelength, λ = 6000 nm = 6000 × 10−9 m
The energy of a signal is given by the relation:
`"E" = ("hc")/lambda`
Where
h = Planck’s constant
= 6.626 × 10−34 Js
c = Speed of light
= 3 × 108 m/s
`"E" = (6.626 xx 10^(-34) xx 3 xx 10^8)/(6000 xx 10^(-9))`
= 3.313 × 10−20 J
But 1.6 × 10−19 J = 1 eV
∴ E = 3.313 × 10−20 J
`= (3.313 xx 10^(-20))/(1.6 xx 10^(-19)) = 0.207 " eV"`
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.
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