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> Learn Additional Exercise 14 > A p-n photodiode is fabricated from a semiconductor...
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A p-n photodiode is fabricated from a semiconductor with band gap of 2.8 eV. Can it detect a wavelength of 6000 nm?

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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 = Where, h = Planck’s constant = 6.626 × 10−34 Js c = Speed of light = 3 × 108 m/s E = 3.313 × 10−20...
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Energy band gap of the given photodiode,Eg= 2.8 eV Wavelength, λ = 6000 nm = 6000 × 10−9m The energy of a signal is given by the relation: E= Where, h= Planck’s constant = 6.626 × 10−34Js c = Speed of light = 3 × 108m/s E = 3.313 × 10−20J But 1.6 × 10−19J = 1 eV ∴E= 3.313 × 10−20J 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. read less
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