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State the reason, why the photodiode is always operated under reverse bias. Write the working principle of operation of a photodiode. The semiconducting material used to fabricate a photodiode has an energy gap of 1.2eV. Using calculations, show whether it can detect light of wavelength of 400 nm incident on it.

 

 
 
 
 
 

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photodiode converts incident light to electric current more effectively at reverse bias.
when the photodiode is illuminated with the photon of energy greater than the energy of semiconductor, electron-hole pairs are generated due to absorption if photon. e-h pair generation takes place near the depletion region. Due to the electric field of the junction electrons and holes are separated before recombination.
The electrons are collected on n side and holes on giving rise to an emf. Now when an external load connected, current flows.
given energy gap = 1.2 e V
E= \frac{hc}{\lambda }= \frac{6\cdot 63\times 10^{-34}\times 3\times 10^{8}}{400\times 10^{-19}}
                 = 4\cdot 9725\times 10^{-19}
                 = \frac{4\cdot 9725eV}{1\cdot 6}= 2\cdot 94eV
 E> E_{g}
Therefore 400 nm can be detected 

Posted by

Safeer PP

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