Draw energy band diagram of p & n type semiconductors. Also write two differences between p and n type semiconductors.

OR


Energy gap in a p – n photodiode is 2.8 eV. Can it detect a wavelength of 6000 nm? Justify your answer.



In P type semiconductor III group element is added as doping element. Impurity added creates vacancy of electrons (holes) called as Acceptor Atom.

In n type semiconductor V group element is added as doping element. Impurity added provides extra electrons and is known as Donor Atom.


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OR


Energy corresponding to wavelength 6000 nm



= 6.6 × 10-34 × 3 × 10⁸ / 6000 × 10-⁹ joule


= 3.3 × 10-2 J


= 3.3 × 10-20 / 1.6 × 10-19= 0.2 eV


The photon energy (E= 0.2eV) of given wavelength is much less than the band gap (Eg = 2.8eV). Hence, it cannot detect the given wave length.


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