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.

Question

Philoid · Accepted 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.

OR

Energy corresponding to wavelength 6000 nm

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

= 3.3 × 10^-²⁰ 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.

Draw energy band diagram of p & n type semiconductors. Also write two differences between p and n type semiconductors.OREnergy gap in a p – n photodiode is 2.8 eV. Can it detect a wavelength of 6000 nm? Justify your answer.

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.