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The concentration of minority carriers in an extrinsic semiconductor under equilibrium is
  • a)
    directly proportional to the doping concentration.
  • b)
    inversely proportional to the doping concentration.
  • c)
    directly proportional to the intrinsic concentration.
  • d)
    inversely proportional to the intrinsic concentration.
Correct answer is option 'B'. Can you explain this answer?
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Explanation:

Extrinsic semiconductors are those in which impurities have been added to the intrinsic semiconductor to increase its electrical conductivity. The impurities added are either trivalent (having three valence electrons) or pentavalent (having five valence electrons) and are called dopants.

Under equilibrium conditions, the concentration of minority carriers in an extrinsic semiconductor is determined by the doping concentration and the intrinsic concentration.

Intrinsic Concentration:

The intrinsic concentration (ni) is the concentration of electrons and holes in an intrinsic semiconductor at a given temperature. It depends on the bandgap and the temperature of the semiconductor. The intrinsic concentration is given by:

ni = √(Nc x Nv) x e^(-Eg/2kT)

where Nc and Nv are the effective densities of states in the conduction and valence bands respectively, Eg is the bandgap, k is the Boltzmann constant, and T is the temperature in Kelvin.

Doping Concentration:

The doping concentration (Nd or Na) is the concentration of dopant atoms in an extrinsic semiconductor. The doping concentration determines the concentration of majority carriers in the semiconductor. Majority carriers are the carriers that are present in large numbers and are responsible for the electrical conductivity of the semiconductor.

Minority Carrier Concentration:

Minority carriers are the carriers that are present in small numbers in the semiconductor. The concentration of minority carriers in an extrinsic semiconductor is determined by the doping concentration and the intrinsic concentration. The concentration of minority carriers is given by:

p = ni^2 / Na (for p-type semiconductor)
n = ni^2 / Nd (for n-type semiconductor)

where p is the concentration of holes (minority carriers) in a p-type semiconductor, n is the concentration of electrons (minority carriers) in an n-type semiconductor, Na is the doping concentration of acceptor atoms in a p-type semiconductor, and Nd is the doping concentration of donor atoms in an n-type semiconductor.

In an extrinsic semiconductor, the concentration of minority carriers is inversely proportional to the doping concentration. This is because as the doping concentration increases, the concentration of majority carriers increases and the concentration of minority carriers decreases. Hence, the correct answer is option B.
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The concentration of minority carriers in an extrinsic semiconductor under equilibrium isa)directly proportional to the doping concentration.b)inversely proportional to the doping concentration.c)directly proportional to the intrinsic concentration.d)inversely proportional to the intrinsic concentration.Correct answer is option 'B'. Can you explain this answer?
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