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Extrinsic Semiconductor - n-type Semiconductor

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Estimated time: 16 minutes
CBSE: Class 12
Maharashtra State Board: Class 11

Definition: Donor Impurity

Since every pentavalent dopant atom donates one electron for conduction, it is called a donor impurity.

CBSE: Class 12
Maharashtra State Board: Class 11

N-type Semiconductor

When a silicon or germanium crystal is doped with a pentavalent impurity such as phosphorus, arsenic, or antimony, an n-type semiconductor is obtained.

  • A dopant atom with 5 valence electrons occupies the position of a silicon atom in the crystal lattice.
  • Four electrons from the dopant form bonds with four neighbouring silicon atoms.
  • The fifth electron remains very weakly bound to its parent atom.
  • Very little energy is needed to free this electron even at room temperature.
  • The required energy is 0.01 eV for germanium and 0.05 eV for silicon.
CBSE: Class 12
Maharashtra State Board: Class 11

Origin of the Name N-Type

  • In this semiconductor, a large number of electrons are present in the conduction band. 
  • Its conductivity is due to negatively charged carriers. 
  • Therefore, it is called an n-type semiconductor.
CBSE: Class 12
Maharashtra State Board: Class 11

Charge Carriers

  • The n-type semiconductor also has a few electrons and holes produced due to thermally broken bonds.
  • The density of conduction electrons, ne, in a doped semiconductor is the total of electrons contributed by donors and thermally generated electrons from the host.
  • The density of holes, nhnh, is only due to the thermal breakdown of some covalent bonds of the host silicon atoms.
  • Some electrons and holes recombine continuously because they carry opposite charges.
  • The number of free electrons exceeds the number of holes.
  • Thus, electrons are the majority carriers, and holes are the minority carriers.
  • Relation: ne >> nh
CBSE: Class 12
Maharashtra State Board: Class 11

Energy Band

  • The free electrons donated by impurity atoms occupy energy levels in the band gap. 
  • These levels are close to the conduction band. 
  • Therefore, these electrons are easily available for conduction. 
CBSE: Class 12
Maharashtra State Board: Class 11

Conductivity of Extrinsic Semiconductors

  • Extrinsic semiconductors are better conductors than intrinsic semiconductors.
  • The conductivity of an extrinsic semiconductor can be controlled by controlling the amount of impurities added.
  • The amount of impurities is expressed as parts per million (ppm), that is, one impurity atom per one million atoms of the host.
CBSE: Class 12
Maharashtra State Board: Class 11

Conductivity of Extrinsic Semiconductors

  1. These are materials doped with pentavalent impurity (donor) atoms.
  2. Electrical conduction in these materials is due to electrons as the majority charge carriers.
  3. The donor atom loses electrons and becomes a positively charged ion.
  4. The number of free electrons is very large compared to the number of holes.
  5. When energy is supplied externally, negatively charged free electrons (majority charge carriers) and positively charged holes (minority charge carriers) are available for conduction.
CBSE: Class 12
Maharashtra State Board: Class 11

Effect of Doping on Electrical Conductivity

  • One cm3 specimen of a metal or semiconductor has of the order of 1022 atoms.
  • In a metal, every atom donates at least one free electron for conduction, so 1 cm3 of metal contains of the order of 1022 free electrons.
  • 1 cm³ of pure germanium at 20°C contains about 4.2 × 1022 atoms, but only 2.5 × 1013 free electrons and 2.5 × 1013 holes.
  • Addition of 0.001% of arsenic donates 1017 extra free electrons in the same volume.
  • The electrical conductivity increases by a factor of 10,000. 
Maharashtra State Board: Class 11

Example

Question: A pure silicon crystal has 4 × 1028 atoms m−3. It is doped with a 1 ppm concentration of antimony. Calculate the number of electrons and holes. Given ni = 1.2 × 1016 m−3

Given

  • Number of silicon atoms = 4 × 1028 m−3
  • Doping concentration = 1 ppm = \[\frac {1}{10^6}\]
  • ni = 1.2 × 1016 m−3

Calculation

Number of Sb atoms: \[\frac{4\times10^{28}}{10^6}=4\times10^{22}\]

As one pentavalent impurity atom donates one free electron:
ne = 4 × 1022 m−3

Number of holes: nh = \[\frac {n_i^2}{n_e}\]

\[n_h=\frac{(1.2\times10^{16})^2}{4\times10^{22}}=3.6\times10^9m^{-3}\]

Answer

  • Number of free electrons: 4 × 1022 m−3
  • Number of holes: 3.6 × 109 m−3
CBSE: Class 12
Maharashtra State Board: Class 11

Key Points: n-type Semiconductor

  • An n-type semiconductor is formed by doping silicon or germanium with a pentavalent impurity.
  • Pentavalent impurities act as donor impurities.
  • The fifth valence electron is weakly bound and can become free easily.
  • Electrons are majority carriers and holes are minority carriers.
  • For an n-type semiconductor, ne >> nh.
  • Donor energy levels lie close to the conduction band.
  • Extrinsic semiconductors are better conductors than intrinsic semiconductors.
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