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Semiconductor Electronics - Intrinsic and Extrinsic Semiconductors

Grade 12CBSEPhysics

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Energy Band Theory: Semiconductors have a forbidden energy gap EgE_g between the Valence Band (VB) and Conduction Band (CB). For semiconductors, Eg<3 eVE_g < 3\text{ eV}. For Silicon, Eg≈1.1 eVE_g \approx 1.1\text{ eV} and for Germanium, Eg≈0.7 eVE_g \approx 0.7\text{ eV}.

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Intrinsic Semiconductors: These are pure semiconductors (e.g., SiSi, GeGe) where the number of free electrons nen_e is equal to the number of holes nhn_h. Thus, ne=nh=nin_e = n_h = n_i, where nin_i is the intrinsic carrier concentration.

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Extrinsic Semiconductors: Formed by adding a small amount of impurity atoms (doping) to intrinsic semiconductors to increase conductivity.

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n-type Semiconductors: Created by doping with pentavalent impurities (Group 15 elements like P,As,SbP, As, Sb). Electrons are the majority charge carriers (ne≫nhn_e \gg n_h). The donor energy level EDE_D lies just below the conduction band.

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p-type Semiconductors: Created by doping with trivalent impurities (Group 13 elements like B,Al,InB, Al, In). Holes are the majority charge carriers (nh≫nen_h \gg n_e). The acceptor energy level EAE_A lies just above the valence band.

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Mass Action Law: In thermal equilibrium, the product of electron and hole concentrations is constant and independent of the amount of doping: nenh=ni2n_e n_h = n_i^2.

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Effect of Temperature: In semiconductors, as temperature TT increases, the number of charge carriers increases, causing the resistivity to decrease and conductivity to increase (negative temperature coefficient of resistance).

📐Formulae

ne=nh=nin_e = n_h = n_i

nenh=ni2n_e n_h = n_i^2

I=Ie+IhI = I_e + I_h

σ=e(neμe+nhμh)\sigma = e(n_e \mu_e + n_h \mu_h)

ρ=1σ=1e(neμe+nhμh)\rho = \frac{1}{\sigma} = \frac{1}{e(n_e \mu_e + n_h \mu_h)}

💡Examples

Problem 1:

A pure silicon crystal has 5×1028 atoms/m35 \times 10^{28} \text{ atoms/m}^3. It is doped by 1 ppm1 \text{ ppm} concentration of Pentavalent Arsenic. If ni=1.5×1016 m−3n_i = 1.5 \times 10^{16} \text{ m}^{-3}, calculate the number of electrons and holes.

Solution:

  1. Doping concentration ND=1 ppmN_D = 1 \text{ ppm} of 5×1028=1106×5×1028=5×1022 m−35 \times 10^{28} = \frac{1}{10^6} \times 5 \times 10^{28} = 5 \times 10^{22} \text{ m}^{-3}.
  2. Since Arsenic is pentavalent, ne≈ND=5×1022 m−3n_e \approx N_D = 5 \times 10^{22} \text{ m}^{-3}.
  3. Using Mass Action Law: nh=ni2ne=(1.5×1016)25×1022=2.25×10325×1022=4.5×109 m−3n_h = \frac{n_i^2}{n_e} = \frac{(1.5 \times 10^{16})^2}{5 \times 10^{22}} = \frac{2.25 \times 10^{32}}{5 \times 10^{22}} = 4.5 \times 10^9 \text{ m}^{-3}.

Explanation:

Because ne≫nhn_e \gg n_h, the resulting semiconductor is an n-type semiconductor.

Problem 2:

The conductivity of an intrinsic semiconductor is given as σi\sigma_i. If it is doped with an acceptor impurity of concentration NAN_A, find the new hole concentration assuming NA≫niN_A \gg n_i.

Solution:

In a p-type semiconductor doped with acceptor impurities, the hole concentration nhn_h is approximately equal to the acceptor concentration because each impurity atom 'accepts' an electron, creating a hole. Therefore, nh≈NAn_h \approx N_A.

Explanation:

In extrinsic semiconductors, the majority carrier concentration is determined almost entirely by the doping concentration at room temperature.

Intrinsic and Extrinsic Semiconductors Revision - Class 12 Physics CBSE