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Semiconductor Electronics - Energy Bands in Solids

Grade 12CBSEPhysics

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

🔑Concepts

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Energy Bands: In a crystal lattice, the interaction between atoms causes the discrete energy levels of isolated atoms to split into closely spaced levels called energy bands.

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Valence Band (VB): The range of energy levels occupied by valence electrons. At 0 K0 \text{ K}, this band is completely filled with electrons.

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Conduction Band (CB): The energy band above the valence band. Electrons in this band are free to move and contribute to electrical conduction.

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Energy Band Gap (EgE_g): The forbidden energy region between the top of the Valence Band (EVE_V) and the bottom of the Conduction Band (ECE_C). No electron can exist in this gap.

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Metals/Conductors: Materials where the VB and CB overlap or the conduction band is partially filled, meaning Eg≈0E_g \approx 0.

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Insulators: Materials with a very large energy gap (Eg>3 eVE_g > 3 \text{ eV}), making it impossible for electrons to jump from VB to CB under normal conditions.

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Semiconductors: Materials with a small energy gap (Eg<3 eVE_g < 3 \text{ eV}). For Silicon (SiSi), Eg≈1.1 eVE_g \approx 1.1 \text{ eV} and for Germanium (GeGe), Eg≈0.72 eVE_g \approx 0.72 \text{ eV}.

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Intrinsic Semiconductors: Pure semiconductors where the number of electrons in the conduction band (nen_e) is equal to the number of holes in the valence band (nhn_h), i.e., ne=nh=nin_e = n_h = n_i.

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Extrinsic Semiconductors: Semiconductors doped with impurities to increase conductivity. In n-type, ne≫nhn_e \gg n_h; in p-type, nh≫nen_h \gg n_e.

📐Formulae

Eg=EC−EVE_g = E_C - E_V

nenh=ni2n_e n_h = n_i^2

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)}

I=Ie+IhI = I_e + I_h

💡Examples

Problem 1:

A semiconductor has an electron concentration of 0.45×1012 m−30.45 \times 10^{12} \text{ m}^{-3} and a hole concentration of 5×1020 m−35 \times 10^{20} \text{ m}^{-3}. Calculate its intrinsic carrier concentration nin_i.

Solution:

Given ne=0.45×1012 m−3n_e = 0.45 \times 10^{12} \text{ m}^{-3} and nh=5×1020 m−3n_h = 5 \times 10^{20} \text{ m}^{-3}. Using the Mass Action Law: ni2=nenhn_i^2 = n_e n_h ni=ne×nhn_i = \sqrt{n_e \times n_h} ni=(0.45×1012)×(5×1020)n_i = \sqrt{(0.45 \times 10^{12}) \times (5 \times 10^{20})} ni=2.25×1032=1.5×1016 m−3n_i = \sqrt{2.25 \times 10^{32}} = 1.5 \times 10^{16} \text{ m}^{-3}

Explanation:

The Mass Action Law states that for a semiconductor in thermal equilibrium, the product of the concentration of free electrons and holes is constant and equal to the square of the intrinsic carrier concentration.

Problem 2:

The energy gap of a Silicon semiconductor is 1.1 eV1.1 \text{ eV}. What is the maximum wavelength of a photon that can excite an electron from the valence band to the conduction band?

Solution:

The energy required to excite an electron is E=EgE = E_g. Given Eg=1.1 eV=1.1×1.6×10−19 JE_g = 1.1 \text{ eV} = 1.1 \times 1.6 \times 10^{-19} \text{ J}. Using the formula: Eg=hcλE_g = \frac{hc}{\lambda} λ=hcEg\lambda = \frac{hc}{E_g} λ=6.63×10−34×3×1081.1×1.6×10−19\lambda = \frac{6.63 \times 10^{-34} \times 3 \times 10^8}{1.1 \times 1.6 \times 10^{-19}} λ≈1.13×10−6 m=1130 nm\lambda \approx 1.13 \times 10^{-6} \text{ m} = 1130 \text{ nm}

Explanation:

To transition an electron across the forbidden gap, the incident photon must have energy at least equal to the band gap EgE_g.

Energy Bands in Solids - Revision Notes & Key Formulas | CBSE Class 12 Physics