#electrical conductivity

Semiconductors, Types and Energy Bands

Which property distinguishes insulators from semiconductors?

**Semiconductor properties** distinguish from conductors and insulators by temperature dependence and doping response. At 0 K intrinsic acts as insulator, conductivity due to thermally generated electron-hole pairs, number of outer electrons 4 for Si/Ge forming covalent bonds, each atom shares electrons, crystal with N atoms has 4N valence electrons, 2N bonds. Insulators have a large energy gap ( E_g > 3 eV ), preventing electron excitation, while semiconductors have a smaller gap ( 0.2 eV to 3 eV ), allowing some conduction. Substituting values gives Energy gap, which matches expected behavio

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Semiconductors, Types and Energy Bands

The resistivity of a semiconductor typically lies in the range:

**Types of semiconductors** elemental Si, Ge group IV with 4 valence electrons, compound GaAs, InP etc. Intrinsic has n_e = n_h, extrinsic doped with pentavalent donors (P, As) gives n-type excess electrons, trivalent acceptors (B, Al) gives p-type excess holes, resistivity range semiconductors 10⁻⁵ to 10⁶ Ω·m vs insulators 10¹¹ Ω·m. Semiconductors have intermediate resistivity between metals and insulators, typically ranging from 10⁻⁵ to 10⁶ Ω m . Substituting values gives 10⁻⁵ to 10⁶ Ω m, which matches expected behaviour for this semiconductor device configuration, confirming doping, depleti

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Semiconductors, Types and Energy Bands

The conductivity of metals is high due to:

**Energy bands in semiconductors** consist of valence band filled at 0 K and conduction band empty, gap E_g small ~1 eV (Si 1.1 eV, Ge 0.7 eV), insulators large gap >3 eV (C diamond 5.4 eV), conductors overlapping. Intrinsic semiconductor at 0 K behaves as insulator because no thermal excitation, at T>0 K electrons jump to conduction band leaving holes, conductivity increases with temperature. Metals have very low resistivity ( 10⁻² to 10⁻⁸ Ω m ) or high conductivity ( 10² to 10⁸ S m⁻¹ ) because of a large number of free electrons available for conduction. Substituting values gives Large numbe

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Semiconductors, Types and Energy Bands

The electrical conductivity of a semiconductor is:

**Types of semiconductors** elemental Si, Ge group IV with 4 valence electrons, compound GaAs, InP etc. Intrinsic has n_e = n_h, extrinsic doped with pentavalent donors (P, As) gives n-type excess electrons, trivalent acceptors (B, Al) gives p-type excess holes, resistivity range semiconductors 10⁻⁵ to 10⁶ Ω·m vs insulators 10¹¹ Ω·m. Semiconductors have intermediate conductivity ( 10⁵ to 10⁻⁶ S m⁻¹ ), between metals ( 10² to 10⁸ S m⁻¹ ) and insulators ( 10⁻¹¹ to 10⁻¹⁹ S m⁻¹ ). Substituting values gives Intermediate to metals and insulators, which matches expected behaviour for this semiconduct

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Temperature Dependence of Resistance and Resistivity

A conductor has a resistivity of \( 7 \times 10^{-8} \, \Omega \text{m} \) and \( \alpha = 4 \times 10^{-3} \, ^\circ\te

**Temperature dependence** of resistance R_t = R₀[1+α(T-T₀)], α temperature coefficient (per °C), R₀ resistance at T₀ (Ω). For metals α positive ≈10⁻³ /°C, resistance increases with temperature because τ decreases due to increased phonon scattering, n nearly constant. Use: rho_t = rho₀ [1 + α (T - T₀)] . Substitute: rho_t = 7 × 10⁻⁸ [1 + 4 × 10⁻³ (85 - 25)] . Calculate: rho_t = 7 × 10⁻⁸ [1 + 0.24] = 7 × 10⁻⁸ × 1.24 = 8.68 × 10⁻⁸ Ω m . Applying I = n e A v_d, R = ρ l/A, R_t = R₀[1+αΔT], Kirchhoff's ΣI=0, ΣV=0, R_eq series/parallel,

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Resistance, Resistivity and Ohm's Law

A conductor has a resistivity of \( 1.2 \times 10^{-7} \, \Omega \text{m} \) and \( \alpha = 4 \times 10^{-3} \, ^\circ\

**Resistance** R = ρ l/A, ρ resistivity (Ω·m), l length (m), A area (m²), ρ = m/(n e² τ) from Drude model, τ average collision time. Ohm's law V = I R holds when ρ constant, independent of V. Volume constant stretching l→2l implies A→A/2, so R' = ρ·2l/(A/2)=4R, resistance quadruples when length doubles at constant volume. Use: rho_t = rho₀ [1 + α (T - T₀)] . Substitute: rho_t = 1.2 × 10⁻⁷ [1 + 4 × 10⁻³ (80 - 20)] . Calculate: rho_t = 1.2 × 10⁻⁷ [1 + 0.24] = 1.2 × 10⁻⁷ × 1.24 = 1.488 × 10⁻⁷ Ω m .

Ref: NCERT > Physics Book > Current Electricity > Resistance, Resistivity and Ohm's Law

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