#material properties

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

Why is carbon an insulator while silicon and germanium are semiconductors?

**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. Carbon (diamond) has a large energy gap (5.4 eV), preventing electron excitation at room temperature, while Si (1.1 eV) and Ge (0.7 eV) have smaller gaps, allowing thermal excitation and conduction. Substituting values gives Carbon has a larger energy gap, which m

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Wheatstone Bridge and Meter Bridge

A conductor of length \( 1 \, \text{m} \) and cross-sectional area \( 1 \times 10^{-6} \, \text{m}^2 \) has a resistance

**Unbalanced bridge** has potential difference between galvanometer nodes, current direction determined by which node higher potential, i.e., if R₁/R₂ > R₃/R₄, left node higher, current flows one way, else opposite. Galvanometer deflection indicates imbalance magnitude. Resistance is given by R = (rho l/A) . Rearrange: rho = (R A/l) . Given: R = 2 Ω , A = 1 × 10⁻⁶ m² , l = 1 m . Substitute: rho = (2 × 1 × 10⁻⁶/1) = 2 × 10⁻⁶ 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, V = ε

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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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Temperature Dependence of Resistance and Resistivity

A platinum wire has a resistance of \( 6 \, \Omega \) at \( 0^\circ \text{C} \) and \( 6.48 \, \Omega \) at \( 100^\circ

**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: R_t = R₀ [1 + α (T - T₀)] . Substitute: 6.48 = 6 [1 + α (100 - 0)] . Solve: 6.48 = 6 + 600α ⇒ 600α = 0.48 ⇒ α = (0.48/600) = 8 × 10⁻⁴ °C⁻¹ . 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, V = ε - I r

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Electric Current, Drift Velocity and Mobility

A copper wire of cross-sectional area \( 4 \times 10^{-7} \, \text{m}^2 \) carries a current of \( 1.2 \, \text{A} \). I

**Mobility** μ = v_d/E = e τ/m, τ relaxation time (s), measures ease of electron drift under field E (V/m). Conductivity σ = n e μ = 1/ρ, linking microscopic τ to macroscopic resistivity, explaining why metals conduct well due to large n and τ. Drift speed: v_d = (I/n e A) . Substitute: v_d = (1.2/8.5 × 10²⁸ × 1.6 × 10⁻¹⁹ × 4 × 10⁻⁷) . Calculate: v_d = (1.2/5.44 × 10³) ≈ 2.21 × 10⁻⁴ m/s . 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, V = ε - I r

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Electric Current, Drift Velocity and Mobility

A conductor has a resistivity of \( 6 \times 10^{-8} \, \Omega \text{m} \) and \( \alpha = 4 \times 10^{-3} \, ^\circ\te

**Mobility** μ = v_d/E = e τ/m, τ relaxation time (s), measures ease of electron drift under field E (V/m). Conductivity σ = n e μ = 1/ρ, linking microscopic τ to macroscopic resistivity, explaining why metals conduct well due to large n and τ. Use: rho_t = rho₀ [1 + α (T - T₀)] . Substitute: rho_t = 6 × 10⁻⁸ [1 + 4 × 10⁻³ (90 - 20)] . Calculate: rho_t = 6 × 10⁻⁸ [1 + 0.28] = 6 × 10⁻⁸ × 1.28 = 7.68 × 10⁻⁸ Ω m . Applying I = n e A v_d, R = ρ l/A, R_t

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Resistance, Resistivity and Ohm's Law

A wire of length \( 3 \, \text{m} \) and resistance \( 6 \, \Omega \) is stretched to \( 6 \, \text{m} \). What is the n

**Ohm's law deviation** at high fields occurs when τ or n vary with E, resistivity ρ = m/(n e² τ) changes, non-ohmic behaviour seen in semiconductors, electrolytes. At moderate fields, linear V-I holds, slope = R. Volume constant: l A = l' A' ⇒ A' = (A/2) . New resistance: R' = (rho l'/A') = (rho (2l)/(A/2)) = 4 (rho l/A) = 4R = 4 × 6 = 24 Ω . 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, V = ε - I r and P = I²R, evaluation yields 24 Ω, consistent

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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 .

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Resistance, Resistivity and Ohm's Law

A wire of length \( 3 \, \text{m} \) and cross-sectional area \( 3 \times 10^{-6} \, \text{m}^2 \) has a resistance of \

**Resistivity** depends on material and temperature, not geometry. For metallic conductor, ρ ≈10⁻⁸ Ω·m for copper. Given ρ=4×10⁻⁸ Ω·m, l=2 m, A=π r², R calculation uses R=ρ l/A. Stretching wire conserves volume V = l A = l' A', so A' = A l/l', new R' = ρ l'²/V. Resistance: R = (rho l/A) . Rearrange: rho = (R A/l) . Given: R = 5 Ω , A = 3 × 10⁻⁶ m² , l = 3 m . Substitute: rho = (5 × 3 × 10⁻⁶/3) = 5 × 10⁻⁶ Ω m . Applying I = n e A v_d, R = ρ l/A,

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