Semiconductor Physics: Carriers, Drift and Diffusion

Energy bands, intrinsic and extrinsic semiconductors, Fermi level, carrier concentrations and mass-action law, mobility, drift and diffusion currents, Einstein relation, generation and recombination.

Every electronic device is built on how electrons and holes behave in a semiconductor. This lesson covers the carrier physics that GATE EC tests directly.

Energy bands

  • Band gap E_g: about 1.12 eV for Si, 0.66 eV for Ge, 1.42 eV for GaAs (at 300 K).
  • Intrinsic semiconductor: n = p = n_i (about 1.5 × 10¹⁰ cm⁻³ for Si at 300 K; 10¹⁰ is often used in problems). n_i rises steeply with temperature: n_i² ∝ T³ e^(−E_g/kT).
  • Extrinsic: donors (P, As) give an n-type material; acceptors (B) give p-type.
  • Direct band gap (GaAs) materials emit light efficiently (LEDs, lasers); indirect (Si) do not.

Carrier concentrations

  • Mass-action law: n p = n_i² (in equilibrium).
  • n-type with N_D ≫ n_i: n ≈ N_D, p = n_i²/N_D.
  • Fermi level position: n = n_i e^((E_F − E_i)/kT), so E_F − E_i = kT ln(n/n_i). In n-type material E_F moves towards the conduction band.
  • Thermal voltage V_T = kT/q ≈ 25.9 mV at 300 K.

Current flow

  • Drift: carriers pushed by a field: J = q(nμ_n + pμ_p)E.
  • Conductivity σ = q(nμ_n + pμ_p); resistivity ρ = 1/σ.
  • Mobility μ falls with doping (impurity scattering) and with temperature (lattice scattering). For Si, μ_n ≈ 1350 and μ_p ≈ 480 cm²/V·s (lightly doped), so electrons move faster.
  • Diffusion: carriers move from high to low concentration: J_n = qD_n dn/dx, J_p = −qD_p dp/dx.
  • Einstein relation: D/μ = kT/q = V_T.

Generation, recombination and the Hall effect

  • Excess carriers decay with lifetime τ; diffusion length L = √(Dτ).
  • Continuity equation combines drift, diffusion, generation and recombination.
  • Hall effect: a current across a magnetic field creates a transverse voltage V_H = BI/(qnt). Its sign shows the carrier type; it measures concentration and mobility.

Worked example

Si doped with N_D = 10¹⁶ cm⁻³, μ_n = 1350 cm²/V·s, n_i = 10¹⁰ cm⁻³.

  • n = 10¹⁶, p = 10²⁰/10¹⁶ = 10⁴ cm⁻³.
  • σ = 1.6 × 10⁻¹⁹ × 10¹⁶ × 1350 = 2.16 S/cm; ρ = 0.463 Ω·cm.

Practice

GATE-styleNAT · numerical answer

A semiconductor has μ_n = 1300 cm²/V·s at 300 K (V_T = 0.026 V). What is the electron diffusion coefficient in cm²/s?

Show answer and explanationAnswer: 33.8

D = μV_T = 1300 × 0.026 = 33.8 cm²/s.

GATE-styleNAT · numerical answer

For n = 10¹⁶ cm⁻³ and n_i = 10¹⁰ cm⁻³ (kT = 0.0259 eV), how far (eV) is E_F above E_i? (Give 3 decimal places.)

Show answer and explanationAnswer: 0.358

E_F − E_i = kT ln(n/n_i) = 0.0259 × ln 10⁶ = 0.0259 × 13.82 = 0.358 eV.

GATE-styleNAT · numerical answer

In the worked example, what is the resistivity in Ω·cm? (Give 3 decimal places.)

Show answer and explanationAnswer: 0.463

ρ = 1/(qnμ_n) = 1/2.16 = 0.463 Ω·cm.

GATE-styleMCQ · one correct answer

Which material is best suited for an LED?

Show answer and explanationAnswer: A

In a direct-gap material, electrons and holes recombine by emitting photons efficiently.