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Semiconductor physics: energy bands, carriers and junctions

Electronics and Communication Engineering (B.E./B.Tech) · Circuits and devices · free study notes for exams, placements and interviews

In short

Semiconductor physics explains why silicon conducts a controllable amount of current: electrons in a crystal may occupy only certain energy bands, a small forbidden gap separates the filled valence band from the empty conduction band, and doping or temperature decides how many mobile electrons and holes exist. Carrier motion by drift and diffusion, and the electrostatics of a pn junction, are the foundation on which every diode, BJT, MOSFET and IC is built. It is the first unit of the Electron Devices paper in every Indian university syllabus and carries a large share of the 8-12 marks of the Electronic Devices section in GATE EC (carrier transport, Poisson and continuity equations, pn junction) as well as in IES/ESE, BARC, ISRO and DRDO papers. Interviews at SCL Mohali, BEL, ISRO, Micron, Intel, AMD, Qualcomm, Texas Instruments, Analog Devices, Applied Materials and the fabs coming up under the India Semiconductor Mission open with exactly these questions.

Explained

From atoms to energy bands

Intrinsic and extrinsic semiconductors

Carrier statistics and the Fermi level

Drift, diffusion and transport

Generation, recombination and excess carriers

The pn junction

Metal-semiconductor and hetero-junctions

Key points

Formulas and examples

f(E) = 1 / [1 + exp((E − EF)/kT)] - occupancy probability; k = 1.38 × 10⁻²³ J/K = 8.62 × 10⁻⁵ eV/K
n = Nc exp(−(Ec − EF)/kT); p = Nv exp(−(EF − Ev)/kT) - cm⁻³, non-degenerate case
np = ni² = Nc Nv exp(−Eg/kT); ni ∝ T1.5 exp(−Eg/2kT)
EF − Ei = kT ln(n/ni) (n-type); Ei − EF = kT ln(p/ni) (p-type) - eV
Charge neutrality: n + Na⁻ = p + Nd⁺ → n = (Nd−Na)/2 + √[((Nd−Na)/2)² + ni²]
σ = q(nμn + pμp) S/cm; ρ = 1/σ Ω·cm; R = ρL/A
J = q(nμn + pμp)E + qDn(dn/dx) − qDp(dp/dx) - A/cm²
D/μ = kT/q = VT = 25.9 mV at 300 K; vd = μE (saturates ≈10⁷ cm/s in Si)
Continuity: ∂n/∂t = (1/q)(∂Jn/∂x) + G − R; Poisson: d²ψ/dx² = −ρ_charge/εs
Δp(t) = Δp(0) e−t/τ; L = √(Dτ) - cm
VH = IB/(q n t); RH = 1/(qn); μH = σ RH - Hall measurement, SI units (t = thickness, m)
Vbi = VT ln(NaNd/ni²) V; Na xp = Nd xn
W = √[2εs(Vbi + VR)(1/Na + 1/Nd)/q]; Emax = 2(Vbi+VR)/W; Cj = εsA/W
Schottky barrier φB = φm − χ (n-type); cut-off wavelength λc(µm) ≈ 1.24/Eg(eV)
Constants: q = 1.6 × 10⁻¹⁹ C; ε₀ = 8.854 × 10⁻¹⁴ F/cm; εs(Si) = 11.7ε₀ = 1.036 × 10⁻¹² F/cm

Example 1 - doped silicon: carriers, resistivity and Fermi level. Si doped with Nd = 10¹⁶ cm⁻³ phosphorus at 300 K; ni = 1.5 × 10¹⁰ cm⁻³, μn = 1200 cm²/V·s at this doping.

Example 2 - compensated sample. Si with Nd = 5 × 10¹⁶ and Na = 2 × 10¹⁶ cm⁻³; at a total impurity of 7 × 10¹⁶ cm⁻³, μn ≈ 1000 cm²/V·s.

Example 3 - temperature and minority carriers. Same n-Si (Nd = 10¹⁶) heated from 300 K to 400 K; take Eg = 1.12 eV constant.

Example 4 - pn junction electrostatics. Abrupt Si junction, Na = 10¹⁷ cm⁻³ (p-side), Nd = 10¹⁶ cm⁻³ (n-side), 300 K, area A = 10⁻⁴ cm².

Example 5 - diffusion and Hall measurement.

Property at 300 KSiGeGaAs
Eg (eV)1.12 (indirect)0.66 (indirect)1.42 (direct)
ni (cm⁻³)~1.5 × 10¹⁰~2.4 × 10¹³~2 × 10⁶
μn / μp (cm²/V·s)1350 / 4803900 / 19008500 / 400
Relative permittivity11.716.012.9
Typical useICs, power, solarIR detectors, historicRF, MMIC, LEDs/lasers
DriftDiffusion
CauseElectric fieldConcentration gradient
ExpressionqnμnE, qpμpEqDn dn/dx, −qDp dp/dx
NeedsApplied biasNon-uniform carrier density
Dominates inMOSFET channel, bulk resistorBJT base, forward-biased diode

Where it is used

Interview questions

Common mistakes

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