Intrinsic and doped semiconductors

Both bands are counted the same way: multiply the density of states by the occupation and integrate. Because E_c - E_F is several kT in a non-degenerate semiconductor, the Fermi function collapses to a Boltzmann exponential, the integral does too, and what survives is n = N_c e^{(-(E_c - E_F)/kT)}, with the entire shape of the band compressed into one effective density of states N_c = 2(2π m*ₑ kT/h²)³⁄² placed at the band edge. Holes give p = Nᵥ e^{(-(E_F - Eᵥ)/kT)}. Multiply, and the Fermi level cancels: np = N_c Nᵥ e^(−E_g/kT). No dopant entered, so that product cannot depend on doping. For an intrinsic sample n = p, hence nᵢ = √(N_c Nᵥ) e^(−E_g/2kT) — the halved exponent is not a convention but the square root of a product. Silicon at 300 K has N_c = 2.8 × 10¹⁹ and Nᵥ = 1.04 × 10¹⁹ cm⁻³, E_g/2kT = 1.12/0.0517 = 21.66, and nᵢ = 1.706 × 10¹⁹ e^(−21.66) = 6.7 × 10⁹ cm^-3. The measured value is 1.0 × 10¹⁰; that factor of 1.5 is a standing reminder that E_g shrinks with temperature and that the density-of-states masses are fitted averages.

Adding donors does not suspend np = nᵢ²; it slides n and p along that hyperbola. The second equation is electrical neutrality, n + Nₐ = p + N_d, counting only ionised impurities. Eliminate p and the positive root is n = (N_d - Nₐ)/2 + √([(N_d - Nₐ)/2]² + nᵢ²). Two limits carry all the work. When the net doping vastly exceeds nᵢ the root collapses to n = N_d - Nₐ; when it does not, n tends to nᵢ. Silicon at 300 K with N_d = 1.0 × 10¹⁵ cm⁻³ is deep in the first limit, five orders clear of nᵢ, so n = 1.0 × 10¹⁵ cm⁻³ and p = nᵢ²/n = 4.5 × 10⁴ cm^-3. Watch what happened to the holes. Doping multiplied n by 10⁵ and divided p by 10⁵, because their product is pinned: the extra electrons recombine away the holes faster than pairs are generated, until the new balance is struck. That suppressed minority density is what a p-n junction leaks and what a bipolar transistor stores, so it is the reason a diode rectifies at all.

Substitute phosphorus for silicon and strip its fifth electron away, and what is left is a fixed +e sitting in a medium of relative permittivity 11.7, binding an electron whose inertia is the conduction-band effective mass rather than mₑ. That is the hydrogen problem with two substitutions: e²/4πε₀ becomes e²/4πε₀ εᵣ, and mₑ becomes m*. Every binding energy is therefore scaled by (m*/mₑ)/εᵣ² and every radius by εᵣ/(m*/mₑ). For silicon that is 13.6 eV × 0.26/11.7² = 26 meV over an orbit of radius 0.0529 nm × 11.7/0.26 = 2.4 nm. Two checks follow at once. That radius spans four lattice constants, which is precisely what licenses treating the crystal as a structureless dielectric; and 26 meV is almost exactly kT = 25.9 meV at 300 K, so, with a conduction band offering far more states than there are donors, the level is emptied thermally and the donor is ionised. Germanium, with εᵣ = 16.0 and m*/mₑ = 0.12, gives 6.4 meV over a 7.1 nm orbit. The estimate is not exact — phosphorus in silicon measures 45 meV — because the effective-mass picture fails inside the donor's own cell, where the potential is no longer a screened Coulomb tail.

Plot ln(n) against 1/T for a doped sample and three stretches appear. Cold, at the right, not all donors have ionised: n = √(N_c N_d/2) e^(−E_d/2kT), a straight line of slope −E_d/2k, and the ionised fraction keeps falling as the sample cools. Warm, through the middle, every donor has given up its electron while nᵢ is still negligible, so n = N_d exactly — flat, and this is the saturation or extrinsic region. Hot, at the left, nᵢ overtakes N_d and the curve turns up with slope −E_g/2k. Where the two knees sit is set by the doping. Silicon at 1.0 × 10¹⁵ cm⁻³ with a 45 meV donor only freezes out below about 70 K, and waits until roughly 600 K for the intrinsic takeover. Raise the doping to 1.0 × 10¹⁸ cm⁻³ and freeze-out climbs to about 230 K while the intrinsic knee leaves the chart. The flat stretch between them is the design window, and it is bought by choosing N_d and E_g together.

σ = e(n μₙ + p μₚ) has two factors, and only one of them is a carrier count. The mobility μ = eτ/m* is set by the scattering time, and the two dominant channels pull opposite ways with temperature. Acoustic-phonon scattering strengthens as the lattice vibrates harder, giving μ_L ∝ T⁻³⁄²; ionised-impurity scattering weakens as carriers move faster past the fixed charges, giving μ_I ∝ T³⁄². Matthiessen adds the rates, 1/μ = 1/μ_L + 1/μ_I, so the mobility peaks where the two cross — high and cold for light doping, lower and hotter for heavy. In lightly doped silicon above about 100 K the phonon term wins outright, and there lies the trap in the saturation region: from 300 K to 400 K, n is stuck at 1.0 × 10¹⁵ cm⁻³ while μₙ falls from 1350 to 877 cm² V⁻¹ s⁻¹, so σ drops 35%. In the intrinsic region the two powers cancel instead — nᵢ carries T³⁄², the phonon mobility carries T⁻³⁄² — so σ goes as e^(−E_g/2kT) with no prefactor left, which is what makes that Arrhenius slope clean, provided you remember it returns E_g/2.

Every line above assumed the Boltzmann tail, which needs E_F at least about 3kT inside the gap — roughly 78 meV in silicon at 300 K. Three things break it. Heavy doping: pinning n at N_d pushes E_F toward E_c, and once N_d approaches N_c = 2.8 × 10¹⁹ cm⁻³ the exponential must give way to a Fermi-Dirac integral, the material is degenerate, and it conducts down to 0 K like a poor metal — which is exactly what ohmic contacts and tunnel diodes are made of. Heavy doping also narrows the gap by tens of meV as the impurity band merges with the band edge, so nᵢ is no longer the pure-crystal number. High temperature: once nᵢ approaches N_d the sample is intrinsic whatever was put into it, and no junction built from it will rectify. And compensation: N_d and Nₐ subtract in the neutrality equation but add in the scattering, so a 1.0 × 10¹⁶ donor crystal holding 9.0 × 10¹⁵ acceptors carries the electrons of a 10¹⁵ sample while scattering like a 1.9 × 10¹⁶ one — which is why carrier density is measured directly, by the Hall effect, and the mobility is then extracted from the resistivity.