University Physics IV · Molecular and Solid-State Physics · 12.8
Intrinsic & Doped Semiconductors
Pure silicon at room temperature carries roughly one mobile electron for every five trillion atoms. Every useful device begins by breaking that number deliberately, with parts per million of phosphorus or boron, and this topic is where you learn what that purchase costs as the temperature moves.
Build the model
Connect the measurement to the mechanism.
A semiconductor is an insulator whose gap is small enough for temperature to matter. Integrate the density of states against the Fermi function; because EF sits several kT inside the gap the Fermi function collapses to its Boltzmann tail, leaving n = Nc e(-(E_c - E_F)/kT) and p = Nᵥ e(-(E_F - Eᵥ)/kT), with Nc and Nᵥ effective densities of states going as T³⁄². Multiply them and EF cancels: np = Nc Nᵥ e(−E_g/kT) = nᵢ², a law of mass action no doping can touch, because no dopant entered the derivation.
That identity is the whole model. Setting n = p gives nᵢ = √(Nc Nᵥ) e(−E_g/2kT), about 10¹⁰ cm⁻³ for silicon at 300 K and nearly thirteen orders below the atom density, because the exponent Eg/2kT is 21.7. Doping does not repeal the identity, it exploits it: a phosphorus atom holds a fifth electron in a hydrogenic orbit that the dielectric constant and the effective mass widen to 2.4 nm and weaken to a few tens of meV, comparable with kT, so at room temperature the donor is ionised, n is pinned at Nd, and the hole density is driven down to nᵢ²/Nd.
The cost is a stack of assumptions — non-degenerate statistics, a Fermi level several kT inside the gap, a dopant orbit wide enough to see the lattice as a dielectric continuum. Raise the doping past 10¹⁹ cm⁻³, or the temperature past the point where nᵢ overtakes Nd, and they fail one by one.
- Simple definition
- An intrinsic semiconductor has equal electron and hole densities fixed only by its gap and its temperature; a doped one has that balance overridden by impurity atoms that supply one carrier type without supplying the other.
- Example
- Pure silicon at 300 K holds n = p = nᵢ = 1.0 × 10¹⁰ cm-3. Add 1.0 × 10¹⁵ cm⁻³ of phosphorus and n rises to 1.0 × 10¹⁵ while p falls to nᵢ²/n = 1.0 × 10⁵ cm⁻³ — the product is untouched.
Silicon: √(2.8 × 10¹⁹ × 1.04 × 10¹⁹) e(−21.66) = 6.7 × 10⁹ cm⁻³, against a measured 1.0 × 1010.
Nc, Nᵥ in cm⁻³; Eg in eV; kT = 25.85 meV at 300 K
That power-law prefactor is real but small beside the exponential: 300 K to 400 K multiplies it by 1.54 and nᵢ by 347.
m*ₑ is the density-of-states mass; both Nc and Nᵥ go as T³⁄²
Doping n-type by a factor 10⁵ raises n by 10⁵ and cuts p by 105. That suppressed minority carrier is what a diode leaks.
thermal equilibrium only, and only while EF stays several kT inside the gap
Collapses to n = Nd - Nₐ when the doping dominates and to n = nᵢ when it does not. Nothing else is needed.
Nd and Nₐ are the ionised donor and acceptor densities, in cm⁻³
26 meV over a 2.4 nm orbit — under kT at 300 K, and four lattice constants wide, which is what licenses the continuum picture.
silicon: εᵣ = 11.7, m*/mₑ = 0.26, a₀ = 0.0529 nm
Through saturation n is flat, so σ follows the mobility alone and falls as the lattice heats.
mobilities in cm² V⁻¹ s⁻¹; silicon at 300 K has μₙ = 1350, μₚ = 480
Where nᵢ comes from, and why the exponent is Eg/2kT
Both bands are counted the same way: multiply the density of states by the occupation and integrate. Because Ec - EF 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 = Nc e(-(E_c - E_F)/kT), with the entire shape of the band compressed into one effective density of states Nc = 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 = Nc Nᵥ e(−E_g/kT). No dopant entered, so that product cannot depend on doping. For an intrinsic sample n = p, hence nᵢ = √(Nc Nᵥ) e(−E_g/2kT) — the halved exponent is not a convention but the square root of a product. Silicon at 300 K has Nc = 2.8 × 10¹⁹ and Nᵥ = 1.04 × 10¹⁹ cm⁻³, Eg/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 Eg shrinks with temperature and that the density-of-states masses are fitted averages.
Doping does not add carriers so much as trade them
Adding donors does not suspend np = nᵢ²; it slides n and p along that hyperbola. The second equation is electrical neutrality, n + Nₐ = p + Nd, counting only ionised impurities. Eliminate p and the positive root is n = (Nd - Nₐ)/2 + √([(Nd - Nₐ)/2]² + nᵢ²). Two limits carry all the work. When the net doping vastly exceeds nᵢ the root collapses to n = Nd - Nₐ; when it does not, n tends to nᵢ. Silicon at 300 K with Nd = 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.
A donor is a hydrogen atom immersed in the crystal
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 0.0529 nm × 11.7/0.26 = 2.4 nm wide. Two checks follow at once. The orbit spans four lattice constants, which is precisely what licenses treating the crystal as a structureless dielectric; and 26 meV sits just under kT = 25.9 meV at 300 K, so 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.
Three regimes, and what fixes the boundaries
Plot ln(n) against 1/T for a doped sample and three stretches appear. Cold, at the right, not all donors have ionised: n = √(Nc Nd/2) e(−E_d/2kT), a straight line of slope −Ed/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 = Nd exactly — flat, and this is the saturation or extrinsic region. Hot, at the left, nᵢ overtakes Nd and the curve turns up with slope −Eg/2k. Where the two knees sit is set by the doping. Silicon at 1.0 × 10¹⁵ cm⁻³ with a 45 meV donor only begins to freeze 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 Nd and Eg together.
Conductivity does not copy the carrier density
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 μL ∝ T⁻³⁄²; ionised-impurity scattering weakens as carriers move faster past the fixed charges, giving μI ∝ T³⁄². Matthiessen adds the rates, 1/μ = 1/μL + 1/μI, 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 Eg/2.
Where the non-degenerate treatment stops
Every line above assumed the Boltzmann tail, which needs EF at least about 3kT inside the gap — roughly 78 meV in silicon at 300 K. Three things break it. Heavy doping: pinning n at Nd pushes EF toward Ec, and once Nd approaches Nc = 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 Nd the sample is intrinsic whatever was put into it, and no junction built from it will rectify. And compensation: Nd 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 and mobility are measured separately, by Hall effect and by resistivity.
Change one variable at a time
Make the relationship visible.
Hold Nd at 10¹⁵ cm⁻³ and drag T from 300 to 400 K: the donors stay fully ionised and nᵢ is still five decades below Nd, so n = Nd exactly — yet the mobility falls from 1350 to 877 and log₁₀ σ drops by 0.19. Then set Nd to 10¹⁸ and the freeze-out knee walks in from the cold right-hand edge.
IONISED DONORS Nd+/Nd1.000
log₁₀ (nᵢ / Nd)-5.18
MOBILITY μₙ1350 cm² V⁻¹ s⁻¹
log₁₀ σ if n = Nd-0.66
Live interpretationIONISED DONORS Nd+/Nd: 1.000. log₁₀ (nᵢ / Nd): −5.18. MOBILITY μₙ: 1350 cm² V⁻¹ s⁻¹. log₁₀ σ if n = Nd: −0.66
Catch the common trap
Explain before calculating.
An n-type silicon sample doped with 1.0 × 10¹⁵ cm⁻³ of phosphorus is heated from 300 K to 400 K. Over that interval its intrinsic density rises from 6.7 × 10⁹ to 2.3 × 10¹² cm-3. What happens to its conductivity?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasySilicon at 300 K is doped with 1.0 × 10¹⁵ cm⁻³ of phosphorus, all of it ionised. Taking nᵢ = 6.7 × 10⁹ cm⁻³, μₙ = 1350 cm² V⁻¹ s⁻¹ and μₚ = 480 cm² V⁻¹ s⁻¹, find the electron and hole densities and the resistivity.
- Phosphorus is a donor, and Nd = 1.0 × 10¹⁵ cm⁻³ stands five orders above nᵢ, so charge neutrality gives n = Nd to better than one part in 10¹⁰: n = 1.0 × 10¹⁵ cm-3.
- The law of mass action fixes the minority carrier: p = nᵢ²/n = (6.7 × 10⁹)²/(1.0 × 10¹⁵) = 4.49 × 10¹⁹/10¹⁵ = 4.5 × 10⁴ cm-3.
- Holes are outnumbered by ten orders of magnitude, so p μₚ = 2.2 × 10⁷ is nothing beside n μₙ = 1.35 × 10¹⁸ in the same units, and the hole term drops out.
- σ = n e μₙ = 1.0 × 10¹⁵ × 1.602 × 10⁻¹⁹ × 1350 = 0.216 S cm⁻¹, so ρ = 1/σ = 4.6 Ω cm.
Answern = 1.0 × 10¹⁵ cm⁻³, p = 4.5 × 10⁴ cm⁻³, ρ = 4.6 Ω cm. Doping two parts in 10⁸ of the silicon atoms has cut the resistivity from about 5 × 10⁵ Ω cm to 4.6.
MediumThe same sample is heated to 400 K. Show that it is still extrinsic, then find its conductivity. Take nᵢ(300 K) = 6.7 × 10⁹ cm⁻³, Eg = 1.12 eV, nᵢ ∝ T³⁄² e(−E_g/2kT), and a phonon-limited mobility falling as T⁻³⁄².
- Scale the prefactor first: (400/300)³⁄² = 1.540.
- Then the exponent. Eg/2kT is 1.12/(2 × 0.025851) = 21.66 at 300 K and 1.12/(2 × 0.034468) = 16.25 at 400 K, so the exponential grows by e(21.66 - 16.25) = e5.41 = 224.
- nᵢ(400 K) = 6.7 × 10⁹ × 1.540 × 224 = 2.3 × 10¹² cm-3. That is 430 times below Nd, so n = Nd still: the sample is extrinsic and squarely in saturation.
- The mobility is not flat, though: μₙ(400 K) = 1350 × (400/300)⁻³⁄² = 1350/1.540 = 877 cm² V⁻¹ s-1.
- σ = n e μₙ = 1.0 × 10¹⁵ × 1.602 × 10⁻¹⁹ × 877 = 0.140 S cm⁻¹, down from 0.216 S cm⁻¹ at 300 K — a 35% fall with the carrier density untouched.
Answernᵢ(400 K) = 2.3 × 10¹² cm⁻³, still 430 times below Nd, so n = 1.0 × 10¹⁵ cm⁻³ and σ = 0.140 S cm⁻¹, 35% below its 300 K value. Saturation flattens the carriers, not the conductivity.
HardAt what temperature does that 1.0 × 10¹⁵ cm⁻³ sample stop being extrinsic — take the boundary as nᵢ = Nd — and what are n and p there? Use nᵢ = 1.706 × 10¹⁹ (T/300)³⁄² e(−E_g/2kT) cm⁻³ with Eg = 1.12 eV.
- Put the condition in base-ten logs: 19.232 + 1.5 log(T/300) - 2822/T = 15, where 2822 K = Eg/(2k ln10) = 1.12/(2 × 8.617 × 10⁻⁵ × 2.3026).
- The exponential dominates, so start by dropping the T³⁄² term: 2822/T = 4.232 gives T = 667 K.
- Restore it and iterate. At 667 K, 1.5 log(667/300) = 0.520, so 2822/T = 4.232 + 0.520 = 4.752 and T = 594 K.
- Again: at 594 K, 1.5 log(594/300) = 0.445, so 2822/T = 4.677 and T = 603 K. One more pass returns 602 K, so the iteration has converged.
- At nᵢ = Nd the neutrality root gives n = Nd(1 + √5)/2 = 1.618 Nd = 1.6 × 10¹⁵ cm⁻³, and p = nᵢ²/n = 0.618 Nd = 6.2 × 10¹⁴ cm-3.
AnswerT = 602 K, about 329 °C. There n = 1.6 × 10¹⁵ cm⁻³ against p = 6.2 × 10¹⁴ cm⁻³, a ratio of only 2.6: the doping asymmetry is gone and no junction cut from this silicon will rectify.