Doping, defects and interfaces

A small addition can alter carrier population

Doping introduces species intended to modify electronic behaviour. Donor and acceptor states can change electron and hole populations, depending on activation and compensation. The added chemical amount is not automatically the mobile-carrier concentration.

Consider two films made with the same number of added donor atoms per volume. One can have more mobile electrons yet conduct less well; another can even have holes as its majority carriers if acceptors compensate its active donors. To understand such outcomes, count the charged centres and mobile carriers separately before calculating current.

Charge carriers and electronic energy bands introduces conduction electrons, valence-band holes and mobility. Here n is the number of mobile conduction electrons per cubic metre, and p the corresponding hole concentration. A hole is an unoccupied electronic state relative to a filled valence band, not a missing lattice atom or a positive dopant ion travelling through the solid.

A donor can release an electron while remaining at its lattice site

A representative substitutional donor in silicon is phosphorus. It has one more valence electron than silicon. In the dilute shallow-donor picture, four electrons participate in the surrounding bonding and an additional electron can occupy a weakly bound donor state. The crystal screens the attractive field of the donor centre. Its binding energy is much smaller than the energy needed to excite a valence electron across the full band gap.

Thermal excitation can transfer that donor-bound electron into extended conduction-band states. The donor centre is then positively charged, D⁺, and the released mobile electron has charge −e. The two changes conserve total charge. The donor atom need not leave its lattice position. A neutral donor D⁰ with its electron still bound is not the same thing as an ionised donor supplying a mobile electron.

A representative acceptor is substitutional boron in silicon. A valence-band electron can occupy the acceptor state, leaving a mobile hole in the valence-band population. The occupied acceptor is negatively charged, A⁻; the hole carries positive charge +e in the carrier description. The acceptor is called ionised even though it has gained an electron. Counting every ionised dopant as a positive charge would therefore give the wrong balance.

The first figure uses electron energy vertically. The shallow donor level E_D lies below the conduction edge E_c; a shallow acceptor level E_A lies above the valence edge E_v. The arrows represent electronic transitions, not atoms moving vertically through a specimen. These are representative dilute states, not a universal rule for every species in every host. Some impurities form deeper states, complexes or other phases instead.

Chemical incorporation does not settle the electronic occupation

Let N_D,chem be the measured or supplied total donor-species concentration. A fraction a_D may occur as electrically active donor centres under the processing and structural conditions of interest. Write N_D,active = a_D N_D,chem. Species at unsuitable sites, in complexes or in precipitates need not supply the same shallow donor state. Calling this processing-related fraction active does not yet say whether its electron is thermally released.

Let N_D⁺ count the active donor centres that are actually ionised. In a simple singly ionisable model, 0 ≤ N_D⁺ ≤ N_D,active ≤ N_D,chem. Define N_A⁻ similarly for ionised acceptors. Their occupation depends on temperature, state energies and the electronic population. The fractions are not independent adjustable knobs in a real equilibrium calculation: ionisation statistics and charge balance must be mutually consistent.

At sufficiently low temperature, many shallow-donor electrons can remain bound: carrier freeze-out. In a suitable higher-temperature dilute regime, nearly complete ionisation may be a good approximation. At still higher temperature, thermally generated electrons and holes can matter strongly. Meanwhile lattice vibrations, charged centres and other defects affect mobility. No temperature-independent rule says that every added atom contributes one mobile carrier.

The calculations below supply a complete-ionisation assumption for the active centres at one common temperature. Inactive donor species are stipulated electrically neutral in these particular accounts. Real inactive complexes can themselves introduce charge or traps, in which case those contributions must be included instead of silently omitted.

A neutral bulk can contain unequal mobile populations

Consider a uniform interior region away from depletion layers and contacts, with no other charged defects in the first model. Count charge relative to the neutral lattice reference. Mobile holes and ionised donors are positive; mobile electrons and ionised acceptors are negative. Bulk charge neutrality therefore gives e[p + N_D⁺ − n − N_A⁻] = 0, or n − p = N_D⁺ − N_A⁻. Define the right-hand side as ΔN, the net ionised-centre concentration.

An n-type population has n > p; a p-type population has p > n. Neither name means that the whole interior is charged with that sign. Fixed ionised centres balance the unequal mobile populations. Compensation means that donor and acceptor charges partly oppose each other in this account. They do not have to destroy or physically remove one another.

Charge neutrality fixes a difference, not the two populations separately. Add a limited thermal-equilibrium relation, np = n_i², for a nondegenerate semiconductor with the same band structure at the chosen temperature. Here n_i is the intrinsic carrier concentration of that reference host at that temperature. Nondegenerate means that carrier occupation is dilute enough for the usual classical carrier-population approximation; it is not a claim of a defect-free material.

The product relation reflects a common equilibrium occupation of the conduction and valence states. In this approximation, shifting that occupation toward more conduction electrons reduces holes so their product remains n_i². It is not an identity for arbitrary illuminated or injected populations. Strong doping that changes the bands or requires degenerate statistics also needs a different treatment. A weak transport probe is assumed not to appreciably disturb the equilibrium populations used here.

Substitute p = n_i²/n into n − p = ΔN. Multiplying by positive n gives n² − ΔN n − n_i² = 0. The positive root is n = [ΔN + √(ΔN² + 4n_i²)]/2, and p = n_i²/n. This expression keeps both populations. Only when positive ΔN is much larger than n_i can n ≈ ΔN and p ≈ n_i²/ΔN be used as a controlled majority-carrier approximation.

Use the charged-centre account before the transport account

Use the invented semiconductor records in the table at 300 K. They are not values for silicon, InGaAs or a commercial film. Both have N_D,chem = 1.00 × 10²¹ m⁻³ and N_A⁻ = 6.00 × 10²⁰ m⁻³. Their active donor fractions differ, and all active centres are stipulated ionised. Supply n_i = 2.00 × 10²⁰ m⁻³ and the nondegenerate equilibrium assumptions above. Inactive species have no extra charge in this model.

Film U has N_D⁺ = 9.00 × 10²⁰ m⁻³, so ΔN = 3.00 × 10²⁰ m⁻³. Its solution is n = 4.00 × 10²⁰ and p = 1.00 × 10²⁰ m⁻³. Check both equations: n − p = 3.00 × 10²⁰, and np = 4.00 × 10⁴⁰ m⁻⁶ = n_i². The positive charge count p + N_D⁺ is 10.0 × 10²⁰ m⁻³, exactly balancing n + N_A⁻.

Film V has N_D⁺ = 3.00 × 10²⁰ m⁻³, so ΔN = −3.00 × 10²⁰ m⁻³. Now n = 1.00 × 10²⁰ and p = 4.00 × 10²⁰ m⁻³. V is p-type despite the same chemical amount of the donor species. Its acceptor charge exceeds its active donor charge; chemical atom counts alone did not give the majority carrier.

Use the low-field drift relation σ = e(nμ_n + pμ_p), with the independently supplied mobilities in the table and the rounded e = 1.60 × 10⁻¹⁹ C. U has electron contribution 6.40 S/m and hole contribution 0.800 S/m, total 7.20 S/m. V gives 3.20 and 6.40 S/m, total 9.60 S/m. Its smaller electron population does not prevent a larger total conductivity.

The change is a supplied comparison, not proof that lowering donor activation causes mobility to rise. Processing and defects can change several inputs. Nor is mobility determined by the neutrality equations: it describes the drift response to a weak field, whereas those equations determine populations under the stated equilibrium model. Four times as many carriers is not automatically four times the current when the other population and mobilities differ.

Defects and interfaces can compete

Traps, grain boundaries, impurities and contacts can scatter or capture carriers. Processing can change which dopants are active and where they reside. Evaluate electrical measurements alongside composition and structure rather than inferring transport solely from a recipe.

Scattering changes a carrier’s directional response while it can remain mobile. Trapping transfers a carrier into a localised state, reducing the mobile population over the relevant observation interval. A trapped carrier can later be thermally released. A defect can also mediate recombination by capturing an electron and a hole in sequence, removing a mobile pair and returning the centre to its previous charge state. Not every capture event is immediate recombination or permanent loss.

Electronic levels associated with a defect depend on the local bonding and charge state, not just whether an atom is labelled an impurity. A gap-state diagram is one useful example; interface and oxide-related trapping can also involve states near or overlapping band energies. Chemical passivation may remove or alter an electrically active state. Such a change must be assessed in its actual structure rather than equating fewer visible defects with a guaranteed mobility.

If a neutral bulk additionally contains occupied negatively charged electron traps with number density N_t⁻, their charge belongs in the balance: p + N_D⁺ = n + N_A⁻ + N_t⁻. Do not subtract the trapped count from a former n while keeping every other equilibrium population fixed. For a separate hypothetical equilibrium record based on U, supply N_t⁻ = 3.00 × 10²⁰ m⁻³ while N_D⁺, N_A⁻ and n_i retain their stipulated values. The equations then give n = p = 2.00 × 10²⁰ m⁻³, not n = 1.00 × 10²⁰ with the old p.

At a real interface the charge need not be locally neutral. An occupied negatively charged boundary can repel electrons and leave a nearby region with fewer mobile electrons and positive fixed donor charge. Its spatial electrostatic potential shifts the band-edge energies, often called band bending. The distribution must satisfy electrostatics and the actual boundary conditions; a uniform bulk-neutrality equation cannot determine that profile or a contact barrier.

A 2016 InGaAs interface study compared gated Hall measurements with capacitance-derived channel charge. Fast trapped charge could contribute to the capacitive response without supplying the same mobile transport. Treating all of it as mobile overestimated carrier density and underestimated the inferred mobility in that analysis. This is specific evidence for separating charge storage from charge transport, not a universal trapped fraction or a performance claim for every interface.

One conductivity value can fit different electronic accounts

Consider a third invented state W at the same n_i and temperature. Let N_D⁺ = N_A⁻ = 6.00 × 10²⁰ m⁻³, so ΔN = 0 and n = p = 2.00 × 10²⁰ m⁻³. With μ_n = 0.175 and μ_p = 0.0500 m²/(V s), its conductivity is also 7.20 S/m. U and W therefore share conductivity while differing in their active-centre balance, electron and hole populations, and mobility. A fit to conductivity alone cannot choose between them.

Different measurements constrain different parts of that account. Composition addresses chemical species and amounts; structural or state-sensitive evidence addresses their sites and electronic states. Temperature-dependent transport can challenge a complete-ionisation or fixed-mobility assumption. Time-dependent charge response can reveal trapping and release, but its sampling interval matters. Agreement with one convenient fit is not independent identification of the microscopic centre.

Hall measurements can add carrier-sensitive information. Simple single-carrier formulas presume a suitable carrier regime, geometry and response model; a two-carrier specimen cannot automatically be reduced to either separate population using such a formula. Hall mobility can differ from drift mobility through a Hall factor. Retain those assumptions rather than using one instrument name as proof that all carrier counts are known.

Contacts, current paths, thickness, sample uniformity, temperature and illumination also matter. Conductors, insulators and semiconductors shows how an extra series contact resistance can change a two-terminal reading without changing bulk conductivity. From measured transport to device behaviour keeps that distinction at component scale. The equilibrium examples here establish neither switching speed, leakage, breakdown nor useful device performance.

Worked examples

Try the reasoning

1. Does more dopant always increase mobility?

   Show the answer and reasoning

   No.

   Additional scattering can reduce it.

2. Can an interface influence carrier transport?

   Show the answer and reasoning

   Yes.

   It may add barriers, states or scattering.

3. In the singly ionisable model, a donor releases one bound electron and an acceptor captures one valence electron. State each fixed charge and the associated mobile carrier. Are D⁺ and A⁻ the particles drifting in the electronic current model?

   Show the answer and reasoning

   D⁺ is fixed positive charge with a released conduction electron; A⁻ is fixed negative charge with a valence-band hole. The dopant centres are not the drifting electronic carriers.

   Each process conserves charge. A hole describes a missing valence-band electron, not a dopant nucleus or lattice vacancy. Ionisation does not always mean losing an electron from the centre.

4. At the same stipulated n_i = 2.00 × 10²⁰ m⁻³, active ionised donors and acceptors are each 8.00 × 10²⁰ m⁻³. There are no other charged states. What are n and p? If μ_n = 0.100 and μ_p = 0.0500 m²/(V s), is conductivity zero?

   Show the answer and reasoning

   n = p = 2.00 × 10²⁰ m⁻³; σ = 4.80 S/m, not zero.

   Equal fixed donor and acceptor charge gives n − p = 0. It does not remove thermal carriers. Their equilibrium product is still n_i². Equal electron and hole concentrations also do not make their conventional drift currents cancel.

5. A different supplied equilibrium record has 9.00 × 10²⁰ active donors per m³, but only two-thirds are ionised; N_A⁻ = 6.00 × 10²⁰ m⁻³ and n_i = 2.00 × 10²⁰ m⁻³. No other charged states occur. Find the populations. May the chemical donor count replace N_D⁺?

   Show the answer and reasoning

   N_D⁺ = 6.00 × 10²⁰ m⁻³, so n = p = 2.00 × 10²⁰ m⁻³. The active or chemical total cannot replace its ionised part.

   This problem supplies a different centre-occupation record, not a temperature-change forecast. In a real specimen, the occupation fractions and carrier populations must satisfy the appropriate ionisation statistics and neutrality simultaneously.

6. Keep V’s supplied carrier populations fixed, but halve both mobilities. What is σ now? Does the unchanged p-type label guarantee that it still exceeds U’s conductivity?

   Show the answer and reasoning

   σ = 4.80 S/m, below U’s 7.20 S/m. V remains p-type by population.

   Each drift contribution is proportional to its own supplied mobility. The majority label describes n versus p, not the total conductivity or a universal scattering law.

7. For W, use n = p = 2.00 × 10²⁰ m⁻³ and μ_p = 0.0500 m²/(V s). Which μ_n gives the same 7.20 S/m as U? Does matching it identify the same doping state?

   Show the answer and reasoning

   μ_n = 0.175 m²/(V s). Equal conductivity does not identify the same centre or carrier populations.

   Solve 7.20/[e × (2.00 × 10²⁰)] = μ_n + 0.0500 = 0.225. W is compensated in the supplied model, unlike U. Additional independent evidence is required to choose between them.

What to carry forward

• Chemical doping and active carriers are distinct.
• Combine transport, structural and chemical evidence.
• A released donor electron leaves positive fixed charge; an occupied acceptor leaves negative fixed charge and a mobile hole.
• Separate total chemical amount, active centres and their actual ionised occupation.
• Neutrality counts all specified positive and negative charges; it does not require n = p in a doped bulk.
• The equilibrium product np = n_i² has a nondegenerate thermal-equilibrium domain; keep minority carriers when the approximation is inadequate.
• Trapping, recombination, scattering and spatial interface charge are different mechanisms. One conductivity cannot identify them all.