Donor and Acceptor Doping

Introduction

An intrinsic semiconductor makes electron–hole pairs by crossing its band gap. A small amount of carefully chosen impurity changes the balance without needing to make a pair for every useful carrier. Donors make electrons the majority carriers; acceptors make holes the majority carriers. The words describe electronic behaviour in a particular host lattice, not a universal property of an element in every compound.

Core explanation

In silicon, a substitutional phosphorus atom has one more valence electron than the Si atom it replaces. In a simple covalent-bond picture, four electrons support bonds and the extra electron is weakly bound to the donor. The donor state lies near the conduction-band edge. Thermal excitation can free the electron into the conduction band, leaving a positively charged ionised donor fixed on a lattice site. Boron has one fewer valence electron than Si. It can accept an electron from the valence band, leaving a mobile hole and a negatively charged ionised acceptor. Neither ionised dopant migrates with each current pulse in an ordinary device; electrons and holes move. MIT's semiconductor lesson covers donor and acceptor doping in this bond-and-band picture.

“Shallow” means a small energy difference from a band edge, not a small physical depth below the wafer surface. A donor activation energy is commonly measured between its level and the conduction band; an acceptor activation energy is measured between its level and the valence band. The hydrogenic effective-mass model explains why host dielectric screening and carrier effective mass can make this binding much weaker than that of an isolated atom. Real impurity levels depend on the host and can instead be deep traps. A dopant that creates a deep level need not produce a large room-temperature free-carrier population. MIT's theory-of-solids lecture discusses shallow hydrogen-like donor states.

Charge neutrality in a simple bulk sample is n + N A⁻ = p + N D⁺ , where N A⁻ and N D⁺ are concentrations of ionised acceptors and donors. If all are ionised and N D > N A, then n − p ≈ N D − N A. With nondegenerate equilibrium mass action np = n i², the majority carrier is roughly n ≈ N D − N A when this difference greatly exceeds n i. Adding a donor to an acceptor-doped sample can compensate holes before it makes the material n-type. The chemical dopant count is not automatically the free-carrier count, especially under freeze-out, compensation or heavy doping.

Temperature separates three qualitative regimes. At low temperature some dopants remain neutral and carriers freeze out. At moderate temperature shallow dopants are mostly ionised, often giving a relatively stable majority concentration. At sufficiently high temperature intrinsic pair generation can dominate. The regime boundaries depend on impurity energy, concentration and host gap. Impurity scattering can also lower mobility as dopant concentration rises, so conductivity does not increase linearly forever with dopant loading.

Step-by-step reasoning

1. Identify the host site and whether the substitution supplies or accepts an electron. 2. Place the impurity level near the appropriate band edge and define its ionisation energy. 3. Determine ionised donor and acceptor counts at the temperature of interest. 4. Solve charge neutrality together with equilibrium mass action where applicable. 5. Check whether carrier mobility, trapping or degeneracy changes the conductivity conclusion.

Visual explanation

Draw a band diagram with a donor level just below E c and an arrow from that level to E c. On a second panel draw an acceptor level just above E v and an arrow from E v into the acceptor. Label the mobile conduction electron and valence hole separately from the stationary ionised dopant sites. Add a temperature axis marking freeze-out, extrinsic and intrinsic regimes.

Real-world analogy

A donor resembles a parked extra bicycle that can be released to a road network with modest effort. An acceptor resembles an empty bicycle rack space that becomes mobile only in the abstract sense as people rearrange bicycles among racks. The analogy distinguishes available carriers from fixed infrastructure but cannot represent band statistics or charge neutrality on its own.

Real-world example

A silicon chip needs adjacent p-type and n-type regions. Controlled boron and phosphorus doping can make those regions while retaining the underlying silicon lattice. If unintended impurities compensate the intended dopant, the electrically active carrier count is lower than the chemical assay suggests. Device fabrication therefore measures both composition and electrical activation rather than treating them as identical.

Why?

Why is a donor's electron easier to free than a valence electron in pure silicon? The weakly bound donor electron need only cross the small separation from impurity level to conduction band, rather than the full valence-to-conduction gap. This is the central advantage of shallow doping. The result still depends on temperature and on whether the impurity state is truly shallow and isolated.

Common misconception

“N-type silicon is negatively charged” is wrong for a neutral bulk sample: mobile electrons are balanced by positively ionised donor sites. “An acceptor releases a hole particle” hides the actual process: the acceptor captures a valence electron, and the resulting empty valence state behaves as a hole.

Worked example

At a temperature where dopants are fully ionised, a silicon sample contains N D = 8.0 × 10¹⁵ cm⁻³ donors and N A = 3.0 × 10¹⁵ cm⁻³ acceptors. Assume n i is much smaller than the net doping. The sample is n-type, with n ≈ N D − N A = 5.0 × 10¹⁵ cm⁻³ . If n i = 1.0 × 10¹⁰ cm⁻³ at that same temperature, p ≈ n i²/n = 2.0 × 10⁴ cm⁻³ . Using N D alone would overestimate electron concentration because the acceptors compensate part of the donors.

Quick check

1. In ordinary boron-doped silicon, which mobile carrier becomes the majority carrier? Answer: Holes; boron acts as an acceptor and captures valence electrons.

Exam focus

Label both the ionised dopant and the mobile carrier with their correct signs. State full-ionisation and nondegenerate assumptions before using net-doping approximations. A shallow impurity level is close in energy to a band edge. Distinguish an element's chemical concentration from the free-carrier concentration.

Advanced insight

At very high dopant concentrations, isolated-impurity pictures break down: wavefunctions overlap, impurity bands may form and band-edge positions can change. Separately, deep impurities may aid recombination by trapping carriers even when they contribute little to equilibrium conductivity. The same dopant can behave differently across host materials because its local bonding and dielectric screening change.

Summary

Donors and acceptors bias the carrier balance through impurity states near band edges. Shallow donors supply conduction electrons; shallow acceptors create valence holes. Ionisation, compensation and temperature determine how many mobile carriers actually appear, while mobility determines how effectively they conduct.

Practice questions

1. What is left on a donor site after its electron enters the conduction band? Answer: A positively ionised donor fixed at its lattice site. 2. Why can a donor-rich sample have fewer electrons than total donor atoms? Answer: Some donors may be neutral, acceptors may compensate them, or other trapping and heavy-doping effects may intervene. 3. An ideal sample has 6 × 10¹⁵ fully ionised donors and 2 × 10¹⁵ fully ionised acceptors per cm³. Estimate its majority concentration. Answer: About 4 × 10¹⁵ electrons cm⁻³ if intrinsic carriers are negligible. 4. Does n-type mean the entire piece has a net negative electrical charge? Answer: No. Positive donor ions balance the mobile-electron excess in neutral bulk material.