9 min readConcept sheet1-mark MCQ and assertion-reason, 2-mark band diagrams, 3-mark n-type versus p-type; also a NEET favourite

Extrinsic semiconductors: n-type and p-type

Doping with a pentavalent atom gives n-type (extra electrons); a trivalent atom gives p-type (extra holes). In both, nₑnₕ = nᵢ² and the crystal stays neutral.

DopingAdding a tiny, controlled amount of impurity (a few parts per million) to a pure semiconductor

n-typePentavalent dopant (As, Sb, P), called a donor; electrons are the majority carriers

p-typeTrivalent dopant (In, B, Al, Ga), called an acceptor; holes are the majority carriers

Law of mass actionnₑnₕ = nᵢ² at thermal equilibrium

Why we dope

Pure silicon has too few carriers to be useful. Adding a tiny, exact amount of impurity, called doping, raises the number of carriers by millions of times. The doped material is an extrinsic semiconductor. The dopant atom is chosen to be about the same size as a silicon atom, so it fits into the crystal without distorting it.

n-type and p-type

n-typeA pentavalent atom (As, Sb, P) uses four electrons in bonds. The fifth electron is held very loosely (about 0.01 eV in Ge, 0.05 eV in Si) and is freed at room temperature. The dopant donates an electron, so it is a donor. Electrons are the majority carriers (nₑ ≫ nₕ). The donor level lies just below the conduction band.

p-typeA trivalent atom (In, B, Al, Ga) makes only three bonds. The missing fourth bond is a hole. The dopant accepts an electron from a neighbour, so it is an acceptor. Holes are the majority carriers (nₕ ≫ nₑ). The acceptor level lies just above the valence band.

Both types are electrically neutral. Every extra free carrier is balanced by a fixed ion: a positive donor ion in n-type, a negative acceptor ion in p-type.

How to draw the band diagrams

n-type: draw the valence band and conduction band with the gap Eg. Draw a dashed line just below the conduction band and label it donor level, Ed. Put many dots (electrons) in the conduction band and a few open circles (holes) in the valence band.
p-type: draw a dashed line just above the valence band labelled acceptor level, Ea. Put many holes in the valence band and a few electrons in the conduction band.
For the lattice: show an As atom (+5) with a loose fifth electron for n-type, and a B or In atom (+3) with one hole for p-type.

Worked example: carriers in doped silicon

Problem

Pure silicon has 5 × 10²⁸ atoms m⁻³ and nᵢ = 1.5 × 10¹⁶ m⁻³. It is doped with arsenic at 2 parts per million. Find nₑ and nₕ, and state whether it is n-type or p-type.

Working

Arsenic atoms = 2 × 10⁻⁶ × 5 × 10²⁸ = 1 × 10²³ m⁻³.
Each arsenic atom gives one free electron, and the thermal electrons (10¹⁶) are negligible, so nₑ ≈ 1 × 10²³ m⁻³.
Law of mass action: nₕ = nᵢ²/nₑ = (1.5 × 10¹⁶)² ÷ 1 × 10²³.
nₕ = 2.25 × 10³² ÷ 10²³ = 2.25 × 10⁹ m⁻³.
Electrons far outnumber holes, so it is n-type.

Answer

nₑ ≈ 1 × 10²³ m⁻³; nₕ ≈ 2.25 × 10⁹ m⁻³; n-type

In real life

Doping is the first step in making every chip, from the camera sensor in a phone to the controller in a two-wheeler. Engineers add exact amounts of boron and phosphorus to silicon to make p-type and n-type regions.

Exam angle

2 marks: “Distinguish between n-type and p-type semiconductors (two points), with a band diagram of each.” n-type: pentavalent donor, electrons majority, donor level just below CB. p-type: trivalent acceptor, holes majority, acceptor level just above VB. Two labelled diagrams.

1 marks: “Name an impurity that makes silicon a p-type semiconductor.” Boron (or indium, aluminium, gallium): a trivalent element.

3 marks: “Silicon (nᵢ = 1.5 × 10¹⁶ m⁻³) is doped so that nₑ = 4.5 × 10²² m⁻³. Find nₕ and state the type of semiconductor.” nₕ = nᵢ²/nₑ = 2.25 × 10³² ÷ 4.5 × 10²² = 5 × 10⁹ m⁻³. Since nₑ ≫ nₕ, it is n-type.

Remember it

Pentavalent gives N-type (a spare electron); trivalent gives P-type (a missing electron, a hole). Always nₑnₕ = nᵢ².