PhysicsCBSE Class 12 · Semiconductor Electronics: Materials, Devices and Simple Circuits

Intrinsic and Extrinsic Semiconductors Class 12: Energy Bands and Doping

Q: What is a hole?

A vacancy left in a covalent bond when an electron escapes. A neighbouring electron can move into it, so the hole moves in the opposite direction and acts like a positive charge carrier.

Q: Why is doping done in very small amounts?

A few parts per million already raise the carrier concentration enormously. Too much impurity would disturb the crystal structure and spoil the semiconductor's properties.

Q: What is the difference between intrinsic and extrinsic semiconductors?

An intrinsic semiconductor is pure, with equal numbers of electrons and holes. An extrinsic semiconductor is doped, so one type of carrier (electrons in n-type, holes in p-type) greatly outnumbers the other.

Q: Why does a semiconductor's resistance fall when heated?

Heating breaks more covalent bonds, creating more electron-hole pairs. The increase in carriers is much larger than the increase in collisions, so resistance falls.

Q: Are energy bands studied mathematically in Class 12?

No. For 2026-27, energy bands in conductors, semiconductors and insulators are treated qualitatively only.

Updated 15 October 2026 · Fact-checked

In a pure (intrinsic) semiconductor like silicon, a small band gap lets a few electrons jump to the conduction band, leaving equal numbers of holes: nₑ = nₕ = nᵢ. Doping makes it extrinsic: pentavalent atoms (P, As, Sb) add electrons to give n-type, trivalent atoms (B, Al, In) add holes to give p-type, with nₑnₕ = nᵢ².

Understand Intrinsic and Extrinsic Semiconductors

Copper conducts well, glass hardly at all, and silicon sits in between, but its conductivity can be raised a million times by adding a tiny trace of another element. The band picture and doping explain how.

Energy bands (qualitative). In a solid, the energy levels of the outer electrons of many atoms spread into bands. The highest band that is full at absolute zero is the valence band; the next one up is the conduction band. The gap between them is the energy band gap $E_{g}$ .

Conductors (metals): the bands overlap (or the conduction band is partly filled), so electrons move freely.
Insulators: a large gap, more than about 3 eV (for example diamond, about 5.4 eV); almost no electrons can cross it.
Semiconductors: a small gap, less than about 3 eV (silicon about 1.1 eV, germanium about 0.7 eV). At 0 K they are insulators, but at room temperature some electrons gain enough thermal energy to cross the gap.

Intrinsic semiconductors. In pure silicon or germanium, each atom forms four covalent bonds with its neighbours. When a bond breaks, an electron becomes free (in the conduction band) and leaves behind a vacancy, a hole, which behaves as a positive charge carrier: a neighbouring electron can jump into it, so the hole moves. In a pure crystal, electrons and holes are created in pairs: $n_{e} = n_{h} = n_{i}$ (the intrinsic carrier concentration). Current is carried by both. Raising the temperature breaks more bonds, so the conductivity of a semiconductor increases with temperature (unlike a metal).

Extrinsic semiconductors (doping). Adding a tiny, controlled amount of impurity (a few parts per million) is called doping. The dopant atom takes the place of a silicon atom in the lattice.

n-type: a pentavalent dopant (phosphorus, arsenic, antimony) has five valence electrons; four form bonds and the fifth is very loosely held and easily freed. The dopant is a donor. Electrons are the majority carriers, holes the minority carriers: $n_{e} ≫ n_{h}$ .
p-type: a trivalent dopant (boron, aluminium, indium, gallium) has only three valence electrons, leaving one bond incomplete: a hole. It can accept an electron, so it is an acceptor. Holes are the majority carriers: $n_{h} ≫ n_{e}$ .
In equilibrium, $n_{e} n_{h} = n_{i}^{2}$ : adding many electrons reduces the number of holes (more recombination), and vice versa.
An extrinsic semiconductor is still electrically neutral: the extra electrons in n-type are balanced by the fixed positive donor ions, and the holes in p-type by fixed negative acceptor ions.

Band diagrams for doped crystals (qualitative). In n-type, the donor energy level lies just below the conduction band, so a little energy lifts donor electrons into it. In p-type, the acceptor level lies just above the valence band, so electrons from the valence band easily move into it, leaving holes behind.

Intrinsic and Extrinsic Semiconductors: formulas and key facts

Intrinsic semiconductor: $n_{e} = n_{h} = n_{i}$; Electrons and holes created in pairs.
Mass action law: $n_{e} n_{h} = n_{i}^{2}$; Holds in thermal equilibrium for doped crystals too.
n-type: $n_{e} \approx N_{D}, n_{h} = \frac{n _{i}^{2}}{N _{D}}$; N_D = donor concentration (P, As, Sb).
p-type: $n_{h} \approx N_{A}, n_{e} = \frac{n _{i}^{2}}{N _{A}}$; N_A = acceptor concentration (B, Al, In, Ga).
Band gaps (approximate): $E_{g} (Si) \approx 1.1 eV, E_{g} (Ge) \approx 0.7 eV, E_{g} (insulator) > 3 eV$; Qualitative classification.
Light that can create electron-hole pairs: $λ \leq \frac{h c}{E _{g}} \approx \frac{1240}{E _{g} ( eV )} nm$; Photon energy must be at least E_g.

How to solve Intrinsic and Extrinsic Semiconductors questions

Use this method for carrier concentration and classification questions.

1Identify the dopant's valency: 5 → donor, n-type; 3 → acceptor, p-type.
2Find the dopant concentration: (number of host atoms per m³) × (doping fraction).
3Majority carriers ≈ dopant concentration (doping is much larger than nᵢ).
4Minority carriers = nᵢ²/(majority carriers).
5Check: majority × minority = nᵢ².
6For band gap questions, compare E_g with about 3 eV, and use λ ≈ 1240/E_g nm for the longest wavelength that can free carriers.
7Remember that doped crystals are electrically neutral.

Intrinsic and Extrinsic Semiconductors in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Effect of doping on carrier numbers

Suppose a chip maker dopes pure silicon (5 × 10²⁸ atoms per m³, nᵢ = 1.5 × 10¹⁶ m⁻³) with phosphorus at one atom per hundred million (10⁸) silicon atoms.

(a) Find the electron concentration. (b) By what factor does it exceed the intrinsic value? (c) Find the hole concentration.

Try it first, then show the solutionSolution

N_D = 5 × 10²⁸/10⁸ = 5 × 10²⁰ m⁻³, so nₑ ≈ 5 × 10²⁰ m⁻³.
Factor = 5 × 10²⁰/1.5 × 10¹⁶ ≈ 3.3 × 10⁴.
nₕ = nᵢ²/nₑ = (1.5 × 10¹⁶)²/5 × 10²⁰ = 4.5 × 10¹¹ m⁻³.

Answer: (a) about 5 × 10²⁰ m⁻³; (b) about 3.3 × 10⁴ times; (c) 4.5 × 10¹¹ m⁻³.

Memory hook: One impurity atom in a hundred million multiplies the carriers by tens of thousands.

Band gap and light

Imagine a light sensor made from a semiconductor with a band gap of 2.4 eV, whose resistance falls when light creates electron-hole pairs.

(a) Find the longest wavelength it responds to. (b) Will it respond to red light (650 nm)? (c) Why would a silicon sensor respond to red light?

Try it first, then show the solutionSolution

λ_max ≈ 1240/2.4 ≈ 517 nm (green).
Red photons have about 1240/650 ≈ 1.9 eV, less than 2.4 eV: no response.
Silicon's band gap is about 1.1 eV, so photons up to about 1127 nm (including red and near infrared) can create pairs.

Answer: (a) about 517 nm; (b) no; (c) silicon's smaller band gap lets red photons create carriers.

Memory hook: A photon must pay the band gap to free a carrier.

Tricks to remember Intrinsic and Extrinsic Semiconductors

Short memory aids. Say each one aloud once, then use it on a question the same day.

PAS for n, BAIG for p
Pentavalent: P, As, Sb give n-type. Trivalent: B, Al, In, Ga give p-type.
Use it when: Choosing dopants.
Donor donates, acceptor accepts
Donors give electrons (n-type); acceptors take electrons, creating holes (p-type).
Use it when: Naming dopants.
Majority times minority = nᵢ²
More of one carrier means fewer of the other.
Use it when: Carrier concentration numericals.
3 eV dividing line
Band gap below about 3 eV: semiconductor; above: insulator; overlap: conductor.
Use it when: Classification.

Intrinsic and Extrinsic Semiconductors: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Holes in n-type germanium · 2 marks · Very short answerBoard level

Germanium has 4.4 × 10²⁸ atoms per m³ and nᵢ = 2.4 × 10¹⁹ m⁻³. It is doped with one arsenic atom per million germanium atoms. Find the electron and hole concentrations.

Try it first, then show the solutionSolution

N_D = 4.4 × 10²⁸ × 10⁻⁶ = 4.4 × 10²² m⁻³ ≈ nₑ. (1)
nₕ = nᵢ²/nₑ = (2.4 × 10¹⁹)²/4.4 × 10²² ≈ 1.3 × 10¹⁶ m⁻³. (1)

Answer: nₑ ≈ 4.4 × 10²² m⁻³; nₕ ≈ 1.3 × 10¹⁶ m⁻³.

Example 2: Conductors, insulators and semiconductors · 3 marks · Short answerBoard level

Distinguish between conductors, insulators and semiconductors on the basis of energy bands. Draw (describe) the band diagram for each.

Try it first, then show the solutionSolution

Conductor: valence and conduction bands overlap (or conduction band partly filled); electrons move freely. Diagram: overlapping bands. (1)
Insulator: wide gap (E_g > 3 eV); conduction band empty; no conduction at ordinary temperatures. Diagram: full lower band, wide gap, empty upper band. (1)
Semiconductor: narrow gap (E_g < 3 eV); some electrons cross at room temperature, leaving holes. Diagram: narrow gap with a few electrons above and holes below. (1)

Answer: Overlap (conductor), wide gap above 3 eV (insulator), narrow gap below 3 eV (semiconductor).

Example 3: Carriers in p-type silicon · 3 marks · Short answerHarder board level

Silicon (5 × 10²⁸ atoms m⁻³, nᵢ = 1.5 × 10¹⁶ m⁻³) is doped with indium at one atom per 5 × 10⁷ silicon atoms. (a) Is it n-type or p-type? (b) Find the majority and minority carrier concentrations.

Try it first, then show the solutionSolution

(a) Indium is trivalent: acceptor, so p-type. (1)
(b) N_A = 5 × 10²⁸/5 × 10⁷ = 10²¹ m⁻³ ≈ nₕ (majority). (1)
nₑ = (1.5 × 10¹⁶)²/10²¹ = 2.25 × 10¹¹ m⁻³ (minority). (1)

Answer: (a) p-type; (b) holes ≈ 10²¹ m⁻³, electrons ≈ 2.25 × 10¹¹ m⁻³.

Example 4: Band gap and wavelength · 2 marks · Very short answerHarder board level

Find the longest wavelength of light that can create electron-hole pairs in (a) silicon (E_g ≈ 1.1 eV) and (b) germanium (E_g ≈ 0.7 eV). Which material can detect longer infrared wavelengths?

Try it first, then show the solutionSolution

(a) λ ≈ 1240/1.1 ≈ 1127 nm; (b) λ ≈ 1240/0.7 ≈ 1771 nm. (1)
Germanium, because its smaller band gap lets lower-energy (longer-wavelength) photons create carriers. (1)

Answer: (a) about 1127 nm; (b) about 1771 nm; germanium.

Example 5: Temperature and conductivity · 5 marks · Long answerHOTS / competency

(a) Explain, using the band picture, why pure silicon behaves as an insulator at 0 K but conducts at room temperature. (2)
(b) Explain why the conductivity of a semiconductor increases with temperature while that of a metal decreases. (2)
(c) Why are semiconductor devices made from doped rather than pure material? (1)

Try it first, then show the solutionSolution

(a) At 0 K the valence band is full and the conduction band empty; there are no free carriers, so it is an insulator. (1)
At room temperature, thermal energy lifts some electrons across the 1.1 eV gap, creating electrons and holes that can carry current. (1)
(b) In a semiconductor, heating greatly increases the number of carriers (more bonds broken), which outweighs the extra collisions. (1)
In a metal the number of free electrons is fixed; heating only increases collisions (shorter relaxation time), so conductivity falls. (1)
(c) Doping gives a controlled, much larger number of chosen carriers (electrons or holes), so the device's behaviour is predictable and depends little on temperature. (1)

Answer: (a) no carriers at 0 K, thermally generated pairs at room temperature; (b) carrier number rises in semiconductors, collisions rise in metals; (c) doping gives controlled majority carriers.

Example 6: Is a doped crystal charged? · 4 marks · Case-basedHarder board level

Read the passage and answer.

A silicon crystal is doped with antimony, so that it has far more free electrons than holes.
(a) Is it n-type or p-type? (1)
(b) Is the crystal negatively charged? Explain. (1)
(c) Where does the donor energy level lie in the band diagram? (1)
(d) What happens to the number of holes compared with pure silicon? Why? (1)

Try it first, then show the solutionSolution

(a) Antimony is pentavalent: n-type. (1)
(b) No. Each extra free electron came from a neutral donor atom, which is left as a fixed positive ion; overall the crystal is neutral. (1)
(c) Just below the bottom of the conduction band. (1)
(d) It falls below the intrinsic value: the many electrons recombine with holes, keeping nₑnₕ = nᵢ². (1)

Answer: (a) n-type; (b) no, it is neutral; (c) just below the conduction band; (d) fewer holes, since nₑnₕ = nᵢ².

Where students lose marks in Intrinsic and Extrinsic Semiconductors

The slips examiners see most often, and what to write instead.

Mistake: Saying n-type material is negatively charged.
Confusing majority carriers with net charge.
Do this: Doped semiconductors are electrically neutral.
Usually costs: 1 mark.
Mistake: Using nₕ = nᵢ in a doped crystal.
Forgetting the mass action law.
Do this: Minority carriers = nᵢ²/majority carriers.
Usually costs: The answer mark.
Mistake: Mixing up donor and acceptor elements.
Memory slip.
Do this: Pentavalent (P, As, Sb) donate electrons; trivalent (B, Al, In, Ga) accept.
Usually costs: 1 mark.
Mistake: Placing donor levels near the valence band.
Reversing the band diagram.
Do this: Donor level just below the conduction band; acceptor level just above the valence band.
Usually costs: 1 mark in a diagram.

Exam tips

Intrinsic and extrinsic semiconductors were set in each year from 2023 to 2026 (for example 2023 55/1/1 Q24(a), 2024 55/1/1 Q10, 2025 55/1/1 Q21 and 2026 55/5/1 Q21).
Draw band diagrams with valence band, conduction band, gap, and donor or acceptor level clearly labelled.
Write nₑnₕ = nᵢ² before carrier-concentration numericals.
State majority and minority carriers whenever a dopant is named.
Energy bands are qualitative only for 2026-27; no band-theory derivations are needed.

Practice questions on Intrinsic and Extrinsic Semiconductors

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
The majority charge carriers in a p-type semiconductor are
(a) electrons (b) holes (c) protons (d) donor ions
Hint
p for positive.
Answer
(b) holes.
Q2 · 1 mark · Very short answerBoard level
Why does pure silicon behave like an insulator at absolute zero?
Hint
Bands at 0 K.
Answer
At 0 K the valence band is completely filled and the conduction band is empty; no electron has the energy to cross the gap, so there are no free carriers.
Q3 · 1 mark · Very short answerHarder board level
In a doped semiconductor, nᵢ = 10¹⁶ m⁻³ and nₑ = 4 × 10²² m⁻³. Find nₕ and state the type.
Hint
nₕ = nᵢ²/nₑ.
Answer
nₕ = 10³²/4 × 10²² = 2.5 × 10⁹ m⁻³; n-type.
Q4 · 1 mark · Very short answerHarder board level
A semiconductor stops responding to light longer than 1000 nm. Estimate its band gap.
Hint
E_g ≈ 1240/λ.
Answer
E_g ≈ 1240/1000 ≈ 1.24 eV.
Q5 · 1 mark · Assertion-reasonHarder board level
Assertion (A): An n-type semiconductor is electrically neutral.
Reason (R): The negative charge of the extra free electrons is balanced by the positive charge of the fixed donor ions.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
Where did the electrons come from?
Answer
(a).
Q6 · 2 marks · Very short answerHOTS / competency
State the type of semiconductor formed when (a) germanium is doped with aluminium, (b) silicon is doped with phosphorus, and (c) silicon is doped with gallium. Name the majority carriers in each.
Hint
Valency of the dopant.
Answer
(a) p-type, holes; (b) n-type, electrons; (c) p-type, holes.
Q7 · 2 marks · Very short answerHOTS / competency
Why is a trivalent impurity called an acceptor? Draw (describe) the position of the acceptor level.
Hint
Incomplete bond.
Answer
A trivalent atom forms only three complete bonds; the fourth bond lacks an electron and can accept one from a neighbouring bond, creating a hole that moves. The acceptor level lies just above the top of the valence band, so valence electrons easily jump into it.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Band gap of silicon?Show answer
About 1.1 eV.
2.Majority carriers in n-type?Show answer
Electrons.
3.Mass action law?Show answer
nₑnₕ = nᵢ².
4.Two acceptor impurities?Show answer
Boron and aluminium (also indium, gallium).
5.Where is the donor level?Show answer
Just below the conduction band.
6.What happens to a semiconductor's conductivity when heated?Show answer
It increases.

Intrinsic and Extrinsic Semiconductors: frequently asked questions

What is a hole?

A vacancy left in a covalent bond when an electron escapes. A neighbouring electron can move into it, so the hole moves in the opposite direction and acts like a positive charge carrier.

Why is doping done in very small amounts?

A few parts per million already raise the carrier concentration enormously. Too much impurity would disturb the crystal structure and spoil the semiconductor's properties.

What is the difference between intrinsic and extrinsic semiconductors?

An intrinsic semiconductor is pure, with equal numbers of electrons and holes. An extrinsic semiconductor is doped, so one type of carrier (electrons in n-type, holes in p-type) greatly outnumbers the other.

Why does a semiconductor's resistance fall when heated?

Heating breaks more covalent bonds, creating more electron-hole pairs. The increase in carriers is much larger than the increase in collisions, so resistance falls.

Are energy bands studied mathematically in Class 12?

No. For 2026-27, energy bands in conductors, semiconductors and insulators are treated qualitatively only.