PhysicsISC Class 12 · Electronic Devices

Semiconductors and Energy Bands: Conductors, Insulators, Intrinsic and Extrinsic Semiconductors

Updated 15 October 2026 · Fact-checked

In a solid, atomic energy levels spread into bands. The valence band and conduction band are separated by a forbidden gap $E_g$. In conductors the bands overlap (no gap); in insulators $E_g$ is large (diamond 5.4 eV); in semiconductors it is small (Si 1.1 eV, Ge 0.7 eV). Doping with pentavalent or trivalent atoms makes n-type or p-type semiconductors.

Understand Semiconductors and Energy Bands

From levels to bands (qualitative). An isolated atom such as copper has sharp levels (1s, 2s, 2p, 3s, ...). When a very large number of atoms come together in a solid, each level splits into a huge number of closely spaced levels, forming an energy band. Bands are separated by ranges of energy that electrons cannot have: forbidden bands (band gaps).

Valence band: the highest band containing the valence electrons; full or nearly full.
Conduction band: the next band above; electrons here are free to move and conduct.
Forbidden gap $E_{g}$ : the energy difference between the top of the valence band and the bottom of the conduction band.

Classification by band diagrams.

Material	Band picture	$E_{g}$	Conduction
Conductor (Cu, Al)	Bands overlap, or conduction band partly filled	No gap	Many free electrons; good conductor
Insulator (diamond, glass)	Large gap	> 3 eV (diamond, carbon, ≈ 5.4 eV)	Practically none
Semiconductor (Si, Ge)	Small gap	Si ≈ 1.1 eV, Ge ≈ 0.7 eV	Poor at 0 K; increases with temperature

At room temperature the thermal energy kT is only about 0.026 eV, but a few electrons in a semiconductor still gain enough energy to cross the small gap; in an insulator almost none can.

Conduction in metals (elementary). Free electrons drift under an electric field. When temperature rises, lattice vibrations increase collisions, so a metal's resistance increases.

Intrinsic semiconductors. Pure Si or Ge. Each atom forms four covalent bonds. Thermal energy breaks some bonds: an electron jumps to the conduction band, leaving a vacancy, a hole, in the valence band. A hole behaves like a positive charge: a neighbouring electron fills it, and the hole moves. In an intrinsic semiconductor $n_{e} = n_{h} = n_{i}$ . At 0 K it behaves as an insulator. As temperature rises, more bonds break, carriers increase, and resistance decreases (negative temperature coefficient).

Extrinsic semiconductors (doping). Adding a tiny amount of impurity (doping) greatly increases conductivity.

n-type: doped with a pentavalent impurity (phosphorus, arsenic, antimony). Four electrons bond; the fifth is loosely held and freed easily. The impurity is a donor. Majority carriers: electrons; minority carriers: holes.
p-type: doped with a trivalent impurity (boron, aluminium, gallium, indium). Only three bonds are completed, leaving a hole. The impurity is an acceptor. Majority carriers: holes; minority carriers: electrons.
Both types are electrically neutral overall: each impurity atom is neutral; when its electron (or hole) moves off, an immobile ion is left, balancing the charge.

Majority and minority carriers. Majority carriers come mainly from doping; minority carriers come from thermal breaking of bonds and are far fewer. A hole is not a proton: it is the absence of an electron in a bond.

Semiconductors and Energy Bands: formulas and key facts

Intrinsic carriers: $n_{e} = n_{h} = n_{i}$; Pure semiconductor.
n-type: $n_{e} ≫ n_{h}, n_{e} \approx N_{D}$; Donor density.
p-type: $n_{h} ≫ n_{e}, n_{h} \approx N_{A}$; Acceptor density.
Longest wavelength to create a pair: $λ_{ma x} = \frac{h c}{E _{g}} = \frac{1240}{E _{g} ( eV )} nm$; Photon energy ≥ gap.
Thermal energy: $k T \approx 0.026 eV at 300 K$; Compare with E_g.
Dopant count: $N_{d o p an t} = \frac{N _{a t o m s}}{ratio}$; e.g. 1 in 10⁷.

How to solve Semiconductors and Energy Bands questions

Most questions are classification and reasoning; a few use $E_{g}$ or doping ratios.

1Classify by $E_{g}$ : none (conductor), small ≈ 1 eV (semiconductor), large > 3 eV (insulator).
2For photon questions: $λ_{ma x} = \frac{1240}{E _{g}}$ nm; shorter wavelengths can create electron-hole pairs.
3For doping: dopant density = atoms per m³ ÷ (1 in N); this is roughly the majority carrier density.
4Pentavalent → n-type (donor, electrons majority); trivalent → p-type (acceptor, holes majority).
5Temperature questions: semiconductor resistance falls; metal resistance rises; give the reason.

Semiconductors and Energy Bands 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.

Why glass is transparent

Suppose you look through a window pane. Glass is an insulator with a band gap of several electronvolts.

Use photon energies of visible light to explain why glass does not absorb it.

Try it first, then show the solutionSolution

Visible photons have energies from $\frac{1240}{700} \approx 1.8$ eV to $\frac{1240}{400} = 3.1$ eV.
These are smaller than glass's band gap, so they cannot lift electrons across it; the light passes through.

Answer: Visible photons are too weak to bridge the gap, so glass transmits them.

Memory hook: No jump, no absorption.

Light sensors in a phone

Imagine the ambient-light sensor in a phone, made of silicon ( $E_{g}$ ≈ 1.1 eV).

Which wavelengths can it detect?

Try it first, then show the solutionSolution

$λ_{ma x} = \frac{1240}{1.1} \approx 1127$ nm.
All visible light (400 to 700 nm) and near infrared up to about 1.1 µm can create electron-hole pairs, so it responds to them.

Answer: Visible and near-infrared light up to about 1127 nm.

Memory hook: Photon energy above the gap makes carriers.

Tricks to remember Semiconductors and Energy Bands

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

P for positive holes, N for negative electrons
p-type: holes majority; n-type: electrons majority.
Use it when: Avoiding the swap (2025 Q20).
Five gives, three takes
Pentavalent donates an electron; trivalent accepts one (creates a hole).
Use it when: Doping questions.
No gap, small gap, big gap
Conductor, semiconductor, insulator.
Use it when: Band diagrams (2025 Q1(B)(vii)).
Heat helps semiconductors
Warmer semiconductor, lower resistance; warmer metal, higher resistance.
Use it when: Temperature questions.

Semiconductors and Energy Bands: solved examples

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

Example 1: Minority carriers in n-type · 1 mark · MCQBoard level

In an n-type semiconductor, the minority charge carriers are:
(a) electrons
(b) holes
(c) protons
(d) donor ions

Try it first, then show the solutionSolution

Electrons are the majority; holes, made by thermal breaking of bonds, are the minority.

Answer: (b).

Example 2: Band gap and light · 2 marks · Short answerBoard level

Gallium arsenide has a band gap of 1.43 eV. Find the longest wavelength of light that can create an electron-hole pair in it, and name the region.

Try it first, then show the solutionSolution

$λ_{ma x} = \frac{1240}{1.43} \approx 867$ nm. (1)
Longer than 700 nm: near infrared. (1)

Answer: ≈ 867 nm, near infrared.

Example 3: Band diagrams · 3 marks · Long answerHarder board level

Draw energy band diagrams of a conductor, an insulator and a semiconductor, and distinguish between them, giving one example and the energy gap of each.

Try it first, then show the solutionSolution

Conductor: valence and conduction bands overlap (no forbidden gap); e.g. copper. (1)
Insulator: filled valence band, empty conduction band, wide gap above about 3 eV; e.g. diamond (≈ 5.4 eV). (1)
Semiconductor: narrow gap, about 1 eV; e.g. silicon (1.1 eV), germanium (0.7 eV); a few electrons cross at room temperature. (1)

Answer: Overlap; wide gap; narrow gap, as described.

Example 4: Doping germanium with indium · 3 marks · Short answerHarder board level

Germanium has $4.4 \times 1 0^{28}$ atoms per m³. It is doped with indium at 1 part in $1 0^{7}$ . (a) Find the number of impurity atoms per m³. (b) What type of semiconductor results? (c) Name the majority and minority carriers and compare the majority density with the intrinsic value $2.4 \times 1 0^{19}$ m⁻³.

Try it first, then show the solutionSolution

(a) $\frac{4.4 \times 1 0 ^{28}}{1 0 ^{7}} = 4.4 \times 1 0^{21}$ m⁻³. (1)
(b) Indium is trivalent, an acceptor: p-type. (1)
(c) Majority: holes (≈ $4.4 \times 1 0^{21}$ m⁻³), about 180 times the intrinsic value; minority: electrons. (1)

Answer: $4.4 \times 1 0^{21}$ m⁻³; p-type; holes majority (≈ 180 × $n_{i}$ ), electrons minority.

Example 5: Why the gap size matters · 3 marks · Short answerHOTS / competency

At 300 K, find kT in eV ( $k = 1.38 \times 1 0^{- 23}$ J/K). Compare it with the band gaps of silicon (1.1 eV) and diamond (5.4 eV), and explain why silicon conducts a little at room temperature but diamond does not.

Try it first, then show the solutionSolution

$k T = \frac{1.38 \times 1 0 ^{- 23} \times 300}{1.6 \times 1 0 ^{- 19}} \approx 0.026$ eV. (1)
$\frac{E _{g}}{k T}$ : silicon ≈ 42; diamond ≈ 210. (1)
Only a very small fraction of electrons have energies many times kT. For a gap of 42kT this tiny fraction is still enough to give some carriers; for 210kT it is effectively zero. (1)

Answer: ≈ 0.026 eV; silicon's gap is small enough for some thermal excitation, diamond's is not.

Example 6: A fire-alarm thermistor · 4 marks · Case-basedHOTS / competency

Suppose a fire alarm uses a semiconductor thermistor whose resistance changes with temperature.
(i) How does its resistance change as it heats up? (1 mark)
(ii) Explain using the band picture. (1 mark)
(iii) How would a copper wire behave instead, and why? (1 mark)
(iv) Could light of 600 nm also create carriers in a silicon thermistor ( $E_{g}$ = 1.1 eV)? (1 mark)

Try it first, then show the solutionSolution

(i) It decreases.
(ii) More electrons gain enough thermal energy to cross the gap, creating more electron-hole pairs (more carriers).
(iii) Copper's resistance rises: its carrier number is fixed, and stronger lattice vibrations cause more collisions.
(iv) Yes: a 600 nm photon has $\frac{1240}{600} \approx 2.07$ eV > 1.1 eV.

Answer: (i) Falls; (ii) more carriers; (iii) copper's rises (more collisions); (iv) yes.

Where students lose marks in Semiconductors and Energy Bands

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

Mistake: Choosing "protons" as carriers in p-type
The ISC 2024 examiners (Q1(A)(vii)) saw candidates confuse holes with protons.
Do this: Holes are vacancies of electrons, not particles in the nucleus.
Usually costs: 1 mark.
Mistake: Answering for a conductor
In 2025 Q1(B)(i) many said resistance increases with temperature, which is true only for metals.
Do this: Semiconductor: resistance decreases with temperature (negative temperature coefficient).
Usually costs: 1 mark.
Mistake: No forbidden band means insulator
The 2025 examiners (Q1(B)(vii)) saw conductors confused with insulators.
Do this: No forbidden gap (overlapping bands) means a conductor (metal).
Usually costs: 1 mark.
Mistake: Mixing up trivalent and pentavalent
Also in 2025 Q20(i).
Do this: Pentavalent → n-type (electrons); trivalent → p-type (holes).
Usually costs: 1 mark.

Exam tips

Label valence band, conduction band and forbidden gap in every band diagram.
Give numerical gaps: Si 1.1 eV, Ge 0.7 eV, diamond about 5.4 eV.
State that doped semiconductors are electrically neutral.
Keep band theory qualitative, as the syllabus requires.

Practice questions on Semiconductors and Energy Bands

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
At absolute zero, a pure semiconductor behaves as:
(a) a conductor
(b) an insulator
(c) a superconductor
(d) an n-type semiconductor
Hint
No thermal energy to cross the gap.
Answer
(b).
Q2 · 1 mark · Short answerBoard level
Name one pentavalent and one trivalent impurity used for doping silicon, and the type each produces.
Hint
Group 15 and group 13.
Answer
Phosphorus or arsenic → n-type; boron or indium → p-type.
Q3 · 1 mark · Assertion-reasonHarder board level
Assertion (A): An n-type semiconductor is electrically neutral.
Reason (R): For every free electron supplied by a donor, a fixed positive donor ion is left in the crystal.
(a) Both true, R explains A. (b) Both true, R does not explain A. (c) A true, R false. (d) A false, R true.
Hint
Charge balance.
Answer
(a).
Q4 · 2 marks · Short answerHarder board level
What is a hole? How does it move?
Hint
Absence of an electron.
Answer
A vacancy in a covalent bond left when an electron is freed; it behaves as a positive charge. A neighbouring bonded electron fills it, creating a new vacancy where it came from, so the hole moves opposite to the electrons.
Q5 · 2 marks · Short answerHarder board level
Find the longest wavelength that can create an electron-hole pair in diamond ( $E_{g}$ = 5.4 eV). Use it to explain why diamond is transparent.
Hint
1240/5.4.
Answer
≈ 230 nm (ultraviolet). Visible photons (400 to 700 nm) have too little energy to cross the gap, so they are not absorbed.
Q6 · 2 marks · Short answerHOTS / competency
Silicon ( $5 \times 1 0^{28}$ atoms m⁻³, $n_{i} = 1.5 \times 1 0^{16}$ m⁻³) is doped with phosphorus at 1 part in $1 0^{8}$ . Find the donor density and how many times the intrinsic carrier density it is.
Hint
Divide by $1 0^{8}$ .
Answer
$5 \times 1 0^{20}$ m⁻³; about $3.3 \times 1 0^{4}$ times $n_{i}$ .
Q7 · 2 marks · Short answerHOTS / competency
How does the resistance of a semiconductor change with temperature, and how does this differ from a metal? Explain both.
Hint
Number of carriers vs collisions.
Answer
Semiconductor: resistance decreases (more electron-hole pairs as bonds break). Metal: resistance increases (carrier number fixed, more collisions with vibrating ions).
Q8 · 4 marks · Case-basedHOTS / competency
Case study: Suppose an engineer chooses a material for solar cells.
(i) What is the longest wavelength silicon ( $E_{g}$ = 1.1 eV) can use? (1 mark)
(ii) Why would an insulator such as diamond be useless? (1 mark)
(iii) Why would a metal also be useless as the absorbing material? (1 mark)
(iv) In a silicon cell, which type is formed by doping with boron? (1 mark)
Hint
1240/1.1.
Answer
(i) ≈ 1127 nm. (ii) Its gap (5.4 eV) is too large: visible sunlight cannot create carriers. (iii) It has no gap: no junction field or separation of light-created carriers is possible, so no emf. (iv) p-type.

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 p-type?Show answer
Holes.
3.Impurity for n-type?Show answer
Pentavalent (donor), e.g. phosphorus or arsenic.
4.Material with no forbidden gap?Show answer
A conductor (metal).
5.Effect of temperature on a semiconductor's resistance?Show answer
It decreases.

Semiconductors and Energy Bands: frequently asked questions

Why is germanium's gap smaller than silicon's important?

More carriers are thermally produced in germanium at room temperature, so it conducts more but also has larger leakage currents.

Is a p-type semiconductor positively charged?

No. It is neutral overall: each mobile hole is balanced by a fixed negative acceptor ion.

Why is doping in such tiny amounts enough?

Even 1 part in 10^8 gives far more carriers than are present intrinsically at room temperature, so conductivity rises enormously.

What does the ISC 2027 syllabus need here?

Qualitative ideas of energy bands and band diagrams to distinguish conductors, insulators and semiconductors, typical gaps (carbon, Ge, Si), electrons and holes, intrinsic and extrinsic semiconductors, doping, donors and acceptors, and majority and minority carriers.

How was this asked recently?

In 1-mark items: ISC 2024 Q1(A)(vii) (majority carriers in p-type), ISC 2025 Q1(A)(v) (n-type), Q1(B)(i) (resistance and temperature) and Q1(B)(vii) (absent forbidden band).