A full lesson: a cartoon movie to start, then the donor atom and its free fifth electron, the acceptor atom and its wandering hole, the mass-action law nenh = ni², the energy-band picture with donor and acceptor levels, why a pinch of impurity makes silicon a million times better at conducting, NEET shortcuts, problems to solve yourself, and 12 MCQs.
A full bus is pure silicon: every seat (bond) taken, nobody can move. One extra rider is a free electron (n-type). One empty seat is a hole that moves backwards as people shift forwards (p-type).
Meera's school bus is packed: every seat taken, and nobody can change places. That is pure silicon near absolute zero: every electron is locked in a covalent bond and nothing can carry a current. Now one extra child squeezes in. There is no seat for him, so he walks freely up and down the aisle. That is the free electron given by a donor atom.
At the next stop, one child gets off. The child behind slides forward into the empty seat, then the next one does the same. The children move forwards, but the empty seat travels backwards. That empty seat is a hole, given by an acceptor atom. Replace just one silicon atom in a million by such an impurity and the conductivity jumps about a million times. This is doping, and it is how every chip, LED and solar cell is made.
"Extra rider means n-type, n for negative carriers. Empty seat means p-type, p for positive holes. But remember: the bus as a whole is still neutral. Nobody brought extra charge from outside."
"If I put a phosphorus atom (5 valence electrons) in place of one silicon atom, does the crystal become negatively charged?"
No. A neutral phosphorus atom brings its own extra proton along with its fifth electron: its core has one more +e than a silicon core. When the fifth electron wanders off, it leaves behind a fixed P⁺ ion. Mobile −e plus fixed +e: the crystal stays neutral.
Line up ten coins or bottle caps in a row with one gap at the right end. Slide the coin next to the gap into it, then the next one, and so on. Watch the gap march to the left while every coin moved one step right. That is exactly how a hole moves through a crystal.
Silicon and germanium are in group 14: four valence electrons, four covalent bonds. A pentavalent atom (group 15: P, As, Sb) of about the same size takes the place of one Si atom. Four of its five electrons bond with the four neighbours. The fifth has no bond to join; it is held only weakly by its parent ion. Tap any Si atom to replace it with a donor, choose the dopant, and change the temperature.
Compare with the band gap: 0.72 eV in Ge, 1.1 eV in Si. Thermal energy at room temperature is kT ≈ 0.026 eV, enough to free almost every donor electron.
Each donor atom gives one free electron and becomes a fixed positive ion. Electrons are the majority carriers: nₑ ≫ nₕ.
The few thermally made holes now meet many more electrons and recombine faster. So the hole density drops below its intrinsic value nᵢ.
"Count the electrons: five come in, four are used in bonds, one is free. That free one does not depend on heating the crystal hard. It is a gift from the donor. That is why the free electron count is set by the doping level, not by the temperature."
A trivalent atom (group 13: B, Al, Ga, In) has only three valence electrons. It bonds with three neighbours; the fourth bond is short of one electron. That vacancy is a hole. A bonded electron next door can jump into it, so the hole appears where that electron was. Press Let an electron hop, or switch on an electric field and watch which way holes and electrons drift.
| Type | Dopant (valency) | Name | Majority | Minority | Fixed ion left |
|---|---|---|---|---|---|
| n-type | P, As, Sb (5) | donor | electrons | holes | positive (+) |
| p-type | B, Al, Ga, In (3) | acceptor | holes | electrons | negative (−) |
"The hole is not a particle that flies in from outside. It is a missing electron in a bond. When the electrons hop one way, the hole goes the other way, so a hole behaves like a positive charge, +e. And once the acceptor has taken its extra electron, it is a fixed negative ion that can never move."
"In a p-type crystal connected to a battery, the holes move towards which terminal?"
Towards the negative terminal, along the field, just like positive charges. The electrons that hop into the holes move towards the positive terminal. Both motions give a conventional current in the same direction.
In thermal equilibrium, electrons and holes are created in pairs and destroyed by recombination. The balance fixes their product, whatever the doping. Slide the doping level and watch the two carrier densities on a log scale. They move like a see-saw balanced at nᵢ.
Valid in thermal equilibrium for pure and doped samples alike. nᵢ depends only on the material and the temperature.
At room temperature almost every dopant atom is ionised, and N_D ≫ nᵢ, so the thermal pairs add almost nothing.
Use nᵢ squared. Silicon with N_D = 5 × 10²² m⁻³: nₕ = (1.5 × 10¹⁶)² / 5 × 10²² = 4.5 × 10⁹ m⁻³.
"On a log scale the law becomes beautiful: log nₑ plus log nₕ is fixed. Whatever one side gains, the other side loses, like a see-saw with its pivot at nᵢ. Push the electrons up a million times and the holes drop a million times."
In the band picture the impurity adds a new energy level inside the forbidden gap. The donor level E_D sits just below the conduction band E_C; the acceptor level E_A sits just above the valence band E_V. Change the temperature and watch where the carriers come from. (Gaps are drawn to show the idea, not to scale.)
Donor electrons need only a tiny jump to reach the conduction band. Ionised donors stay behind as fixed + charges on the E_D line.
Valence electrons jump up to E_A, leaving holes in the valence band. Filled acceptors become fixed − ions.
Very cold: donors hold their electrons. Room temperature: all donors ionised, nₑ ≈ N_D. Very hot: band-to-band pairs swamp the doping and the sample behaves as intrinsic.
"Draw it like this in the exam: dashed line just under the conduction band for n-type, dashed line just over the valence band for p-type. NEET loves to ask exactly where that line is."
A silicon bar 1 cm long with a 1 mm² cross-section is connected to a 1.5 V cell and a small bulb. Silicon has about 5 × 10²⁸ atoms per m³. Choose the dopant type and how many Si atoms share one impurity atom. Conductivity σ = e(nₑμₑ + nₕμₕ), with μₑ = 0.135 and μₕ = 0.048 m²/V·s for silicon.
A boron-doped wafer with a phosphorus-doped skin. The meeting of the two is the p-n junction of the next lesson.
Every transistor in a phone is made of regions doped n or p to exact levels, sometimes one impurity in 10⁸ atoms.
For the same doping, n-type silicon conducts about 2.8 times better than p-type because electrons are more mobile than holes.
"Pentavalent → n" and "trivalent → p". P, As, Sb give n-type; B, Al, Ga, In give p-type.
Minority = nᵢ² ÷ majority. Never divide nᵢ (not squared) by N.
Electrons outnumber holes by the square of N_D/nᵢ. N_D = 10⁶ nᵢ gives a ratio of 10¹².
Acceptors swallow donor electrons first (compensation). The leftover decides the type.
Every mobile carrier has a fixed ion partner. n-type and p-type crystals carry no net charge.
In an extrinsic sample keep only the majority term: σ ≈ e N_D μₑ (n-type) or e N_A μₕ (p-type).
"Three marks are sitting here. Valency decides the type. Neutrality is always true. And the minority density is n i squared over the majority density. Learn these three lines and you will not lose a mark on this topic."
Cover the solution, write the formula, and solve on paper. Open the hint only if you are stuck for more than a minute.
A pure Si crystal has 5 × 10²⁸ atoms m⁻³. It is doped by 1 ppm concentration of pentavalent As. Calculate the number of electrons and holes. Given nᵢ = 1.5 × 10¹⁶ m⁻³.
1 ppm means 1 atom in 10⁶. Each As atom gives one free electron, then use nₑnₕ = nᵢ².
Germanium has nᵢ = 2.5 × 10¹⁹ m⁻³. It is doped with indium to N_A = 6.25 × 10²² m⁻³. Find the electron density. What type is the sample?
Indium is trivalent: acceptor. nₕ ≈ N_A, nₑ = nᵢ² / N_A.
In an n-type Si sample, nₑ = 10²² m⁻³ and nᵢ = 10¹⁶ m⁻³. Find nₕ and the ratio nₑ/nₕ.
nₑ/nₕ = nₑ² / nᵢ².
Find the conductivity of pure Si (nᵢ = 1.5 × 10¹⁶ m⁻³) and of the As-doped Si of Problem 1. Take μₑ = 0.135 and μₕ = 0.048 m²/V·s, e = 1.6 × 10⁻¹⁹ C. By what factor does doping increase σ?
Pure: σ = e nᵢ (μₑ + μₕ). Doped n-type: σ ≈ e N_D μₑ.
A Si sample (nᵢ = 1.5 × 10¹⁶ m⁻³) contains N_D = 4 × 10²² m⁻³ phosphorus and N_A = 10²² m⁻³ boron. Is it n-type or p-type? Find nₑ and nₕ.
Acceptors capture donor electrons first. Net donors = N_D − N_A.
Take e = 1.6 × 10⁻¹⁹ C. Write the formula first, then click.
Score: 0 / 12
| If the question says… | Write this first | The trap |
|---|---|---|
| "doped with As / P / Sb" | pentavalent → donor → n-type | Thinking "n" means "neutral" or that the crystal is negative |
| "doped with B / Al / Ga / In" | trivalent → acceptor → p-type | Calling indium a donor because it is a metal |
| "majority / minority carriers" | n: e⁻ majority · p: holes majority | Forgetting that both carriers are always present |
| "nᵢ = …, doped to N = …, find the other carrier" | minority = nᵢ² / N | Dividing nᵢ (not nᵢ²) by N |
| "charge on an n-type crystal" | zero, it is neutral | Answering "negative" because electrons are majority |
| "position of donor / acceptor level" | E_D just below E_C · E_A just above E_V | Swapping the two levels |
| "both donors and acceptors added" | type set by the larger one, n ≈ |N_D − N_A| | Adding N_D and N_A |
| "conductivity of doped sample" | σ ≈ e N μ(majority) | Using nᵢ for a doped sample |
Put a p-region and an n-region side by side in one crystal and electrons and holes rush across the boundary. Lesson 5 builds the p-n junction, its depletion region and barrier potential from exactly these carriers and fixed ions.
Values used: e = 1.6 × 10⁻¹⁹ C, k = 8.62 × 10⁻⁵ eV/K (kT ≈ 0.026 eV at 300 K), Si: E_g = 1.1 eV, nᵢ = 1.5 × 10¹⁶ m⁻³, 5 × 10²⁸ atoms m⁻³, μₑ = 0.135 and μₕ = 0.048 m²/V·s; Ge: E_g = 0.72 eV. Donor/acceptor energies in Si: P 0.045, As 0.054, Sb 0.039, B 0.045, Al 0.057, In 0.16 eV.
Follow the four steps: understand it, remember it, try it, then prove it with past papers.