Why Doping Is Needed
The conductivity of an intrinsic semiconductor depends strongly on temperature, but at ordinary room temperature it remains very low. With so few free carriers available, no useful electronic device can be built from a pure crystal alone. The conductivity must be raised in a controlled way, and the route to doing so is the addition of impurities.
When a small amount — a few parts per million (ppm) — of a suitable impurity is added to a pure semiconductor, its conductivity increases manifold. The resulting materials are called extrinsic or impurity semiconductors. The deliberate addition of the impurity is called doping, and the impurity atoms are dopants. NIOS notes that the doping fraction can be as small as one impurity atom per $10^{8}$ atoms of the host crystal, yet it dominates the carrier population.
A dopant must not distort the original lattice. It occupies only a few host atom sites, so its atomic size has to be close to that of the host. Silicon and germanium belong to group IV; the dopants are drawn from neighbouring groups — pentavalent atoms from group V or trivalent atoms from group III — to keep the size match.
| Dopant type | Valency | Examples | What it adds | Resulting semiconductor |
|---|---|---|---|---|
| Pentavalent (donor) | 5 | Arsenic (As), Antimony (Sb), Phosphorus (P) | One extra free electron per atom | n-type |
| Trivalent (acceptor) | 3 | Indium (In), Boron (B), Aluminium (Al) | One hole per atom | p-type |
n-type Semiconductor: Pentavalent Donors
Suppose Si or Ge is doped with a pentavalent element such as arsenic. A group V atom carries five valence electrons. When it occupies a lattice site, four of those electrons form covalent bonds with the four neighbouring silicon atoms, exactly as a host atom would. The fifth electron has no bond to join. The four bonding electrons act as part of the effective core of the impurity atom, so the fifth electron is only very weakly held to its parent.
The energy needed to free this loosely bound electron is small — about $0.01\ \text{eV}$ for germanium and $0.05\ \text{eV}$ for silicon. Compare this with the energy needed to lift an electron across the full forbidden gap of the intrinsic crystal: about $0.72\ \text{eV}$ for germanium and $1.1\ \text{eV}$ for silicon. Because the donor electron is so easily released, it is free to move through the lattice even at room temperature. The pentavalent atom thus donates one electron for conduction and is called a donor impurity.
The number of conduction electrons supplied by donors depends strongly on the doping level and is essentially independent of any rise in ambient temperature. By contrast, the electron–hole pairs generated by the host Si atoms themselves increase only weakly with temperature. In a doped n-type crystal, the total electron concentration $n_e$ comes from both donor electrons and the small intrinsic contribution, while the holes $n_h$ come only from intrinsic generation. The abundance of electrons raises the rate at which holes recombine, so the hole population is driven down even further.
With a suitable doping level, the conduction electrons greatly outnumber the holes. Electrons therefore become the majority carriers and holes the minority carriers. Such a material is called an n-type semiconductor, for which
$$n_e \gg n_h$$
p-type Semiconductor: Trivalent Acceptors
A p-type semiconductor is produced by doping Si or Ge with a trivalent impurity such as boron, aluminium or indium. A group III atom has one valence electron fewer than the host. It can form covalent bonds with only three of its four silicon neighbours; the bond with the fourth neighbour is left incomplete, creating a vacancy — a hole.
Because the neighbouring Si atom needs an electron to fill the gap, an electron from a nearby bond can jump in to complete it, leaving a fresh hole at its own former site. The hole therefore moves through the lattice and is available for conduction. When the trivalent atom captures that fourth electron, it becomes effectively a negatively charged core with an associated mobile hole. One acceptor atom yields exactly one hole.
These acceptor-generated holes are in addition to the holes produced intrinsically, while the only source of conduction electrons remains intrinsic generation. The plentiful holes recombine with the small electron population, reducing the intrinsic electron concentration $n_i$ down to a smaller value $n_e$. In a p-type material, therefore, holes are the majority carriers and electrons the minority carriers. The defining relation is
$$n_h \gg n_e$$
New to carriers and the energy gap? Revisit Intrinsic Semiconductor before pushing into doping, mass action and band-level diagrams.
n-type vs p-type at a Glance
The two extrinsic types differ only in the dopant chosen, yet that single choice fixes the valency, the kind of carrier added, the majority and minority carriers, and the position of the impurity energy level. The table below collects the contrasts that NEET questions most often probe.
| Feature | n-type | p-type |
|---|---|---|
| Dopant | Pentavalent — As, Sb, P | Trivalent — B, Al, In |
| Valency of dopant | 5 | 3 |
| Impurity name | Donor | Acceptor |
| Carrier added per atom | One electron | One hole |
| Majority carrier | Electrons | Holes |
| Minority carrier | Holes | Electrons |
| Concentration relation | ne ≫ nh | nh ≫ ne |
| Fixed core left behind | Positive (ionised donor) | Negative (ionised acceptor) |
| Impurity energy level | Donor level $E_D$ just below $E_C$ | Acceptor level $E_A$ just above $E_V$ |
| Net charge of crystal | Neutral | Neutral |
Mass-Action Law and Minority Carriers
Doping does not change the carriers independently of one another. In thermal equilibrium the product of the electron and hole concentrations is fixed at a given temperature by the mass-action law:
$$n_e\, n_h = n_i^{2}$$
Here $n_i$ is the intrinsic carrier concentration of the undoped crystal at that temperature. The relation holds for both intrinsic and extrinsic semiconductors. Its consequence is decisive: if doping raises the majority-carrier concentration sharply, the minority-carrier concentration must fall in exact proportion so that the product stays equal to $n_i^{2}$. This is why an abundance of majority carriers actively suppresses the minority population — the thermally generated minority carriers find majority carriers to recombine with far more readily, so the dopant indirectly lowers the intrinsic concentration of the minority type.
Doping does not make the crystal charged, and minority carriers never vanish
Two confusions are exploited again and again. First, students assume an n-type crystal is negatively charged because it has extra electrons, or that a p-type crystal is positive. It is not: the extra electrons are exactly balanced by fixed positive donor cores, so the crystal stays electrically neutral (and likewise for p-type). Second, students forget that minority carriers still exist. In n-type material holes are few but nonzero, set by $n_h = n_i^{2}/n_e$; in p-type material the electrons satisfy $n_e = n_i^{2}/n_h$.
Remember: doping changes which carrier dominates and the carrier ratio — never the net charge. And $n_e n_h = n_i^2$ always holds in equilibrium.
Charge Neutrality of a Doped Crystal
Although doping floods the crystal with one type of mobile carrier, the material as a whole remains electrically neutral. In n-type silicon each donor that releases an electron is left as a fixed positive ion locked in the lattice; the mobile negative electron and the immobile positive core carry equal and opposite charge. In p-type silicon each acceptor that captures an electron becomes a fixed negative ion, balanced by the positive mobile hole.
NIOS states the point directly: the number of free electrons equals the total number of holes plus positively charged ions, so a semiconductor — intrinsic or doped — is electrically neutral. Overall neutrality holds because the charge of the additional mobile carriers is equal and opposite to that of the ionised cores fixed in the lattice.
The extra electron of an n-type donor is balanced by a fixed positive core; the hole of a p-type acceptor is balanced by a fixed negative core. The mobile carriers move; the ionised cores do not. The sum of all charges is zero.
Donor and Acceptor Energy Levels
Doping reshapes the band structure by introducing extra allowed energy states inside the forbidden gap. In n-type silicon a donor level $E_D$ appears slightly below the bottom $E_C$ of the conduction band. Because the gap between $E_D$ and $E_C$ is tiny, electrons sitting on the donor level need only a very small supply of energy to enter the conduction band. At room temperature most donor atoms are ionised, while very few host Si atoms are — so the conduction band is filled mainly by electrons that came from the donor impurities.
In p-type silicon an acceptor level $E_A$ appears slightly above the top $E_V$ of the valence band. With a small energy supply, an electron from the valence band jumps up to $E_A$ and ionises the acceptor negatively — equivalently, a hole sinks from $E_A$ down into the valence band. At room temperature most acceptor atoms are ionised, so the density of holes in the valence band is governed predominantly by the impurity rather than by thermal generation.
Worked Example: Doping Si with Arsenic
The mass-action law is most often tested through a single numerical step: once the majority concentration is known, the minority concentration follows from $n_h = n_i^{2}/n_e$. NCERT Example 14.2 is the standard template.
A pure Si crystal has $5 \times 10^{28}$ atoms m$^{-3}$. It is doped by 1 ppm of pentavalent arsenic, given $n_i = 1.5 \times 10^{16}$ m$^{-3}$. Find the electron and hole concentrations.
Number of As atoms doped per m$^3$ (1 part per million of host atoms):
$$N_D = \frac{5 \times 10^{28}}{10^{6}} = 5 \times 10^{22}\ \text{m}^{-3}$$
Thermally generated electrons ($n_i \sim 10^{16}$ m$^{-3}$) are negligible beside those from doping, so the electron concentration is essentially the donor concentration:
$$n_e \approx N_D = 5 \times 10^{22}\ \text{m}^{-3}$$
Apply the mass-action law $n_e n_h = n_i^{2}$ to get the minority holes:
$$n_h = \frac{n_i^{2}}{n_e} = \frac{(1.5 \times 10^{16})^{2}}{5 \times 10^{22}} = \frac{2.25 \times 10^{32}}{5 \times 10^{22}} \approx 4.5 \times 10^{9}\ \text{m}^{-3}$$
The crystal is overwhelmingly n-type: electrons outnumber holes by about thirteen orders of magnitude, yet holes never reach zero.
Notice how doping at just 1 ppm raises the electron concentration from $n_i \sim 10^{16}$ to $5 \times 10^{22}$ m$^{-3}$ — a vast increase — while simultaneously pushing the hole concentration far below its intrinsic value. This is the controlled enhancement of conductivity that makes extrinsic semiconductors the foundation of the p-n junction and every device built on it.
Extrinsic semiconductor in one screen
- Doping adds a few ppm of impurity to a pure semiconductor, raising conductivity manifold (NCERT §14.4, NIOS §28.2.2).
- n-type: pentavalent donors (As, Sb, P) add electrons; electrons are majority carriers, $n_e \gg n_h$.
- p-type: trivalent acceptors (B, Al, In) add holes; holes are majority carriers, $n_h \gg n_e$.
- Mass-action law: $n_e n_h = n_i^{2}$ in equilibrium — raising the majority population lowers the minority population.
- Charge neutrality: both n-type and p-type crystals are electrically neutral; mobile carriers balance fixed ionised cores.
- Energy levels: donor level $E_D$ just below $E_C$; acceptor level $E_A$ just above $E_V$.
- Carrier currents: for equal carrier concentrations, n-type carries more current because $\mu_e > \mu_h$ (NEET 2021).