Extrinsic Semiconductors: Doping, n-type & p-type

Doping: The Conductivity Multiplier

Intrinsic conductivity is too feeble for devices. The fix: add a few parts per million (ppm) of a suitable impurity and the conductivity multiplies enormously. The result is an extrinsic (doped) semiconductor; the deliberate addition is doping, the impurity atoms are dopants.

The size rule: the dopant must not distort the lattice — its atoms must be nearly the same size as Si/Ge, occupying only a few original lattice sites. Hence dopants come from the neighbouring columns of group-IV Si/Ge:

• Pentavalent (valency 5): Arsenic (As), Antimony (Sb), Phosphorus (P) — donors.
• Trivalent (valency 3): Indium (In), Boron (B), Aluminium (Al) — acceptors.

Two different dopant families → two entirely different semiconductor types.

n-type: The Electron Donors

Put a pentavalent atom in a Si site: four of its five valence electrons bond with the four Si neighbours; the fifth is only weakly bound (the bonding four screen the core). Freeing it costs merely ~0.01 eV in Ge, ~0.05 eV in Si — compare the full gap of 0.72 eV (Ge) / 1.1 eV (Si). At room temperature that fifth electron is free.

• Each dopant donates one conduction electron → donor impurity.
• Donated electrons depend on doping level, not temperature; intrinsic generation adds a weak temperature-dependent trickle (with equal holes).
• Extra electrons speed up recombination of holes → hole population falls below $n_i$.

Result: electrons majority, holes minority —

$$n_e \gg n_h \qquad \text{(n-type)}$$

p-type: The Hole Makers

Put a trivalent atom in: it bonds with only three neighbours; the fourth bond has a vacancy — a hole. A neighbouring bound electron can jump in, moving the hole outward: one acceptor atom yields one conduction hole (and becomes an effectively negative fixed core when it captures that fourth electron).

Result: holes majority, electrons minority —

$$n_h \gg n_e \qquad \text{(p-type)}$$

Key Point: The crystal stays electrically neutral overall — the mobile carriers' charge is exactly balanced by the immobile ionised dopant cores. And in BOTH types, the abundant majority carriers accelerate recombination of the minority ones: doping indirectly suppresses minority carriers below $n_i$.

Energy Levels of Dopants & the Mass-Action Law

Doping adds new energy states in the gap:

• n-type: donor level $E_D$ sits slightly below $E_C$ — electrons hop from $E_D$ into the conduction band with the tiniest energy supply. At room temperature most donors are ionised while only ~$10^{12}$ Si atoms are (per the intrinsic trickle) — so CB electrons come overwhelmingly from donors.
• p-type: acceptor level $E_A$ sits slightly above $E_V$ — valence electrons easily jump up to $E_A$ (equivalently, holes from $E_A$ 'sink' into the valence band; remember: electrons rise, holes sink when given energy). Most acceptors ionise at room temperature.

The law that rules all doping

In thermal equilibrium, whatever the doping:

$$\boxed{n_en_h = n_i^2}$$

Raise one carrier type and the other must fall in exact proportion — the quantitative form of minority-carrier suppression.

NCERT Example 14.2 — the canonical calculation

Pure Si: $5 \times 10^{28}$ atoms/m³, doped 1 ppm with As; $n_i = 1.5 \times 10^{16}$ m⁻³.

1. Donors: $N_D = \dfrac{5 \times 10^{28}}{10^6} = 5 \times 10^{22}$ m⁻³. Since this dwarfs $n_i$: $n_e \approx 5 \times 10^{22}$ m⁻³.
2. Mass action: $n_h = \dfrac{n_i^2}{n_e} = \dfrac{2.25 \times 10^{32}}{5 \times 10^{22}} \approx 4.5 \times 10^{9}$ m⁻³.

One ppm of dopant boosted electrons a million-fold over $n_i$ and crushed holes by nearly seven orders of magnitude.

[JEE Tip] The mass-action division $n_h = n_i^2/n_e$ is the chapter's most-set numerical — recognise Example 14.2's numbers on sight. [NEET Important] Level positions: $E_D$ just below $E_C$ (n-type); $E_A$ just above $E_V$ (p-type) — swapping them is the classic wrong option.

Solved Examples

Example 1: The canonical doping calculation (NCERT Example 14.2)

Pure Si ($5 \times 10^{28}$ atoms/m³) is doped with 1 ppm pentavalent As. Find $n_e$ and $n_h$ ($n_i = 1.5 \times 10^{16}$ m⁻³).

Solution:

1. Donor density: $N_D = 5 \times 10^{28} \times 10^{-6} = 5 \times 10^{22}$ m⁻³.
2. Electrons: thermal contribution (~$10^{16}$) is negligible beside $N_D$: $n_e \approx 5 \times 10^{22}$ m⁻³.
3. Holes (mass action): $n_h = \dfrac{n_i^2}{n_e} = \dfrac{(1.5 \times 10^{16})^2}{5 \times 10^{22}} = 4.5 \times 10^{9}$ m⁻³.
4. Takeaway: majority set by doping; minority set by $n_i^2/n_{majority}$ — the two-step template of every such numerical.

Example 2: Classify the dopants [Board Rapid]

Classify as donor or acceptor for Si: As, B, P, In, Sb, Al.

Solution:

1. Donors (pentavalent): As, P, Sb — give n-type.
2. Acceptors (trivalent): B, In, Al — give p-type.
3. Takeaway: memory hook — donors from group 15 ('Pentavalent Antimony-Arsenic-Phosphorus'), acceptors from group 13 ('Boron-Aluminium-Indium').

Example 3: Why the fifth electron is nearly free [Board Conceptual]

Explain why freeing the pentavalent dopant's fifth electron takes only ~0.05 eV in Si while the intrinsic gap is 1.1 eV.

Solution:

1. Four of the dopant's electrons are locked in bonds — the fifth sees them as part of the atom's effective core.
2. It orbits a net +1 core through the crystal, loosely bound (a hydrogen-like orbit widened and weakened by the medium).
3. Ionisation energy: ~0.05 eV (Si), ~0.01 eV (Ge) — versus the full band jump of 1.1/0.72 eV.
4. Takeaway: kT at room temperature (~0.026 eV) suffices to ionise essentially ALL donors — but almost no intrinsic bonds.

Example 4: The acceptor's negative core [Conceptual]

Why is the trivalent dopant drawn as a fixed core of ONE negative charge with an associated hole?

Solution:

1. The trivalent atom bonds three neighbours; the fourth bond lacks an electron — a hole.
2. When a neighbouring electron fills that bond, the dopant (now holding an extra electron) becomes effectively negative — and immobile.
3. The hole, released into the lattice, conducts; the -1 core stays.
4. Takeaway: mirror image of the donor's fixed +1 core with its freed electron — and the reason the crystal stays neutral overall.

Example 5: Mass action both ways [JEE Numerical]

A Ge sample ($n_i = 2.4 \times 10^{19}$ m⁻³) is doped so that $n_h = 4 \times 10^{22}$ m⁻³. Find $n_e$ and identify the type.

Solution:

1. Mass action: $n_e = \dfrac{n_i^2}{n_h} = \dfrac{(2.4 \times 10^{19})^2}{4 \times 10^{22}}$.
2. Compute: $n_e = \dfrac{5.76 \times 10^{38}}{4 \times 10^{22}} = 1.44 \times 10^{16}$ m⁻³.
3. Type: $n_h \gg n_e$ → p-type (acceptor-doped).
4. Takeaway: the law works in both directions; the bigger carrier names the type.

Example 6: Doping level for a target [JEE Reverse]

What ppm of donor doping makes $n_e = 10^{22}$ m⁻³ in Si ($5 \times 10^{28}$ atoms/m³)?

Solution:

1. Assume full ionisation: $N_D = n_e = 10^{22}$ m⁻³.
2. Fraction: $\dfrac{10^{22}}{5 \times 10^{28}} = 2 \times 10^{-7}$ = 0.2 ppm.
3. Takeaway: sub-ppm doping controls carrier densities across decades — semiconductor engineering is chemistry at the parts-per-million level.

Example 7: Where are the dopant levels? [NEET Sketch]

Draw/describe the band diagrams of n-type and p-type Si at T > 0 K.

Solution:

1. n-type: donor level $E_D$ a sliver below $E_C$; at room temperature most donors ionised — many CB electrons (mostly from donors), very few VB holes.
2. p-type: acceptor level $E_A$ a sliver above $E_V$; most acceptors ionised — many VB holes, very few CB electrons.
3. Takeaway: 'donor near the ceiling's floor, acceptor near the floor's ceiling' — and remember electrons rise / holes sink on gaining energy.

Example 8: Charge neutrality check [Conceptual]

An n-type sample has $n_e = 10^{22}$ m⁻³ mobile electrons. Why isn't the crystal negatively charged?

Solution:

1. Each donated electron leaves behind an ionised donor core of charge +1 fixed in the lattice.
2. Mobile charge ($-10^{22}$ per m³) is exactly cancelled by immobile core charge ($+10^{22}$).
3. NCERT: 'the crystal maintains an overall charge neutrality'.
4. Takeaway: n-type ≠ negatively charged; p-type ≠ positively charged. Type describes CARRIERS, not net charge — a perennial misconception question.

Example 9: Minority suppression, quantified [JEE Insight]

In Example 14.2, holes fell from $n_i = 1.5 \times 10^{16}$ to $4.5 \times 10^9$ m⁻³. Explain the mechanism.

Solution:

1. Doping floods the crystal with electrons ($5 \times 10^{22}$ m⁻³).
2. Any thermally created hole now meets an electron almost immediately — recombination rate for holes soars.
3. Equilibrium re-balances at $n_h = n_i^2/n_e$ — nearly seven orders below intrinsic.
4. Takeaway: NCERT's line — the dopant 'indirectly helps to reduce the intrinsic concentration of minority carriers'. Doping doesn't just add majority carriers; it actively destroys minority ones.

Example 10: Compound semiconductor twist [JEE Extra]

In GaAs, how can the semiconductor type change WITHOUT conventional dopants?

Solution:

1. Ideal GaAs has Ga : As = 1 : 1.
2. NCERT Points to Ponder: a change in the stoichiometric ratio — Ga-rich (Ga$_{1.1}$As$_{0.9}$) or As-rich (Ga$_{0.9}$As$_{1.1}$) — also changes the carrier type.
3. Defects generally control semiconductor properties in many ways.
4. Takeaway: doping is one route; in compounds, off-stoichiometry is another — a subtle assertion-reason candidate.