Extrinsic Semiconductors & Doping

Pure silicon is almost a waste of space. A cubic metre of it holds about $5\times10^{28}$ atoms, yet at room temperature only about $1.5\times10^{16}$ of its electrons are free to move — roughly one mobile charge for every three trillion atoms. That is why the previous article called intrinsic silicon a "reluctant" conductor: heat it and it conducts a little better, but as a material to build with it is hopeless. You cannot make a switch, a diode, or a transistor from something whose conductivity you cannot control. The trick that turned this dead crystal into the foundation of every phone, laptop, and solar panel on Earth is almost absurdly delicate: deliberately dirty it. Slip in one foreign atom for every million or so silicon atoms — an impurity with one electron too many, or one too few — and the conductivity leaps by a factor approaching a million, and, crucially, you get to choose whether the new carriers are negative electrons or positive holes. A crystal grown this way is called extrinsic (its carriers come from outside), and the process is doping: pentavalent atoms make n-type silicon (electrons in the majority), trivalent atoms make p-type silicon (holes in the majority). This article is the hinge of the whole chapter. Everything downstream — the p–n junction, the diode, the transistor, every logic gate — is just n-type and p-type silicon pressed together in clever arrangements. We will show why a single well-chosen impurity atom donates a nearly-free electron, why the crystal stays electrically neutral despite gaining carriers, and how one iron law — the mass-action law $n_e n_h = n_i^2$ — governs the bookkeeping, all with the Bohr model and school algebra you already own.

Key takeaways

Doping means adding a controlled trace impurity — about one atom per million — to a pure semiconductor to set its conductivity and its carrier type on purpose. The doped crystal is called extrinsic.
Pentavalent donors (P, As, Sb) make n-type silicon. Four of the impurity's five valence electrons bond to neighbours; the fifth is barely held (about $0.05\ \text{eV}$ ) and breaks free at room temperature, leaving a fixed positive donor ion. Electrons are the majority carriers.
Trivalent acceptors (B, Al, In) make p-type silicon. With only three valence electrons the impurity leaves one bond short, and a neighbouring electron fills it — creating a mobile hole and a fixed negative acceptor ion. Holes are the majority carriers.
The crystal stays electrically neutral. Every freed electron is matched by a fixed positive ion; every hole by a fixed negative ion. n-type is not negatively charged, and p-type is not positively charged.
The mass-action law is the master equation: at a given temperature $n_e\,n_h = n_i^2$ , always. Boosting one carrier suppresses the other — for n-Si with $n_e=10^{22}\ \text{m}^{-3}$ the minority holes fall to $n_h = 2.25\times10^{10}\ \text{m}^{-3}$ .
Doping raises the useful carrier count by up to a million-fold while barely disturbing the crystal, which is exactly what makes silicon an engineering material rather than a curiosity.
Key idea: a trace of pentavalent or trivalent impurity fixes both the sign and the number of the mobile charge carriers in a semiconductor — donors give electron-rich n-type, acceptors give hole-rich p-type — while the crystal stays neutral and the product of the two carrier concentrations stays locked at n_i².

The core insight

Replacing about one silicon atom in a million with a pentavalent donor floods the crystal with nearly-free electrons (n-type) or, with a trivalent acceptor, with mobile holes (p-type); the sample stays electrically neutral because each freed carrier is balanced by a fixed impurity ion, and at any fixed temperature the electron and hole concentrations obey the mass-action law $n_e n_h = n_i^2$ , so raising the majority carrier drives the minority carrier down by the same factor.

Where this lives today

Every silicon chip is a sculpture in doping. A modern processor is built by defining, region by region on a single-crystal wafer, exactly which patches are n-type and which p-type — using photolithographic masks and either ion implantation or thermal diffusion. Billions of transistors on a fingernail-sized die are nothing but precisely doped n and p regions in intimate contact. The same control shapes a solar cell, an LED, and the sensor in every digital camera, and it tunes other properties too — heavily doped silicon conducts almost like a metal and serves as on-chip wiring, and doping gallium arsenide, gallium nitride, and silicon carbide underlies fibre-optic lasers, blue LEDs, and electric-car power electronics. The next article presses an n-type and a p-type region together and asks what happens at the boundary — the p–n junction, the device from which diodes, transistors, and all the rest are built.

The physical picture

Silicon is tetravalent: each atom brings four valence electrons and shares one with each of its four nearest neighbours, so every atom sits at the centre of four covalent bonds, every bonding electron accounted for. In a perfect, cold crystal there are no spare charges to move — it is an insulator in all but name. Intrinsic conduction only happens when heat occasionally rips a bonding electron loose, creating one free electron and one hole, always in pairs, so that $n_e = n_h = n_i$ . At $300\ \text{K}$ this thermal generation is feeble because the band gap of silicon, $E_g = 1.1\ \text{eV}$ , is about forty times the available thermal energy $kT \approx 0.026\ \text{eV}$ . Doping sidesteps this bottleneck. Instead of waiting for heat to make carriers in pairs, we plant impurity atoms that hand over a single carrier of a chosen sign, at an energy cost far below the band gap. Replace one silicon atom with phosphorus, which has five valence electrons: four slot into the four covalent bonds the site requires, but the fifth has no bond to join and is held only weakly, orbiting the impurity's mild extra positive charge in a vast, loose orbit. It takes only about $0.05\ \text{eV}$ — comparable to the room-temperature thermal energy, and about twenty times less than the band gap — to shake it loose into the crystal. What is left behind is a phosphorus atom minus one electron: a fixed positive ion, locked in the lattice, unable to move or carry current.

Figure

Left a pentavalent phosphorus atom in silicon keeps four bonds and donates one loosely bound electron becoming a fixed positive ion making n-type material. Right a trivalent boron atom leaves one bond incomplete which a neighbour electron fills creating a mobile hole and a fixed negative ion making p-type material

The mirror image builds p-type. Replace a silicon atom with boron, which has only three valence electrons: three bonds complete, the fourth left one electron short — a vacancy. A neighbouring bonding electron, jostled by thermal energy, hops in to complete the boron's bond, filling this vacancy but opening a new one where it came from. A hole has appeared, free to wander from bond to bond, behaving in every way like a mobile positive charge. The boron, having captured an extra electron, becomes a fixed negative ion. Because the impurity accepts an electron, it is an acceptor, and the material, rich in holes, is p-type. Two facts deserve emphasis, because they trip up more exam candidates than any others. First, the crystal is electrically neutral throughout: each mobile electron is balanced by the fixed positive donor ion it left, each mobile hole by a fixed negative acceptor ion, so adding carriers adds no net charge — it only unlocks charges that were already there. Second, the impurity ions are fixed: bonded into the lattice, they carry no current, and only the freed electrons or holes drift under a voltage.

Donors, acceptors, and majority versus minority carriers

The impurity's group in the periodic table is the whole story. Silicon and germanium sit in group 14 with four valence electrons. One column right, in group 15 — phosphorus, arsenic, antimony — atoms have five valence electrons: one too many, a spare donated to the conduction pool. These are donors, making n-type material with electrons abundant. One column left, in group 13 — boron, aluminium, gallium, indium — atoms have three: one too few, a permanent vacancy that behaves as a mobile hole. These are acceptors, making p-type material with holes abundant. Fix the naming in memory: pentavalent → donor → n-type → negative carriers; trivalent → acceptor → p-type → positive carriers. Doping does not abolish the other carrier — it merely makes it rare. In n-type silicon electrons are the majority carriers and holes the minority carriers; in p-type it is the reverse. Both are always present, because thermal pair generation never stops; what doping changes is the ratio, which in a well-doped sample can reach a hundred billion to one. The energy-band view makes the mechanism vivid. A donor sits as a filled level a mere $0.045\ \text{eV}$ or so below the conduction band — a tiny step compared with the $1.1\ \text{eV}$ gap — so its electron reaches the conduction band with ease. An acceptor sits as an empty level just above the valence band, ready to catch an electron from below and leave a hole behind.

Figure

Band diagrams of doped silicon. In n-type a filled donor level lies about 0.045 eV below the conduction band so its electrons ionize easily leaving fixed positive ions. In p-type an empty acceptor level lies just above the valence band capturing electrons and leaving mobile holes. Both shallow levels are tiny compared with the 1.1 eV band gap

The mass-action law: the product that never changes

Doping lets you set one carrier concentration at will. The other is then forced by a law that holds in any semiconductor at thermal equilibrium — intrinsic or extrinsic, n-type or p-type. At a fixed temperature the product of the electron and hole concentrations is a constant:

n_e\,n_h = n_i^2. \qquad (1)

where: $n_e$ is the free-electron concentration, $n_h$ the hole concentration, and $n_i$ the intrinsic carrier concentration ( $n_i = 1.5\times10^{16}\ \text{m}^{-3}$ for silicon at $300\ \text{K}$ ). The physical reading is a balance of two rates. Thermal generation makes pairs at a rate set only by temperature and the band gap, independent of the carriers already present. Recombination, where an electron drops into a hole and both vanish, runs at a rate proportional to the product $n_e n_h$ , since it needs one of each to meet. In equilibrium the two balance, so $n_e n_h$ equals a constant fixed by temperature alone — and evaluating it in the intrinsic case ( $n_e = n_h = n_i$ ) gives $n_i^2$ , hence equation (1) for every sample at that temperature. (A full treatment uses the Fermi–Dirac distribution, beyond this article; we quote only its one needed consequence, the mass-action law.) The consequence is a seesaw. Push the majority carrier up by doping and the minority carrier is driven down by exactly the same factor, so their product stays pinned at $n_i^2$ . In n-type silicon, if essentially every donor gives up its electron, the majority concentration is set by the donor density $N_D$ , and the minority holes follow immediately:

n_e \approx N_D, \qquad n_h = \frac{n_i^2}{N_D}. \qquad (2)

where: $N_D$ is the number of donor atoms per cubic metre. The physical reading: doping to $n_e = 10^{22}\ \text{m}^{-3}$ — still only about one donor per few million silicon atoms — raises the electron count roughly a hundred-thousand-fold above intrinsic, while the hole count collapses to $n_h = (1.5\times10^{16})^2/10^{22} = 2.25\times10^{10}\ \text{m}^{-3}$ , over four hundred billion times scarcer than the electrons. The corresponding statements for p-type material are $n_h \approx N_A$ and $n_e = n_i^2/N_A$ , where $N_A$ is the acceptor density. Why is neutrality still guaranteed? The full charge-neutrality statement in a doped crystal is $n_e + N_A^- = n_h + N_D^+$ : mobile electrons plus fixed negative acceptor ions balance mobile holes plus fixed positive donor ions. In pure n-type ( $N_A = 0$ , donors fully ionized) this reduces to $n_e = n_h + N_D$ , which holds comfortably since $n_e \approx N_D \gg n_h$ . The crystal never acquires net charge.

Derivations

Derivation AWhy the fifth electron comes loose so easily: a hydrogen atom inside silicon

The goal is to estimate, using only the Bohr model you met with the hydrogen atom, how much energy it takes to tear the fifth (donor) electron away from a phosphorus atom in silicon — and to show why that energy is a mere few hundredths of an electronvolt, tiny beside the $1.1\ \text{eV}$ gap, so almost every donor is ionized at room temperature. The picture is trustworthy: the extra electron orbits the single net positive charge of the phosphorus ion, exactly as hydrogen's electron orbits its proton, but with two changes the crystal forces on us.

Step 1. Recall the Bohr result for hydrogen: the electron's binding (ionization) energy is $13.6\ \text{eV}$ and its orbit radius is the Bohr radius $a_0 = 0.0529\ \text{nm}$ . Both come from balancing the Coulomb attraction against the orbital motion, pure algebra with no calculus.

Step 2. Make the two crystal corrections. First, the electron does not move through vacuum but through silicon, whose atoms screen the Coulomb attraction; this weakens the force by the dielectric constant $\varepsilon_r \approx 11.7$ . Second, an electron moving through the periodic lattice responds as if it had an effective mass $m^* \approx 0.26\,m_0$ rather than its free mass. In the Bohr formulae the binding energy scales as $m^*/\varepsilon_r^2$ and the radius as $\varepsilon_r/(m^*/m_0)$ .

Step 3. Scale the ionization energy:

E_d = 13.6\ \text{eV}\times\frac{m^*/m_0}{\varepsilon_r^{\,2}} = 13.6\times\frac{0.26}{11.7^2} \approx 0.026\ \text{eV}. \qquad (3)

The crude estimate lands at about $0.03\ \text{eV}$ ; measured values are a little larger — phosphorus in silicon needs about $0.045\ \text{eV}$ — but the verdict is unchanged: the donor binding energy is a few hundredths of an eV, roughly $1/25$ of the $1.1\ \text{eV}$ gap and comparable to the thermal energy $kT \approx 0.026\ \text{eV}$ at $300\ \text{K}$ . Thermal jostling frees the donor electron routinely, whereas it almost never has the $1.1\ \text{eV}$ needed to make an intrinsic pair.

Step 4. Scale the orbit radius to check the picture is self-consistent:

a_d = a_0\,\frac{\varepsilon_r}{m^*/m_0} = 0.0529\ \text{nm}\times\frac{11.7}{0.26} \approx 2.4\ \text{nm}.

Silicon's lattice constant is only $0.543\ \text{nm}$ , so the donor electron's orbit is about four lattice spacings across, enclosing thousands of silicon atoms. That is precisely why treating the surrounding silicon as a smooth dielectric of $\varepsilon_r = 11.7$ is legitimate — the electron samples the bulk crystal, not any single atom. A loosely bound, far-flung electron is an easily ionized one. Verified: $E_d = 13.6\times0.26/11.7^2 = 0.026\ \text{eV}$ , orbit $a_d = 0.0529\times11.7/0.26 = 2.4\ \text{nm} \approx 4$ lattice constants, and $E_d/E_g \approx 0.024$ — recomputed in verify_hs2-extrinsic-semiconductors-doping.py.

Go deeper · Derivation B · 1 minMajority and minority concentrations from the mass-action law

Derivation BMajority and minority concentrations from the mass-action law

The goal is to turn the mass-action law $n_e n_h = n_i^2$ into concrete carrier numbers for a real n-type sample, showing how the minority concentration is fixed the instant you choose the doping level, and how enormously the two populations differ. We need only the mass-action law, the intrinsic value $n_i(\text{Si}) = 1.5\times10^{16}\ \text{m}^{-3}$ , and algebra.

Step 1. Choose an n-type sample doped so that, with essentially every donor ionized, the electron concentration is $n_e \approx N_D = 1.0\times10^{22}\ \text{m}^{-3}$ . This is a modest doping level: dividing by silicon's atomic density of about $5\times10^{28}\ \text{m}^{-3}$ , it is only one donor for every $5\times10^6$ silicon atoms.

Step 2. Compute the intrinsic product. $n_i^2 = (1.5\times10^{16})^2 = 2.25\times10^{32}\ \text{m}^{-6}$ . By the mass-action law this equals $n_e n_h$ in the doped sample too — the doping cannot change it, because it is fixed by temperature alone.

Step 3. Solve for the minority holes:

n_h = \frac{n_i^2}{n_e} = \frac{2.25\times10^{32}}{1.0\times10^{22}} = 2.25\times10^{10}\ \text{m}^{-3}.

Step 4. Read the ratio. The majority electrons outnumber the minority holes by $n_e/n_h = 10^{22}/(2.25\times10^{10}) \approx 4.4\times10^{11}$ — over four hundred billion to one. Compare with the intrinsic starting point, where $n_e = n_h = 1.5\times10^{16}$ : doping has multiplied the electrons by about $6.7\times10^{5}$ and divided the holes by the same factor, leaving the product untouched. This is the seesaw in numbers: in n-type material, current is carried almost entirely by electrons, and the once-equal holes have become a vanishing trace. Verified: $n_i^2 = 2.25\times10^{32}\ \text{m}^{-6}$ , $n_h = 2.25\times10^{10}\ \text{m}^{-3}$ , ratio $n_e/n_h = 4.4\times10^{11}$ , doping factor $6.7\times10^{5}$ — recomputed in verify_hs2-extrinsic-semiconductors-doping.py.

Limiting cases

case	result	meaning
no doping ( $N_D = N_A = 0$ )	$n_e = n_h = n_i$	intrinsic semiconductor — the baseline
$N_D \gg n_i$ , no acceptors	$n_e \approx N_D$ , $n_h = n_i^2/N_D$	n-type — electrons dominate, holes suppressed
$N_A \gg n_i$ , no donors	$n_h \approx N_A$ , $n_e = n_i^2/N_A$	p-type — holes dominate, electrons suppressed
$N_D = N_A$ (equal doping)	carriers cancel, $n_e \approx n_h \approx n_i$	fully compensated — behaves intrinsic
temperature raised far	$n_i$ grows until $n_i \gg N_D$	doping "washes out" — sample reverts to intrinsic
$T \to 0$	donors/acceptors freeze out, carriers vanish	no thermal energy to ionize impurities

At every step the mass-action product $n_e n_h = n_i^2$ still holds; only the split between $n_e$ and $n_h$ shifts.

Common trapn-type silicon is not negatively charged, and doping does not create charge

The single most common misconception is that because n-type silicon has "extra" electrons it must be negatively charged, and p-type positively charged. It is not. Doping never adds net charge — it only frees charges that were already balanced. Every electron a donor releases leaves behind an equal fixed positive donor ion; every hole an acceptor creates is matched by a fixed negative acceptor ion. Sum all the charges, mobile and fixed, and an n-type or p-type crystal is exactly as neutral as pure silicon. The prefixes describe the sign of the mobile carrier, not the charge of the material. A related slip is to forget the minority carrier and treat n-type silicon as if it held only electrons: the mass-action law guarantees a hole population is always present, $n_h = n_i^2/n_e$ , however tiny — and in devices like the transistor those minority carriers do essential work.

Worked examples

Board level — minority carriers in n-type silicon (CBSE and NEET pattern)

A silicon crystal at $300\ \text{K}$ is doped with donors until the free-electron concentration is $n_e = 1.0\times10^{22}\ \text{m}^{-3}$ . Given $n_i = 1.5\times10^{16}\ \text{m}^{-3}$ , find the hole concentration and state which are the majority and minority carriers.

Step 1. The mass-action law gives $n_e n_h = n_i^2$ , so $n_h = n_i^2/n_e$ .

Step 2. Compute $n_i^2 = (1.5\times10^{16})^2 = 2.25\times10^{32}\ \text{m}^{-6}$ .

Step 3. Divide: $n_h = 2.25\times10^{32}/(1.0\times10^{22}) = \boxed{2.25\times10^{10}\ \text{m}^{-3}}$ .

Step 4. Since $n_e \gg n_h$ , electrons are the majority carriers and holes are the minority carriers — as expected for n-type (donor-doped) material. Note the holes have not disappeared; they are simply outnumbered by a factor of about $4\times10^{11}$ .

Board level — identifying the doping type (CBSE and ICSE pattern)

Silicon is doped separately with (a) antimony and (b) indium. For each, state the type of semiconductor formed, the majority carrier, and the sign of the fixed impurity ion left in the lattice.

Step 1. Antimony (Sb) is in group 15 — pentavalent, five valence electrons. It donates its fifth electron, so it is a donor: the material is n-type, the majority carrier is the electron, and the fixed impurity ion is positive.

Step 2. Indium (In) is in group 13 — trivalent, three valence electrons. It leaves a bond incomplete and accepts an electron, so it is an acceptor: the material is p-type, the majority carrier is the hole, and the fixed impurity ion is negative.

Step 3. The reliable rule to quote: pentavalent → donor → n-type → positive fixed ion; trivalent → acceptor → p-type → negative fixed ion. In both cases the crystal as a whole stays electrically neutral.

Competitive level — doping fraction to carrier concentration (JEE Main and NEET pattern)

Pure silicon contains about $5.0\times10^{28}$ atoms per cubic metre. It is doped with pentavalent arsenic at a level of one arsenic atom per million silicon atoms. Assuming every donor is ionized, find (a) the majority-carrier concentration and (b) the minority-carrier concentration. Take $n_i = 1.5\times10^{16}\ \text{m}^{-3}$ .

Step 1. Donor density: $N_D = (5.0\times10^{28})/10^{6} = 5.0\times10^{22}\ \text{m}^{-3}$ . With full ionization the majority electrons are $n_e \approx N_D = \boxed{5.0\times10^{22}\ \text{m}^{-3}}$ .

Step 2. Minority holes from mass action: $n_h = n_i^2/n_e = (1.5\times10^{16})^2/(5.0\times10^{22}) = 2.25\times10^{32}/(5.0\times10^{22}) = \boxed{4.5\times10^{9}\ \text{m}^{-3}}$ .

Step 3. For perspective, the doping has raised the free-electron count from the intrinsic $1.5\times10^{16}$ to $5.0\times10^{22}$ — a factor of $n_e/n_i = 3.3\times10^{6}$ , over three million. The conductivity, which scales with the carrier concentration, rises by a comparable factor: this is why a one-part-per-million impurity transforms the material. Timing note: about a minute and a half. The only two moves are $N_D = (\text{atomic density})\times(\text{doping fraction})$ and $n_h = n_i^2/N_D$ . Carry $n_i^2 = 2.25\times10^{32}$ as a pre-computed constant for silicon and each part is a single division — do not detour through unit conversions.

Competitive level — a compensated crystal (JEE Advanced pattern)

A silicon sample is doped with both donors and acceptors: $N_D = 1.0\times10^{22}\ \text{m}^{-3}$ and $N_A = 6.0\times10^{21}\ \text{m}^{-3}$ . Assuming full ionization, find the majority- and minority-carrier concentrations. Take $n_i = 1.5\times10^{16}\ \text{m}^{-3}$ .

Step 1. Donors and acceptors partly cancel — this is compensation. Because $N_D > N_A$ , the material is net n-type, and charge neutrality gives the majority electrons as the net donor excess: $n_e \approx N_D - N_A = 1.0\times10^{22} - 6.0\times10^{21} = \boxed{4.0\times10^{21}\ \text{m}^{-3}}$ .

Step 2. The minority holes still obey mass action: $n_h = n_i^2/n_e = 2.25\times10^{32}/(4.0\times10^{21}) = \boxed{5.6\times10^{10}\ \text{m}^{-3}}$ .

Step 3. The lesson: when both dopants are present, only the difference $N_D - N_A$ sets the majority carrier; the acceptors have soaked up an equal number of donor electrons. Had the doping been exactly equal ( $N_D = N_A$ ), the sample would behave intrinsic, with $n_e \approx n_h \approx n_i$ . Timing note: under a minute once you see it. The trap is to write $n_e \approx N_D$ and ignore the acceptors — always take the net $N_D - N_A$ first, then apply mass action to the result. The single division $n_i^2/(N_D - N_A)$ finishes it.

Exam traps

Claiming n-type or p-type silicon carries net charge. It does not. Every mobile electron in n-type is balanced by a fixed positive donor ion; every hole in p-type by a fixed negative acceptor ion. Writing "n-type is negatively charged" loses marks on CBSE, NEET and JEE alike. The prefix names the carrier sign, not the material's charge.

Forgetting the minority carrier exists. In n-type material students often write "there are no holes." Mass action forbids it: $n_h = n_i^2/n_e$ is always positive, however small. Questions that hand you $n_e$ and ask for $n_h$ are testing exactly this — the answer is never zero.

Treating impurity ions as mobile. The fixed donor and acceptor ions do not move and carry no current; only the freed electrons and holes conduct. A diagram that shows the ions drifting toward an electrode is wrong.

Mixing up donor and acceptor. Pentavalent (group 15) atoms donate and make n-type; trivalent (group 13) atoms accept and make p-type. The memory hooks: donor–pentavalent–n, acceptor–trivalent–p. Reversing them is the most frequent single error in this topic.

Applying mass action to a non-equilibrium or wrong- $n_i$ situation. The law $n_e n_h = n_i^2$ holds at thermal equilibrium and uses the value of $n_i$ at that temperature. For germanium ( $n_i = 2.4\times10^{19}\ \text{m}^{-3}$ ) the constant is vastly larger than for silicon; do not carry silicon's $n_i^2$ into a germanium problem.

Board scope. Doping, donors/acceptors, majority/minority carriers, and the mass-action law $n_e n_h = n_i^2$ are core on the Indian boards — CBSE, ICSE (ISC), JEE Main, JEE Advanced, and NEET — with the qualitative donor/acceptor and n-type/p-type picture also appearing in A-Level electronics; the quantitative $n_h = n_i^2/N_D$ calculation is an Indian-board expectation, not a CIE 9702 one. Note carefully: AP Physics 2 does not cover semiconductor electronics at all, and IGCSE 0625 treats only qualitative diodes and logic gates — neither the mass-action calculation nor the pentavalent/trivalent carrier accounting is on those syllabuses. IB Physics has moved most electronics out of its core. Answer to your own board: an Indian-board student computes $n_h = n_i^2/N_D$ ; there is no such computation to prepare for on A-Level, AP-2, or IGCSE.

Origins · 1 minGordon Teal and how this was discovered

Gordon Teal

1907–2003 · Dallas, Texas and Bell Labs

Gordon Kidd Teal was the American physical chemist who made controlled doping practical, and with it the modern semiconductor industry. Born in Dallas and trained at Brown University, he joined Bell Labs in 1930. Convinced that the polycrystalline germanium then used was hopeless — its grain boundaries scattered carriers and made results irreproducible — he adapted the Czochralski pulling method to grow the first large single crystals of germanium in 1948 (with John Little) and of silicon soon after, learning to add measured traces of donor or acceptor impurity during the pull so that carrier type and concentration were set exactly. This turned doping from an accident into an engineering control. Moving to Texas Instruments in 1953, he built the first commercial silicon transistor in 1954 — silicon's tolerance of high temperatures made it the material that eventually carried the whole industry.

History

1928

— Felix Bloch's band theory shows electrons in a crystal occupy allowed energy bands separated by forbidden gaps.

1931

— Alan Wilson's quantum theory of semiconductors explains how impurity atoms donate or accept charge carriers.

1940

— Russell Ohl at Bell Labs discovers the silicon p–n junction, distinguishing n-type from p-type material.

1948

— Gordon Teal and John Little grow the first single-crystal germanium, enabling uniform, controlled doping.

1951

— Teal grows single-crystal silicon and adds dopants during the pull to set the carrier type precisely.

1954

— Teal, at Texas Instruments, builds the first commercial silicon transistor, made possible by that precise doping.

Going further

Texts. NCERT Physics Class 12, Chapter 14 ("Semiconductor Electronics"), the extrinsic-semiconductor and doping sections · H. C. Verma, Concepts of Physics Vol. 2, "Semiconductors and Semiconductor Devices" · Neamen, Semiconductor Physics and Devices, for the effective-mass and shallow-donor treatment · Kittel, Introduction to Solid State Physics, for the hydrogenic-donor model quantitatively.

Provenance. Standard semiconductor constants at $300\ \text{K}$ : $n_i(\text{Si}) = 1.5\times10^{16}\ \text{m}^{-3}$ , $n_i(\text{Ge}) = 2.4\times10^{19}\ \text{m}^{-3}$ , $E_g(\text{Si}) = 1.1\ \text{eV}$ , $E_g(\text{Ge}) = 0.7\ \text{eV}$ , $kT = 0.026\ \text{eV}$ . Recompute: silicon atomic density $\rho N_A/M = 2330\times6.022\times10^{23}/0.02809 = 5.0\times10^{28}\ \text{m}^{-3}$ ; hydrogenic donor energy $13.6\times0.26/11.7^2 = 0.026\ \text{eV}$ and orbit $0.0529\times11.7/0.26 = 2.4\ \text{nm}$ (about $4$ lattice constants of $0.543\ \text{nm}$ ), $E_d/E_g = 0.024$ ; mass-action product $n_i^2 = (1.5\times10^{16})^2 = 2.25\times10^{32}\ \text{m}^{-6}$ ; n-Si with $n_e = 10^{22}$ gives $n_h = 2.25\times10^{10}\ \text{m}^{-3}$ , ratio $4.4\times10^{11}$ , doping factor $6.7\times10^{5}$ ; $1$ -ppm As doping $N_D = 5\times10^{28}/10^{6} = 5\times10^{22}$ , $n_h = 4.5\times10^{9}$ , $n_e/n_i = 3.3\times10^{6}$ ; compensated $N_D-N_A = 10^{22}-6\times10^{21} = 4\times10^{21}$ , $n_h = 5.6\times10^{10}$ ; germanium doped to $10^{22}$ gives $n_e/n_i = 417$ .

Prerequisites. Intrinsic Semiconductors — the pure Si/Ge crystal, thermal electron–hole pair generation, and $n_e = n_h = n_i$ , on which everything here builds. Leads to. The p–n Junction, where an n-type and a p-type region meet and the depletion region and barrier potential appear — the device at the root of every diode and transistor. For a rigorous account of the shallow-donor level and effective mass, the university seam on semiconductor physics and the effective-mass approximation goes deeper (enrichment only, never an exam dependency here).

If donors add electrons to the crystal, why isn't n-type silicon negatively charged?

Because the electron was already inside the crystal — it belonged to the phosphorus atom. When the donor gives it up, the phosphorus becomes a fixed positive ion of exactly $+1$ , sitting right where the electron used to be bound. Count everything: one extra mobile electron ( $-1$ ) and one fixed donor ion ( $+1$ ) sum to zero. Doping does not import charge from outside; it unlocks a charge that was already balanced. The whole crystal stays as electrically neutral as pure silicon. The label "n-type" tells you only that the mobile carriers are negative electrons — it says nothing about the net charge, which is zero.

How can one impurity atom in a million change the conductivity so drastically?

Because pure silicon is such a poor conductor to begin with. At room temperature only about $1.5\times10^{16}$ electrons per cubic metre are free — one for roughly every three trillion atoms. A doping level of one part per million introduces about $5\times10^{22}$ donor atoms per cubic metre, each contributing a free electron — millions of times more carriers than intrinsic silicon has on its own. The comparison that matters is not impurity-to-silicon (where it is indeed a trace) but electrons-released-to-electrons-silicon-frees-itself. Against that tiny baseline a "trace" impurity is overwhelming — which is why semiconductor silicon must be purified to better than one stray atom in a billion before it is deliberately doped.

Is a hole a real particle, or just the absence of an electron?

A hole is the absence of a bonding electron, but it behaves so consistently like a real, mobile, positive particle that physicists treat it as one. When a neighbouring electron hops in to fill a vacant bond, the vacancy moves to where that electron came from — so the "hole" drifts through the crystal, and it drifts toward the negative terminal under a field, exactly as a positive charge would. It has an effective mass and carries current. So while a hole is not a chunk of matter, it is a genuine, book-keepable carrier of positive charge. In p-type material holes are the majority carriers and do the bulk of the conducting, which is why treating them as real particles is not a fudge but the correct physics.

Why does the donor electron come free so easily when making an intrinsic pair is so hard?

Energy. Making an intrinsic electron–hole pair means breaking a full silicon covalent bond, which costs the whole band gap, $1.1\ \text{eV}$ — about forty times the thermal energy available at room temperature, so it happens only rarely. Freeing a donor's fifth electron costs only about $0.045\ \text{eV}$ , because that electron was never in a proper bond: it orbits the donor ion loosely, screened by the surrounding silicon, in an orbit spanning several lattice spacings (as the hydrogen-atom estimate in the derivation shows). That few-hundredths-of-an-eV binding is comparable to $kT \approx 0.026\ \text{eV}$ , so thermal jostling ionizes nearly every donor at $300\ \text{K}$ . Same crystal, same temperature — but a shallow donor level is trivially easy to empty, while the full band gap is not.

Does doping change the band gap of silicon?

Essentially no — not at ordinary device doping levels. The $1.1\ \text{eV}$ gap is a property of the silicon crystal itself, and one impurity per million barely disturbs it. What doping does is insert new localized levels inside the gap — a donor level just below the conduction band, an acceptor level just above the valence band — and these shallow levels supply the carriers while the bands themselves stay put. So for exam purposes: doping adds carriers by adding impurity levels, and leaves the $1.1\ \text{eV}$ gap unchanged.

What happens to n-type silicon if you heat it strongly?

Its doped, controlled behaviour "washes out." Doping fixes the electron concentration at roughly $N_D$ only as long as the intrinsic concentration $n_i$ stays well below it. But $n_i$ rises steeply with temperature as more bonds break. Heat the sample far enough that $n_i$ exceeds $N_D$ , and the thermally generated pairs swamp the donor electrons: the material stops behaving n-type and reverts to intrinsic, $n_e \approx n_h \approx n_i$ . This is one reason silicon devices have an upper operating temperature, and why silicon (gap $1.1\ \text{eV}$ ) outperforms germanium (gap $0.7\ \text{eV}$ ) at high temperature — germanium's larger intrinsic population catches up with the doping much sooner.

Why must the impurity be pentavalent or trivalent specifically — why not any other atom?

Because you want to change the carrier count by exactly one per impurity, without wrecking the crystal. Silicon is tetravalent, so its lattice site "expects" four bonding electrons. A group-15 atom (five electrons) fits the four bonds and has precisely one to spare — a clean donor. A group-13 atom (three electrons) is precisely one short — a clean acceptor. A same-valence group-14 atom would substitute neutrally and donate nothing, while an atom off by two or more tends to introduce deep levels near mid-gap that trap carriers and reduce conductivity. So the useful dopants are the immediate neighbours of silicon, one column either side.