Extrinsic semiconductors

Take a bar of pure silicon and try to light a bulb with it — nothing much happens. Its conductivity at room temperature is almost nothing, because only a stray handful of covalent bonds break by thermal accident to free an electron. Now take the same silicon and, into every million atoms, drop in just one atom of phosphorus. Suddenly the current jumps by a factor of a million or more. Nothing about the crystal looks different to the eye, no new material has been "added" in any bulk sense — one part in a million is a pinch of salt in an Olympic swimming pool — and yet the electrical behaviour has been transformed. This engineered impurity is called doping, and the doped crystal is an extrinsic semiconductor — the single idea that turns silicon from a scientific curiosity into the material every diode, and eventually every chip, is built from. The trick works because doping does not rely on the slow, temperature-sensitive business of breaking covalent bonds. It hands the crystal carriers directly, atom by atom, wherever the crystal grower chooses to place them. Choose an impurity with one more valence electron than silicon and you get a crystal flooded with spare electrons — an n-type semiconductor. Choose one with one fewer valence electron and you get a crystal flooded with spare "vacancies," or holes, ready to carry positive current — a p-type semiconductor. Both crystals stay perfectly neutral overall; only the type of carrier that dominates has changed. This article builds that picture from the ground up: why a five-electron dopant donates a free electron while a three-electron dopant creates a hole, and why doping raises conductivity so much more effectively than raising the temperature ever could. By the end you should be able to name the majority and minority carriers for any doped crystal, and reason quantitatively about why the extrinsic crystal conducts so much better than the pure one it started from.

Key takeaways

Doping is the deliberate addition of a controlled trace of impurity — about one atom in $10^6$ to $10^8$ — into pure (intrinsic) silicon or germanium to boost and control conductivity.
Pentavalent dopants (P, As, Sb — five valence electrons) give up one spare electron each without creating a matching hole; they are called donors, and the doped crystal is n-type, with electrons as majority carriers.
Trivalent dopants (B, Al, In — three valence electrons) leave one bond incomplete, creating a hole each; they are called acceptors, and the doped crystal is p-type, with holes as majority carriers.
In both cases the crystal stays electrically neutral overall — the fixed donor/acceptor ions are exactly balanced by the mobile carriers they release; only the majority-carrier type differs.
Minority carriers never disappear — a doped crystal still generates thermal electron-hole pairs exactly as an intrinsic crystal does, just in far smaller numbers than the doping-supplied majority carriers.
Because conductivity depends on carrier concentration ( $\sigma = e n \mu$ , quoted from Current Electricity), and doping raises carrier concentration by a factor of a million or more over the thermally generated intrinsic level, extrinsic conductivity vastly exceeds intrinsic conductivity at the same temperature.
Key idea: replacing roughly one host atom in a million with a pentavalent or trivalent impurity injects a controllable, temperature-independent flood of majority carriers (electrons for n-type, holes for p-type) that dwarfs the sparse thermally generated carrier population of pure silicon, while the crystal as a whole stays electrically neutral.

The core insight

Doping silicon or germanium with a pentavalent impurity (a donor) creates an n-type semiconductor whose majority carriers are electrons, while doping with a trivalent impurity (an acceptor) creates a p-type semiconductor whose majority carriers are holes, and in both cases the huge, controllable carrier population supplied by doping — not by heat — is what makes an extrinsic semiconductor conduct so much better than the pure crystal it started from.

Where it fits in the syllabus

This article continues directly from intrinsic semiconductors within Unit IX, "Semiconductor Electronics: Materials, Devices and Simple Circuits," and it is explicitly on the CBSE 2025-26 syllabus at a qualitative level: doping, donor and acceptor impurities, and the resulting n-type and p-type materials with their majority and minority carriers. No formula for carrier concentration, no mass-action law, and no numerical carrier-density syllabus content is prescribed here — those stay qualitative ideas. The one quantitative tool this article borrows is the conductivity relation $\sigma = en\mu$ from the earlier Current Electricity chapter, used only to make the "doping raises conductivity enormously" claim concrete with real numbers, not as new syllabus content. Everything here sets up the very next topic, the p-n junction, which is nothing more than a single crystal grown so that one half is doped p-type and the other n-type.

The physical picture

Silicon and germanium are group-14 elements: every atom has four valence electrons, each shared in a covalent bond with one of four neighbours in the crystal lattice, so every atom's outer shell looks locally "full" with eight shared electrons. Doping means substituting, for one host atom in roughly a million to a hundred million, an atom from a different group of the periodic table — always chosen close enough in size to fit into the lattice without wrecking its structure. A pentavalent impurity (group 15: phosphorus, arsenic, antimony) brings five valence electrons to a site that only needs four for bonding. Four of its electrons complete the four covalent bonds with its silicon neighbours exactly as a silicon atom's electrons would; the fifth has no bond to join. It sits only loosely bound to its parent atom, at an energy just below the conduction band, and typical thermal energy at room temperature is enough to set it free into the conduction band. Each such atom therefore donates one conduction electron without creating a matching hole, so it is called a donor, and the resulting crystal is n-type — electrons are the majority carriers, holes (still trickling in from ordinary thermal generation, as in the intrinsic crystal) are the minority carriers. A trivalent impurity (group 13: boron, aluminium, indium) brings only three valence electrons, one short of the four needed to complete its bonds with four neighbours. One bond is left with a vacancy — a hole — ready to be filled by an electron hopping over from a neighbouring bond. Each such atom therefore accepts an electron (once one arrives) and is called an acceptor; the resulting crystal is p-type — holes are the majority carriers, electrons are the minority carriers.

\sigma = e\left(n_e\,\mu_e + n_h\,\mu_h\right) \qquad (1)

where: $\sigma$ is conductivity (siemens per metre), $e$ is the electronic charge, $n_e$ and $n_h$ are the free-electron and hole concentrations (per cubic metre), and $\mu_e,\mu_h$ are their respective mobilities (drift velocity per unit field), all quoted from the Current Electricity chapter's $I = nAev_d$ and $\mu = v_d/E$ . The physical reading is that both carrier types contribute to current in a semiconductor, unlike a metal where only electrons move — but in an extrinsic crystal one term in equation (1) is completely dominated by the doping-supplied majority carrier, while the other stays at the tiny, thermally set intrinsic level. That imbalance, not any change in mobility, is the entire reason doped silicon conducts so much better than pure silicon. The donor and acceptor picture is usually drawn on an energy-band diagram: the donor level sits just below the bottom of the conduction band (a small hop away for the spare electron), and the acceptor level sits just above the top of the valence band (a small hop for a valence electron to jump up and leave a hole behind).

Figure

Energy band diagrams comparing an intrinsic semiconductor with a wide empty gap, an n type semiconductor with a donor level just below the conduction band supplying free electrons, and a p type semiconductor with an acceptor level just above the valence band accepting electrons and leaving holes behind

Derivation AWhy doping raises conductivity by orders of magnitude

The goal is to show, using nothing beyond the conductivity relation already quoted from Current Electricity, that swapping one silicon atom in a million for a pentavalent donor changes conductivity by roughly the same factor of a million — not by some small correction — because doping supplies carriers directly rather than relying on the rare, thermally activated bond-breaking that intrinsic silicon depends on. We keep every step algebraic and compare only orders of magnitude, exactly as a board-level "explain why" answer should.

Step 1. Write the general conductivity of a semiconductor carrying both electron and hole current: $\sigma = e\left(n_e\mu_e + n_h\mu_h\right)$ , as in equation (1). This holds for intrinsic and extrinsic material alike; only the values of $n_e$ and $n_h$ change between them.

Step 2. Evaluate it for the intrinsic crystal before doping. Here $n_e = n_h = n_i$ , the intrinsic carrier concentration set purely by thermal generation, so $\sigma_i = e\,n_i\left(\mu_e+\mu_h\right)$ . For silicon near room temperature $n_i$ is extremely small compared with the roughly $5\times10^{28}\ \text{atoms/m}^3$ that make up the crystal — only a tiny fraction of bonds are thermally broken at any instant, which is exactly why pure silicon barely conducts.

Step 3. Now dope the crystal n-type with donor concentration $N_D$ (donor atoms per cubic metre). Every donor releases essentially one free electron at room temperature, so the electron concentration jumps to $n_e \approx N_D$ — vastly larger than $n_i$ whenever the doping level exceeds the intrinsic level, which it always does in a practical device. The hole concentration stays pinned near the small intrinsic value (it is, if anything, suppressed further, but that refinement needs the mass-action law and is not required here). Equation (1) then simplifies because the hole term is negligible next to the huge electron term: $\sigma_n \approx e\,N_D\,\mu_e$ .

Step 4. Compare the two conductivities. Since $N_D \gg n_i$ , we get $\sigma_n / \sigma_i \approx N_D / n_i$ , a ratio set entirely by how many donor atoms were put in versus how many carriers heat alone could supply. With a typical doping fraction of one impurity atom per $10^6$ silicon atoms, $N_D \approx (5\times10^{28})/10^{6} = 5\times10^{22}\ \text{m}^{-3}$ — a number that towers over the sparse intrinsic population. This is why "add a whisper of impurity" produces a "jump of a million or more" in conductivity: the ratio of carrier concentrations, not any change in mobility, is doing all the work. Read the answer, not the arithmetic, for the exam: conductivity in a semiconductor is controlled by how many carriers are available, and doping supplies carriers directly and in bulk, sidestepping the slow thermal bond-breaking that limits an intrinsic crystal — hence the enormous jump.

Worked examples

Board level - identifying carrier type from a dopant's valence

A silicon crystal is doped with a small, controlled amount of arsenic (group 15). State, with reasoning based on valence electrons and covalent bonding, whether the result is n-type or p-type, and name the majority and minority carriers.

Step 1. Compare valence electrons: silicon has 4, arsenic has 5.

Step 2. Bond formation: an arsenic atom substituting into the silicon lattice uses four of its five valence electrons to complete covalent bonds with its four silicon neighbours, exactly as a silicon atom would.

Step 3. The fifth electron has no bond to join. It is loosely bound to the arsenic atom, at an energy just below the conduction band, and ordinary thermal energy at room temperature is enough to free it into the conduction band. Arsenic has therefore donated a free electron without creating any matching hole — it is a donor.

Step 4. Since every dopant atom supplies one extra free electron and no extra hole, electrons vastly outnumber the small population of thermally generated holes. Answer: the crystal is n-type, with electrons as the majority carriers and holes as the minority carriers.

Competitive level - estimating the donor concentration and the conductivity boost

A silicon crystal (atomic density $N_{\text{Si}} \approx 5\times10^{28}\ \text{m}^{-3}$ ) is doped with phosphorus at one impurity atom per $10^{7}$ silicon atoms. Taking the intrinsic carrier concentration of silicon near room temperature as $n_i \approx 1.5\times10^{16}\ \text{m}^{-3}$ , estimate (a) the donor concentration $N_D$ , and (b) roughly how many times more conducting the doped crystal is than the pure crystal, treating electron mobility as roughly unchanged by doping.

Step 1. Donor concentration from the doping fraction: $N_D = N_{\text{Si}}/10^{7} = (5\times10^{28})/10^{7} = 5\times10^{21}\ \text{m}^{-3}$ . This is the number of free conduction electrons the doping supplies, since each donor contributes essentially one at room temperature.

Step 2. Compare with the intrinsic carrier concentration: $N_D/n_i = (5\times10^{21})/(1.5\times10^{16}) \approx 3.3\times10^{5}$ . The doped crystal's electron population outnumbers the pure crystal's carrier population by roughly $330{,}000$ times.

Step 3. Since $\sigma_n \approx eN_D\mu_e$ and $\sigma_i \approx e\,n_i(\mu_e+\mu_h) \approx 2e\,n_i\mu_e$ (taking electron and hole mobilities as roughly comparable in order of magnitude for this estimate), the conductivity ratio is close to $\sigma_n/\sigma_i \approx N_D/(2n_i) \approx 1.6\times10^{5}$ . Answer: $N_D \approx 5\times10^{21}\ \text{m}^{-3}$ , and the doped crystal conducts on the order of $10^{5}$ times better than the pure crystal — a jump of roughly a hundred thousand-fold from doping alone. Timing note: under two minutes. The whole calculation is two divisions and a ratio — no exponential temperature dependence, no mass-action law. The one place to slow down is the mobility assumption in Step 3: it is stated as an approximation on purpose, because exact mobility values are not required by this syllabus; the point of the estimate is the order of magnitude, not the last significant figure.

Common traps

Believing an n-type crystal carries a net negative charge (or a p-type crystal a net positive charge) — both stay electrically neutral overall; the fixed donor or acceptor ions exactly balance the mobile carriers they release, and only the majority-carrier type differs.
Mixing up donor and acceptor: pentavalent (group 15) atoms donate a spare electron and give n-type; trivalent (group 13) atoms accept an electron, leaving a hole behind, and give p-type.
Forgetting minority carriers still exist in doped material — doping does not switch off thermal generation, it just makes the doping-supplied majority carrier vastly outnumber the thermally generated minority carrier.
Assuming the doping level is large — it is extremely small, typically one impurity atom per $10^6$ to $10^8$ host atoms, which is exactly what makes conductivity so precisely controllable by the crystal grower.
Treating conductivity as boosted mainly by a change in mobility — the derivation above shows the boost comes almost entirely from the huge increase in majority-carrier concentration, not from carriers suddenly moving faster.
Quoting a mass-action law or a numeric $n_e n_h = n_i^2$ relation as syllabus content here — the CBSE 2025-26 prescription keeps extrinsic semiconductors at a qualitative level; the conductivity estimate above only borrows the already-prescribed $\sigma = en\mu$ from Current Electricity.

Common trapDoped does not mean charged

The single most common mark-losing error in this topic is thinking that adding "extra" electrons (n-type) or "extra" holes (p-type) leaves the crystal with a net electric charge. It does not. Every donor atom that releases a free electron becomes a fixed, immobile positive ion sitting in the lattice; every acceptor atom that creates a hole becomes a fixed, immobile negative ion. The mobile carrier that "escapes" is exactly balanced, charge for charge, by the fixed ion left behind. Count up every electron and every proton in the doped crystal and the total is precisely what it was before doping — the crystal is neutral throughout, from the moment of doping onward. What has changed is not the total charge but which carrier type is free to move and in what abundance.

You should be able to…

Explain why a pentavalent impurity donates a free electron and produces n-type material, using valence electrons and covalent bonding.
Explain why a trivalent impurity creates a hole and produces p-type material, using the same bonding picture.
Name the majority and minority carriers correctly for both n-type and p-type semiconductors.
Explain why a doped semiconductor crystal stays electrically neutral overall despite having an excess of one mobile carrier type.
Use the conductivity relation $\sigma = e(n_e\mu_e + n_h\mu_h)$ , quoted from Current Electricity, to reason (at least in order of magnitude) why doping raises conductivity so much more than raising temperature does.
Sketch or read the donor-level/acceptor-level energy-band picture and connect it to why so little thermal energy is needed to activate a dopant.

Ask the tutor: Want to test yourself with a few "given this dopant's group number, name the type and the majority/minority carriers" flash questions across phosphorus, boron, arsenic, indium, and antimony? Ask your AI tutor to quiz you.

If doping adds atoms with a different number of electrons, why does the crystal not become charged?

Because every dopant atom that releases a mobile carrier is left behind as a fixed ion of the opposite sign, permanently locked into the crystal lattice. A phosphorus donor gives up one free electron and becomes a fixed positive ion; that positive ion's charge is exactly the charge the free electron carried away. Add up every charge in the crystal, mobile and fixed together, and it is precisely zero, just as it was before doping. Only the type of mobile carrier that dominates has changed — not the crystal's overall charge.

Why is the doping level always so tiny, like one atom in a million?

Because that tiny fraction is already enough to swamp the natural, thermally generated carrier population by a factor of a hundred thousand or more, as the competitive-level example above shows. Going any heavier with doping would not buy much extra conductivity for the crystal quality you would sacrifice — heavier doping distorts the lattice more and can introduce unwanted defects. The whole appeal of doping as an engineering tool is that a minuscule, precisely controlled amount of impurity produces a huge, predictable, temperature-stable change in conductivity, which is exactly what a device designer needs.

Do minority carriers ever matter, or can I just ignore them once a crystal is doped?

For this syllabus's qualitative picture and for basic conductivity estimates, majority carriers dominate so completely that minority carriers are safely ignored, as in the worked examples above. But minority carriers are not irrelevant everywhere — once you reach the next article, on the p-n junction, it turns out to be precisely the small population of minority carriers on each side that crosses the junction under reverse bias and sets the (very small) reverse saturation current. So keep the concept in your pocket: negligible for bulk conductivity, but the whole story once you get to junction behaviour.

Is silicon doped with a pentavalent and a trivalent impurity at the same time still one clear type?

In principle you can dope with both donors and acceptors together — real fabrication sometimes does exactly this, called compensation — and what matters then is which population is larger. If donors outnumber acceptors, the net effect is still n-type, with an effective donor concentration equal to the difference of the two; if acceptors outnumber donors, it is net p-type. This compensated-doping detail goes beyond the qualitative CBSE 2025-26 treatment, so for this syllabus you can assume a crystal is doped with one impurity type at a time, giving a clean n-type or p-type result.