Chapter 14 · Semiconductor Electronics: Materials, Devices and Simple Circuits

14.2 Intrinsic and Extrinsic Semiconductors

A pure semiconductor has equal numbers of free electrons and holes. Replace one atom in a million by phosphorus or boron and you get a million times more carriers — mostly electrons or mostly holes, as you choose — while the product $n_en_h$ stays fixed at $n_i^2$.

Silicon for computer chips is refined until fewer than one atom in a billion is a stranger — it is among the purest materials ever made. Then, with great care, engineers put impurities back: a few parts per million of phosphorus in one region, boron in the next. A wafer is bombarded with ions of the chosen element at a precisely controlled dose, and the result is a pattern of regions with very different electrical character. Why go to such trouble to purify a material, only to contaminate it again?

The answer is control. In pure silicon the only carriers are those made by heat, so its conductivity is low and depends strongly on temperature (Section 14.1). Adding the right impurity — doping — fixes the number of carriers at whatever value the designer wants, and also fixes their kind: negative electrons or positive "holes". In this section we look at pure (intrinsic) semiconductors, meet the hole, see how doping produces n-type and p-type material, and learn the rule $n_en_h=n_i^2$ that links the two kinds of carrier.

Intrinsic Semiconductors: Electrons and Holes

In a silicon or germanium crystal each atom sits at the centre of a tetrahedron formed by four neighbours, and shares one of its four valence electrons with each of them: every atom takes part in four covalent bonds, each made of a shared pair of electrons. Drawn flat, the crystal looks like a square grid of atoms joined by double lines (the bonds).

At $0$ K every bond is intact and every valence electron is tied up — the valence band is full and the crystal is an insulator. At room temperature thermal vibrations occasionally break a bond. The electron that escapes is free to wander through the crystal: it is in the conduction band. It leaves behind a bond with one electron missing. That vacancy has a net charge $+e$ relative to its surroundings, and it can move: an electron from a neighbouring bond can jump into it, which fills this vacancy and creates one in the neighbouring bond. The vacancy has moved — in the direction opposite to the jump of the bound electron.

📘 Definition 14.4 — Hole
A hole is a vacancy left in a covalent bond (an empty state in the valence band) when an electron is removed from it. It behaves as a mobile particle with charge $+e$: under an electric field it drifts in the direction of the field, and it carries current in that direction. Thermal breaking of a bond creates an electron–hole pair.
📘 Definition 14.5 — Intrinsic semiconductor
An intrinsic semiconductor is a pure semiconductor crystal, in which all the carriers come from thermally broken bonds. Every free electron leaves exactly one hole, so the electron density $n_e$ and the hole density $n_h$ are equal: $$n_e=n_h=n_i,$$ where $n_i$ is the intrinsic carrier concentration. It depends strongly on the material and the temperature: at $300$ K, $n_i\approx1.5\times10^{16}\ \text{m}^{-3}$ for silicon and $\approx2.4\times10^{19}\ \text{m}^{-3}$ for germanium.

Generation and recombination. Bonds break all the time (generation), and now and then a free electron meets a hole and falls back into it (recombination), releasing the energy $E_g$. At a fixed temperature the two processes balance: $n_i$ is the concentration at which recombination, which is proportional to how often electrons meet holes, exactly cancels generation. Warm the crystal and generation speeds up, so a larger $n_i$ is needed for balance.

Two kinds of current. In an applied field $\mathbf E$, the free electrons drift against $\mathbf E$ and the holes drift along $\mathbf E$. Since the charges are opposite, both currents point along $\mathbf E$ and add: $I=I_e+I_h$. With the mobility $\mu$ of Section 3.1 (drift speed per unit field, $v_d=\mu E$), the current density from each carrier is $j=ne\mu E$, which gives the conductivity.

📐 Result 14.3 — Conductivity of a semiconductor

A semiconductor with electron density $n_e$, hole density $n_h$ and mobilities $\mu_e$, $\mu_h$ has conductivity and resistivity $$\sigma=e\left(n_e\mu_e+n_h\mu_h\right),\qquad \rho=\frac1\sigma.$$ For an intrinsic semiconductor $\sigma=en_i(\mu_e+\mu_h)$. Electrons are more mobile than holes (in silicon at $300$ K, $\mu_e\approx0.135$ and $\mu_h\approx0.048\ \text{m}^2\,\text{V}^{-1}\text{s}^{-1}$), because a hole moves only by bound electrons hopping from bond to bond.

✏️ Example 1 — How many bonds are broken?
Pure silicon has $5.0\times10^{28}$ atoms per cubic metre and $n_i=1.5\times10^{16}\ \text{m}^{-3}$ at $300$ K. (a) How many free electrons and holes are there in a $1.0\ \text{cm}^3$ crystal? (b) Each atom forms four bonds, each shared by two atoms. What fraction of the bonds is broken?
(a) Carriers $N_e=N_h=(1.5\times10^{16})(1.0\times10^{-6})$ $=1.5\times10^{10}$ of each.
(b) Bonds Bonds per cubic metre $=\tfrac12\times4\times5.0\times10^{28}=1.0\times10^{29}$. Each broken bond gives one electron–hole pair, so the fraction is $\dfrac{1.5\times10^{16}}{1.0\times10^{29}}=1.5\times10^{-13}$.
Meaning Only about one bond in seven million million is broken — which is why pure silicon is such a poor conductor.
✏️ Example 2 — The resistivity of pure silicon from its carriers
Using $n_i=1.5\times10^{16}\ \text{m}^{-3}$, $\mu_e=0.135$ and $\mu_h=0.048\ \text{m}^2\,\text{V}^{-1}\text{s}^{-1}$, find (a) the conductivity and resistivity of intrinsic silicon at $300$ K, and (b) the fraction of the current carried by electrons. (c) A field of $100$ V/m is applied: find the drift speeds of the electrons and holes.
(a) σ and ρ $\sigma=en_i(\mu_e+\mu_h)$ $=(1.6\times10^{-19})(1.5\times10^{16})(0.183)$ $\approx4.4\times10^{-4}$ S/m, so $\rho\approx2.3\times10^{3}\ \Omega$ m.
(b) Share Equal densities, so the shares follow the mobilities: $\dfrac{\mu_e}{\mu_e+\mu_h}=\dfrac{0.135}{0.183}\approx0.74$ — electrons carry about $74\%$ of the current, holes $26\%$.
(c) Drift $v_e=\mu_eE=13.5$ m/s (against $\mathbf E$), $v_h=\mu_hE=4.8$ m/s (along $\mathbf E$).
Check The resistivity matches the measured value used in Example 1 of Section 14.1.

Extrinsic Semiconductors: n-type and p-type

📘 Definition 14.6 — Doping and extrinsic semiconductors
Doping is the deliberate addition of a small, controlled amount of a suitable impurity (the dopant) to a pure semiconductor, typically one atom in $10^5$ to $10^{9}$. The doped crystal is an extrinsic semiconductor. The dopant atoms must be close in size to the host atoms, so that they replace host atoms in the lattice without distorting it.

Pentavalent dopants: n-type. Suppose a few silicon atoms are replaced by atoms of a group-15 element — phosphorus (P), arsenic (As) or antimony (Sb) — which have five valence electrons. Four of them form bonds with the four silicon neighbours; the fifth has no bond to join. It is held only weakly by its parent ion, and a tiny energy — about $0.01$ eV in germanium and $0.05$ eV in silicon — sets it free. At room temperature ($k_BT\approx0.026$ eV) practically every dopant atom has given up its extra electron, without creating a hole. The dopant is a donor; it is left behind as a fixed positive ion bound into the lattice.

Trivalent dopants: p-type. A group-13 atom — boron (B), aluminium (Al), gallium (Ga) or indium (In) — has only three valence electrons. It completes three bonds; the fourth bond lacks an electron — a hole. An electron from a neighbouring bond easily moves in (again it costs only about $0.01$–$0.05$ eV), which completes the dopant's bonds and sets the hole free to wander. The dopant is an acceptor; it becomes a fixed negative ion.

📘 Definition 14.7 — n-type and p-type semiconductors
  • An n-type semiconductor is doped with donors (pentavalent atoms). Electrons are the majority carriers and holes the minority carriers: $n_e\gg n_h$.
  • A p-type semiconductor is doped with acceptors (trivalent atoms). Holes are the majority carriers and electrons the minority carriers: $n_h\gg n_e$.
In both, the crystal as a whole remains electrically neutral: the charge of the mobile majority carriers is balanced by the fixed ionised dopants (and the minority carriers).

The band picture of doping. A donor adds an energy level $E_D$ in the gap just below the conduction band; it is so close (about $0.01$–$0.05$ eV below $E_C$) that at room temperature almost all donor electrons are lifted into the conduction band. An acceptor adds an empty level $E_A$ just above the valence band; valence electrons are easily lifted into it, leaving holes in the valence band. A few electron–hole pairs are still made thermally across the full gap $E_g$, which supplies the minority carriers.

✏️ Example 3 — Why the donors are all ionised but the bonds are not
At $300$ K, $k_BT=0.026$ eV. Compare $k_BT$ with (a) the donor ionisation energy in silicon, $0.045$ eV, and (b) the band gap of silicon, $1.1$ eV. Explain why practically every donor gives up its electron while only about one bond in $10^{13}$ breaks.
(a) Donor $\dfrac{0.045}{0.026}\approx1.7$: the ionisation energy is less than twice $k_BT$ — thermal kicks of this size happen all the time.
(b) Gap $\dfrac{1.1}{0.026}\approx42$: a kick of $42\,k_BT$ is exceedingly rare.
Conclusion Because excitation becomes exponentially less likely as the energy needed grows (Section 14.1), nearly all donors are ionised while the full gap is almost never crossed. This is why the carrier density of doped silicon at room temperature is simply set by the doping.
Interactive 1 — Doping Lab

Left: a flat sketch of a silicon crystal (circles are atoms, double lines are bonds). Choose a dopant: a phosphorus atom frees its fifth electron (filled dot), a boron atom leaves a hole (open circle) that moves as neighbouring bound electrons hop in. Now and then heat breaks a bond somewhere, making an electron–hole pair that later recombines. Right: carrier densities on a logarithmic scale for silicon at $300$ K, computed from neutrality and $n_en_h=n_i^2$. Set the dopant density with the slider and watch the two markers move equal distances either side of $n_i$.

22.7
type—
$n_e$ ($\text{m}^{-3}$)—
$n_h$ ($\text{m}^{-3}$)—
$n_en_h$ vs $n_i^2$—

Counting Carriers: the Law of Mass Action

Doping silicon with donors adds a great many electrons. What happens to the holes? Recombination happens when an electron meets a hole, so its rate is proportional to $n_en_h$. Generation of pairs depends only on the temperature (it is the same thermal bond-breaking as in the pure crystal). In equilibrium recombination balances generation, so $n_en_h$ must have the same value as in the pure crystal at that temperature, namely $n_i\times n_i$. Extra electrons from donors therefore mop up holes: the holes become far fewer than in pure silicon.

📐 Result 14.4 — The law of mass action and charge neutrality

In a semiconductor in thermal equilibrium at temperature $T$, intrinsic or doped, $$n_en_h=n_i^2,$$ where $n_i$ is the intrinsic concentration of that material at that temperature. With donor density $N_D$ and acceptor density $N_A$ (all ionised), the crystal is neutral: $$n_e+N_A=n_h+N_D.$$ For an n-type sample with $N_D\gg n_i$: $\ n_e\approx N_D$ and $n_h\approx\dfrac{n_i^2}{N_D}$. For a p-type sample with $N_A\gg n_i$: $\ n_h\approx N_A$ and $n_e\approx\dfrac{n_i^2}{N_A}$.

🧭 Problem-solving strategy — carrier densities in a doped semiconductor
  1. Find the dopant density: (fraction of atoms replaced) × (atoms per m³). If both kinds are present, the net doping $|N_D-N_A|$ counts, and the larger one decides the type.
  2. If the net doping is much larger than $n_i$, set the majority density equal to it.
  3. Find the minority density from $n_en_h=n_i^2$.
  4. Check: the minority density should be far below $n_i$, and the product should equal $n_i^2$. If the net doping is comparable with $n_i$, solve neutrality and mass action together (a quadratic).
  5. For conductivity use $\sigma=e(n_e\mu_e+n_h\mu_h)$; usually only the majority term matters.
✏️ Example 4 — Silicon doped with arsenic
Pure silicon ($5.0\times10^{28}$ atoms per m³, $n_i=1.5\times10^{16}\ \text{m}^{-3}$) is doped with arsenic at one part per million. (a) Find $n_e$ and $n_h$ and the type of the sample. (b) Using $\mu_e=0.135\ \text{m}^2\,\text{V}^{-1}\text{s}^{-1}$, find the conductivity, and compare it with that of pure silicon ($4.4\times10^{-4}$ S/m).
Dopant Arsenic is pentavalent, a donor: $N_D=10^{-6}\times5.0\times10^{28}$ $=5.0\times10^{22}\ \text{m}^{-3}\gg n_i$.
(a) Carriers $n_e\approx5.0\times10^{22}\ \text{m}^{-3}$; $$n_h=\frac{n_i^2}{n_e}=\frac{(1.5\times10^{16})^2}{5.0\times10^{22}}=4.5\times10^{9}\ \text{m}^{-3}.$$ n-type: electrons are the majority carriers.
(b) Conductivity $\sigma\approx en_e\mu_e$ $=(1.6\times10^{-19})(5.0\times10^{22})(0.135)$ $\approx1.1\times10^{3}$ S/m — about $2.5\times10^{6}$ times that of pure silicon. The hole term ($en_h\mu_h\approx3.5\times10^{-11}$ S/m) is negligible.
Check $n_h=4.5\times10^9$ is far below $n_i$, and $n_en_h=2.25\times10^{32}=n_i^2$. ✓ Interactive 1 opens at this doping.
✏️ Example 5 — Germanium doped with indium
Germanium ($n_i=2.4\times10^{19}\ \text{m}^{-3}$ at $300$ K) is doped with indium atoms at a density of $4.0\times10^{22}\ \text{m}^{-3}$. Find the majority and minority carrier densities and name the type.
Dopant Indium is trivalent, an acceptor: $N_A=4.0\times10^{22}\gg n_i$, so the sample is p-type.
Majority $n_h\approx4.0\times10^{22}\ \text{m}^{-3}$.
Minority $n_e=\dfrac{(2.4\times10^{19})^2}{4.0\times10^{22}}$ $=\dfrac{5.76\times10^{38}}{4.0\times10^{22}}$ $\approx1.4\times10^{16}\ \text{m}^{-3}$.
✏️ Example 6 — Donors and acceptors together
A silicon sample ($n_i=1.5\times10^{16}\ \text{m}^{-3}$) contains $3.0\times10^{22}$ donor atoms and $1.0\times10^{22}$ acceptor atoms per cubic metre. Find $n_e$ and $n_h$, and the type.
Idea The acceptors capture electrons from the donors, so they cancel ("compensate") an equal number of donors. Neutrality: $n_e-n_h=N_D-N_A=2.0\times10^{22}$.
Carriers Since $2.0\times10^{22}\gg n_i$: $n_e\approx2.0\times10^{22}\ \text{m}^{-3}$ and $n_h=\dfrac{2.25\times10^{32}}{2.0\times10^{22}}\approx1.1\times10^{10}\ \text{m}^{-3}$. The sample is n-type.
Use This is how chip makers turn an n-type region into p-type: they add more acceptors than there are donors.
Interactive 2 — Electrons and Holes in an Applied Field

A silicon bar between two electrodes at $300$ K. Electrons (filled dots) drift against the field $\mathbf E$, holes (open circles) drift along it, each at speed $\mu E$ — notice the electrons move about three times faster. Both give conventional current in the direction of $\mathbf E$. The number of dots is symbolic (minority carriers are far rarer than shown); the readouts use the real densities. Choose the doping and the field.

100
drift speeds $v_e$, $v_h$—
$j_e$ (A/m²)—
$j_h$ (A/m²)—
$\sigma$ (S/m)—
⚠️ Common mistakes with doped semiconductors

1. "n-type material is negatively charged." No — every free electron came from a neutral donor atom, which is left as a fixed positive ion. The crystal is neutral.

2. Adding $n_i$ to the doping. In n-type silicon $n_e\approx N_D$, not $N_D+n_i$, and the holes are fewer than $n_i$, not equal to it. Use $n_h=n_i^2/n_e$.

3. Mixing up donors and acceptors. Group 15 (P, As, Sb: five electrons) → donor → n-type. Group 13 (B, Al, Ga, In: three electrons) → acceptor → p-type.

4. Thinking holes are particles like positrons. A hole is the absence of a bound electron; its motion is really bound electrons hopping the other way. It simply behaves as a $+e$ carrier.

5. Using $n_en_h=n_i^2$ with $n_i$ at the wrong temperature. $n_i$ rises steeply with $T$, so the product does too.

🚀 Beyond the board syllabus (JEE Main · NEET) — logic gates

Syllabus note: digital electronics is no longer in the NCERT book or the CBSE syllabus, and logic gates are not in JEE Advanced, but they are listed in the NEET and JEE Main syllabi. Every chip made from doped silicon processes information as two voltage levels: high = 1, low = 0. A logic gate is a circuit with one or more inputs and one output whose output level is a fixed function of the input levels, summarised in a truth table. (Gates are built from transistors — also beyond the syllabus — but you only need their truth tables.)

  • NOT (inverter): $Y=\overline A$ — the output is the opposite of the input.
  • OR: $Y=A+B$ — the output is 1 if at least one input is 1.
  • AND: $Y=A\cdot B$ — the output is 1 only if both inputs are 1.
  • NAND: $Y=\overline{A\cdot B}$ — AND followed by NOT; 0 only when both inputs are 1.
  • NOR: $Y=\overline{A+B}$ — OR followed by NOT; 1 only when both inputs are 0.

NAND and NOR are called universal gates: joining the two inputs of a NAND gives NOT, and combinations of NANDs alone can make AND and OR. Two rules (De Morgan's theorems) help with combinations: $\overline{A+B}=\overline A\cdot\overline B$ and $\overline{A\cdot B}=\overline A+\overline B$. For example, feeding $\overline A$ and $\overline B$ into a NAND gate gives $\overline{\overline A\cdot\overline B}=A+B$ — an OR gate.

Check: a truth-table lab
output $Y$—

Exercises

I
Intrinsic semiconductors. $e=1.6\times10^{-19}$ C; for Si $n_i=1.5\times10^{16}\ \text{m}^{-3}$, $\mu_e=0.135$, $\mu_h=0.048\ \text{m}^2\,\text{V}^{-1}\text{s}^{-1}$; for Ge $n_i=2.4\times10^{19}\ \text{m}^{-3}$, $\mu_e=0.39$, $\mu_h=0.19\ \text{m}^2\,\text{V}^{-1}\text{s}^{-1}$ (all at $300$ K).
  • 1 Pure germanium at $300$ K has $2.4\times10^{19}$ free electrons per m³. How many holes per m³ does it have, and why?
  • 2 How many free electrons and how many holes are there in a pure silicon chip of volume $2.0\ \text{mm}^3$?
  • 3 Explain how a hole moves through a crystal, and why it behaves as a positive charge.
  • 4 A field of $50$ V/m is applied to pure silicon. Find the drift speeds of electrons and holes and their directions relative to $\mathbf E$.
  • 5 Find the conductivity and resistivity of pure germanium at $300$ K.
  • 6 In an intrinsic semiconductor electrons and holes drift in opposite directions. Why do their currents add rather than cancel?
  • 7 What two processes go on continuously in a semiconductor at constant temperature, and what does it mean to say they are "in balance"?
1 $2.4\times10^{19}$ holes per m³: in a pure crystal every free electron comes from a broken bond that leaves one hole, so $n_h=n_e$.
2 $(1.5\times10^{16})(2.0\times10^{-9})=3.0\times10^{7}$ electrons and $3.0\times10^{7}$ holes.
3 A bound electron from a neighbouring bond jumps into the vacancy, so the vacancy appears in the neighbouring bond: it moves opposite to the electron's jump. The region with the missing electron has a net charge $+e$, and under a field the vacancy drifts along $\mathbf E$ — exactly as a $+e$ particle would.
4 $v_e=\mu_eE=(0.135)(50)\approx6.8$ m/s, opposite to $\mathbf E$; $v_h=(0.048)(50)=2.4$ m/s, along $\mathbf E$.
5 $\sigma=en_i(\mu_e+\mu_h)$ $=(1.6\times10^{-19})(2.4\times10^{19})(0.58)\approx2.2$ S/m; $\rho\approx0.45\ \Omega$ m.
6 Negative charges moving against $\mathbf E$ and positive charges moving along $\mathbf E$ both transfer positive charge in the direction of $\mathbf E$: both currents point along $\mathbf E$.
7 Generation (bonds broken by thermal energy, creating electron–hole pairs) and recombination (free electrons falling back into holes). In balance, pairs are created and destroyed at equal rates, so $n_e$ and $n_h$ stay constant.
II
Doping. Silicon has $5.0\times10^{28}$ atoms per m³.
  • 1 Which of these dopants make silicon n-type and which p-type: P, B, As, Al, Sb, In, Ga?
  • 2 Why must a dopant atom be similar in size to the host atoms?
  • 3 Explain why an n-type semiconductor is electrically neutral even though it has many more free electrons than holes.
  • 4 Sketch the energy-band diagram of an n-type semiconductor at room temperature, marking $E_C$, $E_V$ and the donor level $E_D$.
  • 5 The extra electron of an arsenic atom in germanium needs about $0.013$ eV to be freed. Compare this with $k_BT$ at $300$ K ($0.026$ eV). Are the donors ionised at room temperature?
  • 6 Silicon is doped with boron at one atom in $10^{6}$. Find the acceptor density.
  • 7 Name the majority and minority carriers in a p-type semiconductor, and say where each comes from.
1 n-type (pentavalent donors): P, As, Sb. p-type (trivalent acceptors): B, Al, In, Ga.
2 It must take the place of a host atom and bond to its neighbours without distorting the crystal; a badly mismatched atom would strain the lattice and create defects.
3 Each free electron came from a neutral donor atom, which is left as a fixed positive ion. Total positive charge (donor ions + holes) equals total negative charge (free electrons).
4 Full valence band (top $E_V$); conduction band (bottom $E_C$) with many electrons; a row of donor levels $E_D$ just below $E_C$ (about $0.01$–$0.05$ eV), mostly emptied; a few holes in the valence band.
5 $0.013/0.026=0.5$: the ionisation energy is only half of $k_BT$, so practically all donors are ionised.
6 $N_A=10^{-6}\times5.0\times10^{28}=5.0\times10^{22}\ \text{m}^{-3}$.
7 Majority: holes, created by the acceptor atoms (and a few by thermal bond breaking). Minority: electrons, created only by thermal bond breaking.
III
Mass action and neutrality. Use $n_i=1.5\times10^{16}\ \text{m}^{-3}$ (Si) and $2.4\times10^{19}\ \text{m}^{-3}$ (Ge) at $300$ K; assume all dopants are ionised.
  • 1 Silicon is doped with $1.0\times10^{21}$ donors per m³. Find $n_e$ and $n_h$.
  • 2 Silicon is doped with $3.0\times10^{22}$ acceptors per m³. Find $n_h$ and $n_e$.
  • 3 Germanium is doped with $4.8\times10^{21}$ donors per m³. Find the hole density.
  • 4 A sample has $n_e=4.5\times10^{22}$ and $n_h=5.0\times10^{9}\ \text{m}^{-3}$. Find $n_i$, identify the semiconductor (Si or Ge) and its type.
  • 5 A silicon sample has $n_h=1.0\times10^{20}\ \text{m}^{-3}$ and no donors. Find $n_e$ and state the type.
  • 6 Silicon contains $5.0\times10^{21}$ donors and $2.0\times10^{21}$ acceptors per m³. Find $n_e$ and $n_h$.
  • 7 In an n-type sample, the donor density is doubled. By what factor do $n_e$ and $n_h$ change?
  • 8 Challenge. Germanium is lightly doped with $N_D=2.4\times10^{19}$ donors per m³ — equal to $n_i$, so the usual approximation fails. Solve $n_e-n_h=N_D$ with $n_en_h=n_i^2$ exactly.
1 $n_e\approx1.0\times10^{21}$; $n_h=\dfrac{2.25\times10^{32}}{1.0\times10^{21}}=2.25\times10^{11}$ $\approx2.2\times10^{11}\ \text{m}^{-3}$ (round half to even).
2 $n_h\approx3.0\times10^{22}$; $n_e=\dfrac{2.25\times10^{32}}{3.0\times10^{22}}=7.5\times10^{9}\ \text{m}^{-3}$.
3 $n_h=\dfrac{(2.4\times10^{19})^2}{4.8\times10^{21}}=1.2\times10^{17}\ \text{m}^{-3}$.
4 $n_i=\sqrt{n_en_h}=\sqrt{2.25\times10^{32}}$ $=1.5\times10^{16}\ \text{m}^{-3}$: silicon, n-type (electrons in the majority).
5 $n_e=\dfrac{2.25\times10^{32}}{1.0\times10^{20}}=2.25\times10^{12}$ $\approx2.2\times10^{12}\ \text{m}^{-3}$; p-type.
6 Net donors $3.0\times10^{21}\gg n_i$: $n_e\approx3.0\times10^{21}$, $n_h=\dfrac{2.25\times10^{32}}{3.0\times10^{21}}=7.5\times10^{10}\ \text{m}^{-3}$ (n-type).
7 $n_e\approx N_D$ doubles; $n_h=n_i^2/n_e$ halves.
8 Substituting $n_h=n_i^2/n_e$: $n_e^2-N_Dn_e-n_i^2=0$, so $n_e=\dfrac{N_D}{2}+\sqrt{\dfrac{N_D^2}{4}+n_i^2}$ $=(1.2+\sqrt{1.44+5.76})\times10^{19}$ $\approx3.9\times10^{19}\ \text{m}^{-3}$, and $n_h=n_e-N_D\approx1.5\times10^{19}\ \text{m}^{-3}$. Check: $n_en_h\approx5.76\times10^{38}=n_i^2$. ✓
IV
Conductivity and applications. Silicon: $\mu_e=0.135$, $\mu_h=0.048\ \text{m}^2\,\text{V}^{-1}\text{s}^{-1}$; $e=1.6\times10^{-19}$ C.
  • 1 Find the conductivity and resistivity of n-type silicon with $N_D=1.0\times10^{22}\ \text{m}^{-3}$.
  • 2 Find the conductivity of p-type silicon with $N_A=1.0\times10^{22}\ \text{m}^{-3}$.
  • 3 What donor density gives n-type silicon a resistivity of $0.10\ \Omega$ m?
  • 4 Find the resistance of a bar of the material in part 1, $2.0$ mm long with a cross-section of $0.50\ \text{mm}\times0.50\ \text{mm}$.
  • 5 Modelling. A chip designer needs a $100$ k$\Omega$ resistor made from a strip of p-type silicon $100\ \mu$m long, $10\ \mu$m wide and $1.0\ \mu$m thick. What acceptor density is needed?
  • 6 At the same doping density, which conducts better, n-type or p-type silicon, and by what factor?
  • 7 A heavily doped n-type silicon sample is warmed from $300$ K to $350$ K. Explain what happens to (a) the majority carrier density, (b) the minority carrier density.
1 $\sigma\approx eN_D\mu_e$ $=(1.6\times10^{-19})(1.0\times10^{22})(0.135)$ $\approx216$ S/m $\approx2.2\times10^{2}$ S/m; $\rho\approx4.6\times10^{-3}\ \Omega$ m.
2 $\sigma\approx eN_A\mu_h$ $=(1.6\times10^{-19})(1.0\times10^{22})(0.048)$ $\approx77$ S/m.
3 $N_D=\dfrac{1}{\rho e\mu_e}$ $=\dfrac{1}{(0.10)(1.6\times10^{-19})(0.135)}$ $\approx4.6\times10^{20}\ \text{m}^{-3}$.
4 $R=\dfrac{\rho L}{A}=\dfrac{(4.63\times10^{-3})(2.0\times10^{-3})}{2.5\times10^{-7}}\approx37\ \Omega$.
5 $A=(10\ \mu\text{m})(1.0\ \mu\text{m})=1.0\times10^{-11}\ \text{m}^2$, so $\rho=\dfrac{RA}{L}$ $=\dfrac{(1.0\times10^{5})(1.0\times10^{-11})}{1.0\times10^{-4}}$ $=1.0\times10^{-2}\ \Omega$ m. Then $N_A=\dfrac{1}{\rho e\mu_h}$ $=\dfrac{1}{(0.010)(1.6\times10^{-19})(0.048)}$ $\approx1.3\times10^{22}\ \text{m}^{-3}$.
6 n-type, because electrons are more mobile: $\mu_e/\mu_h=0.135/0.048\approx2.8$ times better.
7 (a) Practically unchanged: $n_e\approx N_D$, already fully ionised. (b) It increases a lot: $n_h=n_i^2/N_D$ and $n_i$ grows steeply with temperature (so $n_i^2$ grows even faster).
🎯 Exam corner — intrinsic and extrinsic semiconductors BoardJEE MainNEET

E1. Board · 3 marks Distinguish between n-type and p-type semiconductors with respect to (i) the dopant used, (ii) the majority carriers, (iii) their energy-band diagrams.

(i) Dopant n-type: pentavalent (P, As, Sb) — donors. p-type: trivalent (B, Al, Ga, In) — acceptors.
(ii) Carriers n-type: electrons majority, holes minority. p-type: holes majority, electrons minority.
(iii) Bands n-type: donor level $E_D$ just below $E_C$, many electrons in the conduction band. p-type: acceptor level $E_A$ just above $E_V$, many holes in the valence band.
Exam tip Draw both diagrams side by side and mark $E_D$ or $E_A$ with a dashed line inside the gap, close to the right band edge.

E2. Board · 2 marks A pure silicon crystal has $5.0\times10^{28}$ atoms per m³ and $n_i=1.5\times10^{16}\ \text{m}^{-3}$. It is doped with boron at one part in $10^{7}$. Find the electron and hole densities and state the type.

Dopant Boron is an acceptor: $N_A=5.0\times10^{28}\times10^{-7}=5.0\times10^{21}\ \text{m}^{-3}$, so $n_h\approx5.0\times10^{21}\ \text{m}^{-3}$.
Minority $n_e=\dfrac{n_i^2}{n_h}=\dfrac{2.25\times10^{32}}{5.0\times10^{21}}=4.5\times10^{10}\ \text{m}^{-3}$. p-type.

E3. JEE Main An intrinsic semiconductor has $n_i=2.0\times10^{16}\ \text{m}^{-3}$. After doping, the hole density is $4.0\times10^{20}\ \text{m}^{-3}$. The electron density is (A) $4.0\times10^{20}$ (B) $2.0\times10^{16}$ (C) $1.0\times10^{12}$ (D) $5.0\times10^{3}\ \text{m}^{-3}$.

Concept $n_e=\dfrac{n_i^2}{n_h}=\dfrac{4.0\times10^{32}}{4.0\times10^{20}}=1.0\times10^{12}\ \text{m}^{-3}$.
Answer (C). Exam tip: (B) is the trap — doping reduces the minority density below $n_i$.

E4. NEET To make germanium a p-type semiconductor, it should be doped with (A) arsenic (B) antimony (C) indium (D) phosphorus.

Concept p-type needs a trivalent acceptor. As, Sb and P are pentavalent donors.
Answer (C) indium.

E5. JEE Main A semiconductor sample has $n_e=1.0\times10^{19}$ and $n_h=5.0\times10^{18}\ \text{m}^{-3}$, with mobilities $\mu_e=1.6$ and $\mu_h=0.20\ \text{m}^2\,\text{V}^{-1}\text{s}^{-1}$. Its conductivity is (A) $2.56$ S/m (B) $2.72$ S/m (C) $0.16$ S/m (D) $27.2$ S/m.

Concept $\sigma=e(n_e\mu_e+n_h\mu_h)$ $=(1.6\times10^{-19})(1.6\times10^{19}+1.0\times10^{18})$ $=(1.6\times10^{-19})(1.7\times10^{19})=2.72$ S/m.
Answer (B). Exam tip: (A) and (C) are the electron-only and hole-only terms — add both.

E6. NEETJEE Main (Beyond NCERT: logic gates.) The inputs $A$ and $B$ are each passed through a NOT gate, and the two outputs are fed to a NAND gate. The combination behaves as (A) an AND gate (B) an OR gate (C) a NOR gate (D) a NAND gate.

Concept Logic gates are beyond the CBSE and JEE Advanced syllabi but in NEET and JEE Main. $Y=\overline{\overline A\cdot\overline B}=\overline{\overline A}+\overline{\overline B}=A+B$ (De Morgan).
Check Truth table: $A=B=0$ gives $\overline A=\overline B=1$, NAND output $0$; any input $1$ makes one NAND input $0$, output $1$ — the OR table.
Answer (B).
🔬 Research–Teaching NexusResearch-tutored · optional, about 30 minutes

Research question. How pure must silicon be before its behaviour is controlled by the dopants we add rather than by the impurities left in it?

Why it matters. Polysilicon factories in China, Germany, the USA, Malaysia and Japan purify silicon to "solar grade" (about six nines, $99.9999\%$) or "electronic grade" (nine nines or better). Every extra nine costs energy and money. Physics tells you how many nines a device actually needs.

What to do.

  1. Silicon has $5.0\times10^{28}$ atoms per m³. Find the impurity density for purities of $6$, $9$ and $11$ nines (impurity fractions $10^{-6}$, $10^{-9}$, $10^{-11}$), assuming every impurity atom is an electrically active donor or acceptor.
  2. Compare each with $n_i=1.5\times10^{16}\ \text{m}^{-3}$ at $300$ K. Which purities still leave the silicon effectively extrinsic?
  3. A chip region is to be doped deliberately at $10^{22}\ \text{m}^{-3}$. Is $9$-nines silicon pure enough that leftover impurities change the carrier density by less than $1\%$?
  4. What impurity fraction would be needed for silicon to behave intrinsically at $300$ K (say, impurities below $n_i/10$)? Express it as a number of nines.

Tools: Calculator or spreadsheet.

Hand in: A short table, your answers to steps 2–4, and a paragraph on why solar cells can tolerate dirtier silicon than microprocessors (hint: think about the doping levels each uses, and about how many devices on a chip must all work).

Going further. Not all impurities are donors or acceptors: metals such as iron and gold add levels near the middle of the gap and make electrons and holes recombine quickly. Find out why these "lifetime killers" matter so much for solar cells.

Instructor note: impurity densities $5\times10^{22}$, $5\times10^{19}$, $5\times10^{17}\ \text{m}^{-3}$ — respectively $3.3\times10^{6}$, $3.3\times10^{3}$ and $33$ times $n_i$: all still extrinsic. Step 3: $5\times10^{19}/10^{22}=0.5\%$ — yes, just. Step 4: fraction $\le1.5\times10^{15}/5\times10^{28}=3\times10^{-14}$, about $13$–$14$ nines — beyond what industry achieves, which is why practical silicon is never truly intrinsic at room temperature. Solar cells are doped heavily and are large, simple devices, so they tolerate more impurity.

🧪 Self-check quiz

0/5 correct

Q1. In n-type silicon the majority carriers are

Q2. An arsenic atom placed in a silicon crystal acts as

Q3. Silicon with $n_i=1.5\times10^{16}\ \text{m}^{-3}$ is doped so that $n_e=4.5\times10^{22}\ \text{m}^{-3}$. The hole density is

Q4. In an applied electric field, holes in a semiconductor drift

Q5. A piece of p-type germanium, taken as a whole, is

← 14.1 Metals, Semiconductors and Insulators: Energy Bands14.3 The p–n Junction, the Semiconductor Diode and Rectifiers →