In short

A semiconductor is a material whose conductivity lies between a metal's and an insulator's, and — crucially — can be changed by many orders of magnitude by adding controlled trace impurities. The standard example is silicon (Si): atomic number 14, four valence electrons, crystal structure diamond-cubic.

At absolute zero, pure (intrinsic) silicon has a filled valence band of energy states and an empty conduction band, separated by a band gap E_g = 1.12 eV. No charge can flow because every valence state is occupied and no conduction state is. Heat the crystal to 300 K, and thermal energy k_B T = 0.026 eV — much smaller than E_g, but exponentially many electrons still tunnel up thanks to the Boltzmann factor — promoting roughly n_i \approx 1.5 \times 10^{16} electrons per cubic metre into the conduction band and leaving the same number of holes in the valence band. Both carry current: electrons by drift, holes by the motion of the missing-electron pattern in the opposite direction.

Extrinsic silicon is intrinsic silicon with controlled dopants added at concentrations as low as one impurity per ten million silicon atoms. Two choices:

  • n-type: pentavalent donors (P, As, Sb) — one extra valence electron per atom, loosely bound with energy \sim 0.045 eV. At room temperature nearly every donor is ionised. Carrier population is dominated by these electrons; holes are the minority.
  • p-type: trivalent acceptors (B, Al, Ga) — one missing valence electron, creating a state just above the valence band. Acceptors bind a valence electron, leaving a mobile hole. Holes dominate; electrons are the minority.

The law of mass action is exact for non-degenerate semiconductors: n \cdot p = n_i^2, independent of doping level. Pile on electrons and the holes are squeezed away, and vice versa.

Silicon's conductivity rises with temperature (unlike copper's, which falls). That is because the carrier density n_i \propto T^{3/2}\exp(-E_g / 2k_B T) increases explosively, overwhelming the weak \sim 1/T decline in mobility. The chips in every Samsung phone assembled at Sriperumbudur, every Tata Electronics SoC packaged at Dholera, and every ISRO radiation-hardened controller aboard Chandrayaan are built on this single temperature signature — and on the ability to paint n-type and p-type regions onto silicon wafers with nanometre precision.

Take a slab of copper and a slab of glass, both the size of a matchbox. Apply a 1-volt battery across each. Copper passes about 10^6 amps per square metre — enough to melt itself almost instantly. Glass passes about 10^{-12} amps per square metre — a current that would take a hundred thousand years to move one coulomb.

The ratio is 10^{18}. Eighteen orders of magnitude.

Now take a slab of silicon the same size. Pure silicon, no impurities. It passes about 3 \times 10^{-4} amps per square metre — roughly the middle of the eighteen-order-of-magnitude gap, on a logarithmic scale.

That is already strange. Silicon sits in the periodic table right below carbon (which forms diamond, a crystalline insulator) and right above germanium (also semi-conducting). It should be insulating by family resemblance. It isn't quite. And then the strangeness compounds: take that silicon slab, add one phosphorus atom per million silicon atoms — a dopant fraction you could measure with a kitchen balance if your silicon slab weighed a tonne — and the conductivity jumps by a factor of about a thousand. Add one boron atom per million silicons instead, and the conductivity jumps by a thousand again, but now in a way that is different — a way you can tell apart from the phosphorus-doped version by a two-minute experiment with a hot plate and a voltmeter.

The entire semiconductor industry is built on these two tricks — n-type and p-type doping — and on a theoretical framework (the band picture) that explains why they work. Every transistor in every chip you have ever used is a controlled interplay of n-type and p-type regions. Every solar panel. Every LED. Every pulse-oximeter clip that a Mumbai hospital clamps on a COVID patient's finger. The physics of how one atom in a million can change a material's electrical behaviour by three orders of magnitude is what this article is about.

The band picture — from atom to crystal

Start with a single silicon atom, isolated in space. Its 14 electrons fill the 1s, 2s, 2p, 3s orbitals completely; the 3p orbital holds 2 more electrons (silicon's 3s^2 3p^2 valence configuration). The four outer electrons — two in 3s, two in 3p — are the valence electrons. The energy levels are discrete: a specific 3s energy, a specific 3p energy, specific gaps between them. If you put two silicon atoms very close together so their wave functions overlap, each atomic level splits into two — one slightly higher (antibonding), one slightly lower (bonding). Put N = 10^{23} silicon atoms together in a crystal, and each atomic level splits into 10^{23} closely-spaced energy levels, which from our scale look like a continuous band.

From isolated atom to crystal: discrete levels broaden into bandsThree columns showing: (left) discrete 3s and 3p atomic energy levels of a single silicon atom; (middle) levels split when two atoms are brought together; (right) levels broaden into filled valence band and empty conduction band separated by a 1.12 eV band gap in crystalline silicon. 1 atom 3p 3s 2p, 2s, 1s 2 atoms split 3p split 3s crystal (N≈10²³ atoms) conduction band (empty at T=0) valence band (filled at T=0) E_g = 1.12 eV energy E
One silicon atom has discrete $3s$ and $3p$ valence levels. Bring two atoms together and each level splits. In a crystal of $10^{23}$ atoms, the levels broaden into continuous bands. For silicon at $T=0$, the valence band is full and the conduction band empty; the 1.12 eV band gap between them is the key number.

In silicon specifically, the 3s and 3p bands hybridise and then split into a pair of sub-bands. The lower one — the valence band — is exactly filled at T = 0 (four electrons per atom, four states per atom in that band). The upper one — the conduction band — is exactly empty. The gap between them is E_g = 1.12 eV.

Why a filled band carries no current

Here is the idea that makes bands matter. An electron in a crystal, travelling without scattering, moves at a group velocity v_g = (1/\hbar)(dE/dk) determined by the slope of the band dispersion E(k). In a filled band, for every electron at wave-vector +k moving to the right at velocity +v, there is another at -k moving left at -v (the band is symmetric, E(k) = E(-k), in any crystal with inversion symmetry). Sum their currents: exactly zero.

Apply an electric field. The field accelerates every electron — each k state would shift to k + \delta k. But Pauli exclusion says no two electrons can sit in the same state. In a filled band, there is no empty state for any electron to move into. The whole band is stuck. No current.

In contrast, in a partially-filled band — the situation in a metal like copper — the topmost occupied states have empty states right above them in k-space. The field shifts the whole filled region slightly, breaking the +k / -k symmetry, and producing a net current. This is why copper conducts.

The band picture distils the whole electronic structure of a solid into one question: is the highest band filled or partly filled? If filled, look at the gap to the next band. That gap decides everything:

Material Band gap Behaviour
Diamond (C) 5.5 eV insulator
Silicon (Si) 1.12 eV semiconductor
Germanium (Ge) 0.67 eV semiconductor
GaAs 1.43 eV semiconductor (direct gap → LEDs)
Copper 0 (half-filled 4s) metal

The boundary between "semiconductor" and "insulator" is fuzzy — materials with E_g > 3 eV are conventionally called insulators, but they are really just wide-gap semiconductors. Diamond at 300 °C conducts measurably, just as silicon does at -50 °C. The physics is the same; only the numbers differ.

Intrinsic silicon — the thermal excitation picture

Take pure, crystalline silicon. At T = 0, the valence band is full and the conduction band is empty. Lift the temperature. Thermal energy k_B T at 300 K is 0.026 eV — fifty times smaller than the 1.12 eV gap. Classically, no electron can cross the gap. Quantum-mechanically, the tail of the Boltzmann distribution reaches to any energy, with probability proportional to \exp(-E/k_B T). The number of conduction-band electrons at temperature T is

n(T) = \int_{E_c}^{\infty} g_c(E) f(E) \, dE

where g_c(E) is the conduction-band density of states (number of allowed k-states per unit energy) and f(E) = 1/(\exp((E-\mu)/k_B T) + 1) is the Fermi-Dirac occupation. The integral evaluates (for non-degenerate semiconductors, where the Fermi level sits well inside the gap) to

n(T) = N_c \exp\!\left(-\frac{E_c - \mu}{k_B T}\right), \qquad N_c = 2\left(\frac{m_e^* k_B T}{2\pi \hbar^2}\right)^{3/2}

Why: the density of states in a 3D parabolic band goes as (E - E_c)^{1/2}, and the Fermi-Dirac tail is approximately Boltzmann when (E - \mu) \gg k_B T. Their product integrates in closed form to the expression above. N_c is the effective density of states in the conduction band — a characteristic carrier density set by thermal de Broglie wavelength. For silicon at 300 K, N_c \approx 2.8 \times 10^{25}\text{ m}^{-3}. m_e^* is the effective mass — the electron's inertia inside the crystal, typically smaller than the free-electron mass because the periodic potential reshapes the dispersion.

Similarly, the hole population (empty states in the valence band) is

p(T) = N_v \exp\!\left(-\frac{\mu - E_v}{k_B T}\right), \qquad N_v = 2\left(\frac{m_h^* k_B T}{2\pi \hbar^2}\right)^{3/2}

Multiply n and p:

np = N_c N_v \exp\!\left(-\frac{E_c - E_v}{k_B T}\right) = N_c N_v \exp\!\left(-\frac{E_g}{k_B T}\right)

Why: the \mu dependence cancels. The product is a property of the material (band gap, effective masses, temperature) alone — independent of where the Fermi level happens to sit. This is why it holds in doped semiconductors too, as long as they are non-degenerate.

Call this product n_i^2, where n_i is the intrinsic carrier concentration:

\boxed{n_i = \sqrt{N_c N_v} \cdot \exp\!\left(-\frac{E_g}{2 k_B T}\right)}

For silicon at 300 K: E_g / 2k_B T = 1.12 / (2 \times 0.0259) = 21.6, so \exp(-21.6) = 4 \times 10^{-10}. Multiplied by \sqrt{N_c N_v} \approx 2.9 \times 10^{25}\text{ m}^{-3}, this gives n_i \approx 1.5 \times 10^{16}\text{ m}^{-3}. One electron in the conduction band per 3 \times 10^{12} silicon atoms — a vanishingly small fraction, but hugely more than the population in an insulator, because the Boltzmann factor for diamond (E_g = 5.5 eV) is \exp(-5.5/0.052) = 10^{-47}.

Electrons and holes — two carriers, one crystal

When one electron gets thermally lifted from valence to conduction, it leaves behind an empty state in the valence band. That empty state is a hole. The remaining 10^{29} - 1 electrons in the valence band are now free to shuffle one position — any one of them can hop into the empty state, leaving a new hole wherever it came from.

An applied electric field drifts electrons against the field. It also drifts the hole: as electrons nearby hop into the hole from the direction opposite to the field, the hole moves with the field, like a bubble in water moving against the water's net motion. Mathematically, the hole behaves like a positive charge +e with an effective mass m_h^*, travelling in the direction of the field. Both carriers contribute to the current.

In intrinsic silicon, n = p = n_i — every conduction-band electron came from a valence-band hole. Total conductivity:

\sigma_i = e n_i \mu_e + e p \mu_h = e n_i (\mu_e + \mu_h)

For silicon, \mu_e \approx 0.135 \text{ m}^2/\text{V·s}, \mu_h \approx 0.048 \text{ m}^2/\text{V·s}. Plugging in n_i = 1.5 \times 10^{16}\text{ m}^{-3}: \sigma_i \approx 1.6 \times 10^{-19} \times 1.5\times 10^{16} \times (0.135 + 0.048) \approx 4.4 \times 10^{-4}\text{ S/m}. Resistivity \rho_i \approx 2300 \text{ Ω·m}. A metre-long, square-cm bar of pure silicon has resistance \sim 2.3 \times 10^7\text{ Ω} — high enough for a good insulator, low enough to be a semiconductor.

Extrinsic silicon — what one phosphorus atom buys you

Silicon's power comes from what happens when you replace one in a million silicon atoms with a dopant from group 15 (pentavalent) or group 13 (trivalent) of the periodic table.

n-type: phosphorus, a free electron per atom

Silicon crystallises in a tetrahedral lattice: each atom shares four covalent bonds with its four neighbours, using all four of its valence electrons. Now swap one silicon out for phosphorus (valence 5). Phosphorus uses four of its five valence electrons to make the four bonds; the fifth electron has no bond to form. It sits in a hydrogen-like orbital around the phosphorus nucleus, weakly bound.

Treat the fifth electron as a hydrogen-atom problem, but inside silicon. The Coulomb interaction is reduced by the silicon dielectric constant \epsilon_r = 11.7, and the electron's inertia is the effective mass m_e^* = 0.26 m_e. Binding energy scales as (1/\epsilon_r)^2 \cdot (m_e^*/m_e) \cdot 13.6 eV:

E_D = \frac{m_e^*/m_e}{\epsilon_r^2} \cdot 13.6 \text{ eV} = \frac{0.26}{11.7^2} \times 13.6 = 0.026 \text{ eV}

Why: the Bohr model of an electron orbiting a proton has binding 13.6 eV. Put that same Coulomb orbit in a medium with dielectric constant \epsilon_r (screening the proton charge) and replace the free mass with the crystal's effective mass, and the energy scales as m^*/\epsilon_r^2. The experimental phosphorus-in-silicon donor binding is 0.045 eV — a touch higher than the hydrogenic estimate because the actual orbital has some deviation from pure hydrogenic form near the phosphorus core. But the order of magnitude is right, and the physics is simple Bohr.

The ionisation energy 0.045 eV is comparable to k_B T = 0.026 eV at room temperature. At 300 K, essentially every donor is ionised: the fifth electron has enough thermal energy to leave its phosphorus parent and wander free through the silicon lattice, joining the conduction band.

If you dope silicon with N_D = 10^{22}\text{ m}^{-3} phosphorus atoms (one per 5 \times 10^6 silicons), you inject 10^{22}\text{ m}^{-3} electrons into the conduction band at room temperature — a carrier density a million times the intrinsic n_i = 1.5 \times 10^{16}. The conductivity multiplies by the same factor.

The holes, by contrast, are crushed. The law of mass action still holds: np = n_i^2. So

p = \frac{n_i^2}{n} = \frac{(1.5\times 10^{16})^2}{10^{22}} = 2.25 \times 10^{10}\text{ m}^{-3}

— six orders of magnitude below intrinsic. The electrons have become the majority carrier; the holes are the minority carrier.

p-type: boron, a hole per atom

Replace a silicon with boron (valence 3). Boron has only three valence electrons; it makes only three of the four expected bonds. The missing bond is an empty valence state, which any nearby valence electron can hop into — releasing a hole that wanders through the lattice. Boron is an acceptor: it accepts an electron from the valence band, creating a hole.

The binding energy of the hole to the boron acceptor is similar to the donor case — about 0.045 eV, comparable to k_B T — so every boron is effectively ionised at room temperature. If N_A = 10^{22}\text{ m}^{-3} boron atoms are added, p \approx N_A = 10^{22}\text{ m}^{-3} and n = n_i^2/p = 2.25 \times 10^{10}\text{ m}^{-3}.

Interactive: doping level shifts the carrier populationsA schematic of valence and conduction bands with populations of electrons (red dots) and holes (open circles), and a slider to change the donor concentration from intrinsic to heavily n-doped. As you raise N_D, electrons multiply and holes deplete; readouts show n, p, and conductivity. log₁₀(Nₐ or N_D, m⁻³) log₁₀(carrier density, m⁻³) 10 14 18 22 26 10 15 20 25 n_i = 1.5×10¹⁶ electrons n holes p drag the red dot to change doping
Drag the red dot along the horizontal axis to change the donor concentration from $10^{10}$ (essentially intrinsic) to $10^{25}$ per m³ (heavily doped). At low doping, $n = p = n_i$; above $n_i$, electrons pin at $N_D$ and holes are suppressed as $n_i^2 / N_D$. The conductivity readout $\sigma \approx e(n\mu_e + p\mu_h)$ tracks the change. Going from $n_i$ to $10^{22}$ electrons per m³ multiplies conductivity by a million.

The law of mass action, in full

For a non-degenerate semiconductor at thermal equilibrium, regardless of doping type or level:

np = n_i^2

This is a restatement of detailed balance. The rate at which electron-hole pairs are thermally generated (proportional to n_i^2 times temperature-dependent factors) must equal the rate at which they recombine (proportional to the product np). In equilibrium, the two rates balance.

The consequences:

This is the single most-used equation in semiconductor device physics. Knowing the doping type and density, you immediately know both carrier populations.

Why semiconductor conductivity rises with temperature

For a metal like copper, the conductivity is \sigma = ne\mu, and n is temperature-independent (the density of conduction electrons in copper is set by the number of free 4s electrons per atom — roughly one per atom, roughly 10^{29}\text{ m}^{-3}, regardless of temperature). The mobility \mu decreases with temperature as \mu \propto 1/T, because hotter lattice vibrations (phonons) scatter electrons more often. Net: copper's conductivity falls with rising temperature. This is why the heating element in your kettle rises in resistance as it glows red.

For a semiconductor, the carrier density n_i grows exponentially with temperature:

n_i(T) \propto T^{3/2} \exp\!\left(-\frac{E_g}{2 k_B T}\right)

Double the temperature from 300 K to 600 K. The pre-factor T^{3/2} grows by 2^{1.5} = 2.83. The exponential factor grows by \exp(-(1.12)/(2 \times 0.0259 \times 300/600)) \cdot \exp(+E_g/2k_B T_1) = \exp(E_g/(2k_BT_1) - E_g/(2k_BT_2)) = \exp(21.6 - 10.8) = \exp(10.8) \approx 50000. The mobility falls by maybe a factor of 5. Net conductivity grows by 2.83 \times 50000 / 5 \approx 28000.

Why the exponential: the band gap is a fixed barrier. Lifting temperature moves more and more electrons over it at an exponential rate — the Boltzmann tail of the Fermi-Dirac distribution. Each tiny increment in T doubles or triples the carrier count, easily outrunning any power-law cooling from mobility. Kids doing a school experiment with a Ge diode and a hair-dryer watch the current rise tenfold in a few seconds — an inversion of what a copper wire would do.

This is why pure semiconductor parts have to be held at controlled temperature — a CPU runs cooler than 85 °C not just to avoid damage, but because above that, thermal carriers start to wash out the designed doping profile. Above a freezeout low temperature (for silicon, about 100 K), donors are not all ionised; above an intrinsic high temperature (for silicon, about 400 °C), thermal pairs dominate the doping. The "extrinsic" operating regime is a window in between, and it is set by the band gap. This is why gallium nitride (E_g = 3.4 eV) and silicon carbide (E_g = 3.3 eV) are replacing silicon in high-power, high-temperature electronics — their wider gaps push the intrinsic temperature up to several hundred Celsius.

Worked examples

Example 1: Intrinsic carrier density at 300 K

Compute n_i for silicon at 300 K using E_g = 1.12 eV, N_c = 2.8 \times 10^{25}\text{ m}^{-3}, and N_v = 1.04 \times 10^{25}\text{ m}^{-3}. Then compute the conductivity.

Silicon band structure at 300 KSchematic silicon band diagram showing valence band filled (red), conduction band mostly empty with a few electron dots, and 1.12 eV band gap labeled. conduction band (E_c) valence band (E_v) E_g = 1.12 eV few e⁻ few holes
At 300 K, a tiny fraction of valence-band electrons are thermally promoted to the conduction band, leaving holes behind. In intrinsic silicon, $n = p = n_i$.

Step 1. Compute k_B T in eV at 300 K.

k_B T = 8.617 \times 10^{-5} \text{ eV/K} \times 300 \text{ K} = 0.02585 \text{ eV}

Why: converting to eV keeps the arithmetic in units matched to the band gap, so the Boltzmann factor stays clean.

Step 2. Compute the Boltzmann factor.

\frac{E_g}{2 k_B T} = \frac{1.12}{2 \times 0.02585} = 21.66
\exp(-21.66) = 3.91 \times 10^{-10}

Why: the 1/2 in E_g/(2k_BT) comes from the square-root in n_i = \sqrt{N_c N_v} \exp(-E_g/2k_BT) — the effective gap for the geometric mean is half the physical gap.

Step 3. Compute \sqrt{N_c N_v}.

\sqrt{N_c N_v} = \sqrt{2.8 \times 10^{25} \times 1.04 \times 10^{25}} = \sqrt{2.91 \times 10^{50}} = 1.71 \times 10^{25}\text{ m}^{-3}

Step 4. Multiply.

n_i = 1.71 \times 10^{25} \times 3.91 \times 10^{-10} = 6.7 \times 10^{15}\text{ m}^{-3}

Why: a common textbook value is n_i = 1.0 \times 10^{16}\text{ m}^{-3} at 300 K, and a common experimental value is 1.45 \times 10^{16}. Our computed 6.7 \times 10^{15} is a little low because we used simple effective-mass N_c and N_v; the band structure of silicon has multiple valleys in the conduction band, which corrects N_c upward by a factor of 6 (the multiple-valley degeneracy). Using the experimental N_c = 1.04 \times 10^{25} \cdot 6 / (m_h^*/m_e)^{3/2} etc. gives the textbook answer.

Step 5. Compute conductivity.

\sigma_i = e n_i (\mu_e + \mu_h) = 1.6\times 10^{-19} \times 6.7 \times 10^{15} \times (0.135 + 0.048)
\sigma_i = 1.6\times 10^{-19} \times 6.7 \times 10^{15} \times 0.183 \approx 1.96 \times 10^{-4}\text{ S/m}
\rho_i = 1/\sigma_i \approx 5100 \text{ Ω·m}

Why: both electrons and holes carry current; the mobility sum is what matters for bulk conductivity. The resistivity 5000 Ω·m is 10⁹ times larger than copper (1.7×10⁻⁸ Ω·m) and 10^9 times smaller than glass (10^{12} Ω·m). Silicon, indeed, sits in the middle.

Result: n_i = 6.7 \times 10^{15}\text{ m}^{-3} (textbook: \sim 10^{16}), \sigma_i \approx 2 \times 10^{-4}\text{ S/m}, \rho_i \approx 5000\text{ Ω·m}.

What this shows: The intrinsic carrier density in silicon is tiny — one electron per about 10^{13} atoms — and the corresponding resistivity is large, but not absurdly so. A small number of impurity atoms can change this by many orders of magnitude.

Example 2: Doping boosts conductivity by a factor of a thousand

Silicon is doped with phosphorus at a concentration of 10^{22}\text{ m}^{-3} (one phosphorus per 5 \times 10^6 silicons). Assume all donors are ionised at 300 K. Find n, p, \sigma, and the factor by which conductivity has increased relative to intrinsic silicon. Use n_i = 1.5\times 10^{16}\text{ m}^{-3}, \mu_e = 0.135, \mu_h = 0.048.

n-type silicon band diagram with donor levelSilicon band diagram with conduction band (mostly empty), valence band (mostly full), plus a shallow donor level 0.045 eV below the conduction band edge populated with electrons that have mostly ionised into the conduction band. conduction band — n ≈ 10²²/m³ donor E_D, 0.045 eV below E_c + + + valence band — p ≈ 2×10¹⁰/m³ (depleted)
n-type silicon band diagram. Donor atoms sit at a shallow level 0.045 eV below the conduction band edge. At room temperature nearly all have ionised, dumping their electrons into the conduction band.

Step 1. Find the majority-carrier concentration.

Each donor contributes one electron. With N_D = 10^{22}\text{ m}^{-3} fully ionised donors and negligible thermal contribution (since N_D \gg n_i):

n \approx N_D = 10^{22}\text{ m}^{-3}

Why: charge neutrality plus N_D \gg n_i. The full form is n = N_D + p, and p is tiny, so n \approx N_D.

Step 2. Find the minority-carrier concentration using the law of mass action.

p = \frac{n_i^2}{n} = \frac{(1.5 \times 10^{16})^2}{10^{22}} = \frac{2.25 \times 10^{32}}{10^{22}} = 2.25 \times 10^{10}\text{ m}^{-3}

Why: the law of mass action np = n_i^2 holds in equilibrium. Pile on electrons and the holes recombine away until the product is restored.

Step 3. Compute conductivity.

\sigma = e n \mu_e + e p \mu_h = 1.6\times 10^{-19} \left[ 10^{22} \times 0.135 + 2.25 \times 10^{10} \times 0.048 \right]

The hole term is 1.6\times 10^{-19} \times 1.08 \times 10^9 = 1.7 \times 10^{-10}\text{ S/m} — utterly negligible. The electron term is

\sigma \approx 1.6 \times 10^{-19} \times 10^{22} \times 0.135 = 2.16 \times 10^{2} \text{ S/m}

Why: in n-type silicon, electrons are the majority carrier and dominate the conductivity. You can ignore the hole contribution completely.

Step 4. Find the enhancement factor.

\frac{\sigma_{\text{n-type}}}{\sigma_i} = \frac{216}{4.4\times 10^{-4}} = 4.9 \times 10^{5}

Why: adding one phosphorus atom per five million silicons has boosted the conductivity by nearly half a million. The resistivity has fallen from 2300 Ω·m to 4.6 × 10⁻³ Ω·m — four hundred thousand times smaller.

Step 5. Confirm conductivity is dominated by doping, not temperature.

Intrinsic conductivity at 300 K: \sigma_i = 4.4\times 10^{-4}\text{ S/m}. n-type conductivity: 216\text{ S/m}. The ratio is 5\times 10^5, matching Step 4.

Result: Doping with 10^{22}\text{ m}^{-3} phosphorus raises the conductivity from 4.4\times 10^{-4} S/m to 216 S/m — a factor of 5\times 10^5. Resistivity drops from 2300 Ω·m to 4.6 mΩ·m — within a factor of 3 of the resistivity of cast iron.

What this shows: The trick of the semiconductor industry, in one calculation. One dopant atom per five million host atoms — a purity level that sounds impossibly demanding in reverse (the host has to be cleaner than that, to less than 1 ppm of unwanted impurities) — changes conductivity by almost six orders of magnitude. Every transistor, every diode, every solar cell manufactured at Samsung's plant in Sriperumbudur near Chennai, at Tata Electronics' Dholera fab in Gujarat, at Foxconn's Bengaluru facility, relies on this sensitivity.

Common confusions

If you came here to understand what a semiconductor is, why doping matters, and why silicon's conductivity rises with temperature, the above is enough. What follows is for readers who want to see the Fermi level picture, the carrier statistics in full, and the connection to device physics.

Where does the Fermi level sit?

The Fermi level \mu (also called the chemical potential, or E_F at T=0) is the energy at which the occupation probability f(E) = 1/2. For an intrinsic semiconductor, it sits near the middle of the gap. Specifically, using the equality n = p:

N_c \exp\!\left(-\frac{E_c - \mu}{k_B T}\right) = N_v \exp\!\left(-\frac{\mu - E_v}{k_B T}\right)

Taking logs:

-\frac{E_c - \mu}{k_B T} + \ln N_c = -\frac{\mu - E_v}{k_B T} + \ln N_v

Solve for \mu:

\mu = \frac{E_c + E_v}{2} + \frac{k_B T}{2} \ln\!\frac{N_v}{N_c}

Why: the Fermi level lies at the midpoint of the band gap only if N_c = N_v, i.e., if the electron and hole effective masses are equal. For silicon, m_e^* < m_h^* by about 2×, so N_c > N_v, and the Fermi level is slightly below midgap by (k_B T/2) \ln(N_v/N_c) \approx -12 meV at 300 K.

In an n-type semiconductor, doping raises \mu toward E_c. For n = N_D:

\mu = E_c - k_B T \ln(N_c / N_D)

For N_D = 10^{22}\text{ m}^{-3} and N_c = 2.8\times 10^{25}\text{ m}^{-3}: \mu = E_c - 0.026 \ln(2800) = E_c - 0.206 eV. The Fermi level has climbed from near-midgap (0.56 eV below E_c) to 0.21 eV below E_c.

In a p-type semiconductor, doping pushes \mu toward E_v. The two Fermi levels, on opposite sides of the gap, are what create the built-in voltage of a p-n junction — the subject of the next article.

Freeze-out, extrinsic, and intrinsic regimes

For an n-type semiconductor with donor density N_D, the carrier density n(T) has three regimes:

  • Freezeout (low T, k_B T \ll E_D): donors are only partially ionised. Carrier density is set by the Boltzmann factor for donor ionisation:
n(T) \approx \sqrt{N_c N_D / 2} \exp\!\left(-\frac{E_D}{2 k_B T}\right)

Silicon's freezeout temperature is about 100 K for typical doping.

  • Extrinsic (middle T): donors are fully ionised, thermal pair generation is negligible.
n(T) \approx N_D

This is the operating regime of all room-temperature electronics.

  • Intrinsic (high T, n_i > N_D): thermal pair generation overwhelms doping.
n(T) \approx n_i(T)

For silicon at N_D = 10^{22}, this happens above about 400 °C.

A chip has to live in the extrinsic window. Wide-gap semiconductors (SiC, GaN) have wider windows, pushing the intrinsic temperature up to 600–800 °C and enabling power electronics in electric vehicle inverters, solar micro-inverters, and high-temperature industrial control. The Indian Space Research Organisation uses SiC devices in the power conversion units of its launch vehicles.

The Hall effect — how you tell n-type from p-type

Pass a current I through a semiconductor bar along \hat x and apply a magnetic field B along \hat z. The Lorentz force deflects the carriers sideways. If the carriers are electrons moving in -\hat x (to carry current in +\hat x), the magnetic force \vec F = -e\vec v \times \vec B points in -\hat y, pushing electrons to one edge and leaving the other positive.

In a p-type sample, holes move in +\hat x. The magnetic force on a positive charge moving in +\hat x with \vec B in +\hat z is e\vec v \times \vec B, which points in -\hat y. Holes pile up on the same edge electrons would have piled up on — but the edge is now positive, not negative.

The Hall voltage V_H therefore has opposite sign in n-type and p-type samples. Measure V_H across the bar while passing a known current through a known field: the sign tells you the carrier type, and the magnitude gives you the carrier density via V_H = IB / (nte) where t is the bar thickness. This is the standard way labs characterise semiconductor samples. The Hall effect is how Edwin Hall, in 1879, established that current in copper is carried by negatively-charged objects — fifty years before the electron was conceptually in place.

Why the mass-action law fails outside equilibrium

In devices, especially when you shine light on a solar cell or forward-bias a diode, carrier populations are driven out of equilibrium. An excess of electrons \Delta n and holes \Delta p above equilibrium is generated. The product np temporarily exceeds n_i^2.

Out of equilibrium, the law of mass action becomes an inequality: np \ge n_i^2. The system drives toward equilibrium by recombination, at a rate R = (np - n_i^2) / \tau where \tau is the recombination lifetime. For silicon, \tau ranges from microseconds (direct-gap recombination dominated by impurities) to milliseconds (indirect recombination in ultra-pure crystals). The excess carrier lifetime sets the performance ceiling of solar cells — you have roughly \tau microseconds to collect an electron-hole pair before it recombines.

This is why silicon solar cells need to be relatively thick (hundreds of microns) — a photogenerated pair has to diffuse to the junction before recombining, and diffusion length L = \sqrt{D\tau} needs to be at least as long as the cell thickness. Thinner, direct-gap materials (CIGS, CdTe, perovskites) can be much thinner for the same performance.

The effective mass — what the band curvature buys you

Why should the electron in silicon have an effective mass m_e^* = 0.26 m_e, smaller than its free mass? Because in a crystal, the electron's inertia is governed by the curvature of the band dispersion E(k):

\frac{1}{m^*} = \frac{1}{\hbar^2} \frac{d^2 E}{dk^2}

A strongly curved band — a band where E grows fast with k — corresponds to a light effective mass. A flat band corresponds to a heavy effective mass.

In silicon's conduction band, the band curvature near the minimum gives m_e^* \approx 0.26 m_e along one direction and \approx 0.98 m_e along perpendicular directions. The density-of-states effective mass (the geometric mean, weighted by degeneracy) is what enters N_c. Similar story for the valence band, which has multiple sub-bands (heavy hole, light hole, split-off) — the "effective mass" is a summary of a richer band-structure picture.

The effective mass controls everything that involves inertia: drift velocity, cyclotron resonance frequency, Bohr radius of the hydrogenic donor orbital. In GaAs, m_e^* = 0.067 m_e — much lighter than silicon, which is why GaAs transistors are faster for a given doping: electrons accelerate more for a given field.

Why silicon and not germanium?

Germanium is a fine semiconductor. Its band gap is 0.67 eV, its mobility is higher than silicon's, and early transistors at Bell Labs in the 1950s were made from germanium. So why silicon?

Three reasons:

  • Oxide quality: silicon's native oxide SiO₂ is chemically stable, electrically insulating, and forms a smooth, well-passivated interface with the silicon surface. Germanium's native oxide GeO₂ is water-soluble, electrically leaky, and forms a poor interface. This matters for the MOSFET, which is built directly on the oxide-semiconductor interface.

  • Band gap: silicon's 1.12 eV gap keeps intrinsic carrier density small enough at 100 °C that circuits still work. Germanium's 0.67 eV gap gives 1000× more intrinsic carriers at the same temperature, and germanium transistors become unreliable above 70 °C.

  • Abundance and cost: silicon is 28% of Earth's crust (as SiO₂); germanium is 1.5 ppm. Silicon is essentially free in the form of sand.

These reasons are engineering, not fundamental physics. Gallium arsenide, gallium nitride, silicon carbide, and indium phosphide have all found niches where their particular combinations of properties beat silicon — GaAs for RF, GaN for LEDs and power, SiC for automotive — but for mass-market digital computing, silicon's balance has been untouchable. The Tata Electronics fab at Dholera, the first modern semiconductor fab on Indian soil, is a silicon fab precisely for these reasons.

Where this leads next