Conduction of Electricity in Solids
We have learned what a single trapped electron does and how a whole atom is built. Now we group enormous numbers of atoms into a crystal and ask the question that built the modern world: why does copper conduct electricity, diamond refuse to, and silicon sit in between? The answer is one of quantum physics' most powerful ideas. When \(N\) atoms come together, each sharp atomic energy level splits into \(N\) levels so densely packed that they merge into continuous energy bands, separated by forbidden gaps. Whether a solid conducts comes down to one thing: is its highest occupied band full or only partly filled? From that single distinction flow insulators, metals, and semiconductors — and once we learn to dope a semiconductor and join an n-type piece to a p-type piece, we unlock the p-n junction, the seed of the rectifier, the LED, the laser, and the transistor that powers every computer and phone.
- Crystalline solids are classified by three measured quantities: resistivity \(\rho\), temperature coefficient of resistivity \(\alpha = \dfrac{1}{\rho}\dfrac{d\rho}{dT}\), and charge-carrier density \(n\).
- As atoms assemble, sharp atomic levels merge into energy bands split by gaps; an insulator has a full top band and a large gap, a metal a partly filled band whose highest level at \(T=0\) is the Fermi energy \(E_F\) (≈ \(7.0\,\mathrm{eV}\) for copper).
- Conduction depends on the density of states \(N(E) = \dfrac{8\sqrt{2}\,\pi m^{3/2}}{h^3}\,E^{1/2}\), the Fermi–Dirac occupancy \(P(E) = \dfrac{1}{e^{(E-E_F)/kT}+1}\), and their product \(N_o(E) = N(E)\,P(E)\); the Fermi energy follows from \(E_F = 0.121\,\dfrac{h^2}{m}\,n^{2/3}\).
- Semiconductors have a small gap; doping with donors makes n-type (electron) material, with acceptors makes p-type (hole) material.
- A p-n junction builds a depletion zone and contact potential \(V_0\); it acts as a rectifier, a light-emitting diode with \(\lambda = \dfrac{hc}{E_g}\), a junction laser, and — as the MOSFET — the transistor at the heart of every chip.
What Is Physics?
A central goal of modern physics is to answer a deceptively simple question: by what mechanisms does a material conduct electricity, or fail to? The answer is hard because it requires applying quantum physics not to one atom but to an astronomical number of atoms grouped together and interacting. Yet the payoff is immense — mastering the quantum physics of materials is exactly what gave us computers, calculators, phones, and every other solid-state device. Our starting point is to characterize, then explain, the solids that conduct and the ones that do not.
The Electrical Properties of Solids
We study crystalline solids — those whose atoms repeat in a three-dimensional lattice (copper, silicon, and diamond, not wood or glass). Three measurable properties classify them: the resistivity \(\rho\) at room temperature, the temperature coefficient of resistivity \(\alpha = \tfrac{1}{\rho}\,d\rho/dT\), and the number density of charge carriers \(n\) (from the Hall effect). Insulators have enormous resistivity — diamond's exceeds copper's by a factor of about \(10^{24}\). Among non-insulators, two families emerge.
| Property | Unit | Copper (metal) | Silicon (semiconductor) |
|---|---|---|---|
| Resistivity \(\rho\) | \(\Omega\cdot\mathrm{m}\) | \(2\times10^{-8}\) | \(3\times10^{3}\) |
| Temp. coefficient \(\alpha\) | \(\mathrm{K}^{-1}\) | \(+4\times10^{-3}\) | \(-70\times10^{-3}\) |
| Carrier density \(n\) | \(\mathrm{m}^{-3}\) | \(9\times10^{28}\) | \(1\times10^{16}\) |
Compared with a metal, a semiconductor has much higher resistivity, far fewer charge carriers, and — tellingly — a temperature coefficient that is large and negative: its resistivity falls as it warms, the opposite of a metal. The central question of the chapter is why these three numbers differ so wildly between diamond, copper, and silicon.
Energy Levels in a Crystalline Solid
An isolated copper atom stacks its 29 electrons into discrete subshells (\(1s^2 2s^2 2p^6 3s^2 3p^6 3d^{10} 4s^1\)). Bring two atoms close and their outer wave functions overlap; now the Pauli principle applies to the combined system, and each shared energy level must split into two so no two of the 58 electrons share a state.
A typical band spans only a few electron-volts yet contains a vast number of levels. Lower-energy bands are narrower than higher ones, because the core electrons that occupy them lie deep in the atom and their wave functions overlap less, so their levels split less. This band–gap structure is the master key to the electrical behavior of every crystalline solid.
Insulators
For a current to flow, electrons must gain kinetic energy and move into higher levels. In an insulator the highest band that holds any electrons is completely full, and the Pauli principle forbids an electron from moving into an already-occupied level. The electrons are in gridlock. The only escape is to jump the wide gap to the empty band above — but for diamond that gap is \(E_g = 5.5\,\mathrm{eV}\), about \(140\) times the average thermal energy at room temperature, so essentially no electron makes it.
Metals
A metal is defined by a highest occupied band that is only partly filled, so countless empty levels sit just above the filled ones. Apply a field and electrons slide effortlessly into those nearby vacancies — the metal conducts. At absolute zero the electrons fill the band up to a sharp ceiling called the Fermi level, whose energy is the Fermi energy \(E_F\) (about \(7.0\,\mathrm{eV}\) for copper, corresponding to a Fermi speed near \(1.6\times10^{6}\,\mathrm{m/s}\)). Even at \(T=0\) the electrons are not at rest — Pauli stacks them up to \(E_F\).
Carrying out that integral solves for the Fermi energy directly from \(n\):
Semiconductors
A semiconductor has the band structure of an insulator but with a much smaller gap — silicon's is only \(E_g \approx 1.1\,\mathrm{eV}\) versus diamond's \(5.5\,\mathrm{eV}\). That difference is everything: at room temperature a small but real fraction of electrons in the full valence band are thermally kicked across the gap into the empty conduction band, leaving behind empty states called holes.
This explains all three table entries. The carrier density \(n\) is tiny (silicon has fewer carriers than copper by a factor of about \(10^{13}\)), and since \(\rho = m/(e^2 n\tau)\) the resistivity is correspondingly huge. And because warming the crystal promotes more electrons across the gap, \(n\) climbs fast with temperature, making \(\alpha\) negative — resistivity falls as silicon heats up.
Doped Semiconductors
The real power of semiconductors comes from doping — replacing about one silicon atom in \(10^{7}\) with a chosen impurity. Each silicon atom (valence 4) forms four covalent bonds with its neighbors.
The p-n Junction
Join a piece of n-type material to a piece of p-type material and the result — a p-n junction — is the foundation of nearly every semiconductor device. Right at the junction plane, two things happen at once.
That potential is a barrier to further diffusion of majority carriers, but a downhill ride for the few minority carriers, which are swept across to form an opposing drift current. At equilibrium the diffusion and drift currents exactly cancel, so the net current through an isolated junction is zero — leaving a built-in voltage \(V_0\) across a carrier-poor depletion zone.
The Junction Rectifier
Apply an external voltage and the junction conducts in only one direction — it rectifies. In forward bias (battery's positive terminal on the p side) the barrier \(V_0\) is lowered, the depletion zone narrows, and a large diffusion (forward) current flows. In back bias the barrier is raised, the zone widens, and only a tiny drift (back) current leaks through.
Feed it an alternating voltage whose average is zero and it passes only the forward half-cycles, producing an output with a nonzero average. That is exactly what a junction rectifier does in a power supply: it converts alternating current toward direct current. Its circuit symbol is an arrowhead pointing from the p side in the direction of allowed conventional current.
The Light-Emitting Diode, Photodiode, and Junction Laser
When a conduction-band electron drops into a valence-band hole, it releases the gap energy \(E_g\). In silicon that energy mostly becomes lattice heat, but in materials like gallium arsenide it emerges as a photon — light whose color the gap width sets directly.
Push the forward bias even harder and the huge density of electrons facing holes across the thin zone creates a population inversion. A first recombination photon then triggers a cascade of identical stimulated photons, which reflect between the crystal's polished parallel faces — a junction laser, far more coherent and sharply colored than an LED. These tiny lasers read CDs and DVDs and drive fiber-optic communication, usually in the infrared windows (\(1.31\) and \(1.55\,\mu\mathrm{m}\)) where optical fibers absorb least.
The Transistor
A transistor is a three-terminal semiconductor device that controls one current with another. The workhorse of modern electronics is the MOSFET (metal-oxide-semiconductor field-effect transistor). Electrons flow from the source through a thin n-channel to the drain; a voltage on the gate, separated from the channel by an insulating oxide layer, sets up an electric field that widens the channel's depletion zone and chokes off the current.
MOSFETs roughly \(500\,\mathrm{nm}\) long — about the wavelength of yellow light — switch between states at enormous speed, so digital data races through MOSFET circuits. Millions of them, with their resistors and capacitors, are crafted onto a single semiconductor chip to form an integrated circuit — the engine of every computer and phone.
Putting It to Work
Problem. Estimate the probability that, at room temperature (\(300\,\mathrm{K}\)), an electron at the top of diamond's filled band jumps the gap \(E_g = 5.5\,\mathrm{eV}\) into the empty band.
Solution. Use the Boltzmann factor \(P \approx e^{-E_g/kT}\) with \(kT = (8.62\times10^{-5})(300) = 0.0259\,\mathrm{eV}\).
About \(3\) electrons in \(10^{93}\) — yet any real diamond holds fewer than \(10^{23}\) electrons, so in practice none ever crosses. The wide gap makes diamond an extraordinary insulator. Silicon's gap of \(1.1\,\mathrm{eV}\) gives an exponent five times smaller and a probability vastly larger — the whole difference between an insulator and a semiconductor.
Problem. Copper is monovalent, with molar mass \(M = 63.54\,\mathrm{g/mol}\) and density \(8.96\,\mathrm{g/cm^3}\). Find the number density \(n\) of conduction electrons.
Solution. Each atom contributes one electron, so \(n = (\text{density})\,N_A/M\) in SI units.
This matches the \(9\times10^{28}\,\mathrm{m^{-3}}\) quoted in the table. Such a dense sea of mobile electrons is exactly what makes copper an excellent conductor — and silicon, with around \(10^{16}\,\mathrm{m^{-3}}\), lags by some thirteen orders of magnitude.
Problem. Using \(n = 8.5\times10^{28}\,\mathrm{m^{-3}}\) for copper, compute the Fermi energy and confirm the textbook value.
Solution. Apply \(E_F = 0.121\,(h^2/m)\,n^{2/3}\) with \(h^2/m = 4.83\times10^{-37}\,\mathrm{J^2\cdot s^2/kg}\).
The model nails copper's measured \(E_F = 7.0\,\mathrm{eV}\) — a striking success, given that the only input is how many electrons sit in each cubic meter. The matching Fermi speed, \(v_F = \sqrt{2E_F/m}\approx 1.6\times10^{6}\,\mathrm{m/s}\), shows these electrons move fast even at absolute zero.
Problem. At \(T = 800\,\mathrm{K}\), what is the probability that a state (a) \(0.10\,\mathrm{eV}\) above the Fermi energy is occupied, and (b) \(0.10\,\mathrm{eV}\) below it is occupied?
Solution. Use \(P(E) = 1/(e^{(E-E_F)/kT}+1)\) with \(kT = (8.62\times10^{-5})(800) = 0.0690\,\mathrm{eV}\), so the exponent magnitude is \(0.10/0.0690 = 1.45\).
The two add to \(1\) — a symmetry of the Fermi–Dirac function about \(E_F\): a state a given amount above \(E_F\) is as likely to be filled as the mirror state below is to be empty. Only states within a few \(kT\) of \(E_F\) shift at all; the deep Fermi sea is untouched.
Problem. Pure silicon has \(n_0 \approx 10^{16}\,\mathrm{m^{-3}}\) conduction electrons. To raise this by a factor of \(10^{6}\) with phosphorus donors, what fraction of silicon atoms must be replaced? (Silicon: density \(2330\,\mathrm{kg/m^3}\), molar mass \(0.0281\,\mathrm{kg/mol}\).)
Solution. Each phosphorus donates one electron, so we need \(n_P \approx 10^{6}n_0 = 10^{22}\,\mathrm{m^{-3}}\) donors. Compare with the silicon atom density.
Replacing just one silicon atom in five million multiplies the conduction electrons a millionfold. Even so, the doped silicon still has far fewer carriers than copper — by roughly \(10^{7}\) — which is what keeps a semiconductor a semiconductor, not a metal.
Problem. An LED is built from a Ga–As–P material with energy gap \(E_g = 1.9\,\mathrm{eV}\). What is the wavelength of the emitted light?
Solution. A recombining electron drops the gap energy as a photon, so \(\lambda = hc/E_g\).
About \(650\,\mathrm{nm}\) — red light, just as expected for this gap. Widen the gap (different doping) and the photon energy rises, shifting the color toward green and blue; narrow it and the emission moves into the infrared. The gap width is the color dial.
Chapter Summary
Resistivity \(\rho\), temperature coefficient \(\alpha\), and carrier density \(n\) sort solids into insulators, metals, and semiconductors.
Assembling \(N\) atoms splits each level into \(N\), forming dense bands separated by forbidden gaps.
Insulator: full top band, large gap (\(P \approx e^{-E_g/kT}\) tiny). Metal: partly filled band with Fermi energy \(E_F\).
\(N(E)\propto E^{1/2}\), \(P(E)=1/(e^{(E-E_F)/kT}+1)\), \(N_o = NP\), and \(E_F = 0.121\,(h^2/m)\,n^{2/3}\).
Small gap; thermal electrons and holes both conduct; \(\rho\) falls with temperature (\(\alpha<0\)).
Donors → n-type (electron majority); acceptors → p-type (hole majority); tiny doping fractions change \(n\) enormously.
Depletion zone and contact potential \(V_0\); diffusion and drift currents balance; conducts one way → rectifier.
LED (\(\lambda = hc/E_g\)), photodiode, junction laser, and the MOSFET transistor — the building block of chips.
Problems
Take \(h = 6.63\times10^{-34}\,\mathrm{J\cdot s}\), electron mass \(m = 9.11\times10^{-31}\,\mathrm{kg}\), \(hc = 1240\,\mathrm{eV\cdot nm}\), \(N_A = 6.02\times10^{23}\,\mathrm{mol^{-1}}\), and \(k = 8.62\times10^{-5}\,\mathrm{eV/K}\). Use \(E_F = 0.121\,(h^2/m)\,n^{2/3}\), \(P(E) = 1/(e^{(E-E_F)/kT}+1)\), and \(\lambda = hc/E_g\).
- Using Table 41-1 values, by what factor does copper's resistivity differ from silicon's, and by what factor do their carrier densities differ?
- Explain, in terms of band structure, why a metal's resistivity rises with temperature while a semiconductor's falls.
- Estimate the probability that an electron jumps the \(1.1\,\mathrm{eV}\) gap of silicon at room temperature (\(300\,\mathrm{K}\)), and compare it with the diamond result of Example 1.
- Copper is monovalent with molar mass \(63.54\,\mathrm{g/mol}\) and density \(8.96\,\mathrm{g/cm^3}\). Find the number density of conduction electrons.
- Gold is monovalent, molar mass \(197\,\mathrm{g/mol}\), density \(19.3\,\mathrm{g/cm^3}\). Find its conduction-electron number density and its Fermi energy.
- A metal has \(n = 5.9\times10^{28}\,\mathrm{m^{-3}}\). Use \(E_F = 0.121\,(h^2/m)\,n^{2/3}\) to find its Fermi energy in eV.
- Show that the Fermi speed corresponding to copper's \(E_F = 7.0\,\mathrm{eV}\) is about \(1.6\times10^{6}\,\mathrm{m/s}\).
- Evaluate the density of states \(N(E)\) for a metal at \(E = 8.0\,\mathrm{eV}\) (express the answer in \(\mathrm{m^{-3}\,eV^{-1}}\)).
- What is the probability that a state \(0.062\,\mathrm{eV}\) above the Fermi energy is occupied at (a) \(T = 0\,\mathrm{K}\) and (b) \(T = 320\,\mathrm{K}\)?
- A state \(63\,\mathrm{meV}\) above the Fermi level has occupancy probability \(0.090\). What is the occupancy probability of a state \(63\,\mathrm{meV}\) below the Fermi level?
- Show that, for a state at exactly the Fermi energy, the Fermi–Dirac occupancy probability is \(0.5\) at any nonzero temperature.
- For copper (\(E_F = 7.0\,\mathrm{eV}\)) at \(1000\,\mathrm{K}\), find the occupancy probability of a state at \(7.25\,\mathrm{eV}\).
- Why does conduction in a semiconductor involve both electrons and holes, and in what sense does a hole carry charge \(+e\)?
- To increase silicon's conduction-electron density (\(\sim 10^{16}\,\mathrm{m^{-3}}\)) by a factor of \(10^{5}\) with phosphorus, what fraction of silicon atoms must be replaced? (Silicon: density \(2330\,\mathrm{kg/m^3}\), molar mass \(0.0281\,\mathrm{kg/mol}\).)
- Describe how forward bias and back bias change the depletion-zone width and the current in a p-n junction.
- The longest wavelength a certain crystal transmits is \(295\,\mathrm{nm}\) (it is opaque to shorter wavelengths). What is the energy gap between its filled band and the next empty band, in eV?
- What maximum light wavelength will excite an electron across the \(1.1\,\mathrm{eV}\) gap of silicon? In what part of the spectrum does it lie?
- A computer chip \(2.54\,\mathrm{cm}\times2.22\,\mathrm{cm}\) holds about \(3.5\times10^{6}\) square transistors. Ignoring wiring, what is their maximum edge length?