Solids are classified by how they conduct electricity, and the key to understanding why lies in the energy band theory. When atoms come together to form a crystal, the discrete energy levels of individual atoms split and merge into nearly continuous energy bands. The highest band that is completely filled with electrons at $0\,\text{K}$ is the valence band; the next higher band, normally empty, is the conduction band. The gap between them is the forbidden energy gap $E_g$.
The size of $E_g$ decides the class of solid. In a conductor (metal) the valence and conduction bands overlap ($E_g = 0$), so countless free electrons carry current. In an insulator the gap is large ($E_g > 3\,\text{eV}$, e.g. diamond $\approx 6\,\text{eV}$), so almost no electron can jump across. A semiconductor has a small gap ($E_g \approx 1\,\text{eV}$; silicon $1.1\,\text{eV}$, germanium $0.7\,\text{eV}$): at room temperature thermal energy lifts a few electrons into the conduction band, giving modest conductivity that rises with temperature.
A pure semiconductor is an intrinsic semiconductor. When an electron is excited to the conduction band it leaves behind a vacancy called a hole, which behaves as a mobile positive charge. In intrinsic material the electron and hole concentrations are equal: $n_e = n_h = n_i$. Conduction happens by both, but intrinsic conductivity is too small to be useful.
To make it useful we add a tiny, controlled amount of impurity — doping — producing an extrinsic semiconductor. Doping silicon (a tetravalent atom) with a pentavalent donor (P, As, Sb) gives an n-type semiconductor: four electrons bond and the fifth is loosely held, becoming a free electron. Here electrons are the majority carriers and holes the minority carriers. Doping with a trivalent acceptor (B, Al, In) gives a p-type semiconductor: one bond is incomplete, creating a hole, so holes are majority and electrons minority. In every doped sample the law of mass action holds: $n_e\, n_h = n_i^2$. Each doped crystal stays electrically neutral overall.
Joining a p-type and an n-type region in a single crystal forms a p-n junction. Near the boundary, electrons diffuse from n to p and holes from p to n, recombining and leaving behind immobile charged ions — negative acceptor ions on the p-side, positive donor ions on the n-side. This carrier-free zone is the depletion region (width $\sim 10^{-6}\,\text{m}$). The exposed ions set up an internal field and a barrier potential $V_B$ ($\approx 0.3\,\text{V}$ for Ge, $\approx 0.7\,\text{V}$ for Si) that opposes further diffusion until equilibrium.
Applying an external voltage biases the junction. In forward bias (p to +, n to minus) the applied field opposes the barrier: the depletion width shrinks, the barrier lowers, and once $V > V_B$ a large current of majority carriers flows. In reverse bias (p to minus, n to +) the field adds to the barrier: the depletion width grows and only a tiny reverse saturation current (from minority carriers, in the microamp range) flows. The V-I characteristic is therefore strongly non-linear and one-directional — the diode conducts well only one way. Beyond a critical reverse voltage the junction undergoes breakdown.
The energy gap of silicon is $1.1\,\text{eV}$. Find the maximum wavelength of light that can excite an electron from the valence band to the conduction band. (Take $hc = 1240\,\text{eV nm}$.)
Solution- An electron is excited when the photon energy equals at least $E_g$: $E = \dfrac{hc}{\lambda} \ge E_g$.
- Maximum wavelength corresponds to minimum photon energy $E = E_g = 1.1\,\text{eV}$.
- $\lambda_{max} = \dfrac{hc}{E_g} = \dfrac{1240\,\text{eV nm}}{1.1\,\text{eV}}$.
- $\lambda_{max} \approx 1127\,\text{nm}$.
Answer: The maximum wavelength is about $1127\,\text{nm}$ (in the infrared), so silicon responds to infrared and visible light.
A silicon sample is doped with phosphorus. State the type of semiconductor formed, the majority and minority carriers, and whether the crystal is charged.
Solution- Phosphorus is pentavalent (5 valence electrons); silicon is tetravalent.
- Four electrons form covalent bonds; the fifth is loosely bound and becomes a free electron — the impurity is a donor.
- This gives an n-type semiconductor: electrons are the majority carriers, holes the minority carriers.
- Each donor contributes a free electron but also a fixed positive ion, so the crystal stays electrically neutral overall.
Answer: n-type; majority carriers are electrons, minority carriers are holes; the crystal as a whole is electrically neutral.
In a sample of n-type silicon at room temperature, the intrinsic carrier concentration is $n_i = 1.5\times10^{16}\,\text{m}^{-3}$ and the electron concentration is $n_e = 5\times10^{22}\,\text{m}^{-3}$. Find the hole concentration.
Solution- The law of mass action: $n_e\, n_h = n_i^2$.
- $n_h = \dfrac{n_i^2}{n_e} = \dfrac{(1.5\times10^{16})^2}{5\times10^{22}}$.
- $n_i^2 = 2.25\times10^{32}\,\text{m}^{-6}$.
- $n_h = \dfrac{2.25\times10^{32}}{5\times10^{22}} = 4.5\times10^{9}\,\text{m}^{-3}$.
Answer: The minority (hole) concentration is $4.5\times10^{9}\,\text{m}^{-3}$, far smaller than $n_e$, confirming n-type behaviour.
The barrier potential of a silicon p-n junction is $0.7\,\text{V}$ and the depletion width is $1\,\mu\text{m}$. Estimate the electric field across the depletion region.
Solution- The barrier potential is set up across the depletion width $d$.
- The (average) electric field is $E = \dfrac{V_B}{d}$.
- $E = \dfrac{0.7\,\text{V}}{1\times10^{-6}\,\text{m}}$.
- $E = 7\times10^{5}\,\text{V m}^{-1}$.
Answer: The internal electric field is about $7\times10^{5}\,\text{V m}^{-1}$, directed from the n-side to the p-side, opposing diffusion.
A semiconductor has resistance that decreases as temperature rises, while a metal's resistance increases. Explain why, using band theory.
Solution- In a semiconductor only a few electrons cross the small gap $E_g$ at room temperature; raising the temperature gives more electrons enough energy to jump into the conduction band.
- The number of free charge carriers therefore rises rapidly with temperature, increasing conductivity (decreasing resistance).
- In a metal the carrier number is essentially fixed; raising temperature increases lattice vibrations, scattering electrons more.
- More scattering reduces carrier mobility, so a metal's resistance increases with temperature.
Answer: A semiconductor's resistance falls with temperature because carrier number grows; a metal's rises because increased scattering lowers mobility while carrier number stays fixed.
A p-n junction diode carries a forward current of $10\,\text{mA}$ when forward-biased at $0.7\,\text{V}$, and a reverse saturation current of $5\,\mu\text{A}$. Compare the forward and reverse resistances at these operating points.
Solution- Static (DC) resistance is $R = \dfrac{V}{I}$ at the operating point.
- Forward: $R_f = \dfrac{0.7\,\text{V}}{10\times10^{-3}\,\text{A}} = 70\,\Omega$.
- Reverse (say at $5\,\text{V}$): $R_r = \dfrac{5\,\text{V}}{5\times10^{-6}\,\text{A}} = 1\times10^{6}\,\Omega = 1\,\text{M}\Omega$.
- Ratio $\dfrac{R_r}{R_f} = \dfrac{10^{6}}{70} \approx 1.4\times10^{4}$.
Answer: Forward resistance $\approx 70\,\Omega$, reverse resistance $\approx 1\,\text{M}\Omega$ — about $14000$ times larger, which is why a diode conducts essentially in one direction.