Past exam of the condensed matter physics course of the University of Oxford / 2014 / 3

An intrinsic semiconductor has its carrier populations set primarily by thermal excitation across the band gap; in the ideal pure, neutral case its conduction-electron and hole densities are equal, $n=p=n_i$. An extrinsic semiconductor has carrier populations significantly controlled by donor or acceptor impurities, producing predominantly n-type semiconductor or p-type semiconductor behavior. Chemical purity alone does not fix conductivity independently of temperature and gap size.

The positive charge carrier mobility is the magnitude of mean drift speed per unit weak electric field. For electrons ${\mathbf{v}}_e=-{\mu }_e\mathbf{E}$, for holes ${\mathbf{v}}_h={\mu }_h\mathbf{E}$, and its SI unit is ${\mathrm{m}}^2{\mathrm{V}}^{-1}{\mathrm{s}}^{-1}$. In an isotropic relaxation-time model, $\mu =|q|\tau /m^{\ast }$.

A band effective mass characterizes how an electron wave packet accelerates in an energy band, rather than replacing its vacuum rest mass. In one dimension and its tensor generalization,These follow by differentiating $v_i={\hslash }^{-1}\partial E/\partial k_i$ and using $\hslash {\dot{k}}_j=F_j$. Effective masses can be anisotropic and electron band curvature can be negative.

An electron hole is a missing electron in an otherwise almost filled valence band. A completely filled band has no net transport current: its velocities cancel over the Brillouin zone. Describing a few missing states as mobile excitations is therefore simpler than following almost every occupied electron state.

Near an isotropic valence maximum, write the electron dispersion as $E_v(\mathbf{k})=E_v(0)-{\hslash }^2{k}^2/(2m_h)$ with $m_h>0$. The electron curvature mass is negative there. Removing that electron creates an excitation with momentum $\hslash {\mathbf{k}}_h=-\hslash \mathbf{k}$ and energy measured upward from the filled reference,ThusThe sign refers to the positive excitation energy of a missing electron near a maximum, not to negative vacuum mass. A general maximum gives the corresponding positive hole-curvature tensor.^[ref]

Removing an electron of charge $-e$ leaves charge $+e$ relative to the filled-band reference. This also reproduces the current. If the missing electron state had velocity ${\mathbf{v}}_e(\mathbf{k})$, subtracting its contribution from the zero current of a filled band givesThe hole dispersion with ${\mathbf{k}}_h=-\mathbf{k}$ gives ${\mathbf{v}}_h({\mathbf{k}}_h)={\mathbf{v}}_e(\mathbf{k})$, soLikewise $\hslash \dot{\mathbf{k}}=-e\mathbf{E}$ becomes $\hslash {\dot{\mathbf{k}}}_h=+e\mathbf{E}$. Positive hole charge and positive hole effective mass consistently describe drift along the electric field, while the valence electrons themselves drift through an almost full band. A hole is a quasiparticle, not a proton moving through the crystal.

In the simple transport model, ${\mu }_e=e{\tau }_e/m_e^{\ast }$ and ${\mu }_h=e{\tau }_h/m_h^{\ast }$. Conduction bands in many common semiconductors have lighter transport masses than the dominant valence hole bands. For comparable momentum-relaxation times, this gives larger electron mobility. It is a common trend, not a theorem: band anisotropy, light and heavy hole populations and different scattering times can alter the comparison. The hole's mass is not the mass of an atom, and crystal translational symmetry does not make its mobility infinite.

For the intrinsic germanium sample, electrons and holes both contribute positive conventional conductivity:ThereforeThis is the density of each species; the combined carrier number density is $2n_i$. Assigning separate densities from the two mobilities would violate intrinsic charge neutrality.

A monovalent FCC metal supplies one conduction electron per atom and four atoms per conventional cell. HenceWith a one-carrier conductivity $\sigma =n_{m}e{\mu }_m$,This is about $84$ times smaller than the stated germanium electron mobility. Nevertheless the metal is vastly more conductive because its carrier density is about $3.7\times 10^9$ times larger. Resistivity depends on both density and mobility; low resistivity alone does not establish high mobility or long scattering time.

Neglect the battery's internal resistance, contact drops and the very small resistance of laboratory-length metal leads. The germanium resistance and field areTaking the positive direction along the electric field, the carrier drift velocities areTheir conventional current densities add: $j_e=e{n}_i{\mu }_e{E}_g\simeq 266.7\mathrm{A}{\mathrm{m}}^{-2}$ and $j_h\simeq 133.3\mathrm{A}{\mathrm{m}}^{-2}$. Thus $I=(j_e+j_h)A_g=0.400\mathrm{mA}$ is the same total current through either metal lead. The current continuity between a metal and an intrinsic semiconductor givesThe metal electrons move opposite its conventional current. This extremely small drift speed is compatible with large microscopic Fermi velocities; drift is the small imbalance of the occupied momentum distribution.

The wire lengths were not specified. Their resistance per metre is ${\rho }_m/A_m=0.034\Omega {\mathrm{m}}^{-1}$, so neglecting them relative to $5000\Omega$ is justified for an ordinary apparatus. If finite total wire length $L_w$ is retained, the exact ideal-contact circuit expression is $I=2/(5000+0.034L_w)\mathrm{A}$ with $L_w$ in metres; all three drift magnitudes above scale with $I/(0.400\mathrm{mA})$. Thus the approximation and its missing geometric input are explicit.