Past exam of the condensed matter physics course of the University of Oxford / 2012 / 3

Assume thermal equilibrium, nondegenerate carriers, and isotropic parabolic band dispersions, with the supplied effective masses interpreted as density-of-states masses. Take one effective valley with two electron spin states for each band; any additional valley multiplicity must instead be absorbed into the masses or explicit degeneracy factors.

Let the conduction band edge be $E_c$, the valence band edge $E_v$, and $E_g=E_c-E_v$. The spin-summed conduction density of states per volume isUsing the nondegenerate limit of the Fermi-Dirac distribution and the substitution $x=(E-E_c)/(k_{B}T)$ givesThe factor $\sqrt{\pi }/2$ in the supplied integral produces the normalization. Counting missing valence band electrons similarly givesMultiplication cancels the chemical potential, establishing the semiconductor mass-action law:The result does not require an intrinsic semiconductor: semiconductor doping changes $n$ and $p$ separately while preserving this product under the stated equilibrium and nondegeneracy assumptions.

A substitutional group-V atom such as phosphorus has one more valence electron than germanium and acts as a donor impurity, giving an n-type semiconductor. Approximate its outer electron by a hydrogenic donor state with screened Coulomb attraction and effective mass $m_e^{\ast }$. The ionization energy is the positive binding-energy magnitudeIts effective Bohr radius of a hydrogenic donor is about $3.80\mathrm{nm}$, much larger than an atomic core, consistent with a shallow extended bound state.

At $300\mathrm{K}$, $k_{B}T=25.9\mathrm{meV}$, while the mean three-dimensional classical translational kinetic energy is $3k_{B}T/2=38.8\mathrm{meV}$. Both exceed the donor binding energy, so substantial thermal ionization is expected, unlike excitation across the much larger $0.661\mathrm{eV}$ host band gap. A precise ionized fraction also depends on the chemical potential, dopant density and donor charge-state degeneracy; comparing two energy scales alone does not prove complete ionization at arbitrary concentration.

With the supplied fixed band gap and effective masses, the intrinsic carrier concentration isNumerically,For uncompensated, approximately ionized donor impurities, charge neutrality and the semiconductor mass-action law requireThere is no abrupt extrinsic-to-intrinsic transition. An order-of-magnitude lower crossover density is thereforeRobust extrinsic dominance requires $N_D\gg n_i$; at the nominal equality, $n=1.618N_D$ and $p=0.618N_D$, so intrinsic contributions are already appreciable. For a quantitative tolerance $p/n\le \eta$, the minimum ionized dopant density is $n_i(1-\eta )/\sqrt{\eta }$. For example, $\eta =0.1$ would require $N_D\ge 2.85n_i$ at the upper temperature.

Using the nominal crossover estimate and the complete-ionization approximation at room temperature, $n\simeq N_D$ andFor the twofold neutral-donor degeneracy, a consistency calculation gives about $91.3\%$ ionization at $300\mathrm{K}$, raising this minority density to $1.30\times 10^{15}{\mathrm{m}}^{-3}$, about $9.6\%$ above the complete-ionization estimate. Stronger extrinsic dominance uses more donors and correspondingly fewer electron holes. Actual germanium has temperature-dependent band parameters and multiple valleys; the numerical estimate consistently uses the fixed effective data supplied rather than adding an unsupplied multiplicity.