Past exam of the condensed matter physics course of the University of Oxford / 2017 / 4

Assume thermal equilibrium, a three-dimensional nondegenerate semiconductor, one parabolic conduction extremum and one parabolic valence extremum, with two spin states in each. Use the supplied effective masses as scalar density-of-states parameters. With conduction energy $E=E_c+\epsilon$, state counting gives the per-volume density of statesSince $E_c-\mu \gg k_{B}T$, replace the Fermi-Dirac distribution by its Boltzmann factor. Using $x=\epsilon /(k_{B}T)$ and the supplied integral,The electron hole occupation is $1-f(E)$ in the valence band. Applying the same integration below $E_v$ givesTheir product cancels the chemical potential, giving the semiconductor mass-action lawThus, for masses expressed in ordinary mass units,The labels $B,C$ in this last box are exponents, unrelated to the magnetic field or Curie constant above. Valley degeneracy or multiple valence bands must be included in the effective density of states, or absorbed into suitable density-of-states effective masses, not counted twice. Real silicon has anisotropic multiple valleys, so its conductivity mass is not automatically its density-of-states mass; the numerical answers below implement the paper's supplied scalar-mass model.^[ref]

The formula is not an identity for arbitrarily degenerate carriers. For a concrete counterexample, take $\mu -E_c=10k_{B}T$ while holes remain dilute. Fermi occupation gives $n/N_c\simeq 24.1$, whereas the Boltzmann replacement would give $e^{10}\simeq 22026$; the predicted product then differs by about three orders of magnitude. Separate electron and hole quasi-Fermi levels out of equilibrium would also prevent the cancellation used above.

For an undoped neutral intrinsic semiconductor, $n=p$. Equating the two expressions from a and taking logarithms givesand hence the intrinsic-semiconductor chemical potentialTherefore $\boxed{\alpha =\frac{3}{4}\ln (m_h^{\ast }/m_e^{\ast })}$. The logarithm is temperature-independent for fixed parabolic effective masses. The chemical potential is exactly at midgap only if the two effective density of states values coincide; for $m_h^{\ast }>m_e^{\ast }$ it moves above midgap. Additional valley multiplicities would add their logarithmic contribution.

A substitutional phosphorus atom supplies one more valence electron than the silicon atom it replaces. In the shallow hydrogenic donor state model, the remaining positive impurity attracts that extra electron through the screened Coulomb potentialRescaling the hydrogen atom binding energy by the effective mass and dielectric screening gives the ionization energyThe donor level lies at $E_c-E_D^{\mathrm{bind}}$; its binding magnitude is positive. The associated effective Bohr radius of a hydrogenic donor is $a_B^{\ast }={ϵ}_r(m_e/m_e^{\ast })a_0\simeq 2.38\mathrm{nm}$, consistent with a shallow state extended over several lattice spacings.

At $1000\mathrm{K}$, $k_{B}T=0.08617\mathrm{eV}$, approximately $3.3$ times this binding energy. Dilute donors are consequently expected to be largely ionized, rather than frozen out. The actual ionized fraction also depends on the chemical potential, available conduction states and donor charge-state degeneracy; comparing $k_{B}T$ to the binding energy alone is not a normalized ionization probability. At sufficiently high temperature, intrinsic electron-hole production can dominate the total carriers even though donors are ionized. Band anisotropy and short-range impurity corrections limit the numerical hydrogenic estimate.

At $300\mathrm{K}$, $k_{B}T=0.0258520\mathrm{eV}$. With the scalar effective masses and two-spin convention of a, the effective density of states values areThe intrinsic carrier concentration follows from $n=p=n_i$:From b, $\alpha =(3/4)\ln (0.39/0.26)=0.304099$. Taking $E_v=0$ as a reference,It lies $7.862\mathrm{meV}$ above the middle of the band gap, because the hole effective density of states is larger. Both band edges are more than twenty $k_{B}T$ away, validating the nondegenerate semiconductor approximation.

Take uncompensated donor impurities and initially approximate them as ionized. Charge neutrality then gives $n-p=N_d$, while the semiconductor mass-action law gives $np=n_i^2$. Solving the quadratic without prematurely discarding minority carriers,Since $N_d\gg n_i$, the hole density isUse the conduction population to locate the chemical potential:It moves toward the conduction band but remains about $10.4k_{B}T$ below it, so the mobile electrons are nondegenerate. Despite the donor binding energy being comparable with $k_{B}T$, the large number of available conduction states makes ionization almost complete at this dilute doping. For a singly occupied two-spin donor, $N_d^{+}/N_d=[1+2e^{(\mu -E_D)/(k_{B}T)}{]}^{-1}$, with $E_D=E_c-0.02584\mathrm{eV}$. The inferred value is $0.99984$, confirming the approximation to much better than one percent. Including this small correction raises the hole density only to about $4.5824\times 10^{10}{\mathrm{m}}^{-3}$.