Past exam of the condensed matter physics course of the University of Oxford / 2022 / 3

Let $\beta =1/(k_{B}T)$, with $k_B$ the Boltzmann constant, $T$ the absolute temperature, and $\mu$ the common equilibrium chemical potential. The conduction band minimum and valence band maximum are ${ϵ}_c$ and ${ϵ}_v$; their separation is the band gap $E_g={ϵ}_c-{ϵ}_v$. The positive parameters $m_e,m_h$ in the supplied density of states functions are the band effective masses of electrons and electron holes, not necessarily the bare electron mass.

The electron occupation is the Fermi-Dirac distribution $f(ϵ)=[e^{\beta (ϵ-\mu )}+1{]}^{-1}$. The appropriate statistical factors are $f_c=f$ for electrons and $f_v=1-f$ for electron holes. In a nondegenerate semiconductor, ${ϵ}_c-\mu \gg k_{B}T$ and $\mu -{ϵ}_v\gg k_{B}T$, so the Boltzmann approximation givesFor the conduction band, set $u=ϵ-{ϵ}_c$ and then $\xi =\beta u$. The supplied integral yieldsFor the valence band, set $u={ϵ}_v-ϵ$ to obtainThe effective density of states factors $N_c,N_v$ include the two spin states already present in the supplied density of states functions. Multiplication cancels the chemical potential and gives the semiconductor mass-action lawHere $n_i=\sqrt{N_c{N}_v}{e}^{-E_g/(2k_{B}T)}$ is the intrinsic carrier concentration. Equality $n=p=n_i$ additionally requires an electrically neutral intrinsic semiconductor. The product law also applies to an equilibrium doped semiconductor while both populations remain nondegenerate. The derivation assumes parabolic energy bands, the stated spin degeneracy, and a single common chemical potential.

For a single impurity level in a crystal with allowed occupations zero and one and equal occupied/empty statistical weights, the Fermi-Dirac distribution givesSince $n_{\mathrm{full}}=N_D{f}_D$ and $n_{\mathrm{empty}}=N_D(1-f_D)$,The same result follows from the two grand canonical ensemble weights $1$ and $e^{-\beta ({ϵ}_D-\mu )}$.

The equal-statistical-weight convention is necessary for the printed formula when $n_{\mathrm{full}}$ and $n_{\mathrm{empty}}$ count whole donor impurities. More generally, donor charge-state degeneracy gives $n_{\mathrm{full}}/n_{\mathrm{empty}}=(g_{\mathrm{full}}/g_{\mathrm{empty}})e^{\beta (\mu -{ϵ}_D)}$. An ordinary singly occupied donor with two spin orientations and a nondegenerate empty state has ratio $2e^{\beta (\mu -{ϵ}_D)}$; at $\mu ={ϵ}_D$ its occupied-to-empty ratio is two, a concrete counterexample to treating the printed expression as universal. The following part uses the paper's ratio-one convention.

Assume uncompensated donor impurities, negligible intrinsic electron holes, and a nondegenerate conduction band. Then charge neutrality equates the conduction electron number density to the ionized-donor number density:The low-temperature condition relative to the band gap supports neglect of intrinsic carriers; one also needs $n_i\ll n$ and the Boltzmann approximation for the conduction electrons. Combining part (b)(i) with the effective density of states result givesEliminate the chemical potential by writing $e^{\beta (\mu -{ϵ}_D)}=(n/N_c)e^{\beta ({ϵ}_c-{ϵ}_D)}$. With the donor binding energy $E_i={ϵ}_c-{ϵ}_D>0$,Writing $A(T)=N_c{e}^{-\beta E_i}$, the physical root of the resulting quadratic equation isThis lies between zero and $N_D$. At semiconductor carrier freeze-out, $A\ll N_D$, it becomes $n\simeq \sqrt{N_D{N}_c}{e}^{-E_i/(2k_{B}T)}$; at essentially complete ionization, $A\gg N_D$, it tends to $N_D$. Including donor charge-state degeneracy would replace $A$ by $A{g}_{\mathrm{empty}}/g_{\mathrm{full}}$.

There are three regimes. At the lowest temperatures, most donor impurities retain their electrons; thermal ionization gives semiconductor carrier freeze-out. The result from part (b) isThis produces the approximately straight descending low-temperature tail on the logarithmic scale plot against $1/T$. The activation energy in the exponential is $E_i/2$, because the chemical potential adjusts with donor ionization.

At intermediate temperatures, the donor impurities are essentially fully ionized while intrinsic excitation remains small. Charge neutrality gives $n\simeq N_D$, producing the extrinsic plateau, roughly from $20$ to $40\mathrm{K}$ in the figure.

At higher temperatures, excitation from the valence band to the conduction band creates electron-hole pairs. The intrinsic carrier concentration grows as $n_i\propto T^{3/2}{e}^{-E_g/(2k_{B}T)}$ for constant parabolic band effective masses and constant band gap. With ionized donor impurities, charge neutrality and the semiconductor mass-action law giveThus $n\simeq n_i+N_D/2$ when $n_i\gg N_D$. This explains the renewed rapid rise and the eventual intrinsic branch; that branch is governed by the band gap, rather than the donor binding energy.

The plateau in both source plots is approximately $10^{18.5}{\mathrm{m}}^{-3}$, soThese are graphical estimates, not high-precision fitted material parameters. To extract activation energies consistently with part (a), retain the temperature powers in the effective density of states factors.

On the semiconductor carrier freeze-out tail, put $x=1/T$. ThenFor example, the right-hand graph gives roughly ${\log }_{10}[n/({\mathrm{m}}^{-3})]\simeq 17.7$ at $x=0.10{\mathrm{K}}^{-1}$ and $13.0$ at $x=0.30{\mathrm{K}}^{-1}$. Subtract the known prefactor contribution before interpreting the slope:Thus $\boxed{E_i\simeq 9\mathrm{meV}}$, to the accuracy of the drawing. Ignoring the weak prefactor gives about $9.3\mathrm{meV}$, a similar rough estimate.

For the intrinsic branch, useThe left-hand graph gives $n(100\mathrm{K})\simeq 2.5\times 10^{21}{\mathrm{m}}^{-3}$ and $n(500\mathrm{K})\simeq 1.5\times 10^{24}{\mathrm{m}}^{-3}$, both far above $N_D$. ThereforeHere $k_B=8.6173\times 10^{-5}\mathrm{eV}{\mathrm{K}}^{-1}$. Omitting the $T^{3/2}$ prefactor over this wide temperature interval would instead give approximately $0.14\mathrm{eV}$, so it is not a negligible correction for the band-gap extraction from intrinsic carrier activation. The source graph warrants an approximate narrow band gap, rather than several significant figures.