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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.

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.

With equal constant electron mobility and hole mobility $\mathcal{M}$, the sheet electrical conductivity is $\sigma =e\mathcal{M}(n+p)$. In the intrinsic regime, $n+p\simeq 2n_i\propto T{e}^{-E_g/(2k_{B}T)}$, soIt decreases steeply with increasing $1/T$ on the high-temperature side. There is a small logarithmic curvature rather than a perfectly straight activation line.

In the fully ionized extrinsic regime, charge neutrality and the semiconductor mass-action law give $n-p=n_d$, $np=n_i^2$, henceThis gives a nearly horizontal region passing through $\sigma (200\mathrm{K})=0.01{\Omega }^{-1}$. At lower temperatures semiconductor carrier freeze-out decreases the ionized donor density and bends the curve downward again.

To locate that bend realistically, include donor charge-state degeneracy. For a neutral donor with two spin states and an empty donor with one state, the ionized density isSolve $n-n_i^2/n=n_d^{+}$ for the mobile density. In the asymptotic freeze-out regime, where $p$ is negligible and $n\ll n_d$,Its activation scale $E_B/2\simeq 3.14\mathrm{meV}$ is much smaller than the intrinsic scale $E_g/2=180\mathrm{meV}$.

Within the specified interval the low-temperature downturn is modest: $N_c(30\mathrm{K})\simeq 2.81\times 10^{10}{\mathrm{cm}}^{-2}$, and the partial-ionization equation gives $n(30\mathrm{K})\simeq 0.65n_d$. The sample is not yet deep in the asymptotic freeze-out limit at 30 K. Calibrating the mobility to the stated 200 K value gives representative valuesThe donor statistical weights were not explicitly supplied, so they are a stated assumption for these illustrative curve details; a one-state donor approximation changes the low-temperature rounding but not the three physical regimes or their activation scales.

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 $A{g}_{\mathrm{empty}}/g_{\mathrm{full}}$.

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.