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The number of dopant sites per volume. For a roughly homogeneous three-dimensional distribution, the characteristic site separation is of order $n^{-1/3}$. This counts introduced sites rather than automatically counting free conduction band electrons: semiconductor carrier freeze-out or compensation can make the mobile carrier density much smaller.

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

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.

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.

A completely filled valence band carries no net current: its occupied crystal-momentum states have velocities that cancel. Near a valence-band maximum, an almost full band is conveniently described by the few missing electrons, or electron holes. Removing an electron of charge $-e_0$, with $e_0>0$, changes the charge relative to the filled band by $+e_0$. This is a quasiparticle description of electronic motion, not a mobile positive ion or a Positron.

For an isotropic parabolic maximum, ${ϵ}_v(\mathbf{k})=E_v-{\hslash }^2{k}^2/(2m_h^{\ast })$. Assign the missing-electron state momentum ${\mathbf{k}}_h=-{\mathbf{k}}_e$ and hole energyIt has positive effective mass $m_h^{\ast }$. Its velocity equals the velocity of the missing electron state, and its positive-charge current replaces the negative of that state's electronic current. Semiclassical motion is thus $\hslash {\dot{\mathbf{k}}}_h=+e_0(\mathbf{E}+{\mathbf{v}}_h\times \mathbf{B})$. With momentum relaxation time ${\tau }_h$, the low-field mobility is ${\mu }_h=e_0{\tau }_h/m_h^{\ast }$ andIf electrons also contribute, $\sigma =e_0(p{\mu }_h+n{\mu }_e)$: the two species' longitudinal conductivities add.

A Hall bar separates carrier concentration from mobility. Apply current density $j_x$, magnetic field $B_z$, and measure transverse electric field $E_y$. For one isotropic hole population, transverse force balance gives $E_y=v_x{B}_z$, soFor thickness $t$, the Hall-voltage magnitude is $|V_H|=IB/(e_{0}pt)$. A longitudinal four-terminal resistance measurement supplies $\sigma$ independently. Use ohmic contacts, known conducting thickness, controlled temperature and low enough current to avoid self-heating. Reversing magnetic field removes even-in-field longitudinal offsets from contact misalignment; reversing current also helps reject thermoelectric offsets. These are central controls, rather than merely taking one transverse voltage reading.^[ref]

The simple inference assumes one carrier type and a known Hall scattering factor. More generally $R_H=r_H/(e_{0}p)$, so $\sigma R_H$ measures Hall mobility rather than automatically the drift mobility. With two carrier populations at weak field,Its sign alone need not identify the more numerous carrier. A Hall experiment also cannot determine $m_h^{\ast }$ and ${\tau }_h$ separately: their ratio sets mobility. cyclotron resonance supplies an independent mass measurement by sweeping frequency or magnetic field through $\omega =e_{0}B/m_c$ and detecting resonant absorption. The measured cyclotron effective mass can depend on band anisotropy and need not equal every curvature mass; the isotropic parabolic model identifies them. Its linewidth provides a scattering-time estimate only with a specified damping model.

semiconductor doping generates holes by adding acceptor impurities. In silicon, a substitutional group-III atom such as boron provides one fewer valence electron than silicon. A shallow acceptor level lies slightly above the valence-band edge: accepting a valence electron makes the dopant negatively ionized and leaves a mobile hole in the valence band. The impurity stays fixed in the lattice. charge neutrality and the nondegenerate equilibrium mass-action law giveFor fully ionized, uncompensated acceptors in the extrinsic regime, $p\simeq N_A$ and $n\simeq n_i^2/N_A$; with fully ionized donor compensation, $p\simeq N_A-N_D$ while holes remain dominant. The Fermi level moves towards the valence band. At low temperature, semiconductor carrier freeze-out suppresses mobile electron holes; at high temperature intrinsic electron-hole generation can dominate. Increasing dopant density also changes scattering, so a conductivity increase alone cannot distinguish increased carrier density from changed mobility. Temperature-dependent Hall and conductivity measurements separate these effects, subject to the carrier and scattering qualifications above.