Past exam of the condensed matter physics course of the University of Oxford / 2019 / 3

The Boltzmann carrier density of a two-dimensional parabolic band can be derived directly. Use one isotropic parabolic conduction band, $E(\mathbf{k})=E_c+{\hslash }^2{k}^2/(2m_e^{\ast })$, with two spin states per wave vector. In the nondegenerate limit the Fermi-Dirac distribution reduces to its Boltzmann factor. The number per area is thereforeThe integral uses ${\int }_0^{\infty }k{e}^{-a{k}^2}dk=1/(2a)$. ThusEquivalently, the two-dimensional density of states is constant, $D_c=m_e^{\ast }/(\pi {\hslash }^2)$, and $n=N_c{e}^{-\beta (E_c-\mu )}$ with $N_c=D_c{k}_{B}T$. These expressions include twofold spin degeneracy and one valley. With a general total band degeneracy $g_c$, replace the factor $1/\pi$ by $g_c/(2\pi )$; extra valleys or occupied confinement subbands would change the coefficient.

Near a parabolic valence band maximum $E_v$, the missing-electron effective mass is $m_h^{\ast }>0$. The same two-dimensional state counting givesMultiplication eliminates the chemical potential, giving the semiconductor mass-action lawFor an intrinsic semiconductor, charge neutrality imposes $n=p=n_i$, soThe intrinsic chemical potential is not exactly midgap when the two effective masses differ: ${\mu }_i=(E_c+E_v)/2+(k_{B}T/2)\log (m_h^{\ast }/m_e^{\ast })$. The exponential suppression still justifies the assumed nondegenerate regime here.

At $300\mathrm{K}$, the two-dimensional effective density of states values areUsing $E_g=0.36\mathrm{eV}$ and $k_{B}T\simeq 0.02585\mathrm{eV}$ givesFor the doped sample, estimate the intrinsic-to-extrinsic crossover by $n_i(T_{\ast })\simeq n_d=10^9{\mathrm{cm}}^{-2}$. With the masses and gap held fixed, $n_i(T)=AT{e}^{-B/T}$, whereSolving $A{T}_{\ast }{e}^{-B/T_{\ast }}=10^9{\mathrm{cm}}^{-2}$ givesIt is a smooth carrier crossover, not a thermodynamic phase transition. Donors are essentially ionized near this temperature because their binding energy, calculated next, is much smaller than $k_B{T}_{\ast }$. At substantially higher temperatures intrinsic pairs dominate; below the crossover and above appreciable donor freeze-out, the majority density approaches $n_d$.

Take the screened Coulomb potential of a singly ionized donor as $V(r)=-\kappa /r$, with $\kappa =e^2/(4\pi {ϵ}_0{ϵ}_r)$ and $r=|\mathbf{r}-{\mathbf{r}}_0|$. The electron's motion is two-dimensional, but its electrostatic interaction is the ordinary three-dimensional $1/r$ interaction restricted to the layer; it is not a logarithmic two-dimensional electrostatic potential.

For the rotationally symmetric bound state, the two-dimensional radial Schrödinger equation isFor $\psi =A{e}^{-br}$, the radial Laplacian is $(b^2-b/r)\psi$. Cancel the $1/r$ term to obtainThus the positive two-dimensional hydrogenic donor binding energy is four times the three-dimensional effective Rydberg:The decay length is $1/b=a_B^{\ast }/2\simeq 15.3\mathrm{nm}$. Normalization, if desired, follows from $1=2\pi |A{|}^2{\int }_0^{\infty }r{e}^{-2br}dr=\pi |A{|}^2/(2b^2)$, giving $A=b\sqrt{2/\pi }$ up to a constant phase. This ideal result neglects finite well thickness, dielectric mismatch at the interfaces and short-range impurity corrections.

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.