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

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 $4.5824\times 10^{10}{\mathrm{m}}^{-3}$.