Exercises — Fermi level and Fermi-Dirac distribution

WHAT: substitute $E=E_F$ so $x=0$. WHY: the definition of $E_F$ is "the energy where $f=\frac{1}{2}$", so this is a sanity check. $f=\frac{1}{1+e^0}=\frac{1}{1+1}=\frac{1}{2}.$ No $T$ appears once $x=0$, so the answer is $\frac{1}{2}$ at every temperature. On the figure, the curve passes through the point $(E_F,\frac{1}{2})$ no matter how steep or smeared it is.

WHAT: multiply $k_B=8.617\times 10^{-5}$ eV/K by each temperature. WHY: $k_{B}T$ is the width of the fuzzy region; you need it before any numeric $f$. $k_{B}T(300)=8.617\times 10^{-5}\times 300=0.02585\text{ eV}\approx 0.0259\text{ eV}.$ $k_{B}T(77)=8.617\times 10^{-5}\times 77=0.006635\text{ eV}\approx 6.6\text{ meV}.$ Colder → smaller $k_{B}T$ → sharper step. This is the "freezing stadium" from the parent note.

WHY it works by sign: $f=\frac{1}{1+e^x}$ is a decreasing function of $x=(E-E_F)/k_{B}T$.

• (a) $E<E_F\Rightarrow x<0\Rightarrow e^x<1\Rightarrow$ denominator $<2\Rightarrow f>\frac{1}{2}$. Mostly full.
• (b) $E>E_F\Rightarrow x>0\Rightarrow e^x>1\Rightarrow$ denominator $>2\Rightarrow f<\frac{1}{2}$. Mostly empty. Read this straight off the shaded regions of the figure.

WHAT: here $x=+2$. WHY: the energy is given directly in $k_{B}T$ units, so no need for eV. $f=\frac{1}{1+e^2}=\frac{1}{1+7.389}=\frac{1}{8.389}=0.1192\approx 11.9\%.$ About $88\%$ empty — states a couple of $k_{B}T$ above $E_F$ are largely vacant.

WHAT: convert energy to $x$ first. WHY: the formula eats $x$, not raw eV. $x=\frac{+0.05}{0.0259}=\mathrm{1.931.}$ $f=\frac{1}{1+e^{1.931}}=\frac{1}{1+6.895}=\frac{1}{7.895}=0.1267\approx 12.7\%.$

WHAT: empty probability $=1-f$. WHY: a state is either occupied ($f$) or not ($1-f$); the two must sum to 1. $1-f=1-0.1267=0.8733\approx 87.3\%.$ This "$1-f$" is the hole occupation probability — remember it, it drives L4.

WHAT: use the inversion $E-E_F=k_{B}T\ln \left(\frac{1-f}{f}\right)$. WHY: we're given $f$ and asked for position — the reverse direction. $\frac{1-f}{f}=\frac{0.1}{0.9}=0.1111,\ln (0.1111)=-\mathrm{2.197.}$ $E-E_F=0.0259\times (-2.197)=-0.0569\text{ eV}.$ Negative → below $E_F$. The state sits $0.0569$ eV below the Fermi level. Makes sense: mostly filled ($f=0.9>\frac{1}{2}$) means below the water line.

WHAT: compute both terms with $x=+2$ and $x=-2$. WHY: verifying the symmetry that pairs "filled above" with "empty below". $f(E_F+2k_{B}T)=\frac{1}{1+e^2}=\mathrm{0.11920.}$ $f(E_F-2k_{B}T)=\frac{1}{1+e^{-2}}=\frac{1}{1.13534}=\mathrm{0.88080.}$ Sum $=0.11920+0.88080=1.00000$. ✓ Exactly 1 — the curve is antisymmetric about $(E_F,\frac{1}{2})$.

WHY compare: the parent note says Boltzmann is valid when $x\gg 1$; this quantifies "how far is far enough". At $x=1$: exact $f=\frac{1}{1+e^1}=\frac{1}{3.718}=0.26894$; approx $e^{-1}=0.36788$. Relative error $=\frac{0.36788-0.26894}{0.26894}=0.368=36.8\%$ — bad. At $x=4$: exact $f=\frac{1}{1+e^4}=\frac{1}{55.598}=0.017986$; approx $e^{-4}=0.018316$. Relative error $=\frac{0.018316-0.017986}{0.017986}=0.0183=1.83\%$ — good. Lesson: at $x=1$ the "+1" in the denominator still matters a lot; by $x=4$ it's negligible. See the figure below where the two curves peel apart near $E_F$ and merge far above it.

WHY combine: electrons live at $E_C$, holes live at $E_V$; this ties the distribution to the two carrier types of Intrinsic and extrinsic semiconductors. $k_{B}T=0.0259$ eV. (a) $x_C=+0.55/0.0259=21.24$. $f(E_C)=\frac{1}{1+e^{21.24}}=\frac{1}{1+1.68\times 10^9}=5.96\times 10^{-10}.$ Vanishingly small — the conduction band is nearly empty (as expected for an insulator-like intrinsic material). (b) $x_V=-0.55/0.0259=-21.24$. $f(E_V)=\frac{1}{1+e^{-21.24}}=1-5.96\times 10^{-10}.$ Hole probability $=1-f(E_V)=5.96\times 10^{-10}.$ (c) They are equal: $f(E_C)=1-f(E_V)$. That's the antisymmetry (L3.2) applied at $\Delta =0.55$ eV. Equal electron-top and hole-bottom probabilities is exactly why mid-gap $E_F$ gives $n=p$ (charge neutrality). See Chemical potential and equilibrium.

WHAT: compute each $f$, then divide. WHY: ratios like this appear when comparing occupation across a small energy window. $x_A=-0.05/0.0259=-1.931$, $x_B=+1.931$. $f(A)=\frac{1}{1+e^{-1.931}}=\frac{1}{1.14497}=\mathrm{0.87338.}$ $f(B)=\frac{1}{1+e^{1.931}}=\frac{1}{7.89528}=\mathrm{0.12666.}$ $\frac{f(A)}{f(B)}=\frac{0.87338}{0.12666}=\mathrm{6.895.}$ Notice $6.895=e^{1.931}$ — the ratio equals $e^{(E_B-E_A)/k_{B}T}$ when both are near $E_F$? Only approximately; here it's exact because $f(A)=1-f(B)$ and $\frac{1-f}{f}=e^x$. A neat consistency check.

WHY Boltzmann: $x_0=21.24\gg 4$, so $f(E_C)\approx e^{-x}$ safely (L3.3 justified this). We want $e^{-x_1}=10e^{-x_0}$, i.e. $x_0-x_1=\ln 10=2.3026$. $x_1=21.24-2.3026=\mathrm{18.937.}$ Since $x=(E_C-E_F)/k_{B}T$ with the numerator fixed at $0.55$ eV: $k_B{T}_1=\frac{0.55}{18.937}=0.029045\text{ eV}\Rightarrow T_1=\frac{0.029045}{8.617\times 10^{-5}}=337.1\text{ K}.$ Factor on $k_{B}T$: $0.029045/0.02585=1.1235$. So raising $T$ from $300$ K to about $337$ K multiplies the conduction occupation ~$10\times$. Small temperature change, big occupation change — the exponential is fierce.

WHY limits: "$T=0$" cannot be plugged in directly ($k_{B}T=0$ divides by zero inside the exponent), so we take the limit $T\to 0^{+}$. Write $x=(E-E_F)/k_{B}T$.

• Below $E_F$ ($E<E_F$): numerator negative, and $k_{B}T\to 0^{+}$ makes $x\to -\infty$. Then $e^x\to 0$, so $f\to \frac{1}{1+0}=1$. Fully occupied.
• Above $E_F$ ($E>E_F$): numerator positive, $x\to +\infty$, $e^x\to \infty$, so $f\to \frac{1}{\infty }=0$. Empty.
• Exactly at $E_F$ ($E=E_F$): the numerator is exactly zero for every $T$, so $x=0$ at all $T$, giving $f=\frac{1}{2}$ throughout the limit. The single point $(E_F,\frac{1}{2})$ survives; the two limits above approach from either side, producing the discontinuous step with the mid-value pinned at $\frac{1}{2}$. The figure below shows three curves ($T$ high → low) collapsing onto the step, all pivoting through $(E_F,\frac{1}{2})$.