Fermi level and Fermi-Dirac distribution: Worked examples

This is the "drill every case" companion to the parent note. We will hit every kind of situation the Fermi–Dirac formula can throw at you: energies above, below and exactly at the Fermi level; the two temperature extremes ( $T = 0$ and $T \to \infty$ ); the "hole" (empty-state) side; a real-world device question; and an exam-style twist. Every symbol is re-earned here, so you can start from line one.

Recall The one formula we keep reusing

$f (E) = \frac{1}{1 + e ^{(E - E_{F}) / k_{B} T}}$

$f (E)$ is a probability between $0$ and $1$ that a state at energy $E$ holds an electron.
$E_{F}$ is the Fermi level: the energy where $f = \frac{1}{2}$ .
$k_{B} T$ is the "thermal energy unit". At $T = 300$ K, $k_{B} T \approx 0.0259$ eV.
The only quantity that matters is the dimensionless ratio $x = \frac{E - E _{F}}{k _{B} T}$ : energy measured in $k_{B} T$ steps, counted from $E_{F}$ .

Let me name that ratio once and for all, because every example below is really just "compute $x$ , then compute $\frac{1}{1 + e ^{x}}$ ."

Figure — Fermi level and Fermi-Dirac distribution

Look at the figure: the curve is a smooth "S" (an S-curve, technically a sigmoid) that drops from $1$ on the left to $0$ on the right, passing through exactly $\frac{1}{2}$ at $x = 0$ . Every example is just picking one point on this one curve.

The scenario matrix

Here is every distinct class of problem. Each later example is tagged with the cell(s) it covers.

Cell	Situation	What's special	Covered by
A	$E > E_{F}$ (state above)	$x > 0$ , $f$ small	Ex 1
B	$E < E_{F}$ (state below)	$x < 0$ , $f$ near $1$ ; ask for empty = $1 - f$	Ex 2
C	$E = E_{F}$ exactly	$x = 0$ , $f = \frac{1}{2}$ always	Ex 3
D	$T = 0$ K (degenerate limit)	step function, $x \to \pm \infty$	Ex 3
E	$T \to \infty$ (hot limit)	$x \to 0$ , $f \to \frac{1}{2}$ everywhere	Ex 3
F	Inverse problem: given $f$ , find $E - E_{F}$	invert the S-curve	Ex 4
G	Boltzmann tail valid? ( $x ≫ 1$ )	compare exact vs approx	Ex 5
H	Antisymmetry check across $E_{F}$	$f (+ Δ) + f (- Δ) = 1$	Ex 6
I	Real-world word problem (device, temperature effect)	pick $T$ , compute change	Ex 7
J	Exam twist: "electron at $E_{F}$ ?" / no-state trap	probability vs. count; needs Density of states	Ex 8

Example 1 — State above the Fermi level (Cell A)

Worked example

$E$ is $2 k_{B} T$ above $E_{F}$ Find the occupation probability $f$ when $E - E_{F} = 2 k_{B} T$ .

Forecast: Above $E_{F}$ means mostly empty. Guess: is $f$ closer to $0.1$ or $0.4$ ? Write your guess down before reading on.

Step 1 — Build $x$ . $x = \frac{E - E _{F}}{k _{B} T} = \frac{2 k _{B} T}{k _{B} T} = 2$ . Why this step? The formula only ever sees $x$ ; the actual eV value and $T$ cancel into this clean number.

Step 2 — Evaluate the exponential. $e^{2} = 7.389$ . Why this step? The denominator is $1 + e^{x}$ ; we need $e^{x}$ first.

Step 3 — Plug in. $f = \frac{1}{1 + 7.389} = \frac{1}{8.389} = 0.1192$ . Why this step? Direct substitution into $f = \frac{1}{1 + e ^{x}}$ .

Answer: $f \approx 0.119$ , i.e. about a 12% chance the state is occupied — so $\approx 88%$ empty. (Did your forecast beat "closer to $0.1$ "?)

Verify: Sanity — $x > 0$ demands $f < \frac{1}{2}$ . Indeed $0.119 < 0.5$ . ✓ And it's not tiny, because $2 k_{B} T$ is only "a little" above $E_{F}$ ; you have to go several $k_{B} T$ up before $f$ collapses to near zero.

Example 2 — State below the Fermi level, ask for EMPTY (Cell B)

Worked example A state

$0.05$ eV below $E_{F}$ at $300$ K — chance it is empty A state sits $0.05$ eV below $E_{F}$ . Temperature $T = 300$ K so $k_{B} T = 0.0259$ eV. Find the probability the state is empty (this "empty state below the top of a band" is a hole).

Forecast: Below $E_{F}$ = mostly full, so empty should be a small number. Bigger or smaller than 10%?

Step 1 — Build $x$ with the correct sign. Below means $E - E_{F} = - 0.05$ eV, so $x = \frac{- 0.05}{0.0259} = - 1.931.$ Why this step? The sign is the whole game — forget the minus and you'd compute a full state as empty. "Below" → negative $x$ .

Step 2 — Occupation. $e^{- 1.931} = 0.1449$ , so $f = \frac{1}{1 + 0.1449} = \frac{1}{1.1449} = 0.8734$ . Why this step? First get how full it is; the empty probability is the leftover.

Step 3 — Empty probability. $1 - f = 1 - 0.8734 = 0.1266$ . Why this step? A state is either filled ( $f$ ) or empty ( $1 - f$ ); the two must sum to $1$ .

Answer: $\approx 12.7%$ chance empty. So even $0.05$ eV below $E_{F}$ , roughly one state in eight is vacant — these vacancies are the holes that carry current. See Carrier concentration n and p.

Verify: Units — $x$ is (eV)/(eV) = dimensionless ✓. Sign check — below $E_{F}$ gave $f > \frac{1}{2}$ ✓. Add-up check — $f + (1 - f) = 0.8734 + 0.1266 = 1.000$ ✓.

Example 3 — The three extreme limits at once (Cells C, D, E)

Worked example

$E = E_{F}$ ; then $T = 0$ ; then $T \to \infty$ Take a state and examine it in three regimes.

Forecast: Which knob makes $f$ independent of energy: freezing to $T = 0$ , or heating to $T \to \infty$ ? (Trap: many say $T = 0$ .)

(C) Exactly at the Fermi level, any $T$ . Here $E = E_{F}$ so $x = 0$ , giving $e^{0} = 1$ and $f = \frac{1}{1 + 1} = \frac{1}{2} .$ Why this step? $x = 0$ kills the temperature dependence — that's why $E_{F}$ is defined as the half-occupation energy: it's the one point that's $\frac{1}{2}$ at every temperature.

(D) Absolute zero, $T = 0$ K. Now $k_{B} T \to 0$ , so $x = \frac{E - E _{F}}{k _{B} T} \to \pm \infty$ depending on the sign of $E - E_{F}$ :

$E < E_{F}$ : $x \to - \infty$ , $e^{x} \to 0$ , $f \to 1$ (full).
$E > E_{F}$ : $x \to + \infty$ , $e^{x} \to \infty$ , $f \to 0$ (empty).

This is the step function — the sharp cliff in the figure below. Why this step? At $T = 0$ there is no thermal energy to lift any electron above $E_{F}$ , so filling is perfectly sharp. This is the degenerate limit.

(E) Infinitely hot, $T \to \infty$ . Now $k_{B} T \to \infty$ , so for any finite $E - E_{F}$ , $x \to 0$ , giving $f \to \frac{1}{2}$ everywhere. Why this step? Infinite thermal energy makes every state equally likely to be filled or empty — the "S" flattens into a horizontal line at $\frac{1}{2}$ .

Answer: $f = \frac{1}{2}$ at $E_{F}$ always; a sharp step at $T = 0$ ; a flat $\frac{1}{2}$ line at $T \to \infty$ . So it's the hot limit that erases energy dependence — forecast trap sprung.

Verify: At $T = 0$ plug a state $0.1$ eV below $E_{F}$ : $x \to - \infty \Rightarrow f = 1$ exactly ✓. Continuity check — as $T$ rises from $0$ , the step must smear but still pass through $(E_{F}, \frac{1}{2})$ , which both limits respect ✓.

Example 4 — Inverse problem: given $f$ , where is the state? (Cell F)

Worked example A state is 25% occupied — how far above

$E_{F}$ ? Given $f = 0.25$ at $T = 300$ K ( $k_{B} T = 0.0259$ eV), find $E - E_{F}$ .

Forecast: $f < \frac{1}{2}$ means the state is above $E_{F}$ . Above by more or less than $k_{B} T$ ?

Step 1 — Invert the formula algebraically. Start from $f = \frac{1}{1 + e ^{x}}$ . Flip: $\frac{1}{f} = 1 + e^{x} \Rightarrow e^{x} = \frac{1}{f} - 1 = \frac{1 - f}{f} .$ Why this step? We want $x$ alone; the S-curve is monotonic so it has exactly one inverse (the logit).

Step 2 — Take the logarithm. $x = ln (\frac{1 - f}{f})$ . Why this step? $ln$ is the tool that undoes the exponential — it answers "what power of $e$ gives this?"

Step 3 — Plug numbers. $\frac{1 - 0.25}{0.25} = \frac{0.75}{0.25} = 3$ , so $x = ln 3 = 1.0986$ . Why this step? Pure substitution.

Step 4 — Convert $x$ back to energy. $E - E_{F} = x k_{B} T = 1.0986 \times 0.0259 = 0.02845$ eV. Why this step? $x$ is in " $k_{B} T$ units"; multiplying restores eV.

Answer: $E - E_{F} \approx + 0.0285$ eV — the state is about one $k_{B} T$ above $E_{F}$ .

Verify: Feed $x = 1.0986$ forward: $e^{1.0986} = 3.000$ , $f = \frac{1}{1 + 3} = 0.25$ ✓. Sign check — $f < \frac{1}{2}$ gave $x > 0$ (above) ✓.

Example 5 — Is the Boltzmann shortcut safe here? (Cell G)

Worked example Compare exact

$f$ with $e^{- x}$ at $x = 5$ and at $x = 1$ The parent note says: when $E - E_{F} ≫ k_{B} T$ you may drop the " $1$ " and use the Boltzmann approximation $f \approx e^{- x}$ . Test how good it is at $x = 5$ (far) and $x = 1$ (close).

Forecast: At which $x$ does dropping the $1$ cost you a big error?

Step 1 — Far case $x = 5$ . Exact: $f = \frac{1}{1 + e ^{5}} = \frac{1}{1 + 148.41} = 6.693 \times 1 0^{- 3}$ . Approx: $e^{- 5} = 6.738 \times 1 0^{- 3}$ . Why this step? When $e^{x}$ is huge, the " $+ 1$ " barely changes $1 + e^{x}$ , so the approximation nearly matches.

Step 2 — Relative error at $x = 5$ . $\frac{0.006738 - 0.006693}{0.006693} = 0.0067 = 0.67%$ . Excellent. Why this step? Relative error tells us whether the shortcut is device-grade accurate.

Step 3 — Close case $x = 1$ . Exact: $f = \frac{1}{1 + e ^{1}} = \frac{1}{3.718} = 0.2689$ . Approx: $e^{- 1} = 0.3679$ . Relative error $= \frac{0.3679 - 0.2689}{0.2689} = 0.368 = 37%$ . Terrible. Why this step? Here $e^{x} = 2.72$ is not huge, so ignoring the $1$ in $1 + e^{x}$ is a big lie.

Answer: Boltzmann is superb far above $E_{F}$ ( $x = 5$ : $< 1%$ off) and useless close to it ( $x = 1$ : $37%$ off). Rule of thumb: trust it once $x ≳ 3$ . This is the backbone of non-degenerate Intrinsic and extrinsic semiconductors math.

Verify: $x = 5$ error $0.67% < 1%$ ✓; $x = 1$ error $\approx 37% > 10%$ ✓ — the shortcut's validity flips exactly as claimed.

Example 6 — Antisymmetry about $E_{F}$ (Cell H)

Worked example Show

$f (E_{F} + Δ) + f (E_{F} - Δ) = 1$ for $Δ = 2 k_{B} T$ Verify the filling/emptying mirror symmetry using $Δ$ equal to $2 k_{B} T$ (so $x = \pm 2$ ).

Forecast: Will the two probabilities add to exactly $1$ , or just approximately?

Step 1 — Above point, $x = + 2$ . $f_{+} = \frac{1}{1 + e ^{2}} = \frac{1}{8.389} = 0.1192$ . Why this step? This is "chance a state $2 k_{B} T$ above $E_{F}$ is filled."

Step 2 — Below point, $x = - 2$ . $f_{-} = \frac{1}{1 + e ^{- 2}} = \frac{1}{1.1353} = 0.8808$ . Why this step? This is "chance a state $2 k_{B} T$ below $E_{F}$ is filled."

Step 3 — Add them. $f_{+} + f_{-} = 0.1192 + 0.8808 = 1.0000$ . Why this step? If the sum is exactly $1$ , then chance-above-filled = chance-below-empty ( $1 - f_{-}$ ), which is the electron–hole symmetry.

Answer: Exactly $1$ . The curve has point symmetry about $(E_{F}, \frac{1}{2})$ : rotate the S-curve $18 0^{\circ}$ about that centre and it lands on itself.

Verify: Algebra guarantees it: $\frac{1}{1 + e ^{x}} + \frac{1}{1 + e ^{- x}} = \frac{1}{1 + e ^{x}} + \frac{e ^{x}}{e ^{x} + 1} = \frac{1 + e ^{x}}{1 + e ^{x}} = 1$ for all $x$ , not just $x = 2$ ✓.

Example 7 — Real-world word problem: heating a device (Cell I)

Worked example Same state, warmed from 300 K to 400 K

A trap state sits $0.15$ eV above $E_{F}$ . Its occupation matters for leakage current. At $T = 300$ K, $k_{B} T = 0.0259$ eV; a chip heats to $T = 400$ K, where $k_{B} T = 0.0345$ eV. Find $f$ at both temperatures and the factor by which occupation rises.

Forecast: Heating populates higher states, so $f$ should rise. By roughly $\times 2$ , or much more?

Step 1 — Cold case, $300$ K. $x_{300} = \frac{0.15}{0.0259} = 5.792$ ; $e^{5.792} = 327.6$ ; $f_{300} = \frac{1}{328.6} = 3.043 \times 1 0^{- 3}$ . Why this step? Establish the baseline leakage occupation.

Step 2 — Hot case, $400$ K. $x_{400} = \frac{0.15}{0.0345} = 4.348$ ; $e^{4.348} = 77.31$ ; $f_{400} = \frac{1}{78.31} = 1.277 \times 1 0^{- 2}$ . Why this step? Same energy gap, but bigger $k_{B} T$ shrinks $x$ → higher occupation.

Step 3 — Ratio. $\frac{f _{400}}{f _{300}} = \frac{0.01277}{0.003043} = 4.20$ . Why this step? Device engineers care about the multiplier — a $100$ K rise more than quadruples occupation of this state.

Answer: $f_{300} \approx 0.30%$ , $f_{400} \approx 1.28%$ , a $\times 4.2$ jump. This exponential sensitivity is why hot chips leak so much more current.

Verify: Boltzmann cross-check (both $x > 3$ , so valid): ratio $\approx \frac{e ^{- 4.348}}{e ^{- 5.792}} = e^{5.792 - 4.348} = e^{1.444} = 4.24$ , matching $4.20$ within rounding ✓. Sign — hotter gave larger $f$ ✓.

Example 8 — Exam twist: "is there an electron at $E_{F}$ ?" (Cell J)

Worked example Intrinsic silicon:

$f (E_{F}) = \frac{1}{2}$ , so half an electron sits at $E_{F}$ ? True or false? In intrinsic (undoped) silicon at $300$ K, $E_{F}$ lies near the middle of the band gap — a region with no allowed states. A student argues: " $f (E_{F}) = \frac{1}{2}$ , so on average half an electron occupies $E_{F}$ ." Judge the claim.

Forecast: Is the flaw in the probability $f$ , or in the assumption that a state exists there?

Step 1 — Confirm $f (E_{F}) = \frac{1}{2}$ . $x = 0 \Rightarrow f = \frac{1}{2}$ . The student's probability is correct. Why this step? Don't dismiss what's right — the arithmetic is fine.

Step 2 — Recall what $f$ actually multiplies. The real electron count is $n = \int g (E) f (E) d E,$ where $g (E)$ is the density of states: how many states exist per unit energy. See Density of states. Why this step? $f$ is only the chance a state is filled if a state exists. No state ⇒ nothing to fill.

Step 3 — Apply to the band gap. Inside the gap $g (E) = 0$ . So the contribution at $E_{F}$ is $g (E_{F}) f (E_{F}) = 0 \times \frac{1}{2} = 0$ electrons — regardless of $f$ . Why this step? Zero times a half is zero: the missing factor is $g$ , not $f$ .

Answer: False. $f (E_{F}) = \frac{1}{2}$ is true as a probability, but there is no state at $E_{F}$ in the gap, so zero electrons live there. $f$ is a conditional probability; occupancy needs $g (E)$ too. This is exactly the parent note's steel-manned mistake — see also Chemical potential and equilibrium.

Verify: Dimensional logic — $f$ is dimensionless (0–1), $g$ has units states·eV $^{- 1}$ ; only their product integrated over $E$ yields a count. In the gap $g = 0 \Rightarrow$ contribution $0$ ✓.

Recap: reading the matrix off one curve

Recall What every cell reduces to

Compute $x = \frac{E - E _{F}}{k _{B} T}$ ::: then $f = \frac{1}{1 + e ^{x}}$ ; everything else is a special case of this. Sign of $x$ for a state below $E_{F}$ ::: negative → $f > \frac{1}{2}$ (mostly full). Occupation exactly at $E_{F}$ at any $T$ ::: $\frac{1}{2}$ . Which limit flattens $f$ to $\frac{1}{2}$ everywhere ::: $T \to \infty$ (hot), not $T = 0$ . When is $f \approx e^{- x}$ trustworthy ::: when $x ≳ 3$ (error $< 1%$ ); fails badly near $E_{F}$ . Why $f (E_{F}) = \frac{1}{2}$ does NOT mean an electron sits at $E_{F}$ ::: $f$ is only occupation probability; count needs $g (E)$ , which is $0$ in the gap.

Connections

Fermi level and Fermi-Dirac distribution — the parent this drill supports.
Density of states — the $g (E)$ factor Example 8 needs.
Carrier concentration n and p — holes = $1 - f$ (Ex 2) integrated over the valence band.
Intrinsic and extrinsic semiconductors — Boltzmann tail (Ex 5) powers non-degenerate carrier math.
Pauli exclusion principle — why $f$ caps at $1$ in the first place.
Band gap and conduction/valence bands — where $E_{F}$ sits (Ex 8).
Chemical potential and equilibrium — $E_{F}$ as the flat electrochemical potential.