Compare band gaps - conductor - semiconductor - insulator: Visual walkthrough

Before any algebra, let us agree on the cast of characters, because every symbol here must be earned before it is used.

Definition The words we will need (each is a picture, not just a letter)

Energy $E$ — how much "climbing power" an electron has. Higher up on our diagrams = more energy. Measured in electron-volts (eV): one eV is the energy one electron gains falling through one volt. A tiny unit, perfect for atoms.
Valence band (VB) — the crowded lower shelf where electrons sit glued to atoms. Its top edge is called $E_{v}$ .
Conduction band (CB) — the empty upper shelf where an electron is free to roam and carry current. Its bottom edge is $E_{c}$ .
Band gap $E_{g} = E_{c} - E_{v}$ — the forbidden height between the two shelves. No electron is allowed to rest inside it.
Temperature $T$ — measured in kelvin (K). Room temperature $\approx 300$ K.
$k$ (Boltzmann's constant) — the exchange rate between temperature and energy: $k \approx 8.617 \times 1 0^{- 5} eV/K$ . So $k T$ is an energy: at 300 K, $k T \approx 0.0259$ eV. Think of $k T$ as the typical energy a thermal kick can deliver to one electron.

Keep that last idea burning: $k T$ is the size of a random thermal shove. The whole story is a race between how tall the gap is ( $E_{g}$ ) and how hard heat can shove ( $k T$ ).

Step 1 — Draw the two shelves and the forbidden gap

WHAT. We sketch the energy landscape: a full lower shelf (VB), an empty upper shelf (CB), and the forbidden gap between them.

WHY. Everything about conduction comes down to getting an electron from the full shelf onto the empty shelf, because only a partly-filled band carries current (a totally full band has equal left- and right-movers → zero net flow). So we must first see the height an electron has to climb.

PICTURE.

Figure — Compare band gaps - conductor - semiconductor - insulator

In the figure, the height of the orange strip is $E_{g}$ . The electron at the bottom shelf has to gain at least $E_{g}$ of energy to reach the empty shelf. Hold that image — the rest is just counting how many electrons win that climb.

Step 2 — Ask "how likely is a state up top to be occupied?" (Fermi–Dirac)

WHAT. We introduce the rule that tells us the probability a spot at energy $E$ actually holds an electron: $f (E) = \frac{1}{1 + e ^{(E - E_{F}) / k T}}$

WHY this tool and not another? Electrons are antisocial — no two can share the same quantum state (Pauli exclusion). A dice-roll (classical) probability ignores that. The Fermi–Dirac distribution is the correct occupancy rule for such particles, so we start there rather than a plain Boltzmann guess. See Fermi–Dirac Distribution for its full birth.

Term by term:

$f (E)$ — a number between $0$ and $1$ : the chance the state at energy $E$ is filled.
$E_{F}$ — the Fermi level, the energy where $f = \frac{1}{2}$ (a state is a coin-flip). It is our reference "sea level."
$E - E_{F}$ — how far above sea level our state sits.
$e^{(E - E_{F}) / k T}$ — the exponential that punishes states far above $E_{F}$ . When $E - E_{F}$ is large, this blows up, and $f$ shrinks toward $0$ .

PICTURE.

Notice the S-shape: below $E_{F}$ almost everything is filled ( $f \approx 1$ ), above $E_{F}$ it falls off. The steepness of that fall is set by $k T$ — warmer means a gentler, more spread-out fall, so more high states get filled.

Step 3 — Place the Fermi level in the middle of the gap

WHAT. For a pure (intrinsic) semiconductor, $E_{F}$ sits near the middle of the gap. So a state at the bottom of the conduction band ( $E = E_{c}$ ) is a distance $E_{c} - E_{F} \approx \frac{E _{g}}{2}$ above sea level.

WHY. In an intrinsic crystal, every electron that jumps up leaves behind an empty spot (a hole) below. Electrons up and holes down must balance in number. That balance forces the coin-flip line $E_{F}$ to land roughly halfway up the gap (see Intrinsic vs Extrinsic Semiconductors and Doping and the Fermi Level for why doping later moves it). This is the origin of the famous factor of 2 — remember it.

Term by term:

$E_{c}$ — bottom edge of the empty shelf (where new free electrons appear).
$E_{F} \approx (E_{c} + E_{v}) /2$ — mid-gap sea level.
$E_{c} - E_{F} \approx E_{g} /2$ — the climb an electron really faces, which is half the full gap, because $E_{F}$ meets it halfway.

PICTURE.

Step 4 — The gap is huge next to a thermal kick, so simplify

WHAT. Compare the two energies in the exponent. At room $T$ , $k T \approx 0.026$ eV, while $E_{c} - E_{F} \approx E_{g} /2$ is hundreds of meV. So $\frac{E - E _{F}}{k T} ≫ 1 ⟹ e^{(E - E_{F}) / k T} ≫ 1.$

WHY. When a number is enormous, adding $1$ to it changes nothing. So the " $+ 1$ " in the Fermi–Dirac denominator becomes dead weight and we can drop it: $f (E) = \frac{1}{1 + e ^{(E - E_{F}) / k T}} \approx \frac{1}{e ^{(E - E_{F}) / k T}} = e^{- (E - E_{F}) / k T} .$

This is the Boltzmann approximation — Fermi–Dirac's simple tail.

Term by term:

We threw away the $1$ because $e^{(E - E_{F}) / k T}$ dwarfs it (error under $1$ part in a million for a real gap).
The result $e^{- (E - E_{F}) / k T}$ is a clean decaying exponential: the higher the state, the rarer it is filled.

PICTURE.

The two curves (true Fermi–Dirac and the Boltzmann tail) sit right on top of each other out in the conduction band — that is what "drop the +1" looks like.

Step 5 — Count the electrons: fold in the density of states

WHAT. Probability alone is not a number of electrons. We need how many seats exist at each energy in the CB — that is the density of states, $g (E)$ . Multiply "seats available" by "chance a seat is filled," then add up over the whole conduction band: $n_{i} = \int_{E_{c}}^{\infty} g (E) f (E) d E .$

WHY. A high shelf could have a low occupancy but many seats, or vice-versa. Only seats × occupancy, summed, gives the true head-count. The integral sign $\int \dots d E$ just means "slice the CB into thin energy layers and total them up." (Full construction of $g (E)$ lives in Density of States.)

Term by term:

$g (E)$ — number of available states per unit energy at height $E$ (seats per shelf-slice).
$f (E) \approx e^{- (E - E_{F}) / k T}$ — the chance each seat is taken (Step 4).
$\int_{E_{c}}^{\infty}$ — sum from the bottom of the empty shelf upward.

PICTURE.

The shaded sliver in the figure is the product $g (E) f (E)$ — the actual number of electrons in each thin layer. It is squeezed into a narrow band just above $E_{c}$ , because $g$ grows upward but $f$ crashes downward — their product is a small bump.

Step 6 — Do the sum: out pops the master formula

WHAT. Carrying out that integral (the standard result — the $g (E)$ part contributes a temperature-dependent prefactor) gives: $n_{i} = N_{c} N_{v} e^{- E_{g} /2 k T}$

WHY. The integral neatly separates into (i) a prefactor $N_{c} N_{v}$ that collects all the "how many seats" bookkeeping, and (ii) the surviving exponential $e^{- E_{g} /2 k T}$ that carries the physics of the climb. Because $E_{g}$ lives in an exponent, it dominates everything.

Term by term:

$n_{i}$ — intrinsic carrier concentration (free electrons per unit volume in pure material).
$N_{c}, N_{v}$ — effective density of states in CB and VB: think "effective number of seats" on each shelf. $N_{c} N_{v}$ is their geometric average.
$e^{- E_{g} /2 k T}$ — the star. The exponent is a ratio: (half the gap you must climb) ÷ (size of a thermal kick). Bigger gap → more negative exponent → exponentially fewer carriers. The $/2$ is the mid-gap Fermi level from Step 3.

PICTURE.

Because the vertical axis is logarithmic, $ln n_{i}$ falls as a straight line against $1/ T$ , with slope $- E_{g} /2 k$ . Every straight line in the figure is a different material; steeper line = wider gap.

Step 7 — Edge case A: the metal, where the gap is zero

WHAT. Set $E_{g} = 0$ (bands overlap): $e^{- E_{g} /2 k T} = e^{0} = 1.$

WHY. With no climb, the exponential suppression vanishes. Carriers are present at every temperature — the head-count no longer depends on heat at all.

Consequence (the important twist): since carrier count is fixed, a metal actually gets more resistive when heated — extra lattice vibrations (phonons) scatter the already-present electrons. That is a mobility story, not a carrier-count story (see Conductivity and Mobility and Temperature Dependence of Resistance in Metals).

PICTURE.

The metal's line is flat (carriers constant); the semiconductor's line climbs steeply with temperature. Same equation, opposite behaviour, decided entirely by $E_{g}$ .

Step 8 — Edge case B: the insulator, where the gap is enormous

WHAT. For diamond, $E_{g} \approx 5.5$ eV. Put it against Silicon ( $E_{g} = 1.12$ eV) at 300 K, taking the ratio so prefactors cancel: $\frac{n _{i}^{Si}}{n _{i}^{dia}} = e^{- (E_{g}^{Si} - E_{g}^{dia}) /2 k T} = e^{(5.5 - 1.12) / (2 \times 0.02585)} = e^{84.7} \approx 1 0^{37} .$

WHY. This shows the exponential's brutality: a gap only ~5× larger produces 37 orders of magnitude fewer carriers. There is no "insulator equation" — it is the same formula, just a taller climb.

Term by term:

Numerator vs denominator differ only in $E_{g}$ ; the shared $N_{c} N_{v}$ cancels.
$84.7$ in the exponent → $84.7/ ln 10 = 84.7/2.303 \approx 36.8$ , i.e. $\approx 1 0^{37}$ .

PICTURE.

Three materials share the identical straight-line law; only the slope ( $- E_{g} /2 k$ ) differs, splitting them into conductor / semiconductor / insulator.

Step 9 — Sanity check: heat the semiconductor and watch carriers explode

WHAT. Ratio of $n_{i}$ for Si heated from 300 K to 600 K: $\frac{n _{i} ( 600 )}{n _{i} ( 300 )} = exp [\frac{E _{g}}{2 k} (\frac{1}{300} - \frac{1}{600})] .$

WHY. Taking the ratio at two temperatures cancels the prefactor and isolates the exponential's temperature response — a direct test of the formula.

Term by term:

$\frac{E _{g}}{2 k} = \frac{1.12}{2 \times 8.617 \times 1 0 ^{- 5}} = 6499 K$ — the gap re-expressed as a temperature.
$(\frac{1}{300} - \frac{1}{600}) = 0.001667 K^{- 1}$ .
Product $= 6499 \times 0.001667 = 10.8$ , so the factor is $e^{10.8} \approx 5 \times 1 0^{4}$ .

Doubling temperature multiplies free carriers by ~50,000×. This is why semiconductors conduct better when hot — the opposite of metals (Step 7).

The one-picture summary

Read it left to right: two shelves + a gap → occupancy rule (Fermi–Dirac) → mid-gap Fermi level gives the half-gap climb → drop the +1 → multiply by seats and sum → out falls $n_{i} = N_{c} N_{v} e^{- E_{g} /2 k T}$ → the single exponential splits all three material classes.

Recall Feynman retelling of the whole walkthrough

Picture a building with a crowded ground floor (valence band) and an empty top floor (conduction band). Only kids on the empty top floor can push carts around and make electricity. The gap is the height of the jump between floors.

How many kids make it up? First we need the rule for whether any given upstairs spot is occupied — that's the Fermi–Dirac curve, an S-shaped "how full is this height" graph. The coin-flip line (Fermi level) sits smack in the middle of the gap, so a kid really only has to jump half the gap to reach the first upstairs spot — that's where the sneaky "divide by 2" comes from.

Way up top the odds are tiny, so the fancy curve simplifies into a plain fading exponential. Multiply "odds a spot is filled" by "how many spots exist," add them all up, and you get the head-count: $n_{i} = N_{c} N_{v} e^{- E_{g} /2 k T}$ .

The whole personality of a material hides in that exponent. Zero gap (metal): the exponential is just $1$ , kids always upstairs — always conducts. Small gap (silicon): a few make it, and heating gives everyone more jumping energy, so carriers explode with temperature. Huge gap (diamond): almost nobody jumps — an insulator. One equation, three worlds, decided by the height of a jump sitting in an exponent.

Recall

Why is the exponent $E_{g} /2 k T$ and not $E_{g} / k T$ ? ::: Because the intrinsic Fermi level sits near mid-gap, so a carrier only climbs about $E_{g} /2$ on average. What does $k T$ physically represent? ::: The typical energy of a random thermal kick; at 300 K it is about 0.0259 eV. Why can we drop the "+1" in Fermi–Dirac out in the conduction band? ::: Because $E - E_{F} ≫ k T$ there, so $e^{(E - E_{F}) / k T} ≫ 1$ and the +1 is negligible. What two pieces multiply inside the carrier-count integral? ::: The density of states $g (E)$ (seats available) and the occupancy $f (E)$ (chance a seat is filled). For a metal ( $E_{g} = 0$ ), what does the exponential become and what then sets resistance? ::: $e^{0} = 1$ ; carrier count is fixed, so resistance is set by phonon scattering (mobility), which rises with temperature.

Connections

Parent: Compare band gaps
Fermi–Dirac Distribution
Density of States
Intrinsic vs Extrinsic Semiconductors
Doping and the Fermi Level
Conductivity and Mobility
Temperature Dependence of Resistance in Metals