Band gap and its meaning for conductivity: Visual walkthrough

We assume you know only: energy is measured in a number, temperature is "how much random jiggling," and higher means harder to reach. Everything else we draw.

Step 1 — The energy ladder: where can an electron sit?

WHAT. Picture a vertical axis. Up = more energy. On this axis, electrons in a crystal are only allowed to sit in certain bands (thick allowed stripes) and are forbidden to sit in the gap between two of them.

WHY. Before we can talk about "an electron jumping the gap," we must first draw the two places it jumps between. The bottom stripe is packed full of electrons; the top stripe is empty.

PICTURE. Look at the figure. The red band at the bottom is the valence band — full of electrons. Its top edge is labelled $E_{v}$ . The band above is the conduction band, empty, with bottom edge $E_{c}$ . The white strip between them is where no electron is ever allowed to be.

Figure — Band gap and its meaning for conductivity

Each symbol here is just a mark on the ladder. Nothing is happening yet — we have only set the stage.

Step 2 — Why a full band cannot carry current

WHAT. We ask: if the valence band is completely full of electrons, can it conduct? Answer: no. So current must come from lifting an electron up into the empty band above.

WHY. Current means charge drifting one way. For an electron to drift, it must nudge into a slightly higher-energy neighbouring state. In a packed band every neighbouring seat is taken — nowhere to nudge. This is the whole reason the gap matters: conduction lives entirely in the empty band up top.

PICTURE. Left half: a full row of seats, every one occupied (red dots), arrows blocked — a traffic jam. Right half: one electron has been promoted across the gap into the empty band; now it has empty seats all around and can move (red arrow drifts).

So the number of moving carriers is exactly the number of electrons we manage to promote across $E_{g}$ . The rest of the page computes that number.

Step 3 — What decides whether one electron is up or down? Temperature

WHAT. Heat is random energy. At temperature $T$ , every electron gets random "kicks." A rare, big kick can throw an electron up across the gap. We need the chance of a kick big enough.

WHY. We can't track one electron; we track the probability that a state at energy $E$ is occupied. The tool that gives "chance of having energy $E$ at temperature $T$ " is the Boltzmann factor — see Fermi level and Fermi-Dirac distribution. We choose it because it is precisely the function that answers "how likely is a rare high-energy state given random thermal jiggling?"

PICTURE. A curve of occupation probability $f$ versus energy $E$ . Near the bottom, $f \approx 1$ (seats full). As we climb, the red curve decays: the higher the seat, the less likely it is filled.

Read the fraction as "climb divided by kick size." If the climb is many times the kick, the exponent is a big negative number and $f$ is tiny.

Step 4 — Where is the reference level $E_{F}$ ? Dead centre of the gap

WHAT. The formula above needs a reference energy $E_{F}$ , the Fermi level. For a pure (intrinsic) semiconductor it sits exactly in the middle of the gap.

WHY. $E_{F}$ is the energy at which a seat is 50/50 likely to be filled. For every electron promoted up into the conduction band, an empty seat — a hole — is left down in the valence band. By that symmetry the 50/50 point balances halfway between $E_{v}$ and $E_{c}$ . This single geometric fact is where the 2 will come from.

PICTURE. The ladder again, with a dashed red line for $E_{F}$ drawn precisely midway. Two equal braces: from $E_{F}$ up to $E_{c}$ is $E_{g} /2$ ; from $E_{v}$ up to $E_{F}$ is also $E_{g} /2$ .

This is the heart of the whole page: the electron does not climb the full gap $E_{g}$ ; from mid-gap it only climbs $E_{g} /2$ .

Step 5 — Plug the mid-gap climb into the tail

WHAT. We now put the climb from Step 4, $E_{c} - E_{F} = E_{g} /2$ , into the Boltzmann tail from Step 3.

WHY. The number of electrons $n$ that reach the conduction band is proportional to the chance of a seat there being filled, $f (E_{c})$ . So we just evaluate $f$ at $E = E_{c}$ .

PICTURE. The decay curve of Step 3, with a vertical red drop marking $E = E_{c}$ ; the height where it lands on the curve is $n$ . The horizontal distance from $E_{F}$ to $E_{c}$ is labelled $E_{g} /2$ .

Step 6 — From carriers to conductivity

WHAT. Conductivity $σ$ (how easily current flows) is proportional to the number of carriers times how freely each one moves. The "free-moving" part changes only slowly with $T$ , so the exponential dominates.

WHY. More carriers $\Rightarrow$ more current for the same push. Since $n$ carries the exponential, $σ$ inherits it. (The mobility factor is developed in Conductivity, mobility and drift current.)

PICTURE. $σ$ plotted against $T$ : a red curve that rises steeply as temperature climbs — the signature semiconductor shape.

Step 7 — Cover the edge cases (so no scenario surprises you)

WHAT. Push the formula to its extremes and read off each limit.

WHY. A derivation you trust must survive its own boundaries. We test $E_{g} \to 0$ , $E_{g}$ huge, and $T \to 0$ .

PICTURE. Three mini-panels of the ladder: (a) bands overlapping, gap zero — a metal; (b) an enormous gap — an insulator; (c) $T = 0$ , no thermal kicks at all.

The one-picture summary

Everything above, compressed: the ladder on the left feeds the decay curve on the right; the mid-gap climb $E_{g} /2$ is the single distance that becomes the exponent; the exponent becomes the rising $σ (T)$ curve.

Recall Feynman retelling (plain words)

Draw a ladder. Fill the bottom shelf completely with electrons and leave the top shelf empty. A packed shelf can't shuffle — no current — so the only way to get current is to throw an electron up to the empty shelf across a forbidden gap. How high is the throw? Not the full gap: the 50/50 "waterline" (the Fermi level) sits exactly halfway up, so from the waterline the throw is only half the gap. Now, how often does a random heat-kick reach that high? The Boltzmann rule says the chance dies off exponentially with (height ÷ kick-size), and here the height is $E_{g} /2$ and the kick-size is $k_{B} T$ . So the number of electrons that make it is $e^{- E_{g} / (2 k_{B} T)}$ , and conductivity, riding on that number, is $σ \propto e^{- E_{g} / (2 k_{B} T)}$ . Make the gap bigger and the throw becomes hopeless (insulator); shrink it to nothing and everyone's already up top (metal); cool it toward absolute zero and the kicks vanish so nobody climbs. One picture, one waterline, one factor of two.

Recall Check yourself

Why is the exponent $E_{g} /2$ and not $E_{g}$ ? ::: Because the Fermi level sits mid-gap, so an electron only climbs $E_{c} - E_{F} = E_{g} /2$ to reach the empty band. Where does the minus sign come from? ::: From the Boltzmann tail — a higher climb means a smaller occupation chance, so the exponent is negative. What happens to $σ$ as $T \to 0$ ? ::: The kick size $k_{B} T \to 0$ , the exponent $\to - \infty$ , so $n \to 0$ ; the semiconductor becomes a perfect insulator.

Connections

Fermi level and Fermi-Dirac distribution
Intrinsic vs extrinsic semiconductors
Doping and carrier concentration
Conductivity, mobility and drift current
Temperature dependence of resistivity in metals