Energy bands - valence band and conduction band: Visual walkthrough

This is the parent note's central result, drawn out one picture at a time. We start with a single number line of energies and end with the intrinsic carrier formula — never using a symbol before we have earned it.

Step 1 — Draw the energy axis and the two bands

WHAT. We put electron energy on a vertical axis (higher = more energetic). We shade two allowed regions and leave a forbidden strip between them.

WHY. Everything in band theory is a statement about which energies an electron is allowed to have. Before counting jumps we must name the floor it jumps from and the ceiling it lands on. The parent note built these bands from splitting atomic levels; here we just need their edges.

PICTURE. Look at the figure. The blue block is the valence band (VB) — full of electrons at the bottom. The green block is the conduction band (CB) — empty. The dark strip is the band gap: no allowed states live there.

Figure — Energy bands - valence band and conduction band

Step 2 — Ask the real question: how likely is a state occupied?

WHAT. Pick any single energy level $E$ . We ask: what fraction of the time is it filled by an electron? Call that fraction $f (E)$ , a number between $0$ (never) and $1$ (always).

WHY. We cannot count electrons in the CB without a rule for how full each CB state is. That rule is the occupation probability. We introduce it now because every later step multiplies it against a count of states.

PICTURE. The curve slides from $1$ (low energies, always full) down through a halfway point to $0$ (high energies, always empty). The halfway point sits at a special energy $E_{F}$ .

Step 3 — Simplify: far above $E_{F}$ the exponential dominates

WHAT. A conduction-band state sits at $E \geq E_{c}$ , and in a semiconductor $E_{c}$ is many $k_{B} T$ above $E_{F}$ . When $E - E_{F} ≫ k_{B} T$ , the " $1 +$ " in the denominator is negligible next to the huge exponential.

WHY. We simplify because the exact Fermi-Dirac curve is clumsy to integrate, and in the region that matters (well above $E_{F}$ ) it collapses to a clean Boltzmann exponential. This is the single approximation that makes the whole derivation solvable by hand.

PICTURE. Zoom into the far-right tail of the same curve. The true curve (yellow) and the pure exponential (red dashed) lie right on top of each other once we are a few $k_{B} T$ past $E_{F}$ .

Step 4 — You cannot count electrons without counting seats

WHAT. $f (E)$ is only a probability per state. To get an actual number of electrons we must multiply by how many states exist at each energy — the density of states $g (E)$ .

WHY. A 90%-full row of $2$ seats holds fewer people than a 10%-full row of $1000$ seats. Probability alone is blind to how many seats there are. We need seats × fill-fraction.

PICTURE. Near a band edge the number of available seats grows like a square-root curve as you move into the band — few seats right at $E_{c}$ , more just above it.

Step 5 — Multiply and sum: electrons in the conduction band

WHAT. Electrons in the CB = (seats at each energy) × (chance each is filled), summed over all CB energies. Summing over a continuous axis is an integral.

WHY. We use an integral because energy is continuous — there is no "next" state to add one at a time. The integral adds up a smear of infinitely thin slices $g_{c} (E) f (E) d E$ .

PICTURE. The shaded product curve (seats × probability) peaks just above $E_{c}$ : at the very edge there are no seats, far up the exponential kills it, so the electrons pile up in a bump. The area of that bump is $n$ .

Step 6 — The hole side: empty seats in the valence band

WHAT. By identical logic, the number of empty valence states (a missing electron = a hole) is a mirror image, measured downward from $E_{v}$ .

WHY. We must count holes too, because every electron that jumped up left a hole behind. Both carry current. Covering only electrons would count half the physics. See Holes as charge carriers.

PICTURE. Flip the Step-5 picture upside down. The hole bump sits just below $E_{v}$ ; its area is $p$ , the hole density.

Step 7 — Multiply $n$ and $p$ : the $E_{F}$ cancels

WHAT. Multiply the electron count by the hole count. Watch the Fermi level — the one quantity we did not know — vanish.

WHY. In a pure (intrinsic) crystal we do not know $E_{F}$ in advance, so a formula that still contains it is useless. Multiplying $n \times p$ combines the two exponents so that $+ E_{F}$ and $- E_{F}$ annihilate, leaving only $E_{g} = E_{c} - E_{v}$ , which we do know.

PICTURE. Two exponent bars, one labelled $- (E_{c} - E_{F})$ and one $- (E_{F} - E_{v})$ , snap together end to end; the $E_{F}$ pieces cancel and the surviving length is exactly $- (E_{c} - E_{v}) = - E_{g}$ .

Step 8 — Intrinsic case: every electron leaves exactly one hole

WHAT. In a pure crystal, each promoted electron creates one hole, so the counts are equal: $n = p = n_{i}$ .

WHY. This charge-balance fact turns the product $n p$ into a single unknown. If $n = p = n_{i}$ , then $n p = n_{i}^{2}$ , so $n_{i} = n p$ — take the square root of Step 7.

PICTURE. One arrow lifts a blue electron from VB to CB, leaving a red hole exactly where it was. The tally on each side reads the same number, $n_{i}$ .

Edge & degenerate cases

The one-picture summary

Read it left to right: a band diagram (seats) → times the exponential tail (fill chance) → integrate into $n$ and $p$ → multiply so $E_{F}$ dies → square-root for the intrinsic case → the boxed formula.

Recall Feynman retelling (plain words)

Picture a tall filled bottom shelf (valence) and an empty top shelf (conduction) with a forbidden air-gap of height $E_{g}$ between them. Warmth throws electrons upward, but the chance of a throw reaching height $E_{g}$ shrinks exponentially — that is the $e^{- E_{g} / (k_{B} T)}$ feeling. But throws alone don't count: you also need to know how many landing spots (seats) exist up top, and how many empty spots the electron left behind below. Multiply seats by throw-chance and add up all the slices — that gives the electrons up top ( $n$ ) and the holes down below ( $p$ ). When you multiply those two totals, the one thing you never knew — the exact "half-full line" $E_{F}$ — cancels itself out, and you're left with only the gap height and the temperature. Since every electron up leaves exactly one hole down in a pure crystal, both counts are the same number $n_{i}$ , and taking the square root of the product hands you $n_{i} = N_{c} N_{v} e^{- E_{g} / (2 k_{B} T)}$ . The little $2$ downstairs is just the gap energy being shared between the electron you made and the hole you left.

Recall

Why does $E_{F}$ drop out of the product $n p$ ? ::: The electron exponent carries $+ E_{F}$ and the hole exponent carries $- E_{F}$ ; adding them cancels $E_{F}$ , leaving only $- E_{g}$ . Where does the factor of 2 in the final exponent come from? ::: The square root (from $n = p = n_{i}$ ) halves the exponent of $e^{- E_{g} / (k_{B} T)}$ ; physically the gap is split between making one electron and one hole. What must be true for the Boltzmann simplification in Step 3? ::: The band edge must sit many $k_{B} T$ above $E_{F}$ (non-degenerate), so the " $1 +$ " in Fermi-Dirac is negligible.

Connections

Energy bands - valence band and conduction band — the parent this walkthrough derives.
Fermi level and Fermi-Dirac distribution — supplies $f (E)$ and $E_{F}$ used in Steps 2–3.
Density of states — supplies $g (E)$ , $N_{c}$ , $N_{v}$ used in Steps 4–5.
Intrinsic and extrinsic semiconductors — where the degenerate edge case matters.
Holes as charge carriers — the mirror count of Step 6.