Donor - acceptor energy levels in the band gap: Worked examples

Before any numbers, meet the symbols that run the whole show.

Definition The band edges

$E_{C}$ , $E_{V}$ , and the gap $E_{g}$ Inside a solid, electrons are only allowed to have certain energies. These allowed energies clump into two blocks with a forbidden "no-go" strip between them:

$E_{V}$ = valence band edge: the energy at the top of the lower, mostly-full block (where bound bonding electrons live). Picture a floor.
$E_{C}$ = conduction band edge: the energy at the bottom of the upper, mostly-empty block (where free, current-carrying electrons live). Picture a ceiling.
$E_{g} = E_{C} - E_{V}$ = band gap: the height of the forbidden strip between floor and ceiling. For silicon $E_{g} = 1.12 eV$ .

An electron sitting on the floor ( $E_{V}$ ) must be lifted a full $E_{g}$ to reach the ceiling ( $E_{C}$ ) and become free — unless a dopant places a stepping-stone level partway up.

The scenario matrix

Cell	Case class	What changes	Example
A	Donor, room temp	electron climbs $E_{C} - E_{D}$	Ex 1
B	Acceptor, room temp	electron climbs $E_{A} - E_{V}$	Ex 2
C	Sign / position check	is a given level donor or acceptor?	Ex 3
D	Deep level (large $Δ E$ )	ionization no longer easy + traps	Ex 4
E	Low temperature (freeze-out limit)	$k T \to$ small, carriers vanish	Ex 5
F	High temperature (intrinsic limit)	gap itself starts to ionize	Ex 6
G	Hydrogen-model estimate (derive the ~meV)	why the level is shallow	Ex 7
H	Word problem (real device choice)	pick the right dopant	Ex 8
I	Exam twist (compensation)	both dopants at once	Ex 9

Each example below is labelled with the cell it fills. Together they cover every row.

Example 1 — Cell A: a donor at room temperature

Forecast: guess before reading — will $45 meV$ beat a $26 meV$ nudge, or lose to it? Jot down "yes/no".

Write the ionization energy. $Δ E = E_{C} - E_{D} = 45 meV$ . Why this step? For a donor the carrier that matters is the loosely-held electron climbing up into the conduction band, so the relevant gap is measured from $E_{D}$ up to $E_{C}$ .
Write the thermal budget. $k T = 8.617 \times 1 0^{- 5} eV/K \times 300 K = 25.9 meV$ . Why this step? We need a yardstick to say whether $45 meV$ is "small". $k T$ is that yardstick.
Form the ratio. $\frac{Δ E}{k T} = \frac{45}{25.9} \approx 1.74$ . Why this step? A pure ratio strips units and tells us how many "nudges" tall the climb is.

What figure s01 shows: two horizontal bands fill the top ( $E_{C}$ , red) and bottom ( $E_{V}$ , blue) of an energy axis. A dashed yellow donor line $E_{D}$ sits just under the red ceiling. A green double arrow between $E_{D}$ and $E_{C}$ is labelled " $45 meV$ ". On the left, a short white double arrow drops down from $E_{C}$ by only $26 meV$ — the $k T$ bracket. The key visual: the yellow gap ( $45$ ) is only a little taller than the white $k T$ bracket ( $26$ ), so the climb is roughly one-and-three-quarter nudges. That "barely taller than one nudge" picture is the whole meaning of shallow.

Figure — Donor - acceptor energy levels in the band gap

Interpret. $1.74$ is small (of order one), not $50$ . So many electrons do make the jump — the level is shallow and effectively ionized at room temperature.

Verify: $25.9 \times 1.74 = 45.1 meV$ ✓ recovers the input. Units: $meV / meV$ = dimensionless ✓.

Example 2 — Cell B: an acceptor at room temperature

Forecast: the number will look identical to Ex 1 — why? Predict which carrier moves.

Write the ionization energy. $Δ E = E_{A} - E_{V} = 45 meV$ . Why this step? An acceptor grabs an electron up from the valence band into the level $E_{A}$ . That electron climbs from $E_{V}$ to $E_{A}$ , leaving a hole behind — so the relevant gap is measured upward from $E_{V}$ .
Ratio. $\frac{45}{25.9} \approx 1.74$ , same yardstick as before. Why this step? Same $k T$ , same climb size.

What figure s02 shows: the same band picture as s01, but now the dashed yellow acceptor line $E_{A}$ sits just above the blue floor $E_{V}$ . A green double arrow of height " $45 meV$ " measures $E_{A} - E_{V}$ . A small white arrow marks an electron hopping up from $E_{V}$ into $E_{A}$ , and a red label near the floor marks the empty seat — the hole — left behind. It is s01 flipped upside-down: the action happens at the floor, and the freed carrier is a hole rather than an electron.

Interpret. Fully ionized at room temperature; each ionized acceptor leaves a mobile hole.

Verify: identical arithmetic to Ex 1, $25.9 \times 1.74 = 45.1$ ✓.

Example 3 — Cell C: donor or acceptor? Read the position

Forecast: near which edge does it sit — top or bottom?

Locate it in the gap. $E_{X} - E_{V} = 0.05 eV$ , while $E_{C} - E_{X} = 1.12 - 0.05 = 1.07 eV$ . Why this step? A level's role is decided by which band edge it hugs. Hugging $E_{V}$ ⇒ acceptor; hugging $E_{C}$ ⇒ donor.
Decide. It sits far from $E_{C}$ ( $1.07 eV$ ) but close to $E_{V}$ ( $0.05 eV$ ) → acceptor. Why this step? Only a level near $E_{V}$ can cheaply pull an electron out of the valence band.
Ionization energy. $E_{A} - E_{V} = 0.05 eV = 50 meV$ .

Verify: $0.05 + 1.07 = 1.12 eV$ = the full gap ✓ — the two distances must add to the gap.

Example 4 — Cell D: a deep level, ionization is hard (and it traps carriers)

Forecast: the gap grew ~12×. Guess: does ionization drop 12×, or far more?

Shallow factor. $e^{- 45/25.9} = e^{- 1.74} \approx 0.176$ . Why this step? This factor is the "fraction that can jump." We built it in the third definition.
Deep factor. $Δ E = 0.55 eV = 550 meV$ , so $e^{- 550/25.9} = e^{- 21.2} \approx 6.0 \times 1 0^{- 10}$ . Why this step? Same formula, bigger climb — but the exponential punishes size multiplicatively.
Ratio of the two. $\frac{0.176}{6.0 \times 1 0 ^{- 10}} \approx 2.9 \times 1 0^{8}$ . Why this step? This is the whole point of the exponential (see the third definition): a 12× larger gap makes the level roughly 300 million times less ionized, not 12× less.

Verify: $e^{- 21.2} / e^{- 1.74} = e^{- 19.46} \approx 3.5 \times 1 0^{- 9}$ , so the inverse is $\approx 2.9 \times 1 0^{8}$ ✓.

Example 5 — Cell E: freeze-out, the low-temperature limit

Forecast: we dropped $T$ by 10×. Guess the fate of the carriers.

New thermal budget. $k T = 8.617 \times 1 0^{- 5} \times 30 = 2.585 \times 1 0^{- 3} eV = 2.585 meV$ . Why this step? Colder crowd, weaker jiggles. The yardstick shrank 10×.
New ratio. $\frac{45}{2.585} \approx 17.4$ . Why this step? Now the climb is ~17 nudges tall — far into the "hard" regime.
Occupation factor. $e^{- 17.4} \approx 2.8 \times 1 0^{- 8}$ : almost nothing jumps. Why this step? The exponential collapses. Electrons fall back onto their donors — this is carrier freeze-out.

What figure s03 shows: a graph of $k T$ (blue curve, in meV) rising with temperature $T$ along the horizontal axis. A dashed yellow horizontal line marks the fixed $45 meV$ donor energy. Two dots sit on the blue curve: a green dot at $300 K$ (where $k T \approx 26 meV$ , comfortably in the "easy" zone) and a red dot at $30 K$ (where $k T \approx 2.6 meV$ , far below the yellow line). The visual message: as you slide left (cool down), the blue $k T$ curve drops under the yellow level, so the nudge can no longer reach it — carriers freeze out.

Verify: ratio $17.4 = 45/2.585$ ✓, and it is exactly $10 \times$ the room-temperature ratio $1.74$ since $k T \propto T$ ✓.

Example 6 — Cell F: high temperature, the intrinsic limit

Forecast: will it be a few hundred K, or thousands?

Set the condition. $k T = E_{g} /4 = 1.12/4 = 0.28 eV$ . Why this step, and why the factor $1/4$ ? We want the temperature at which heat can excite electrons straight across the full gap $E_{g}$ in appreciable numbers. The number that actually controls that is the factor $e^{- E_{g} /2 k T}$ (the intrinsic carrier density carries a half-gap in the exponent, because an electron and its hole share the cost). "Appreciable" means that exponent should be small, of order a few — say $E_{g} /2 k T \approx 2$ , i.e. $k T \approx E_{g} /4$ . So the $1/4$ is not arbitrary: it is $\frac{1}{2}$ (from sharing the gap) times $\frac{1}{2}$ (from "exponent of about 2 is where things get going"). It marks the rough temperature where intrinsic conduction begins to swamp the dopants.
Solve for $T$ . $T = \frac{0.28}{8.617 \times 1 0 ^{- 5}} \approx 3250 K$ . Why this step? Rearranging $k T = 0.28$ . That is far above silicon's melting point (~1687 K), which tells you doping dominates over the whole usable temperature range — a real, reassuring result.

Verify: $8.617 \times 1 0^{- 5} \times 3250 = 0.280 eV$ ✓.

Example 7 — Cell G: the hydrogen model — deriving "why ~tens of meV"

Forecast: hydrogen binds at $13.6 eV$ . Silicon must weaken it hugely. Guess by how much.

Recall the scaling law — and where it comes from. $E_{b} = 13.6 eV \times \frac{m ^{*} / m _{0}}{ε _{r}^{2}}$ . Why this step? Two physical changes turn a real hydrogen atom into a donor-electron-orbiting-a-fixed-+1-core inside silicon:
- The core is screened. In a vacuum the +1 core pulls with the full Coulomb force. Inside silicon the sea of bonding electrons partly cancels that pull — the medium screens it by the dielectric constant $ε_{r}$ . The force weakens by $ε_{r}$ , and because the hydrogen energy goes as (charge $\times$ charge), which now carries two screened factors, the energy scales as $1/ ε_{r}^{2}$ .
- The electron is lighter. Moving through the silicon lattice, the electron responds as if it had a reduced effective mass $m^{*}$ . The hydrogen binding energy is directly proportional to the orbiting mass, so it scales by $m^{*} / m_{0}$ . Both effects shrink the binding. What figure s04 shows: a "before / after" pair — on the left a tight hydrogen orbit (small radius, deep $13.6 eV$ well); on the right a fat, loose orbit around the +1 donor core inside a shaded silicon medium, with a shallow well labelled " $\sim$ tens of meV". The green arrows point from the two shrinking factors ( $\div ε_{r}^{2}$ and $\times m^{*} / m_{0}$ ) to the loosened orbit. The picture makes it visual: screen the charge and lighten the electron, and the orbit balloons while the binding collapses.

Plug in. $E_{b} = 13.6 \times \frac{0.26}{11. 7 ^{2}} = 13.6 \times \frac{0.26}{136.9}$ . Why this step? Just arithmetic in the scaling law.
Evaluate. $= 13.6 \times 1.899 \times 1 0^{- 3} = 0.0258 eV = 25.8 meV$ . Why this step? Same order as the measured $45 meV$ — the crude model explains the shallowness without any fitting. (The factor-2 gap is because real dopants aren't perfect hydrogen atoms.)

Verify: $13.6 \times 0.26/136.89 = 0.02583 eV$ ✓, which is $\sim 0.6 \times$ the real $45 meV$ — right order of magnitude ✓.

Example 8 — Cell H: word problem, pick the dopant

Forecast: name your pick before the numbers.

"Many free electrons" ⇒ donor. Rules out gold as a source (it's a trap). Phosphorus donates electrons. Why this step? Cell C logic: we need a level near $E_{C}$ that releases electrons.
Check phosphorus at 200 K. $k T = 8.617 \times 1 0^{- 5} \times 200 = 17.2 meV$ ; ratio $= 45/17.2 = 2.61$ . Why this step? Verify the shallow level is still nearly ionized when cooled. $2.61$ is still order-one → still mostly ionized.
Check gold at 200 K. ratio $= 550/17.2 = 32.0$ , factor $e^{- 32} \approx 1.3 \times 1 0^{- 14}$ → essentially dead. Why this step? Confirms the deep level fails completely.
Decide: phosphorus. Shallow donor stays ionized across the whole $200$ – $300 K$ range.

Verify: $45/17.2 = 2.62$ ✓ and $550/17.2 = 31.98$ ✓.

Example 9 — Cell I: exam twist — compensation

Forecast: do the two dopings add, subtract, or cancel? Guess the sign of the leftover.

Ionize everything. All donors give up an electron: $N_{D}^{+} = 1 \times 1 0^{16} cm^{- 3}$ . All acceptors grab one: $N_{A}^{-} = 3 \times 1 0^{15} cm^{- 3}$ . Why this step? Both levels are shallow (Ex 1/2 logic), so at room temperature the Boltzmann factor is order-one and essentially every dopant is ionized.
Compensation — the electrons and holes cancel first. Each acceptor's empty seat greedily swallows a donor electron before that electron ever reaches the conduction band. So the net free electrons are the excess of donors over acceptors: $n \approx N_{D} - N_{A} = 1 \times 1 0^{16} - 3 \times 1 0^{15} = 7 \times 1 0^{15} cm^{- 3} .$ Why this step? Charge neutrality: the fixed positive donor ions ( $N_{D}^{+}$ ) must be balanced by fixed negative acceptor ions ( $N_{A}^{-}$ ) plus mobile electrons. Rearranged, the mobile electrons equal the difference of the dopings, not their sum — the dopings partially cancel.
Identify the majority carrier. Since $N_{D} > N_{A}$ , the leftover mobile charge is negative (electrons), so electrons are the majority carrier. Why this step? Whichever dopant wins the subtraction sets the sign of the surviving free charge.
State the type. Majority carriers are electrons ⇒ the sample is n-type. Why this step? "n-type" literally means negative majority carriers (electrons); "p-type" means positive (holes). The winning dopant's sign names the type.

Verify: $1 0^{16} - 3 \times 1 0^{15} = 7 \times 1 0^{15}$ ✓, and $N_{D} > N_{A}$ so the sample is n-type ✓.

Recall Quick self-test

Donor ionization energy is measured from ::: $E_{D}$ up to $E_{C}$ , i.e. $E_{C} - E_{D}$ Room-temperature $k T$ in meV ::: about $25.9 meV$ at $300 K$ Why a deep level barely ionizes ::: the Boltzmann factor $e^{- Δ E / k T}$ collapses exponentially with $Δ E$ What a deep midgap level does to carriers ::: acts as a recombination centre, cutting the carrier lifetime $τ$ When does the $e^{- Δ E / k T}$ shortcut break ::: in degenerate (very heavily doped) material where $E_{F}$ enters a band In a compensated sample with $N_{D} > N_{A}$ , net electrons ::: $N_{D} - N_{A}$ , and the sample is n-type The hydrogen model predicts shallow binding because ::: light effective mass and large dielectric screening shrink $13.6 eV$ to tens of meV

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