Electron-hole pair generation: Exercises — Hardware deep dive

Before we start, we agree on exactly one toolbox of numbers, so no symbol appears un-earned.

Definition The constants we will use (nothing more)

$E_{g}$ = band gap = smallest energy (in electron-volts, eV) an electron must gain to jump from bound (valence band) to free (conduction band). Si: $1.12$ , Ge: $0.66$ , GaAs: $1.42$ eV. See Energy Bands & Band Gap.
$k$ = Boltzmann constant $= 8.617 \times 1 0^{- 5} eV/K$ . It converts temperature into a typical thermal energy: at temperature $T$ (in kelvin), a jiggling atom carries roughly $k T$ of energy.
$T$ = absolute temperature in kelvin (K). Room temperature $\approx 300$ K.
$h c = 1240 eV\cdotnm$ . This single combined constant lets us turn a photon's wavelength $λ$ (in nanometres) into its energy via $E = \frac{h c}{λ}$ . Why bundle $h$ and $c$ ? Because we never need them apart here — one number, one job.
$n_{i}$ = intrinsic carrier concentration = number of free electrons (or holes) per cm³ in a pure crystal at temperature $T$ . Formula: $n_{i} = N_{c} N_{v} e^{- E_{g} / (2 k T)}$ .

Look at the map of the ladder before climbing:

Level 1 — Recognition

Recall Solution L1.1

Rule: an EHP forms only when the supplied energy is at least $E_{g}$ . For Si, $E_{g} = 1.12$ eV.

(a) $0.5 eV < 1.12 eV$ → no. Too weak to lift an electron across the gap.
(b) $2.0 eV \geq 1.12 eV$ → yes. Plenty of energy; the extra $0.88$ eV becomes heat.
(c) Cooling removes thermal energy → no (in fact cooling reduces the number of pairs). Generation needs energy in.

Recall Solution L1.2

False. A hole is the absence of an electron in the valence band — a bookkeeping label for the collective shuffle of many bound electrons; it merely behaves like a $+ q$ carrier.

WHAT: find $λ$ at the threshold where photon energy exactly equals the gap. WHY: longer wavelength ⇒ lower energy ( $E = h c / λ$ , so bigger $λ$ = smaller $E$ ). The longest usable wavelength is the one whose energy has just dropped to $E_{g}$ . $λ_{ma x} = \frac{h c}{E _{g}} = \frac{1240 eV\cdotnm}{0.66 eV} \approx 1879 nm$ So Ge absorbs out to ~1.88 µm — deeper into the infrared than Si (1107 nm), because its smaller gap needs less energy.

Recall Solution L2.2

$E = \frac{h c}{λ} = \frac{1240}{550} \approx 2.25 eV$ $2.25 eV \geq 1.42 eV$ → yes, green light generates EHPs in GaAs.

Level 3 — Analysis

Recall Solution L3.1

$λ_{ma x}^{GaAs} = \frac{1240}{1.42} \approx 873 nm$ Ranking (using $λ_{ma x} = h c / E_{g}$ ):

GaAs: $E_{g} = 1.42$ eV → $873$ nm
Si: $E_{g} = 1.12$ eV → $1107$ nm
Ge: $E_{g} = 0.66$ eV → $1879$ nm

Ge reaches deepest (1879 nm). Reason: smallest gap ⇒ smallest threshold energy ⇒ even low-energy (long-wavelength) photons clear the bar. This is the same inverse relation as L2, now used to order three materials.

Recall Solution L3.2

Look at the band diagram:

The Fermi level $E_{F}$ (the energy where a state is 50% likely occupied — see Fermi-Dirac Distribution & Fermi Level) sits near the middle of the gap in an intrinsic crystal. An electron only has to climb from $E_{F}$ up to the conduction-band edge — a distance of $E_{g} /2$ , not the full $E_{g}$ . Boltzmann occupation $\propto e^{- Δ E / k T}$ with $Δ E = E_{g} /2$ gives $e^{- E_{g} / (2 k T)}$ . The full $e^{- E_{g} / k T}$ belongs to the product $n p = n_{i}^{2}$ : multiply two halves and the exponents add back to a whole gap. That is why $n_{i} = N_{c} N_{v} e^{- E_{g} / k T} = N_{c} N_{v} e^{- E_{g} / (2 k T)}$ .

Level 4 — Synthesis

Recall Solution L4.1

WHAT: take a ratio so $N_{c} N_{v}$ cancels. $\frac{n _{i} ( 400 )}{n _{i} ( 300 )} = exp [- \frac{E _{g}}{2 k} (\frac{1}{400} - \frac{1}{300})]$ First, $\frac{E _{g}}{k} = \frac{1.12}{8.617 \times 1 0 ^{- 5}} \approx 12998 K$ . $\frac{1}{400} - \frac{1}{300} = \frac{3 - 4}{1200} = - \frac{1}{1200}$ $exponent = - \frac{12998}{2} \cdot (- \frac{1}{1200}) = \frac{12998}{2400} \approx 5.416$ $\frac{n _{i} ( 400 )}{n _{i} ( 300 )} = e^{5.416} \approx 225$ So a modest $100$ K rise multiplies the intrinsic carriers by ~225×. This is why raw semiconductors are so temperature-sensitive, motivating Doping — Donors & Acceptors.

Recall Solution L4.2

Tool: mass-action law $n p = n_{i}^{2}$ , valid in equilibrium even after doping (generation always makes pairs, so the product stays pinned). $p = \frac{n _{i}^{2}}{n} = \frac{( 1.5 \times 1 0 ^{10} ) ^{2}}{1 0 ^{16}} = \frac{2.25 \times 1 0 ^{20}}{1 0 ^{16}} = 2.25 \times 1 0^{4} cm^{- 3}$ Interpretation: adding electrons pushes holes down by a factor of ~ $1 0^{12}$ . Holes are now the tiny minority carrier. See Intrinsic vs Extrinsic Semiconductors.

Level 5 — Mastery

Recall Solution L5.1

WHY this formula: at equilibrium nothing accumulates, so pairs are destroyed (recombined) exactly as fast as they are born (generated): $G = R = r n p = r n_{i}^{2}$ . See Recombination & Carrier Lifetime. $G_{t h} = r n_{i}^{2} = (2 \times 1 0^{- 10}) \times (1.5 \times 1 0^{10})^{2}$ $= (2 \times 1 0^{- 10}) \times (2.25 \times 1 0^{20}) = 4.5 \times 1 0^{10} pairs cm^{- 3} s^{- 1}$

Recall Solution L5.2

Step 1 — photon energy of the signal: $E = \frac{h c}{λ} = \frac{1240}{1310} \approx 0.947 eV$ Step 2 — detection rule: the material generates an EHP only if $E \geq E_{g}$ , i.e. only if $E_{g} \leq 0.947$ eV. Step 3 — check each:

Ge: $E_{g} = 0.66 \leq 0.947$ → detects ✅ (also matches its long 1879 nm cutoff).
Si: $E_{g} = 1.12 > 0.947$ → transparent, no pair, cutoff 1107 nm is shorter than 1310 nm ❌.
GaAs: $E_{g} = 1.42 > 0.947$ → transparent ❌.

Answer: only Ge. This is exactly why Ge / InGaAs (not Si) is used for 1310/1550 nm fibre detectors — the signal photon must clear the gap. Ties together L1 (threshold), L2 (energy from $λ$ ), L3 (cutoff ordering).

Recall Self-test checklist (hide & recite)

Threshold to make an EHP? ::: energy $\geq E_{g}$ .
Photon energy from wavelength? ::: $E = h c / λ$ , $h c = 1240$ eV·nm.
Cutoff wavelength? ::: $λ_{ma x} = h c / E_{g}$ (inverse to gap).
Why exponent $E_{g} /2 k T$ ? ::: Fermi level mid-gap; climb half the gap (= square-root of $e^{- E_{g} / k T}$ ).
Mass-action law? ::: $n p = n_{i}^{2}$ , fixed at given $T$ .
Equilibrium generation? ::: $G_{t h} = r n_{i}^{2}$ .

Connections

Energy Bands & Band Gap
Intrinsic vs Extrinsic Semiconductors
Doping — Donors & Acceptors
Recombination & Carrier Lifetime
Fermi-Dirac Distribution & Fermi Level
Photodiodes & Solar Cells
Drift & Diffusion Currents

Exercises — Electron-hole pair generation