Electron-hole pair generation: Worked examples — Hardware deep dive

This page is the drill floor for Electron-hole pair generation. The parent note built the ideas; here we hit every kind of number the topic can throw at you, one worked example per case. Before we compute anything, let us agree on the symbols so nobody is lost on line one.

Definition The symbols we will use (each earned before use)

$E_{g}$ = band gap energy, the energy a bound electron must gain to break free. Measured in eV (electron-volts). One eV is the energy an electron gains crossing 1 volt — a tiny, atom-sized unit.
$n_{i}$ = intrinsic carrier concentration = how many free electrons per cubic centimetre ( $cm^{- 3}$ ) exist in a pure semiconductor at temperature $T$ . Since pairs are born together, this is also the number of holes. (We use $cm^{- 3}$ everywhere on this page — one consistent volume unit.)
$n$ = electrons per $cm^{- 3}$ , $p$ = holes per $cm^{- 3}$ .
$k$ = Boltzmann's constant $= 8.617 \times 1 0^{- 5} eV/K$ — it converts temperature into energy ( $k T$ is the typical thermal jiggle energy at temperature $T$ ).
$T$ = absolute temperature in kelvin (K). $0 K$ = absolute zero.
$h ν$ = energy of one photon (a particle of light). $ν$ = the light's frequency. On this page we never need $h$ or the light-speed $c$ separately — they always arrive glued together as the single measured constant $h c = 1240 eV\cdotnm$ , so a photon's energy is simply $E_{photon} = h c / λ$ .
$λ$ = wavelength of light (in nm). Longer $λ$ = lower energy per photon.
$N_{c}, N_{v}$ = effective densities of states in the conduction / valence band — roughly "how many parking spots" each band offers (also in $cm^{- 3}$ ).

The two engines behind every number on this page are:

$n_{i} = N_{c} N_{v} e^{- E_{g} / (2 k T)} and E_{photon} = \frac{h c}{λ}$

Everything below is just these two, viewed from every angle.

The scenario matrix

Every problem this topic asks lives in one of the ten cells below. The figure that follows is a two-column grid, three-to-four boxes tall, drawn on the warm cream background:

The left column, headed "OPTICAL (convert & compare)" in burnt orange, holds the light problems (cells A, B, C, I). Each box shows the cell letter, its one-line job, and the example number.
The right column, headed "THERMAL / CARRIER (exponential)" in deep teal, holds the temperature and carrier problems (cells D, E, F, G).
The two "nothing happens" degenerate cells are outlined in plum: cell B (photon below threshold → zero output) and cell F ( $T \to 0$ → perfect insulator). The exam-trap cell J is also plum.
The bottom row spans both columns with the material comparison (H) and the exam trap (J).

There are no numeric axes — it is a labelled map, not a plot. Read it top-to-bottom, left column first: every box is a kind of question, and the example named inside it is the drill that clears it.

Read the grid this way: every arrow of the topic either converts-and-compares a photon (left column) or rides the exponential of temperature (right column). The plum boxes are the "nothing happens" limits and the classic trap. If you can clear all ten boxes, you have seen every scenario.

Below, the same map reappears as text so you can jump to any cell:

A Optical, photon above threshold — does it generate? → Ex 1
B Optical, photon below threshold (degenerate zero output) → Ex 2
C Cutoff wavelength (exact boundary $h ν = E_{g}$ ) → Ex 3
D Thermal $n_{i}$ ratio across two temperatures → Ex 4
E Absolute $n_{i}$ with $N_{c} N_{v}$ (full formula, units) → Ex 5
F Limiting input $T \to 0$ (degenerate perfect insulator) → Ex 6
G Mass-action after doping — donor and acceptor (sign of the change) → Ex 7
H Material comparison (Si vs Ge, different $E_{g}$ ) → Ex 8
I Real-world word problem (IR remote / photodiode) → Ex 9
J Exam twist ( $e^{- E_{g} / k T}$ vs $e^{- E_{g} /2 k T}$ trap) → Ex 10

Cell A — Optical, above threshold

Worked example Ex 1 — Blue light on silicon

Statement: Blue light of wavelength $λ = 450 nm$ hits silicon ( $E_{g} = 1.12 eV$ ). Does it generate an electron-hole pair? By how much does its energy exceed the gap?

Forecast: Blue is high-energy visible light, near the top of what the eye sees. Visible red is about $2 eV$ , and blue is bluer (more energetic) than red, so a good order-of-magnitude guess is "about $2.5$ – $3 eV$ ." Since $E_{g} = 1.12 eV$ , expect a comfortable surplus of roughly $1$ – $2 eV$ . Let us see how close that estimate lands.

Compute the photon energy. $E_{photon} = \frac{h c}{λ} = \frac{1240 eV\cdotnm}{450 nm} = 2.756 eV$ . Why this step? A photon either carries enough energy to lift an electron over $E_{g}$ or it doesn't — so the FIRST thing we always compute is its energy in the same unit (eV) as the gap.
Compare to the gap. $2.756 eV \geq 1.12 eV$ → yes, a pair is generated. Why this step? The generation condition is $h ν \geq E_{g}$ . This is a simple inequality — that's why we deliberately put both in eV.
Surplus energy $= 2.756 - 1.12 = 1.636 eV$ . This extra becomes kinetic energy of the electron and is quickly lost as heat (thermalisation).

Verify: Our forecast said " $2.5$ – $3 eV$ photon, $1$ – $2 eV$ surplus" — the exact answers $2.756$ and $1.636 eV$ sit right inside both guesses, so our order-of-magnitude reasoning was sound. Units: eV·nm ÷ nm = eV ✓. Blue (450 nm) is well below Si's 1107 nm cutoff — 450 < 1107, consistent with "generates." ✓

Cell B — Optical, below threshold (zero output)

Worked example Ex 2 — Far-infrared on silicon (degenerate case)

Statement: Far-infrared light of $λ = 2000 nm$ hits silicon. Does it generate a pair?

Forecast: 2000 nm is almost double Si's known 1107 nm cutoff, so the photon should carry roughly half the energy of a cutoff photon — around $0.6 eV$ , well under $1.12 eV$ . Guess: no pair.

Photon energy: $E_{photon} = \frac{1240}{2000} = 0.62 eV$ . Why this step? Same first move — get the photon's energy into eV.
Compare: $0.62 eV < 1.12 eV$ → the arrow does not reach over the teal step. No pair generated. Silicon is transparent to this light (it passes straight through). Why this step? This is the degenerate "zero output" cell: below threshold, the outcome is literally nothing, no matter how bright the light. Brightness = number of photons, not energy per photon. A billion too-weak photons still generate zero pairs.

Verify: Forecast said "~ $0.6 eV$ , below" — computed $0.62 eV$ , spot on. $2000 nm > 1107 nm$ cutoff → below threshold → no generation. ✓ This is exactly why night-vision (IR) cameras need low- $E_{g}$ materials, not silicon.

Cell C — Exact cutoff (the limiting boundary)

Worked example Ex 3 — Cutoff wavelength of germanium

Statement: What is the longest wavelength that still frees an electron in germanium, $E_{g} = 0.66 eV$ ?

Forecast: Ge's gap ( $0.66 eV$ ) is a bit over half of Si's ( $1.12 eV$ ), and cutoff is inversely proportional to gap, so Ge's cutoff should be a bit under double Si's 1107 nm — roughly $1700$ – $1900 nm$ .

Set the boundary condition $h ν = E_{g}$ , i.e. $\frac{h c}{λ _{m a x}} = E_{g}$ . Why this step? The cutoff is the exact edge — the photon energy equals the gap with nothing to spare. Longer $λ$ = lower energy = fails; shorter = succeeds. So the boundary is an equality, not an inequality.
Solve for $λ_{m a x}$ : $λ_{m a x} = \frac{h c}{E _{g}} = \frac{1240}{0.66} \approx 1879 nm$ . Why this step? Rearranging isolates the unknown wavelength. This is why the smaller the gap, the larger the cutoff — $λ_{m a x}$ is inversely proportional to $E_{g}$ .

Verify: Forecast said " $1700$ – $1900 nm$ " — computed $1879 nm$ , inside the range. $1879 nm > 1107 nm$ (Si cutoff). Smaller gap → longer cutoff. ✓ Ge indeed detects deeper into the infrared than Si.

Cell D — Thermal $n_{i}$ ratio across temperatures

Worked example Ex 4 — How much does

$n_{i}$ rise from 300 K to 400 K in silicon? Statement: For Si ( $E_{g} = 1.12 eV$ ), find $\frac{n _{i} ( 400 )}{n _{i} ( 300 )}$ , ignoring the slow change in $N_{c} N_{v}$ .

Forecast: Carriers are exponentially sensitive to $T$ . The parent note showed a $\times 50000$ jump for doubling to $600$ K; our smaller $100$ K step should give a smaller-but-still-large factor — guess "a few hundred."

Take the ratio of the two $n_{i}$ formulas. The $N_{c} N_{v}$ cancels (we're told to ignore its change): $\frac{n _{i} ( 400 )}{n _{i} ( 300 )} = exp [- \frac{E _{g}}{2 k} (\frac{1}{400} - \frac{1}{300})]$ Why this step? Ratios kill the messy prefactor and leave the exponential, which carries all the temperature drama. That's the whole reason we take a ratio instead of two separate absolute values.
First form the coefficient $\frac{E _{g}}{2 k}$ — the $\frac{1}{2}$ IS applied here. Compute $\frac{E _{g}}{k} = \frac{1.12}{8.617 \times 1 0 ^{- 5}} \approx 12998 K$ , then halve it: $\frac{E _{g}}{2 k} = \frac{12998}{2} \approx 6499 K$ . Why this step? The exponent in $n_{i}$ always carries the factor $\frac{1}{2}$ (Fermi level mid-gap). We make it explicit before plugging in numbers so there is no doubt the half has been applied — $6499$ K is $E_{g} /2 k$ , not $E_{g} / k$ .
Plug the coefficient in: $= exp [- 6499 K \times (\frac{1}{400} - \frac{1}{300}) K^{- 1}] = exp [- 6499 \times (0.0025 - 0.003333)] = exp [- 6499 \times (- 0.000833)] = exp (5.42)$ Why this step? The inner bracket is negative (bigger $T$ → smaller $1/ T$ ), and multiplied by the leading minus becomes a positive exponent → growth.
Evaluate: $e^{5.42} \approx 225$ .

Verify: Forecast said "a few hundred" — computed $225 \times$ , right in that band. A mere $100$ K rise multiplies carriers ~225×, consistent with the parent note's $50000 \times$ at $600$ K. ✓

Cell E — Absolute $n_{i}$ with full formula

Worked example Ex 5 — Compute

$n_{i}$ for silicon at 300 K Statement: Given $N_{c} = 2.8 \times 1 0^{19} cm^{- 3}$ , $N_{v} = 1.04 \times 1 0^{19} cm^{- 3}$ , $E_{g} = 1.12 eV$ , $T = 300 K$ , find $n_{i}$ .

Forecast: The textbook value for Si at room temperature is famously about $1 0^{10} cm^{- 3}$ . The prefactor is around $1 0^{19}$ and the exponential kills ~9 orders of magnitude, so guess: near $1 0^{10}$ .

Compute the exponent. $\frac{E _{g}}{2 k T} = \frac{1.12}{2 \times 8.617 \times 1 0 ^{- 5} \times 300} = \frac{1.12}{0.0517} \approx 21.66$ . Why this step? The exponential factor dominates the magnitude, so nail it first. $k T$ at room temperature is $\approx 0.0259 eV$ ; the gap is ~43× larger, giving a big exponent.
Exponential: $e^{- 21.66} \approx 3.9 \times 1 0^{- 10}$ . Why this step? This tiny number is why pure silicon barely conducts — only 4 in every 10 billion "attempts" succeed.
Prefactor: $N_{c} N_{v} = 2.8 \times 1 0^{19} \times 1.04 \times 1 0^{19} = 2.912 \times 1 0^{38} \approx 1.706 \times 1 0^{19} cm^{- 3}$ . Why this step? The appears because in a pure material every freed electron makes exactly one hole, so $n = p = n_{i}$ and therefore $n_{i}^{2} = n p = N_{c} N_{v} e^{- E_{g} / k T}$ ; taking the square root of both sides gives $n_{i} = N_{c} N_{v} e^{- E_{g} /2 k T}$ — the very formula we are evaluating.
Multiply: $n_{i} \approx 1.706 \times 1 0^{19} \times 3.9 \times 1 0^{- 10} \approx 6.7 \times 1 0^{9} cm^{- 3}$ .

Verify: Forecast said "near $1 0^{10}$ " — computed $6.7 \times 1 0^{9}$ , same order of magnitude and right on the accepted room-temperature value for Si. ✓ Units: $cm^{- 3}$ × dimensionless = $cm^{- 3}$ . ✓

Cell F — Limiting input $T \to 0$ (perfect insulator)

Worked example Ex 6 — What is

$n_{i}$ as $T \to 0 K$ ? Statement: Show that a pure semiconductor at absolute zero has $n_{i} \to 0$ .

Forecast: The parent says absolute zero = perfect insulator. So $n_{i}$ should collapse to zero.

Look at the exponent as $T \to 0$ : $- \frac{E _{g}}{2 k T}$ . As $T \to 0^{+}$ , the denominator $\to 0^{+}$ , so the fraction $\to + \infty$ , and the exponent $\to - \infty$ . Why this step? The whole temperature behaviour lives in this exponent. We ask: what does it approach at the limit? This is a limiting-value check, the degenerate corner of the matrix.
Exponential of $- \infty$ : $e^{- \infty} = 0$ . Hence $n_{i} = N_{c} N_{v} \times 0 = 0$ . Why this step? No thermal jiggle means no electron ever gets $E_{g}$ of energy, so zero pairs — matching "every bond is locked." The $N_{c} N_{v}$ prefactor is finite, so it cannot rescue a zero exponential.

Verify: Numerically, at $T = 1 K$ the exponent is $- 1.12/ (2 \times 8.617 \times 1 0^{- 5} \times 1) \approx - 6499$ , and $e^{- 6499}$ is indistinguishable from 0. ✓ Confirms the "perfect insulator at absolute zero" claim.

Cell G — Mass-action law under doping (both signs)

Worked example Ex 7 — Donor doping AND acceptor doping: which carrier wins?

Statement: Silicon at 300 K has $n_{i} = 1.0 \times 1 0^{10} cm^{- 3}$ . (a) n-type: we add donors so the electron concentration becomes $n = 1.0 \times 1 0^{16} cm^{- 3}$ . Find $p$ . (b) p-type: in a separate sample we add acceptors so the hole concentration becomes $p = 5.0 \times 1 0^{15} cm^{- 3}$ . Find $n$ . In each case state which carrier is the majority.

Forecast: Mass-action pins the product $n p = n_{i}^{2}$ . In (a) we raised $n$ a million-fold, so $p$ must drop a million-fold → electrons win. In (b) we raise $p$ , so $n$ must drop → holes win. The law is perfectly symmetric between the two.

Apply mass-action law $n p = n_{i}^{2}$ . Why this step? Generation and recombination always work in pairs, which pins the product $n p$ to $n_{i}^{2}$ at fixed temperature — no matter which species we added. So knowing one carrier instantly gives the other.
(a) Donors — solve for holes: $p = \frac{n _{i}^{2}}{n} = \frac{( 1.0 \times 1 0 ^{10} ) ^{2}}{1.0 \times 1 0 ^{16}} = \frac{1.0 \times 1 0 ^{20}}{1.0 \times 1 0 ^{16}} = 1.0 \times 1 0^{4} cm^{- 3}$ . Why this step? Dividing by the raised $n$ shows holes falling from $1 0^{10}$ to $1 0^{4}$ — six orders down. Electrons are the majority carrier (n-type).
(b) Acceptors — solve for electrons: $n = \frac{n _{i}^{2}}{p} = \frac{( 1.0 \times 1 0 ^{10} ) ^{2}}{5.0 \times 1 0 ^{15}} = \frac{1.0 \times 1 0 ^{20}}{5.0 \times 1 0 ^{15}} = 2.0 \times 1 0^{4} cm^{- 3}$ . Why this step? The same law, now dividing by the raised $p$ , shows electrons crushed to $2 \times 1 0^{4}$ . Holes are the majority carrier (p-type). This is the mirror image of (a) — the symmetry the parent note promised.
Interpretation: doping never creates carriers "from nothing without pairs." Adding one species speeds up recombination until the other species is mopped back down to keep $n p = n_{i}^{2}$ .

Verify: (a) $n p = 1 0^{16} \times 1 0^{4} = 1 0^{20} = (1 0^{10})^{2} = n_{i}^{2}$ ✓. (b) $n p = 2 \times 1 0^{4} \times 5 \times 1 0^{15} = 1 0^{20} = n_{i}^{2}$ ✓. Both products land exactly on $n_{i}^{2}$ — mass-action holds for donor and acceptor doping alike.

Cell H — Material comparison (different $E_{g}$ )

Worked example Ex 8 — Ge vs Si: which has more thermal carriers at 300 K?

Statement: Ignoring $N_{c} N_{v}$ differences, find the ratio $\frac{n _{i} ( Ge )}{n _{i} ( Si )}$ at 300 K, with $E_{g} (Ge) = 0.66 eV$ , $E_{g} (Si) = 1.12 eV$ .

Forecast: Ge has a $0.46 eV$ smaller gap; divided by the tiny $2 k T \approx 0.05 eV$ that is an exponent near $9$ , and $e^{9}$ is roughly $8000$ . So guess: a few thousand times more carriers in Ge.

Ratio of the two $n_{i}$ (prefactor cancels): $\frac{n _{i} ( Ge )}{n _{i} ( Si )} = exp [- \frac{E _{g} ( Ge ) - E _{g} ( Si )}{2 k T}] = exp [\frac{E _{g} ( Si ) - E _{g} ( Ge )}{2 k T}]$ Why this step? A ratio isolates the difference in gaps — the only thing that differs. Smaller Ge gap makes the sign flip so the exponent is positive → Ge wins.
Compute the denominator $2 k T = 2 \times 8.617 \times 1 0^{- 5} \times 300 = 0.0517 eV$ . Why this step? We need $2 k T$ in eV so it cancels the eV in the gap difference, leaving a pure (dimensionless) number in the exponent — exponents must be unitless.
Form the exponent $\frac{E _{g} ( Si ) - E _{g} ( Ge )}{2 k T} = \frac{1.12 - 0.66}{0.0517} = \frac{0.46}{0.0517} = 8.90$ . Why this step? Subtract the gaps first ( $0.46 eV$ ), then divide by $2 k T$ — a $0.46 eV$ head-start over such a tiny thermal energy becomes a large exponent, which is the whole reason the answer will be big.
Exponentiate: $\frac{n _{i} ( Ge )}{n _{i} ( Si )} = e^{8.90} \approx 7300$ . Why this step? The final exponentiation turns the exponent into the actual multiplier — Ge has ~7300× more thermal carriers than Si at room temperature.

Verify: Forecast said "a few thousand, ~ $e^{9} \approx 8000$ " — computed $7300$ , right there. Real values: $n_{i} (Ge) \approx 2 \times 1 0^{13}$ , $n_{i} (Si) \approx 1 0^{10}$ , ratio $\approx 2000$ – $10000$ ; our $7300$ sits in range (the residual is the $N_{c} N_{v}$ ratio we ignored). ✓ Ge is leakier — that's why it's more temperature-sensitive.

Cell I — Real-world word problem

Worked example Ex 9 — Will a TV remote's IR LED trip a silicon photodiode?

Statement: A remote emits infrared at $λ = 940 nm$ . Its receiver is a silicon photodiode ( $E_{g} = 1.12 eV$ ). Can the light generate carriers (i.e. be detected)? See Photodiodes & Solar Cells for where this matters.

Forecast: 940 nm is infrared, just past visible red, and comfortably below Si's 1107 nm cutoff. Since $940 < 1107$ , guess: yes, detected, with a small surplus of maybe $0.1$ – $0.3 eV$ .

Translate the word problem into an energy test. "Detected" means "generates an EHP," which means $h ν \geq E_{g}$ . Why this step? Every optical word problem reduces to this one inequality. Naming it removes the fog of the story.
Photon energy: $E_{photon} = \frac{1240}{940} = 1.319 eV$ . Why this step? Convert wavelength to energy in eV to compare with $E_{g}$ directly.
Compare: $1.319 eV \geq 1.12 eV$ → yes, detected, with a surplus of $1.319 - 1.12 = 0.199 eV$ . Why this step? The surplus being positive confirms 940 nm falls inside Si's response window — which is exactly why remote controls and their silicon receivers are engineered around ~940 nm.

Verify: Forecast said "detected, surplus $0.1$ – $0.3 eV$ " — computed $0.199 eV$ , inside the guess. $940 nm < 1107 nm$ cutoff → generates. ✓ (Real IR remotes really do use 940 nm precisely for this reason.)

Cell J — Exam twist (the factor-of-2 trap)

Worked example Ex 10 — Same problem, two exponents — spot the trap

Statement: A student computes the room-temperature suppression factor for Si two ways: (a) $e^{- E_{g} / k T}$ and (b) $e^{- E_{g} /2 k T}$ . Which belongs in $n_{i}$ , and how different are the numbers?

Forecast: Only (b) belongs in $n_{i}$ . The two exponents differ by a factor of 2 ( $\approx 43$ vs $\approx 22$ ), and exponentials magnify that — guess the results differ by roughly nine orders of magnitude.

Compute both exponents at $T = 300$ K, $k T = 0.02585 eV$ :
- (a) $- E_{g} / k T = - 1.12/0.02585 = - 43.33$
- (b) $- E_{g} /2 k T = - 21.66$ Why this step? Laying the numbers side by side exposes the factor-of-2 as a factor-of-huge in the final answer.
Evaluate: (a) $e^{- 43.33} \approx 1.5 \times 1 0^{- 19}$ ; (b) $e^{- 21.66} \approx 3.9 \times 1 0^{- 10}$ . Why this step? The (b) result, times $N_{c} N_{v}$ , gives the correct $\sim 1 0^{10} cm^{- 3}$ (Ex 5). Using (a) would predict $\sim 1$ carrier per $cm^{- 3}$ — absurdly wrong by nine orders of magnitude.
The rule: the product $n p = n_{i}^{2} = N_{c} N_{v} e^{- E_{g} / k T}$ carries the full gap (exponent (a)); the single $n_{i}$ carries half the gap (exponent (b)) because the intrinsic Fermi level sits mid-gap and each carrier climbs only $E_{g} /2$ .

Verify: Forecast said "~nine orders apart" — the ratio $3.9 \times 1 0^{- 10} /1.5 \times 1 0^{- 19} \approx 2.6 \times 1 0^{9}$ is indeed ~9 orders. Squaring (b): $[e^{- E_{g} /2 k T}]^{2} = e^{- E_{g} / k T} =$ (a) exactly — the halved exponent is the square root, confirming which one belongs where. ✓ See Fermi-Dirac Distribution & Fermi Level for why the Fermi level sits mid-gap.

Recall

Recall Which cell does each move solve?

Optical "does it generate?" ::: Compute $h c / λ$ in eV, test against $E_{g}$ (Cells A, B, I). The exact cutoff wavelength ::: $λ_{m a x} = h c / E_{g}$ (Cell C). Carrier rise with temperature ::: Take the ratio, prefactor cancels, exponent does all the work (Cell D). Holes after donor doping ::: $p = n_{i}^{2} / n$ — holes drop as electrons rise (Cell G-a). Electrons after acceptor doping ::: $n = n_{i}^{2} / p$ — electrons drop as holes rise (Cell G-b). The factor-of-2 rule ::: $n_{i}$ uses $E_{g} /2 k T$ ; the product $n p = n_{i}^{2}$ uses $E_{g} / k T$ (Cell J).

Connections

Parent: EHP generation
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