Recombination and generation mechanisms: Exercises

Before we start, one shared picture of the machinery we keep reusing:

Figure — Recombination and generation mechanisms

Everything below is just this diagram — carriers falling from the upper band to the lower one by three routes — turned into numbers.

Level 1 — Recognition

Can you pick the right formula and plug in?

Exercise 1.1 — Net rate vs. gross rate

A silicon sample at equilibrium has recombination rate $R = 5 \times 1 0^{20} cm^{- 3} s^{- 1}$ and generation rate $G = 5 \times 1 0^{20} cm^{- 3} s^{- 1}$ . What is the net recombination rate $U$ , and is the sample above, below, or at equilibrium?

Recall Solution 1.1

The net rate is defined as $U = R - G$ . $U = 5 \times 1 0^{20} - 5 \times 1 0^{20} = 0 cm^{- 3} s^{- 1}$ $U = 0$ means the sample is exactly at equilibrium. Notice $R$ and $G$ are both huge and nonzero — there is a busy traffic of pairs forming and dying — but they cancel. The only thing we can measure is the difference $U$ . See Law of Mass Action: at equilibrium $n p = n_{i}^{2}$ , which is precisely the condition $U = 0$ .

Exercise 1.2 — Radiative lifetime

An n-type direct-gap semiconductor has $B = 1.0 \times 1 0^{- 10} cm^{3} / s$ and $n_{0} = 5 \times 1 0^{16} cm^{- 3}$ . Find the low-injection minority-carrier lifetime $τ$ .

Recall Solution 1.2

Low-injection radiative lifetime in n-type material is $τ = \frac{1}{B n _{0}}$ (the "majority concentration" is $n_{0}$ ). $τ = \frac{1}{( 1.0 \times 1 0 ^{- 10} ) ( 5 \times 1 0 ^{16} )} = \frac{1}{5 \times 1 0 ^{6}} = 2.0 \times 1 0^{- 7} s = 200 ns$ Why $n_{0}$ and not $p_{0}$ ? A hole (the minority carrier) is surrounded by a sea of electrons; its chance of meeting one per second is set by how many electrons there are — the majority count. That is why lifetime shrinks as doping rises.

Exercise 1.3 — Which mechanism?

Match each situation to its dominant recombination channel (Radiative / SRH / Auger): (a) a GaAs LED at moderate current, (b) a dirty silicon wafer with many defects, (c) a silicon region doped to $1 0^{20} cm^{- 3}$ .

Recall Solution 1.3

(a) Radiative — GaAs is direct-gap, so band-to-band photon emission is fast and efficient (that is the whole point of an LED; see LEDs and Lasers). (b) SRH — defects create mid-gap trap levels, the "stepping stones" that dominate in impure/indirect material. (c) Auger — very heavy doping means carriers are extremely dense, so the three-carrier Auger process ( $\propto n_{0}^{2}$ ) takes over.

Level 2 — Application

Can you run the standard calculations end-to-end?

Exercise 2.1 — Exponential decay of excess carriers

$Δ p (0) = 4 \times 1 0^{15} cm^{- 3}$ is injected into a bar with lifetime $τ = 3 μ s$ , then the source is switched off. Find $Δ p$ after $t = 9 μ s$ , and the time for $Δ p$ to fall to one-tenth of its initial value.

Recall Solution 2.1

With no field and no external generation, the Continuity Equation collapses to $\frac{d Δ p}{d t} = - \frac{Δ p}{τ}$ , whose solution is the pure exponential $Δ p (t) = Δ p (0) e^{- t / τ}$ . $Δ p (9 μ s) = 4 \times 1 0^{15} e^{- 9/3} = 4 \times 1 0^{15} e^{- 3} = 4 \times 1 0^{15} (0.0498) = 1.99 \times 1 0^{14} cm^{- 3}$ For one-tenth: set $e^{- t / τ} = 0.1 \Rightarrow t = τ ln 10 = 3 μ s \times 2.303 = 6.91 μ s$ . Why an exponential? The rate of loss is proportional to how much excess is left — the more you have, the faster you lose it. That "rate ∝ amount" rule is the defining property of exponential decay.

Exercise 2.2 — SRH in low injection

An n-type Si sample ( $n_{0} = 1 0^{16} cm^{- 3}$ ) has traps with $σ_{p} = 1 0^{- 15} cm^{2}$ , $v_{t h} = 1 0^{7} cm/s$ , $N_{t} = 1 0^{13} cm^{- 3}$ . Find the minority-hole SRH lifetime $τ_{p}$ .

Recall Solution 2.2

In low-injection n-type, the SRH formula reduces (because $n \approx n_{0}$ dominates the denominator) to $U \approx Δ p / τ_{p}$ with $τ_{p} = \frac{1}{σ _{p} v _{t h} N _{t}}$ $τ_{p} = \frac{1}{( 1 0 ^{- 15} ) ( 1 0 ^{7} ) ( 1 0 ^{13} )} = \frac{1}{1 0 ^{5}} = 1.0 \times 1 0^{- 5} s = 10 μ s$ Why $τ_{p} \propto 1/ N_{t}$ ? Each trap is an independent catch site. Double the traps, double the catch rate, halve the lifetime — the cleaner your crystal, the longer carriers survive.

Exercise 2.3 — Auger lifetime

Heavily doped n-type Si: $C_{n} = 2.8 \times 1 0^{- 31} cm^{6} / s$ , $n_{0} = 1 0^{19} cm^{- 3}$ . Find the Auger lifetime.

Recall Solution 2.3

In n-type low injection the Auger channel gives $τ_{A ug er} = \frac{1}{C _{n} n _{0}^{2}}$ . $τ_{A ug er} = \frac{1}{( 2.8 \times 1 0 ^{- 31} ) ( 1 0 ^{19} ) ^{2}} = \frac{1}{( 2.8 \times 1 0 ^{- 31} ) ( 1 0 ^{38} )} = \frac{1}{2.8 \times 1 0 ^{7}} = 3.57 \times 1 0^{- 8} s \approx 36 ns$ Why the square of $n_{0}$ ? Auger needs three particles at once: a recombining electron–hole pair plus a third carrier to take the energy. The third carrier's abundance scales with doping, and the pair's meeting also scales with electron density — hence $n_{0}^{2}$ .

Level 3 — Analysis

Can you read the structure of a formula and predict behaviour?

Exercise 3.1 — Why mid-gap traps kill lifetime

For the SRH denominator in high injection ( $n = p = Δ n$ ), $D = τ_{p} (Δ n + n_{1}) + τ_{n} (Δ n + p_{1}), n_{1} p_{1} = n_{i}^{2},$ with $τ_{n} = τ_{p} = τ_{0}$ . Show $D$ is minimised (so $U$ maximised) when $n_{1} = p_{1} = n_{i}$ , i.e. the trap sits at mid-gap $E_{t} = E_{i}$ .

Recall Solution 3.1

With $τ_{n} = τ_{p} = τ_{0}$ , drop the constant $τ_{0} \cdot 2Δ n$ part; the tunable piece is $τ_{0} (n_{1} + p_{1})$ . We must minimise $n_{1} + p_{1}$ subject to the fixed product $n_{1} p_{1} = n_{i}^{2}$ . By the AM–GM inequality, for a fixed product the sum is smallest when the two terms are equal: $n_{1} + p_{1} \geq 2 n_{1} p_{1} = 2 n_{i}^{2} = 2 n_{i},$ with equality iff $n_{1} = p_{1} = n_{i}$ . Since $n_{1} = n_{i} e^{(E_{t} - E_{i}) / k T}$ and $p_{1} = n_{i} e^{(E_{i} - E_{t}) / k T}$ , equality means $E_{t} - E_{i} = 0$ , i.e. $E_{t} = E_{i}$ (mid-gap). Physical meaning: a mid-gap trap is equally good at grabbing electrons and holes, so once it catches one carrier it very likely catches the opposite one before re-emitting — the event completes. An off-centre trap catches one type but re-emits it, wasting the event.

Exercise 3.2 — The $n p - n_{i}^{2}$ driver

A silicon sample is illuminated so that $n = 2 \times 1 0^{16}$ , $p = 1 \times 1 0^{16} cm^{- 3}$ , with $n_{i} = 1.5 \times 1 0^{10} cm^{- 3}$ . Is the net process recombination or generation, and by roughly what factor does $n p$ exceed $n_{i}^{2}$ ?

Recall Solution 3.2

Every net rate ( $U_{r a d}$ , $U_{S R H}$ , $U_{A ug er}$ ) carries the same numerator $n p - n_{i}^{2}$ . $n p = (2 \times 1 0^{16}) (1 \times 1 0^{16}) = 2 \times 1 0^{32} cm^{- 6}, n_{i}^{2} = (1.5 \times 1 0^{10})^{2} = 2.25 \times 1 0^{20} cm^{- 6}$ Since $n p ≫ n_{i}^{2}$ , $n p - n_{i}^{2} > 0$ ⇒ net recombination (the crystal is trying to shed excess pairs). The ratio: $\frac{n p}{n _{i}^{2}} = \frac{2 \times 1 0 ^{32}}{2.25 \times 1 0 ^{20}} = 8.9 \times 1 0^{11}$ So $n p$ exceeds equilibrium by nearly twelve orders of magnitude — a massively driven, recombining system.

Level 4 — Synthesis

Can you combine multiple channels and reason about limits?

Exercise 4.1 — Parallel lifetimes

A material has $τ_{r a d} = 100 ns$ , $τ_{S R H} = 20 ns$ , $τ_{A ug er} = 500 ns$ . Find the total effective lifetime, and name the dominant channel.

Recall Solution 4.1

Channels act in parallel — each independently removes carriers, so rates add: $\frac{1}{τ} = \frac{1}{τ _{r a d}} + \frac{1}{τ _{S R H}} + \frac{1}{τ _{A ug er}} = \frac{1}{100} + \frac{1}{20} + \frac{1}{500} ns^{- 1}$ $= 0.010 + 0.050 + 0.002 = 0.062 ns^{- 1} \Rightarrow τ = \frac{1}{0.062} = 16.1 ns$ The total is smaller than any single lifetime, and dominated by the fastest channel — SRH (20 ns), whose reciprocal $0.050$ is the biggest term. Why reciprocals add? Think of three drains in a bathtub: each drains water at its own rate, and their flow rates add. The tub empties in the reciprocal of the summed rates — always faster than any drain alone.

Exercise 4.2 — Crossover doping between SRH and Auger

In n-type Si, SRH gives constant $τ_{S R H} = 1 μ s$ while Auger gives $τ_{A ug er} = 1/ (C_{n} n_{0}^{2})$ with $C_{n} = 2.8 \times 1 0^{- 31} cm^{6} / s$ . At what doping $n_{0}$ does Auger start to dominate (i.e. $τ_{A ug er} = τ_{S R H}$ )?

Recall Solution 4.2

Set the two lifetimes equal: $\frac{1}{C _{n} n _{0}^{2}} = τ_{S R H} \Rightarrow n_{0}^{2} = \frac{1}{C _{n} τ _{S R H}} = \frac{1}{( 2.8 \times 1 0 ^{- 31} ) ( 1 0 ^{- 6} )} = \frac{1}{2.8 \times 1 0 ^{- 37}} = 3.57 \times 1 0^{36}$ $n_{0} = 3.57 \times 1 0^{36} = 1.89 \times 1 0^{18} cm^{- 3}$ Above roughly $2 \times 1 0^{18} cm^{- 3}$ , Auger wins because it grows like $n_{0}^{2}$ while SRH stays flat. This is exactly why heavily doped emitters in Solar Cells and diodes suffer Auger-limited lifetimes.

Level 5 — Mastery

Can you build the answer from equilibrium principles with no formula handed to you?

Exercise 5.1 — Derive $τ$ for high-injection radiative recombination

Start from $U_{r a d} = B (n p - n_{i}^{2})$ . In high injection ( $Δ n = Δ p ≫ n_{0}, p_{0}$ ) in an intrinsic-like slab, find how the effective lifetime $τ_{e f f} \equiv Δ n / U$ depends on $Δ n$ . Contrast with the low-injection constant lifetime.

Recall Solution 5.1

Write $n = n_{0} + Δ n$ , $p = p_{0} + Δ n$ . Then $n p - n_{i}^{2} = (n_{0} + p_{0}) Δ n + Δ n^{2} .$ High injection: $Δ n ≫ n_{0}, p_{0}$ , so the linear term is negligible and $n p - n_{i}^{2} \approx Δ n^{2}$ . Hence $U_{r a d} \approx B Δ n^{2}, τ_{e f f} = \frac{Δ n}{U _{r a d}} = \frac{Δ n}{B Δ n ^{2}} = \frac{1}{B Δ n} .$ So $τ_{e f f} \propto 1/Δ n$ — the lifetime shrinks as you inject harder. Contrast (low injection, n-type): $n p - n_{i}^{2} \approx n_{0} Δ n$ , giving $U \approx B n_{0} Δ n$ and $τ = 1/ (B n_{0})$ , a constant. The switch from constant $τ$ to $τ \propto 1/Δ n$ is the fingerprint of entering high injection — a key check in Minority Carrier Diffusion and laser threshold analysis.

Exercise 5.2 — Full net rate from balance, then a numeric lifetime

An n-type GaAs LED region has $B = 7.2 \times 1 0^{- 10} cm^{3} / s$ , $n_{0} = 2 \times 1 0^{18} cm^{- 3}$ , and a competing SRH lifetime $τ_{S R H} = 50 ns$ . (a) Find the radiative lifetime. (b) Combine with SRH to get the total lifetime. (c) Compute the internal radiative efficiency $η = τ / τ_{r a d}$ (fraction of recombinations that emit a photon).

Recall Solution 5.2

(a) Radiative low-injection lifetime: $τ_{r a d} = \frac{1}{B n _{0}} = \frac{1}{( 7.2 \times 1 0 ^{- 10} ) ( 2 \times 1 0 ^{18} )} = \frac{1}{1.44 \times 1 0 ^{9}} = 6.94 \times 1 0^{- 10} s = 0.694 ns .$ (b) Parallel combination: $\frac{1}{τ} = \frac{1}{0.694 ns} + \frac{1}{50 ns} = 1.440 + 0.020 = 1.460 ns^{- 1} \Rightarrow τ = 0.685 ns .$ (c) Radiative efficiency: $η = \frac{τ}{τ _{r a d}} = \frac{1/ τ _{r a d}}{1/ τ} = \frac{1.440}{1.460} = 0.986 = 98.6%.$ Interpretation: because $τ_{r a d}$ (0.69 ns) is far shorter than $τ_{S R H}$ (50 ns), almost every recombination goes down the photon path. That is precisely why heavily doped, high-quality direct-gap GaAs makes bright, efficient LEDs (see LEDs and Lasers). If instead SRH were the fast channel, $η$ would collapse — the fate of a silicon LED.

Wrap-up recall

Recall One-line answers

Driver of every net rate ::: $n p - n_{i}^{2}$ Low-injection n-type radiative lifetime ::: $τ = 1/ (B n_{0})$ Worst trap position for lifetime ::: mid-gap, $E_{t} = E_{i}$ How channels combine ::: $1/ τ = \sum 1/ τ_{i}$ (rates add, in parallel) High-injection radiative lifetime scaling ::: $τ_{e f f} \propto 1/Δ n$ Auger lifetime scaling with doping ::: $τ_{A ug er} \propto 1/ n_{0}^{2}$