Worked examples — Recombination and generation mechanisms

Before we start, one reminder of the master driver you'll see in every single example:

Recall The one quantity behind all three mechanisms

What drives net recombination in radiative, SRH, and Auger alike? ::: the excess $np-n_i^2$

So that every symbol below is earned, let's restate all three mechanism formulas up front — exactly as the parent note built them. Notice how the same driver $np-n_i^2$ sits in every numerator:

Notation used everywhere below (all built in the parent note): $n,p$ = electron and hole concentrations (per ${\text{cm}}^3$); $n_0,p_0$ = their equilibrium values; $n_i$ = intrinsic concentration; $\Delta p$ = excess holes; $\tau$ = minority-carrier lifetime; $U=R-G$ = net recombination rate; $n_1,p_1$ = the SRH reference concentrations defined just above.

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The scenario matrix

Every recombination problem is one cell of this grid. Read it as: pick a mechanism × pick a regime × pick an edge case.

#	Case class	The distinguishing feature	Covered by
A	Net recombination, $U>0$	$np>n_i^2$ (carriers injected)	E1, E2, E4
B	Net generation, $U<0$	$np<n_i^2$ (carriers depleted, e.g. reverse-bias depletion region)	E3
C	Equilibrium / zero input	$np=n_i^2\Rightarrow U=0$ (yet $R,G\ne 0$)	E1 sanity
D	Low injection ($\Delta p\ll n_0$)	$\tau$ = constant material property	E1, E2
E	High injection ($\Delta n\approx \Delta p\gg n_0$)	$\tau$ shifts, becomes nonlinear	E4
F	Trap position sweep ($E_t$ vs $E_i$)	mid-gap = lifetime killer	E5
G	Parallel channels — which $\tau$ wins	$1/\tau =\sum 1/{\tau }_i$, smallest wins	E6
H	Real-world word problem	LED vs solar-cell design choice	E7
I	Exam twist — degenerate/limit	zero trap density, $\Delta p\to 0$, $t\to \infty$	E8

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E1 — Net recombination in low injection (cells A, C, D)

Forecast: guess before computing — is $U$ positive or negative? And is $\Delta p=10^{12}$ "low injection" compared to $n_0=10^{16}$? (Answer: yes, four orders smaller.)

1. Equilibrium hole concentration. By the Law of Mass Action, $p_0=n_i^2/n_0=(10^{10}{)}^2/10^{16}=10^4{\text{cm}}^{-3}$. Why this step? We need $p_0$ to write the disturbed concentrations honestly. Notice $p_0=10^4$ is utterly dwarfed by $\Delta p=10^{12}$ — the minority population is dominated by the injection.

2. Disturbed concentrations. $n=n_0+\Delta p=10^{16}+10^{12}\approx 10^{16}$; $p=p_0+\Delta p=10^4+10^{12}\approx 10^{12}$. Why this step? Recombination needs an electron to meet a hole, so we need both actual populations.

3. Full net rate. $np-n_i^2=(10^{16})(10^{12})-(10^{10}{)}^2=10^{28}-10^{20}\approx 10^{28}$. So $U_{rad}=10^{-14}\times 10^{28}=10^{14}{\text{cm}}^{-3}{\text{s}}^{-1}$. Why this step? The $n_i^2=10^{20}$ term is eight orders below $10^{28}$ — it's the "equilibrium traffic" that nearly cancels; the excess is what survives.

4. Lifetime. $\tau =\frac{1}{B{n}_0}=\frac{1}{10^{-14}\times 10^{16}}=10^{-2}\text{s}=10\text{ms}$. Why this step? In low injection $U\approx B{n}_0\Delta p=\Delta p/\tau$, so $\tau =1/(B{n}_0)$ — a constant (cell D), independent of $\Delta p$.

5. Cross-check the approximation. $\Delta p/\tau =10^{12}/10^{-2}=10^{14}{\text{cm}}^{-3}{\text{s}}^{-1}$ — identical to step 3.

Verify: Units of $U$: $[{\text{cm}}^3/\text{s}]\cdot [{\text{cm}}^{-3}][{\text{cm}}^{-3}]={\text{cm}}^{-3}{\text{s}}^{-1}$ ✓ (pairs lost per volume per time). Cell C sanity: set $\Delta p=0$ → $np=n_0{p}_0=n_i^2$ → $U=0$, yet $R=B{n}_0{p}_0\ne 0$ still runs. The system is busy but balanced.

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E2 — Exponential decay of the excess (cell D, time domain)

Forecast: $25\text{ms}$ is $2.5$ lifetimes — expect a drop to roughly $e^{-2.5}\approx 8\%$.

1. Which equation governs decay? With no field and no external generation, the Continuity Equation collapses to $\frac{d\Delta p}{dt}=-\frac{\Delta p}{\tau }$. Why this step? Every drift/diffusion/generation term vanishes; only recombination removes carriers. This is the defining ODE of lifetime.

2. Solve it. The solution is $\Delta p(t)=\Delta p(0)e^{-t/\tau }$. Why this step? Exponential is the unique function whose rate of loss is proportional to how much is left — a constant fractional loss per unit time.

3. Plug in. $\Delta p(25\text{ms})=10^{12}{e}^{-25/10}=10^{12}{e}^{-2.5}=10^{12}(0.0821)=8.21\times 10^{10}{\text{cm}}^{-3}$.

4. The $1/e$ time. $\Delta p=\Delta p(0)/e$ when $t/\tau =1$, i.e. $t=\tau =10\text{ms}$. Why this step? This is the physical meaning of lifetime — the time to decay to $\approx 37\%$.

Verify: At $t=\tau$, $e^{-1}=0.368$, so $\Delta p=3.68\times 10^{11}$, indeed $1/e$ of start ✓. Sanity: $25\text{ms}>10\text{ms}$ so answer must be below $3.68\times 10^{11}$; $8.21\times 10^{10}$ is ✓.

The figure below plots this decay so you can see the "one lifetime per $1/e$" rule and locate both marked points (the $1/e$ crossing and the $t=25\text{ms}$ value from step 3):

Read the red curve: each horizontal step of one $\tau =10\text{ms}$ multiplies the height by $e^{-1}\approx 0.37$. The black dashed lines mark the $1/e$ crossing (step 4); the red dot at $t=25\text{ms}$ is the $8.21\times 10^{10}$ answer from step 3.

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E3 — Net generation, $U<0$ (cell B — depletion region)

Forecast: carriers are missing, so nature should create pairs — expect $U<0$ (net generation).

1. Set the depleted concentrations. Take $n=p=0$ (fully swept). Why this step? This is the degenerate edge case $np<n_i^2$ — the opposite of injection.

2. Numerator. $np-n_i^2=0-(10^{10}{)}^2=-10^{20}{\text{cm}}^{-6}$. Negative ⇒ net generation (cell B). Why this step? The same driver as always, but now the sign flips because we're below equilibrium.

3. Denominator (mid-gap, $n_1=p_1=n_i$): ${\tau }_p(n+n_1)+{\tau }_n(p+p_1)={\tau }_0(0+n_i)+{\tau }_0(0+n_i)=2{\tau }_0{n}_i$.

4. Combine. $U=\frac{-n_i^2}{2{\tau }_0{n}_i}=\frac{-n_i}{2{\tau }_0}=\frac{-10^{10}}{2\times 10^{-6}}=-5\times 10^{15}{\text{cm}}^{-3}{\text{s}}^{-1}$. Why this step? Defining the generation lifetime ${\tau }_g=2{\tau }_0$, we get $U=-n_i/{\tau }_g$ — the standard thermal generation current source of reverse leakage.

Verify: Sign is negative ✓ (generation). Magnitude cross-check: $n_i/(2{\tau }_0)=10^{10}/(2\cdot 10^{-6})=5\times 10^{15}$ ✓. Units: ${\text{cm}}^{-3}/\text{s}$ ✓. This is exactly the physics of reverse-bias leakage — the region manufactures pairs to refill the vacuum.

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E4 — High injection: lifetime shifts (cell E)

Forecast: in high injection the excess outnumbers $n_0$; guess ${\tau }_{eff}$ gets shorter (more carriers meet).

1. General net rate. $U=B(np-n_i^2)$ with $n=n_0+\Delta p$, $p=p_0+\Delta p\approx \Delta p$. Expand: $np-n_i^2=(n_0+\Delta p)\Delta p=n_0\Delta p+\Delta p^2$ (dropping $n_i^2$, tiny). Why this step? We keep the $\Delta p^2$ term — that's exactly what low-injection dropped, and it's what dominates at high injection.

2. Low injection $\Delta p=10^{12}\ll n_0=10^{15}$: $U\approx B{n}_0\Delta p$, so ${\tau }_{eff}=\frac{\Delta p}{B{n}_0\Delta p}=\frac{1}{B{n}_0}=\frac{1}{10^{-14}\cdot 10^{15}}=0.1\text{s}$. Why this step? Constant lifetime — this is the familiar $1/(B{n}_0)$ (cell D).

3. High injection $\Delta p=10^{17}\gg n_0$: now $\Delta p^2$ dominates, $U\approx B\Delta p^2$, so ${\tau }_{eff}=\frac{\Delta p}{B\Delta p^2}=\frac{1}{B\Delta p}=\frac{1}{10^{-14}\cdot 10^{17}}=10^{-3}\text{s}=1\text{ms}$. Why this step? Here ${\tau }_{eff}$ depends on $\Delta p$ — it is not a fixed constant. This is the nonlinear regime the "lifetime is constant" mistake forgets.

Verify: Ratio ${\tau }_{low}/{\tau }_{high}=0.1/10^{-3}=100$. Independent check: ${\tau }_{high}/{\tau }_{low}=(1/B\Delta p)/(1/B{n}_0)=n_0/\Delta p=10^{15}/10^{17}=10^{-2}$ ✓. Lifetime shrank 100× — high injection does shorten it, matching the forecast.

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E5 — Trap-position sweep: mid-gap is the killer (cell F)

Forecast: the parent note says mid-gap wins. Expect the mid-gap denominator smaller ⇒ larger $U$.

1. Numerator (same for both). $np-n_i^2=(10^{15}{)}^2-(10^{10}{)}^2=10^{30}-10^{20}\approx 10^{30}$. Why this step? Trap position only affects the denominator; isolate what changes.

2. Mid-gap denominator. ${\tau }_0(n+n_1)+{\tau }_0(p+p_1)=10^{-6}(10^{15}+10^{10})+10^{-6}(10^{15}+10^{10})\approx 2\times 10^{-6}\times 10^{15}=2\times 10^9$. So $U_{mid}=10^{30}/(2\times 10^9)=5\times 10^{20}{\text{cm}}^{-3}{\text{s}}^{-1}$.

3. Shallow-trap denominator. ${\tau }_0(n+n_1)+{\tau }_0(p+p_1)=10^{-6}(10^{15}+10^{15})+10^{-6}(10^{15}+10^5)=10^{-6}(2\times 10^{15})+10^{-6}(10^{15})=3\times 10^9$. So $U_{shallow}=10^{30}/(3\times 10^9)=3.33\times 10^{20}{\text{cm}}^{-3}{\text{s}}^{-1}$. Why this step? The large $n_1\approx 10^{15}$ inflates the denominator — the trap emits captured electrons back before a hole arrives, wasting the event.

4. Compare. $U_{mid}/U_{shallow}=5/3.33=1.5$ — mid-gap recombines 50% faster here. Why this step? Confirms mid-gap traps are the "lifetime killers" — same product $n_1{p}_1=n_i^2$ but their sum is minimized at $n_1=p_1$.

Verify: $n_1{p}_1$ must equal $n_i^2$ in both cases: mid-gap $10^{10}\cdot 10^{10}=10^{20}$ ✓; shallow $10^{15}\cdot 10^5=10^{20}$ ✓. $U_{mid}>U_{shallow}$ ✓, matching the forecast.

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E6 — Which channel wins? Parallel lifetimes (cell G)

Forecast: heavy doping ($10^{19}$) → the parent note flags Auger ($\propto n_0^2$). Guess Auger dominates.

1. Auger lifetime. ${\tau }_{Aug}=\frac{1}{C_n{n}_0^2}=\frac{1}{10^{-31}\times (10^{19}{)}^2}=\frac{1}{10^{-31}\times 10^{38}}=\frac{1}{10^7}=10^{-7}\text{s}=0.1\mu s$. Why this step? At $10^{19}$, $n_0^2=10^{38}$ is enormous — Auger scales with density squared, so this is where it bites.

2. List all three rates ($1/{\tau }_i$): $1/{\tau }_{rad}=100{\text{s}}^{-1}$; $1/{\tau }_{SRH}=10^6{\text{s}}^{-1}$; $1/{\tau }_{Aug}=10^7{\text{s}}^{-1}$. Why this step? Channels run in parallel — rates add, not times.

3. Total. $\frac{1}{\tau }=100+10^6+10^7=11,010,100{\text{s}}^{-1}\approx 1.1\times 10^7{\text{s}}^{-1}\Rightarrow \tau =\frac{1}{11,010,100}=9.08\times 10^{-8}\text{s}\approx 91\text{ns}$. Why this step? The sum is dominated by the largest term ($10^7$, Auger) — the fastest path controls total lifetime.

4. Name the winner. Auger (${\tau }_{Aug}=0.1\mu s=100\text{ns}$) sets the scale; the total ($91\text{ns}$) sits just below it because SRH ($10^6{\text{s}}^{-1}$) adds a further $\sim 10\%$ to the rate. Radiative ($100{\text{s}}^{-1}$) is utterly negligible.

Verify: $\tau \le \min ({\tau }_i)={\tau }_{Aug}=100\text{ns}$, and indeed $91\text{ns}<100\text{ns}$ ✓ (parallel lifetime is always below the smallest). Common mistake avoided: adding times would give $10\text{ms}+1\mu s+0.1\mu s\approx 10\text{ms}$ — wildly wrong.

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E7 — Real-world design choice (cell H, word problem)

Forecast: LEDs need a fast radiative path. GaAs radiative is $14\text{ns}$ vs Si $10\text{ms}$ — expect GaAs efficiency near 1, Si near 0.

1. GaAs. $1/{\tau }_{rad}=1/14\text{ns}=7.14\times 10^7{\text{s}}^{-1}$; $1/{\tau }_{nr}=10^6{\text{s}}^{-1}$. ${\eta }_A=\frac{7.14\times 10^7}{7.14\times 10^7+10^6}=\frac{7.14\times 10^7}{7.24\times 10^7}=0.986$. Why this step? Radiative rate is 71× the non-radiative one — almost every recombination emits a photon.

2. Si. $1/{\tau }_{rad}=1/10\text{ms}=100{\text{s}}^{-1}$; $1/{\tau }_{nr}=10^6{\text{s}}^{-1}$. ${\eta }_B=\frac{100}{100+10^6}=\frac{100}{1.0001\times 10^6}=1.0\times 10^{-4}$. Why this step? Radiative rate is 10 000× slower — SRH/phonon path swallows the energy as heat.

3. Decision. Choose GaAs: ${\eta }_A\approx 99\%$ vs ${\eta }_B\approx 0.01\%$. Roughly $10^4$× brighter. Why this step? This is precisely why LEDs are built from direct-gap GaAs/GaN, not Si — the same reason Si excels instead in Solar Cells, where a long lifetime helps carriers reach the contacts.

Verify: Both $\eta \in [0,1]$ ✓. Ratio ${\eta }_A/{\eta }_B=0.986/10^{-4}\approx 9.9\times 10^3$ — GaAs is ~$10^4$× more efficient, exactly why Si can't make LEDs. Consistency with the LED/solar-cell contrast in the parent note ✓.

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E8 — Degenerate limits & exam twist (cell I)

Forecast: no traps → SRH is switched off (${\tau }_{SRH}\to \infty$), so radiative alone sets $\tau$. As $t\to \infty$ all excess must vanish.

1. SRH lifetime with no traps. ${\tau }_{SRH}=\frac{1}{\sigma v_{th}{N}_t}$. As $N_t\to 0$, ${\tau }_{SRH}\to \infty$, so its rate $1/{\tau }_{SRH}\to 0$. Why this step? No stepping-stone defects means the trap-assisted channel simply doesn't exist — the degenerate zero-input case.

2. Total lifetime. $\frac{1}{\tau }=\frac{1}{{\tau }_{rad}}+\underset{0}{\underbrace{\frac{1}{{\tau }_{SRH}}}}=\frac{1}{{\tau }_{rad}}$, so $\tau ={\tau }_{rad}=\frac{1}{B{n}_0}=\frac{1}{7.2\times 10^{-10}\times 10^{17}}=1.39\times 10^{-8}\text{s}\approx 14\text{ns}$. Why this step? When one parallel rate vanishes, it drops silently out of the sum — the remaining channel takes over. (Matches parent example E1.)

3. Long-time limit. $\Delta p(t)=\Delta p(0)e^{-t/\tau }$; as $t\to \infty$, $e^{-t/\tau }\to 0$, so $\Delta p\to 0$. Why this step? The system must return to equilibrium $np=n_i^2$. Any finite $\tau$ guarantees full decay — no residual excess survives.

Verify: $\tau =14\text{ns}$ matches the parent's GaAs result ✓. Limit check: plug $N_t=0$ into $1/{\tau }_{SRH}=\sigma v_{th}{N}_t=0$ ✓. And ${\lim }_{t\to \infty }{e}^{-t/\tau }=0$ for any $\tau >0$ ✓ — the crystal always wins in the end.

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Matrix coverage check

Recall Did we hit every cell?

A (net recomb) → E1,E2,E4 · B (net gen) → E3 · C (equilibrium) → E1 · D (low inj) → E1,E2 · E (high inj) → E4 · F (trap position) → E5 · G (parallel $\tau$) → E6 · H (word problem) → E7 · I (degenerate/limit) → E8. All nine covered ✓

Related builds: Minority Carrier Diffusion (where these lifetimes feed the diffusion length), Continuity Equation (the ODE behind E2), and Solar Cells (where E5's lifetime killers cost you efficiency).