Visual walkthrough — Recombination and generation mechanisms

We build only the radiative result in full (it is the cleanest), then show in pictures how the very same "distance-from-equilibrium" driver $np-n_i^2$ reappears in SRH and Auger.

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Step 1 — Meet the two players: an electron and a hole

WHAT. We place one electron up top and one hole down below.

WHY. Recombination is literally the electron dropping into the hole — the seat gets filled, both "vanish" as free carriers. Before we can count how often this happens, we must picture what is happening once.

PICTURE. In the figure, the blue dot (electron) sits in the conduction band, the red circle (hole) sits in the valence band, and the yellow arrow is the drop that annihilates them, releasing energy $\approx E_g$ as a photon (the wavy yellow line).

Recall Why is the released energy

$\approx E_g$? Because the electron falls from the bottom of the conduction band to the top of the valence band ::: the vertical height of that fall is exactly the gap $E_g$, and energy is conserved, so the photon carries away $\approx E_g$.

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Step 2 — Counting meetings: why the rate is proportional to $np$

WHAT. We want the recombination rate $R$ = number of these drops per second per unit volume.

WHY a product? For a drop to happen, an electron must find a hole in the same place. Think of a dance floor: blue dancers and red dancers wandering randomly. The number of blue–red pairings per second is proportional to (how many blues) $\times$ (how many reds). Double the blues → double the pairings; double the reds → double again. That "multiply the two crowds" logic is exactly a product.

PICTURE. The left panel has a sparse floor (few dots → few crossings marked yellow). The right panel doubles both crowds → the crossings more than double, growing like the product $n\times p$.

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Step 3 — The crystal is never still: generation runs too

WHAT. Heat alone keeps kicking electrons up across the gap — that is generation $G$, the opposite of the drop in Step 1.

WHY we cannot ignore it. At room temperature the crystal is warm. Warmth = vibration = random energy kicks. Some kicks are big enough to launch an electron across $E_g$, creating a fresh electron–hole pair. This runs all the time, whether or not we disturb the crystal.

PICTURE. Yellow up-arrows (generation, thermal kicks) and blue down-arrows (recombination) on the same band diagram — two opposing traffic streams.

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Step 4 — Equilibrium: the busy stalemate pins down $G_0$

WHAT. At thermal equilibrium (no light, no injected current, just sitting warm) the up-traffic and down-traffic exactly cancel.

WHY this lets us solve for $G_0$. "Cancel" means $R_0=G_0$. But we already know $R_0=B{n}_0{p}_0$ from Step 2, using the equilibrium crowd sizes $n_0,p_0$. So:

$G_0=R_0=B{n}_0{p}_0.$

Now bring in the Law of Mass Action, which says at equilibrium the two crowds always multiply to the same fixed number: $n_0{p}_0=n_i^2,$ where $n_i$ = the intrinsic carrier concentration (the crowd size in a perfectly pure crystal). Therefore:

PICTURE. A balance scale: left pan "recombination $B{n}_0{p}_0$", right pan "generation $G_0$", perfectly level, labelled $=B{n}_i^2$.

Recall Why is

$G_0$ still $B{n}_i^2$ after we disturb the crystal? Because $G_0$ depends only on temperature and $E_g$ (Step 3), which we do not change by shining light or injecting current ::: so the generation number stays frozen at its equilibrium value $B{n}_i^2$ even when $n$ and $p$ shoot up.

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Step 5 — Disturb it: the net rate $U=R-G$

WHAT. Now shine light or inject current. Crowds rise to $n$ and $p$ (bigger than $n_0,p_0$). Recombination speeds up to $R=Bnp$, but generation stays frozen at $G_0=B{n}_i^2$. The net downward flow is:

$U=R-G_0=Bnp-B{n}_i^2.$

WHY subtract. $R$ tells us drops per second; $G_0$ tells us kicks-up per second. What actually reduces the excess is the difference. If $R=G_0$ the excess neither grows nor shrinks — $U=0$, back to stalemate.

PICTURE. A number line for the product $np$. A dot at $n_i^2$ is the "rest" point. When we push the dot right ($np>n_i^2$), a blue arrow $U>0$ pushes it back left (net recombination). Push it left ($np<n_i^2$), a green arrow $U<0$ pushes it right (net generation). The formula is a restoring force.

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Step 6 — Every case: above, below, and exactly at equilibrium

We must cover all signs of the driver $np-n_i^2$ so no reader hits an unshown scenario.

Situation	$np$ vs $n_i^2$	Sign of $U$	Physical meaning
Injected carriers (light on)	$np>n_i^2$	$U>0$	net recombination: excess pairs removed
At rest (dark, no current)	$np=n_i^2$	$U=0$	stalemate: $R=G_0$, nothing net
Depleted region (reverse bias)	$np<n_i^2$	$U<0$	net generation: crystal creates pairs to refill

WHY the depleted case matters. In the reverse-biased PN Junction Diode, carriers are swept out so $np$ drops below $n_i^2$. Then $U$ goes negative — the same one formula now describes generation, no new equation needed. This is the origin of reverse-bias generation current.

PICTURE. Three mini band diagrams side by side: (left) crowded bands, blue down-arrow dominating; (middle) balanced; (right) sparse bands, yellow up-arrow dominating.

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Step 7 — Low injection: from $U$ to a clean lifetime $\tau$

WHAT. Take n-type material: electrons are the majority, so $n_0\gg p_0$. Inject a small excess $\Delta p$ (with $\Delta n=\Delta p$, since light makes pairs). Write $n=n_0+\Delta p$, $p=p_0+\Delta p$ and expand:

= (n_0+p_0)\Delta p + \Delta p^2.$$ **WHY drop terms.** ==Low injection== means the injected excess is tiny next to the majority crowd: $\Delta p \ll n_0$. So $\Delta p^2$ is negligible next to $n_0\Delta p$, and $p_0\ll n_0$. Only $n_0\Delta p$ survives: $$U_{rad} \approx B\,n_0\,\Delta p \;=\; \frac{\Delta p}{\tau},\qquad \boxed{\tau=\frac{1}{Bn_0}}.$$ - $\tau$ = the ==minority-carrier lifetime==: how long, on average, an injected hole lasts before recombining. - It is a **constant** here — set by material ($B$) and doping ($n_0$), *not* by how much we injected. Plug $U=\Delta p/\tau$ into the [[Continuity Equation]] with no field or extra light and you get $\frac{d\Delta p}{dt}=-\Delta p/\tau$, whose solution is the exponential decay $\Delta p(t)=\Delta p(0)e^{-t/\tau}$ — the workhorse of [[Minority Carrier Diffusion]]. **PICTURE.** The parabola $U=B[(n_0+p_0)\Delta p+\Delta p^2]$ plotted; near the origin it hugs the straight line $Bn_0\Delta p$ (the low-injection approximation). The two curves peel apart only at large $\Delta p$ — that peeling-apart *is* high injection. > [!example] Numbers, so the picture is concrete > GaAs: $B=7.2\times10^{-10}\,\text{cm}^3/\text{s}$, $n_0=10^{17}\,\text{cm}^{-3}$. > $$\tau=\frac{1}{Bn_0}=\frac{1}{(7.2\times10^{-10})(10^{17})}=1.39\times10^{-8}\ \text{s}\approx 14\ \text{ns}.$$ > A short ns lifetime = fast light emission — exactly why GaAs powers [[LEDs and Lasers]]. --- ## Step 8 — The same driver in the other two channels **WHAT.** Look at the parent's SRH and Auger formulas and hunt for the driver: $$U_{SRH}=\frac{\overbrace{np-n_i^2}^{\text{same driver!}}}{\tau_p(n+n_1)+\tau_n(p+p_1)}, \qquad U_{Auger}=(C_n n + C_p p)\big(\overbrace{np-n_i^2}^{\text{same driver!}}\big).$$ **WHY this is the punchline.** The factor $np-n_i^2$ appears in **all three**. Why? Because *every* mechanism must obey the same law: at equilibrium $np=n_i^2$ forces $U=0$. That single requirement stamps the driver onto all of them. What differs is only the **prefactor** — constant $B$ (radiative), a denominator with traps (SRH), or a carrier-density factor $C_nn+C_pp$ (Auger). Because the channels act in parallel, their rates add, so the lifetimes combine as reciprocals — the key result relied on by [[Solar Cells]]: $$\frac{1}{\tau}=\frac{1}{\tau_{rad}}+\frac{1}{\tau_{SRH}}+\frac{1}{\tau_{Auger}}.$$ **PICTURE.** Three boxes each showing the shared driver $np-n_i^2$ entering, times a different prefactor cartoon (photon / trap-ladder / third-carrier), all funnelling into $U_{total}$. --- ## The one-picture summary Everything on one canvas: two crowds → product $np$ → subtract the frozen equilibrium $n_i^2$ → multiply by a prefactor → net rate $U$ → in low injection collapses to $\Delta p/\tau$ → decays exponentially. > [!recall]- Feynman retelling — the whole walkthrough in plain words > Picture a two-storey building. Kids upstairs (electrons) sometimes slide down a fireman's pole into an empty seat downstairs (a hole) — that's a **recombination**, and it flashes a light (a photon). How many slides per second? You need a kid *and* an empty seat in the same spot, so it's (kids) times (seats): the product $np$, scaled by how slippery the pole is, $B$. But the building is warm, so heat keeps flinging kids back *upstairs* at a steady pace $G_0$ — and at rest the up-flinging and down-sliding exactly match, which pins $G_0=Bn_i^2$. When we crank up the crowd by shining light, sliding speeds up but flinging stays the same, so the *net* downward flow is $U=B(np-n_i^2)$: current-crowd product minus its resting value. That leftover $np-n_i^2$ is just "how far from calm we are," and it shows up in every recombination story because every story must go quiet at rest. If we only nudge things a little (low injection), the messy formula straightens into $U=\Delta p/\tau$, and the excess simply fades away like $e^{-t/\tau}$. That $\tau$ — the lifetime — is what decides how fast a diode switches, how bright an LED glows, and how much of the sun a solar cell keeps. --- > [!recall]- Rapid self-test > 1. Why product $np$ not sum $n+p$? → if either crowd is zero, no meeting is possible; only a product vanishes. ::: > 2. Why is $G_0$ frozen at $Bn_i^2$ under injection? → it depends only on $T$ and $E_g$, which we don't change. ::: > 3. What is the sign of $U$ in a reverse-biased depletion region? → negative ($np<n_i^2$): net generation. ::: > 4. Which term did low injection let us drop? → $\Delta p^2$ (and $p_0$), leaving $Bn_0\Delta p$. ::: > 5. Why does $np-n_i^2$ appear in all three mechanisms? → equilibrium demands $U=0$, forcing that exact driver on every channel.