Atomic layer deposition (ALD): Visual walkthrough — Hardware deep dive

This is the mathematical heart of ALD. The chemistry behind it is Langmuir surface kinetics; keep that link in your pocket.

Step 1 — Draw the surface as a row of parking spots

WHAT. Before any equation, picture the surface. It is not a smooth sheet — it is a finite set of reactive spots. For Al₂O₃ growth those spots are surface -OH groups (hydroxyl groups sticking up). Call the number of spots per square centimetre $N_{0}$ .

WHY. Everything self-limiting comes from one fact: there is a finite number of spots. If spots were infinite, the reaction would never run out and ALD would behave like ordinary CVD. So we must start by counting spots.

PICTURE. Look at the grid below. Each square is one reactive site. Green = still free (a spot a precursor molecule can grab), orange = already reacted (dead, cannot be used again).

Step 2 — Define coverage $θ$ : the one number that tracks progress

WHAT. We do not want to track every spot individually — that's millions per cm². Instead we track one fraction:

θ = \frac{number of reacted spots}{total spots} = \frac{N _{reacted}}{N _{0}}

$θ$ (Greek "theta") is the fraction of sites already reacted.
The numerator counts orange squares; the denominator is all squares.

WHY. A fraction is dimensionless and lives on a fixed scale. Whatever happens, $θ$ is trapped between two meaningful ends:

θ = 0 (nothing reacted yet) θ = 1 (every spot used up)

That fixed range $0 \leq θ \leq 1$ is the whole story: the reaction starts at 0 and cannot pass 1. The ceiling at 1 is where self-limiting comes from.

PICTURE. The bar below fills from left (all green, $θ = 0$ ) to right (all orange, $θ = 1$ ). Watch how "free spots left" is exactly the un-filled part, $1 - θ$ .

Step 3 — Write the reaction rate: fuel × opportunity

WHAT. How fast do new reactions happen? A reaction needs two things to coincide:

A precursor molecule arriving at the surface, and
A free spot waiting for it.

We multiply "how much precursor" by "how many free spots":

\frac{d θ}{d t} = how eager k how much gas P how many free spots (1 - θ)

Term by term, right where it sits:

$\frac{d θ}{d t}$ = the rate of change of coverage — how fast orange spreads per second (units: per second, since $θ$ is dimensionless). (This is a derivative; see the box on why we need one.)
$k$ = a rate constant — bundles how reactive this chemistry is. For the whole product $k P$ to come out in per second (matching the left side), $k$ must carry units of per second per unit pressure. Fix a reference pressure unit once and stick to it: here we use the pascal (Pa), so $k$ has units $s^{- 1} Pa^{- 1}$ and $P$ is measured in Pa. (Use Torr consistently instead and $k$ would be $s^{- 1} Torr^{- 1}$ — the number for $k$ changes but the physics does not.)
$P$ = partial pressure of the precursor gas, in Pa — more gas in the chamber ⇒ molecules arrive faster.
$(1 - θ)$ = free-spot fraction from Step 2 — the fuel gauge.

WHY a derivative and not just a ratio? Because coverage changes over time and we want to know the instantaneous speed at each moment, which is precisely what a derivative $d θ / d t$ measures — the slope of the coverage-vs-time curve at one instant. A plain ratio would only give an average; we need the moment-by-moment law.

WHY the product? If either factor is zero, nothing happens: no gas ( $P = 0$ ) ⇒ no reactions; no free spots ( $θ = 1$ so $1 - θ = 0$ ) ⇒ no reactions. Multiplication is the only simple operation that makes both conditions necessary.

PICTURE. Two dials feed one output. Turn up $P$ (left dial), reactions speed up. But as $θ$ climbs, the $(1 - θ)$ dial turns itself down — the reaction chokes its own fuel supply.

Step 4 — Separate the variables (untangle $θ$ from $t$ )

WHAT. We have an equation mixing $θ$ and $t$ . To solve it we gather all $θ$ -stuff on one side and all $t$ -stuff on the other:

\frac{d θ}{d t} = k P (1 - θ) ⟹ \frac{d θ}{1 - θ} = k P d t

What we did: divided both sides by $(1 - θ)$ and multiplied both sides by $d t$ .

Left side: everything about spots ( $θ$ ).
Right side: everything about time ( $t$ ), with the constants $k P$ riding along. (This grouping only works because $k$ and $P$ are constant during the pulse, as we argued in Step 3.)

WHY. This trick is called separation of variables. Once each side depends on only one letter, we can integrate each side by itself — turning a rate law into an actual formula for $θ$ . It's the standard key for any equation of the form $d y / d t = (stuff) \cdot (function of y)$ .

PICTURE. Think of it as sorting a mixed pile into two bins — a "θ bin" and a "t bin" — so each can be totalled separately.

Step 5 — Integrate both sides (add up all the tiny changes)

WHAT. Now sum every tiny step. We keep the starting coverage general — call it $θ_{0}$ , the coverage already present at time 0 — so nothing is lost if the surface wasn't perfectly bare:

\int_{θ_{0}}^{θ} \frac{d θ ^{'}}{1 - θ ^{'}} = \int_{0}^{t} k P d t^{'}

The left integral collects all the little coverage increments, from the starting value $θ_{0}$ up to the current $θ$ .
The right integral collects all the little time increments.
$θ^{'}$ and $t^{'}$ are just dummy variables — placeholders that sweep from the lower to the upper limit.

Do each integral:

[- ln (1 - θ^{'})]_{θ_{0}}^{θ} = k P [t^{'}]_{0}^{t} ⟹ - ln (1 - θ) + ln (1 - θ_{0}) = k P t

which tidies to

ln \frac{1 - θ _{0}}{1 - θ} = k P t

WHY the natural log appears. The left integrand is $\frac{1}{1 - θ ^{'}}$ , and the antiderivative of "one over a linear thing" is a logarithm (specifically $- ln (1 - θ^{'})$ ). That is why $ln$ shows up here and not, say, a square root — it is dictated purely by the shape of the $(1 - θ)$ term. The natural log $ln$ is the inverse of $e^{(\cdot)}$ , which is exactly what lets us undo it in the next step.

PICTURE. The area under the $1/ (1 - θ)$ curve from $θ_{0}$ to $θ$ equals the flat rectangle area $k P t$ on the right. As $θ \to 1$ , the left area blows up — foreshadowing that reaching full coverage takes forever.

Step 6 — Solve for $θ$ : the exponential is forced

WHAT. Undo the logarithm. Exponentiate both sides, then rearrange:

\frac{1 - θ _{0}}{1 - θ} = e^{k P t} ⟹ 1 - θ = (1 - θ_{0}) e^{- k P t} ⟹ θ (t) = 1 - (1 - θ_{0}) e^{- k P t}

This is the general solution for any starting coverage $θ_{0}$ . The parent note's boxed result is just the bare-surface case $θ_{0} = 0$ :

θ_{0} = 0 ⟹ θ (t) = 1 - e^{- k P t}

Reading the general formula piece by piece:

$(1 - θ_{0})$ = the free-spot fraction we started with — the pool available to be filled.
$e^{- k P t}$ starts at 1 (when $t = 0$ : $e^{0} = 1$ ) and shrinks toward 0 as time grows.
So $θ = 1 - (1 - θ_{0}) (shrinking thing)$ climbs from $θ_{0}$ toward 1.

WHY $e$ and not some other curve? Because the free spots decay at a rate proportional to how many are left — and the one function whose rate of decrease equals itself is $e^{- x}$ . The exponential is not a choice; it is the unique solution of "the more you've done, the slower you go."

PICTURE. The saturation curve. Steep at first (many free spots), bending over, then crawling toward the ceiling $θ = 1$ . Dashed guides mark the famous checkpoints (drawn for the bare-surface case $θ_{0} = 0$ ).

Step 7 — Edge and degenerate cases (never leave a gap)

WHAT. Check the corners so no scenario surprises you. Use the general solution $θ (t) = 1 - (1 - θ_{0}) e^{- k P t}$ .

Case $t = 0$ (before any dose). $θ = 1 - (1 - θ_{0}) e^{0} = 1 - (1 - θ_{0}) = θ_{0}$ . The surface starts at whatever coverage it already had — correct.

Case $P = 0$ (no precursor in the chamber). Then $k P t = 0$ for all $t$ , so $θ = 1 - (1 - θ_{0}) = θ_{0}$ forever. No gas, no growth — the coverage just stays put.

Case $k = 0$ (dead chemistry, even with gas present). Even if $P > 0$ , if the reaction constant $k = 0$ then again $k P t = 0$ and $θ$ is frozen at $θ_{0}$ for all time. So the "nothing happens" behaviour needs both $P > 0$ and $k > 0$ ; either one being zero freezes the surface. Worth noting because " $P > 0$ " alone does not guarantee growth.

Case $t \to \infty$ with $k P > 0$ (dose forever). Only when the product $k P$ is strictly positive does $e^{- k P t} \to 0$ , giving $θ \to 1$ . Coverage saturates and stops — the mathematical fingerprint of ALD. (If $k P = 0$ , per the two cases above, there is no approach to 1 at all — the limit is just $θ_{0}$ .)

The real-world ceiling — steric hindrance and $θ_{ma x}$ . Bulky ligands (the CH₃ groups on TMA) physically block neighbouring spots, so the surface can never fill past some fraction $θ_{ma x} < 1$ . Define:

To put this into the physics we change the driving term. The free-spot fraction is no longer $(1 - θ)$ but the fraction of still-reachable spots, $(1 - \frac{θ}{θ _{ma x}})$ , which correctly hits zero when $θ = θ_{ma x}$ . The rate law becomes:

\frac{d θ}{d t} = k P (1 - \frac{θ}{θ _{ma x}})

Running the identical separate-integrate-solve steps (starting from a bare surface, $θ_{0} = 0$ ) gives:

θ (t) = θ_{ma x} (1 - e^{- k P t / θ_{ma x}})

Same shape as before, just a lower ceiling $θ_{ma x}$ and a rescaled time constant. This is why one cycle is sub-monolayer, not a full monolayer.

PICTURE. Three curves on one axis: the ideal ceiling at 1, a realistic ceiling at $θ_{ma x} \approx 0.4$ , and the flat $k P = 0$ line hugging $θ_{0}$ .

Step 8 — From coverage to thickness (why we cared)

WHAT. One cycle deposits a slab whose thickness is the growth per cycle:

GPC = \frac{θ _{ma x} N _{0}}{N _{atoms/vol}} d = GPC \times (# cycles)

$θ_{ma x} N_{0}$ = reacted spots per cm² per cycle.
Divide by the atomic density $N_{atoms/vol}$ to convert areal atoms into a thickness.
Multiply by cycle count for the total film $d$ .

WHY this is the payoff. Because $θ_{ma x}$ is fixed by chemistry and geometry — not by time — every cycle lays down the same thickness. Stack $N$ identical slabs and you get digital, angstrom-level thickness control, verified by ellipsometry and impossible with line-of-sight sputtering.

Worked check. Need 5 nm = 50 Å with GPC = 1.0 Å/cycle: $N = 50/1.0 = 50$ cycles. And coverage after a 2 s pulse at $k P = 0.5 s^{- 1}$ (bare surface): $θ = 1 - e^{- 1} = 0.63$ .

The one-picture summary

The whole derivation on one canvas: a finite grid of spots → coverage $θ$ → self-throttling rate $k P (1 - θ)$ → separate → integrate (log appears) → solve ( $e$ appears) → the saturating curve → stack cycles for thickness.

Recall Feynman retelling — say it in plain words

Picture a wall of parking spots. Each second, a molecule grabs a free spot; the chance of a grab depends on how much gas is around ( $P$ ) times how many spots are still empty ( $1 - θ$ ). During one pulse the temperature and gas pressure are held steady, so those two knobs ( $k$ and $P$ ) stay fixed and we can treat their product as one number. Early on, lots of empty spots, so grabs happen fast. As the lot fills, empty spots get rare, so grabbing slows — the process throttles itself. Mathematically, "the emptier fraction shrinks in proportion to itself" is exactly the rule that produces an exponential, so the empty fraction decays like $e^{- k P t}$ and the filled fraction climbs to a ceiling as $θ = 1 - (1 - θ_{0}) e^{- k P t}$ (from any starting coverage $θ_{0}$ ). In the real world, bulky molecules block their neighbours, so the ceiling $θ_{ma x}$ sits below a full monolayer. Because that ceiling is the same every cycle, you get identical slabs — count the cycles, get the thickness. That is ALD's whole magic in one sentence.

Recall Quick self-test

Answer before revealing. Why does the rate law contain $(1 - θ)$ ? ::: Because only free spots can react; a reacted spot cannot react again, so the driving force is the free-spot fraction $1 - θ$ . Why may we treat $k$ and $P$ as constant while integrating? ::: The pulse is at fixed temperature (so $k$ is constant) and the precursor pressure is held steady (so $P$ is constant) for the duration of the pulse. Why does a natural log appear during the derivation? ::: Integrating $\frac{1}{1 - θ}$ gives a logarithm — dictated by the "one over a linear term" shape. What is the general solution for nonzero starting coverage $θ_{0}$ ? ::: $θ (t) = 1 - (1 - θ_{0}) e^{- k P t}$ , which reduces to $1 - e^{- k P t}$ when $θ_{0} = 0$ . Does $P > 0$ alone guarantee growth? ::: No — you also need $k > 0$ ; if either $k$ or $P$ is zero the coverage stays frozen at $θ_{0}$ . Why is one cycle sub-monolayer? ::: Steric hindrance from bulky ligands caps coverage at $θ_{ma x} < 1$ .

Connections

Parent: ALD overview — this page derives its boxed saturation result
Surface Chemistry & Adsorption — the Langmuir kinetics we used in Step 3
Chemical Vapor Deposition (CVD) — no $(1 - θ)$ brake, so no self-limiting
Physical Vapor Deposition (Sputtering) — line-of-sight; contrast the conformality
High-k Gate Dielectrics — where 50-cycle Al₂O₃/HfO₂ films go
DRAM Capacitor Fabrication — why over-dosing helps deep trenches
Thin-Film Thickness Metrology (Ellipsometry) — measuring the GPC we derived