Atomic layer deposition (ALD): Worked examples — Hardware deep dive

Before any symbol is used, here is the vocabulary, each anchored to a picture you'll meet below.

See Surface Chemistry & Adsorption for the Langmuir kinetics that birth $θ (t) = 1 - e^{- k P t}$ , and the Hinglish parent if you want the plain-language story first. Everything here deepens 4.3.14 Atomic layer deposition (ALD) (Hinglish)'s parent — never contradicts it.

The scenario matrix

Every ALD numeric question falls into one of these cells. Each example below is tagged with the cell it covers.

Cell	What makes it distinct	Example
C1 Thickness ↔ cycles	plain $d = GPC \times N$ , both directions	Ex 1
C2 Short dose (under-saturation)	small $k P t$ , film NOT saturated	Ex 2
C3 Long dose (over-dose)	large $k P t$ , diminishing returns	Ex 3
C4 Zero / degenerate input	$t = 0$ , $P = 0$ , $N = 0$ — nothing happens	Ex 4
C5 Limiting behaviour	$t \to \infty$ ceiling; % to reach 99%	Ex 5
C6 Sub-monolayer + steric	GPC < one monolayer, ligand blocking	Ex 6
C7 Word problem: deep trench	conformality, diffusion time to bottom	Ex 7
C8 Exam twist: temperature/decomposition	outside the ALD window, law breaks	Ex 8

C1 — Thickness ↔ cycles (both directions)

Worked example Example 1 — count the cycles, then invert it

(a) You need a 8 nm Al₂O₃ layer. GPC = 1.0 Å/cycle. How many cycles? (b) You accidentally ran 120 cycles at GPC = 1.1 Å/cycle. How thick is the film in nm?

Forecast: Guess — is (a) around 8, 80, or 800 cycles? Jot it down before reading on.

Convert units so they match. $8 nm = 80 \overset{˚}{A}$ (since $1 nm = 10 \overset{˚}{A}$ ). Why this step? GPC is quoted in Å/cycle; you cannot divide nm by Å/cycle without a unit clash.
Divide thickness by GPC. $N = 80 \overset{˚}{A} \div 1.0 \overset{˚}{A} /cycle = 80$ cycles. Why this step? Thickness stacks linearly, so $d = GPC \times N \Rightarrow N = d / GPC$ .
For (b), multiply the other way. $d = 1.1 \overset{˚}{A} /cycle \times 120 cycles = 132 \overset{˚}{A} = 13.2 nm$ . Why this step? Same law, run forward instead of inverted.

Answer: (a) 80 cycles; (b) 13.2 nm.

Verify: Plug back — $80 \times 1.0 = 80 \overset{˚}{A} = 8 nm$ ✓. Units: $\frac{A ˚}{cycle} \times cycle = \overset{˚}{A}$ ✓. This is the everyday task behind High-k Gate Dielectrics targeting.

C2 — Short dose: the film is NOT saturated yet

Worked example Example 2 — under-dosing

$k P = 0.4 s^{- 1}$ . You pulse for only 1 s. What fraction of sites reacted, and did you saturate?

Forecast: More than 90%? Around 60%? Under 40%? Guess first.

Compute the dimensionless dose $k P t = 0.4 \times 1 = 0.4$ . Why this step? $θ$ depends only on the product $k P t$ — the "number of clock ticks" of dosing. That single number decides everything.
Feed it to the law $θ = 1 - e^{- k P t} = 1 - e^{- 0.4}$ . Why this step? This is the self-limiting formula from the parent note; it is the only equation you need.
Evaluate $e^{- 0.4} \approx 0.670$ , so $θ \approx 1 - 0.670 = 0.330$ .

Answer: == $θ \approx 0.33$ == — only a third of sites reacted. Not saturated.

Verify: Since $k P t = 0.4 < 1$ , we are on the steep left part of the s01 curve — coverage well below the ceiling, exactly as the geometry predicts. A film built this way would be thin and non-conformal: the bottom of a trench (which sees less precursor) would get even less. Fix = dose longer (see Ex 3).

C3 — Long dose: over-dosing and diminishing returns

Worked example Example 3 — is a 6 s pulse worth it?

Same $k P = 0.4 s^{- 1}$ . Compare coverage after 2 s, 6 s, and 12 s. How much extra do the last 6 seconds buy?

Forecast: Does 12 s give roughly double the coverage of 6 s? Guess yes/no.

Three doses. $k P t = 0.8, 2.4, 4.8$ respectively. Why this step? Same lump $k P t$ ; only $t$ changes.
Evaluate the law. $θ (2) = 1 - e^{- 0.8} \approx 0.551$ . $θ (6) = 1 - e^{- 2.4} \approx 0.909$ . $θ (12) = 1 - e^{- 4.8} \approx 0.992$ .
Compare the gains. From 2→6 s coverage rose by $0.358$ ; from 6→12 s it rose by only $0.083$ . Why this step? This quantifies "diminishing returns" — the extra 6 s adds an eighth as much as the first 4 s.

Answer: Last 6 s add only ~0.08 of coverage while doubling precursor use.

Verify: All three points sit on the flattening right shoulder of the s01 curve. This is exactly why recipes over-dose only slightly — enough to guarantee saturation (Ex 5), not to waste gas. Contrast with Chemical Vapor Deposition (CVD), where longer time keeps piling on material with no ceiling.

C4 — Zero and degenerate inputs (nothing happens)

Worked example Example 4 — the "does nothing" cases

Evaluate coverage / thickness for each degenerate input: (a) pulse time $t = 0$ ; (b) partial pressure $P = 0$ (valve shut); (c) cycle count $N = 0$ .

Forecast: All three give the same answer. What is it?

(a) $t = 0$ . $θ = 1 - e^{- k P \cdot 0} = 1 - e^{0} = 1 - 1 = 0$ . Why this step? $e^{0} = 1$ ; a zero-length pulse reacts with nobody. Fresh surface stays fresh.
(b) $P = 0$ . Then $k P t = 0$ for any $t$ , so again $θ = 1 - e^{0} = 0$ . Why this step? No molecules arrive ⇒ no reaction, no matter how long you wait. This is the sanity floor of the model.
(c) $N = 0$ . $d = GPC \times 0 = 0 \overset{˚}{A}$ . Why this step? No cycles ⇒ no stacked layers ⇒ zero film.

Answer: All three give 0 (no reaction, no growth).

Verify: The law must return the fresh-surface state $θ = 0$ whenever any driving quantity is zero — a model that grew film with the valve shut would be wrong. It passes. Note $P = 0$ and $t = 0$ are physically different ("shut valve" vs "instant pulse") yet mathematically identical because only the product $k P t$ enters.

C5 — Limiting behaviour: the ceiling and how long to reach it

Worked example Example 5 — how long to hit 99% saturation?

With $k P = 0.5 s^{- 1}$ : (a) what is the ceiling as $t \to \infty$ ? (b) what pulse time reaches $θ = 0.99$ ?

Forecast: Is the 99% time about 2 s, 5 s, or 20 s? Guess.

(a) Take the limit. As $t \to \infty$ , $e^{- k P t} \to 0$ , so $θ \to 1$ . Why this step? The exponential decays to nothing; the surface fills completely (or to $θ_{ma x}$ once steric limits kick in — see Ex 6). This ceiling is the mathematical fingerprint of self-limiting growth.
(b) Invert the law for a target. Set $0.99 = 1 - e^{- k P t}$ , so $e^{- k P t} = 0.01$ . Why this step? We want the time that produces a chosen coverage — so we solve for $t$ , undoing the exponential with a logarithm (the tool that answers "what exponent gives this value?").
Take the natural log. $- k P t = ln (0.01) = - 4.605$ , so $t = 4.605/0.5 = 9.21 s$ .

Answer: (a) ceiling == $θ \to 1$ ==; (b) == $t \approx 9.2 s$ == to reach 99%.

Verify: Plug back: $θ (9.21) = 1 - e^{- 0.5 \times 9.21} = 1 - e^{- 4.605} = 1 - 0.01 = 0.99$ ✓. Geometrically this is the far-right tail of s01. Because $ln$ grows so slowly, chasing the last 1% costs disproportionate time — the reason "100% saturation" is a target you approach, never truly touch. This informs how Thin-Film Thickness Metrology (Ellipsometry) checks whether saturation was really reached.

C6 — Sub-monolayer growth with steric hindrance

Worked example Example 6 — why GPC is less than a full monolayer

A perfect Al₂O₃ monolayer would add ≈3.0 Å. Bulky TMA ligands block neighbours so only a fraction $θ_{ma x} = 0.40$ of sites react per cycle. (a) What GPC results? (b) How many cycles for 6 nm?

Forecast: Is GPC closer to 3.0 Å or 1.2 Å per cycle?

Scale the monolayer by the reachable fraction. $GPC = θ_{ma x} \times 3.0 \overset{˚}{A} = 0.40 \times 3.0 = 1.2 \overset{˚}{A} /cycle$ . Why this step? Only the sites that actually react contribute thickness; steric hindrance from big ligands physically shadows the rest. "Atomic layer" means layer-by-layer control, not one full layer per cycle — the common mistake from the parent note.
Convert the target. $6 nm = 60 \overset{˚}{A}$ . Why this step? Match units to GPC (Å/cycle).
Divide. $N = 60 \div 1.2 = 50$ cycles.

Answer: (a) ==GPC = 1.2 Å/cycle==; (b) 50 cycles.

Verify: $50 \times 1.2 = 60 \overset{˚}{A} = 6 nm$ ✓. The result (1.2 Å) lands right in the parent note's stated 0.9–1.2 Å/cycle band for Al₂O₃, confirming realistic steric fractions ( $θ_{ma x} \sim 0.3$ – $0.5$ ) reproduce measured GPC. This is the physics behind High-k Gate Dielectrics.

C7 — Word problem: coating a deep trench

Worked example Example 7 — will the trench bottom get coated?

A DRAM trench has aspect ratio 50:1. Suppose precursor takes about $t_{diff} = 1.5 s$ to diffuse from top to bottom, and the reaction clock is $k P = 0.5 s^{- 1}$ everywhere it arrives. You dose for 10 s. Does the bottom saturate?

Forecast: Yes or no — will the bottom reach ≥99%?

Find effective reaction time at the bottom. The bottom only starts reacting after precursor arrives: $t_{eff} = 10 - 1.5 = 8.5 s$ . Why this step? Self-limiting reactions saturate every surface to the same $θ_{ma x}$ — the only question is whether molecules reach the bottom with enough dwell time left.
Coverage at the bottom. $θ_{bottom} = 1 - e^{- 0.5 \times 8.5} = 1 - e^{- 4.25} \approx 0.986$ . Why this step? Same law; the bottom is just a delayed copy of the top.
Compare to the 99% target. $0.986$ is a hair under 0.99 — extend the dose slightly to be safe. Why this step? High-aspect-ratio features are the classic reason to over-dose (Ex 3): you pay for diffusion, not for extra thickness.

Answer: Bottom ≈ 98.6% covered after 10 s — nearly saturated; a slightly longer pulse guarantees ~100% step coverage.

Verify: $θ (8.5) = 1 - e^{- 4.25}$ ; $e^{- 4.25} \approx 0.0143$ , so $θ \approx 0.986$ ✓. This is exactly why ALD beats line-of-sight Physical Vapor Deposition (Sputtering) in trenches: sputtered atoms fly straight and shadow the bottom, but ALD's self-limiting chemistry saturates every wall given enough dwell time.

C8 — Exam twist: stepping outside the ALD window

Worked example Example 8 — the temperature trap

An exam states: "At $250^{\circ} C$ , GPC = 1.0 Å/cycle. A student raises the temperature to $400^{\circ} C$ and finds the film grows to 300 Å in 50 cycles. Compute the apparent GPC and explain." Is the film still ALD?

Forecast: Does apparent GPC stay at 1.0 Å/cycle, or jump?

Compute apparent GPC. $GPC_{app} = 300 \overset{˚}{A} \div 50 cycles = 6.0 \overset{˚}{A} /cycle$ . Why this step? Same $d = GPC \times N$ rule, inverted for GPC.
Compare to the true ALD value. $6.0$ Å/cycle is 6× the window value of 1.0 — impossibly large for a self-limiting sub-monolayer process. Why this step? Inside the ALD window GPC is flat vs temperature; a jump signals the law no longer applies.
Diagnose. At $400^{\circ} C$ the precursor thermally decomposes — it now deposits like Chemical Vapor Deposition (CVD), continuously and non-self-limiting. The reaction stopped being surface-referee'd. Why this step? The parent note's "ALD window" says: too hot ⇒ decomposition ⇒ CVD-like growth ⇒ no saturation, no true GPC.

Answer: ==Apparent GPC = 6.0 Å/cycle==, and no — it is no longer ALD but CVD-mode decomposition; the self-limiting law $θ = 1 - e^{- k P t}$ does not hold there.

Verify: $6.0/1.0 = 6 \times$ the window GPC — a self-limiting process cannot exceed roughly one monolayer (~3 Å) per cycle, so any GPC well above that flags a broken window. The trap is trusting $d = GPC \times N$ blindly; it only holds inside the ALD window.

Recall Which cell is which? (click to reveal)

"8 nm needs how many cycles?" ::: C1 — plain thickness↔cycles. " $θ$ after a 1 s pulse, $k P = 0.4$ ?" ::: C2 — short dose / under-saturation. "Time to reach 99% coverage?" ::: C5 — limiting behaviour, invert with $ln$ . "Does the trench bottom saturate?" ::: C7 — conformality word problem. "GPC jumped to 6 Å/cycle at 400 °C?" ::: C8 — outside the ALD window (CVD mode).

Connections

Surface Chemistry & Adsorption — Langmuir kinetics behind $θ = 1 - e^{- k P t}$
Chemical Vapor Deposition (CVD) — what ALD becomes outside its window (Ex 8)
Physical Vapor Deposition (Sputtering) — line-of-sight, fails the trench of Ex 7
DRAM Capacitor Fabrication — the deep-trench word problem (Ex 7)
High-k Gate Dielectrics — target-thickness examples (Ex 1, 6)
Thin-Film Thickness Metrology (Ellipsometry) — verifying the GPC you computed