Exercises — Atomic layer deposition (ALD)

(c) is ALD. The deciding feature is that the precursors are pulsed sequentially and never together, with a purge between them.

• (a) is Chemical Vapor Deposition (CVD) — simultaneous flow, gas-phase reaction, no self-limiting stop.
• (b) is Physical Vapor Deposition (Sputtering) — line-of-sight, physical (not chemical) transport.

Pulse A → Purge → Pulse B → Purge. (Mnemonic from the parent: "Please Purge, Please Purge.")

False. Each half-reaction is self-limiting — once the surface sites are used up, $\theta \to {\theta }_{max}$ and the reaction stops. Longer pulses beyond saturation add nothing. Thickness is set by cycle count, per $d=\text{GPC}\times N$.

Convert first so units match: $6\text{nm}=60\text{A˚}$ (since $1\text{nm}=10\text{A˚}$). $N=\frac{d}{\text{GPC}}=\frac{60\text{A˚}}{1.2\text{A˚/cycle}}=50\text{cycles.}$ Answer: 50 cycles.

Rearrange the thickness law for GPC: $\text{GPC}=\frac{d}{N}=\frac{108\text{A˚}}{120\text{cycles}}=0.90\text{A˚/cycle.}$ Answer: 0.90 Å/cycle — right in the typical Al₂O₃ band of 0.9–1.2 Å/cycle.

Plug into the saturation law: $\theta =1-e^{-kPt}=1-e^{-0.5\times 2}=1-e^{-1}=1-0.3679=\mathrm{0.632.}$ Answer: $\theta \approx 0.63$ (about 63 % of sites reacted). Look at the saturation figure: at $t=1/kP$ the curve has always climbed to exactly $1-1/e\approx 0.63$, regardless of the number's size.

Solve the saturation law for $t$. Start from $\theta =1-e^{-kPt}$, so $e^{-kPt}=1-\theta$, and taking the natural log (the operation that undoes the exponential): $t=\frac{-\ln (1-\theta )}{kP}.$ With $\theta =0.95$ and $kP=0.5{\text{s}}^{-1}$: $t=\frac{-\ln (0.05)}{0.5}=\frac{2.996}{0.5}=5.99\text{s}\approx 6\text{s}.$ Meaning: going from 63 % to 95 % coverage took the pulse from 2 s to 6 s — triple the time for the last third of the sites. The exponential approach means the final empty sites are the rarest for an incoming molecule to hit. That's why real recipes over-dose: to guarantee the stubborn last sites saturate.

Sputtering is line-of-sight: atoms travel in straight lines, so the trench opening shadows the bottom — the top gets thick, the bottom gets starved. Look at the trench figure, left panel. ALD is chemically self-limiting. Even though precursor reaches the deep bottom slowly (diffusion down a narrow 40:1 hole), every surface still reacts only until its sites are consumed, then stops at the same ${\theta }_{max}$ everywhere. The top cannot "over-grow" because it, too, saturated and stopped. Give the molecules enough dose/purge time to diffuse to the bottom and you get ~100 % conformal step coverage. Right panel of the figure.

Both count 50 cycles, but Lab Y is no longer doing ALD. Without purges, leftover precursor A is still in the chamber when B is pulsed, so A and B react in the gas phase — this is Chemical Vapor Deposition (CVD)-like growth: rough surface, particle contamination, and loss of conformality in the trench. Mechanism: the purge is what enforces "surface-only" reaction. Remove it and you lose the self-limiting referee. Lab X gets a smooth, conformal 5 nm film; Lab Y gets an uneven, possibly thicker, defect-ridden layer.

(a) $8\text{nm}=80\text{A˚}$, so $N=\frac{80\text{A˚}}{1.0\text{A˚/cycle}}=80\text{cycles.}$ (b) Total time $=N\times t_{\text{cycle}}=80\times 4=320\text{s}=5.33\text{min}.$ Answer: 80 cycles, ≈ 5.3 minutes.

Design for the worst case (the bottom). Using $t=-\ln (1-\theta )/kP$ with $\theta =0.99$, $kP=0.20{\text{s}}^{-1}$: $t=\frac{-\ln (0.01)}{0.20}=\frac{4.605}{0.20}=23.0\text{s}.$ Top coverage at that same 23 s but with its faster $kP=0.80{\text{s}}^{-1}$: ${\theta }_{\text{top}}=1-e^{-0.80\times 23.0}=1-e^{-18.4}\approx \mathrm{1.0000.}$ Answer: a ≈ 23 s pulse. The top is at essentially 100 % (over-dosed, which is harmless — it simply stopped at ${\theta }_{max}$), while the bottom just reaches the required 0.99. This is exactly why deep-trench ALD uses long doses.

• 150–250 °C (flat): the ALD window. Growth is purely surface-controlled and self-limiting, so GPC is independent of temperature. Run here — it's the reproducible regime.
• Above 250 °C (rising): thermal decomposition. The precursor breaks down on its own like CVD, adding uncontrolled extra growth — GPC climbs, self-limiting is lost. (Ligand desorption can also spoil coverage here.)
• Below 150 °C (rising): precursor condensation / incomplete reaction. Precursor physically condenses (multilayers) or reacts too sluggishly, so apparent GPC rises and varies. Why flat = true ALD: if growth depended on time, flux, or temperature you would not get a plateau. A flat GPC proves the surface itself is stopping every pulse — the mathematical fingerprint $\theta \to {\theta }_{max}$ of self-limiting chemistry (rooted in the Langmuir picture from Surface Chemistry & Adsorption).

$\text{fraction}=\frac{\text{GPC}}{\text{monolayer thickness}}=\frac{1.1\text{A˚}}{2.5\text{A˚}}=\mathrm{0.44.}$ Answer: about 0.44 of a monolayer per cycle — sub-monolayer, as expected. Cause: steric hindrance. The bulky ligands (e.g. the three methyl groups on TMA) physically occupy space and block neighbouring surface sites, so not every site can react in one pulse. "Atomic layer" in the name refers to layer-by-layer control, not one complete monolayer per cycle.

Substitute $t^{\star }=1/(kP)$ into $\theta (t)=1-e^{-kPt}$: $\theta (t^{\star })=1-e^{-kP\cdot \frac{1}{kP}}=1-e^{-1}=1-0.3679=\mathrm{0.6321.}$ The $kP$ cancels — the result is a universal constant, independent of pressure or rate constant. Diagnostic use: to find an unknown $kP$, find the pulse length at which coverage (or measured film) reaches 63.2 % of its saturated value; that pulse length is $1/(kP)$. So $kP=1/t^{\star }$. It converts a hard curve-fit into reading off a single characteristic time — the same "time constant" trick used for any exponential.