Chemical vapor deposition (CVD): Worked examples — Hardware deep dive

We only use two governed quantities from the parent, and we re-earn them here so nothing is assumed:

Related vault topics: Arrhenius Equation, Boundary Layer (Fluid Dynamics), Physical Vapor Deposition (PVD), Step Coverage & Conformality, Thin Film Metrology, Epitaxy, Thermal Oxidation.

The scenario matrix

Every CVD rate problem is one (or a blend) of these cells. The examples below are tagged with the cell they cover.

#	Cell class	Distinguishing input	What it tests
C1	Reaction-limited	$k_{s} ≪ h_{g}$ (low $T$ , or low $P$ )	$F \approx k_{s} C_{g}$ ; strong $T$ -dependence
C2	Transport-limited	$h_{g} ≪ k_{s}$ (high $T$ , high $P$ )	$F \approx h_{g} C_{g}$ ; flow/pressure sets rate
C3	Crossover / comparable	$k_{s} \approx h_{g}$	full series formula, neither term dropped
C4	Degenerate: $k_{s} \to 0$	dead chemistry (too cold, poisoned surface)	$F \to 0$
C5	Degenerate: $h_{g} \to \infty$	infinitely fast delivery (vacuum limit, tiny $P$ )	$F \to k_{s} C_{g}$ (pure reaction)
C6	Arrhenius extraction	two $(T, v)$ data points	solve for $E_{a}$ using $k_{s} = k_{0} e^{- E_{a} / k T}$
C7	Pressure lever	change $P$ , use $D \propto 1/ P$ , $h_{g} \propto D$	why LPCVD improves uniformity
C8	Word problem (real fab)	aspect-ratio trench fill	conformality vs line-of-sight
C9	Exam twist	data that looks like one regime but isn't	reading the diagnostic signs correctly

The limiting behaviour figure below shows how the single formula slides across all these cells as we change one knob ( $k_{s}$ , controlled by temperature).

Figure — Chemical vapor deposition (CVD)

Read that curve left to right: at low $T$ the flux tracks the steep exponential ( $k_{s}$ small, cell C1); at high $T$ it flattens onto the horizontal ceiling $h_{g} C_{g}$ (cell C2); the bend in between is the crossover (cell C3). Every example lands somewhere on this curve.

Worked examples

Example 1 — Reaction-limited (cell C1)

Step 1 — Exact series formula. $F = \frac{k _{s} h _{g}}{k _{s} + h _{g}} C_{g} = \frac{( 2.0 ) ( 40 )}{2.0 + 40} (3.0 \times 1 0^{16})$ Why this step? We are told both resistances, so we can compute the true flux instead of an approximation — the exact value is our reference.

Step 2 — Evaluate the prefactor. $\frac{80}{42} = 1.9048 cm/s \Rightarrow F = 1.9048 \times 3.0 \times 1 0^{16} = 5.714 \times 1 0^{16} cm^{- 2} s^{- 1}$ Why this step? Multiplying the effective velocity by concentration gives molecules per area per second (units check below).

Step 3 — Approximation and error. $F_{approx} = k_{s} C_{g} = 2.0 \times 3.0 \times 1 0^{16} = 6.0 \times 1 0^{16}$ $error = \frac{6.0 - 5.714}{5.714} \times 100% = 5.0%$ Why this step? In cell C1 we throw away the $h_{g}$ resistance; the leftover error is exactly $k_{s} / (k_{s} + h_{g})$ fraction, here $\approx 4.8%$ of the true.

Verify: Units: $[cm/s] \cdot [cm^{- 3}] = cm^{- 2} s^{- 1}$ = flux ✓. Since $k_{s} / h_{g} = 0.05 ≪ 1$ , we are firmly reaction-limited, matching the small 5% correction. ✓

Example 2 — Transport-limited (cell C2)

Step 1 — Exact flux. $F = \frac{( 500 ) ( 10 )}{500 + 10} (3.0 \times 1 0^{16}) = \frac{5000}{510} (3.0 \times 1 0^{16}) = 2.941 \times 1 0^{17}$ Why this step? Same series law, different regime — we must confirm which resistance dominates by computing, not guessing.

Step 2 — Compare to the transport approximation $h_{g} C_{g}$ . $h_{g} C_{g} = 10 \times 3.0 \times 1 0^{16} = 3.0 \times 1 0^{17}$ The true flux is $98%$ of this ceiling. Why this step? It proves $h_{g}$ is the bottleneck: the fast chemistry is "waiting" for gas to arrive.

Step 3 — Double the temperature-driven $k_{s}$ to $1000$ . $F^{'} = \frac{( 1000 ) ( 10 )}{1010} (3.0 \times 1 0^{16}) = 2.970 \times 1 0^{17}$ Change $= (2.970 - 2.941) /2.941 = 1.0%$ . Why this step? Doubling chemistry moved the rate by only $1%$ — the hallmark of transport limitation, and the reason high- $T$ CVD is nearly $T$ -independent.

Verify: $F$ approaches but never exceeds $h_{g} C_{g} = 3.0 \times 1 0^{17}$ (a resistance in series can only lower the flux) ✓. Weak $T$ -response confirms cell C2. ✓

Example 3 — Crossover, comparable resistances (cell C3)

Step 1 — Plug equal values. $F = \frac{( 20 ) ( 20 )}{20 + 20} (3.0 \times 1 0^{16}) = \frac{400}{40} (3.0 \times 1 0^{16}) = 10 \times 3.0 \times 1 0^{16} = 3.0 \times 1 0^{17}$ Why this step? This is the one cell where you may drop neither term — the crossover is where both physics matter.

Step 2 — Compare to a single-resistance estimate. $k_{s} C_{g} = 20 \times 3.0 \times 1 0^{16} = 6.0 \times 1 0^{17}, F = \frac{1}{2} (6.0 \times 1 0^{17})$ Why this step? Equal series resistances double the total resistance, halving the flux — a clean sanity anchor.

Verify: $\frac{k _{s} h _{g}}{k _{s} + h _{g}} = \frac{k _{s}^{2}}{2 k _{s}} = \frac{k _{s}}{2}$ when $k_{s} = h_{g}$ ✓. So $F = \frac{1}{2} k_{s} C_{g}$ exactly. ✓

Example 4 — Degenerate limits $k_{s} \to 0$ and $h_{g} \to \infty$ (cells C4, C5)

Step 1 (a) — Take $k_{s} \to 0$ in the series formula. $F = \frac{k _{s} h _{g}}{k _{s} + h _{g}} C_{g} k_{s} \to 0 \frac{0 \cdot h _{g}}{0 + h _{g}} C_{g} = 0$ Why this step? If molecules arrive but never stick, flux into the film is zero regardless of delivery. Physically: a dead surface grows nothing.

Step 2 (b) — Take $h_{g} \to \infty$ . $F = \frac{k _{s} h _{g}}{k _{s} + h _{g}} C_{g} = \frac{k _{s}}{k _{s} / h _{g} + 1} C_{g} h_{g} \to \infty k_{s} C_{g} = 2.0 \times 3.0 \times 1 0^{16} = 6.0 \times 1 0^{16}$ Why this step? Divide top and bottom by $h_{g}$ first so the limit is obvious — the delivery resistance $1/ h_{g}$ vanishes, leaving pure reaction control (cell C5 is C1's ideal endpoint).

Verify: (a) $F = 0$ : a series circuit with an infinite resistance ( $1/ k_{s} \to \infty$ ) carries no current ✓. (b) $F = 6.0 \times 1 0^{16}$ : matches the $F_{approx}$ of Example 1, the reaction-limited ceiling ✓.

Example 5 — Arrhenius extraction (cell C6)

Step 1 — Take the log ratio to cancel $A$ . $ln \frac{v _{2}}{v _{1}} = - \frac{E _{a}}{k} (\frac{1}{T _{2}} - \frac{1}{T _{1}})$ Why this step? We don't know the prefactor $A$ ; dividing the two equations and taking $ln$ makes it disappear, isolating $E_{a}$ .

Step 2 — Plug numbers. $ln 2 = 0.6931, \frac{1}{923} - \frac{1}{873} = - 6.204 \times 1 0^{- 5} K^{- 1}$ Why this step? Both temperature reciprocals are needed; the difference is negative (higher $T$ has smaller $1/ T$ ), which will cancel the minus sign.

Step 3 — Solve. $E_{a} = \frac{k l n 2}{- ( \frac{1}{T _{2}} - \frac{1}{T _{1}} )} = \frac{( 8.617 \times 1 0 ^{- 5} ) ( 0.6931 )}{6.204 \times 1 0 ^{- 5}} = 0.9627 eV$ Why this step? Rearranging isolates $E_{a}$ ; the two minus signs cancel to give a positive activation energy, as it must be.

Verify: Substitute back: $A e^{- E_{a} / (k T_{2})} / A e^{- E_{a} / (k T_{1})} = e^{- E_{a} (1/ T_{2} - 1/ T_{1}) / k}$ . With $E_{a} = 0.9627$ this gives $e^{0.693} = 2.0$ = $v_{2} / v_{1}$ ✓. $E_{a} \approx 0.96 eV$ is physically sane (real poly-Si is $1.5$ – $2 eV$ ). ✓

Example 6 — Pressure lever, LPCVD (cell C7)

Step 1 — Scale $h_{g}$ . $h_{g, new} = h_{g, old} \times 100 = 8 \times 100 = 800 cm/s$ Why this step? $h_{g} \propto D \propto 1/ P$ , so a $100 \times$ pressure drop multiplies $h_{g}$ by $100$ . This is the whole reason LPCVD exists.

Step 2 — Identify the new regime. Now $k_{s} = 200 ≪ h_{g} = 800$ , so we are reaction-limited (cell C1). Why this step? The bottleneck flipped from delivery to chemistry; in this regime thickness no longer follows gas-flow non-uniformity → better uniformity.

Step 3 — Flux ratio (note $C_{g}$ scales with $P$ in real life, but the problem fixes $C_{g}$ , so we compare only the resistance prefactors). $\frac{F _{new}}{F _{old}} = \frac{\frac{k _{s} h _{g, new}}{k _{s} + h _{g, new}}}{\frac{k _{s} h _{g, old}}{k _{s} + h _{g, old}}} = \frac{\frac{200 \cdot 800}{1000}}{\frac{200 \cdot 8}{208}} = \frac{160}{7.692} = 20.8$ Why this step? The exact series prefactors tell the truth: raising a bottleneck resistance helps a lot, but once we cross into reaction limitation the remaining $k_{s}$ caps the gain.

Verify: New effective velocity $160 cm/s$ is below the ceiling $k_{s} = 200$ ✓ (reaction-limited endpoint). Ratio $20.8 < 100$ ✓ — you never get the full pressure factor because the other resistance now dominates. ✓

Example 7 — Real-fab word problem: trench fill (cell C8)

Step 1 — Aspect ratio. $AR = \frac{depth}{width} = \frac{1.0}{0.20} = 5$ Why this step? Fill difficulty scales with AR; AR = 5 is deep and narrow, the classic voiding regime.

Step 2 — Half-angle seen from the bottom. $tan θ = \frac{half-width}{depth} = \frac{0.10}{1.0} = 0.10 \Rightarrow θ = arctan (0.10) = 5.7 1^{\circ}$ Why this step? $tan θ = opposite / adjacent$ on the right triangle whose vertical leg is the depth and horizontal leg is the half-width (see figure). A tiny $θ$ means the bottom is starved of line-of-sight flux.

Step 3 — Conclusion. Because $θ \approx 6^{\circ}$ is minuscule, PVD flux at the bottom is a small fraction of the mouth flux → the overhang closes first → void. CVD gas reaches all walls and reacts everywhere ( $F$ set by local $k_{s}, h_{g}$ , not geometry) → conformal fill. Why this step? This is exactly why tungsten plugs (from $WF_{6} + H_{2}$ ) are CVD, not sputtered — see Step Coverage & Conformality.

Verify: $tan (5.7 1^{\circ}) = 0.100$ ✓; AR $= 5$ ✓. The narrow cone quantifies the line-of-sight shadowing that dooms PVD in deep holes. ✓

Example 8 — Exam twist: misleading diagnostics (cell C9)

Step 1 — Match each response to its regime.

Strong response to flow / pressure → transport-limited (rate set by $h_{g}$ ).
Strong response to temperature → reaction-limited (rate set by $k_{s} = k_{0} e^{- E_{a} / k T}$ ).

Why this step? The parent's two diagnostic signatures are the only reliable classifier; you read whichever knob moves the rate most.

Step 2 — Apply. Flow moved rate $40%$ ; temperature only $8%$ . The dominant lever is flow → transport-limited (cell C2). Why this step? The small $8%$ temperature effect is the residual $T$ -dependence of $h_{g}$ (diffusivity rises weakly with $T$ ), not the exponential Arrhenius signature.

Step 3 — Fix the reasoning. The student confused "any temperature effect" with "reaction control." Only a strong exponential $T$ -response (like the rate doubling over $50 K$ in Example 5) signals reaction limitation. Why this step? Guards against the exact trap in the parent's mistake callout.

Verify: In Example 5, $50 K$ doubled the rate ( $+ 100%$ ) → reaction-limited. Here $50 K$ gave only $+ 8%$ → not reaction-limited ✓. Flow sensitivity ( $+ 40%$ ) is the transport-limited fingerprint ✓.

Recall Self-test (hidden)

Which cell has $F = \frac{1}{2} k_{s} C_{g}$ ? ::: C3 — the crossover, where $k_{s} = h_{g}$ . As $h_{g} \to \infty$ , what does $F$ approach? ::: $k_{s} C_{g}$ — the pure reaction-limited ceiling (cell C5). Strong response to gas flow but weak to $T$ means which regime? ::: Transport-limited (cell C2). Why does dropping pressure 100× raise $h_{g}$ 100×? ::: $h_{g} \propto D \propto 1/ P$ , so $P$ down 100× → $D$ up 100× → $h_{g}$ up 100×. A deep narrow trench (AR = 5) voids under which deposition method and why? ::: PVD — line-of-sight flux starves the bottom (tiny visible angle) while the mouth pinches shut.