Chemical vapor deposition (CVD): Exercises — Hardware deep dive

Before we start, let me build the one master equation everything below leans on — not just state it — so no symbol is used unnamed.

First, why "two resistances in series"?

Picture a single reactant molecule that must complete a two-stage journey to become part of the film:

Stage A — delivery. It has to cross the boundary layer, the thin blanket of nearly-still gas hugging the wafer. Crossing it is a slow diffusion trudge.
Stage B — reaction. Once it arrives at the surface, it has to actually react (decompose and bond into the solid). That chemistry takes its own finite time.

These two stages happen one after the other, so the molecule's total delay is the sum of the two delays — exactly like water flowing through two narrow pipes joined end to end: the total resistance to flow is the sum. This is the mechanical reason the resistances add in series.

Now the electrical picture that makes it quantitative:

Figure — Chemical vapor deposition (CVD)

Let us derive it in three honest steps.

Step 1 — write each stage's flux (what and why). Delivery flux is proportional to the concentration it must cross: $F_{deliver} = h_{g} (C_{g} - C_{s}) .$ Why this form? Bigger gas-to-surface difference $C_{g} - C_{s}$ ⇒ steeper "hill" for diffusion ⇒ faster crossing. $h_{g}$ is the proportionality: the gas-phase mass-transfer coefficient.

Reaction flux is proportional to how much reactant is actually sitting at the surface: $F_{react} = k_{s} C_{s} .$ Why this form? More reactant present ( $C_{s}$ ) ⇒ more reacts per second. $k_{s}$ is the surface reaction rate constant.

Step 2 — steady state couples them (what and why). In steady state nothing piles up at the surface: everything delivered is consumed. So the two fluxes must be equal, and that common value is the flux $F$ : $h_{g} (C_{g} - C_{s}) = k_{s} C_{s} = F .$ Why? Conservation — the surface is not a storage tank.

Step 3 — eliminate the hidden $C_{s}$ (what and why). Solve the equality for the surface concentration: $C_{s} = \frac{h _{g}}{h _{g} + k _{s}} C_{g},$ then substitute into $F = k_{s} C_{s}$ : $F = \frac{k _{s} h _{g}}{k _{s} + h _{g}} C_{g} = \frac{C _{g}}{\frac{1}{k _{s}} + \frac{1}{h _{g}}}$ Why the right-hand form? Because $\frac{1}{k _{s}} + \frac{1}{h _{g}}$ is literally the sum of two series resistances $R = R_{react} + R_{deliver}$ , and $F = C_{g} / R$ is Ohm's law with concentration as voltage. The coupling between mass transfer and surface reaction is now explicit: the smaller of $k_{s}, h_{g}$ (the larger resistance) dominates the sum.

And the temperature knob:

How $h_{g}$ depends on pressure and temperature

$h_{g}$ is set by the gas diffusivity $D$ — the "spreadiness" of the gas, i.e. how fast molecules random-walk through it (units $cm^{2} / s$ ). A larger $D$ means faster crossing of the boundary layer, so $h_{g} \propto D$ . From kinetic theory: $D \propto \frac{T ^{α}}{P} (α \approx 1.5 - 2), hence h_{g} \propto \frac{T ^{α}}{P} .$ So two honest nuances:

Pressure: $h_{g} \propto 1/ P$ . Lower pressure ⇒ larger $D$ ⇒ larger $h_{g}$ . This is the lever LPCVD pulls.
Temperature: $h_{g}$ is not flat — it rises mildly as $T^{1.5}$ or so. But this is a gentle power-law climb, utterly dwarfed by the exponential climb of $k_{s}$ . So relative to $k_{s}$ , $h_{g}$ looks nearly flat, which is why it becomes the ceiling at high $T$ .

Figure s02 is your map for the whole page: blue = $h_{g}$ (delivery, mild $T^{1.5}$ rise), yellow = $k_{s}$ (chemistry, exponential explosion), red = the actual growth rate $\propto \frac{k _{s} h _{g}}{k _{s} + h _{g}}$ — it hugs whichever curve is lower. The green knee is where control passes from chemistry (left) to delivery (right).

Level 1 — Recognition

Recall Solution

(a) CVD — the reaction happens on all exposed surfaces (conformal), so the hole fills from walls and bottom. See Step Coverage & Conformality. (b) PVD — PVD is line-of-sight; atoms mostly land on the top rim and seal it, trapping a void.

Recall Solution

Mass-transport resistance $= 1/ h_{g}$ (crossing the boundary layer) and surface-reaction resistance $= 1/ k_{s}$ (the chemistry). Total resistance $R = \frac{1}{h _{g}} + \frac{1}{k _{s}}$ , and $F = C_{g} / R$ .

Recall Solution

Reaction-limited ( $k_{s} ≪ h_{g}$ ) is temperature sensitive because $k_{s}$ obeys Arrhenius (exponential in $T$ ). Mass-transport-limited ( $h_{g} ≪ k_{s}$ ) is flow/pressure sensitive because it depends on delivery ( $h_{g} \propto T^{1.5} / P$ ) — strongly on $P$ , only mildly on $T$ .

Level 2 — Application

Recall Solution

$k_{eff} = \frac{100 \times 25}{100 + 25} = \frac{2500}{125} = 20 cm/s .$ Units check: $\frac{( cm/s ) ( cm/s )}{cm/s} = cm/s$ ✓ — same units as $k_{s}$ and $h_{g}$ . Since $h_{g} = 25 ≪ k_{s} = 100$ , transport is the bottleneck ( $k_{eff} \approx h_{g}$ ). We are mass-transport-limited. Notice $k_{eff} = 20$ sits just below $h_{g} = 25$ , confirming delivery is the choke point.

Recall Solution

$k_{eff} = \frac{100 \times 50}{100 + 50} = \frac{5000}{150} = 33. \overline{3} cm/s .$ Ratio $= \frac{100/3}{20} = \frac{5}{3} \approx 1.667$ . So halving pressure (doubling $h_{g}$ ) raised the rate by $\times 1.67$ — a sub-linear response, because we are moving out of transport limitation toward the reaction-limited plateau set by $k_{s}$ .

Recall Solution

$t = \frac{120 nm}{8 nm/min} = 15 min .$

Level 3 — Analysis

Recall Solution

Flow-insensitive ⇒ not transport-limited. Strongly $T$ -sensitive ⇒ chemistry ( $k_{s}$ ) controls ⇒ reaction-limited. For uniformity you want this regime: since delivery ( $h_{g}$ , which varies with local flow) is not the bottleneck, small flow non-uniformities across the wafer do not print into thickness. So run low pressure (LPCVD) to keep $h_{g}$ high and stay here.

Recall Solution

Take the log of the ratio to kill the unknown $A$ : $ln \frac{v _{2}}{v _{1}} = - \frac{E _{a}}{k} (\frac{1}{T _{2}} - \frac{1}{T _{1}}) .$ Left side: $ln 2 = 0.6931$ . Bracket: $\frac{1}{923} - \frac{1}{873} = 0.0010834 - 0.0011455 = - 6.206 \times 1 0^{- 5} K^{- 1}$ . So $E_{a} = \frac{0.6931 \times ( 8.617 \times 1 0 ^{- 5} )}{6.206 \times 1 0 ^{- 5}} \approx 0.962 eV .$ (Real poly-Si is $\approx 1.5$ – $2$ eV; the method is what matters.)

Recall Solution

The rate is $\frac{k _{s} h _{g}}{k _{s} + h _{g}} C_{g}$ . As $T \to$ high, $k_{s} \to \infty$ (exponential) while $h_{g}$ climbs only mildly ( $T^{1.5}$ ), so $\frac{k _{s} h _{g}}{k _{s} + h _{g}} \to h_{g}$ . The chemistry becomes so fast it's no longer the bottleneck — gas can only be delivered at rate $h_{g}$ . The rate rides the gentle $h_{g}$ ceiling. That's the "flat" top (mildly rising, not exactly flat): a series system can never beat its weakest link.

Level 4 — Synthesis

Recall Solution

Pick LPCVD (low-pressure CVD). Reasoning chain:

Uniformity requires the reaction-limited regime (flow non-uniformity must not print into thickness).
Reaction-limited means $k_{s} ≪ h_{g}$ , so we need a large $h_{g}$ .
Since $h_{g} \propto D \propto 1/ P$ , lowering pressure raises $h_{g}$ . LPCVD (low $P$ ) does exactly this.
Bonus: at low $P$ the mean free path is long, gas fills the tube evenly, so a batch of 200 wafers stacked vertically all see the same chemistry-limited rate. APCVD (atmospheric-pressure CVD) at $P = 1 atm$ gives small $h_{g}$ → transport-limited → thickness tracks flow → non-uniform. LPCVD wins.

Recall Solution

Use PECVD (plasma-enhanced CVD). A plasma (energetic electrons/ions) supplies the activation energy that heat would otherwise provide, so the surface reaction proceeds at $< 400^{\circ} C$ — safely below the aluminum limit. In terms of Arrhenius: the plasma effectively lowers the barrier $E_{a}$ the reaction must climb, letting $k_{s}$ be large even at low $T$ .

Recall Solution

(a) Aspect ratio $= depth / width = 0.9/0.15 = 6$ . (b) In reaction-limited growth, delivery is not the bottleneck, so even the slow-to-reach trench bottom gets essentially the same reactant concentration as the top ( $C_{s} \approx C_{g}$ everywhere). Growth rate $\approx k_{s} C_{g}$ uniformly → conformal. In transport-limited growth (or PVD), reactant is consumed near the opening faster than it can diffuse deep; the bottom is starved → the top grows faster → mouth pinches off → void. Hence high-aspect trenches demand reaction-limited CVD.

Level 5 — Mastery

Recall Solution

(a) $k_{eff} = \frac{k _{s} h _{g}}{k _{s} + h _{g}} = \frac{5 \times 20}{5 + 20} = \frac{100}{25} = 4 cm/s .$ (b) $F = k_{eff} C_{g} = 4 \times (3 \times 1 0^{16}) = 1.2 \times 1 0^{17} cm^{- 2} s^{- 1} .$ (c) $v = \frac{F}{N} = \frac{1.2 \times 1 0 ^{17}}{5 \times 1 0 ^{22}} = 2.4 \times 1 0^{- 6} cm/s .$ Convert: $2.4 \times 1 0^{- 6} cm/s = 2.4 \times 1 0^{- 6} \times 1 0^{7} nm/s = 24 nm/s = 1440 nm/min .$ (d) $k_{s} = 5 ≪ h_{g} = 20$ , and $k_{eff} = 4 \approx k_{s}$ : reaction-limited.

Recall Solution

Need $\frac{k _{s} ( T _{2} )}{k _{s} ( T _{1} )} = 2$ : $ln 2 = - \frac{E _{a}}{k} (\frac{1}{T _{2}} - \frac{1}{T _{1}}) \Rightarrow \frac{1}{T _{2}} = \frac{1}{T _{1}} - \frac{k l n 2}{E _{a}} .$ $\frac{k l n 2}{E _{a}} = \frac{( 8.617 \times 1 0 ^{- 5} ) ( 0.6931 )}{1.8} = 3.318 \times 1 0^{- 5} K^{- 1} .$ $\frac{1}{T _{2}} = \frac{1}{873} - 3.318 \times 1 0^{- 5} = 1.14548 \times 1 0^{- 3} - 3.318 \times 1 0^{- 5} = 1.11230 \times 1 0^{- 3} .$ $T_{2} = 899.0 K$ (about $626^{\circ} C$ ). A modest $26 K$ rise doubles the chemistry — that steep sensitivity is the reaction-limited signature.

Recall Solution

(a) $k_{eff} = \frac{k _{s} h _{g}}{k _{s} + h _{g}} = \frac{h \cdot h}{h + h} = \frac{h ^{2}}{2 h} = \frac{h}{2} .$ (b) New $k_{eff} = \frac{h \cdot 1.1 h}{h + 1.1 h} = \frac{1.1 h ^{2}}{2.1 h} = \frac{1.1}{2.1} h = 0.5238 h .$ Rise $= \frac{0.5238 - 0.5}{0.5} = 0.0476 = 4.76%.$ (c) A $10%$ change in one resistance moved the rate by only $\approx 4.8%$ — less than half. In the intermediate regime both resistances matter, so tugging on only one gives diminishing returns. The lesson: when $k_{s} \approx h_{g}$ , you must move both knobs (raise $T$ to lift $k_{s}$ and lower $P$ to lift $h_{g}$ ) to get a strong response — or, better, deliberately push the process to one side (usually reaction-limited) so a single, predictable knob controls it.

Recall Solution

Oxidation vs CVD: thermal oxidation reacts oxygen with the wafer's own silicon ( $Si + O_{2} \to SiO_{2}$ ), so it eats into the substrate. CVD brings both atoms from the gas ( $SiH_{4}$ , $WF_{6}$ , etc.), depositing on top without consuming the wafer. Epitaxy: epitaxial CVD needs the deposited atoms to line up with the crystal lattice of the substrate. That alignment requires (i) an atomically clean surface (no oxide) so atoms bond to the true lattice, and (ii) high $T$ so arriving atoms have enough mobility to hop to correct lattice sites before getting buried. Ordinary CVD tolerates disorder (polycrystalline/amorphous films); epitaxy does not.

Recall One-line master check

Growth rate always follows the weakest link: $v = \frac{C _{g}}{N} \cdot \frac{k _{s} h _{g}}{k _{s} + h _{g}}$ . Low $T$ → chemistry weak → reaction-limited (uniform, $T$ -sensitive). High $T$ → delivery weak → transport-limited (mildly $T$ -rising, flow/pressure-sensitive).

First, why "two resistances in series"?

How hg​ depends on pressure and temperature

Level 1 — Recognition

Level 2 — Application

Level 3 — Analysis

Level 4 — Synthesis

Level 5 — Mastery

How $h_{g}$ depends on pressure and temperature