Physical vapor deposition (PVD - sputtering): Exercises

Before we start, one symbol table so nothing appears unearned:

Definition Symbols used on this page

$M_{1}$ — mass number of the incident ion (the marble you shoot), e.g. Argon $= 40$ .
$M_{2}$ — mass number of the target atom (the ball you knock loose).
$E$ — kinetic energy of the incident ion, measured in electron-volts (eV) — the energy one electron gains crossing a 1-volt gap.
$γ$ (gamma) — the fraction of the ion's energy that can be handed to a target atom in a head-on hit. A pure number between 0 and 1.
$U_{s}$ — surface binding energy: how much energy an atom needs to break free of the target surface, in eV.
$E_{t h}$ — threshold energy: the smallest ion energy that can eject even one atom.
$Y$ — sputter yield: atoms ejected per incident ion.
$J$ — ion flux: ions hitting each unit area of target per second.
$R$ — deposition rate: how fast the film grows.

Figure — Physical vapor deposition (PVD - sputtering)

The figure above plots $γ$ as the two masses change. Notice the peak sits exactly where $M_{1} = M_{2}$ — matched masses transfer the most energy, just like two identical billiard balls where the cue ball stops dead and hands over all its energy. We will use this picture repeatedly.

Level 1 — Recognition

Recall Solution L1.1

(c) momentum transfer. The ion acts like a marble; it drives a collision cascade through a few atomic layers, and statistically a surface atom gets kicked backwards out of the solid. No melting (that's evaporation) and no chemistry (that's Chemical Vapor Deposition).

Recall Solution L1.2

The target is the cathode, held at a large negative voltage. Negative attracts the positive Ar⁺ ions, which is exactly the acceleration that gives them enough energy to sputter.

Recall Solution L1.3

False. The magnets trap electrons (which are light and easily steered by a magnetic field) in a racetrack near the target — see Plasma Physics. Sputtered atoms are neutral and heavy; the field barely affects them. The magnets boost ionization, not atom steering.

Level 2 — Application

Recall Solution L2.1

$γ = \frac{4 ( 40 ) ( 64 )}{( 40 + 64 ) ^{2}} = \frac{10240}{10816} = 0.947$ WHAT we did: plugged masses into the match formula. WHY: $γ$ tells us the ceiling fraction of ion energy Cu atoms can receive. WHAT IT LOOKS LIKE: on the s01 curve, $M_{2} = 64$ against $M_{1} = 40$ is near — but not at — the peak, so $γ \approx 0.95$ , slightly below the perfect $1.0$ .

Recall Solution L2.2

$E_{t h} = \frac{U _{s}}{γ} = \frac{3.5}{0.947} \approx 3.7 eV$ WHY divide: an atom only escapes if the delivered energy $γ E$ exceeds $U_{s}$ . Setting $γ E = U_{s}$ and solving for $E$ gives the smallest ion energy that just barely works. In practice we run at hundreds of eV — far above this — to sit in the productive linear-yield regime.

Recall Solution L2.3

$R \propto Y \cdot J$ . Yield fixed, flux $\times 3$ , so $\frac{R _{n e w}}{R _{o l d}} = \frac{Y \cdot 3 J}{Y \cdot J} = 3.$ Deposition rate triples. The magnetron changed the ions-per-second term, not the atoms-per-ion term.

Level 3 — Analysis

Recall Solution L3.1

Compute each: $γ_{N e} = \frac{4 ( 20 ) ( 28 )}{4 8 ^{2}} = \frac{2240}{2304} = 0.972$ $γ_{A r} = \frac{4 ( 40 ) ( 28 )}{6 8 ^{2}} = \frac{4480}{4624} = 0.969$ $γ_{K r} = \frac{4 ( 84 ) ( 28 )}{11 2 ^{2}} = \frac{9408}{12544} = 0.750$ Neon wins narrowly ( $γ = 0.972$ ) because $M_{1} = 20$ sits closest to $M_{2} = 28$ — look at the s01 peak: efficiency is maximal at matched mass. Krypton is heavier and mismatched, so its $γ$ drops to $0.75$ . Analysis caveat: real recipes still favor Ar because it's cheap and inert; $γ$ isn't the only criterion, but on transfer efficiency alone the closest mass wins.

Recall Solution L3.2

Higher pressure = more gas atoms per volume = shorter mean free path (average distance an atom flies before hitting a gas molecule). Sputtered atoms now collide many times crossing to the wafer, so they (i) lose kinetic energy and (ii) arrive from random directions. Low-energy, randomly-arriving atoms don't pack tightly → porous, weakly-adhering film. The magnetron's whole point is to sustain the plasma at lower pressure so atoms fly straight and hit hard — denser film.

Recall Solution L3.3

Under DC, Ar⁺ ions land on the insulating surface and their positive charge has nowhere to drain (the insulator won't conduct it away). The surface charges positive, repels incoming Ar⁺, and the discharge dies. RF at 13.56 MHz flips polarity fast: on the negative half-cycle Ar⁺ ions arrive and sputter; on the positive half-cycle light, mobile electrons rush in and neutralize the accumulated charge. Net effect: sputtering continues indefinitely on an insulator.

Level 4 — Synthesis

Recall Solution L4.1

(a) Sputter — TiN is high-melting; momentum transfer works without melting, and gives good adhesion/stress control (see Thin Film Deposition). (b) TiN is conductive, so DC suffices and is cheaper. (RF would only be needed if the target were insulating.) (c) Magnetron — traps electrons, densifies the plasma, gives high deposition rate at low voltage/pressure and low substrate heating. (d) Low pressure — long mean free path ⇒ energetic, directional arrival ⇒ dense film with good step coverage. Each choice traces to one physical lever: melting-point → sputtering; conductivity → DC; ionization efficiency → magnetron; mean free path → low pressure.

Recall Solution L4.2

$γ \frac{E}{U _{s}} = 0.962 \times \frac{300}{3.4} = 0.962 \times 88.24 = 84.9$ Interpretation: the ion delivers about $0.962 \times 300 = 288.6$ eV to a surface atom — roughly $85$ times the binding energy. That huge overshoot is why real sputtering (hundreds of eV) sits comfortably in the productive regime, nowhere near the $\sim 3.5$ eV threshold. (This is a trend factor, not the literal atom count — real $Y$ includes a proportionality constant of order $1 0^{- 2}$ .)

Level 5 — Mastery

Recall Solution L5.1

Sputtering ejects atoms from the surface, and it does so via a collision cascade that must reach back up to the surface. At low energy the cascade sits in the top atomic layers — perfect. As energy climbs, the ion penetrates deeper before depositing its momentum. Past some point the cascade's energy is buried too far below the surface: it can't propagate back up to kick surface atoms loose, and instead just heats/implants the bulk (this is the regime Sputter Etching and ion implantation exploit). So yield saturates then declines — not because energy is wasted, but because it's deposited in the wrong place (too deep to help ejection). The optimum energy keeps the cascade shallow.

Recall Solution L5.2

Step 1 — clean metal: charge drains freely, stable DC plasma, high rate. Step 2 — thin oxide patches form: those patches are insulating, so Ar⁺ charge accumulates locally; the field there rises until micro-arcs discharge violently — arcing, which spits droplets and ruins the film. Step 3 — full oxide skin: behaves like the L3.3 insulator — charge can't drain, local regions repel ions, rate collapses and becomes unstable. Rescue: pulsed-DC briefly reverses polarity each cycle so electrons neutralize the built-up positive charge before it can arc; RF does the same continuously. Both restore charge balance on the insulating regions — exactly the mechanism that lets RF sputter true insulators. The unifying principle: any surface that can't drain its acquired positive charge will quench or arc under steady DC; you must periodically supply electrons.

Recall One-line self-check

Why does matched ion/target mass give the highest $γ$ ? ::: Because in a head-on elastic hit between equal masses the projectile stops dead and transfers all its energy — the $\frac{4 M _{1} M _{2}}{( M _{1} + M _{2} ) ^{2}}$ ratio equals 1 exactly when $M_{1} = M_{2}$ .