Chemical mechanical planarization (CMP): Worked examples

This page is a drill sheet. The parent note CMP gave you two tools:

$R R = k_{p} P v and ∣ v ∣ = ω d (when ω_{p} = ω_{w} = ω)$

Here $R R$ is the removal rate (how many nanometres of material vanish per second), $k_{p}$ the Preston coefficient (units $Pa^{- 1}$ , it swallows all the chemistry), $P$ the pressure (force per area, units Pascals), $v$ the relative sliding speed between wafer and pad (metres/second), $ω$ an angular speed (radians/second), and $d$ the offset between the wafer centre and the pad centre.

Before we start: think of $R R$ as a speed and the film you must remove as a distance. Every one of these problems is secretly "distance ÷ speed = time", dressed up.

The scenario matrix

Every CMP homework/exam question is one of these cells. We work an example for each.

Cell	What makes it tricky	Example
A. Direct forward	Given $P, v, k_{p}$ → find $R R$ and time	Ex 1
B. Inverse solve	Given a target time → solve for $P$ (or $v$ )	Ex 2
C. Unit landmine	Inputs in rpm, cm, kPa — convert first	Ex 3
D. Zero / degenerate	$v = 0$ (no spin) or $d = 0$ (wafer centred on pad)	Ex 4
E. Multi-layer stack	Two films with two different $k_{p}$ (selectivity, endpoint)	Ex 5
F. Trade-off / limiting	Double $P$ vs double $v$ ; what does Preston vs defects say	Ex 6
G. Word problem (geometry)	Velocity varies across a single-spin disk, then the two-spin fix	Ex 7
H. Exam twist	$R R$ depends on local pressure that changes as feature dishes	Ex 8

Now the worked cases. Guess the answer at each Forecast before reading on.

Worked example Example 1 — Cell A: direct forward

Oxide CMP. $k_{p} = 1.5 \times 1 0^{- 13} Pa^{- 1}$ , $P = 25 kPa$ , $v = 0.5 m/s$ . Remove a $600 nm$ oxide layer. Find $R R$ and the time $t$ .

Forecast: Will the rate be a few nm/s or a few hundred? Will the job take seconds or minutes?

Convert $P$ to SI. $25 kPa = 25000 Pa = 2.5 \times 1 0^{4} Pa$ . Why this step? Preston multiplies $k_{p}$ (in $Pa^{- 1}$ ) by $P$ ; the Pascals must cancel or the answer is garbage.
Apply Preston. $R R = (1.5 \times 1 0^{- 13}) (2.5 \times 1 0^{4}) (0.5) = 1.875 \times 1 0^{- 9} m/s$ . Why this step? This is the removal-rate law from the parent note.
Read in friendly units. $1.875 \times 1 0^{- 9} m/s = 1.875 nm/s$ (since $1 nm = 1 0^{- 9} m$ ). Why this step? Film thicknesses live in nm; matching units makes the next division clean.
Time = distance/rate. $t = \frac{600 nm}{1.875 nm/s} = 320 s$ . Why this step? Rate is constant, so total time is simply how many "nm-worth of seconds" you need.

Verify: Units of $R R$ : $Pa^{- 1} \cdot Pa \cdot m/s = m/s$ ✓. Multiply back: $1.875 nm/s \times 320 s = 600 nm$ ✓.

Worked example Example 2 — Cell B: inverse solve for pressure

Same slurry/tool as Ex 1 ( $k_{p} = 1.5 \times 1 0^{- 13} Pa^{- 1}$ , $v = 0.5 m/s$ ). Your boss wants the $600 nm$ oxide gone in exactly $200 s$ . What pressure $P$ do you need? Is that wise?

Forecast: You need it faster than Ex 1 (320 s → 200 s). Should $P$ go up or down, and by roughly what factor?

Required rate. $R R_{need} = \frac{600 nm}{200 s} = 3 nm/s = 3 \times 1 0^{- 9} m/s$ . Why this step? Fix the target first; everything else is solving Preston backwards for the unknown.
Invert Preston. $P = \frac{R R}{k _{p} v} = \frac{3 \times 1 0 ^{- 9}}{( 1.5 \times 1 0 ^{- 13} ) ( 0.5 )}$ . Why this step? Only $P$ is unknown, so isolate it algebraically.
Compute. $P = \frac{3 \times 1 0 ^{- 9}}{7.5 \times 1 0 ^{- 14}} = 4 \times 1 0^{4} Pa = 40 kPa$ .
Judge it. That's $40/25 = 1.6 \times$ the Ex-1 pressure. Higher $P$ → more dishing/scratches (Archard Wear Law says worn volume rises with load). Why this step? An exam wants the number and the engineering caveat — Preston is symmetric in $P$ and $v$ but defects are not (see Ex 6).

Verify: Plug back: $R R = (1.5 \times 1 0^{- 13}) (4 \times 1 0^{4}) (0.5) = 3 \times 1 0^{- 9} m/s$ → $t = 600/3 = 200 s$ ✓.

Worked example Example 3 — Cell C: unit landmine (velocity from rpm & cm)

Platen and carrier both spin at $ω = 60 rpm$ ; wafer-centre offset $d = 10 cm$ . With $k_{p} = 2 \times 1 0^{- 13} Pa^{- 1}$ and $P = 30 kPa$ , find $v$ and $R R$ .

Forecast: rpm sounds fast; will $v$ be near 0.1, 0.5, or 5 m/s?

rpm → rad/s. $ω = 60 \times \frac{2 π}{60} = 2 π \approx 6.283 rad/s$ . Why this step? "Revolutions per minute" is not SI. One revolution is $2 π$ radians; per minute means divide by 60 s.
cm → m. $d = 10 cm = 0.10 m$ . Why this step? Preston/velocity formulas are in metres; a stray factor of 100 is the classic exam trap.
Matched-speed velocity. Because $ω_{p} = ω_{w} = ω$ , the wafer-radius terms cancel and $∣ v ∣ = ω d = 6.283 \times 0.10 = 0.6283 m/s$ . Why this step? This is the uniformity result from the parent note — the same $v$ everywhere on the wafer.
Removal rate. $R R = (2 \times 1 0^{- 13}) (3 \times 1 0^{4}) (0.6283) = 3.770 \times 1 0^{- 9} m/s = 3.77 nm/s$ .

Verify: Dimensional check on step 3: $rad$ is dimensionless, so $rad/s \times m = m/s$ ✓. Sanity: $2 π \approx 6.28$ , times 0.1 ≈ 0.63 ✓.

Worked example Example 4 — Cell D: zero & degenerate inputs

Two failure/edge cases on the same tool. (a) The platen jams: $v = 0$ . (b) An intern mounts the wafer dead-centre on the pad, $d = 0$ , while both still spin at $ω$ .

Forecast: In which case does nothing get polished? Is (a) and (b) the same outcome or different?

Case (a), $v = 0$ . $R R = k_{p} P (0) = 0$ . Why this step? No sliding means abrasive particles never drag across the surface — no wear events, no matter how hard you press. Pure static load ≠ polishing.
Case (b), $d = 0$ with matched spin. $∣ v ∣ = ω d = ω \cdot 0 = 0$ , so again $R R = 0$ . Why this step? When the wafer centre sits exactly on the pad's spin axis and speeds match, the whole wafer's sliding cancels to zero — the wafer just co-rotates like a passenger on a merry-go-round it's glued to.
Contrast them. (a) is no relative motion of the pad; (b) is no offset. Different causes, identical Preston result: $R R = 0$ . Why this step? Recognising two physical roads to the same math is exactly what "cover every degenerate case" means.
The escape from (b). Real tools also give the carrier a small oscillating sweep across the pad, so effective $d \neq = 0$ on average and slurry keeps refreshing. Why this step? Shows the design fix, not just the failure.

Verify: Both cases give $R R = 0$ ; any positive thickness would need infinite time ( $t = Δ h /0 \to \infty$ ), consistent with "never removed."

Worked example Example 5 — Cell E: two-layer stack (selectivity / endpoint)

Damascene Process copper CMP. First you polish $500 nm$ of copper overburden, then you must clear a $30 nm$ tantalum barrier underneath. $P = 20 kPa$ , $v = 0.5 m/s$ . Preston coefficients: $k_{p, Cu} = 3 \times 1 0^{- 13} Pa^{- 1}$ , $k_{p, Ta} = 0.5 \times 1 0^{- 13} Pa^{- 1}$ . Find total time and the selectivity $S = R R_{Cu} / R R_{Ta}$ .

Forecast: The barrier is thin (30 nm) but hard (small $k_{p}$ ). Will clearing it take a negligible or a surprisingly large slice of total time?

Convert $P$ to SI. $20 kPa = 20000 Pa = 2 \times 1 0^{4} Pa$ . Why this step? This sheet's own Cell C warning: $k_{p}$ is in $Pa^{- 1}$ , so pressure must be in Pascals before any multiplication.
Copper rate & time. $R R_{Cu} = (3 \times 1 0^{- 13}) (2 \times 1 0^{4}) (0.5) = 3 \times 1 0^{- 9} m/s = 3 nm/s$ . $t_{Cu} = 500/3 = 166.7 s$ . Why this step? Each layer has its own $k_{p}$ , so each gets its own Preston evaluation.
Barrier rate & time. $R R_{Ta} = (0.5 \times 1 0^{- 13}) (2 \times 1 0^{4}) (0.5) = 0.5 \times 1 0^{- 9} m/s = 0.5 nm/s$ . $t_{Ta} = 30/0.5 = 60 s$ . Why this step? Same $P, v$ ; only the material coefficient changes — that's the whole point of selectivity.
Total time. $t = 166.7 + 60 = 226.7 s$ . Why this step? The two polishes happen in sequence, so times add.
Selectivity. $S = \frac{R R _{Cu}}{R R _{Ta}} = \frac{k _{p, Cu}}{k _{p, Ta}} = \frac{3 \times 1 0 ^{- 13}}{0.5 \times 1 0 ^{- 13}} = 6$ . Why this step? Because $P$ and $v$ are shared, they cancel in the ratio — selectivity is purely a chemistry/material property. High $S$ means the barrier acts as a natural stop for endpoint detection.

Verify: A 30 nm barrier eating 60 s of a 227 s job (≈26%) is not negligible — the slow barrier dominates far beyond its thickness, exactly why selectivity matters. Ratio check: $500/166.7 = 3$ , $30/60 = 0.5$ , so $S = 3/0.5 = 6$ ✓.

Worked example Example 6 — Cell F: double

$P$ vs double $v$ (trade-off / limiting) Baseline $R R_{0} = 3 nm/s$ . Marketing wants $2 \times$ throughput. Option ①: double $P$ . Option ②: double $v$ . What does Preston predict, and what breaks?

Forecast: Preston says both give $6 nm/s$ — so are they truly interchangeable?

Preston, both options. $R R = k_{p} P v$ is linear in each, so doubling either factor gives $2 R R_{0} = 6 nm/s$ . Mathematically identical. Why this step? Establish that the rate formula cannot distinguish them.
Defect physics of ①. Contact stress scales with $P$ . From Archard Wear Law, deeper penetration → more scratches; the soft pad also bends further into wide lines → dishing grows with load. Why this step? Preston hides these; the exam tests whether you know the hidden cost.
Defect physics of ②. Higher $v$ keeps contact stress fixed (good) but can fling slurry off / starve the interface → poorer uniformity and heating. Why this step? Every knob has a downside; naming ②'s failure mode shows balanced judgement.
Engineering call. Prefer a modest $v$ bump plus a better oxidiser (raise $k_{p}$ via Slurry Chemistry and Colloids) — that gets speed without extra mechanical stress. Why this step? $k_{p}$ is the "free" throughput lever because it doesn't raise load.

Verify: $2 \times 3 = 6$ for both options numerically ✓; the tiebreaker is defects, not the number — matching the parent note's "Preston treats them symmetrically, defect physics does not."

Worked example Example 7 — Cell G: word problem with geometry (single vs matched spin)

Only the platen spins at $ω = 5 rad/s$ ; the carrier is held still ( $ω_{w} = 0$ ). Set up a coordinate line running through both the pad centre and the wafer centre. Let $d = 12 cm$ be the wafer-centre offset (its distance from the pad centre) and $R = 15 cm$ the wafer radius. A general point on the wafer, measured along this line from the wafer centre, is called $r$ and ranges from $- R$ (the wafer edge nearest the pad centre) to $+ R$ (the edge farthest from it). For any such point, let $ρ$ be its distance from the pad centre — so $ρ = ∣ d + r ∣$ . Find the sliding speed at $r = - R$ and $r = + R$ . Then say what matched spin ( $ω_{w} = ω$ ) fixes.

Figure below: the large white circle is the pad, spinning about its centre (yellow dot). The blue circle is the wafer, its centre offset by $d$ . The two pink dots are the near edge ( $r = - R$ ) and far edge ( $r = + R$ ); the pink arrows are their sliding-speed vectors — notice the far-edge arrow is much longer, because that point sits farther from the pad's spin axis.

Forecast: With only the platen spinning, do inner and outer edges of the wafer see the same speed or different speeds?

Speed of any point on a spinning pad. A point at distance $ρ$ from the pad centre moves at $∣ v ∣ = ω ρ$ (rigid rotation: farther out = faster). Why this step? On a single rigid spinning disk, every point traces a circle about the axis; the bigger the circle (radius $ρ$ ), the more distance covered per second, so speed grows linearly with $ρ$ . This is precisely the "outer points move faster" effect that ruins uniformity if left uncorrected.
Near edge, $r = - R$ : $ρ = ∣ d + r ∣ = ∣12 + (- 15) ∣ = ∣ - 3∣ = 3 cm = 0.03 m$ . Speed $= ω ρ = 5 \times 0.03 = 0.15 m/s$ . Why this step? This wafer point actually lands on the other side of the pad axis ( $d + r < 0$ ); only the magnitude $ρ$ sets the speed, so we take the absolute value.
Far edge, $r = + R$ : $ρ = ∣ d + r ∣ = ∣12 + 15∣ = 27 cm = 0.27 m$ . Speed $= ω ρ = 5 \times 0.27 = 1.35 m/s$ . Why this step? Compare the two edges to expose the spread.
The non-uniformity. $1.35$ vs $0.15 m/s$ — a 9× difference. By Preston $R R \propto v$ , the far edge polishes 9× faster → badly non-flat wafer. Why this step? Translates the velocity spread into the real defect (removal non-uniformity).
The fix: matched spin. Set $ω_{w} = ω$ . Then the wafer-radius ( $r$ ) terms cancel and every point sees $∣ v ∣ = ω d = 5 \times 0.12 = 0.60 m/s$ , independent of $r$ . Why this step? Shows the exact reason tools rotate carrier and platen together — uniform $v$ ⇒ uniform $R R$ ⇒ Photolithography gets a flat surface within its Depth of Focus (Optics).

Verify: Ratio $1.35/0.15 = 9 = (d + R) / (d - R) = 27/3$ ✓. Matched-spin value $5 \times 0.12 = 0.6 m/s$ , independent of $r$ ✓.

Worked example Example 8 — Cell H: exam twist (pressure redistributes as a feature dishes)

A wide copper line and the neighbouring oxide start coplanar. Model: the pad carries total average pressure $P_{0} = 20 kPa$ . Because copper is softer, the pad presses it harder — say local pressure on copper is $P_{Cu} = 1.5 P_{0}$ and on oxide $P_{ox} = 0.7 P_{0}$ (the extra load shifts to the softer region). Using $k_{p}$ equal for both ( $= 2 \times 1 0^{- 13} Pa^{- 1}$ ) and $v = 0.5 m/s$ , how much faster does the copper recede? This is why dishing happens.

Forecast: Will the copper recess grow or shrink over time relative to oxide?

Local pressures in SI. $P_{Cu} = 1.5 \times 20 kPa = 30 kPa = 3 \times 1 0^{4} Pa$ ; $P_{ox} = 0.7 \times 20 kPa = 14 kPa = 1.4 \times 1 0^{4} Pa$ . Why this step? Convert to Pascals before Preston, exactly as in Ex 1 and Ex 5 — never feed kPa into a $Pa^{- 1}$ coefficient.
Local rates. $R R_{Cu} = k_{p} P_{Cu} v = (2 \times 1 0^{- 13}) (3 \times 1 0^{4}) (0.5) = 3 \times 1 0^{- 9} m/s = 3 nm/s$ . $R R_{ox} = (2 \times 1 0^{- 13}) (1.4 \times 1 0^{4}) (0.5) = 1.4 \times 1 0^{- 9} m/s = 1.4 nm/s$ . Why this step? Preston is a local law — feed it the local pressure, not the average.
Dishing growth. Copper sinks faster than oxide by $3 - 1.4 = 1.6 nm/s$ . After $10 s$ the copper centre sits $1.6 \times 10 = 16 nm$ below the oxide — that dip is dishing. Why this step? Turns the rate gap into a measurable defect depth.
Why it self-worsens then self-limits. As copper dishes, the pad bends in and the pressure imbalance can grow (worse dishing) — but once copper drops enough, the pad may lift off the copper centre, cutting $P_{Cu}$ toward zero and slowing further loss. Why this step? Every "cover the limiting behaviour" answer needs the eventual saturation, not just the runaway.

Verify: $R R_{Cu} = 3$ , $R R_{ox} = 1.4$ nm/s; gap $1.6$ nm/s; $\times 10 s = 16 nm$ dishing ✓. Same $k_{p}, v$ means the rate ratio equals the pressure ratio: $3/1.4 = 30/14 \approx 2.143$ ✓.

All eight cells of the scenario matrix are now covered: direct (Ex 1), inverse (Ex 2), unit conversion (Ex 3), degenerate zeros (Ex 4), multi-layer selectivity (Ex 5), the $P$ -vs- $v$ trade-off (Ex 6), the geometry word problem (Ex 7), and the local-pressure exam twist (Ex 8). Whatever the exam throws, it is one of these — or a combination of two.

Recall Quick self-test across the matrix

Which cell has $R R = 0$ for two different physical reasons? ::: Cell D — either $v = 0$ (platen jammed) or $d = 0$ (wafer centred, matched spin). In Ex 5, why does selectivity $S$ not depend on $P$ or $v$ ? ::: They are shared by both materials and cancel in the ratio $R R_{Cu} / R R_{Ta} = k_{p, Cu} / k_{p, Ta}$ . Preston says doubling $P$ and doubling $v$ are identical — what breaks the tie? ::: Defect physics: high $P$ causes scratches/dishing; high $v$ risks slurry starvation.