Exercises — Chemical mechanical planarization (CMP)

(a) Slurry chemistry — oxidizers / pH buffers react with the top material to form a soft film. Without it you only scratch. (b) Abrasive particles (silica, ceria, alumina) — they supply the "cut rate." (c) The pad — its stiffness sets whether it bridges over valleys (global planarity) or bends into every dip (only local). See Slurry Chemistry and Colloids for the particle side.

Unit check: ${\text{Pa}}^{-1}\cdot \text{Pa}\cdot \text{m/s}=\text{m/s}$. ✓

Step 1 — rate (why: Preston directly gives $RR$). $RR=k_{p}Pv=(1.5\times 10^{-13})(4\times 10^4)(0.5)=3.0\times 10^{-9}\text{m/s}=3.0\text{nm/s}.$ Step 2 — time (why: constant rate ⇒ time = distance / rate). $t=\frac{\Delta h}{RR}=\frac{600\text{nm}}{3.0\text{nm/s}}=\boxed{200\text{s}}.$

Step 1 — convert to SI (why: Preston/geometry need $\text{rad/s}$, not rpm). $\omega =50\times \frac{2\pi }{60}=5.236\text{rad/s}.$ Step 2 — matched-speed result (why: with ${\omega }_p={\omega }_w$ the wafer-radius terms cancel, $|v|=\omega d$ everywhere). $v=\omega d=5.236\times 0.10=\boxed{0.524\text{m/s}}.$

Step 1 — required rate. $RR=\frac{500\text{nm}}{100\text{s}}=5.0\text{nm/s}=5.0\times 10^{-9}\text{m/s}.$ Step 2 — invert Preston (why: $k_p$ is the only unknown). $k_p=\frac{RR}{Pv}=\frac{5.0\times 10^{-9}}{(2.5\times 10^4)(0.7)}=\boxed{2.86\times 10^{-13}{\text{Pa}}^{-1}}.$

A wafer point at position $\vec{r}$ (measured from the wafer center) sits at $\vec{d}+\vec{r}$ from the platen center. Its velocity relative to the platen is $\vec{v}={\omega }_p(\hat{z}\times (\vec{d}+\vec{r}))-{\omega }_w(\hat{z}\times \vec{r}).$ Why two terms: the platen carries the point along at ${\omega }_p$; the carrier spins the point about the wafer center at ${\omega }_w$ — subtract because we want relative motion. Set ${\omega }_p={\omega }_w=\omega$: $\vec{v}=\omega (\hat{z}\times \vec{d})+\omega (\hat{z}\times \vec{r})-\omega (\hat{z}\times \vec{r})=\omega (\hat{z}\times \vec{d}).$ The two $\vec{r}$ terms cancel exactly, leaving $\vec{v}=\omega (\hat{z}\times \vec{d})$, whose magnitude is $|v|=\omega d$ — independent of $\vec{r}$. Every point moves at the same speed (see figure below). Uniform $v$ ⇒ uniform $RR$ ⇒ flat removal. This is why real tools match speeds.

With ${\omega }_w=0$: $\vec{v}=\omega (\hat{z}\times (\vec{d}+\vec{r}))$, magnitude $|v|=\omega |\vec{d}+\vec{r}|$.

• Center ($\vec{r}=0$): $|v_c|=\omega d$.
• Edge, aligned with $d$ ($|\vec{d}+\vec{r}|=d+R$): $|v_e|=\omega (d+R)$. Ratio (why: it exposes the non-uniformity that matched speeds cure): $\frac{v_e}{v_c}=\frac{d+R}{d}=1+\frac{R}{d}.$ For a $300\text{mm}$ wafer ($R=0.15\text{m}$) at $d=0.10\text{m}$: ratio $=1+1.5=\boxed{2.5}$. Edge polishes $2.5\times$ faster than center → a dished, non-uniform wafer. That severe variation is exactly why the carrier is driven, not left idle.

Since $RR=k_{p}Pv$ and only $v$ is free, $v$ must have doubled. Why this route is safer: doubling $v$ leaves contact stress (∝ $P$) unchanged, so scratch and dishing risk (which grow with load) do not increase — dishing is a pressure-driven, pad-bending effect. Doubling $P$ instead would double the mechanical stress and worsen dishing/erosion (relevant in the Damascene Process where wide copper lines dish). The trade-off cost of higher $v$ is possible slurry starvation (slurry flung off), a uniformity — not a stress — problem.

Step 1 — required rate. $RR=\frac{1.0\times 10^{-6}}{90}=1.111\times 10^{-8}\text{m/s}.$ Step 2 — velocity from geometry. $\omega =45\times \frac{2\pi }{60}=4.712\text{rad/s}$, so $v=\omega d=4.712\times 0.12=0.5655\text{m/s}.$ Step 3 — solve Preston for $P$ (why: $P$ is the only free knob left). $P=\frac{RR}{k_{p}v}=\frac{1.111\times 10^{-8}}{(2.5\times 10^{-13})(0.5655)}=7.86\times 10^4\text{Pa}=78.6\text{kPa}.$ Step 4 — verdict. $78.6\text{kPa}>P_{\text{max}}=35\text{kPa}$ ⇒ NOT safe. You cannot hit 90 s this way; either lower the target rate (longer time), raise $v$ (faster $\omega$ or larger $d$), or improve $k_p$ (better chemistry). Chemistry is the cleanest fix — it raises $RR$ with no extra mechanical stress.

(a) $S=\frac{6}{0.5}=\boxed{12}.$ (b) Barrier lost in $10\text{s}$: $0.5\text{nm/s}\times 10\text{s}=\boxed{5\text{nm}}.$ Why it's a natural stop: when polishing breaks through Cu into the barrier, the rate suddenly drops by a factor of $12$. The wafer "self-brakes," giving a wide time window before you over-thin the barrier — the friction/reflectance change also feeds Interconnects and Metallization endpoint detectors. Low selectivity (say $S\approx 1.2$) would keep eating downward at nearly full speed, leaving no margin.

(a) Use $\delta =cPw$ with $P=3\times 10^4\text{Pa}$.

• $w=100\mu \text{m}=1.0\times 10^{-4}\text{m}$: $\delta =(4.0\times 10^{-8})(3\times 10^4)(1.0\times 10^{-4})=1.2\times 10^{-7}\text{m}=\boxed{120\text{nm}}.$
• $w=1\mu \text{m}=1.0\times 10^{-6}\text{m}$: $\delta =(4.0\times 10^{-8})(3\times 10^4)(1.0\times 10^{-6})=1.2\times 10^{-9}\text{m}=\boxed{1.2\text{nm}}.$ Wide lines dish $100\times$ more — the pad has more span to sag into. (b) Spec: $\delta =cPw\le 20\text{nm}$. Solve for $w$: $w\le \frac{\delta }{cP}=\frac{2.0\times 10^{-8}}{(4.0\times 10^{-8})(3\times 10^4)}=1.667\times 10^{-5}\text{m}=\boxed{16.7\mu \text{m}}.$ Wider lines need lower $P$ or dummy-fill patterning to survive the spec.

Step 1 — pressure ceiling from dishing (why: dishing is the tighter, defect constraint). $P_{\text{max}}=\frac{\delta }{cw}=\frac{1.5\times 10^{-8}}{(4.0\times 10^{-8})(2.0\times 10^{-5})}=1.875\times 10^4\text{Pa}=18.75\text{kPa}.$ Step 2 — required rate. $RR=\frac{1.0\times 10^{-6}}{120}=8.333\times 10^{-9}\text{m/s}.$ Step 3 — needed velocity at $P=P_{\text{max}}$ (why: with $P$ pinned, $v$ is the last knob). $v=\frac{RR}{k_{p}P}=\frac{8.333\times 10^{-9}}{(2.5\times 10^{-13})(1.875\times 10^4)}=1.778\text{m/s}.$ Step 4 — convert $v=\omega d$ to rpm. $\omega =\frac{v}{d}=\frac{1.778}{0.12}=14.81\text{rad/s}\Rightarrow \omega =14.81\times \frac{60}{2\pi }=\boxed{141.5\text{rpm}}.$ Verdict: feasible only if the tool reaches ~142 rpm without slurry starvation. Otherwise the honest engineering answer is: raise $k_p$ (better chemistry) so the same rate needs less speed — the recurring lesson that chemistry is the free lunch.

Preston in words ::: Removal rate = Preston coefficient × pressure × relative velocity, i.e. $RR=k_{p}Pv$. Why match platen and carrier speeds ::: So the wafer-radius terms cancel and $v=\omega d$ is uniform → uniform, flat removal. What sets the pressure ceiling ::: The defect (dishing/erosion) limit, not the rate equation.