Extreme UV (EUV) lithography — Hardware notes

WHY do we need EUV at all?

WHAT is the problem? Chip features (transistor gates, wires) keep shrinking below 20 nm. The smallest feature you can print (the resolution) is set by the wavelength $\lambda$ of the light you use. Old lithography used $\lambda = 193$ nm (ArF excimer laser). You cannot print a 15 nm feature cleanly with 193 nm light without heroic tricks.

HOW do we know resolution depends on $\lambda$ ? From the Rayleigh criterion for the minimum resolvable half-pitch:

$\text{CD} = k_1 \frac{\lambda}{\text{NA}}$

Let's derive why this form appears rather than just quote it.

Formula Deriving the resolution limit from diffraction

First principle: A photomask is a diffraction grating with pitch $p$ . Light passing through bends into diffraction orders at angles $\theta$ given by the grating equation $p\sin\theta_m = m\lambda .$ Why: To reproduce the pattern, the lens must capture at least the 0th and 1st diffracted orders — if it only catches the 0th, no interference fringe forms and the image is blank. So the lens must accept angle $\theta_1$ where $p\sin\theta_1 = \lambda$ . The lens's ability to collect angle is the numerical aperture $\text{NA} = n\sin\theta_{\max}$ . Setting $\sin\theta_1 = \text{NA}$ (marginal capture): $p_{\min} = \frac{\lambda}{\text{NA}} .$ Half of the pitch is the printed feature (Critical Dimension). With a process factor $k_1$ (bundling everything real-world: partial coherence, resist, tricks; theoretical floor $k_1 = 0.25$ ): $\boxed{\text{CD} = k_1\frac{\lambda}{\text{NA}}}$

Takeaway (the 80/20): To shrink CD you can (1) lower $\lambda$ , (2) raise NA, or (3) lower $k_1$ . EUV attacks the biggest lever, $\lambda$ — dropping from 193 nm → 13.5 nm.

Worked example Why the jump to EUV is worth the pain

With DUV: $\lambda=193$ nm, NA $\approx 1.35$ (water immersion), $k_1\approx0.28$ : $\text{CD} = 0.28 \times \frac{193}{1.35} \approx 40 \text{ nm}$ Why this step? Plugging the best real DUV numbers gives the single-exposure floor ≈ 40 nm — you need multi-patterning (2–4 masks) to go below. With EUV: $\lambda=13.5$ nm, NA $=0.33$ , $k_1\approx0.4$ : $\text{CD} = 0.4\times\frac{13.5}{0.33} \approx 16\text{ nm}$ Why this step? Even with a smaller NA and worse $k_1$ , the 14× shorter wavelength wins overwhelmingly — one exposure prints what DUV needs 3–4 masks for.

WHY is EUV so hard? (Everything absorbs 13.5 nm)

WHAT changes at 13.5 nm? The photon energy is $E = \frac{hc}{\lambda} = \frac{1240\text{ eV·nm}}{13.5\text{ nm}} \approx 92\text{ eV}.$ That is ionizing radiation — it kicks electrons off atoms. So:

No lenses. Glass and every transparent material absorbs EUV. You must use reflective mirrors, not refractive lenses.
No air. Air (N₂, O₂, H₂O) absorbs EUV over centimeters. The entire beam path must be vacuum.
No normal mirrors. Ordinary metal mirrors reflect only ~a few % at 13.5 nm. You need Bragg multilayer mirrors.

Formula Why the mirrors are stacks of ~40 Mo/Si layers (Bragg reflection)

A single interface reflects almost nothing at EUV. But if you stack many thin bilayers of Molybdenum + Silicon, each interface reflects a tiny amount, and these reflections add in phase when the Bragg condition holds: $2d\cos\theta = m\lambda$ Why: waves reflected from layers spaced $d$ apart travel an extra path $2d\cos\theta$ ; constructive interference needs this to equal a whole wavelength. Solving for $\lambda=13.5$ nm near-normal incidence gives $d\approx 7$ nm bilayers. Stacking ~40 of them, the small reflections sum to a peak reflectivity of only about 70% per mirror.

Consequence: with ~10 mirrors in the system, throughput $\approx 0.7^{10}\approx 3\%$ . Most of the light is lost, which is why EUV needs a brutally powerful source.

HOW is 13.5 nm light generated? (Laser-Produced Plasma)

WHAT is the source? Tin (Sn) droplets are hit by a high-power CO₂ laser, creating a plasma at ~200,000 °C that emits EUV.

HOW does it work (the clever double-pulse):

A stream of molten tin droplets (~25 µm) falls at ~50,000/second.
A weak pre-pulse flattens each droplet into a pancake (better target).
A powerful main pulse (CO₂ laser, ~kW) vaporizes it into Sn plasma.
Sn ions (Sn⁹⁺–Sn¹⁴⁺) emit a line at exactly 13.5 nm.
A collector mirror gathers the EUV toward the scanner.

Why tin? Its multiply-ionized states happen to have strong emission right at 13.5 nm, matching the Mo/Si mirror reflectivity peak. A cosmic coincidence that made EUV feasible.

The full light path (WHAT / WHY each part)

Stage	What	Why
Source	Sn plasma, CO₂ laser	Only way to get intense 13.5 nm
Collector	multilayer ellipsoidal mirror	focuses EUV, must be in vacuum
Illuminator	multilayer mirrors	shapes beam onto mask
Mask	reflective, patterned absorber on Mo/Si	EUV can't pass through a mask
Projection optics	6 aspheric mirrors	de-magnify pattern 4× onto wafer
Wafer	EUV photoresist	records the pattern

Forecast-then-Verify

Feynman

Recall Explain to a 12-year-old (click to reveal)

Imagine you're painting tiny dots with a spray can. If your nozzle is fat, your dots are fat and blurry. To paint tinier dots you need a finer nozzle. In chip-making, "light" is the spray, and its wavelength is how fat the nozzle is. Normal light has a fat nozzle. EUV light has a super-thin nozzle (14× thinner!) so it paints teeny transistors. The catch: this special light is so delicate that ordinary air eats it and normal mirrors swallow it — so the whole machine is a vacuum box with special stacked mirrors, and to make the light we shoot a laser at falling tin droplets to make a tiny star hotter than the Sun. Wild, but that's how your phone chip gets built.

Connections

Rayleigh Criterion and Resolution
Deep UV (DUV) Immersion Lithography
Multi-patterning Techniques
Bragg Reflection and Multilayer Mirrors
Numerical Aperture
Photoresist Chemistry
Moore's Law and Feature Scaling
Photon Energy and Wavelength

Flashcards

What wavelength does EUV lithography use?

13.5 nm (photon energy ≈ 92 eV, in the soft-X-ray-like regime)

Why must EUV optics be reflective (mirrors) rather than lenses?

EUV at 13.5 nm is absorbed by all transparent materials (glass), so refraction is impossible; only reflection works.

Why must the entire EUV beam path be under vacuum?

Air (O₂, N₂, H₂O) strongly absorbs 13.5 nm light over short distances, so any gas would kill the beam.

State the Rayleigh resolution formula and define each term.

CD = k₁·λ/NA; CD = critical dimension, k₁ = process factor (~0.25 floor), λ = wavelength, NA = numerical aperture.

Derive why the lens must capture the 1st diffraction order.

The mask acts as a grating (p sinθ = mλ); capturing only the 0th order gives no interference fringes → blank image. You need ≥0th and 1st orders to reconstruct the pattern.

How is EUV light generated?

Laser-Produced Plasma: a CO₂ laser vaporizes falling molten tin (Sn) droplets into a ~200,000°C plasma that emits 13.5 nm.

Why is tin used as the EUV source material?

Multiply-ionized Sn (Sn⁹⁺–Sn¹⁴⁺) has strong emission lines at exactly 13.5 nm, matching the Mo/Si mirror reflectivity peak.

What are EUV mirrors made of and why stacks?

Alternating Mo/Si bilayers (~40 pairs, d≈7 nm) forming a Bragg reflector; single interfaces reflect ~nothing, but many add in phase (Bragg: 2d cosθ = mλ) → ~70% reflectivity.

Why does EUV throughput suffer badly?

Each mirror reflects only ~70%; with ~10 mirrors, total ≈ 0.7¹⁰ ≈ 3%, wasting most photons and demanding a high-power source.

Is an EUV photomask transmissive or reflective?

Reflective — a patterned absorber on a Mo/Si multilayer; EUV bounces off it because it can't pass through.

What is High-NA EUV and its effect on CD?

EUV scanners with NA raised from 0.33 to 0.55; since CD ∝ 1/NA, this shrinks CD by ~40% (e.g., 16 nm → ~10 nm).

Compute single-exposure CD for EUV with k₁=0.4, λ=13.5nm, NA=0.33.

CD = 0.4 × 13.5/0.33 ≈ 16 nm.

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho, chip banane ka basic funda simple hai: transistor jitne chote print karne hain, light ki wavelength utni hi choti chahiye. Purani lithography 193 nm deep-UV use karti thi, lekin ab features 15-20 nm se bhi neeche jaa rahe hain. Resolution ka formula hai CD = k₁·λ/NA — matlab CD chota karna hai to sabse bada lever hai λ ko chota karna. Isliye EUV aaya jo 13.5 nm wavelength use karta hai, yaani 14 guna chhoti — game changer!

Lekin problem yeh hai ki 13.5 nm light ki energy ~92 eV hoti hai, yeh soft X-ray jaisi hai. Yeh har cheez me absorb ho jaati hai — glass, hawa, sab kuch. Isliye lenses use nahi kar sakte (glass light kha jaata hai), sirf reflective Mo/Si multilayer mirrors chalte hain jo Bragg condition (2d cosθ = mλ) se kaam karte hain. Pura machine vacuum me hota hai kyunki hawa bhi light absorb karti hai. Aur mask bhi transmissive nahi, reflective hota hai — light us par se bounce hoti hai.

Light banana bhi kamaal ka hai: pighle hue tin ke droplets giraate hain, unpar taaqatwar CO₂ laser maarte hain, jisse ~200,000°C ka plasma banta hai jo exactly 13.5 nm emit karta hai. Tin isliye kyunki uske multiply-ionized states ka emission theek 13.5 nm par hai — Mo/Si mirror ke peak se match! Har mirror sirf ~70% reflect karta hai, aur ~10 mirrors ke baad sirf ~3% light bachti hai, isliye source bahut powerful chahiye. Yaad rakhna: "13 Tin Vampires Mirror Reflectively" — 13.5nm, Tin source, Vacuum, Reflective mirrors. Yeh technology aaj ke sabse advanced chips (5nm, 3nm nodes) banane ke liye essential hai.

Test yourself — Semiconductor Fabrication

Connections