Extreme UV (EUV) lithography — Hinglish notes

4.3.8 · Hardware › Semiconductor Fabrication

EUV ki zaroorat kyun hai?

Problem kya hai? Chip features (transistor gates, wires) 20 nm se neeche shrink hote ja rahe hain. Sabse chhoti feature jo aap print kar sakte ho (resolution) woh light ki wavelength $λ$ se decide hoti hai. Puraani lithography mein $λ = 193$ nm (ArF excimer laser) tha. Aap 193 nm light se 15 nm feature cleanly print nahi kar sakte bina bahut mushkil tricks ke.

Hume kaise pata ki resolution $λ$ par depend karta hai? Rayleigh criterion se, minimum resolvable half-pitch ke liye:

$CD = k_{1} \frac{λ}{NA}$

Chaliye isko sirf quote karne ki jagah derive karte hain ki yeh form kyun aata hai.

Formula Diffraction se resolution limit derive karna

Pehla principle: Ek photomask ek diffraction grating hai jiska pitch $p$ hai. Light iske through jaate waqt diffraction orders mein bend hoti hai, angles $θ$ par, grating equation se: $p sin θ_{m} = mλ .$ Kyun: Pattern ko reproduce karne ke liye, lens ko kam se kam 0th aur 1st diffracted orders pakadne chahiye — agar sirf 0th pakde, toh koi interference fringe nahi banega aur image blank hogi. Toh lens ko angle $θ_{1}$ accept karna chahiye jahan $p sin θ_{1} = λ$ . Lens ki angle collect karne ki ability hai numerical aperture $NA = n sin θ_{m a x}$ . $sin θ_{1} = NA$ set karte hain (marginal capture): $p_{m i n} = \frac{λ}{NA} .$ Pitch ka aadha hissa printed feature hai (Critical Dimension). Ek process factor $k_{1}$ ke saath (jisme sab real-world cheezein bundle hain: partial coherence, resist, tricks; theoretical floor $k_{1} = 0.25$ ): $CD = k_{1} \frac{λ}{NA}$

Takeaway (the 80/20): CD shrink karne ke liye aap (1) $λ$ kam kar sakte ho, (2) NA badha sakte ho, ya (3) $k_{1}$ kam kar sakte ho. EUV sabse bade lever, $λ$ par attack karta hai — 193 nm se 13.5 nm par drop karke.

EUV itna mushkil kyun hai? (Har cheez 13.5 nm absorb karti hai)

13.5 nm par kya badalta hai? Photon energy hai: $E = \frac{h c}{λ} = \frac{1240 eV\cdotnm}{13.5 nm} \approx 92 eV .$ Yeh ionizing radiation hai — yeh atoms se electrons kick karta hai. Toh:

Koi lens nahi. Glass aur har transparent material EUV absorb karta hai. Aapko refractive lenses ki jagah reflective mirrors use karne padte hain.
Koi hawa nahi. Hawa (N₂, O₂, H₂O) EUV ko centimeters mein absorb kar leti hai. Poora beam path vacuum mein hona chahiye.
Ordinary mirrors nahi. Aam metal mirrors 13.5 nm par sirf ~kuch % reflect karte hain. Aapko Bragg multilayer mirrors chahiye.

Formula Mirrors kyun ~40 Mo/Si layers ke stack hote hain (Bragg reflection)

Ek single interface EUV par almost kuch reflect nahi karta. Lekin agar aap Molybdenum + Silicon ke bahut se patले bilayers stack karo, toh har interface thoda sa reflect karta hai, aur yeh reflections phase mein add hote hain jab Bragg condition poori hoti hai: $2 d cos θ = mλ$ Kyun: $d$ distance par layers se reflect hoti waves $2 d cos θ$ extra path travel karti hain; constructive interference ke liye yeh poori wavelength ke barabar hona chahiye. $λ = 13.5$ nm near-normal incidence ke liye solve karne par $d \approx 7$ nm bilayers milte hain. ~40 stack karne par, chhoti reflections milkar ek peak reflectivity tak pohonchti hain jo sirf 70% per mirror hai.

Consequence: system mein ~10 mirrors ke saath, throughput $\approx 0. 7^{10} \approx 3%$ . Zyaadatar light waste ho jaati hai, isliye EUV ko bahut powerful source ki zaroorat hai.

13.5 nm light kaise generate hoti hai? (Laser-Produced Plasma)

Source kya hai? Tin (Sn) droplets ko ek high-power CO₂ laser maarta hai, jisse ~200,000 °C par ek plasma banta hai jo EUV emit karta hai.

Yeh kaise kaam karta hai (clever double-pulse):

Molten tin droplets (~25 µm) ki ek stream ~50,000/second ki speed se girती hai.
Ek kamzor pre-pulse har droplet ko pancake shape mein flatten karta hai (better target).
Ek powerful main pulse (CO₂ laser, ~kW) isko Sn plasma mein vaporize kar deta hai.
Sn ions (Sn⁹⁺–Sn¹⁴⁺) exactly 13.5 nm par ek line emit karte hain.
Ek collector mirror EUV ko scanner ki taraf gather karta hai.

Tin kyun? Iske multiply-ionized states ki strong emission exactly 13.5 nm par hoti hai, jo Mo/Si mirror reflectivity peak se match karti hai. Yeh ek cosmic coincidence hai jisne EUV ko feasible banaya.

Poora light path (WHAT / WHY har part)

Stage	Kya hai	Kyun
Source	Sn plasma, CO₂ laser	Intense 13.5 nm pane ka sirf yahi tarika
Collector	multilayer ellipsoidal mirror	EUV focus karta hai, vacuum mein hona chahiye
Illuminator	multilayer mirrors	beam ko mask par shape karta hai
Mask	reflective, patterned absorber on Mo/Si	EUV mask ke through nahi ja sakta
Projection optics	6 aspheric mirrors	pattern ko wafer par 4× de-magnify karta hai
Wafer	EUV photoresist	pattern record karta hai

Forecast-then-Verify

Feynman

Recall 12-saal ke bachche ko samjhao (click to reveal)

Socho tum ek spray can se tiny dots paint kar rahe ho. Agar tumhara nozzle mota hai, toh tumhare dots mote aur blurry honge. Tinier dots paint karne ke liye tumhe thinner nozzle chahiye. Chip-making mein, "light" spray hai, aur uski wavelength nozzle ki motai hai. Normal light ka nozzle mota hai. EUV light ka nozzle super-thin hai (14× thinner!) toh yeh tiny transistors paint karta hai. Catch yeh hai: yeh special light itni nazuk hai ki aam hawa ise kha jaati hai aur normal mirrors ise nigal jaate hain — toh poori machine ek vacuum box hai special stacked mirrors ke saath, aur light banane ke liye hum girte hue tin droplets par laser maarte hain jisse Sun se bhi zyada garam ek tiny star banta hai. Wild hai, lekin isi tarah tumhare phone ka chip banta hai.

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

EUV lithography mein kaunsi wavelength use hoti hai?

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

EUV optics reflective (mirrors) kyun hone chahiye, lenses ki jagah?

EUV 13.5 nm par saare transparent materials (glass) absorb kar lete hain, toh refraction possible nahi; sirf reflection kaam karta hai.

Poora EUV beam path vacuum mein kyun hona chahiye?

Hawa (O₂, N₂, H₂O) 13.5 nm light ko short distances mein strongly absorb kar leti hai, toh koi bhi gas beam ko khatam kar degi.

Rayleigh resolution formula batao aur har term define karo.

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

Derive karo ki lens ko 1st diffraction order kyun pakadna chahiye.

Mask ek grating ki tarah kaam karta hai (p sinθ = mλ); sirf 0th order pakadne par koi interference fringes nahi bante → blank image. Pattern reconstruct karne ke liye ≥0th aur 1st orders chahiye.

EUV light kaise generate hoti hai?

Laser-Produced Plasma: ek CO₂ laser girte hue molten tin (Sn) droplets ko vaporize karke ~200,000°C plasma banata hai jo 13.5 nm emit karta hai.

EUV source material ke roop mein tin kyun use hota hai?

Multiply-ionized Sn (Sn⁹⁺–Sn¹⁴⁺) ke emission lines exactly 13.5 nm par hain, jo Mo/Si mirror reflectivity peak se match karte hain.

EUV mirrors kis cheez ke bane hote hain aur stacks kyun?

Alternating Mo/Si bilayers (~40 pairs, d≈7 nm) jo ek Bragg reflector banaate hain; single interfaces ~kuch reflect nahi karte, lekin bahut saare phase mein add hote hain (Bragg: 2d cosθ = mλ) → ~70% reflectivity.

EUV throughput itna bura kyun hota hai?

Har mirror sirf ~70% reflect karta hai; ~10 mirrors ke saath, total ≈ 0.7¹⁰ ≈ 3%, zyaadatar photons waste ho jaate hain aur high-power source ki zaroorat padti hai.

EUV photomask transmissive hai ya reflective?

Reflective — ek Mo/Si multilayer par patterned absorber; EUV iske off bounce hoti hai kyunki yeh through nahi ja sakti.

High-NA EUV kya hai aur CD par uska kya effect hai?

EUV scanners jisme NA 0.33 se 0.55 tak badhayi gayi hai; kyunki CD ∝ 1/NA, yeh CD ~40% shrink karta hai (e.g., 16 nm → ~10 nm).

EUV ke liye k₁=0.4, λ=13.5nm, NA=0.33 ke saath single-exposure CD compute karo.

CD = 0.4 × 13.5/0.33 ≈ 16 nm.