Yield, defect density, and binning: Visual walkthrough

4.3.21 · D2 · Hardware › Semiconductor Fabrication › Yield, defect density, and binning

Step 1 — Hum kya count kar rahe hain: rectangle par dust

WHAT. Poore wafer ko random killer specks se chidka hua imagine karo, aur ek single die par zoom karo.

WHY. Kisi bhi math se pehle, hume us event ke baare mein bilkul clear hona chahiye jise hum dhundh rahe hain: "is ek rectangle ne koi speck catch nahi kiya." Baaki sab machinery hai us ek probability ko compute karne ke liye.

PICTURE. Figure mein, orange dots defects hain jo wafer par bikhre hain. Teal rectangle hamaari die hai. Sirf ek sawaal poochho: kya koi dot teal ke andar land kiya?

Figure — Yield, defect density, and binning

Step 2 — Defect density $D_{0}$ : process kitna dirty hai?

WHAT. Hum "poora process kitna dusty hai" ko ek single number $D_{0}$ mein compress karte hain. Agar kisi region ka area $A$ hai, to woh region defects ka average count pakadta hai

$λ = defects per cm^{2} D_{0} \times cm^{2} of one die A .$

Units padho: $\frac{defects}{cm ^{2}} \times cm^{2} = defects$ . To $λ$ ek die par defects ka ek pure expected count hai. Hum ise $λ$ (Greek "lambda") naam dete hain kyunki yeh woh ek number hai jis par poori derivation pivot karti hai.

Multiply kyun karte hain? Density times area hamesha "is region mein kitna stuff hai" hota hai — same tarah jaise (log per km²) × (km²) = log. $D_{0}$ ek density hai; $A$ region hai; unka product count hai.

PICTURE. Figure mein usi dusty wafer par do dies hain alag-alag sizes ki. Badi die zyada dots overlap karti hai — to uska $λ = D_{0} A$ bada hai. Same $D_{0}$ , bada $A$ , zyada expected defects.

Step 3 — Die ko bahut saare tiny boxes mein slice karo

WHAT. Die ko $n$ equal tiny squares mein kaato, jahan $n$ bahut bada hai. Har box itna chota hai ki usme zyada se zyada ek defect aa sake — ya to usme ek speck hai ya nahi.

WHY aisa karte hain? Ek tiny yes/no box kuch aisa hai jis par hum plain counting se reason kar sakte hain. Hum ek mushkil sawaal ("poori die par kitne specks?") ko $n$ aasaan coin-flip sawaalon mein badal dete hain ("kya is box mein ek speck hai?"). Yahi woh trick hai jo hamare liye $e$ manufacture karegi.

Ek box ke hit hone ki chance. Die $n$ boxes mein spread hue $λ$ defects average karti hai, to har box $λ / n$ defects average karta hai. Kyunki ek box zyada se zyada ek hold karta hai, woh average hi uski hit hone ki probability hai:

$p = \frac{λ}{n} (tiny, kyunki n bahut bada hai) .$

PICTURE. Die ab $n$ chote cells ka ek grid hai. Kuch cells orange shade hain (hit), zyaatar empty hain (clean). Har cell ek independent coin flip hai jo "defect" land karti hai tiny probability $p = λ / n$ ke saath.

Step 4 — Ek die survive karti hai sirf tab jab HAR box clean ho

WHAT. Die tab jeeti hai jab saare $n$ boxes empty hon. Ek box clean hone ki probability $(1 - p)$ hai. Kyunki boxes independent hain (specks independently land karte hain), hum multiply karte hain:

$Y = box 1 clean (1 - p) \times (1 - p) \times \dots = (1 - p)^{n} = (1 - \frac{λ}{n})^{n} .$

Term by term:

$1 - p$ = probability ki ek box defect se bach jaye.
exponent $n$ = hume saare $n$ ko bachna hai, to hum $n$ copies multiply karte hain.
$p = λ / n$ substitute karne par single number $λ$ ke terms mein boxed expression milta hai.

WHY multiply karte hain? Independent events ke liye "AND" matlab unki probabilities multiply karo. "Box1 clean AND box2 clean AND … AND box $n$ clean" → $n$ equal factors ka product → ek power.

PICTURE. Figure mein $n$ green "clean" tiles line up hain; survival chance product left to right run karta hai. Agar koi single tile orange flip ho, poora product ek dead die ki taraf collapse ho jaata hai.

Step 5 — Boxes ko infinitely fine hone do: $e$ ka janam

WHAT. Hamare boxes ek approximation the (ek defect in principle ek box share kar sakta tha). Picture ko exact banane ke liye, boxes ko zero size tak shrink karo — matlab $n \to \infty$ . Hume yeh limit chahiye

$Y = lim_{n \to \infty} (1 - \frac{λ}{n})^{n} .$

WHY limit? Limit woh tool hai jo jawab deta hai "jaise main approximation ko perfect banata hoon, yeh value kya approach karti hai?" Yahan yeh artificial grid ko remove karta hai aur sach mein continuous answer deta hai.

Ek fact jo hume chahiye. Ek famous limit hai jo number $e \approx 2.71828$ ko define karta hai:

$lim_{n \to \infty} (1 + \frac{x}{n})^{n} = e^{x} .$

Yahan $x = - λ$ . Plug in karne par:

$Y = e^{- λ} = e^{- D_{0} A}$

Result ka symbol check:

$e$ — woh special growth constant jo yeh limit hamare haath mein deta hai.
$- λ = - D_{0} A$ — ek negative exponent, kyunki zyada expected defects matlab kam survival. Zyada dust → chota $Y$ .

PICTURE. Figure mein $(1 - λ / n)^{n}$ plot hai $n = 1, 2, 4, 10, 50$ ke liye jo $e^{- λ}$ par ek dashed horizontal line ki taraf climb kar raha hai. Tum literally staircase ko exponential value par settle hote dekhte ho.

Step 6 — $Y$ vs area ka shape: kyun bade dies ko saza milti hai

WHAT. $Y = e^{- D_{0} A}$ ko $A$ mein ek curve ke roop mein dekho ( $D_{0}$ fixed ke saath). Yeh $A = 0$ par $Y = 1$ se start hota hai aur decay karta hai, zero ko kabhi touch nahi karta.

WHY exponential, $1/ A$ nahi? Kyunki area double karna matlab survival factor ko khud se multiply karna:

$e^{- D_{0} (2 A)} = (e^{- D_{0} A})^{2} .$

Do guna bada die half yield nahi karta — yield squared ho jaati hai. 1 se neeche ke number ko square karna use halving se kaafi zyada tezi se shrink karta hai. Yahi poori wajah hai ki giant GPU dies yield bleed karti hain.

PICTURE. Plum curve $e^{- D_{0} A}$ ek dotted " $1/ A$ " curve ke saath jo log galti se expect karte hain. Dekho kaise sachi curve tezi se dive karti hai aur dono sharply disagree karte hain — classic mistake exposed.

Step 7 — Edge aur degenerate cases (formula sab mein survive karna chahiye)

Hum $Y = e^{- D_{0} A}$ ko har extreme par test karte hain taaki koi reader kabhi unseen scenario na dekhe.

Case	Input	$Y = e^{- D_{0} A}$ deta hai	Sanity
Perfect process	$D_{0} = 0$	$e^{0} = 1$	Har die jeeti hai ✅
Zero-area die	$A = 0$	$e^{0} = 1$	Ek point koi dust nahi pakadta ✅
Filthy / huge	$D_{0} A \to \infty$	$e^{- \infty} = 0$	Kuch survive nahi karta ✅
Chota $λ$	$D_{0} A ≪ 1$	$\approx 1 - D_{0} A$	≈ "kam defects, chota loss" ✅
Defects clump karte hain	(Poisson zyada pessimistic)	$(1 + \frac{D _{0} A}{α})^{- α}$ use karo	$e^{- D_{0} A}$ mein reduce hota hai jab $α \to \infty$ ✅

WHY aakhri do important hain.

Chote $λ$ ke liye, exponential ka opening slice $e^{- λ} \approx 1 - λ$ hai: intuitive "roughly $λ$ fraction lose karo." Mature nodes ke liye accha gut-check.
Clustering ke liye, real dust clumps mein aati hai, to kuch dies bahut saare defects absorb karti hain aur zyada dies clean rehti hain jo pure randomness predict karta hai usse — negative-binomial model (dekho Poisson distribution) $Y$ raise karta hai. Jab uska clustering knob $α \to \infty$ (clumping vanish) ho jaata hai to yeh hamare $e^{- D_{0} A}$ par wapas collapse ho jaata hai, confirm karta hai ki Poisson no-clustering limit hai.

PICTURE. Teen miniature panels: $D_{0} = 0$ (all-green die), $A \to 0$ (ek dot, hamesha clean), aur clustered dust (orange dots ek corner mein bunched baki ko spare karte hue).

Ek-picture summary

Poori derivation ek single canvas par: die → tiny boxes → "all clean" product → limit → $e^{- D_{0} A}$ curve, edge anchors marked ke saath ( $A = 0 \Rightarrow Y = 1$ , bada $A \Rightarrow Y \to 0$ ).

Recall Feynman: walkthrough plain words mein

Ek chip ek rectangle hai; killer dust specks wafer par randomly land karti hain. Is chance ko find karne ke liye ki hamaari chip saari dust se bache, hum ise ek zillion tiny boxes mein kaatate hain, har ek itna chota ki usme sirf ek speck aa sake ya nahi. Har box $1 - λ / n$ chance se clean rehta hai, jahan $λ = D_{0} A$ chip par expected specks ka average number hai. Chip tab jeeti hai jab har box clean ho, to hum woh chance khud se $n$ baar multiply karte hain: $(1 - λ / n)^{n}$ . Ab boxes ko infinitely tiny banao ( $n \to \infty$ ) — aur yeh exact expression $e^{- λ}$ ki famous definition hai. To $Y = e^{- D_{0} A}$ seedha nikal aata hai. Kyunki yeh exponential hai, chip ka area double karna survival halve nahi karta — use square karta hai, yahi wajah hai ki huge chips itni aasaani se marti hain. $D_{0} = 0$ ya $A = 0$ set karo aur $Y = 1$ milta hai (kuch maarne wala nahi); dust badhaao aur $Y \to 0$ . Agar dust clumps mein aaye, kuch chips mess absorb kar leti hain aur zyada survive karti hain jo pure randomness kehta hai usse — yahi negative-binomial correction hai, aur yeh quietly hamare formula mein wapas turn ho jaata hai jab clumping disappear ho.

Recall

Yield derivation mein number $e$ kahan aata hai? ::: Limit $lim_{n \to \infty} (1 - λ / n)^{n} = e^{- λ}$ se jab die ko infinitely many tiny clean-or-not boxes mein slice kiya jaata hai. Boxes par survival multiply kyun kiya jaata hai? ::: Die tab jeeti hai jab SAARE boxes clean hon, aur independent "AND" events apni probabilities multiply karte hain. Die area double karne par yield ka kya hota hai? ::: Survival factor square ho jaata hai: $e^{- D_{0} (2 A)} = (e^{- D_{0} A})^{2}$ — halving se kaafi bura. $λ$ kya hai aur kya yeh 1 se zyada ho sakta hai? ::: Ek die par defects ka expected count, $λ = D_{0} A$ ; haan, yeh 1 se zyada ho sakta hai — yeh count hai, probability nahi.

Connections

Parent — Yield, defect density, and binning
Poisson distribution — tiny-boxes limit hi Poisson $P (0)$ hai; yeh page ise haath se derive karta hai.
Process node scaling — chote nodes $A$ aur $D_{0}$ change karte hain, $Y$ ko shift karte hain.
Chip economics and cost per transistor — $Y$ cost per good die mein feed hota hai.
Chiplets and MCM — ek badi die ko choti pieces mein split karna Step 6 mein dikhaye exponential area penalty se ladta hai.