Yield, defect density, and binning: Worked examples

4.3.21 · D3 · Hardware › Semiconductor Fabrication › Yield, defect density, and binning

Tumne parent page Yield, defect density, and binning par formulas dekhe hain. Ab hum har tarah ke numbers ke saath unhe grind karenge. Goal yeh hai: is page ke baad, koi bhi exam ya real-world case tumhe surprise nahi kar sakta, kyunki tum us shape ki problem pehle hi solve karke dekh chuke hoge.

Hum sirf teen tools use karte hain, sab parent page par bane hain:

Gross dies per wafer $DPW \approx \frac{π r ^{2}}{A} - \frac{π \cdot 2 r}{2 A}$ — kitne rectangles us round disk par fit hote hain.
Poisson yield $Y = e^{- D_{0} A}$ — chance ki ek die mein zero killer defects hain.
Clustered (negative-binomial) yield $Y = (1 + \frac{D _{0} A}{α})^{- α}$ — same idea jab dust clump karta hai.

Neeche sab kuch sirf yahi teen hain, corners mein push kiye hue.

The scenario matrix

Har yield problem ko is table mein ek point ki tarah socho. Har row ek class of input hai jo topic tumhare samne rakh sakta hai; last column us example ka naam batata hai jo use nail karta hai.

Cell	Kya cheez ise tricky banati hai	Covered by
Baseline	ordinary $D_{0}$ , ordinary $A$	Ex 1
Big-die punishment	area badhti hai → yield exponentially collapse hoti hai	Ex 2
Zero / degenerate input	$D_{0} = 0$ (perfect fab) ya $A \to 0$ (tiny die)	Ex 3
Limiting behaviour	$α \to \infty$ : clustered model zaroor Poisson ban jaata hai	Ex 4
Clustering vs Poisson	finite $α$ yield ko $e^{- D_{0} A}$ se upar uthata hai	Ex 5
Economics word problem	dollars per good die, real wafer cost	Ex 6
Reverse / solve-for-input	ek target yield diya gaya hai, allowed $D_{0}$ nikalo	Ex 7
Binning ladder	survivors ko speed grades mein baanto → revenue	Ex 8
Chiplet twist (exam)	small dies vs ek bada die, same silicon	Ex 9

Shuru karne se pehle units ka ek note: $D_{0}$ hai defects per cm², area $A$ cm² mein hai, toh $λ = D_{0} A$ ek pure number hai (defects). Yahi wajah hai ki yeh ek exponent mein baithh sakta hai — tum sirf ek bare number ko exponentiate kar sakte ho, kabhi bhi aisi cheez ko nahi jo units carry karti ho.

Ex 1 — Baseline (ordinary case)

Worked example Cell: Baseline

Ek 300 mm wafer ( $r = 15$ cm), die area $A = 1 cm^{2}$ , defect density $D_{0} = 0.5 /cm^{2}$ . Yield, gross dies, aur good dies nikalo.

Forecast: andaza karo — kya aadhe se zyaada dies kaam karenge? Kya 700 se zyaada ya kam gross dies honge?

Step 1 — expected defects per die. $λ = D_{0} A = 0.5 \times 1 = 0.5$ . Yeh step kyun? $Y = e^{- D_{0} A}$ ko pehle pure number $λ$ chahiye; yeh ek die par killers ki average count hai.

Step 2 — yield. $Y = e^{- 0.5} = 0.6065$ , lagbhag 61%. Yeh step kyun? $e^{- λ}$ Poisson probability hai zero defects ki — sirf yahi survivors hain.

Step 3 — gross dies. Naive $\frac{π r ^{2}}{A} = π \cdot 225 = 706.86$ . Edge loss $\frac{π \cdot 2 \cdot 15}{2} = 66.64$ . Toh $DPW \approx 640$ . Yeh step kyun? Half-cut dies ka rim dead area hai; ise subtract karna hume over-counting se rokta hai.

Step 4 — good dies. $640 \times 0.6065 \approx 388$ . Yeh step kyun? Good = gross $\times$ fraction jo survive karti hai.

Verify: $e^{- 0.5} \approx 0.6065$ $\frac{1}{2}$ aur $1$ ke beech hai ✓ (aadhe se zyaada survive karte hain, forecast se match karta hai). $706.86 - 66.64 = 640.2$ ✓, dimensionless (cm²/cm² cancel ho jaata hai). Good dies $388 < 640$ ✓.

Ex 2 — Big-die punishment

Worked example Cell: Big-die punishment

Same $D_{0} = 0.5$ , lekin ek GPU jisme $A = 4 cm^{2}$ hai. Iska yield Ex 1 se compare karo.

Forecast: area ×4 ho gayi. Kya yield quarter (×0.25) tak giregi, ya aur bura?

Step 1 — naya $λ$ . $λ = 0.5 \times 4 = 2$ . Kyun? Chaar guna area chaar guna expected defects sweep karta hai.

Step 2 — yield. $Y = e^{- 2} = 0.1353$ , lagbhag 13.5%. Kyun? Same zero-defect rule; exponent bas chaar guna ho gaya.

Step 3 — compare. Ex 1 tha $e^{- 0.5} = 0.6065$ . Ratio $\frac{0.1353}{0.6065} = 0.223$ , roughly ek paanchwa hissa, ek chautha nahi. Kyun? Kyunki $e^{- 2} = (e^{- 0.5})^{4}$ — yield area mein exponential hai, toh ×4 area survival factor ko 4th power tak uthata hai, ise kisi bhi linear guess se zyaada crush karta hai.

Verify: $(0.6065)^{4} = 0.1353$ ✓ — confirm karta hai ki yield ratios multiply hote hain, subtract nahi hote. Forecast (×0.25 guess karna) bahut optimistic tha — asli giravat zyaada steep hai. Dekho Process node scaling aur Chip economics and cost per transistor kyun yeh real design choices drive karta hai.

Ex 3 — Zero aur degenerate inputs

Worked example Cell: Zero / degenerate input

(a) Ek hypothetically perfect fab: $D_{0} = 0$ . (b) Ek infinitely tiny die: $A \to 0$ . Kaun sa yield?

Forecast: ek perfect fab 100% dena chahiye. Ek vanishing die koi defect collect nahi kar sakta — woh bhi 100%?

Step 1 — perfect fab. $λ = 0 \times A = 0$ , toh $Y = e^{0} = 1 = 100%$ . Kyun? Koi defects kahin bhi nahi → har die saaf hai. Formula yeh earn karta hai: $e^{0} = 1$ exactly.

Step 2 — vanishing die. Jaise $A \to 0$ , $λ = D_{0} A \to 0$ , toh $Y = e^{- D_{0} A} \to e^{0} = 1$ . Kyun? Ek point-sized die kisi speck ke liye koi target area present nahi karta. Yield perfect hone ki taraf jaata hai.

Step 3 — opposite extreme. Jaise $A \to \infty$ (ek die poore wafer jitni badi), $λ \to \infty$ aur $Y = e^{- \infty} \to 0$ . Kyun? Ek enormous die almost certainly kam se kam ek killer pakad lega.

Verify: $e^{0} = 1$ ✓ aur $e^{- λ} \to 0$ jaise $λ \to \infty$ ✓. Dono limits behave karte hain, aur do "perfect" cases ke forecast se match karte hain jabki teesra expose karte hain (giant die → 0).

Ex 4 — Limiting behaviour: clustered → Poisson

Worked example Cell: Limiting behaviour

Clustered model lo $Y = (1 + \frac{D _{0} A}{α})^{- α}$ jisme $D_{0} A = 2$ hai. Ise $α = 1, 10, 1000$ par compute karo aur dikhao ki yeh Poisson $e^{- 2} = 0.1353$ ke paas aata hai.

Forecast: jaise clustering kamzor hoti hai ( $α$ bada), kya clustered yield Poisson value ki taraf girna chahiye ya usse door jaana chahiye?

Step 1 — $α = 1$ . $Y = (1 + 2)^{- 1} = 1/3 = 0.3333$ . Kyun? Heavy clustering: defects kuch dies par pile ho jaate hain, baaki ko spare karte hain → high yield.

Step 2 — $α = 10$ . $Y = (1 + 0.2)^{- 10} = (1.2)^{- 10} = 0.1615$ . Kyun? Kamzor clustering, random scatter ke zyaada paas → yield Poisson ki taraf girta hai.

Step 3 — $α = 1000$ . $Y = (1 + 0.002)^{- 1000} = 0.1354$ . Kyun? Almost koi clustering nahi; identity $lim_{α \to \infty} (1 + x / α)^{- α} = e^{- x}$ kick in karta hai.

Verify: target $e^{- 2} = 0.13534$ . Sequence $0.3333 \to 0.1615 \to 0.1354$ iske upar march karta hai ✓. Yeh parent page ka sanity anchor hai: fancy model zaroor simple model mein collapse hona chahiye jab clumping vanish ho jaaye. Dekho Poisson distribution.

Ex 5 — Clustering vs Poisson (same die, do models)

Worked example Cell: Clustering vs Poisson

Ek GPU jisme $D_{0} A = 2$ hai. Pure Poisson yield ko clustered model se compare karo jisme $α = 2$ hai. Kaun sa zyaada hai, aur kitna?

Forecast: clustering defects ko kam dies par push karti hai — kya good fraction upar jaana chahiye ya neeche?

Step 1 — Poisson. $Y_{P} = e^{- 2} = 0.1353$ (13.5%). Kyun? Perfectly uniform scatter assume karte hue baseline.

Step 2 — clustered. $Y_{C} = (1 + 2/2)^{- 2} = 2^{- 2} = 0.25$ (25%). Kyun? $α = 2$ matlab moderate clumping; clumps damage concentrate karte hain, zyaada dies spare karte hain.

Step 3 — ratio. $\frac{0.25}{0.1353} = 1.85$ — surviving dies ki almost double count. Kyun? Real fabs naive Poisson se upar yields report karte hain exactly kyunki dust clumps karta hai; yahi wajah hai ki fabs practice mein negative-binomial use karte hain.

Verify: $2^{- 2} = 0.25$ ✓ aur $0.25 > 0.1353$ ✓ (clustering help karti hai, forecast se match karta hai). Yeh cost per good die ko directly change karta hai — Poisson estimate ke wasted silicon ka aadha.

Ex 6 — Economics word problem

Worked example Cell: Economics word problem

Ek finished 300 mm wafer ki cost $15{,}000 hai. Ex 1 ke numbers use karte hue ( $DPW = 640$ , $Y = 0.6065$ ), cost per good die nikalo. Phir ek naive "cost ÷ gross dies" se compare karo.

Forecast: kya true cost per good die $23 (=15000/640) ke paas hogi ya noticeably zyaada?

Step 1 — good dies. $DPW \times Y = 640 \times 0.6065 = 388.2$ . Kyun? Tum sirf working dies bechte ho; dead ones bhi tumhara paisa khaate hain.

Step 2 — cost per good die. $\dfrac{15000}{388.2}=\$ 38.64 $. * K y u n ? *$ \text{Cost}{\text{good}}=\dfrac{\text{Cost}{\text{wafer}}}{\text{DPW}\times Y}$ — poore wafer ki price survivors mein divide hoti hai.

Step 3 — naive comparison. Yield ignore karte hue: $\dfrac{15000}{640}=\$ 23.44 $. * K y u n ? * P e na l t y d ik ha t ahai :$ 38.64/23.44=1.65\times $z y aa d a, e x a c tl y$ 1/Y=1/0.6065=1.649$.

Verify: $1/0.6065 = 1.649$ aur $23.44 \times 1.649 = 38.65$ ✓ (rounding). Units: dollars/die throughout ✓. Forecast "noticeably higher" sahi hai — yield cost ko $1/ Y$ factor se inflate karti hai. Dekho Chip economics and cost per transistor.

Ex 7 — Reverse problem: input ke liye solve karo

Worked example Cell: Reverse / solve-for-input

Tumhare product ko kam se kam 80% yield chahiye. Die area $A = 1 cm^{2}$ . Tumhare process mein sabse bada defect density $D_{0}$ kaun sa ho sakta hai (Poisson)?

Forecast: 80% alive rakhne ke liye, kya $D_{0}$ 0.5 se kaafi kam hona chahiye (woh value jo sirf 61% deti thi)?

Step 1 — target likho. $0.80 = e^{- D_{0} \cdot 1}$ . Kyun? Yield formula ko requirement ke barabar set karo aur ulta solve karo.

Step 2 — exponential undo karo. Natural log lo: $ln 0.80 = - D_{0}$ . Logarithm woh tool hai jo $e^{(\cdot)}$ ko undo karta hai — yeh poochta hai "kaun sa exponent yeh number deta hai?" Yeh tool kyun? $D_{0}$ ek exponent ke andar trapped hai; sirf $ln$ ise free karta hai.

Step 3 — solve karo. $D_{0} = - ln 0.80 = 0.2231 /cm^{2}$ . Kyun? $ln 0.80 = - 0.2231$ , positive density paane ke liye negate karo.

Verify: $e^{- 0.2231} = 0.800$ ✓. Sanity: $0.2231 < 0.5$ , toh Ex 1 se cleaner process — higher yield chahne ke consistent ✓. Forecast se match karta hai. Yeh Photolithography / yield-ramp target hai jo ek fab chase karta hai.

Ex 8 — Binning ladder (survivors ko product ladder mein convert karna)

Worked example Cell: Binning ladder

388 good dies mein se, testing (dekho Wafer testing and probe) max-clock distribution dikhata hai: 25% 5.0 GHz hit karte hain, 50% 4.5 GHz hit karte hain, 25% sirf 4.0 GHz hit karte hain. Prices: $500 / $300 / $180. Total revenue aur average price per die nikalo.

Forecast: kya average middle price ($300) ke paas hogi ya premium tier se upar pull hogi?

Step 1 — har bin count karo. Premium $0.25 \times 388 = 97$ ; middle $0.50 \times 388 = 194$ ; low $0.25 \times 388 = 97$ . Kyun? Binning ek identical die design ko measured speed se SKUs mein split karta hai — same silicon, different grade.

Step 2 — revenue per bin. $97 \times 500 = 48, 500$ ; $194 \times 300 = 58, 200$ ; $97 \times 180 = 17, 460$ . Kyun? Har grade apni price par bikta hai; revenue = count × price.

Step 3 — totals. Total $=48{,}500+58{,}200+17{,}460=\$ 124{,}160 $. A v er a g e$ =124160/388=$320.0$. Kyun? Average price per good die measure karta hai ki binning ne value kitni recover ki.

Verify: $97 + 194 + 97 = 388$ ✓. Average $320 middle price $300 se thoda upar hai ✓ — premium tier se pull hua, forecast se match karta hai. Binning slow dies scrap karne se better hai: iske bina woh 97 low-tier dies ($17,460) lost ho jaate.

Ex 9 — Chiplet twist (exam-style)

Worked example Cell: Chiplet twist

Same silicon, do strategies $D_{0} = 0.5$ par: (A) ek monolithic die $A = 4 cm^{2}$ ; (B) chaar chiplets of $A = 1 cm^{2}$ each, baad mein joined. Us silicon ke fraction compare karo jo ek working part mein end up hota hai (assembly loss ignore karo).

Forecast: barabar total area ke chaar small dies vs ek bada die — kaun kam silicon waste karta hai?

Step 1 — monolithic yield. $Y_{A} = e^{- 0.5 \times 4} = e^{- 2} = 0.1353$ . Kahin bhi ek defect saare 4 cm² ko kill kar deta hai. Kyun? Ek killer defect poore large die ko doom kar deta hai.

Step 2 — per-chiplet yield. Har 1 cm² chiplet: $Y_{B} = e^{- 0.5} = 0.6065$ . Kyun? Ab ek defect sirf apna 1 cm² tile kill karta hai, apne neighbours ko nahi.

Step 3 — fraction of good silicon. Monolithic: area ka 13.5% survive karta hai. Chiplets: har tile 60.65% par independently survive karta hai, toh silicon area ka 60.65% good tiles yield karta hai. Kyun? Splitting defects ko isolate karta hai — yahi Chiplets and MCM ka core argument hai.

Verify: $e^{- 2} = 0.1353$ vs $e^{- 0.5} = 0.6065$ ✓, aur $0.6065 > 0.1353$ by a factor of $4.48$ ✓. Note karo $(e^{- 0.5})^{4} = e^{- 2}$ : chance ki saare chaar chiplets good hain monolithic yield ke barabar hai — lekin chiplets ke saath tum good ones rakhte ho set scrap karne ki jagah. Forecast confirm: splitting bahut kam silicon waste karta hai.

Figure — Yield, defect density, and binning

Recall

Recall Ek hi saanson mein nine leans

Baseline → plug in karo. Bada die → exponentiate down. Zero input → $e^{0} = 1$ . Infinite $α$ → Poisson mein collapse karo. Finite $α$ → yield upar jaati hai. Money → good dies se divide karo. Reverse → $ln$ lo. Binning → survivors ko priced grades mein split karo. Chiplets → area split karo defects isolate karne ke liye.

Recall Reveal drills

Yield 80% chahiye, $A = 1$ — max $D_{0}$ ? ::: $D_{0} = - ln 0.80 = 0.223 /cm^{2}$ Monolithic 4 cm² yield at $D_{0} = 0.5$ ? ::: $e^{- 2} = 0.135$ Chaar 1 cm² chiplets, har ek ki yield? ::: $e^{- 0.5} = 0.607$ Cost per good die agar wafer $15k, 640 gross, 61% yield? ::: $38.6 Clustered yield $D_{0} A = 2, α = 2$ ? ::: $2^{- 2} = 0.25$

Connections

Parent: Yield, defect density, and binning
Poisson distribution — $e^{- D_{0} A}$ ke peeche zero-defect probability.
Wafer testing and probe — woh speed distribution measure karta hai jo binning feed karta hai (Ex 8).
Chiplets and MCM — Ex 9 ka defect-isolation win.
Chip economics and cost per transistor — jahaan cost-per-good-die (Ex 6) land karta hai.