Yield, defect density, and binning: Exercises — Hinglish deep dive

4.3.21 · D4 · Hardware › Semiconductor Fabrication › Yield, defect density, and binning

Intuition Is page ko kaise use karein

Yeh problems ek ladder ki tarah chadhte hain. L1 sirf check karta hai ki aap sahi formula pehchante ho. L2 mein aapko numbers cleanly plug karne hote hain. L3 mein aapko compare karna ya reason karna hota hai ki kyun ek situation doosre se better hai. L4 kai ideas ko ek economic decision mein stitch karta hai. L5 open-ended mastery hai jahan aap ek design choice defend karte ho.

Yahan use kiya gaya har symbol parent note mein build kiya gaya tha: the parent topic. Shuru karne se pehle alphabet ka quick refresher:

$A$ = ek die ka area (cm² mein, taaki numbers chhote rahein).
$D_{0}$ = average fatal defects per cm² — process kitna dirty hai.
$λ = D_{0} A$ = ek die par land hone wale defects ki expected number.
$Y$ = yield = fraction of dies jo kaam karti hain (jinmein zero fatal defects hain).
$DPW$ = gross dies per wafer (defects se pehle kitne rectangles fit hote hain).
$α$ = clustering parameter (defects kitna clump karte hain).

Teen formulas jinpar hum rely karenge, sab parent mein derived hain: $Y_{Poisson} = e^{- D_{0} A}, Y_{cluster} = (1 + \frac{D _{0} A}{α})^{- α}, Cost_{good} = \frac{Cost _{wafer}}{DPW \times Y} .$

Level 1 — Recognition

Recall Solution 1.1

Humein KYA chahiye: $λ$ bas "average defects per die" hai. Parent ne isse $λ = D_{0} A$ ki tarah build kiya — density (per area) times area ek pure count deta hai. YEH kyun aur $Y$ kyun nahi: question mean count par ruk jaata hai. Yield baad mein aati hai, yeh poochh kar ki "is count ke zero hone ki kya chance hai?" $λ = D_{0} A = 0.4 \times 2 = 0.8 .$ Units cancel ho jaati hain: $\frac{defects}{cm ^{2}} \times cm^{2} = defects$ . Ek pure number, jaise ek count hona chahiye.

Recall Solution 1.2

Answer: (a) $e^{- D_{0} A}$ . Exactly $k$ events ki Poisson probability hai $P (k) = \frac{λ ^{k} e ^{- λ}}{k !}$ . $k = 0$ set karein: $\frac{λ ^{0} e ^{- λ}}{0 !} = e^{- λ} = e^{- D_{0} A}$ . (b) sirf $e^{- λ}$ ki Taylor series ke pehle do terms hain aur $D_{0} A > 1$ hone par negative ho jaata hai — probability ke liye impossible. (c) jab $D_{0} A \to 0$ tab blow up karta hai, 1 se upar yields deta hai — yeh bhi impossible hai.

Level 2 — Application

Recall Solution 2.1

Step 1 — mean: $λ = D_{0} A = 0.8$ (1.1 se). Step 2 — $e^{- λ}$ mein plug karein: Kyun? Yield = zero defects ki probability. $Y = e^{- 0.8} = 0.4493 \approx 44.9% .$ Toh har 100 dies mein se jo aap print karte ho, lagbhag 45 survive karti hain.

Recall Solution 2.2

Step 1 — radius: diameter $200 mm = 20 cm$ , isliye $r = 10 cm$ . Step 2 — naive count (circle area ÷ die area): $\frac{π r ^{2}}{A} = \frac{π \times 100}{0.5} = \frac{314.159}{0.5} = 628.3.$ Step 3 — edge-loss term (partial dies ki rim): $\frac{π 2 r}{2 A} = \frac{π \times 20}{1} = \frac{62.83}{1} = 62.83.$ Step 4 — subtract karein: $628.3 - 62.8 = 565.5 \Rightarrow 565$ dies. Subtract kyun? Round wafer, square dies — curved edge par ring mein adhi-katI dies hoti hain jo use nahi ho saktin.

Recall Solution 2.3

Step 1 — IS die ke liye yield: $λ = 0.4 \times 0.5 = 0.2$ , isliye $Y = e^{- 0.2} = 0.8187 \approx 81.9%.$ Chhoti die → defects ke liye chhota target → high yield. Step 2 — good dies: $DPW \times Y = 565 \times 0.8187 = 462.6 \approx 462$ good dies. Step 3 — cost per good die: Good se kyun divide karein, gross se kyun nahi? Aap poore wafer ka paisa dete ho chahe kitni bhi dies hon; sirf survivors paisa kamaate hain. $Cost_{good} = \frac{5000}{462.6} = $10.81 .$

Level 3 — Analysis

Recall Solution 3.1

Step 1 — dono likhein: $Y_{S} = e^{- 0.5 \cdot 1} = e^{- 0.5}$ , $Y_{B} = e^{- 0.5 \cdot 4} = e^{- 2}$ . Step 2 — exponents relate karein: $e^{- 2} = e^{- 0.5 \times 4} = (e^{- 0.5})^{4}$ . Yeh kyun kaam karta hai? Exponent laws: $e^{- D_{0} A}$ mein $A$ ko 4 se multiply karna 4 ko power ki tarah bahar nikaalata hai, factor ki tarah nahi. Yahi poora reason hai ki bade chips ke liye yield kyun collapse hoti hai. Step 3 — numbers: $Y_{S} = e^{- 0.5} = 0.6065$ (≈61%), $Y_{B} = e^{- 2} = 0.1353$ (≈13.5%). Interpretation: area ko chaar guna karna yield ko ek-chauthai nahi karta (yeh luck se $\approx 15%$ hota, lekin mechanism alag hai) — yeh survival factor ko 4th power tak raise kar deta hai. Figure mein steep exponential dekho: ek badi die ek bada target hai, aur area ka har extra bit death chance ko multiply karta hai.

Figure — Yield, defect density, and binning

Recall Solution 3.2

Poisson baseline: $Y = e^{- 2} = 0.1353$ . Clustered: $Y = (1 + \frac{2}{α})^{- α}$ .

$α = 1$ : $(1 + 2)^{- 1} = 3^{- 1} = 0.3333$ → 33.3%.
$α = 4$ : $(1 + 0.5)^{- 4} = 1. 5^{- 4} = 0.1975$ → 19.8%.
$α \to \infty$ : $e^{- 2} = 0.1353$ par collapse → 13.5% (sanity anchor). Trend: chhota $α$ = heavy clustering = defects kuch dies par pile ho jaate hain, baaki ko spare karte hain → highest yield. Jaise $α$ badhta hai, clustering fade hoti hai aur yield pessimistic Poisson floor ki taraf sink hoti hai. Dekho ki curve dashed Poisson line ki taraf neeche slide karta hai.

Level 4 — Synthesis

Recall Solution 4.1

Plan Mono. DPW: $\frac{π \cdot 225}{6} - \frac{π \cdot 30}{12} = \frac{706.86}{6} - \frac{94.25}{3.4641} = 117.81 - 27.21 = 90.6 \to 90$ . Yield: $λ = 0.3 \times 6 = 1.8$ , $Y = e^{- 1.8} = 0.1653$ . Good products = $90 \times 0.1653 = 14.88 \to 14$ . Cost per product $= \frac{8000}{14.88} = \boxed{\$ 537.6}$.

Plan Chiplet. DPW (per chiplet, $A = 3$ ): $\frac{706.86}{3} - \frac{94.25}{6} = 235.62 - 38.48 = 197.1 \to 197$ . Yield per chiplet: $λ = 0.3 \times 3 = 0.9$ , $Y = e^{- 0.9} = 0.4066$ . Good chiplets = $197 \times 0.4066 = 80.1 \to 80$ . Good products (har ek ko 2 chiplets chahiye) $= 80/2 = 40$ . Cost per product $= \frac{8000}{40} = \boxed{\$ 200}$.

Recommendation: Plan Chiplet. $200 vs $537.6 per working product par yeh kaafi sasta hai, kyunki area split karne se har chiplet ka $λ$ 1.8 se 0.9 par aa jaata hai aur yield $λ$ mein exponential hai: do chhote dies aksar ek bade die se kahin zyada baar survive karte hain. Yahi argument Chiplets and MCM ke peechhe hai. (Real life mein packaging/interconnect cost add hoti hai — Chip economics and cost per transistor dekho.)

Level 5 — Mastery

Recall Solution 5.1

Har core probability $q = 1 - p = 0.9$ se good hai. Good cores ki number binomial distribution follow karti hai: $P (k good) = (k 8) q^{k} p^{8 - k}$ . Binomial kyun, Poisson kyun nahi? Humare paas fixed, chhoti number of independent yes/no trials hain (8 cores), jo binomial setting hai — Poisson bahut saare rare trials ki limit hai.

(a) Perfect 8-core: sabhi cores good. $P (8) = q^{8} = 0. 9^{8} = 0.4305 \to 43.0% .$ (b) 6-core bin = exactly 6 ya 7 good (8-good top bin ne already claim kar liye): $P (7) = (7 8) q^{7} p^{1} = 8 \times 0. 9^{7} \times 0.1 = 8 \times 0.4783 \times 0.1 = 0.3826.$ $P (6) = (6 8) q^{6} p^{2} = 28 \times 0. 9^{6} \times 0.01 = 28 \times 0.5314 \times 0.01 = 0.1488.$ $P (6 or 7) = 0.3826 + 0.1488 = 0.5314 \to 53.1% .$ (c) Point: harvesting ke bina, sirf 43% dies (perfect wale) bikenge — baaki "failed" hain. Binning un 53.1% ko rescue karta hai jinmein 6–7 good cores hain: kharaab cores fuse off karo aur unhe saste 6-core SKU ki tarah becho. Total bikne wala fraction 43% se $43.0 + 53.1 = 96.1%$ tak jump karta hai. Yahi recovered revenue ka reason hai ki Wafer testing and probe sirf pass/fail nahi, per-block health measure karta hai — yeh decide karta hai ki har die kis bin mein jaati hai.

Recall Solution 5.2

Effect 1 (yield help karta hai): chhota area. Transistors shrink karne se $A$ shrink hota hai, aur $Y = e^{- D_{0} A}$ badhta hai jaise $A$ girta hai. Achha. Effect 2 (yield hurt karta hai): naye node par zyada $D_{0}$ . Ek brand-new node immature hota hai — tighter feature sizes ka matlab hai ab chhote specks bhi fatal count hote hain, aur process tune nahi hua hai, isliye $D_{0}$ spike karta hai. Kyunki $Y = e^{- D_{0} A}$ product $D_{0} A$ par depend karta hai, $D_{0}$ mein bada jump $A$ ki drop ko overwhelm kar sakta hai. Verdict: node ki life ke shuruaat mein, badhta $D_{0}$ aksar jeetta hai → yield dip hoti hai (dreaded early "yield valley"), phir recover hoti hai jaise Photolithography aur defect control mature hoti hai (the "yield ramp") aur $D_{0}$ girta hai. Parent ka mistake-callout — " $D_{0}$ ek fixed material constant nahi hai" — yahi story hai: $D_{0}$ ek maturity number hai jo time ke saath move karta hai.

Recall Ek-line self-test

Solutions cover karo aur Exercise 3.1 ki identity $Y_{B} = (Y_{S})^{4}$ sirf exponent laws se re-derive karo. $Y_{B} = e^{- D_{0} (4 A)} = (e^{- D_{0} A})^{4} = (Y_{S})^{4}$ ::: kyunki exponent ko 4 se multiply karna poore factor ko 4th power tak raise karne ke barabar hai.

Connections

Yield, defect density, and binning — woh parent jisko yeh drill karta hai.
Poisson distribution — $e^{- D_{0} A}$ ka origin.
Chiplets and MCM — L4 area-splitting payoff.
Chip economics and cost per transistor — jahan cost-per-good-die rehti hai.
Process node scaling — L5 ka $D_{0}$ -vs- $A$ tension.