Yield, defect density, and binning: Worked examples

You have met the formulas on the parent page Yield, defect density, and binning. Now we grind through every kind of number they can be fed. The goal: after this page, no exam or real-world case can surprise you, because you will have already seen that shape of problem solved.

We only use three tools, all built on the parent page:

Gross dies per wafer $DPW \approx \frac{π r ^{2}}{A} - \frac{π \cdot 2 r}{2 A}$ — how many rectangles fit on the round disk.
Poisson yield $Y = e^{- D_{0} A}$ — chance a die has zero killer defects.
Clustered (negative-binomial) yield $Y = (1 + \frac{D _{0} A}{α})^{- α}$ — same idea when dust clumps.

Everything below is just these three, pushed into corners.

The scenario matrix

Think of every yield problem as a point in this table. Each row is a class of input the topic can throw at you; the last column names the example that nails it.

Cell	What makes it tricky	Covered by
Baseline	ordinary $D_{0}$ , ordinary $A$	Ex 1
Big-die punishment	area grows → yield collapses exponentially	Ex 2
Zero / degenerate input	$D_{0} = 0$ (perfect fab) or $A \to 0$ (tiny die)	Ex 3
Limiting behaviour	$α \to \infty$ : clustered model must become Poisson	Ex 4
Clustering vs Poisson	finite $α$ raises yield above $e^{- D_{0} A}$	Ex 5
Economics word problem	dollars per good die, real wafer cost	Ex 6
Reverse / solve-for-input	given a target yield, find allowed $D_{0}$	Ex 7
Binning ladder	split survivors into speed grades → revenue	Ex 8
Chiplet twist (exam)	small dies vs one big die, same silicon	Ex 9

A note on units before we start: $D_{0}$ is defects per cm², area $A$ is in cm², so $λ = D_{0} A$ is a pure number (defects). That is why it can sit in an exponent — you can only exponentiate a bare number, never something carrying units.

Ex 1 — Baseline (the ordinary case)

Worked example Cell: Baseline

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

Forecast: guess — will more than half the dies work? Will there be more or fewer than 700 gross dies?

Step 1 — expected defects per die. $λ = D_{0} A = 0.5 \times 1 = 0.5$ . Why this step? $Y = e^{- D_{0} A}$ needs the pure number $λ$ first; it is the average count of killers on one die.

Step 2 — yield. $Y = e^{- 0.5} = 0.6065$ , about 61%. Why this step? $e^{- λ}$ is the Poisson probability of zero defects — the only survivors.

Step 3 — gross dies. Naive $\frac{π r ^{2}}{A} = π \cdot 225 = 706.86$ . Edge loss $\frac{π \cdot 2 \cdot 15}{2} = 66.64$ . So $DPW \approx 640$ . Why this step? The rim of half-cut dies is dead area; subtracting it stops us over-counting.

Step 4 — good dies. $640 \times 0.6065 \approx 388$ . Why this step? Good = gross $\times$ fraction that survive.

Verify: $e^{- 0.5} \approx 0.6065$ is between $\frac{1}{2}$ and $1$ ✓ (more than half survive, matching forecast). $706.86 - 66.64 = 640.2$ ✓, dimensionless (cm²/cm² cancels). Good dies $388 < 640$ ✓.

Ex 2 — Big-die punishment

Worked example Cell: Big-die punishment

Same $D_{0} = 0.5$ , but a GPU with $A = 4 cm^{2}$ . Compare its yield to Ex 1.

Forecast: area went up ×4. Does yield drop to a quarter (×0.25), or worse?

Step 1 — new $λ$ . $λ = 0.5 \times 4 = 2$ . Why? Four times the area sweeps up four times the expected defects.

Step 2 — yield. $Y = e^{- 2} = 0.1353$ , about 13.5%. Why? Same zero-defect rule; the exponent just quadrupled.

Step 3 — compare. Ex 1 was $e^{- 0.5} = 0.6065$ . Ratio $\frac{0.1353}{0.6065} = 0.223$ , roughly one fifth, not one quarter. Why? Because $e^{- 2} = (e^{- 0.5})^{4}$ — yield is exponential in area, so ×4 area raises the survival factor to the 4th power, crushing it harder than any linear guess.

Verify: $(0.6065)^{4} = 0.1353$ ✓ — confirms yield ratios multiply, they don't subtract. Forecast (guessing ×0.25) was too optimistic — the real drop is steeper. See Process node scaling and Chip economics and cost per transistor for why this drives real design choices.

Ex 3 — Zero and degenerate inputs

Worked example Cell: Zero / degenerate input

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

Forecast: a perfect fab should give 100%. A vanishing die can't collect any defect — also 100%?

Step 1 — perfect fab. $λ = 0 \times A = 0$ , so $Y = e^{0} = 1 = 100%$ . Why? No defects anywhere → every die is clean. The formula earns this: $e^{0} = 1$ exactly.

Step 2 — vanishing die. As $A \to 0$ , $λ = D_{0} A \to 0$ , so $Y = e^{- D_{0} A} \to e^{0} = 1$ . Why? A point-sized die presents no target area for a speck to land on. Yield tends to perfect.

Step 3 — the opposite extreme. As $A \to \infty$ (a die the size of the whole wafer), $λ \to \infty$ and $Y = e^{- \infty} \to 0$ . Why? An enormous die is almost certain to catch at least one killer.

Verify: $e^{0} = 1$ ✓ and $e^{- λ} \to 0$ as $λ \to \infty$ ✓. Both limits behave, and match the forecast for the two "perfect" cases while exposing the third (giant die → 0).

Ex 4 — Limiting behaviour: clustered → Poisson

Worked example Cell: Limiting behaviour

Take the clustered model $Y = (1 + \frac{D _{0} A}{α})^{- α}$ with $D_{0} A = 2$ . Compute it at $α = 1, 10, 1000$ and show it approaches the Poisson $e^{- 2} = 0.1353$ .

Forecast: as clustering weakens ( $α$ big), should the clustered yield fall toward the Poisson value or rise away from it?

Step 1 — $α = 1$ . $Y = (1 + 2)^{- 1} = 1/3 = 0.3333$ . Why? Heavy clustering: defects pile on few dies, sparing many → high yield.

Step 2 — $α = 10$ . $Y = (1 + 0.2)^{- 10} = (1.2)^{- 10} = 0.1615$ . Why? Weaker clustering, closer to random scatter → yield drops toward Poisson.

Step 3 — $α = 1000$ . $Y = (1 + 0.002)^{- 1000} = 0.1354$ . Why? Almost no clustering; the identity $lim_{α \to \infty} (1 + x / α)^{- α} = e^{- x}$ kicks in.

Verify: target $e^{- 2} = 0.13534$ . Sequence $0.3333 \to 0.1615 \to 0.1354$ marches down onto it ✓. This is the sanity anchor from the parent page: the fancy model must collapse to the simple one when clumping vanishes. See Poisson distribution.

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

Worked example Cell: Clustering vs Poisson

A GPU with $D_{0} A = 2$ . Compare pure Poisson yield to the clustered model with $α = 2$ . Which is higher, and by how much?

Forecast: clustering pushes defects onto fewer dies — should the good fraction go up or down?

Step 1 — Poisson. $Y_{P} = e^{- 2} = 0.1353$ (13.5%). Why? Baseline assuming perfectly uniform scatter.

Step 2 — clustered. $Y_{C} = (1 + 2/2)^{- 2} = 2^{- 2} = 0.25$ (25%). Why? $α = 2$ means moderate clumping; clumps concentrate damage, sparing more dies.

Step 3 — ratio. $\frac{0.25}{0.1353} = 1.85$ — nearly double the surviving dies. Why? Real fabs report yields above naive Poisson exactly because dust clumps; this is why fabs use negative-binomial in practice.

Verify: $2^{- 2} = 0.25$ ✓ and $0.25 > 0.1353$ ✓ (clustering helps, matching forecast). This directly changes cost per good die — half the wasted silicon of the Poisson estimate.

Ex 6 — Economics word problem

Worked example Cell: Economics word problem

A finished 300 mm wafer costs $15{,}000. Using Ex 1's numbers ( $DPW = 640$ , $Y = 0.6065$ ), find the cost per good die. Then compare to a naive "cost ÷ gross dies".

Forecast: will the true cost per good die be near $23 (=15000/640) or noticeably higher?

Step 1 — good dies. $DPW \times Y = 640 \times 0.6065 = 388.2$ . Why? You only sell working dies; the dead ones still cost you.

Step 2 — cost per good die. $\dfrac{15000}{388.2}=\$ 38.64 $. * W h y ? *$ \text{Cost}{\text{good}}=\dfrac{\text{Cost}{\text{wafer}}}{\text{DPW}\times Y}$ — the whole wafer's price divided across survivors.

Step 3 — naive comparison. Ignoring yield: $\dfrac{15000}{640}=\$ 23.44 $. * W h y ? * S h o w s t h e p e na l t y :$ 38.64/23.44=1.65\times $m or e, e x a c tl y$ 1/Y=1/0.6065=1.649$.

Verify: $1/0.6065 = 1.649$ and $23.44 \times 1.649 = 38.65$ ✓ (rounding). Units: dollars/die throughout ✓. The forecast "noticeably higher" is right — yield inflates cost by the factor $1/ Y$ . See Chip economics and cost per transistor.

Ex 7 — Reverse problem: solve for the input

Worked example Cell: Reverse / solve-for-input

Your product needs at least 80% yield. Die area $A = 1 cm^{2}$ . What is the largest defect density $D_{0}$ your process may have (Poisson)?

Forecast: to keep 80% alive, must $D_{0}$ be well under 0.5 (the value that gave only 61%)?

Step 1 — write the target. $0.80 = e^{- D_{0} \cdot 1}$ . Why? Set the yield formula equal to the requirement and solve backwards.

Step 2 — undo the exponential. Take natural log: $ln 0.80 = - D_{0}$ . The logarithm is the tool that undoes $e^{(\cdot)}$ — it asks "what exponent gives this number?" Why this tool? $D_{0}$ is trapped inside an exponent; only $ln$ frees it.

Step 3 — solve. $D_{0} = - ln 0.80 = 0.2231 /cm^{2}$ . Why? $ln 0.80 = - 0.2231$ , negate to get a positive density.

Verify: $e^{- 0.2231} = 0.800$ ✓. Sanity: $0.2231 < 0.5$ , so a cleaner process than Ex 1 — consistent with wanting higher yield ✓. Matches forecast. This is the Photolithography / yield-ramp target a fab chases.

Ex 8 — Binning ladder (turning survivors into a product ladder)

Worked example Cell: Binning ladder

Of 388 good dies, testing (see Wafer testing and probe) shows the max-clock distribution: 25% hit 5.0 GHz, 50% hit 4.5 GHz, 25% hit only 4.0 GHz. Prices: $500 / $300 / $180. Find total revenue and average price per die.

Forecast: will the average sit closer to the middle price ($300) or get pulled up by the premium tier?

Step 1 — count each bin. Premium $0.25 \times 388 = 97$ ; middle $0.50 \times 388 = 194$ ; low $0.25 \times 388 = 97$ . Why? Binning splits one identical die design into SKUs by measured speed — same silicon, different grade.

Step 2 — revenue per bin. $97 \times 500 = 48, 500$ ; $194 \times 300 = 58, 200$ ; $97 \times 180 = 17, 460$ . Why? Each grade sells at its own price; revenue is count × price.

Step 3 — totals. Total $=48{,}500+58{,}200+17{,}460=\$ 124{,}160 $. A v er a g e$ =124160/388=$320.0$. Why? Average price per good die measures how well binning recovered value.

Verify: $97 + 194 + 97 = 388$ ✓. Average $320 sits just above the middle price $300 ✓ — pulled up by the premium tier, matching forecast. Binning beats scrapping the slow dies: without it those 97 low-tier dies ($17,460) would be lost.

Ex 9 — Chiplet twist (exam-style)

Worked example Cell: Chiplet twist

Same silicon, two strategies at $D_{0} = 0.5$ : (A) one monolithic die $A = 4 cm^{2}$ ; (B) four chiplets of $A = 1 cm^{2}$ each, later joined. Compare the fraction of silicon that ends up in a working part (ignore assembly loss).

Forecast: four small dies vs one big die of equal total area — which wastes less silicon?

Step 1 — monolithic yield. $Y_{A} = e^{- 0.5 \times 4} = e^{- 2} = 0.1353$ . A defect anywhere kills all 4 cm². Why? One killer defect dooms the whole large die.

Step 2 — per-chiplet yield. Each 1 cm² chiplet: $Y_{B} = e^{- 0.5} = 0.6065$ . Why? A defect now only kills its own 1 cm² tile, not its neighbours.

Step 3 — fraction of good silicon. Monolithic: 13.5% of area survives. Chiplets: each tile survives independently at 60.65%, so 60.65% of the silicon area yields good tiles. Why? Splitting isolates defects — this is the core argument for Chiplets and MCM.

Verify: $e^{- 2} = 0.1353$ vs $e^{- 0.5} = 0.6065$ ✓, and $0.6065 > 0.1353$ by a factor of $4.48$ ✓. Note $(e^{- 0.5})^{4} = e^{- 2}$ : the chance all four chiplets are good equals the monolithic yield — but with chiplets you keep the good ones instead of scrapping the set. Forecast confirmed: splitting wastes far less silicon.

Figure — Yield, defect density, and binning

Recall

Recall The nine leans in one breath

Baseline → plug in. Bigger die → exponentiate down. Zero input → $e^{0} = 1$ . Infinite $α$ → collapse to Poisson. Finite $α$ → yield rises. Money → divide by good dies. Reverse → take $ln$ . Binning → split survivors into priced grades. Chiplets → split area to isolate defects.

Recall Reveal drills

Yield needed 80%, $A = 1$ — max $D_{0}$ ? ::: $D_{0} = - ln 0.80 = 0.223 /cm^{2}$ Monolithic 4 cm² yield at $D_{0} = 0.5$ ? ::: $e^{- 2} = 0.135$ Four 1 cm² chiplets, each yield? ::: $e^{- 0.5} = 0.607$ Cost per good die if 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 — the zero-defect probability behind $e^{- D_{0} A}$ .
Wafer testing and probe — measures the speed distribution that feeds binning (Ex 8).
Chiplets and MCM — the defect-isolation win of Ex 9.
Chip economics and cost per transistor — where cost-per-good-die (Ex 6) lands.