Session 4 · Exercises 1, 2 and 3

Wafer die cost calculator

The area is 8 times larger, but the cost is 20 times higher. A big chip loses twice: once on yield and once at the wafer edge. Lay dies out on a wafer and count them yourself, then change the defect density and how strongly defects cluster to see how the cost per die moves.

① Dies on the wafer and cost per die

GDPW ≈ (πd²/4)/A − πd/√(2A)　·　Y = e^−AD₀　·　cost per die = W / (GDPW × Y)　·　A₅₀ = ln 2 / D₀

Die area A (square)

Defect model (α for negative binomial)

Grid origin, horizontal offset

Grid origin, vertical offset

Defect density D₀

Wafer price W (assumed value)

Count at this offset
Mean over 400 offsets
√(2A) formula
√A formula
Yield Y
Good dies = √(2A) formula × Y
Cost per die
Cost relative to a 100 mm² die
A₅₀ = ln 2 / D₀ (Poisson yield 50%)

Counted die (all four corners inside the circle) Die hit by a defect Die not counted Defect ＋ wafer center · ● grid origin

Where you place the grid also changes the count (66 vs. 71 dies at 800 mm²). Putting one die corner at the wafer center gives 648, 148 and 68 dies at 100, 400 and 800 mm², almost the same as the mean. The 20× cost at 8× the area is the product of three factors: 8× from area, 2.0× from yield and 1.2× from edge loss.

② Where Poisson comes from, and what happens when defects cluster

(1 − λ/n)ⁿ → e^−λ (n → ∞)　·　Y = (1 + AD₀/α)^−α　·　Var(D) = D₀²/α

(a) Splitting a die into n cells

Mean defects per die λ = AD₀

Number of cells n (log scale)

(1 − λ/n)ⁿ
Limit e^−λ
Fraction of the limit

(b) Poisson vs. negative binomial at the same AD₀

Die area A

Defect density D₀

Clustering α

λ = AD₀
Poisson yield
Negative binomial yield
Difference
Var(D) = D₀²/α
Standard deviation/mean = 1/√α

Splitting into just ten cells already gets you to 97% of the answer. The stronger the clustering (the smaller α), the higher the yield, even though the expected total number of defects is the same. With α = 1 the standard deviation of the defect density equals the mean; with α = 5 it is about 45% of the mean. Yield is a question of "how many dies are contaminated", not "how many defects there are".

Check against the solution tables ()

Values from the Session 4, Section 0 table, Section 5, and the solutions to Exercises 1–3 in Self-study appendix A, recomputed with the formulas of this figure. Solution values are rounded. "Rounding diff" marks cases where the solution multiplied or divided by rounded intermediate values (yield 0.905, good dies 96.1 and 29.1) and so differs from the exact value in the last digit.
Item	Computed	Solution

What this calculation leaves out

Edge exclusion and scribe line width are set to 0. Real wafers have both, so a direct count would also come out lower than this.
Good dies and cost are computed as in the solution to Exercise 2, by multiplying the GDPW approximation (√(2A)) by the yield. For 100–800 mm² dies this formula gives 1–5% fewer dies than a direct count.
Strictly, the A in λ = AD₀ is the critical area, not the die area. Using the die area as is folds the effect of the critical area into the density, a simplification.
e^−AD₀ counts defects only. Parametric yield, the subject of speed and power binning, is not explained by this formula.
Cost per die is the silicon cost only. HBM, packaging and test come on top of it. The wafer price of USD 15,000 is an assumed value that the text introduces with "suppose".