Die per wafer and yield calculator

Enter the wafer diameter, die size and defect density to see how many dies fit on a wafer, how many are good, and what each good die costs. Try making the die larger and watch the cost climb.

Inputs

Results

Die area
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Gross dies per wafer (grid count)
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Formula estimate
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Yield modelYieldGood diesCost per good die
Poisson–––
Murphy–––
Negative binomial–––

Wafer map (one random defect pattern)

Good Bad · Good dies in this example: –

The expected number of defects is scattered uniformly at random, and any die hit by at least one defect is marked bad (the same assumption as the Poisson model). Real defects cluster, so real yield is usually somewhat higher.

How dies per wafer is counted

Dies are laid out on a grid, separated by scribe lines that the dicing saw cuts through. The outer few millimeters of the wafer are not used because film thickness and flatness are less controlled there (edge exclusion), so only dies that fit completely inside the usable circle count. This calculator tries four grid placements and reports the best count. The common estimate DPW ≈ πr²/S − πd/√(2S), where S is the die area including scribe and d is the usable diameter, subtracts the partial dies lost at the edge and usually comes out slightly below the grid count.

Yield models compared

With die area A and defect density D₀, the Poisson model, which assumes defects land completely at random, gives yield Y = e^(−A·D₀). Real defects tend to cluster in parts of the wafer, so at the same average density fewer dies are actually lost. The Murphy model Y = ((1 − e^(−A·D₀)) / (A·D₀))² and the negative binomial model Y = (1 + A·D₀/α)^(−α) account for this. A large clustering factor α approaches the Poisson result; a small α means strongly clustered defects.

Why big dies get expensive fast

A larger die reduces the number of dies per wafer in proportion to area, loses a larger share at the wafer edge, and is more likely to catch a defect. For example, on a 300 mm wafer with 100 µm scribe lines, 3 mm edge exclusion and D₀ = 0.1 per cm², a 10 × 10 mm die gives 612 dies at about 90% Poisson yield, or about 554 good dies. At 20 × 20 mm it drops to 148 dies at about 67% yield, about 99 good dies. At 10,000 dollars per wafer, the cost per good die rises from about 18 to about 101 dollars: four times the area, about 5.6 times the cost. This is a key reason designs are increasingly split into smaller chiplets.

Limits of this estimate

These are first-order estimates for understanding the trade-offs. Real die counts depend on how exposure shots are stepped across the wafer, where test structures sit, and how the notch area is handled. Defect density varies widely with process maturity, layer count and design rules: around 0.1 per cm² is often quoted for a mature high-volume process, and several times that during ramp-up. Yield loss from parametric variation, as opposed to random defects, is not included.

Practice yield questions

With the same defect density, what generally happens to yield when chip area is increased?

  1. AIt decreases
  2. BIt increases
  3. CIt does not change
Show answer and explanation

Answer: A. It decreases

In the Poisson model, yield is Y = exp(−D×A), where D is defect density and A is die area. With randomly landing defects, a larger die is more likely to contain at least one, so yield falls exponentially with area. This is one reason chiplets, which split a large die into several small ones, attract attention: small dies can be tested and only good ones combined.

In the Poisson model Y = exp(−D × A), with defect density D = 0.1 /cm² and chip area A = 1 cm², approximately what is the yield?

  1. AAbout 50%
  2. BAbout 90%
  3. CAbout 99%
Show answer and explanation

Answer: B. About 90%

D × A = 0.1 × 1 = 0.1, so Y = exp(−0.1) ≈ 0.905, about 90%. When D×A is small, Y ≈ 1 − D×A, so the average number of defects per die approximates the failure fraction. At the same defect density, a 5 cm² die yields only exp(−0.5) ≈ 61%. About 50% corresponds to D×A ≈ 0.7 and about 99% to D×A ≈ 0.01.

FAQ

How many 10 mm × 10 mm dies fit on a 300 mm wafer?

About 600, assuming 100 µm scribe lines and 3 mm edge exclusion. The exact count shifts with the scribe width, edge exclusion and grid placement.

Should I use the Poisson or the negative binomial yield model?

The negative binomial model, which accounts for defect clustering, usually matches real fab data better and is widely used in industry. The Poisson model is simpler and gives a conservative, lower yield for large dies.

How is cost per good die calculated?

Divide the cost of one processed wafer by the number of good dies it yields (gross dies × yield). A finished product also adds test, packaging and final-test costs and their own yields.

Related terms

Yield Poisson yield model Chiplet Dicing

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