Question bank — Yield, defect density, and binning

True — $D_0$ is a property of the process/wafer, not the die. What changes with die size is $\lambda =D_{0}A$, the expected count per die, not the density itself.

False — yield is $Y=e^{-D_{0}A}$, so doubling $A$ squares the survival factor: $e^{-2\lambda }=(e^{-\lambda }{)}^2$. A 60% yield becomes 36%, not 30%.

False in general — a chip can carry a defect in a redundant/harvestable block and still work as a lower SKU. "Fatal defect" (kills the die) is the yield-relevant category, not "any defect."

False — ${\lim }_{\alpha \to \infty }(1+D_{0}A/\alpha {)}^{-\alpha }=e^{-D_{0}A}$, so it collapses back to Poisson. Infinite $\alpha$ means "no clustering," i.e. perfectly uniform defects — exactly the Poisson assumption.

True for finite $\alpha$ — clumping piles defects onto fewer dies, sparing more of the rest, so the good fraction rises. Equality is only reached in the limit $\alpha \to \infty$.

False — that ignores the round rim. Rectangular dies straddling the curved edge are lost, so we subtract an edge-loss term; $\pi r^2/A$ is an upper estimate only.

False — speed and harvest binning use the same die and mask. Sorting happens after test; the ladder of SKUs comes from variation and fusing, not from new designs.

True — you pay for the wafer regardless of survivors. Yield changes cost per good die by changing the denominator $\text{DPW}\times Y$, not the wafer price.

The formula is right but the conclusion is absurd — a real die has irreducible minimum area, and shrinking also raises $D_0$ (finer features are harder to print, see Process node scaling). The limit is mathematical, not physical.

Wrong — the tunable input is $D_0$, a process-maturity metric that starts high on a new node and falls during the yield ramp as the fab learns. Yield changes over the product's life.

Wrong — it is usually the same die with two defective cores fused off (harvesting). One mask, many SKUs.

Random does not mean uniform. Real dust clumps, creating spatial clustering; that's precisely why pure Poisson under-predicts real yield and the negative-binomial correction exists.

It omits yield. You must divide by good dies $=\text{DPW}\times Y$, because the dead dies still cost wafer area but earn nothing.

It is an estimate. $\sqrt{2A}$ approximates an average die edge length over orientation and $2\pi r$ is written as $\pi \cdot 2r$ loosely — the whole correction is engineering-grade, not a derivation.

A trap — clustering only changes the statistical distribution of a given defect population; it doesn't remove defects. The real goal is lowering $D_0$. Clustering is a modelling reality, not a lever to pull.

Poisson describes counts of rare, independent events at a fixed average rate over a region, which is exactly "how many random defects fall on this die area." See Poisson distribution.

A die works only if it has zero fatal defects; any $k\ge 1$ kills it. So the working fraction is precisely $P(k=0)=e^{-\lambda }$.

Yield is exponential in area, so large $A$ crushes the survivor fraction super-linearly, and each survivor must pay for the wafer share of many dead neighbours. This is a key motivation for Chiplets and MCM.

Process variation makes "working" chips genuinely differ in speed/power/functional units; binning turns that continuous spread into a product ladder, so a chip that just misses the top grade is sold instead of scrapped.

Each small chiplet has small $A$, so $e^{-D_{0}A}$ stays high; you assemble only good chiplets, avoiding the exponential penalty of one giant monolithic die. (Details: Chiplets and MCM.)

Cost per good die $={\text{Cost}}_{\text{wafer}}/(\text{DPW}\times Y)$, and $Y=e^{-D_{0}A}$ responds exponentially to $D_0$ — small density improvements can multiply the number of good dies, outweighing linear wafer savings. See Chip economics and cost per transistor.

Poisson assumes one fixed rate everywhere; real defects vary in density across the wafer, and $\alpha$ encodes how much that density itself fluctuates (the clustering). One number ($\lambda$) can't capture both mean and spread of the local rate.

$Y=e^0=1$, i.e. 100% — no defects means every die survives, as the formula demands.

$Y=e^{-D_{0}A}\to 0$ — an infinitely large die is almost certain to catch at least one fatal defect. This is the limiting form of the "big-die punishment."

$Y=(1+D_{0}A/\alpha {)}^{-\alpha }\to 1$ as $\alpha \to 0^{+}$ — all defects pile onto a vanishing fraction of dies, so almost every die is spared. It's the opposite extreme from the Poisson $\alpha \to \infty$ limit.

No — edge loss dominates when few dies fit, so the correction term is a large fraction of the total and DPW is very uncertain; the estimate is meant for wafers holding many dies.

Yield being 1 says every die is functional, but binning can still sort them by speed/power variation — functional yield and speed grade are independent axes. See Wafer testing and probe.

Not necessarily — equal $Y$ means equal $\lambda =D_{0}A$, so the fab with the larger die must have the lower $D_0$ to compensate. Yield alone can't rank process cleanliness without knowing $A$.

Yes — only fatal defects (those that actually kill a die) count toward $Y$; many measured particles are benign or land in non-critical areas, so the yield-relevant $D_0$ is typically lower than the raw particle count.