Tue Dec 5, 2017
Determining alignment tolerances between directly aligned layers is, well, direct. These tolerances will be set by equipment capability1. However, unless you are planning to measure O(n²) values, indirect alignment tolerances require a little calculation. Unfortunately, I have seen this done incorrectly, using the root-mean-square (RMS) value.
The problem with using the RMS is determining whether or not the direct alignment tolerances are linked to the alignment variance. If the alignment tolerances were set at a multiple of the standard deviations, then using the RMS will work. This is linked to the Bienaymé formula. However, if the alignment tolerances were determined in some other manner, then using the RMS will be incorrect. One way this might happen is if you have a significant rate of photo rework, and so your distribution will not be normal.
As a worse-case scenario, assume that the alignment is uniformly distributed2 within the alignment tolerance. In this case, the variance will be b²/3. To keep things simple, let’s use a tolerance of ±1μm.
In the above figure, you can see that total tolerance provides the smallest calculated tolerance when the number of links in the alignment is limited. The estimates based on 3σ is larger. The difference is due to the kurtosis in the uniform distribution; there are no tails so 3×σ is overly conservative. As the number of links increases, the central limit starts to take effect, and the total tolerance becomes overly conservative. The pattern for estimates based on 6σ are the same, but the total tolerance remain less than the statistically calculated tolerance until the number of links becomes quite high3.
For those who continue to use RMS on their absoluate tolerances without knowing the actual distribution, they could be grossly underestimating the process capability. In the worse case of a uniform distribution, they are essentially running their photo tolerances at 1.7σ. I don’t know too many plants that would be confortable with that.
More or less. Alignment tolerance can be affected by dimensional instability of wafers, degradation of alignment marks, topography… ↩︎
Instead of a uniform distribution, one could consider a triangular distribution, sometimes known as the “lack of knowledge” distribution. The variance in this case is b²/6. This doesn’t really change the following comments, except that it becomes more efficient to use the 3σ or 6σ limits sooner. ↩︎
The total tolerance remains less than the statistical tolerance until the number of links reaches 12! ↩︎