Comment by ajb
3 days ago
So they massively reduce the area lost to defects per wafer, from 361 to 2.2 square mm. But from the figures in this blog, this is massively outweighed by the fact that they only get 46222 sq mm useable area out of the wafer, as opposed to 56247 that the H100 gets - because they are using a single square die instead of filling the circular wafer with smaller square dies, they lose 10,025 sq mm!
Not sure how that's a win.
Unless the rest of the wafer is useable for some other customer?
It's a win because they have to test one chip, and don't have to spend resources on connecting the chiplets. The latter costs a lot (though it has other advantages). I suspect that a chiplet-based device with total 900k cores would just be not viable due to the size constraints.
If their routing around the defects is automated enough (given the highly regular structure), it may be a massive economy of efforts on testing and packaging the chip.
Why does it have to be a square? There’s no need to worry about interchangeable third-party heat sink compatibility. Is it possible to make it an irregular polygon instead of square?
Additional wafer area would be a marginal increase in performance (+~20% core core best case) but increases the complexity of their design, and requires they figure out how to package/connect/house/etc. a non-standard shape. A wafer scale chip is already a huge tech risk, why spend more novelty budget on nonessential weirdness?
Why does their chip have to be rectangular, anyways? Couldn't they cut out a (blocky) circle too?
You need a rectilinear polygon that tessellates, and has the fewest sides possible to minimize the number of cuts necessary. And it would probably help the cutting if the shape is entirely convex, so that cuts can overshoot a bit without damaging anything.
That suggests a rectangle is the only possible shape.
If it's just one chip per wafer, why even bother cutting?
Why does it need to tessellate if there's only one chip per wafer?
Rather I wonder why do they even need to cut the extra space, instead of putting something there. I suppose that the structure of the device is highly rectangular from the logical PoV, so there's nothing useful to put there. I suspect smaller unrelated chips can be produced on these areas along the way.
I've never cut a wafer, but I assume cutting is hard and single straight lines are the easiest.
I wonder if you could… just not cut the wafer at all??
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The cost driver for fabbing out wafers is the number of layers and the number of usable devices per wafer. Higher layer count increases cost and tends to decrease yield, and more robust designs with higher yields increase usable devices per wafer. If circles or other shapes could help with either of those, they would likely be used. Generally the end goal is to have the most usable devices per wafer, so they'll be packed as tightly as possible on the wafer so as to have the highest potential output.
Right, but they're making just one usable device per wafer already.
Is the wafer itself so expensive? I assume they don't pattern the unused area, so the process should be quicker?
> I assume they don't pattern the unused area
I’m out of date on this stuff, so it’s possible things have changed, but I wouldn’t make that assumption. It is (used to be?) standard to pattern the entire wafer, with partially-off-the-wafer dice around the edges of the circle. The reason for this is that etching behavior depends heavily on the surrounding area — the amount of silicon or copper whatever etched in your neighborhood affects the speed of etching for you, which effects line width, and (for a single mask used for the whole wafer) thus either means you need to have more margin on your parameters (equivalent to running on an old process) or have a higher defect right near the edge of the die (which you do anyway, since you can only take “similar neighborhood” so far). This goes as far as, for hyper-optimized things like SRAM arrays, leaving an unused row and column at each border of the array.
All the process steps are limited by wafers for hour. Lithography (esp EUV) might be slightly faster, but that's not 30% of total steps, since you generally have deposit and etch/implant for every lithography step.
It's close to a dead loss in process cost.
> I assume they don't pattern the unused area, so the process should be quicker?
The primary driver of time and cost in the fabrication process is the number of layers for the wafers, not the surface area, since all wafers going through a given process are the same size. So you generally want to maximize the number of devices per wafer, because a large part of your costs will be calculated at the per-wafer level, not a per-device level.
Yes, but isn't a big driver of layer costs the cost of the machines to build those layers?
For patterning, a single iteration could be (example values, no actual values used, probably only ballpark accuracy) on a 300M$ EUV machine with 5-year write off cycle, patterns on average 180 full wafers /hour. Excluding energy usage and service time, each wafer that needs full patterning would cost ~38$. If each wafer only needed half the area patterned, the lithography machine might only spend half its usual time on such a wafer, and that could double the throughput of the EUV machine, halving the write-off based cost component of such a patterning step.
Given that each layer generally consists of multiple patterning steps, a 10-20% reduction in those steps could give a meaningful reduction in time spent in the machines whose time spend on the wafer depends on the used wafer area.
This of course doesn't help reduce time in polishing or etching (and other steps that happen with whole wafers at a time), so it won't be as straightforward as % reduction in wafer area usage == % reduction in cost, but I wouldn't be surprised if it was a meaningful percentage.
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Yes, but my understanding is that the wafer is exposed in multiple steps, so there would still be less exposure steps? Probably insignificant compared to all the rest though. (Etching, moving the wafer, etc.)
EDIT: to clarify - I mean the exposure of one single pattern/layer is done in multiple steps. (https://en.wikipedia.org/wiki/Photolithography#Projection)
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There's also no reason they couldn't pattern that area with some other suitable commodity chips. Like how sawmills and butchers put all cuts to use.
Often those areas are used for test chips and structures for the next version. They are effectively free, so you can use them to test out ideas.
Good question. I think the wafer has a cost per area which is fairly significant, but I don't have any figures. There has historically been a push to utilise them more efficiently, eg by building fabs that can process larger wafers. Although mask exposure would be per processed area, I think that there are also some proportion of processing time which is per wafer, so the unprocessed area would have an opportunity cost relating to that.
AIUI Wafer marginal cost is lower than you'd expect. I had $50k in my head, quick google indicates[1] maybe <$20k at AAPL volumes? Regardless seems like the economics for Cerebras would strongly favor yield over wafer area utilization.
[1] https://www.tomshardware.com/tech-industry/tsmcs-wafer-prici...
They probably pattern at least next nearest neighbors for local uniformity. That’s just litho though. The rest of the process is done all at once on the wafer
It’s a win if you can use the wafer as opposed to throwing it away.
A win is a manufacturing process that results in a functioning product. Wafers, etc. aren't so scarce as to demand every mm2 be used on every one every time.