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100x defect tolerance: How we solved the yield problem

cerebras.ai

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Re: 100x defect tolerance: How we solved the yield problem

#71

Neat. What about power density? An H100 has a TDP of 700 watts (for the SXM5 version). With a die size of 814 mm^2 that's 0.86 W/mm^2. If the cerebras chip has the same power density, that means a cerebras TDP of 37.8 kW. That's a lot. Let's say you cover the whole die area of the chip with water 1 cm deep. How long would it take to boil the water starting from room temperature (20 degrees C)? amount of water = (die…

Minor correction, the keynote video says ~20 kW

Re: 100x defect tolerance: How we solved the yield problem

#72
post #65
post #56

Earlier quoted context omitted.

I had been under the impression that Nvidia had done something similar here, but they did not talk about deploying the space saving design and instead only talked about the server rack where all of the chips on the mega wafer normally are. https://www.sportskeeda.com/gaming-tech/what-nvlink72-nvidia...

That shield is just a prop that looks nothing like the real product. The NVL72 rack doesn't use any wafer-scale-like packaging.

It would be nice if they made it real. The cost savings from not needing so much material should be fantastic.

Re: 100x defect tolerance: How we solved the yield problem

#73

Neat. What about power density? An H100 has a TDP of 700 watts (for the SXM5 version). With a die size of 814 mm^2 that's 0.86 W/mm^2. If the cerebras chip has the same power density, that means a cerebras TDP of 37.8 kW. That's a lot. Let's say you cover the whole die area of the chip with water 1 cm deep. How long would it take to boil the water starting from room temperature (20 degrees C)? amount of water = (die…

The machine that actually holds one of their wafers is almost as impressive as the chip itself. Tons of water cooling channels and other interesting hardware for cooling.

Re: 100x defect tolerance: How we solved the yield problem

#74

Neat. What about power density? An H100 has a TDP of 700 watts (for the SXM5 version). With a die size of 814 mm^2 that's 0.86 W/mm^2. If the cerebras chip has the same power density, that means a cerebras TDP of 37.8 kW. That's a lot. Let's say you cover the whole die area of the chip with water 1 cm deep. How long would it take to boil the water starting from room temperature (20 degrees C)? amount of water = (die…

The enthalpy of vaporization of water (at standard pressure) is listed by Wikipedia[1] as 2.257 kJ/g, so boiling 462 grams would require an additional 1.04 MJ, adding 26 seconds. Cerebras claims a "peak sustained system power of 23kW" for the CS-3 16 Rack Unit system[2], so clearly the power density is lower than for an H100.

[1] https://en.wikipedia.org/wiki/Enthalpy_of_vaporization#Other... [2] https://cerebras.ai/product-system/

Re: 100x defect tolerance: How we solved the yield problem

#75
post #2

I think this is an important step, but it skips over that 'fault tolerant routing architecture' means you're spending die space on routes vs transistors. This is exactly analogous to using bits in your storage for error correcting vs storing data. That said, I think they do a great job of exploiting this technique to create a "larger"[1] chip. And like storage it benefits from every core is the same and you don't nee…

"While I continue to believe that many people are going to collectively lose trillions of dollars ultimately pursuing "AI" at this stage"

Can you please explain more why you think so ?

Thank you.

Re: 100x defect tolerance: How we solved the yield problem

#77
This is a strange blog post. Their tables say:

Cerebras yields 46225 * .93 = 43000 square millimeters per wafer

NVIDIA yields 58608 * .92 = 54000 square millimeters per wafer

I don't know if their numbers are correct but it is a strange thing for a startup to brag that it is worse than a big company at something important.

Re: 100x defect tolerance: How we solved the yield problem

#78
post #2

I think this is an important step, but it skips over that 'fault tolerant routing architecture' means you're spending die space on routes vs transistors. This is exactly analogous to using bits in your storage for error correcting vs storing data. That said, I think they do a great job of exploiting this technique to create a "larger"[1] chip. And like storage it benefits from every core is the same and you don't nee…

Any thoughts on why they are disabling so many cores in their current product? I did some quick noodling based on the 46/970000 number and the only way I ended up close to 900,000 was by assuming that an entire row or column would be disabled if any core within it was faulty. But doing that gave me a ~6% yield as most trials had active core counts in the high 800,000s

Re: 100x defect tolerance: How we solved the yield problem

#79
post #40

Understanding that there's inherent bias by them being competitors of the other companies, but still this article seems to make some stretches. If you told me you had an 8% core defect rate reduced 100x, I'd assume you got to close to 99% enablement. The table at the end shows... Otherwise. They also keep flipping between cores, SMs, dies, and maybe other block sizes. At the end of the day I'm not very impressed. The…

I think you're missing the point. The comparison is not between 93% and 92%. The comparison is between what they're getting (93%) and what you'd get if you scaled up the usual process to the core size they're using (0%). They are doing something different (namely: a ~whole wafer chip) that isn't possible without massively boosting the intra-chip redundancy. (The usual process stops working once you no longer have any extra dies to discard.)

> Despite having built the world’s largest chip, we enable 93% of our silicon area, which is higher than the leading GPU today.

The important part is building the largest chip. The icing on the top is that the enablement is not lower. Which it would be without the routing-to-spare-cores magic sauce.

And the differing terminology is because they're talking about differing things? You could call an SM a core, but it kind of contains (heterogeneous) cores itself. (I've no idea whether intra-SM cores can be redundant to boost yield.) A die is the part you break off and build a computer out of, it may contain a bunch of cores, a wafer can be broken up into multiple dies but for Cerebras it isn't.

If NVIDIA were to go and build a whole-wafer die, they'd do something similar. But Cerebras did it and got it to work. NVIDIA hasn't gotten into that space yet, so there's no point in building a product that you can't sell to a consumer or even a data center that isn't built around that exact product (or to contain a Balrog).

Re: 100x defect tolerance: How we solved the yield problem

#80
post #2

I think this is an important step, but it skips over that 'fault tolerant routing architecture' means you're spending die space on routes vs transistors. This is exactly analogous to using bits in your storage for error correcting vs storing data. That said, I think they do a great job of exploiting this technique to create a "larger"[1] chip. And like storage it benefits from every core is the same and you don't nee…

Any thoughts on why they are disabling so many cores in their current product? I did some quick noodling based on the 46/970000 number and the only way I ended up close to 900,000 was by assuming that an entire row or column would be disabled if any core within it was faulty. But doing that gave me a ~6% yield as most trials had active core counts in the high 800,000s

They did mention that they stash extra cores to enable the re-routing. Those extra cores are presumably unused when not routed in.
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