4 ms·

> Competition with "mature nodes" is different in the sense that here one company (like TSMC) will be "competing" with itself. No, why? Lots of foundries out t

> Competition with "mature nodes" is different in the sense that here one company (like TSMC) will be "competing" with itself. No, why? Lots of foundries out there that don't compete on leading edge nodes. > Apple can't increase the price of their product arbitrarily. So if the price per performance increases, the new iPhone would be slower than the old one. Maybe people will still buy it if it has better power efficiency, but that is a trade-off which has its limits. That doesn't address what I said -- that price per transistor is not the single limiting factor. I never said any performance increase is worth infinite value.

cubefox 3y ago
> No, why? Lots of foundries out there that don't compete on leading edge nodes The discussion was about Moore's law ("scaling down"), which is only about the leading edge.
- throwawaylinux 3y ago
  The discussion moved to talking about is competing with mature nodes, where you said they'd only be competing with themselves (which doesn't seem to be right).
  - cubefox 3y ago
    Well, it's right for the leading edge. The point is that they will stop scaling chips before technical limits are reached.
    - throwawaylinux 3y ago
      > Well, it's right for the leading edge. No, the point was that the leading edge will stop advancing when it can no longer compete economically with companies that sell mature nodes. Which is obvious, but I was just being similarly reductive. > The point is that they will stop scaling chips before technical limits are reached. Sure, like most commercial ventures, cost becomes prohibitive at some point and overtakes benefits. As it always has been.
      - cubefox 3y ago
        > No, the point was that the leading edge will stop advancing when it can no longer compete economically with companies that sell mature nodes. The "mature node" here will just be the one before the first node that doesn't make economic sense anymore. And this will be from a leading edge company, most likely TSMC. Say, if 2nm is too expensive compared to 3nm, then TSMC will stick with 3nm.
        throwawaylinux 3y ago
        Yep.

cubefox 3y ago

> No, why? Lots of foundries out there that don't compete on leading edge nodes The discussion was about Moore's law ("scaling down"), which is only about the leading edge.

throwawaylinux 3y ago
The discussion moved to talking about is competing with mature nodes, where you said they'd only be competing with themselves (which doesn't seem to be right).
- cubefox 3y ago
  Well, it's right for the leading edge. The point is that they will stop scaling chips before technical limits are reached.
  - throwawaylinux 3y ago
    > Well, it's right for the leading edge. No, the point was that the leading edge will stop advancing when it can no longer compete economically with companies that sell mature nodes. Which is obvious, but I was just being similarly reductive. > The point is that they will stop scaling chips before technical limits are reached. Sure, like most commercial ventures, cost becomes prohibitive at some point and overtakes benefits. As it always has been.
    - cubefox 3y ago
      > No, the point was that the leading edge will stop advancing when it can no longer compete economically with companies that sell mature nodes. The "mature node" here will just be the one before the first node that doesn't make economic sense anymore. And this will be from a leading edge company, most likely TSMC. Say, if 2nm is too expensive compared to 3nm, then TSMC will stick with 3nm.
      - throwawaylinux 3y ago
        Yep.