4 ms·
> major power savings with integrated memory. M1's integrated memory is completely generic LPDDR5. It's the same stuff the rest of the industry uses, there's n
by kllrnohj 4y ago
> major power savings with integrated memory.
M1's integrated memory is completely generic LPDDR5. It's the same stuff the rest of the industry uses, there's no power savings here (and it's not any more "integrated" than AMD & Intel have been doing for half a decade or more, either)
The primary interesting thing about the M1/M2's memory is the width of the memory bus, which is how it gets such huge bandwidth numbers. But if anything that's costing power, not saving it.
- titzer 4y agoIt takes a ton of power to drive a memory bus across a motherboard.
- kllrnohj 4y agoWhich the M1/M2 is doing as well... It's just soldered RAM, not on-die RAM. Completely off-the-shelf LPDDR modules that are shipping in dozens of other laptops. Absolutely nothing special about the RAM at all.
- titzer 4y agoDescribing it as "soldered" is not right. It's on the same package. https://en.wikipedia.org/wiki/Apple_M1 https://en.wikipedia.org/wiki/Apple_M1. There's no multi-centimeter hundred-plus-line bus to drive charges through. Pushing signals through those big long wires takes juice.
- kllrnohj 4y ago> It's on the same package. Which means it's still over an external interconnect, which is what actually "takes juice" > Pushing signals through those big long wires takes juice. lol no it doesn't. That's a pretty minuscule amount of resistance.
- titzer 4y agoResistance is directly proportional to length. Yeah it saves a lot to move the RAM closer by putting it on package. Not sure why you are still trying to argue about basic EE.
- kllrnohj 4y agoMath it out then. How much resistance is that few centimeters of PCB trace costing & what does that translate to in terms of milliwatts? Resistance is proportional to length, yes, but that doesn't make it significant in the given application.