3 ms·

These days the 3.3v rail and -3.3v rail are practically vestigial (iirc the major consumer on those rails was floppy disk drives...) and 5V is mostly used to dr

by paulmd 4y ago

These days the 3.3v rail and -3.3v rail are practically vestigial (iirc the major consumer on those rails was floppy disk drives...) and 5V is mostly used to drive SSDs and USB devices. Almost everything runs off 12V these days - which is the whole point behind the 12VO standard, just make everything use 12V and step it down at the point-of-consumption for the handful of things that still need something else. So basically I wouldn't worry about motherboard compatibility due to 5v, 3.3v, and -3.3v rails - there just isn't much of anything still being done on those rails, and the things that do use it pretty much are the same across motherboards (because they're things that are plugged into the motherboard, not the motherboard itself).

mschuster91 4y ago
> and 5V is mostly used to drive SSDs and USB devices. The most important 5V rail is 5VSB - it powers wake-on-LAN and on some systems wake-via-bluetooth. The interesting question is, is it more efficient to always let 12V->5V regulators run for 5VSB or to use an extremely small separate regulator?
- glowingly 4y ago
  To be fair, something is stepping it down from AC, and many modern PSUs step everything down from a common DC rail using DC-DC conversion (i.e, 12V, 5V, 3.3V, etc are all generated from a source DC voltage). Then things like the CPU and GPU have long moved off of the 3.3V rails to harness the 12V rail for DC-DC conversion in various VRM stages, down to <<2V. My main appreciation for ATX12VO (12 volts only) is the MB can choose to right size the 5V and 3.3V rails appropriately. M.2 slots are driven from 3.3V only (even if M.2 devices step it up or down for their specific uses). A board with 4 M.2 slots will have a different potential 3.3V power demand than a board with only 2 slots. Typically, a decent board will have 1 M.2 2230 A or E slot for WiFi, and 1-2 M.2 2280/22110 M (and rarely B) slots for M.2 storage. PCIe x16 lanes will have 9W of 3.3V, as part of their 75W total power (the remaining 66W is 12V). As noted earlier, 5V is heavily I/O driven. A board with 15 USB ports can expect to need more 5V than a stingy board with only 8 (both are accounting for front I/O headers). I'd rather the MB have to right-size the 3.3V and 5V rails, than forcing every PSU to include copious amounts of 3.3 and 5 just to cover every edge case. Sure, it's a tiny bit of potential efficiency. It's also passing a bit of cost from the PSU to the board itself. Sadly, it does seem ATX12VO's main impact will be lost. OEMs are going to continue with their custom PSUs with unique connectors and the DIY vendor + consumer market has largely turned their backs to this.
- paulmd 4y ago
  > The interesting question is, is it more efficient to always let 12V->5V regulators run for 5VSB or to use an extremely small separate regulator? To complicate this further, full system power-off is not the only time you want to do wake-on-lan... if you want to wake from system sleep then you have to be running the 12V rail anyway, because that 12V rail drives the VRMs which drive Vcore and Vmem. No 12V rail means you lose your memory and CPU state... As such I don't think you can really frame this as "having a separate 5V regulator with the 12v rail turned off", or at least that use-case must exist in addition to the deep-sleep use-case, it cannot supplant it entirely. Anyway though, power-supply 5V rails are usually much much larger than they need to be, and that has higher loss inside a transformer. Even if you make that a solid-state thing inside the transformer, moving power at 5V is less efficient than moving it at 12V and stepping it down at the point-of-consumption. So actually the answer is probably that it's more efficient to always run a step-down regulator.

by paulmd 4y ago

mschuster91 4y ago

> and 5V is mostly used to drive SSDs and USB devices. The most important 5V rail is 5VSB - it powers wake-on-LAN and on some systems wake-via-bluetooth. The interesting question is, is it more efficient to always let 12V->5V regulators run for 5VSB or to use an extremely small separate regulator?

glowingly 4y ago
To be fair, something is stepping it down from AC, and many modern PSUs step everything down from a common DC rail using DC-DC conversion (i.e, 12V, 5V, 3.3V, etc are all generated from a source DC voltage). Then things like the CPU and GPU have long moved off of the 3.3V rails to harness the 12V rail for DC-DC conversion in various VRM stages, down to <<2V. My main appreciation for ATX12VO (12 volts only) is the MB can choose to right size the 5V and 3.3V rails appropriately. M.2 slots are driven from 3.3V only (even if M.2 devices step it up or down for their specific uses). A board with 4 M.2 slots will have a different potential 3.3V power demand than a board with only 2 slots. Typically, a decent board will have 1 M.2 2230 A or E slot for WiFi, and 1-2 M.2 2280/22110 M (and rarely B) slots for M.2 storage. PCIe x16 lanes will have 9W of 3.3V, as part of their 75W total power (the remaining 66W is 12V). As noted earlier, 5V is heavily I/O driven. A board with 15 USB ports can expect to need more 5V than a stingy board with only 8 (both are accounting for front I/O headers). I'd rather the MB have to right-size the 3.3V and 5V rails, than forcing every PSU to include copious amounts of 3.3 and 5 just to cover every edge case. Sure, it's a tiny bit of potential efficiency. It's also passing a bit of cost from the PSU to the board itself. Sadly, it does seem ATX12VO's main impact will be lost. OEMs are going to continue with their custom PSUs with unique connectors and the DIY vendor + consumer market has largely turned their backs to this.
paulmd 4y ago
> The interesting question is, is it more efficient to always let 12V->5V regulators run for 5VSB or to use an extremely small separate regulator? To complicate this further, full system power-off is not the only time you want to do wake-on-lan... if you want to wake from system sleep then you have to be running the 12V rail anyway, because that 12V rail drives the VRMs which drive Vcore and Vmem. No 12V rail means you lose your memory and CPU state... As such I don't think you can really frame this as "having a separate 5V regulator with the 12v rail turned off", or at least that use-case must exist in addition to the deep-sleep use-case, it cannot supplant it entirely. Anyway though, power-supply 5V rails are usually much much larger than they need to be, and that has higher loss inside a transformer. Even if you make that a solid-state thing inside the transformer, moving power at 5V is less efficient than moving it at 12V and stepping it down at the point-of-consumption. So actually the answer is probably that it's more efficient to always run a step-down regulator.