According to the wiki page on flash memory, flash chips have built-in charge pumps to step up the 1.8v supply to a higher voltage to program/erase the cells. This wastes a lot of power and is apprarently a common failure mode. The wiki page mentions the idea of using a single shared boost converter for chips with a separate Vpp pin, but apparently this is not very common anymore. Besides, this still seems like only a half-solution; I recall reading that this voltage is on the order of 5-10 volts, a voltage range that is abundantly available on PCs.

For SATA SSDs, one could use the 12V rail if available, and fall back to a boost converter drawing power from the 5V rail; for standard PCI-e cards, there's a guaranteed 12V supply available. (m.2 appears to only have a 3.3v supply, which begs the question of why, given what appear to be obvious advantages of higher voltage supplies for flash)

I understand single power supply is convenient, but surely a couple more layers would be worth it for the efficiency and performance(?) benefits. And economies of scale doesn't seem like a good explanation, given there is a pretty large market for high-performance flash.

So, why are single-power-supply flash chips so ubiquitous?

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    \$\begingroup\$ How do you provide a 12V supply in an unpowered USB enclosure? \$\endgroup\$ May 7, 2017 at 23:24
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    \$\begingroup\$ You don't; that's what having a boost converter fallback would be for. If you're putting a SATA SSD in a USB enclosure, chances are you don't care too much about pushing the boundaries of write performance though. And AFAIK there is no such thing as an unpowered USB enclosure for standard PCIe cards. Anyways, even 5V and a boost converter would be better than 1.8V and a charge pump \$\endgroup\$
    – user371366
    May 7, 2017 at 23:27
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    \$\begingroup\$ do you want to break backwards compat or split a universal part into internal and external models? what's the advantage for doing so? \$\endgroup\$
    – dandavis
    May 8, 2017 at 4:21
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    \$\begingroup\$ 12V might be abundant in PCs but in smartphones and tablets it's a lot harder to come by. As they're such a huge market for storage now, it's quite likely that market forces are making such chips cheaper and therefore getting them used more widely in other applications like SSDs. Whatever your views on design, the answer to these questions is usually "market forces". \$\endgroup\$
    – Finbarr
    May 8, 2017 at 8:53
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    \$\begingroup\$ There were about 1.5 billion smartphones sold last year (Gartners). I don't know how many data centres there are in the world, but... \$\endgroup\$
    – Finbarr
    May 8, 2017 at 11:07

2 Answers 2


Three reasons really - TLDR: Reliability, and Compatibility

  • SSDs use 1.8V in their logic circuits to prevent interference.

AFAIK, The 12V level for programming and erasing applies to NOR flash cells, I'm not certain what the voltage level for NAND and VNAND flash is, as they use a different method to program and erase data. (Why does NAND erase only at block-level and not page level?). The voltage MUST be higher than Vccq, but not necessarily 12V.

  • Flash Cells are unreliable, and must use a controller to be usable.

An SSD is just Flash Cells with a controller that emulates a hard drive. By directly piping in a 12V line past the controller to the logic side of the device, you're really playing with fire. Flash memory is basically designed to fail predictably, and the controller is always maintaining the device, cleaning up messes, and sanitizing data streams. While the proper voltage rail could be supplied from the power input, it's more reliable to use localized voltage regulation within the controller's circuit space.

  • and finally, even modern computers don't have reliable voltage rails

The motherboard has a lot of voltage regulators itself because of this to provide clean power levels to the more sensitive components of the motherboard. Many PSUs have a limited number of 12V rails, and quite often on cheaper models there will be only 1. This can cause voltage dips as well as surges, the most I've seen in practice is 12V ±1V, but flash cells are extremely sensitive to voltage levels, and relying on external power regulation could end in disaster should something jump on/off the same rail and pull the voltage up or down. This would actually be common since an erase or write operations mean the computer is busy and will be pulling additional power compared to idle.

  • \$\begingroup\$ surely even just using a higher voltage to feed the regulated charge pumps built in to the flash chips would help reduce strain and improve performance though? \$\endgroup\$
    – user371366
    Jun 6, 2019 at 5:33

There is no 12V in smartphones, and it would take unnecessarily much board space while you can just put it in silicon.

That a 3.5" SSD has a lot of boards space is just because it conforms to the standard. If you look at an M2 SSD, board space becomes scarce again.

In short: why waste board space if you don't have to?

  • \$\begingroup\$ how does changing up the power routing slightly consume more board space? running a line from an existing pin on an existing connector doesn't add any additional footprints to the board \$\endgroup\$
    – user371366
    Jun 6, 2019 at 5:32
  • \$\begingroup\$ @dn3s Adding additional converters increases component count. \$\endgroup\$
    – Jeroen3
    Jun 6, 2019 at 6:15
  • \$\begingroup\$ and adding a charge pump doesn't? \$\endgroup\$
    – user371366
    Aug 9, 2022 at 16:32

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