2

Context

So I have 2 SDXC cards of identical storage space by Lexar (Lexar 1000X) and Sandisk (Sandisk Extreme Plus). They have a small variance in terms of storage space <~1%. Functionally, they work the same on my Sony a6300 and I've had no problems with either one.

Question

I've noticed that the Lexar takes about 5 seconds or less to format on my camera (regardless of how full it is) but the Sandisk takes about 30 seconds to 1 minute. I can see the number of images I can store vary slightly between brands but is there a reason why it takes forever to format one over another?

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Your cards are not in the same speed category.

The Lexar 1000x belongs to the UHS-II category while the Sandisk Extreme Plus belongs to UHS-I (UHS = Ultra High Speed, more information about class/category on Wikipedia: https://en.wikipedia.org/wiki/Secure_Digital). UHS-I cards have a theoretical limit of 104MB/s vs 312MB/s for UHS-II.

In real life, this test (http://alikgriffin.com/best-fuji-x-t1-memory-cards-uhs-ii-speed-tests) shows that the UHS-II Lexar 1000x (32GB) card have a writing speed of about 145.0 MB/s, compared to 89 MB/s for the slower UHS-I Sandisk Extreme Plus. Those values are given for images transfer (RAW+JPEG and JPEG). Formatting involves a different data flow and the real writing speeds while formatting are probably different than those displayed here (...and can differ a lot !).

One question might interests you : How can I know what speed card to get for my camera?

  • 1
    The major counterpoint is that it assume the camera supports UHS-II. I think the Nikon D500 does but the Sony doesn't. Wouldn't that mean both cards revert to UHS-I speeds? – unsignedzero Sep 19 '16 at 17:24
  • Good remark, I just checked... and the Sony A6300 doesn't support UHS-II => back to square one ? – Olivier Sep 19 '16 at 17:32
  • I really can't say. I don't have any other non UHS-I camera to test the card either. I doubt it's a dud since it works just fine though. – unsignedzero Sep 19 '16 at 17:36
  • UHS-II cards are compatibles with USH-I readers, just like USB 3 and USB 2, but then you should probably have USH-I performances... – Olivier Sep 19 '16 at 18:36
  • *UHS but yes I have that. I don't have a UHS-II reader since the UHS-I speed works just fine for me. However, I format the cards on my camera which is my speed problem, should have mentioned that. – unsignedzero Sep 20 '16 at 5:28
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In-camera formatting involves writing data to the flash card. Whenever a flash card writes new data, it must first erase an existing flash page and then write new data to it. The flash card maintains some extra pages (usually pre-erased) to make this faster. However, if there are not enough pre-erased pages to handle a series of writes, performance slows way down while it erases pages.

In some cases, it may even need to evict existing data so that it spreads the erasures out over more flash pages, because you can only erase a flash page so many times.

For flash cards that start out full, there are many fewer free pages than there were when the flash card was new, which is one reason why flash cards tend to be slower the second time around. (The other reason has to do with the way signals are encoded in each flash cell, but this tends to be a much more gradual degradation in performance.)

Either way, there are only a certain number of spares. The more spares, the faster the card can write new data.

If you know whether your flash card's NAND flash erases to all zeros or all ones, you could ostensibly write all zeros or ones across the card so that no block will require an erase when you write data to it next. This is because a flash page only has to be erased when bits change in one direction and not when they change in the opposite direction. (The details depend on the flash part and on how the controller is configured.) Of course, realistically, you can never be sure, so this is probably not the best approach.

Another, better solution is to send the right SD card commands (CMD32, CMD33, and CMD38 with appropriate delays in between) to erase all of the flash pages, assuming your flash card supports those commands. This will dramatically improve performance. However, depending on whether the flash card does the erasure synchronously or asynchronously, it could take a long time.

It is possible that the SanDisk card supports the SD erase commands, and that your camera is using it, and that this is the reason for the extra time. If so, the SanDisk card will be much faster than the Lexar after reformatting.

It is also possible that the Lexar card supports the SD erase commands, and that your camera has used them to wipe deleted files, resulting in more free flash pages ready for new data, thus making the format faster because it doesn't have to wait for a slow erasure. In that case, the Lexar card will be much faster after formatting

Obviously none of this really tells us for sure what's going on—you'd probably need a proper bus analyzer to be certain—but maybe it at least gives you a vague idea of some of the issues that might be involved.

And it could be something else entirely, such as the SanDisk card doing more wear leveling than the Lexar, or something else that I've forgotten.

Either way, I wouldn't worry about it.

  • Yeah I'm not worrying too much since the card works after the format and that's what matters. I'm more asking out of curiosity if there's an explanation. I thought formatting wrote a new page file system and assumed all cells are dirty? – unsignedzero Sep 21 '16 at 2:24
  • AFAIK, unless the camera sends an erase command to tell the card to release the flash pages, the flash pages that were in use before are still in use as far as the flash card's controller chip is concerned. The controller in the card doesn't know that you're formatting the card; a block write is a block write. (Well, that's not strictly true—I seem to recall some research papers about flash controllers actually trying to understand FAT and/or EXFAT and trying to figure out whether filesystem changes meant that pages could be freed, but... I'm not sure if they exist in the wild.) – dgatwood Sep 22 '16 at 6:16

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