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I recently picked up a Crucial M500 240GB SSD (20nm NAND) and I'm trying to figure out the best way to partition it. Currently, I'm using fdisk -cu starting at sector 2048.

I believe the nand page size is 16KB.

I cannot find anywhere what the nand erase block size is for it.

Does anyone know the answer to this or general advice on partitioning this particular series of SSDs?

  • 1
    Opening the drive and Googling the part numbers on the NAND chips may be needed. – LawrenceC Mar 14 '14 at 3:48
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This information is sometimes published in SSD manufacturer specs, but other times it's not there, especially for CF or SD memory cards. Short of using Google to search for someone else who has done the research, you can try to estimate it yourself using FlashBench. Download it here: https://github.com/bradfa/flashbench

This tool does random reads on an SSD and will plot a table showing the read times. (You should have done some writes to the SSD already, because reading all-erased pages is often simulated by the controller chip.) By looking for breaks in the time by block size, you can infer what the erase block size is. Here is a sample from the README:

== Guess erase block and page sizes ==

''flashbench -a <device>''

This is a simple read-only test doing small reads across boundaries of various sizes. Example:

$ sudo ./flashbench -a /dev/mmcblk0  --blocksize=1024
align 134217728 pre 735µs       on 1.08ms       post 780µs      diff 324µs
align 67108864  pre 736µs       on 1.05ms       post 763µs      diff 300µs
align 33554432  pre 722µs       on 1.04ms       post 763µs      diff 294µs
align 16777216  pre 727µs       on 1.05ms       post 772µs      diff 302µs
align 8388608   pre 724µs       on 1.04ms       post 768µs      diff 299µs
align 4194304   pre 741µs       on 1.08ms       post 788µs      diff 317µs
align 2097152   pre 745µs       on 950µs        post 811µs      diff 171µs
align 1048576   pre 745µs       on 945µs        post 807µs      diff 169µs
align 524288    pre 743µs       on 936µs        post 799µs      diff 165µs
align 262144    pre 746µs       on 948µs        post 809µs      diff 171µs
align 131072    pre 737µs       on 935µs        post 804µs      diff 165µs
align 65536     pre 735µs       on 925µs        post 796µs      diff 159µs
align 32768     pre 735µs       on 925µs        post 800µs      diff 157µs
align 16384     pre 745µs       on 911µs        post 781µs      diff 148µs
align 8192      pre 785µs       on 808µs        post 725µs      diff 53.3µs
align 4096      pre 784µs       on 788µs        post 779µs      diff 5.85µs
align 2048      pre 787µs       on 793µs        post 789µs      diff 4.65µs

This shows the access times to do two 1024 byte reads around the boundaries of power-of-two aligned blocks. Reading at the end of a 128 MB unit takes around 735 microseconds, reading the last block of this unit together with the first block of the next one takes about 1080 microseconds and reading the first two blocks in a 128 MB unit takes around 780 microseconds.

The most interesting number here is the last one, the difference between the second number and the average of the first and the third is 324 microseconds. These numbers all stay roughly the same for all units between 4 MB and 128 MB.

However, from 2 MB down to 16 KB, the last column has a much lower value. This indicates that whatever the memory card does on a 4 MB boundary does not happen at other boundaries. The educated guess here is that 4 MB is the erase block size, also called the segment or allocation unit size. This erase blocksize will need to be used in other tests following this one.

Similarly, both 16 KB and 8 KB boundaries are special. The logical explanation for this is that the card has 8 KB pages, but can use multi-plane accesses to read two 8 KB pages simultaneously.

Some cards only show a clear pattern using accesses with certain block sizes, other cards do not show any pattern, which means that the numbers need to be determined differently.

Also, cards that were never fully written may show a different behaviour because access times on pre-erased segments are different from those that have been written.

1

Another attempt is to align on a boundary that is a multiply of any practical block size.

With this concept, it is most common to align on 1MB boundary, so it won't matter if the block size is 4 or 16kb; all of these will be mulitplys of 2 and below 1M so aligning on that boundary will fit them all.

However, applying this concept depends on what you are aligning; loosing 1MB at the start of a mass storage device is totally acceptable while loosing that many times in a different scenario might not be.

0

The erase block size has no bearing on alignment, and the M500 supports garbage collection, so performance is not an issue. Please refer to the 2nd page of this PDF from micron's site which will help you determine erase block size based on the NAND used in your M500.

as far as alignment advice, please check out this fantastic Superuser Post.

Here's the screenshot of the page: enter image description here

  • 2
    So in this case, where in this diagram would the erase block size be given? – hbogert Oct 25 '16 at 13:12

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