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idata
Community Manager
1,040 Views

X25-M(V) erase block size?

On the Internet I've found two different oppinions about erase block size of Intel's X25-M (and X25-V) SSDs: 128kB and 512kB

128kB:

http://techreport.com/articles.x/15433

Intel gives an example in which a host system generates a 4KB write request that, thanks to a drive's 128KB erase block size, actually incurs a 128KB NAND write. Dividing the NAND write size by the request size yields the amplification factor, which is 32 in this case. Intel says the X25-M's write-amplification factor is extremely low at 1.1, while "traditional" SSDs have much higher amplification factor of 20.

512kB:

http://www.anandtech.com/show/2614/3

Group a bunch of cells together and you've got a page. A page is the smallest structure you can program (write to) in a NAND flash device. In the case of most MLC NAND flash each page is 4KB. A block consists of a number of pages, in the Intel MLC SSD a block is 128 pages (128 pages x 4KB per page = 512KB per block = 0.5MB). A block is the smallest structure you can erase. So when you write to a SSD you can write 4KB at a time, but when you erase from a SSD you have to erase 512KB at a time.

http://www.xbitlabs.com/articles/storage/display/intel-x25m-ssd_2.html

For the X25-M this block is as large as 128 pages or 512 kilobytes or half a megabyte. As a result, if there is a request to erase (or rewrite) one page, the drive has to erase 128 pages.

and then again, even on this forum I've found both 128kB and 512kB info:

128kB:

http://communities.intel.com/message/28730# 28730

512kB:

http://communities.intel.com/message/68942# 68942

So, can someone officially say: is the size of erase block on X25-M(V) 128kB or 512kB ?

Thx.

1 Reply
MBall5
New Contributor II
124 Views

Hello, I would like to know that unfortunately Intel cannot disclose the size of the block used at the Intel Solid State Drives, I suggest you to use the default size if the hardware provides de option to set the block size.

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