1 Reply Latest reply on Oct 4, 2010 12:49 PM by Mauriceb@Intel

    X25-M(V) erase block size?

    trx

      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.