Re: Lifetime of flash memory

[linux-os (Dick Johnson)]

On Mon, 27 Mar 2006, Sergei Organov wrote:

> "linux-os \(Dick Johnson\)" <linux-os@xxxxxxxxxxxx> writes:
> [...]
>> CompactFlash(tm) like SanDisk(tm) has very good R/W characteristics.
>
> Try to write 512-byte sectors in random order, and I'm sure write
> characteristics won't be that good.
>
>> It consists of a connector that exactly emulates an IDE drive connector
>> in miniature, an interface controller that emulates and responds to
>> most IDE commands, plus a method of performing reads and writes using
>> static RAM buffers and permanent storage in NVRAM.
>
> Are you sure they do have NVRAM? What kind of NVRAM? Do they have backup
> battery inside to keep NVRAM alive?
>

NVRAM means [N]on-[V]olatile-[RAM]. Any of many types, currently NAND flash.
No battery required.

> [...]
>
>> Note that the actual block size is usually 64k, not the 512 bytes of a
>> 'sector'. Apparently, some of the data-space on each block is used for
>> relocation and logical-to-physical mapping.
>
> Wrong. AFAIK, first disks had FLASH with 512b blocks, then next
> generation had 16K blocks, and currently most of cards have 128K
> blocks. Besides, each page of a block (64 pages * 2K for 128K block) has
> additional "system" area of 64 bytes. One thing that is in the system
> area is bad block indicator (2 bytes) to mark some blocks as bad on
> factory, and the rest could be used by application[1] the same way the
> rest of the page is used. So physical block size is in fact 64 * (2048 +
> 64) = 135168 bytes.
>
> Due to FLASH properties, it's a must to have ECC protection of the data
> on FLASH, and AFAIK 22-bits ECC is stored for every 256 bytes of data,
> so part of that extra memory on each page is apparently used for ECC
> storage taking about 24 bytes out of those 64. I have no idea how the
> rest of extra memory is used though.
>

Huh? There is no ECC anywhere nor is it required. The flash RAM is
the same kind of flash used in re-writable BIOS, etc. It requires
that an entire page be erased (all bits set high) because the
write only writes zeros. The write-procedure is a byte-at-a-time
and results in a perfect copy being written for each byte. This
procedure is hidden in devices that emulate hard-disks. The
immediate read/writes are cached in internal static RAM and
an ASIC manages everything so that the device looks like an
IDE drive.

> BTW, the actual block size could be rather easily found from outside, --
> just compare random access write speed against sequential write speed
> using different number of 512b sectors as a write unit. Increase number
> of sectors in a write unit until you get a jump in random access write
> performance, -- that will give you the number of sectors in the block.
>

Huh? The major time is the erase before the physical write, the entire
physical page needs to be erased. That's why there is static-RAM buffering.
It is quite unlikely that you will find a page size using any such
method.

> [1] By application here I mean the code that works inside the CF card
> and deals with the FLASH directly. This memory is invisible from outside
> of CF card.
>
> -- Sergei.
>

Cheers,
Dick Johnson
Penguin : Linux version 2.6.15.4 on an i686 machine (5589.42 BogoMips).
Warning : 98.36% of all statistics are fiction, book release in April.
_


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