K9F5608U0B-YIB0000: 528 byte in page, or 512?
A.L. Klyutchenya
asoneofus at nm.ru
Tue Apr 1 06:02:14 EST 2003
Dear sir,
Thank you for quick answer.
1 Апрель 2003 07:36, Charles Manning написал:
> This is NAND flash. Do not try to treat it as NOR flash.
> Use it with mtd/nand and use YAFFS or JFFS2 file system.
I want to write the first image, for booting system. I create
this image (with kernel, MTD & Co). I have a possibility his
write. Herewith: I have an physical access to page: size
256+256+16, and block on 32 pages. While NO MTD and NO kernel!
> You must write NAND in whole pages, not on a per-byte basis.
I understand that page it is necessary to write, but what bring
the first image: chain on 512+512.... or 512+16 free + 512+16
free...
As even so write? :-) (to hrdware-level)
> On Tue, 01 Apr 2003 10:07, A.L. Klyutchenya wrote:
> > I try to use the microcircuits K9F5608U0B-YIB0000 from
> > Samsung. Herewith, size of the page size of this chip is 512
> > + 16 bytes (528). When I write the image of the disk on
> > chip, me it is necessary to write all bytes consecutively?
> > Or only 512 on each page, but remained 16 not to use?
--
ВсехБлаг!
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/ | / / А. Л. Клютченя
/ /| | / / (asoneofus)
/ /_| |/ / mail: asoneofus at kde.ru
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