I always asked myself why the pages from the mentioned DataFlash chips, are not(standard) binary sized, but have eight more bytes every 256 bytes?
Is it a physical/production side effect, or are those bytes meant for storing some "˜file system' data, what's the use for it?
I'm using these chips in combination with a FAT file system, for storing logged data, and thus use/need "˜binary' ordered page/buffer byte addresses. Now as people know, for those that use these chips, that to obtain a "˜binary' page and buffer byte addresses, some left and right shifting is needed on the page and buffer byte address bits, to gain access to the correct bytes. Not really a big deal, but for one, the extra bytes are not used anyway, second, one cannot use the "˜continuous read' commands in a binary way, thus when reading sequential data, each time a page has been read, one must stop reading, send a new address command(and shifting address bits again), and start reading again, a lot of overhead when reading big sequential files on one DF chip.
Fortunately, there is the possibility to order the page/buffer addresses in a "˜binary' way(although, once programmed for "˜binary' addresses, you cannot go back), then no(special) bit shifting is needed to obtain the correct page/buffer byte address, and one can use the "˜continuous read' commands in a "˜binary' way(which speeds up the transfers a lot, Dean could use this in he's LUFA projects, when using only one DF chip), it also simplifies the coding.
This is what I've done, programmed the DataFlash for binary use, the only drawback is that I lose , in my case(AT45DB642D), 256KB of flash space.
So what's the reason why these DataFlash chips use, standard, non binary sized pages?
