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The driver RAM requirement depends on flash parameters such as block size and run-time configurations such as sector size. For a particular instance, a general formula can give an approximate:

No Format
    if (secs_per_blk < 255) {
        temp1 = ceil(blk_cnt_used / 8) + (blk_cnt_used * 1);
    } else {
        temp1 = ceil(blk_cnt_used / 8) + (blk_cnt_used * 2);
    }
    if (sec_cnt < 65535) {
        temp2 = sec_cnt * 2;
    } else {
        temp2 = sec_cnt * 4;
    }
    temp3 = sec_size;
    TOTAL = temp1 + temp2 + temp3;

where

secs_per_blk

The number of sectors per block.

...

The sector size configured for the device, in octets.

 

secs_per_blk and sec_cnt can be calculated from more basic parameters :

No Format
    secs_per_blk = floor(blk_size / sec_size);
    sec_cnt      = secs_per_blk * blk_cnt_used;

where

blk_size

The size of a block on the device, in octets

 

Take as an example a 16-Mb NOR that is entirely dedicated to file system usage, with a 64-KB block size, configured with a 512-B sector. The following parameters describe the format :

No Format
    blk_cnt_used = 32;
    blk_size     = 65536;
    sec_size     = 512;
    secs_per_blk = 65536 / 512 = 128;
    sec_cnt      = 128 * 32    = 4096;

and the RAM usage is approximately

No Format
    temp1 = (32 / 8) + (32 * 2) = 68;
    temp2 = 4096 * 2 = 8192;
    temp3 = 512;
    TOTAL = 68 + 8192 + 512 = 8772;

In this example, as in most situations, increasing the sector size will decrease the RAM usage. If the sector size were 1024-B, only 5188-B would have been needed, but a moderate performance penalty would be paid.