I was looking at the serial eeprom and wondering what else it contained (apart from the obvious VIN)
Spotted that the last 2048 bytes are a mirror image of the first 2048 bytes.
Then I got a bit distracted and realised just what
this post was trying to explain about checksums in the serial eeproms
Interesting read.
However my own assumption is that the pages in my example are 32 bytes in size and the last 2 bytes are the checksum. The checksum is calculated by adding the first 30 bytes together and then negating the sum.
Example 1(0x01 + 0x03 + 0xDC + 0x9D + 0x56 + 0x20 + 0x04 + 0xDD + 0xBE +
0x01 + 0x05 + 0x6F + 0xB1 + 0x48 + 0xD7 + 0xB3 + 0xE8 + 0xF5 +
0x2D + 0x13 + 0x9E + 0xBC + 0x09 + 0x6A + 0x52 + 0x52 + 0x02 +
0x1C + 0x03 + 0x09) = 0x0B42
0x0B42 = 0000 1011 0100 0010
Negate this (flip the ones and zeros)
0000 1011 0100 0010
1111 0100 1011 1101
1111 0100 1011 1101 =
0xF4BD (last two bytes)
Example 2(0x05 + 0x03 + 0x38 + 0x50 + 0x30 + 0x39 + 0x30 + 0x37 + 0x31 +
0x31 + 0x35 + 0x42 + 0x20 + 0x20 + 0x20 + 0x20 + 0x20 + 0x20 +
0x00 + 0x00 + 0x00 + 0x00 + 0x00 + 0x00 + 0x00 + 0x00 + 0x00 +
0x00 + 0x00 + 0x00) = 0x02F9
0x02F9 = 0000 0010 1111 1001
Negate this (flip the ones and zeros)
0000 0010 1111 1001
1111 1101 0000 0110
1111 1101 0000 0110 =
0xFD06 (last two bytes)
Picture shows what I'm trying to explain!
There is a LOT of information and answers in this forum. If I look at some of the queries I posted, I'm cringing because they are the same questions everyone asks time and again,
and they've already been answered! Sorry guys