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brianko
11-11-2012, 08:07 PM
OK, here's a question: You have a 16K calibration going into a 64K EEPROM. There are 4 16K offsets to choose from (0000, 4000, 8000, C000). GM uses the 4000 offset. If you use another offset, will the computer look for the first 16K block that is not empty? Reason I ask is I noticed an EEPROM from another tuner is using the C000 offset rather than the 4000 offset.

RobertISaar
11-11-2012, 10:37 PM
if you're throwing it into a 64KB PROM, then you could just stack the BIN 4 times and make a 64KB BIN out of it and it will work.

otherwise, offseting a 16KB BIN starting at either 4000 or C000 will work. reason being that when using a 64KB PROM, the A15 pin is connected to 5V, so the PROM is basically mirrored from 0000-7FFF into 8000-FFFF. you can thank GM for that, why they chose to purposely disable the ability to use 64KB PROMs with older ECMs, i'll never know. it's a pain to try and enable the other data.

1project2many
11-12-2012, 07:15 AM
GM uses the 4000 offset.
Yes, but no. The processor uses the last two addresses in the calibration to find the address for the beginning of the code. This means the calibration has to fit into the space that ends with the last address in the chip, $FFFF. But using $C000 works because the processor cannot actually access addresses higher than $7FFF. A request for $FFFF is the same as $7FFF.