
Originally Posted by
skimask
I thought PBP handled that part for you. 'cause with a generic serial eeprom, the address is something like $a0, and that's how you address it...$a0, without the left shift, PBP handles the R/W bit.
But then again, we don't know exactly what kind of secret magic device is being addressed here...
$A0 is the LOWER 7 bit's of the address, shifted left one place. The DEVICE address being sent out and 'seen' by the reciever is %01010000, or $50. Check the specs of I2C, not just EEPROMS. The device address is specified as a 7 bit address, not 8. But since we are adding the R/W bit to the 8 bit address byte, we have to left shift the DEVICE address one place.
The confusion comes in dealing with 2 different data "specs" or "types" combined into one 8 bit Byte being send. Then suming the two values to "use" one value. That "one" value is not the device address.
You ever do this: (I know I have)
Gang 8 64KBit EEproms and use the device address (bit 4:1) by just rolling the long address over directly into the device address. It won't work. Trust me, just shift the address left once to embed the address into the device address byte.
Last edited by JD123; - 2nd April 2008 at 21:52.
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