hankshone
- 10th June 2009, 16:36
I am trying to use 24AA1025 for data storage.
Word address <ADD> and word variable <DA>. I am trying to save DA into the EEPROM memory at a bandwidth of 50.
I read through all the datasheets and the forum but still have some problems and confusions...
1. When I was trying to access the second 512KBs of this chip, it says in the datasheet I need to change the control byte from $A0 to $A8. I did that but it seems that there is no response. The first segment, I mean when the control byte is $A0, everything works fine. I saw some previous programs showing that when accessing the second 512KB segment, the address(ADD)should start over from 0, I did that, with still no clue.
2. Another confusion is when ADD is defined as a word variable, the largest address is 65535. As the first segment of 24AA1025 has 64000 bytes, so where does it go when ADD is 64000-65535 when control byte is $A0?
I just don't know how to access the second 512KB of this chip.....
Any suggestion is greatly appreciated..
Word address <ADD> and word variable <DA>. I am trying to save DA into the EEPROM memory at a bandwidth of 50.
I read through all the datasheets and the forum but still have some problems and confusions...
1. When I was trying to access the second 512KBs of this chip, it says in the datasheet I need to change the control byte from $A0 to $A8. I did that but it seems that there is no response. The first segment, I mean when the control byte is $A0, everything works fine. I saw some previous programs showing that when accessing the second 512KB segment, the address(ADD)should start over from 0, I did that, with still no clue.
2. Another confusion is when ADD is defined as a word variable, the largest address is 65535. As the first segment of 24AA1025 has 64000 bytes, so where does it go when ADD is 64000-65535 when control byte is $A0?
I just don't know how to access the second 512KB of this chip.....
Any suggestion is greatly appreciated..