Post by thread:[EE] Calculate the last location of 24LC32 eeprom

Hi Stupid question, that I don't find on the datasheet. How do I calculate the last location in the eeprom. Hi and Low address. datasheet say's: 24LC32 is a 4K x 8 (32K bit) I think the first address is HI-0x00 LOW-0x00 last address I would calculate: (4 x 1024) - 1 = 4095 -> HI-0x0F LOW-0xFF is this correct? Stefano Vanzo

Stefano Vanzo wrote: > How do I calculate the last location in the eeprom. > Hi and Low address. > > datasheet say's: > 24LC32 is a 4K x 8 (32K bit) Yep -- (32 x 1024) bits, or ((32 x 1024) / 8) = 4096 bytes = 4Kbytes. > I think the first address is HI-0x00 LOW-0x00 That's right. > last address I would calculate: > > (4 x 1024) - 1 = 4095 -> HI-0x0F LOW-0xFF Exactly right. 4096 bytes total, minus 1 (address 0x000 is the first addressable byte), which is address 4095. 4095 in hex is 0xFFF, which means the high byte is 0x0F, and the low byte is 0xFF. -- Phil. spam_OUTpiclistTakeThisOuTphilpem.me.uk http://www.philpem.me.uk/

Stefano Vanzo wrote: > How do I calculate the last location in the eeprom. > Hi and Low address. > > datasheet say's: > 24LC32 is a 4K x 8 (32K bit) 4096 - 1 = 4095 = FFFh ******************************************************************** Embed Inc, Littleton Massachusetts, http://www.embedinc.com/products (978) 742-9014. Gold level PIC consultants since 2000.

Thanks Philip Pemberton ha scritto: {Quote hidden}> Stefano Vanzo wrote: > >> How do I calculate the last location in the eeprom. >> Hi and Low address. >> >> datasheet say's: >> 24LC32 is a 4K x 8 (32K bit) >> > > Yep -- (32 x 1024) bits, or ((32 x 1024) / 8) = 4096 bytes = 4Kbytes. > > >> I think the first address is HI-0x00 LOW-0x00 >> > > That's right. > > >> last address I would calculate: >> >> (4 x 1024) - 1 = 4095 -> HI-0x0F LOW-0xFF >> > > Exactly right. 4096 bytes total, minus 1 (address 0x000 is the first > addressable byte), which is address 4095. 4095 in hex is 0xFFF, which means > the high byte is 0x0F, and the low byte is 0xFF. > >