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'[EE] Calculate the last location of 24LC32 eeprom'
2008\11\18@145400 by

Hi
Stupid question, that I don't find on the datasheet.

How do I calculate the last location in the eeprom.

datasheet say's:
24LC32 is a 4K x 8 (32K bit)

I think the first address is HI-0x00 LOW-0x00

(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.
>
> 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.
piclistphilpem.me.uk
http://www.philpem.me.uk/
Stefano Vanzo wrote:
> How do I calculate the last location in the eeprom.
>
> datasheet say's:
> 24LC32 is a 4K x 8 (32K bit)

4096 - 1 = 4095 = FFFh

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Thanks

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