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It doesn't say on the data-sheet, but I'm curious where i can find if Microchip uses IEC or SI as standard for there memory sizes.

For example:

http://ww1.microchip.com/downloads/en/DeviceDoc/21754M.pdf

The Microchip Technology Inc. 24AA512/24LC512/ 24FC512 (24XX512*) is a 64K x 8 (512 Kbit) Serial Electrically Erasable PROM, capable of operation across a broad voltage range (1.7V to 5.5V)….

My question is:

By IEC (binary) its

(512*1024)/8 = 65.536 bytes

but according to SI its

(512*1000)/8 = 64.000 bytes

Anyone?

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    \$\begingroup\$ Think about how addressing has to work inside these memories. It's going to be a power of 2. Therefore "64K x 8" means 65536 bytes of 8 bits each. \$\endgroup\$ – Olin Lathrop Dec 16 '13 at 17:13
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I thought SI could be distinguished by the lowercase 'k'? In any case, because of the structure of address lines, memory is almost always a power of two in size. So it's 65536 bytes.

(Hard disk manufacturers have taken to using SI as it produces larger numbers for the same number of storage units, but this hasn't spread to RAM/EEPROM yet)

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  • \$\begingroup\$ That's right, there's a bit of a table of alternatives at en.wikipedia.org/wiki/Kibibyte for byte sizes although I haven't seen KiB used much in practice. \$\endgroup\$ – PeterJ Dec 16 '13 at 11:13

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