M24C04-F DNF5 I2C address

MMici.1 · ‎2025-09-28

Dear All,

can someone, please, specify the meaning of the bit 1 as 'A8' in the I2C address of the M24C04-FMH6TG?

I can't find any explanation of this value.

Thank you

Peter BENSCH · ‎2025-09-29

The M24C04 contains 512 (2^9) bytes, which require 9 bits for addressing. Since an address byte only has 8 bits (A0...A7), the 9th bit (A8) must also be transferred, for which bit b1 in the device select code is used.

Does it answer your question?

Regards
/Peter

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View solution in original post

Peter BENSCH · ‎2025-09-29

The M24C04 contains 512 (2^9) bytes, which require 9 bits for addressing. Since an address byte only has 8 bits (A0...A7), the 9th bit (A8) must also be transferred, for which bit b1 in the device select code is used.

Does it answer your question?

Regards
/Peter

In order to give better visibility on the answered topics, please click on Accept as Solution on the reply which solved your issue or answered your question.

MMici.1 · ‎2025-10-01

Of course, thanks :D.