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Visitor II
September 28, 2025
Solved

M24C04-F DNF5 I2C address

  • September 28, 2025
  • 2 replies
  • 295 views

Dear All,

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

MMici1_0-1759081896628.png

I can't find any explanation of this value.

 

Thank you

    This topic has been closed for replies.
    Best answer by Peter BENSCH

    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

    2 replies

    Technical Moderator
    September 29, 2025

    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

    MMici.1Author
    Visitor II
    October 1, 2025

    Of course, thanks :D.