I`m planning to use the 24C64 EEPROM with my Raspberry Pi.

According to the datasheet the EEPROM "Single Supply Voltage" is 4.5 to 5.5 V.

I read somewhere that 5V will destroy the I2C pins from the Pi.

Is this right and what do I have to do to use the EEPROM without destroying my Pi?

  • there are plenty of EEPROM chips that support 3v3 instead of 5v, use them.
    – lenik
    Commented Oct 7, 2014 at 13:57

1 Answer 1


To use it from the Pi you'd need to supply power from the 5V and ground pins.

Check that the chip does not actively drive the SDA line (it shouldn't drive the SDA line with 5V, it should let the line float to the Pi's pull-up to 3V3).

As long as the chip doesn't actively drive SDA it should be quite safe to use with the Pi.

  • So i should just connect Vcc to 5V / GND to Vss / SDA to SDA and SCL to SCL and everything should work fine ?
    – Tim
    Commented Oct 7, 2014 at 17:33
  • That is NOT what I said. At a minimum I'd check the voltage on the SDA output of the chip. If you only want to read from the chip then just power it from the 3.3V pin. It only needs a higher voltage if you want to write to the chip. But the specs say it needs 0.7*VCC as the logic high. 0.7*3 is 3.5V which suggests that the Pi floating to 3.3V wouldn't be seen as high. However there is a way around that problem. If this is all nonsense to you then you probably ought to buy a 3.3V EEPROM.
    – joan
    Commented Oct 7, 2014 at 18:09
  • So if i power the chip with 3.3V it is possible to read, but if i want to write to the chip i would at least need 3.4 V power ?
    – Tim
    Commented Oct 7, 2014 at 18:32
  • There is a difference between the chip power supply voltage and the voltage needed to be seen as logic high and logic low by the chip. To write to the chip you need to take both into account (as well as the write select pin). This is detailed in the linked specs. I suggest you try re-reading the specs.
    – joan
    Commented Oct 7, 2014 at 18:48
  • You definitely know more than me, but in your previous post you wrote 0.7 times 3 is 3.5 V which is wrong (it is 2.1). Did you make a mistake and what you said is wrong or am i wrong ? Also why exactly are you multiplying the Input High Voltage with 3 ?
    – Tim
    Commented Oct 7, 2014 at 19:05

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