I have a chip that operates on 1.8V Vcc and has a digital interface I2C that also obviously outputs 1.8V signals and receives 1.8V signals. The datasheet says anything above 2V on VCC or digital input pins is fatal.

I want to interface this to a 3.3V micro controller dev board (which has its own

I have one 1.8V regulator that can output up to 1A. I also have some 2N4403 and 2N4401 and some resistors at hand.Is there a way that I can create a 1.8V to 3.3V and 3.3V to 1.8V with some passive components?

I have +5V/2A, +3.3V/1A and 1.8V/1A supply rails on my bread board.

If not, I would assume I will need a transciver or logic translator chip? Can you recomment me a few if possible?


I found the circuit below, but I guess 1.8V is not enough to switch mosfets on/off.... is there any trick I can use with just my NPN/PNP jelly bean transistors?

enter image description here


1 Answer 1


The trick you can exploit with I2C is that it is open collector. This simplifies translating voltage somewhat.

But rather than raiding your parts bin, consider using a translation IC. Here’s a bunch from NXP: https://www.nxp.com/products/interfaces/ic-spi-serial-interface-devices/ic-voltage-level-translators.

  • \$\begingroup\$ Thanks for your answer. How can I exploit open collector? \$\endgroup\$
    – DEKKER
    Commented Mar 13, 2022 at 21:52
  • 1
    \$\begingroup\$ The fact that I2C only sinks current makes it easier to deal with for voltage translation. Put the pull-up in the 3.3V domain and it covers the translated-down 1.8V too. \$\endgroup\$ Commented Mar 13, 2022 at 23:39

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