I've looked at articles detailing I2C level shifting and referencing the below figure I have a question. enter image description here

If the gates of the FETs were driven by 2.5V and everything else remained the same, would the level shifting still work?

I would think a low from the 3.3V side would propagate but a low from the 5V would not get through.

At what low-side voltage or gate voltage would this circuit not work?


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    \$\begingroup\$ Did you notice the body diodes of the FETs? \$\endgroup\$ Sep 2, 2016 at 5:27

1 Answer 1


The FETs activate when their gate voltage, relative to the source, is sufficiently above the gate threshold voltage. (The threshold voltage is specified for a very small drain current, such as 250 µA; here, we need more current for the pull-up resistors. The gate voltage we want is the one for which the datasheet specifies RDS(on).)

The low-voltage side can activate the FETs by driving low the source.

The high-voltage side can activate the FETs by driving low the source through the body diode, which adds about 0.7 V.

So this circuit will work with a gate voltage of 2.5 V as long as the FETs fully switch on at 1.8 V.


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