Help with high voltage low current on gpio pin

[cdb101]

I use this voltage regulator to convert a 12 volts battery to 5 volts output.
It include a enable pin that can be set "high" (>1.4V) to turn the regulator on or "low" (0V) to turn it off. So far so good.

I discover that the regulator board pull the enable pin high through a 200k resistor if left unconnected. This way the regulator do it's job if enable pin is not used. I measure a voltage of 6.14volts on this pin and in my application it is connected to one uC GPIO pin through a 1k resistor. Obviously this voltage exceed the 3.3v gpio pin rating although it only represent 62uA of current for the pin to sink. Strange enough the high voltage do not destroy my uC pin and I don't know what to think or do about this?

It's true that at program startup, the digital pin is quickly turn to "low" thus sinking this 62uA current and dropping the voltage to 0 which effectively cut the regulator 5v output. Does 6.14v be a problem or there is so little current involve that voltage do not really matter in this case?

I can probably include a voltage divider to further drop the 6.14v to about 3v (adding the green 1k resistor in the drawing below). Another option migth be to change the digital pin mode from output to input_pulldown and combine this to the voltage divider? Any comments appreciated!
Thanks

[LarryD]

An open Drain logic level MOSFET could be used.

[cdb101]

This is a solution I will consider. On the other hand, this make the enable pin pointless since I could use a logic mosfet as a relay directly and toggle the 5v in and out.

[cdb101]

@cdb101

Does 6.14v be a problem or there is so little current involve that voltage do not really matter in this case?

I put a voltage divider with 10 k resistor. The enable pin voltage drop to 1.14 and the current sourced by the pin can be calculated to be around 56 uA.

Thus the esp pin is exposed to a very brief 56 uA @ 6.14v worst case scenario during transient. Should I care?

Alternatively, adding a 10k resistor to ground just like the figure in post #1 also worked. With a 10k, the voltage available on the En pin is closer to 3.3v. It is better in term of making sure the enable pin is really enable compare to the initial 1K suggestion.