Max reverse INPUT voltag/current for gpio set to OUTPUT

Question

What would happen to a gpio pin set as high output if the pin already see a voltage higher than the MCU's voltage? Lets say 3.3V vs 3.4V

I will probably solve this with a diode and using one extra gpio pin, but I got curious. How will the internal circuitry of the gpio react to a small over-voltage?

For the long story: I have two signals controlling the same point for a power enable pin. One gpio from a battery powered esp32 at 3V to 3.6V range. The other is from a voltage divider from USB 5V, and this will also have a voltage range, since every usb port is not the same. The resistance is almost 100k, so the current is probably too low for anything to happen anyway, or is it? The MCU also want to measure the usb voltage to boot in to other code if the device is powered from usb instead of the battery, so a diode for that pin does not work. I am short on space, so it would be neat if these two things could be connected on one gpio pin.

On-button turns the circuit on, and the MCU can decide on its own when to turn off. With USB connected, the circuit is always on

Link to Falstad circuit.

Answer 1

What would happen to a gpio pin set as high output if the pin already see a voltage higher than the MCU's voltage? Lets say 3.3V vs 3.4V

^{I'm assuming the GPIO is set to push-pull output (Open drain, a.k.a. Hi-Z, is another output config).}

The output section of a push-pull output GPIO is a CMOS totem-pole buffer.

When the output is driven the pin will be connected to either rail (depending on whether the output is HI or LO) through a relatively low resistance which is the buffer transistor's equivalent resistance (The actual value depends on the chip but is usually less than 50 Ohms).

So, if the output is HIGH the external network and the output buffer will form an adder (a.k.a. summer). As you might expect, the net voltage will be dominated by the side having the lowest output resistance, in your case the MCU methinks. So the pin voltage will still be 3V (for 3V MCU supply voltage and 4.75~5.25V USB voltage).

But...

A non-zero current (in the order of microamps, you can calculate by yourself) will flow into the pin, through the high-side equivalent resistance (GPIO output driver MOSFET), out of the VDD pin and back to the source (battery, in your case). If the current coming out of the battery (i.e. load current) is already that low (or even lower) then the reverse current will win and the difference current will flow back to the battery. So, if the battery is a non-rechargeable one then this could be a problem, otherwise the USB port will recharge your battery with very low current.

If the GPIO output is set to zero whilst the external network is still connected then a current set by the external network's equivalent resistance will flow into the pin.

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One other thing to consider here is the start up i.e. the GPIO's default (or reset) config. For some MCUs the GPIO pin is set to Hi-Z input at power-on reset and it'll stay such until it's configured for something else. So, if the external network will be present at power-up you may want to check the default config. If it's input then, as long as the voltage seen by the pin is within the ABS max ratings, it's fine.