RE: Why GPIO can be used as power supply pin?

andlier — Tue, 03 Nov 2015 10:06:17 GMT

You're "backpowering" the chip through the protection diode in the GPIO-pin. (google "backpowering mcu" for more on that.)

Even if it works, this is definitely not the way the chip is intended to be operated, so the datasheet can't be trusted and you can easily end up with brown-outs etc. depending on what your code does because of the higher impedance path through the gpio protection circuitry.

There is a "diode" or protection circuit to the Vdd rail from the GPIO pin. This is there to protect the GPIO input from over-voltage conditions in case the GPIO pin voltage goes above Vdd. In your case Vdd is not powered, so effectively at 0V and you apply a voltage at the GPIO which will flow through that protection diode. A Nordic employee would have to chime in on the long term effects of continuous current through that diode.
System reset is performed when waking up from off mode. So you'd have to init again. There is an option to retain RAM blocks in off-mode, but then the init-code would have to write retained values to the registers again.
Not 100% sure what you are suggesting. Typically you should have Vdd applied to the chip at all times and either use the internal comparator/ADC to sample the 3.5V or a divided down version of it if Vdd is lower than 3.5V. Or you can use a GPIO pin together with an external voltage comparator.