Decoupling Capacitor on GPIO instead of connected directly to battery

Question

I am building a device that is for most of the time in sleep mode. When woken up, I am sending some BLE signals and use a LED. To support the battery I have a 2u2F capacitor. But because my battery is so small and I want the device to have a long runtime I am concerned about the leakage currents of the capacitor. Especially because I would like to put a bigger capacitor than 2u2F into the circuit for supporting the battery.

An idea and question now are: Because I don't need the decoupling capacitor during the sleep mode I wanted to put it on a GPIO and when the device is awake I could switch the GPIO to VCC.

Is this a possibility?

Answer 1

It would not work due to many reasons.

You only assume you don't need a decoupling capacitor in sleep mode, but since the device is capable of waking up from sleep mode, there must be something running in order to the device to detect when to wake up.

Also the capacitor would be quite useless on a GPIO pin. The GPIO pin can't be thought that it is like a zero ohm relay contact between IO pin and VDD. The MCU would need to wake up and start running before you can control the pin high to charge the capacitor, so if the capacitor is needed to start up then there is no capacitor. The IO pin current to charge a directly connected capacitor would be limited only by the output stage transistor and would most likely exceed absolute maximum limits for the current. And if there is a drop in the supply, the capacitor won't be directly connected to the supply but via the IO stage transistor or via IO stage protection diode so it could not work as a bypass cap.