I'm designing a pcb with a 3.3v mcu and some 5.5++V modules on it.
My microcontroller is doing the power-management, turning the 5.5V supply on/off. But I'm not sure how to turn the "high" voltage supply on/off from the mcu(GPIO).
The 5.5V is from a battery and is powering some modules(GSM/GPS) with internal regulators, so supplier encouraged "raw" battery-voltage so they can consume the current they need.
I would like to be able to power these modules off completely(ultra-low-power system) when they aren't in use, often hours/days at a time.
The mcu is also powered from the battery, but via its own 3.3v LDO reg, and should always be powered on.
My knowledge in analog electronics is basic and limiting :p
How to I enable the power-supply to one part of the circuit using my 3.3V mcu?
Is a MOSFET the way to go? I would like to avoid a relay and other large components.
EDIT: Link for the Datasheet(GSM)
The power-off current is 40µA, and would almost triple my idle current consumption if left on.
Best Answer
Depending on how much load current you need to supply the following circuit should work: -
If you need less than 2A this FET will work.
At 2A it'll dissipate about 0.3W so a little heatsinking in Cu may be advised. It's good for up to 20V but I wouldn't run it higher than 12V. Also this was the first FET I came across so there will probably be cheaper devices.