2
\$\begingroup\$

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.

\$\endgroup\$
  • \$\begingroup\$ I've recently built a couple of similar systems, and the standby consumption on the GSM module when "off" is a few microamps, enough that I've not worried about it and left it directly connected to the battery. Could you link datasheets for the modules please? \$\endgroup\$ – pjc50 Mar 28 '13 at 12:47
  • \$\begingroup\$ My whole circuit in idle mode shouldn't use more than max 20µA, and is currently doing so. When my circuit is active(gsm on), i'm not too worried about the current consumption, as this is once a day perhaps. Its the idle-current I need as low as possible \$\endgroup\$ – Jeffa Mar 28 '13 at 12:55
  • \$\begingroup\$ You say the micro needs 3.3V and the module 5.5V, where is this all ultimately coming from? You mention a battery, but failed to tell us its voltage. \$\endgroup\$ – Olin Lathrop Mar 28 '13 at 13:02
  • \$\begingroup\$ The MCU is an energyMicro 1.8 - 3.3V mcu. It can't handle more voltage. The GSM module has it's lowest operating voltage on 5V, and the manufacturer recommends an even higher voltage. The battery is yet to be decided, but would have a voltage higher than 5 volts, but not much more. Perhaps 2cell Lipo \$\endgroup\$ – Jeffa Mar 28 '13 at 13:06
6
\$\begingroup\$

Depending on how much load current you need to supply the following circuit should work: -

enter image description here

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.

\$\endgroup\$
  • 1
    \$\begingroup\$ That looks good - leakage of 1ua. GSM current profile is quite spiky but the device seems to have its own regulator which should smooth it a bit. \$\endgroup\$ – pjc50 Mar 28 '13 at 13:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.