simple power switching circuit

Nov 11, 2007 6 Replies

Hello,



I am working on a small project using a pic microcontroller that I want to have run for months at a time. I can put the pic to sleep to conserve power but I would also like to remove power to the peripherals I am adding (i2c eeprom, etc). What I want to do is use one of the pic output pins (high) to power up the devices when I need to use them. This seems like it should be pretty easy, can anybody point me to a simple circuit to do this or recommend a good way to do this.



thanks !


Look for a logic Mosfet transistor that can be gate driven via an output of the PIC. A high side P-channel config should do it. That would be the S (Source) to the + rail, G (gate) via a parasitic resistor to an output of the pic. (D) drain will supply the peripherals. The pic's output will have to go low to turn on the fet. Select a fet that has full turn on at 50% or less of the rail voltage you're using.. That's about the drift of it.. P.S. Make sure you have the output of the pic programmed for pull up.

"I\'m never wrong, once i thought i was, but was mistaken" Real Programmers Do things like this. http://webpages.charter.net/jamie_5

Thank you.

Is there a mosfet which will do the same thing you describe except with the gate being pulled high to turn on ? All the pins on the pic will powerdown I think when I put it to sleep, plus I would like a circuit that draws no power when the pic is sleeping. I'm not sure how much power the mosfet draws when 'turned off' with the pull up resister.

Yes, it's very practical. What kind of peripheral (voltage/current) do you need to power up/down ? That will influence the design.

Graham

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If using a mosfet to switch the load I'd suggest an n-channel type with a high side driver chip. IR do lots of high side drivers and I think Microchip even has some.

Graham

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Out of curiosity, what's the advantage of using an n-channel type?

It helps the manufacturer sell an extra high-side driver chip.

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