microcontrollermosfetswitches
I want to control the power supply of a sensor with a microcontroller, so I can turn it off when it isn't being used. My goal is to minimize power consumption on a battery-powered device.
Both the microcontroller and sensor are running on 3.3 V.
Can I use a single MOSFET directly driven by the controller? Or is there a better method?
atmega328p micro controller and GY 61 accelerometer sensor are used.
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.
I don't see an overly-compelling reason to boost up the voltage to 5V. Unless you're doing some intense time-critical math heavy calculations, I doubt 16MHz will buy you any advantage over 8MHz for the application you're describing. Single-purpose hardware can get a lot done at 8MHz.
A 3.3V linear regulator with a dropout voltage of 0.3V or less should give you pretty good performance using a 3.7V LiPo. Take a look at this graph I lifted from adafruit's website:
Let's assume your battery will be fairly small. Say 500mAh. So the 250mA load represents the red curve (0.5C). If you keep the battery between 0% and 80% discharge capacity, the battery voltage will be between 4.2 and 3.6V. A linear regulator will see efficiencies between 80% and 90%. If you can keep your current draw below 250mA, you'll see even better numbers.
One possible difficulty when working with 3.3V is parts selection. You'll need to make sure the transistors and op-amps you choose work at that voltage. If, for example, you plan to use an N-channel MOSFET to switch on an air pump that draws a few hundred milliamps from the battery, the Vgs(th) of the FET will need to be quite a bit lower than 3.3V. Otherwise its Vgs(on) will be too high to allow the pump to turn on fully.
Best Answer
Probably- one method would be to use a P-channel MOSFET as a high-side switch to control the 'sensor' power. You might have to add a pull-down or otherwise ensure that the output of the 'sensor' was close to 0V with no power- if it is allowed to float around the MCU power consumption might increase.
There are many inexpensive logic-level MOSFETs that will work fine from -3.3V gate voltage with guaranteed very low Rds(on) and reasonable guaranteed leakage.
simulate this circuit – Schematic created using CircuitLab
You may have to ensure that the 'sensor' supply has some series impedance or that the MOSFET does not switch too rapidly if the 'sensor' has large bypass capacitance- otherwise it could glitch the MCU power when it turns on and cause problems.