I'm currently designing a device that will be powered by batteries, I'm planning to make the MCU sleep and only wake up a few times a day to read sensor values and go to sleep again, one of the sensors is a GPS, It will be used once a day only a for a few seconds (not every time the device wakes up), the GPS I want to use can be put to standby mode but won't be more efficent to just cut de power with a transistor controlled by the MCU and only power it when is going to be used (very rarely) than using the software based power saving features of the device? I guess I'm missing something here because if it was the case the device would probably include this feature from the factory.

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    \$\begingroup\$ the PMOS is connected backwards for what you want. \$\endgroup\$
    – Big6
    Apr 2, 2020 at 20:14

1 Answer 1


but won't be more efficent to just cut de power with a transistor controlled by the MCU?

Start looking at the datasheet!

If standby mode consumes no more power than the leakage of your FET, then the FET is pointless.

There's probably no leverage to be had here because of your 1-day interval, but a GPS receiver takes time to lock on to enough satellites to get a measurement. So you don't just turn it on for a moment (unless your "moment" is way longer than mine). You need to turn it on, wait for it to acquire, and then get a position -- and during that time it'll be consuming power.

I could conceive of a well-built GPS device that could maintain lock on the satellites in some level of standby, and only do the rest of the processing when it's "on". For such a device, there would be some interval (almost certainly less than a day) where there could be power advantages to holding it in standby.

But -- only if the power savings of standby is great enough, and only if the time to acquire a position is significantly less. You'd need to look at the data sheet for times and power levels (or measure them yourself), then do the math to see if there's a savings.


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