I have a circuit that is off almost all the time, when a button is pressed an MCU will be brought outside of sleep mode perform a log and go back to sleep. This device should run on 3 1.5V AAA batteries giving more than 3.3V.

How can I regulate the voltage when I wake up from sleep and have the regulator not waste energy when device is not used?

  • \$\begingroup\$ Depends what causes you to wake up from sleep... \$\endgroup\$
    – vicatcu
    Commented Oct 29, 2017 at 21:50
  • \$\begingroup\$ @vicatcu A button press brings the MCU out of sleep mode. The button generates an asynchronous external interrupt request by changing the logic state of one of the MCU's pins. \$\endgroup\$
    – user34920
    Commented Oct 29, 2017 at 21:59
  • \$\begingroup\$ Yeah but you have to supply the MCU 3.3v before it can receive interrupt, right? \$\endgroup\$ Commented Oct 29, 2017 at 22:04
  • \$\begingroup\$ @user3528438 Yes, that's the main part of the question :) \$\endgroup\$
    – user34920
    Commented Oct 29, 2017 at 22:07
  • \$\begingroup\$ does the mcu need to be on during "sleep" our could power be removed completely in principle \$\endgroup\$
    – vicatcu
    Commented Oct 29, 2017 at 22:13

1 Answer 1


If you use a CMOS flip/flop, the 'button press' can SET the flip/flop, which can enable a low-dropout regulator like TPS706 . The regulator turns on, your MCU goes through power-on operations, does its thing, and turns the power off by issuing a RESET to the flip/flop. When in 'shutdown', that TPS706 takes under 150 nA.

This is usually needless, though: many MCU chips have a variety of on-chip functions to go into sleep mode. A trio of alkaline AAA cells has roughly 1200 mAh of capacity, so a modest drain (like, 30 uA) might be supportable. 30 uA will drain the cells in a bit over 4 years, and 150 nA will take longer (so long that battery shelf-life dominates).


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