Say I have a 12 V battery powered device, with a logic and control board that take 5V sourced from the 12V supply, what is the solution to powering up the device from standby while not unnecessarily draining the battery while in standby?

I expect I do not want a constant 12>5V downconverter operating full time to power the logic and respond to a power on request, so what is the solution to enabling standby in a system such as this?

  • \$\begingroup\$ What do you mean under 'respond to a power on request'? Sounds like that could be a simple power switch. \$\endgroup\$
    – NStorm
    Commented Sep 3, 2021 at 12:31
  • \$\begingroup\$ I don't think there is an alternative. Also, beware that for ultra-low currents you can get all sorts of problems (If your buck converter is aimed at hundreds of ma, pulling 1ma will be VERY inefficient, not even sure if it will work with every chip). (I assume standby means deep sleep, but powered on) \$\endgroup\$
    – Ilya
    Commented Sep 3, 2021 at 12:32
  • \$\begingroup\$ Having a LDO and down converter with EN signal. I do think there are already made ICs for that. \$\endgroup\$ Commented Sep 3, 2021 at 12:36
  • \$\begingroup\$ You'll indeed have to clarify if "standby" is "power off", no current needed at all or if it is "deep sleep", where a lot of things are turned off, but there are still a few things that need current to monitor for an external event (timer, pin level change...), and possibly even sensors, probably drawing anywhere between a fraction of a µA and a few milliamps. You'll probably also need to specify how much peak current you need, and possibly the duty cycle (how long on and off). \$\endgroup\$
    – jcaron
    Commented Sep 3, 2021 at 12:39
  • \$\begingroup\$ It would be a 'deep sleep'. The system would still need a 3.3/5 V supply, but at essentially no current (enough to drive an RTC say) \$\endgroup\$
    – J Collins
    Commented Sep 3, 2021 at 12:41

1 Answer 1



simulate this circuit – Schematic created using CircuitLab

This uses an LDO which allows Vout to be higher than the set voltage without drawing any current while the SMPS is regulating. The LDO can have Iq = 1uA if you want.

LDO's using BJT's are Emitter Follower Darlington's which pullup but if driven higher than the regulated output , draw no current and they cannot pull-down with a single ended transistor.

MOSFET LDO's are similar but Common source outputs.

When the SMPS buck regulators are disabled, it should draw no current from input or output.

I chose 4.5 because that should still drive the onboard 3.3V regulator (check) but not interact or fight with SMPS regulator. The onboard LDO might draw more than the RTC.

  • \$\begingroup\$ Could you explain this in more detail? There is a 4.5 V regulator with a switching 5V power supply? \$\endgroup\$
    – J Collins
    Commented Sep 3, 2021 at 12:50

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