I want to find the most efficient way to power an STM32 microcontroller from a lipo (3.7V nominal).

I have calculated that @ 3.3V, the current consumption is approximately:

  • in run mode:1.75mA
  • in idle mode: 200uA
  • in shutdown mode: 0.6uA

One of the smallest buck boost converters I could find, LTC3532, has the following efficiency curve:

enter image description here

and its quiescent current is at best 50uA.

As a result, it would not be very efficient (~75%) to power the uC even from this small converter when the uC is in idle, or even worse completely inefficient in shutdown mode.

I also considered using an LDO, but again, at best I could find quiescent currents in the range of several microamps, thus not being efficient for the shutdown mode.

Is there any way to provide such low powers efficiently and not directly from the battery?

I also found this quote on a forum

The biggest impact for increasing efficiency to an LDO, apart from Iq, will be to reduce your input voltage to the LDO. Often this is done by using a switching regulator before the LDO.

What is the point of this? I think I understand the theory, where with the switching regulator you can get Vin of LDO to be as close to Vout+Vdrop as possible to have maximum efficiency, but why not use a switching regulator directly then?

EDIT: As mentioned by W5VO in the comment, I could find a lower Iq LDO than the several microamps mentioned above. So is the best way to go to use an LDO at 3.3V with the lowest possible quiescent current and dropout voltage, and only exploit the energy of the LiPo from ~4.2V down to 3.3V+Vdo?

  • 2
    \$\begingroup\$ Using the online catalogs, you can filter/sort by Iq. A word of warning, if the regulator can be "shut down", the Iq listed will be when the regulator is "off". First in-stock part I found had a "typical" on Iq of 95nA, so I think your searching needs a bit of refinement. \$\endgroup\$
    – W5VO
    Apr 15, 2022 at 18:41
  • 1
    \$\begingroup\$ Yes, that is correct. So the best way to go is to use an LDO at 3.3V with the lowest possible quiescent current and dropout voltage, and only exploit the energy of the LiPo from ~4.2V down to 3.3V+Vdo? \$\endgroup\$
    – NickG
    Apr 15, 2022 at 18:48
  • 1
    \$\begingroup\$ It sure is the simplest... and probably the cheapest. An ideal LDO has an efficiency of 78% converting a 4.2V source to 3.3V - food for thought. \$\endgroup\$
    – W5VO
    Apr 15, 2022 at 18:55
  • \$\begingroup\$ @W5VO you should use the average voltage of the batter, in my opinion. That means 3.7 V. 3.3 / 3.7 = 89 percent (without accounting for Iq). \$\endgroup\$
    – user57037
    Apr 16, 2022 at 18:44

3 Answers 3


I would just use an LDO. I would like to point out a few things.

The efficiency of most switching regulators under very light load is not much better than an LDO (or may not be better at all). It is very common to use linear regulators for very light loads, even when Vin / Vout > 10 or 20. Things like laptop supplies or other AC-DC converters often have a small, low quiescent current linear regulator running directly off of rectified mains voltage, because this makes a lot of things easier.

LDOs often continue to operate (they do not shut down) after they reach the drop-out voltage. This means that you can discharge the battery more deeply than 3.3 V + VDO. Perhaps you can continue operating until the output of the LDO is 3 V or 2.7 V, depending on what circuitry you are using. If I remember correctly, the STM32 processors do not need 3.3 V. They can operate substantially below 3.3 V. But double-check the datasheet for your processor and any other ICs you have on the board.

Finally, a warning. If you are measuring battery voltage using an ADC, and using VCC as the ADC voltage reference, then you will eventually run into a problem once the LDO is in drop-out where you can't correctly measure the battery voltage any more. Essentially, your reference and your battery will be going down together and it will look like the voltage is not changing. I have had this problem before. See here if you want to understand more about that issue: Why does this lithium battery have such a flat discharge curve?

You can work around this by using the internal ADC reference (which is less than 3.3 V) when you measure battery voltage.

Good luck and have fun.


Look for low quiescent current Buck converters. Here is one (TPS6274x) from TI that has a quiesent current as low as 320nA and most likely works for the voltage input range you have and output voltage level.


Just to add to @mkeith's excellent answer:

Think about when the efficiency matters and how it affects your design. You'll want to do a power budget that includes the amount of time you expect to be in each mode, then multiply by the efficiency of the switchmode at each mode.

Depending on the amount of time you're in high current run modes like 1.75mA, the efficiency at that point may have a greater or lesser effect on the total system energy consumption. For instance 90% efficiency at 1.75mA means you lose 0.175mA as heat (roughly), but at 0.6uA and 10% efficiency you lose only 0.54uA as head (roughly). These aren't quite correct as you need to take into account quiescent current and how that affects you budget and the fact that a switchmode supply is an energy converter. But it's close.

Knowing where you're losing your energy allows you to focus on how to make the most bang-per-buck.



Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.