I've question similar to another SO question, with some minor differences, which I'll explain below:

I'm designing a circuit which should work with 3 AND 4 AA batteries. Idea is to create two models, one with slightly longer life than the other one (33% more). The circuit uses an STM8L, and would periodically "wake up" to perform some periodic stuff, and then go back to sleep, typical of certain battery powered devices. I'm planning to use a voltage regulator of 3.3 V output. I want to know which kind of regulator would make more sense (LDO vs switching). My use case doesn't require any RF application, so bit of output noise is acceptable (meaning switching regulator should be just fine).

With 3 AA battery system, based on the discharge characteristics, "95% + charge" would be available upto 2.8-2.9V. So, a LDO regulator, will be giving on average 3.8V battery regulated output 3.3V, which makes it around 85% efficient. This is putting it on par with most switching regulators out there. Please correct if wrong.

With 4 AA battery system, based on the discharge characteristics, regulator will be on average 4.5V battery regulated output 3.3V, which is <75 % efficient. Now we are getting into that territory where switching regulator would most definitely start to shine.

Please let me know if my above understanding is on the correct path. Should I use a LDO/switching regulator or design the two circuits separately one with LDO other with switching regulator? Any other suggestion also welcome.

  • 1
    \$\begingroup\$ "3 AND 4 AA batteries" or '3 OR 4 AA batteries"? \$\endgroup\$
    – Transistor
    Oct 25, 2020 at 12:49
  • \$\begingroup\$ Yes I thought about it while writing, and made sure its clear in the description. \$\endgroup\$
    – Ouroboros
    Oct 25, 2020 at 12:53
  • \$\begingroup\$ So a device which has 2 models : one with 3 AA and another with 4 AA \$\endgroup\$
    – Ouroboros
    Oct 25, 2020 at 12:53
  • \$\begingroup\$ Without a switching regulator, the runtime will be almost identical with either 3 or 4 batteries in series. \$\endgroup\$
    – winny
    Oct 25, 2020 at 13:00
  • \$\begingroup\$ You really shouldn't try to consider this in isolation of detail of what is being powered. In many cases, you can use a voltage lower than 3v3 with or without a regulator, in others allowed range may run higher meaning you don't need a regulator. And if your device sleeps rather than being in constant operation, then you need to worry about regulator quiescent current. Low power design is an art requiring comprehensive attention, one question in isolation isn't really meaningful. \$\endgroup\$ Oct 25, 2020 at 14:37

1 Answer 1


As winny said, with a linear regulator the input current is equal to the output current (plus quiescent current tax) so if you use a linear reg, the reason why 4 AAs will last a bit longer than 3 AAs will only be due to the last bit of the discharge curve below 1.1V per cell being exploited.

So a switcher makes a lot more sense. You need one with low quiescent current, especially in sleep mode. Here's an example.

Note if you have, say a radio transmitter that uses quite a bit of current especially in pulses, the actual dropout of your LDO and its transient response at very low dropout may be worse than a switcher.

  • \$\begingroup\$ Buying a 8usd switching regulator is out of question. Are there cheap (less than 1 usd) options out there? \$\endgroup\$
    – Ouroboros
    Oct 25, 2020 at 18:50
  • \$\begingroup\$ There are plenty of cheaper options. TI/AD parametric search can filter and sort by quiescent current, you'll find what you need. Since you didn't say how much peak current you needed, I figured you'd have to search anyway so I picked the luxury model as an example ;) \$\endgroup\$
    – bobflux
    Oct 25, 2020 at 19:39
  • \$\begingroup\$ My peak current is 100ma. \$\endgroup\$
    – Ouroboros
    Oct 26, 2020 at 4:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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