I'm building a board (simple toy) that needs two power voltages. The first part has MCU and sensors and can use anything from 3.3 to 2.6 volts. The second part has an audio amplifier and servo motors; it needs 5 to 5.5 volts. The total current is ~50mA on average, with short spikes to 80mA when the speaker is active.

I currently use 4 AA cells and two linear regulators (5V and 3V3), but this has unwanted drawbacks:

  1. lowering 6V to 3V3 wastes almost 50% of energy;
  2. it only works with fresh batteries. As soon as the power pack voltage drops below 5V, the first regulator stops working, and the 5V part becomes underpowered.

My next best idea is to use 3 AA cells, a boost DC-DC regulator to power the 5V part, and a linear regulator to get 2V8. This way, I can use batteries until dry; they should go about 1V each in the end.

Is it a good design, or is there a better one? This is one of my first boards, and I don't have a good intuition, even less knowledge.

  • \$\begingroup\$ Better voltage regulators are an option too. Check out murata brand. They're expensive compared to a linear, but they're easy to use, drop-in replacement, and best of all they have nice datasheets. There are lots of other brands, and also non branded switchers you can get. Ones with datasheets are nice because you can size them to put your load in a better part of the efficiency spectrum. \$\endgroup\$
    – K H
    Feb 9, 2021 at 3:23

1 Answer 1


I think 3 AA cells is the best option, like you say. In toys powered by AA batteries, in order to avoid electrolyte leakage problems, it is good if the battery current is very low (like 15uA or less) when the batteries are totally dead. If the device turns off with an actual switch that disconnects the battery, then that is great. The current will be zero when the device is off.

But if there is some kind of "sleep mode" where the device turns off automatically after not being used, then you need to take care that the sleep mode current drops to a very low value when the batteries are drained.

The second big problem is avoiding reboot loops. The battery voltage will drop under load, and then rebound when the load is removed. So sometimes if you start up the device, it will begin to boot up, but then the load will cause the battery voltage to crash and it will shut down, then the voltage goes up and it reboots, etc.

The best way to avoid this is to have startup code which executes very early and prevents the device from booting when battery voltage is less than around 1.1 or 1.15 volts per cell (you can play with that threshold). During operation, you can keep running until 0.9 or 0.8V per cell before low battery shutdown. But on startup, the voltage must be at least 1.1 (or whatever). This combination will help prevent any kind of reboot loop from happening. Again, you may need to adjust the numbers a bit. But the main idea is that the min startup voltage should be higher than the min run voltage.

I know most of this is not what you asked about. But you did say you are new to this.

  • \$\begingroup\$ Thank you, this is indeed much more that I asked about or expected to learn! \$\endgroup\$
    – blaze
    Feb 10, 2021 at 2:16
  • \$\begingroup\$ It also wouldn't be a bad idea, if you have the option, of switching to a single lithium ion battery. But you may have perfectly good reasons for sticking with AA and I am not trying to talk you out of it. \$\endgroup\$
    – user57037
    Feb 10, 2021 at 3:10
  • \$\begingroup\$ What kind of lithium battery you have in mind? My main design criteria are size, price, and availability. AA batteries are ubiquitous and cheap, flat 3xAA holder fits on the device frame. I briefly considered using 18500, but it's too bulky, both in length and diameter. \$\endgroup\$
    – blaze
    Feb 10, 2021 at 18:28
  • \$\begingroup\$ I was thinking of one 18650. But if the 18500 is too large, then 18650 is much too large. AA works well. That is why it is so popular. \$\endgroup\$
    – user57037
    Feb 10, 2021 at 18:37

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