I am still new to electronics and I am looking for a low noise and efficient way to provide 5V and 3.3V from AA batteries. My circuit requires 3.3V/200mA 10% of the time and less than 1mA the other 90%. 5V is 99% <1mA, 1% 50mA.

Here is how I am currently approching this problem :

I could use 4xAA and step-down converters but given my circuit requirements the efficiency would be horrible. The noise would also be terrible.

I could use 4xAA and LDOs, but steping down to 3.3V from 6V doesn't sound like the good way to do it.

So I could use 3xAA (4.5V) with a LDO to provide 3.3V and a charge pump to provide 5V, but then :

  • the LDO will fail under 3.4V, wasting around 20% of the AA batteries. So maybe I should put the LDO after the 5V charge pump... or find a way to do that only when the LDO fails...
  • charge pump arn't exactly low noise. Maybe I should use an LDO after it to get rid of the noise...

I'm starting to have more questions than answers. What is the correct way to solve this beginner electronics engineering problem ?

  • 1
    \$\begingroup\$ Welcome, Ron. You could use 3 AA's. Use an LDO or a buck (switch mode regulator) for 3.3V. Not all switchers are noisy. You have to decide on the tradeoff. Some buck converters transition to a different mode when the load current is low, to save power. Please also note that the useful range of an AA battery is around 1.6V to 0.9 or even 0.8V per cell. So if you use only 3 AA batteries, your 3.3V rail will drop out before you fully discharge the batteries. But if you use 4AA batteries, then the LDO seems too wasteful (and it may get warm). \$\endgroup\$
    – user57037
    Commented Mar 24, 2015 at 1:35
  • \$\begingroup\$ Sometimes for critical audio functions, people will use a switch mode regulator followed by an LDO. In this case, the job of the LDO is basically to filter out the switch-mode noise. Obviously if you do this, the switcher output would need to be higher than the LDO. \$\endgroup\$
    – user57037
    Commented Mar 24, 2015 at 1:38
  • \$\begingroup\$ How is your system switching between high current and low current modes of operation? 200mA/3.3V and 5V/50mA is a fair amount of power consumption out of AA batteries, what do you expect for battery life? How complex can your system get? For optimum battery life, you certainly want to use a switch mode regulator, the question is how you partition things. A block diagram would be helpful. \$\endgroup\$
    – rfdave
    Commented Mar 24, 2015 at 1:39
  • \$\begingroup\$ Thank you @mkeith. I understand that with 3xAA and an LDO my 3.3V rail will drop out before I fully discharge This is what i meant by "wasting around 20% of the AA batteries", which is a rough order of magnitude estimate based on the AA battery discharge profiles I was able to find. \$\endgroup\$
    – Rom Ain
    Commented Mar 24, 2015 at 2:05
  • \$\begingroup\$ @mkeith, I do not have critical audio functions but I do have RF. I have for instance a PN532 (13.56MHz NFC) with pcb "antenna". I don't know how critical it is regarding the audio functions you were thinking of, but it seems that I do have to be careful about the noise. \$\endgroup\$
    – Rom Ain
    Commented Mar 24, 2015 at 2:15

1 Answer 1


Low noise switching regulators do exist; by controlling slew rate of a switching element you can make it pretty quiet, down to ~100-200 uV. For example, Linear's LT1533, LT1534. Also, Application note 70, from Linear as well.

  • \$\begingroup\$ Thank you for pointing out Linear AN70. So, ok, I could operate from two 1,5V Alkaline cells and use two ultralow noise step-up converters to provide 3.3V and 5V. However, the efficiency of step-up converters is pretty bad for really light load current. As the device will spend most of its time doing nothing, I'm afraid the lack of efficiency will really hinder battery life. \$\endgroup\$
    – Rom Ain
    Commented Mar 24, 2015 at 3:43
  • \$\begingroup\$ There is a sample schematic somewhere in the appnotes showing how to put this controller to sleep. You'll be getting some ripple but since you will be sleeping too this won't be noticed. \$\endgroup\$ Commented Mar 24, 2015 at 3:47
  • \$\begingroup\$ As far as I understand it : sleep mode = bypass mode = no noise + minimal quiescent current + unregulated. If I use 2xAA, it would mean providing something between 1.6V-3V to my chips, which is way out of specs. \$\endgroup\$
    – Rom Ain
    Commented Mar 24, 2015 at 13:53

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