• My device requires ~4.5 V.
  • I'm using 4 battery packs connected in parallel
  • Each battery pack contains 3 metal-encased non-rechargeable alkaline AA batteries.
  • I used voltmeter and verified that each battery pack individually provides ~4.5 V.
  • I have also used voltmeter to verify that all battery packs are correctly connected in parallel.
  • All the batteries are of the same nominal voltage, the same brand and product name (Varta Longlife).
  • (It shouldn't matter here, but for completeness: The batteries are used as a backup to the main power supply, which supplies about 0.5V more than the batteries. Both the power supply and the batteries are connected via a Shottky diode [1N5822 40V/3A DO201] to prevent "charging" each other.)

The device runs fine on the battery packs, but sometimes, obviously when a load spike comes, the device suddenly switches off. I measured that the device, when connected to the power supply, draws 1.6 A during spikes. Normally it draws about 300-500 mA.

Question 1: Shall I keep adding more battery packs in parallel? Will it help or does it only increase the lifetime, but not max current/power?

Question 2: If I use a battery with higher mAh rating, will it help, or does it only increase the lifetime?

As a side note: I am also going to try top-notch lithium AA battery instead of alkaline to see if it handles the spikes.

Thank you.

EDIT: The backup is not supposed to be used much, just very rarely, during a power outage (which we haven't had in years).

UPDATE: I've tried the best lithium AA batteries on the market, replaced the alkaline batteries, and found out that they didn't handle the spikes, either. Moreover, they supply ~5.5V instead of the ~4.7V that the alkaline batteries do. So that's not nice, either, because that is higher voltage than that of the power supply. I'm thinking about using the capacitors or the suggested buck converter with 12 batteries in parallel. The thing is, I have no experience with capacitors and I don't even know what the circuit should look like. As for buck converter I don't know anything about them, either, and am a bit worried that they would introduce more complexity to the system, possibly increasing the chance of failure. So adding more battery packs in parallel, and changing to D-type batteries as some suggested, still seems to me to be the simplest (least complex) solution, which I feel capable of designing and building myself.

UPDATE 2: 4x3 D-type batteries (instead of the AA-type batteries I original used) did help and the device no longer shuts down during high-drain events.

  • 3
    \$\begingroup\$ You could try to connect a couple of 10uF capacitors in parallel to cover the current spikes. \$\endgroup\$
    – Swedgin
    Jun 12, 2020 at 7:18
  • 3
    \$\begingroup\$ Indeed supply decoupling capacitors should help. If 10 uF isn't enough, there is no harm by trying 1000 uF and see if that does the job. Connect the capacitor close to the components / circuit that draws the current spike so that the loop Capacitor-circuit is small. \$\endgroup\$ Jun 12, 2020 at 7:36
  • 1
    \$\begingroup\$ You'd be better off using D cells. \$\endgroup\$
    – user16324
    Jun 12, 2020 at 11:04
  • \$\begingroup\$ What device is this? Is it a finished product supplied for a third part, or it was designed and built for yourself? \$\endgroup\$
    – mguima
    Jun 12, 2020 at 14:43
  • \$\begingroup\$ Thanks for all the helpful comments, guys. I will definitely try the suggested solution with capacitors. The device is a Samsung Galaxy phone with the internal battery removed. I just kept the CB that was connected to the original battery and I'm powering the phone via the pins of the CB (not via its USB connector). This fools the phone into thinking it has a real battery (without which it refuses to run). I'm doing this because I need this phone to be permanently connected to a charger and that causes the Lithium battery to inflate. \$\endgroup\$
    – deLock
    Jun 12, 2020 at 14:58

4 Answers 4


At 1.6A you're shorting the battery. Maximum load is about 500mA, this is already <1 hour. Internal resistance is about 0.3 Ohm, at 1.6A sets you back half a volt per cell. So, for one pack with 4.5V (real 4.2V ) you go down to 2.7V in real conditions.

If equally divided over 4 packs this 1.6A is maybe fine (0.4A per pack), but if one cell fails, your pack cascades into failure. In theory with 4 packs you have 0.1 Ohm internal resistance, which is about 120mV drop during you worst case condition. For the entire pack that is about a drop to 3.72V. (4.2 - (0.12*4) when full. In theory.

Can you retry and measure the current of each pack vs the supplied voltage? This should give you insight in the paralleling behavior.

I also suggest rethinking this backup solution. Either

  • Put all the 12 cells in series and use a buck converter. (less current, same power, better power point)
  • Use C or D-cells.
  • Use protected lithium cells eg: 14500 (in series).
  • Maybe even consider lead acid if that fits.

Now you're abusing the poor AA's.

  • 2
    \$\begingroup\$ I prefer your answer. Much better, overall. +1. Of course, my first thought to the OP is to just say, "sure, feel free to add thousands of AA, all soldered together in a massive, parallel heap." ;) But I had to restrain myself. \$\endgroup\$
    – jonk
    Jun 12, 2020 at 7:55
  • \$\begingroup\$ @Jeroen3 Thanks so much for all the helpful advice and tips! I was thinking about using a buck converted. Would you use this one for those purposes? meanwell-web.com/en-gb/… \$\endgroup\$
    – deLock
    Jun 12, 2020 at 15:34
  • \$\begingroup\$ For now I've decided to give D-cells a try. It's the simplest, fastest and cheapest solution for me. I'll try with Energizer D Max (if it doesn't work, I'll try several D packs in parallel) and will report back. \$\endgroup\$
    – deLock
    Jun 12, 2020 at 17:14
  • 1
    \$\begingroup\$ @deLock Isn't that quite big, and 10A when 2A should be fine? I'd say get TMDC 10-2411 one or SKM10A-05. \$\endgroup\$
    – Jeroen3
    Jun 12, 2020 at 20:15
  • 1
    \$\begingroup\$ So I've tested 4x3 D-type batteries Energizer D Max (instead of the AA-type batteries I original used) and they did help -- the device no longer shuts down during high-drain events (3x3 were not enough). \$\endgroup\$
    – deLock
    Jun 26, 2020 at 12:26

Yes, adding more battery packs in parallel will share the load and have ability to deliver larger currents and survive the peaks. But it's not necessarily the best option.

What about using D cells? They are bigger, and besides larger capacity they can also deliver more current.

Always read the available datasheet to get a general idea about current they can deliver. If not available for the particular brand you're using, at least find datasheet for that battery type to provide a general idea.

Lithium cells in general can handle higher currents.

  • \$\begingroup\$ Thank you for the advice. Yes, I've decided to give D-cells a try. It's the simplest, fastest and cheapest solution for me. I'll try with Energizer D Max (if it doesn't work, I'll try several D packs in parallel) and will report back. \$\endgroup\$
    – deLock
    Jun 12, 2020 at 17:15
  • \$\begingroup\$ So I've tested 4x3 D-type batteries Energizer D Max (instead of the AA-type batteries I original used) and they did help -- the device no longer shuts down during high-drain events (3x3 were not enough). \$\endgroup\$
    – deLock
    Jun 26, 2020 at 12:26

During a 1.6 Amp spike each of your four battery packs will supply 400 mA, so the voltage from three alkaline cells will be about 3.5 volts. You will lose another 0.35 volts across the diode. You need to identify the minimum voltage requirement of your device.

Adding more battery packs will reduce the current demand from each pack, so it would help, but probably not enough. Putting a diode on each pack would also reduce the drop across the diode from 0.35 volts to 0.25 volts.

Using Lithium batteries would work much better. The voltage from three cells should be 4.2 volts and with a separate diode on each pack the voltage should only drop to 4.0 volts during a 1.6 Amp spike.

If your device can tolerate 6 volts you would be better using a 6 volt supply and four alkaline cells in each pack.

  • 1
    \$\begingroup\$ Thanks for the explanation. And, I've just tried lithium batteries but unfortunately, they didn't help. And, no, I cannot exceed 4.5V from the batteries, because they need to be below the 5V supplied by the main power supply. \$\endgroup\$
    – deLock
    Jun 12, 2020 at 17:17

Considering your current requirements, number, size and cost of batteries, as well as their capacity and as a consequence lifetime, I would definitely go for Li-ion.

First of all, there's plenty of choice in terms of size and capacity. They can hold more charge really well. Besides, you can get a tiny charging-discharging PCB for them. The one from powerbank will do. Google powerbank charger pcb or module. Of course, it will output 5V, If your circuit is ok with 5V, good for you, if not, you can throw in some voltage converter or modify the charger board's boost circuit (most likely replacing one resistor near boost converter).

  1. Make sure your li-ion can support current you may need. Look at 18650 batteries first.
  2. Make sure powerbank board supports 2A/2.1A output

Even if you assemble this whole thing from pre-made modules (charger/boost converter etc) it will still take less space than your army of AA. Besides, you'll be able to charge your li-ion without taking them out.


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