I'm tinkering with some DC motors, and have been able to drive them with a 1.2A wall wart that produces a variable 3-12v. I want to drive the motors with 24v+. Would I be able to get what I need by soldering together 16 AA batteries in series? Or would internal resistance foil my plans?

Would it be better to hook up three 9v batteries in series?

Some more details: I'm working towards building an autonomous quadcopter with an arduino for the controller and misc surplus parts. At this point I'm just seeing if I can push enough voltage thru the four small motors I've got to even get them to lift their own weight. I found the motors for $1.50 each at a local surplus store. I'm willing to throw $2 of AA at a one-shot feasability test before spending real money on li-po and/or ESC or other 'real' kit.


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    \$\begingroup\$ If you mean the small rectangle 9V (PP3), your better off with AA's The AA has lower internal resistance. \$\endgroup\$
    – Spoon
    Dec 27 '13 at 22:22
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    \$\begingroup\$ Small dry cells replacement become expensive fast. Consider two small motorcycle batteries, they are rechargeable. \$\endgroup\$ Dec 27 '13 at 22:26
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    \$\begingroup\$ Since you mentioned soldering, note that soldering to batteries can overheat and damage the batteries if not done quickly with proper technique. \$\endgroup\$
    – Kevin Reid
    Dec 28 '13 at 3:26
  • \$\begingroup\$ AA better than PP3 9V. | Cheap AA very limited in current capability | Put laptop supply + 12V wallwart in series for 31V nominal (!!!). What are motors rated at? Using a meter to measure current highly advisable. 1 Ohm in series drops 1V per Amp. 0.1 Ohm in series drops 0.1V per amp. | Measuring voltage at motor wires shows you how much "droop" is happening. | Presumably motors are wired in parallel. | Aa cells give ~ 1.5V when new but rapidly droop to say 1.3V each then fall to about 1V when flat. | ... \$\endgroup\$
    – Russell McMahon
    Dec 28 '13 at 19:49
  • \$\begingroup\$ ... As Optionparty notes - almost any 12V lead acid x 2 in series will give solider 24V than other sources. | Boost converter a good idea as long as it can supply required current. \$\endgroup\$
    – Russell McMahon
    Dec 29 '13 at 0:39

Would I be able to get what I need by soldering together 16 AA batteries in series? Or would internal resistance foil my plans?

Probably it would at least make your goal difficult, if we assume alkaline batteries. Let's take some numbers from an Energizer application note, which says that the series resistance for a AA might be around 200mΩ. This value will increase as the battery is drained, and also depends on temperature and frequency, so see the application note for all the details.

If we use that 200mΩ number, then with 16 batteries in series, the total series resistance of the batteries is

$$ 200\mathrm m\Omega \cdot 16 = 3.2 \Omega $$

If you were to draw 1A from this, the voltage drop across the battery's internal resistance would be (by Ohm's law):

$$ 1\mathrm A \cdot 3.2\Omega = 3.2 \mathrm V $$

Assuming a nominal 1.5V across each cell, the output voltage won't be 24V as you'd expect, it will be:

$$ 24\mathrm V - 3.2 \mathrm V = 20.8 \mathrm V $$

While this might work for a short while (pretty soon the batteries will drain, and the internal resistance will rise), it's not terribly efficient. The power lost in the batteries, through their internal resistance is:

$$ 3.2 \mathrm V \cdot 1 \mathrm A = 3.2 \mathrm W $$

The power delivered to the load (your motor) is:

$$ 20.8 \mathrm V \cdot 1 \mathrm A = 20.8 \mathrm W $$

The efficiency is thus:

$$ \frac{20.8 \mathrm W}{20.8 \mathrm W + 3.2 \mathrm W} \approx 87\% $$

And again remember, these are for fresh batteries. It gets far worse as the batteries discharge, even before they are dead.

While 87% efficiency might work, it won't work well. You are making a thing designed to fly. Flight is a lot of work, and an inefficient system means you have to carry more battery weight. The problem is compounding, because more weight means you also need more thrust, which requires more energy, thus more weight. While you can in theory make it fly, it might end up being pretty huge.

Would it be better to hook up three 9v batteries in series?

Probably not. 9V batteries achieve their high voltage in a small package by containing internally several cells in series. Knowing that an alkaline cell (any kind, AA, A, C, D...) has a voltage of about 1.5V, we can infer that a 9V battery is 9V/1.5V = 6 alkaline cells in series. However, each of these cells is much smaller than a AA, and each will have a higher internal resistance.

In order to decrease the internal resistance, you could do a couple things. The first would be to wire batteries in parallel. Two AA cells in parallel have the same voltage, but twice the stored energy, and half the effective resistance as one AA cell. So, you could make 16 pairs of parallel AA cells, then wire those 16 pairs in series. However, this arrangement also has twice the weight, which is really not good in your situation (flight).

The better solution would be to use a different battery chemistry. Alkaline batteries are good because they are cheap, and that's about it. In terms of internal resistance, they are poor. They also have not great energy density, meaning per unit of weight, they contain less energy than you could have for the same weight of some other chemistry of battery. Again, low energy density is really bad for your situation.

Among the other chemistries you might consider are:

With lithium polymer batteries and chargers being widely available from RC hobby suppliers, and their superb energy density, I'd recommend them first for your situation.

  • \$\begingroup\$ Fantastic answer, with bonus points for the xckd reference. This is a very clear explanation including math of all the relevant information needed. LiPo is the eventual goal, but I needed a cheap solution to see if the motor, while tethered to the batteries on the floor, could lift itself. I was able to use Russell's comments above to feel safe in putting two wall warts in series, measuring the voltage under load and sadly seeing that the stupid motor can't even lift itself with the prop I've got. Next step, lighter motor, bigger prop, or both. Thanks! \$\endgroup\$
    – Rob Fagen
    Feb 26 '14 at 20:31
  • \$\begingroup\$ @RobFagen Sounds about right. Weight is really key. I don't know the relevant maths for atmospheric craft, but there's a problem in spaceflight called the Tsiolkovsky rocket equation. For some reduction in weight of your craft, you get an exponential increase in the distance it can travel. Since you like XKCD, the current what-if is also relevant. Though you don't have a rocket, you still have a \$\Delta V\$ requirement, working against the constant acceleration of gravity. \$\endgroup\$
    – Phil Frost
    Feb 26 '14 at 20:40
  • \$\begingroup\$ @Phil Frost - excellent answer. The IR of the energizer battery you quote was measured at a discharge of 500 mA. It has been demonstrated in alkaline batteries that the IR increases as the discharge rate increases. Considering this, your example with the 1 amp disharge, the IR would be higher than the 200 milliohms and so the efficiency would be even less than 87%. How much less, I don't know, but your example shows that 16 AA alkaline batteries in series is certainly getting near the limit as efficiency is getting low. \$\endgroup\$
    – Filek
    Mar 2 '14 at 8:13
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    \$\begingroup\$ @RobFagen I think the secret to electric RC planes (and later the drones) was the invention of extremely light-weight motors containing neo supermagnets. Perhaps buy cheap drone replacement motors from RC catalogs? \$\endgroup\$
    – wbeaty
    Dec 12 '20 at 12:08

IIRC you can’t get much more than about an amp from a primary AA (although it must be 40 years since I tried that) but you can get a few amps from rechargeable AAs. However, 18650 cells can be good for 40A (choose carefully) and each one will provide over 4v when fully charged, so the weight penalty isn’t that bad - 45g per cell vs 23g for an alkaline AA, so in fact you’d be far better off since you can use about one third the number of cells to achieve the same voltage and you would have significantly more power available. Take care though, shorting lithium cells will result in smoke and flames, although most pre-assembled packs have some protection built in.


Why not use an (RC battery) and a (boost converter)?

How big are these DC motors? Maybe use a car battery and a really beefy boost converter.

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    \$\begingroup\$ 16x 400mAh batteries in series does not give you 6400mAh - it will still be 400mAh \$\endgroup\$
    – Andy aka
    Dec 27 '13 at 22:48
  • \$\begingroup\$ Removed the incorrect information. Sorry about that. \$\endgroup\$
    – jmunsch
    Dec 27 '13 at 22:51

As you've mentioned this is just for a test then connecting 16 in series won't be a problem as long as you don't have any 'dud' cells in the chain. The series resistance will just limit the overall current to the same as a single cell. I just looked at the first Zinc Carbon AA cell I came across with a datasheet and according to that it could deliver around 800mA in 15 second per minute pulse mode.

Also consider some of the other options mentioned such as boost converters (quite cheap modules are available on e-bay) as you my find something with a variable output voltage valuable during testing. You could also consider using a bench power supply, of course much more expensive but very useful and a good long-term investment.

You haven't mentioned the current drawn by the motor but whatever power source you choose you'll need to make sure it can adequately deliver enough current for your tests. Otherwise apart from possible damage the voltage may dip. You should measure the voltage while testing to make sure it remains at the level you're expecting.


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