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Let me preface with an apology for asking what's probably a really basic question. A quick search couldn't turn up an answer to a question in the exact same situation.

I have a DC motor (12V) that, according to the seller's specs, has a free current of 0.5A, a no-load current of 0.4A, and a stall current of 11.5A. I'm understanding this as the motor drawing up to 11.5A for heavy loads (yes, it's amps, not milliamps). Max power is labelled as 14 Watts.

I'm planning to run it using 8 1.5V alkaline AA batteries, wired in series. While that's 12V, intuition says AA batteries might not be able to provide 11.5A (the manufacturer's spec sheet does not have anything about max current).

Would doing so damage the batteries (or possibly the motor, but I can't think of a reason the motor would get damaged)? If it does, would switching to D cells help or is that still not enough? Or, can I switch to 9V batteries even though the motor wants 12V?

For what it's worth, I plan on running this motor using a RPi controlling a MOSFET which would switch the connection to the batteries on and off (and possibly regulate the wattage to change the speed, though I'm not sure if MOSFETs can do that)? Also, I've tested the motor with a single 1.5V D cell and a 9V cell; both are able to start the motor (which I currently don't have a load to attach to).

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    \$\begingroup\$ AA alkaline will not give 10A. High loads will shorten life disproportionately. Motor MAY start poorly if start current is excessively limited. LiIon cells rated appropriately are probably a better choice if possible. \$\endgroup\$
    – Russell McMahon
    Commented Oct 30, 2021 at 23:53
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    \$\begingroup\$ I doubt anything will get damaged unless the motor stays stalled for an extended period. Then the batteries will get very hot, and of course they will discharge very rapidly. I doubt the motor will get damaged, but it could heat up a bit. Alkaline batteries may not be capable of delivering 11.5 Amps into a short circuit. But NiMH AA size batteries probably can deliver a lot more than 11.5 Amps to a short circuit (but maybe not to a stalled motor). NiMH batteries can get damaged from short circuits. \$\endgroup\$
    – user57037
    Commented Oct 31, 2021 at 7:37
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    \$\begingroup\$ Note that most DC motors are designed to work at loads well below the stall load. They overheat if stalled for more than a second or so. The stall current usually happens only as a very short transient when the motor is initially energized. The current at the expected load should be more relevant than the stall current for the purpose of choosing the batteries. \$\endgroup\$ Commented Oct 31, 2021 at 9:50
  • \$\begingroup\$ Are you stuck on AA batteries? A car SLA battery is designed to deliver a lot of current for a short time, anything from 100A to over 1kA. You might find an "alarm" or "UPS" grade of SLA battery that is already 12V and can suit your requirements better. \$\endgroup\$
    – Criggie
    Commented Nov 2, 2021 at 3:21
  • \$\begingroup\$ Are you able to please measure the current draw under normal operating circumstances and load? It will be less than 11.5 A. \$\endgroup\$
    – Criggie
    Commented Nov 2, 2021 at 3:23

6 Answers 6

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There is no way you'll get a sustainable 11.5A from any normal 1.5V AA alkaline cell that I know of, due to their internal resistance, which typically seems to be about 150mΩ at room temperature.

The highest current you can expect from such a cell is under a short circuit condition, where 0V appears at its terminals, and its entire EMF of the cell is presented across its internal resistance. This would be a current of:

$$ I = \frac{V}{R} = \frac{1.5V}{0.15Ω} = 10A $$

You can't get any power from the cell in those circumstances because there's no potential difference across its terminals, so short-circuit current isn't a very useful specification. Rather you would be more interested in the greatest power you can expect from a cell, which would be when the connected load also has 150mΩ resistance. In that case, the cell has half its EMF (0.75V) across its own internal resistance, and the remaining 0.75V at its terminals, across the load. See Maximum Power Transfer Theorem. It that condition, the current through the load, and the corresponding power delivered to it is:

$$ \begin{aligned} I &= \frac{0.75V}{0.15Ω} = 5A \\ \\ P &= IV = 5A \times 0.75V = 3.8W \end{aligned} $$

With 8 of those cells in series, the maximum power they can make available to you would be \$8 \times 3.8 = 30W\$, at a voltage of \$8 \times 0.75V = 6V\$.

Compare that with the power demanded by your stalled motor, to begin spinning:

$$ P = 12V \times 11.5A = 140W $$

Another way to look at this is from the motor's perspective. In a stall condition, the motor represents a near short circuit across the battery of cells. You can estimate the impedance of the motor windings:

$$ R_{STALL} = \frac{12V}{11.5A} = 1\Omega $$

Your battery of 8 cells has a total internal resistance of:

$$ R_{BATT} = 8 \times 0.15\Omega = 1.2\Omega $$

The circuit with a stalled motor across that battery looks like this:

schematic

simulate this circuit – Schematic created using CircuitLab

Those 5.5A may or may not be enough to get the motor started, overcoming friction in the motor and its mechanical load, but it's a big ask from the cells.

How much would the a cell heat up with 5.5A through its internal resistance? The power dissipated in each cell would be:

$$ P = I^2R = (5.5A)^2 \times 0.15\Omega = 4.5W $$

That's quite a lot of power, and the resulting heating could quite easily damage the cell if sustained for more than a few seconds. I can only state this qualitatively, because I have no idea of the heat capacity of the cell's materials, and only you can estimate how long such conditions will prevail.

Another consideration is how long can the cells actually provide 5.5A? Internal resistance rises as the cells discharge, and to get an idea of what 1Ω across 8 cells would do over time, you have to turn to the cell's datasheet. Duracell's datasheet contains a number of graphs from which you can get an idea of what to expect from a cell. This is one of them:

enter image description here

As you can see here, a cell might still produce 1V, under a load of 1A after about 1 hour. From that I will infer (very approximately) that the internal resistance has risen to:

$$ R = \frac{V}{I} = \frac{1.5V - 1.0V}{1A} = 0.5\Omega $$

in a space of one hour, or in other words at a rate of about 0.5Ω per hour. At 6 times that current, I'll guess that the cell's internal resistance will rise 6 times faster, at 3Ω per hour. This is a terribly imprecise estimation, but it serves to give us some idea, at least. I'll also guess that once the cell reaches 0.2Ω (an increase of 0.05Ω) internal resistance it will be useless to you, because it will no longer be able to provide enough stall current to get the motor started.

That gives you a useful service life for that cell of:

$$ \frac{0.05}{3} \times 60min = 1min $$

Of course this assumes a constant current draw of over 5A, which is not going to be the case. Again, without more information, I'm flying blind, but you get the idea.

Of course the normal running current of the motor will be either 0.4A with no load, or \$\frac{14W}{12V} = 1.2A\$ at full load. I'll let you work out for youself what that means for your cells.

Replacing AA cells with D cells will make quite a difference, since they have lower internal resistance, and obviously more energy. Another approach you might take is to connect two batteries of 8 cells in parallel, which will drop the combined internal resistance to 0.4Ω half of what it was. Even better, just use a bunch of lithium ion/polymer cells in series to obtain near 12V, or a single 12V sealed lead-acid battery, both of which have significantly lower internal resistance, solving your problem outright.

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    \$\begingroup\$ I'd go with Ds too if I really wanted alkalines (perhaps for shelf life). BUT: some cheap primary cells and many cheap rechargeables are really AAs in D casings (check the weight) so it's worth getting alkalines from a reputable brand. \$\endgroup\$
    – Chris H
    Commented Oct 31, 2021 at 17:59
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    \$\begingroup\$ Thanks for the answer. When you suggest lithium ion/polymer cells to obtain 12V, is there any significant chance of damage due to an accidental short circuit or something similar? I've heard many bad stories of Li-Ion/LiPo batteries catching on fire. I'm not terribly experienced with EE, so I'm trying to reduce my risks. \$\endgroup\$
    – A. Owl
    Commented Nov 1, 2021 at 2:04
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    \$\begingroup\$ @A.Owl Yes, there's a risk, of course. But that's the price you pay for anything that can deliver the kind of currents and power that you require. It's not a particular failing of the cells, being "dangerous" because they are some type or other. If Superman hurts you, it's not because he's a bad guy, it's because you pissed off a guy who can hurt you. \$\endgroup\$ Commented Nov 1, 2021 at 3:02
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    \$\begingroup\$ @Simon Fitch LiIon are more like Lex Luther. While shorting any cell is a bad idea, with LiIon it's badder than most. That said, LiIon are safe enough if you charge them correctly,( which is ESSENTIAL) and dont short them. LiFePO4 lithium cells are much safer if it matter. 4 X LiFePO4 AA would make an excellent 12V supply. Charging should be done correctly but is less demanding than foe LiIon. \$\endgroup\$
    – Russell McMahon
    Commented Nov 1, 2021 at 4:44
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    \$\begingroup\$ @RussellMcMahon Ha! I was trying to avoid making the Lion the villain in this story! How about this analogy: if you poke a hamster, you might get bitten, but if you poke a Lion you get seriously messed up. \$\endgroup\$ Commented Nov 1, 2021 at 6:20
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Here is the performance of a Duracell alkaline AA cell. At 1A its voltage drops very quickly and at 2A it begins at about 1.1V for a moment: AA

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Alkaline cells are generally non-rechargeable, so when they are discharged that’s the end of their life. Drawing a large current will shorten their life considerably but is unlikely to do anything beyond that; they won’t explode or burst into flame (unlike lithium cells). Whether the motor is ok to short is another matter, best to look at the datasheet for the motor itself, but then it’s unlikely that an AA pack will deliver more than a couple of amps. C or D cells will deliver more but probably less than 11A. Also check your FET - with 3.3V on the gate it may not be fully switched on at high current, which would cause rapid overheating. If you choose a FET with a threshold voltage of about 1V you should be ok. You’ll want a freewheel diode too, as motors are somewhat inductive and Will generate electricity if they’re switched off when rotating.

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Alkaline cells have only simple failure modes when overloaded. They are single use, so "damage" is nor exactly defined for them.

  • They can overheat.For this to happen, you need to overload/short them for substantial amount of time (like, say, 30 seconds). The result may be leaking and/or outgasing with pretty much limited probability of catching fire.

  • Their voltage may (briefly) drop lower than what your control circuity tolerates. A common problem when you try to drive a motor that draws a lot when starting. A simple combination of a diode and a capacitor may protect your control circuit from transient voltage drops.

  • They may profoundly underperform. For AA cell, you may get, say, 2Ah at 0.5A load and something like 0.3Ah at 1.5A load.

p.s. be aware that different alkaline cells have quite different properties in regard to high currents, depending on the brand, model, storage time and temperature. Even different lots may behave differently.

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A real life experience: While building a large battery pack (perhaps 50 cells) with AA alkaline batteries, there was a short. Enough heat was generated in the center of the battery pack which caused the batteries to vent and ooze green goo. Fortunately, a towel was nearby which enable me to toss the assembly outdoors. Needed to air out the room for an hour.

To answer your question, it depends on how your battery pack is constructed. High current can create an unpleasant situation if the cells are surrounding each other.

For that sort of current, you're better off with lead acid (gel cell, car, or motorcycle battery) or a LiPo battery, all rechargeable. Alkaline batteries aren't suited for your application due to the current requirements and cost (unless it's a one time experiment).

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    \$\begingroup\$ Since you're suggesting LiPo, it should be noted that it's very easy to damage lipo cells in a way that can be very dangerous. If you accidentally short it, it'll vent noxious gases and potentially ignite. Lead-acid is a bit safer to work with, but should still be treated with respect. \$\endgroup\$
    – Hearth
    Commented Nov 1, 2021 at 3:11
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500mA is kind of high for AA batteries, they won't giver optimum performance there.

The 14A stall current however is not a problem, so long as the switches and wiring are up-to the task. the AAs will give whatever they can ans the motor will spin up on the availalbele current until its draw reduces to something more sensible.

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