Lets say I have a motor which draws 25 amps and the battery I'm using can only provide 20 amps. I would assume this would cause a drop in voltage and poorer performance of the motor, but is this potentially dangerous or could the battery be damaged?

Another scenario, the motor again draws 25 amps and this time the battery is able to provide 35 amps, however the motors starting current is 420 amps which is significantly more amps than the battery could provide. My guess is this would cause a temporary drop in voltage which would eventually stabilize, but would this scenario be potentially dangerous or could the batteries be damaged in the process?

Thanks for your time and help!

  • 1
    \$\begingroup\$ You're right that voltage droops. Such a thing naturally squanders some of the energy stored in the battery; what happens beyond that is not answerable without specifics - type of battery, safety features, if the motor is loaded at low RPM or if the load really only becomes apparent at higher speed due to either the nature of the load or a clutch, etc. \$\endgroup\$ Oct 29, 2020 at 19:22
  • 2
    \$\begingroup\$ The second case happens all the time, every day, for short durations. The starter motor on a vehicle does exactly this. It pulls 100s of amps for a short period of time from the car battery. Yes the voltage dips, no the battery isn't damaged. Whether something gets damaged depends on several things, mostly heat generated in either the motor and/or the battery. \$\endgroup\$
    – Aaron
    Oct 29, 2020 at 19:36
  • \$\begingroup\$ @Aaron Vehicle batteries are rated for that. Other batteries may not be : that's why they have a short term max discharge rating (aka "CCA" cold cranking amps) as well as a continuous discharge rating. ShinyWhaleFood : if you're pushing batteries to their limits, measure their temperature and shut down at 50 or 60C. \$\endgroup\$
    – user16324
    Oct 29, 2020 at 20:14
  • \$\begingroup\$ @BrianDrummond Right. I wasn't sure if including CCA would be relevant since only automotive batteries seem to carry that specification. \$\endgroup\$
    – Aaron
    Oct 29, 2020 at 20:17
  • \$\begingroup\$ @Aaron other batteries also carry some kind of high-current spec. In Li-Ion world, it is common to see a battery rated 5C, 20C, 30C and it can have two ratings - for short-term and for long term (as if 30C discharge can be long term, ...) \$\endgroup\$
    – fraxinus
    Oct 30, 2020 at 7:42

2 Answers 2


I'll answer your first scenario only, because I think it's more interesting. Let's dissect:

the battery I'm using can only provide 20 amps

But what that really means? What would happen if you try to pull 21 amps? If it is a smart battery, some sort of overcurrent protection might kick in. If yes, then you have the answer - the motor will just not run.

OK, let's say it's a plain chemical battery — no fuses, no BMS. The 20 amps figure likely is printed on the battery, this is its rated current. It's important to realize that

  1. Ratings like that have certain requirements to be met, e.g. temperature, state-of-charge, etc.
  2. Such ratings are not hard limits, they can be exceeded given the right circumstances. But you need to know what you are doing.

By exceeding the rated current you'd run into some issues, which the manufacturers tried to guard you from. Likely the battery will heat more (power lost through internal resistance is proportional to the square of the current. Your example is at 125% of rated current, so heating power is 156% of the nominal that was deemed acceptable by the designers). Will the battery tolerate it is not clear cut. Heating takes some time (the battery has thermal inertia). For a short time ­— e.g. a minute — the heat rise is likely acceptable. Or you may compensate the extra heat by supplying ample cooling (a water cooling jacket if you really insist).

Also, consider that a battery — even at full rated current — still needs to be a decent voltage source, so its voltage cannot sag too deeply. Let's imagine a specific example: it's a 12V battery, and the designers decided 5% voltage sag at rated current is good. 5% of 12V is 0.6V; this is at 20 amps, so 0.03 ohms of internal resistance. If you were to short the battery terminals with a heavy copper bar, that 12V/0.03 ohms means 400 amps will flow (sadly the batteries don't behave that predictably, but the figures here aren't too wrong, so let's stick with them). This partially answers your #2 scenario - the battery will likely be able to start your motor.

I think the biggest problem in your #1 scenario is actually precisely with the starting current. By using an undersized battery you increase the chance that the battery could be incapable of supplying the starting current. If the motor has any load attached, it may fail to turn (just like how a car may not start due to its battery being too old). This would lead to a pretty bad scenario, in which the motor does not turn, but consumes a ton of current. If that situation is left unchecked, it can get really ugly. The possible outcomes I can think of:

  1. the mildest one. Everything gets hot, but no fire or melting, the battery just depletes very quickly. As we assumed it has no protection, it gets depleted down to 0V. For most chemistries (and especially Li-Ion) the battery will be ruined, by a chemical process which happens inside it below certain voltages. The motor in this case is unaffected.
  2. the motor windings get so hot that their insulation melt, or the windings break open. In this case the motor is ruined.
  3. the battery may get so hot that it bursts into flames or explodes. Pretty much worst case scenario.

So is it potentially dangerous? Yes. Could the battery be damaged? Yes.

  • \$\begingroup\$ Thanks for the advice, I was planning to use four unprotected 18650 batteries in parallel to power my motor. They should provide about 15 volts altogether (3.7 * 4 = 14.8), and each battery is rated for 35 amps so they should be able to power my motor which draws 25 amps at full load. Alternatively, I can use a different model of the motor which draws less amps but more voltage. In this case I would use seven of the same model 18650 in series for 26 (3.7 * 7 = 25.9) volts. This time the motor would only draw 13amps and 230 at startup. Which do you recommend, thank you so much for the help! \$\endgroup\$ Oct 29, 2020 at 21:17
  • \$\begingroup\$ Both scenarios should work. There are other things that may tip your decision either way. The 7-cell solution would have more power and more runtime, if that's important to you. But obviously it would weigh and cost more. It's your call here, really, but from EE standpoint both options work. \$\endgroup\$
    – anrieff
    Oct 29, 2020 at 22:36
  • 8
    \$\begingroup\$ #1 is actually a worst case scenario for Li-Ion. If you do something stupid like hook up a Li-Ion cell with no protection, you will burn something down. A deep discharge for Li-Ion is extremely dangerous, because it looks fine until you recharge it, then the crystals formed during deep discharge will puncture the cell due to heat expansion, creating a short, and burn. Li-Ion cannot be deep discharged. \$\endgroup\$
    – Nelson
    Oct 30, 2020 at 8:48
  • \$\begingroup\$ @Nelson meaning that deep discharging a lithium cell is perfectly fine if I make sure to never ever use it again? \$\endgroup\$ Oct 30, 2020 at 16:52
  • 3
    \$\begingroup\$ @JohnDvorak, yes it will not combust during the discharge. The real problem is: there'd be no way to know it was deep-discharged. Apart from voltage, but not everyone checks that every time (do you?) and you may even be able to slowly recharge it to a voltage that seems "normal". Then it will look just like any normal depleted battery. It won't look physically crooked. Basically, a ticking time bomb. For this reason alone is why it's much better to use protected cells. \$\endgroup\$
    – anrieff
    Oct 30, 2020 at 19:45

Both scenarios are potentially hazardous. In the first case, the internal resistance in the battery will probably reduce the current sufficiently to prevent damage to the motor, but the battery may be damaged. In the second case, the internal battery resistance will not reduce the motor current sufficiently tp prevent damage to the motor. Some small motors may not draw so much current when they start, but those motors are not very efficient because their internal resistance is wasting power during normal operation.


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