I have two very basic motors, basically this one: amazon

Motor specifications:
Power: 150W
Volts: 12-24V
Amps: 4.42 @ 12V, 4.76 @ 24V
Speed: 3300 RPM @ 12V, 5500 RPM @ 24V

The project is a robot consisting of these two motors and 3 servos. But the main concern is these two motors since they will be running at full speed (but not high load) most of the time.

I need to be able to run these two motors for at least 2h 30min. I'm trying to figure out what kind of battery do I need but I find mixed answers between forums and formulas.

I'm looking at Lithium-Ion or LiPo batteries. I have a 5000mAh, 6S, 50C LiPo (that I'm a bit afraid to use) and option to get 24V 6Ah. As I understand 6Ah means that the battery could supply 6A for an 1 hour so one option is to have two of these batteries. I'm assuming this would give me about 1h+ so I would need a 4 pack.

Otherwise I likely need to rethink this project with gear/belt setup an 12V 30W DC motors.

How do I determine the battery needed for these two motors.

I'm also confused by this formula that I found:

Total Current Draw = 2 * 4.67A = 9.34A
Operational Time = 2.5 hours
Battery Voltage = 24V
Battery Capacity (Ah) = (9.34A) * (2.5 hours) / 24V ≈ 0.977 Ah

Which is kinda odd. This would mean my 6Ah Lithium-Ion battery for a cordless hand drill could power the two motors for nearly 6h..

Another formula I found gives a much smaller run time:
300W / 24V = 12.5A
6Ah / 12.5A = 0.48 minutes.

Not sure which one is correct.

  • \$\begingroup\$ " at Lithium-Ion or LiPo batteries". What you call "LiPo" IS a Li-ion battery. They use Li-ion cells in a pouch package but they are Li-ion cells nonetheless. \$\endgroup\$ Aug 31, 2023 at 23:32
  • \$\begingroup\$ @DavideAndrea thanks for the explanation. The reason I mentioned -Ion is because they are super cheap/easy to get. \$\endgroup\$ Aug 31, 2023 at 23:41
  • \$\begingroup\$ I added the calculations but I'm not sure which one is correct. \$\endgroup\$ Aug 31, 2023 at 23:42
  • \$\begingroup\$ Without any voltage regulation, will the motors run properly as the 6S battery voltage runs down to 18V? Do you have a battery protection circuit to disconnect the battery's load when the voltage drops to 18V? Super cheap batteries are fakes. Instead of battery chemicals they have flour for its weight. THE SPECS ARE LIES. \$\endgroup\$
    – Audioguru
    Sep 1, 2023 at 0:53

3 Answers 3


The problem with the calculations is that Ah does not specify voltage. This is actually a problem in a lot of other contexts as well. What you are really storing in a battery is power, traditionally measured in Watt-hours. For comparison, your electric bill is generally measured in kWh = kilowatt-hours = 1,000 Watt-hours.

However, for whatever reasons, a lot of batteries are rated as storing a particular number of Ah = Amp-hours, or mAh = milli-Amp-hours. I think the whole mAh thing is a bit crazy and more for marketing than anything else - 5,000 mAh sounds like more than 5 Ah. That number (whether mAh or Ah) is only meaningful if you also have the voltage. Multiply voltage x Ah to get Watt-hours. Technically you get VAh, and they are not always the same, but for a rough calculation they're close enough for most purposes.

Let's examine each of these equations:

Total Current Draw = 2 * 4.67A = 9.34A
Operational Time = 2.5 hours
Battery Voltage = 24V
Battery Capacity (Ah) = (9.34A) * (2.5 hours) / 24V ≈ 0.977 Ah

The first line should probably be 4.76A based on the earlier specs. So that's really 9.52A. But the real problem is in the last line. Ah is not A * h / V. It is A * h. So the actual Ah is 9.52A x 2.5h = 23.8 Ah. If you use 24V batteries (or 6V batteries, 4 of them together) then the voltage is the same and you need 23.8 Ah of batteries to last 2.5 hours.

Look at the second equation:

300W / 24V = 12.5A
6Ah / 12.5A = 0.48 minutes.

The first line is basically correct. W = V x A, so W / V = A. 300W is the rated total power of 2 motors (150W each), so the result is 12.5A of current draw. Except that it isn't exactly correct because the 150W rated power for each motor is substantially more than 4.76A x 24V = 114W. Which could be for all sorts of reasons, among them that the motor might draw more at startup but a nominal current draw of 4.76A when running.

The second line should be hours, not minutes. Ah/A = h. 0.48 hours = 28.8 minutes.

But we now have two equations with different meanings. The first equations says "the battery capacity to run for 2.5 hours is 23.8 Ah @ 24V". The second equation says "A 6A hour battery pack at 24V (same as motor, by definition) will run for 28 minutes".

Extrapolating, we can take the second equation and multiply to get 2.5 hours. 2.5 hours / 0.48 hours = 5.2 x 6A = 31.2. So the second equation says you need 31.2 Ah of battery to last 2.5 hours. The first equations says 23.8 Ah. That is a bit off, but not by much. Because if you divide the initial current, 9.52A, used in equation 1 by the initial current, 12.5A, use in equation 2, you get 0.76. Multiply that by the result of equation 2 and you get 31.2 x 0.76 = 23.7 - the same as equation 1, except for some rounding errors.

In other words, once you get the equations right they really say the same thing except for starting with a different starting point (max. W vs. a very exact A).

  • \$\begingroup\$ Thank you very much for the very detailed explanation. Seems like I won't be able to run this off batteries for any sensible price. I will have to rethink the project with simpler motors and some kind of gear system. \$\endgroup\$ Sep 1, 2023 at 1:09
  • \$\begingroup\$ If you have something that has extremely low load then yes, a geared solution may work. But if you have to move something of any significant wait then it really doesn't matter whether you start with a fast motor and slow it down or a slow motor and use it at its natural speed - the work done, which requires electrical energy to make the motor do the work - is the same. \$\endgroup\$ Sep 1, 2023 at 1:14

Running one of the motors at 24V, unloaded, will draw 4.8A. This will have to continue for 2h30m (2.5h). The required battery capacity for one motor will be:

$$ Q = 4.8A \times 2.5h = 12Ah $$

Note: The units are consistent, \$A \times h = Ah\$. Voltage is not considered here, since then \$V\$ would also have to show up in the answer's units somewhere.

The power rating of the motor is 150W at full load. This will give you some idea of the current that the motor will draw under maximum mechanical load:

$$ I_{MAX} = \frac{P}{V} = \frac{150W}{24V} = 6.3A $$

I have assumed that this power rating is electrical input power, not mechanical output power. If this power is mechanical output power, then you must account for motor efficiency to find the corresponding electrical input power. Refer to the motor's documentation.

At full load, then, the required battery capacity will be:

$$ Q_{MAX} = 6.3A \times 2.5h = 16Ah $$

I have assumed that the batteries are 24V. If each battery pack is only 12V, you must connect two such batteries in series, for 24V. Doing so will not change the capacity (in Ah), since being in series they both pass the same current. For instance, with two 6Ah, 12V batteries, they can still pass 6A, and each can still maintain this condition for 1h.

However, at 24V, each Ampere of current delivers 24W of power, in contrast to the 12W per Ampere available from a 12V source. Therefore, even though it seems counter-intuitive that combining batteries in series doesn't change overall charge capacity, it does increase the amount of energy available per unit of charge.

To obtain 16Ah of capacity, from batteries of only 6Ah capacity, they must be connected in parallel, such that the total 6.3A is shared between them. In the next diagram, I am treating each pair of 12V batteries as a single 24V battery. Therefore, there are \$N=3\$ 24V batteries here:


simulate this circuit – Schematic created using CircuitLab

With \$N\$ individual batteries of \$Q_{IND}[Ah]\$ capacity, connected in parallel, you have total capacity:

$$ Q_{TOT} = N \times Q_{IND} $$


$$ N = \frac{Q_{TOT}}{Q_{IND}} = \frac{16Ah}{6Ah} = 2.7 $$

You can't have a fractional number of batteries, so round that up to \$N=3\$.

With the motor running under no load, which we calculated above would need 12Ah of battery capacity to keep going for 2.5h, you have:

$$ N = \frac{12Ah}{6Ah} = 2.0 $$

You should still probably use \$N=3\$ batteries, since their capacity will reduce with age and discharge/recharge cycles.

As a sanity check, we can find the longevity of this setup in terms of energy. Start by calculating the total energy contained in an individual fully charged battery. You'll notice there we do involve voltage now. While it wasn't strictly necessary to consider voltage before (it was sufficient to deal with charge, currents and time), if we wish to make our calculations in terms of energy and power, then voltage must enter the equations.

If you are using six 12V 6Ah batteries, each fully charged battery stores:

$$ E_{IND}[Wh] = Q_{IND} \times V = 6Ah \times 12V = 72Wh $$

The total amount of energy available from all six batteries is:

$$ E_{TOT} = E_{IND} \times N = 72Wh \times 6 = 430Wh $$

If you are using three 24V 6Ah batteies, you'll find the energy stored is the same:

$$ E_{IND}[Wh] = Q_{IND} \times V = 6Ah \times 24V = 144Wh $$

The total amount of energy available from all three batteries is:

$$ E_{TOT} = E_{IND} \times N = 144Wh \times 3 = 430Wh $$

Convert those 430Wh to Joules, to be compatible with our power calculations below (because the units of power are joules per second \$[W] = \left[\frac{J}{s}\right]\$:

$$ E_{TOT[J]} = E_{TOT[Wh]} \times 3600 = 1.5MJ $$

Drawing 150W (that's 150 joules per second) from this energy source, we can expect the source to be depleted after time \$T\$:

$$ T = \frac{1.5MJ}{150W} = 10000s \approx 160min \approx 2.8h $$

All that is for one motor. To obtain the same run-time for two motors, you need twice as much energy stored, which means twice as many batteries:


simulate this circuit



simulate this circuit

In terms of charge capacity, this works out as follows: each 24V battery has 6Ah capacity. Connecting six such batteries in parallel increases total capacity:

$$ Q_{TOT} = N \times Q_{IND} = 6 \times 6Ah = 36Ah $$

At a load of \$I=2\times 6.3A = 12.6A\$, this compound battery will last:

$$ T = \frac{Q_{TOT}}{I} = \frac{36Ah}{12.6A} \approx 2.8h $$

Alternatively use two 12V car batteries in series, each with a capacity of over 50Ah. Even a pair of small 50Ah batteries gets you a run time of:

$$ T = \frac{50Ah}{2 \times 6.3A} \approx 4h $$


The maximum current draw for two motors at 24 V and fully loaded will be

$$ 2\times 4.76~A = 9.52~A $$

For 2.5 hours that will be $$ 2.5~h\times 9.52~A = 23.8~Ah $$

You're going to want a safety margin, say 20%. $$ 120\% \times 23.8~Ah = 28.56~Ah $$

That's going to be your worst case battery capacity requirement for a 20 % safety factor.

If not running at full load you can reduce that relative to the load, the best thing to do would be to measure the motor current under the expected load and plug that number into the formulas.

You also have to understand that the Ah rating does not tell you what the maximum current you can draw from a battery is. A 6 Ah battery is not likely to supply 600 A for 36 seconds as simply multiplying the time and current might suggest, there will be a rating for maximum current draw and you need to take that into account.

Update - Using the measured load current from your comment we get this:

  • Run time = 180 minutes = 10800 seconds
  • Throw rate = Every 5 seconds
  • Max load time per throw = 0.5 seconds
  • Total throws = \$10800s / 5s = 2160\$
  • Max load time = \$2160\times 0.5s = 1080s\$
  • Idle time = \$10800s - 1080s = 9720s\$
  • Max load current = 3.5 A
  • Idle current = 0.45 A
  • Max load consumption (Ah) = \$\frac{1080s\times 3.5~A}{3600~s/h} = 1.05~Ah\$
  • Idle consumption (Ah) = \$\frac{9720s\times 0.45~A}{3600~s/h} = 1.215~Ah\$
  • Total Ah = \$1.05~Ah + 1.215~Ah = 2.265~Ah\$

With a 20% safety factor that's 2.718 Ah, so if my math is right, with a 6 Ah battery you should have just over twice the capacity you need. Even if the numbers are off a bit you should probably have plenty of leeway. I would think the next thing to do would be to build a prototype and test it, updating the data with further measurements.

  • \$\begingroup\$ thanks for the explanation it seems it will be nearly impossible to run these two motors for 2.5h on batteries since I can't find any for any reasonable price that can deliver that kind of capacity. \$\endgroup\$ Sep 1, 2023 at 1:07
  • \$\begingroup\$ Even 1h runtime comes quite a bit on the expansive side I might have to find new motors and maybe use a gearbox system. \$\endgroup\$ Sep 1, 2023 at 1:08
  • \$\begingroup\$ @SterlingDuchess Do you know how much current they will actually draw in your use-case? The 4.76 A rating is full load, no load is 0.5 A, you're going to be somewhere in between, you need to find out where before you can know what size battery you need. \$\endgroup\$
    – GodJihyo
    Sep 1, 2023 at 1:14
  • \$\begingroup\$ thank you for the clarification. I think I completely messed up, I'm building a tennis ball shooter so the load wont be constant, in fact the motor will mostly be no-load. Balls will fire every 5s which means it will be under load every 5s for an extremely short time. So I looked up how to wire the motor to test the load. At 24V the motor needs 2A to start, then has 0.45A no-load @ max RPM. I tried gripping the shaft strongly and softly. Under load I applied it reached 3.5A. \$\endgroup\$ Sep 1, 2023 at 11:07
  • \$\begingroup\$ This means that I would need to supply 0.45 for majority of time and 2.5-3.5A every 5 seconds for a timeframe <0.5s (duration of contact with the ball). This means that in 180m session I would need to provide 2.5-3.5A for total of 36 minutes, and 0.45A for 144 minutes. But that 36 minutes is not continuous. So I assume that my 24 6Ah battery could last me some time. \$\endgroup\$ Sep 1, 2023 at 11:11

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