I have a motor attached to a foam wheel (made of insulation material). The wheel is VERY light, yet the motor I am using cannot handle the torque required to run it, resulting in the motor creating a burning smell. This is an 18V motor that I connected to a 12V 2.2 A supply. If I lower the current further, wouldn't the speed decrease along with the torque? Or is voltage related to speed and torque related to current? I am not sure if this motor can handle the load, but I want to know if there is anything I can do to give it a chance.

I know this question has been asked, but it was about a brushless motor. This is a brushed motor, so it might work differently.

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    \$\begingroup\$ I'm a little surprised that the motor would even attempt to turn with that much undervoltage. \$\endgroup\$ Commented Aug 25, 2015 at 2:59
  • \$\begingroup\$ How much voltage is the supply actually producing? If the motor is trying to draw more than 2.2 A, the supply will either automatically reduce it's output voltage to limit the current to 2.2 A or it will produce more current, get hot and the voltage will be reduced somewhat because of the overload. How is the wheel connected to the motor? Unless it is directly attached to and supported by the motor shaft, the additional mechanism requires torque. What is the rated motor current and speed? \$\endgroup\$
    – user80875
    Commented Aug 25, 2015 at 4:00
  • \$\begingroup\$ @Ignacio - many motor ratings are much less critical in many respects than you might expect. At voltages lower than rated (downto maybe 20%) it'll just run at commensurately lower speed. Below that point the stall current (V/R) simply doesn't produce enough torque to turn it over. This motor actually has speed specs at 12V and 9V in addition to 18V. \$\endgroup\$
    – user16324
    Commented Aug 25, 2015 at 10:52

2 Answers 2


You're looking at it slightly wrong : voltage affects speed, but it's closer to say that torque affects current than vice-versa.

So run the motor unloaded : at 12V it'll run at about 2/3 the speed at 18V, that being the speed at which it generates enough back EMF to cancel most of the driving voltage. (Datasheet says 15200 vs 24000 at 18V).

The rest of the driving voltage (maybe 10 or 20%) is dropped across the motor's winding resistance, resulting in two effects : enough torque to turn the unloaded motor that fast, and heating the motor, by I^2R watts. You can't measure that voltage directly, but you can measure the resistance and current, and I recommend you do. I'm guessing about 1 ohm, and the spec gives a no-load current of 400mA, so 0.16W in the windings (and 4.8W total, so the rest is lost in friction in brushes and windage).

Now add a little load. As you add a torque load, the motor will slow, so the back EMF will reduce, and the current will increase until it supplies the torque you need. This is why I say torque affects current... Efficiency also drops too, because that extra current creates more heat. You don't want to use the motor below 75% of its unloaded speed - maybe 60% for short periods then cool it off.

Now add your foam wheel. You haven't told us but I'm guessing it's quite large, and simply spinning, you're not running it across the floor. How fast does it run? You haven't told us, but I'm guessing its wind resistance is acting as a huge airbrake, so a few hundred RPM.

Which is pretty much stalling the motor. It can't generate any back EMF, and virtually all of that 12V is developed across the winding resistance. So I=12A, power = 144W wasted as heat. Fortunately your PSU isn't up to that, so you may still have a working motor...

You need a high torque, low speed motor. Or, more likely, gearing. This will increase the motor speed and reduce the torque demand to let the motor drive the wheel efficiently.

  • \$\begingroup\$ +1 for gearing. It will trade torgue for speed. That way the motor can run within its specifications while turning the fan even faster than before. An alternative solution with the same end-result would be a V-belt. \$\endgroup\$
    – Philipp
    Commented Aug 25, 2015 at 14:46
  • \$\begingroup\$ +1 for the breakdown of how torque affects current. Very illustrative, thanks! \$\endgroup\$
    – U007D
    Commented Dec 20, 2016 at 23:43
  • \$\begingroup\$ @Brian, can you please explain how for a given load, voltage affects speed? I'm looking for a conceptual answer rather than a mathematical one. \$\endgroup\$
    – penguin99
    Commented Dec 13, 2019 at 7:15
  • \$\begingroup\$ @Noorav :follow through he answer. First consider no-load speed - then consider how speed reduces as you add load. \$\endgroup\$
    – user16324
    Commented Dec 13, 2019 at 16:49

In general for permanent magnet motors the torque produced will be proportional to the current flowing through armature and the no-load speed will be proportional to the voltage.

As you increase the load on the motor the speed will reduce but even this can be modeled as the effective voltage being reduced because of the drop across the armature resistance so reduce the effective voltage that the motor sees.

When you say you have a light foam wheel - are you attempting to propel a vehicle with it or is the wheel in free air? What diameter is the wheel?

If you are trying to drive a vehicle with the motor you will need a gear reduction to increase the torque (and reduce the speed).


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