I ordered a 36V 500W brushed DC motor but unfortunately a 24V one came. I'm faced with the dilemma of returning it and waiting for a month before getting a refund, then ordering a new one and waiting about a month again for it to arrive, totaling 3 months of waiting (with the current one), OR run it beyond its ratings.

I understand that increasing the voltage will increase the rotation speed proportionally. It is currently rated at 2700 rpm @ 24V and ~27A drawn. If I increase the voltage to 36V then rpm will rise to about 4000 and to maintain the 500W power output I will need to supply ~14A.

Questions that arise are:

  1. Are my assumptions correct in the first place?
  2. Is it possible for construction damage to occur when running a motor beyound its rated rotation speed?
  3. Is efficiency compromised in such setup.
  • 3
    \$\begingroup\$ You can't control both the voltage and current supplied to your motor independently of each other. You only get to control one of those, and the other is determined by the motor load. \$\endgroup\$ – brhans Sep 4 at 17:07
  • \$\begingroup\$ @brhans yeah I felt like something is off but I wasn't sure what. \$\endgroup\$ – php_nub_qq Sep 4 at 17:09
  • \$\begingroup\$ It's not an option to run it beyond it's ratings if it burns up instantly. \$\endgroup\$ – Voltage Spike Sep 4 at 17:47
  • \$\begingroup\$ Bursting force on the rotor rises as speed squared. How much excess strength is there in the rotor construction? \$\endgroup\$ – Neil_UK Sep 4 at 19:29
  • 1
    \$\begingroup\$ @php_nub_qq How strong is the rotor? Of course, that question was rhetorical. They've tested it to the speed they've specified. Your safety margin disappears not as the overspeed, but with the square of the overspeed. So my real question is do you feel lucky, punk? The standard method for finding the maximum speed of a motor is to increase the speed until the rotor disintegrates, then back off a bit. \$\endgroup\$ – Neil_UK Sep 5 at 4:43

It is probably no issue to run a motor beyond its voltage rating.
Electronically seen, you might exceed the break down voltage between two adjecent copper wires in the armature, but it is quite unlikely at this low voltage.
Mechanically seen, as voltage is proportional to rotational speed, a higher speed burdens the bearings more. Contineously running a motor at higher speed will not directly damage the bearings, but the bearings will wear out faster.

Running a motor beyond its current rating will heat up the motor. You can temporarily run it beyond its current rating provided you don't exceed the thermal limits. If the motor becomes too hot, the isolation of the copper wires of the armature degrades or even melts, causing shorts between the windings, which on its turn increases the motor current causing a thermal runaway. The point when this happens depends on the cooling of the motor housing, ambient temperature, etc, so, not easily to predict.

You cannot do simple power calculations with motors: when the motor is running with no load, the current will be low, so the input / output power will be < 500W. When stalling the motor, the motor voltage will be close to zero and the the current high, but still the power will be way lower than 500W.

Read this answer to see how the power and efficiency are related to the torque and only have a local maximum.

To make useful assumptions for a motor, you need its torque vs speed / current / power / efficiency graphs (as in the linked answer above) AND/OR its physical constants (speed constant, torque constant etc)

  • \$\begingroup\$ As I don't have credentials for this, the next assumption is in a comment: The wear of the bearings will be rather in the order from 10000 hours of operation degraded to 7000 hours than from 10000 hours to 3 hours. \$\endgroup\$ – Huisman Sep 4 at 19:00
  • \$\begingroup\$ Fortunately for me I live in an area where a lot of companies produce bearings and is really easy to find good quality bearings on low prices. Having in mind that in my situation the only thing I'm risking is to shorten bearing life I'm keen on taking my chances with this motor. Only thing that's making me think about it is if efficiency will drop at higher rpm but I'm guessing that would be really hard to tell without graphs. \$\endgroup\$ – php_nub_qq Sep 4 at 19:05
  • \$\begingroup\$ Efficiency will drop as frictional losses increase with speed. It will also drop because a 24V motor requires more current than a 36V motor, resulting in higher copper losses. But I wouldn't bother about efficiency as the efficiency is zero at the mentioned working points (no load speed and stall) and somewhere between zero and the best possible number that the manufacturer managed to push out the motor under the optimal circumstances and that is presented in the datasheet to make the datasheet more irresistable. \$\endgroup\$ – Huisman Sep 4 at 20:49
  • \$\begingroup\$ With some restraint (because the mechanical approach regarding friction is incorrect) I would like to point to engineering.stackexchange.com/questions/26270/… where you can find an elaboration regarding working points of a motor. It may be quite possible your rated 500W motor will not be able to output 500W because your required torque / speed deviates from the rated torque / rated speed \$\endgroup\$ – Huisman Sep 4 at 21:07
  • \$\begingroup\$ it's probably no issue to run a motor beyond its voltage rating Erm, you might mention the strength required to hold the rotor together goes as the speed squared. There will be a margin, but, how much? How to find the maximum speed? Increase the voltage until the rotor bursts, then back off a bit! \$\endgroup\$ – Neil_UK Sep 5 at 4:47

I think you have 2 options, lower your supply voltage to 24V or return the motor and get the one you want. A motor has voltage and speed ratings for a purpose. One to limit current to prevent the wire from overheating and melting the insulation and a speed limit to operate within the limitations of the bearings.

  • \$\begingroup\$ So basically, I could get stronger bearings (which are pretty cheap locally I live in a place with many bearing factories) and this could allow me to run the motor at higher speeds? Fortunately they are even really easy to replace on this motor. \$\endgroup\$ – php_nub_qq Sep 4 at 17:22
  • \$\begingroup\$ It is also the rpm limits of the bearings not just load. I would rather try a 24v power supply before changing the bearings. \$\endgroup\$ – Rob B. Sep 4 at 17:29
  • \$\begingroup\$ That means I'll have to run in the 30s of amps which will quadruple my I2R losses, I would very much like to keep amperage as low as possible that's why I'm looking at those extreme solutions. \$\endgroup\$ – php_nub_qq Sep 4 at 17:33
  • \$\begingroup\$ Remember that a motor is like an inductor and as rpm (frequency) increase so does the impedance. So Z = 2pi * f * L. So the impedance increases with rpm and current decreases. A 36V motor is wound differently than a 24V motor. You really haven't said much about the expected load. \$\endgroup\$ – Rob B. Sep 4 at 17:35
  • \$\begingroup\$ You stated you ordered a 500W motor at 36V. That puts your power supply at slightly over 3A. At 24V that >4.5A How do you get 27 amps? Did you mean 5000 Watts \$\endgroup\$ – Rob B. Sep 4 at 17:41

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