I am trying to find a BLDC servo motor for a pump. I have calculated the required speed, but while the manufacturer was good in getting back to me with the torque requirements, they did not give much detail. So, I only have a ballpark figure for worst-case the torque.

My instinct is to pick a motor with a much higher rated torque than quoted. My thinking goes that the higher torque rating basically means higher current carrying capacity, which means larger windings. If the motor is used up to its limits, it will work OK because its rated for them. If used well below its limits, the larger body will dissipate heat more efficiently, and the windings will have a lower resistance - so if anything it will use less power a smaller one operating at its limits.

So, rather than try to estimate the torque, I should just buy the largest motor that fits with my other constraints.

Have I assessed this correctly? Are there any downsides to oversizing a BLDC motor?

Related: Aside from cost, is there any disadvantage to using thicker wire than required?

  • \$\begingroup\$ A good assessment, I think. Note that the BLDC controller must be able to start the bigger motor with its lower winding resistance. \$\endgroup\$
    – user16324
    Jan 10, 2016 at 13:23
  • \$\begingroup\$ "My thinking goes that the higher torque rating basically means higher current carrying capacity, which means larger windings" Not quite true... Higher torque could be via high Kt \$\endgroup\$
    – user16222
    Jan 10, 2016 at 13:31
  • \$\begingroup\$ Experienced mechanical engineers have told me that you ideally want to match the moment of inertia of the motor itself to the moment of inertial of the load applied to the motor in the correct ratio. Seems to me that picking an arbitrarily larger motor than the application requires would violate this consideration. \$\endgroup\$ Jan 10, 2016 at 14:25
  • 2
    \$\begingroup\$ With regard to my above comment it may be informative to read this white paper on Load/Motor Inertial Mismatch. diequa.com/download/articles/inertia.pdf \$\endgroup\$ Jan 10, 2016 at 14:35
  • \$\begingroup\$ Thanks JonRB and Brian for those good points! @Michael, Thank you for that link! This is exactly the kind of knowledge gap I was worried about. I'll give that paper another couple of reads to make sure, though it seems I should be OK so long as its not too far out, since my pump speed will very rarely change/my application doesn't require much bandwidth. \$\endgroup\$
    – sebf
    Jan 17, 2016 at 15:20

2 Answers 2


BLDC with higher torque rating will differ in the size, thus it will have larger inertia, making the motor less prone to the fast dynamic change of speed, position.
If your driver can fed the motor with less than rated current the overall torque will be smaller than rated. The only constraint is the match between driver and motor - the driver has to have the capabilty to drive the motor within the range of winding inductance.


In general, electric motors are designed for maximum efficiency near rated speed and load. However, the peak efficiency is often slightly below the rated load because motors are generally selected to be somewhat oversized. Also, the efficiency change with load is generally smallest near the rated speed and load. With a moderately oversized motor, say 20%, there is likely little or no reduction in efficiency, while with a greatly oversized motor, say more than 30%, the reduction in efficiency may be significant. Bear in mind that the efficiency curve must drop to zero as the load drops from rated load to zero. That is caused by the fact that there is no output power at zero load but some power is required to keep the motor turning even when there is no load. You should examine the efficiency curve for your specific motor, but I would expect that the curve is relatively flat near rated load and gradually drops more quickly as the load decreases.


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