I know that motor speed depends on applied voltage, both for DC and brushless motors: I applied 12V to a 6V motor and it started running very fast... until it burnt! :-)

Questions are actually two:

  1. how much voltage can I apply over the rated voltage without damaging the motor?
  2. why?

About question 2: I think it depends on maximum heat the windings can dissipate, which depends on current flowing into them... but then I got lost: power dissipated in a winding is I^2 * R, but how much is I? Is it V/R? But I tried plotting the following formula and it results in absurd values...

Power dissipated in motor:

  • \$ P_1 = \frac {V_1^2}R \$ (nominal condition)
  • \$P_2 = \frac{(V_1 + V_2)^2}R \$ (overvolting)

This results in:

\$P_2 = P_1 + \frac{2*V_1V_2 + V_2^2}R\$

But this gives impossibile results...

  • \$\begingroup\$ Asking for "applying voltage" probably means to the circuitry that drives the motor, which will have an absolute max rating in its datasheet, beyond which all bets are of. \$\endgroup\$
    – PlasmaHH
    Jul 11, 2016 at 21:52

2 Answers 2


In asking the question, you seem to have assumed the brushed or brushless design of the motor is not important. That is not true. A brushless motor requires an electronic controller. The effect on the motor in that case is as much determined by the controller design as it is by the motor itself. You have another answer that addresses that.

For a permanent magnet brushed motor, at the voltage level in question, the effect of doubling the voltage is mostly a question of the effect of doubling the speed. With no load connected to the motor, will probably be ok at twice the rated voltage and twice the rated speed. The current drawn by the motor is primarily determined by the load, not the resistance of the motor. The current is given by I = (V - E) / R, where V is the supply voltage, E is the back EMF (proportional to speed) generated in the motor and R is the winding resistance. If there is no load, E is practically equal to V when the motor is running at full speed. It is only reduced by the slight load due to the friction of the bearings and commutator and the air drag on the moving parts of the motor. If there is a load, doubling the speed may increase the load dramatically thus increasing the current and increasing the power developed in the resistance of the winding. The load will particularly increase dramatically if there is something like a propellor connected to the motor.

Even without a load, it is possible that doubling the speed may increase the heating due to internal friction and air drag to the point that the motor fails. In an inexpensive motor, vibration could also be a factor.


Your power-supply can be higher than the motor's continuous voltage rating, but it cannot be higher than the insulation breakdown voltage rating. On top of that, you need to be sure that you never exceed the motor's peak current, maximum velocity or temperature rating.

To put it simply, you need to run closed-loop if you want your power supply to exceed your motor's voltage rating. A PWM current-loop will typically keep your average voltage below the motor's rated voltage. It will not prevent you from exceeding the motor's maximum speed, so you need to monitor that separately. If you allow the motor's peak current, then you need to either limit the time that peak current is used, or monitor the motor's temperature (which I try to always do, anyway). The motor is not only an expensive part of the system, it is also the part that can't normally be repaired.


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