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I have a 1100KV brushless DC motor (turnigy d3548/4) rated maximum voltage 19V i want to run it at 25V assuming low load as i mostly need the higher rpm !

whats are the effects of this higher voltage on the motor electrically(coils insulation) and mechanically ?

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    \$\begingroup\$ 1100 KV (kelvin volts? I assume you mean kV, kilo-volt) is a 1.1 MV (million volt) motor and you want to run it at 25 V? Welcome to EE.SE. Units and symbols have particular exact meanings. Please modify your question to clarify and provide links or extracts for any relevant data sheets. \$\endgroup\$
    – Transistor
    Mar 18, 2016 at 10:26
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    \$\begingroup\$ @transistor KV refers to the rpm constant of a brushless motor - it is the number of revolutions per minute that the motor will turn when 1V (one Volt) is applied with no load attached to the motor \$\endgroup\$
    – HandyHowie
    Mar 18, 2016 at 10:46
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    \$\begingroup\$ @transistor Probably you never looked at motor spec., it menas 1100rpms/1V, 19V is cca. 21kRPM final speed \$\endgroup\$
    – Marko Buršič
    Mar 18, 2016 at 10:57
  • \$\begingroup\$ It may fall apart due to centrifugal force. \$\endgroup\$
    – Marko Buršič
    Mar 18, 2016 at 10:58
  • \$\begingroup\$ There was no link to a spec. :^) \$\endgroup\$
    – Transistor
    Mar 18, 2016 at 11:01

2 Answers 2

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If you want to run the motor at a higher speed, the most serious mechanical problem is the rotor bursting, though you will also have higher windage losses, and there is also the potential for unbalance vibrations.

Windage load

These rise as the cube of the speed. As you say the motor will have low load, you may be able to accommodate the extra power to overcome this loss without overheating the motor. The cooling will improve at higher speeds as well.

Rotor bursting

The bursting force on the rotor rises as the square of the speed. This means a 50% increase in speed results in 1.5^2 = 2.25 times the bursting force.

With an outrunner motor, the rotor consists of permanent magnets, and an iron return path, that is, a composite structure. How much strength margin has the manufacturer built into it?

Before I used a motor at significantly higher speed than rated, I would spin the motor (either electrically, or passively by using an electric drill or something) to at least 2x the intended maximum speed. This would give a strength margin of 4x.

Unbalance vibrations

A motor does not need to be as accurately balanced at low speed, so it may not be well enough balanced for a higher speed. Run it through and a little above your intended speed to check.

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  • \$\begingroup\$ @OleksandrR. I understand we don't share the same view of the precautionary principle, which is fine, this is a public forum, speech is free. I'm not sure whether you disagree with 'test it faster than you intend to run it' in principle, or the factor of 2x in speed = 4x in force in particular? I look forward to reading a more useful answer. Please feel free to remove your upvote, and post your own answer to this question. \$\endgroup\$
    – Neil_UK
    Mar 19, 2016 at 8:36
  • \$\begingroup\$ I don't disagree with your exhortation to be cautious. I just think that when the test is done and the motor almost certainly doesn't suffer rapid catastrophic mechanical failure, the result will not be very useful and could even lead to a false sense of security that it can operate normally in this condition. In reality, the bearing life will be reduced significantly, so the motor will likely fail anyway, but after some indeterminate time. Your answer is accepted, so there is no point in my posting another. I will just delete the comments, since OP seems to be satisfied with your account. \$\endgroup\$
    – Oleksandr R.
    Mar 19, 2016 at 13:21
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We have a pump application where we run a 14v rated brushless motor at 24v without issues. The core issue is really total watts, within reason. A motor basically becomes a large dc dc converter anyway, for sure under pwm conditions.

You'll have to keep rpms within spec of course.

Some approximate values from this setup at the same load and rpms are -

24v @ 0.90 Amps

20v @ 1.05 Amps

16v @ 1.25 Amps

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  • \$\begingroup\$ well this means it draws less current at same load meaning no extra heat generated that would make the coil isolation fail ? do u go to full throttle at 24V setting with no problems ? i'd appreciate if u explain further what u mean by "for sure under pwm conditions" \$\endgroup\$
    – MTF_09
    Mar 18, 2016 at 12:29
  • \$\begingroup\$ Those numbers suggest you are stalling the blades and shifting less air at higher speeds. OTOH if you're changing the blade configuration to achieve the same delivered power (or volume or thrust) at different speeds, they look more reasonable. But I'm guessing at those power levels you're not operating anywhere near 20000 rpm!. \$\endgroup\$
    – user_1818839
    Mar 18, 2016 at 13:00
  • \$\begingroup\$ This looks like you are supplying a higher voltage to the controller, but running the motor itself under the same conditions of speed, load etc, so the controller is doing its buck stepdown thang into the motor windings. The OP specifically asks about higher speed \$\endgroup\$
    – Neil_UK
    Mar 18, 2016 at 13:48
  • \$\begingroup\$ Within the precision of your figures, the motor is dissipating the same power in all 3 cases. In turn, this means that the pump is operating at the same RPM and load and suggests, as Neil_UK stated, that the motor controller is absorbing the voltage differences, but the motor is essentially being driven at the same power. I hope you don't think you're getting better pump output at the higher voltage. \$\endgroup\$
    – WhatRoughBeast
    Mar 18, 2016 at 14:58
  • \$\begingroup\$ Its a pump, running at 4000rpm. Not sure why this deserves a downvote anyway. It is real life figures from years of operation. \$\endgroup\$
    – Erik Friesen
    Mar 18, 2016 at 16:07

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