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You always learn in school that you should never short terminals of a battery because the wires overheat from the large current.

If you look at a brushless motor, you'll see that its nothing but coils of wire. So why don't these motors "short" the power leads? How is this any different than shorting terminals of a battery? How is the current being regulated when the motor is operating?

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    \$\begingroup\$ Brushless motors are not connected directly across the terminals of a battery. If they were, they would indeed burn up. A BLDC controller is used to produce an AC waveform (a somewhat complicated process due to the need for feedback with motor position) which sees a significant impedance from the inductance of the windings as well as the back EMF from the rotor's motion; this prevents undesirably large currents from flowing. \$\endgroup\$ – pericynthion Dec 23 '14 at 22:50
  • \$\begingroup\$ I've shorted one with a 9V battery. \$\endgroup\$ – user54268 Dec 24 '14 at 5:30
  • \$\begingroup\$ Why don't inductors "short"? \$\endgroup\$ – Nick T Dec 24 '14 at 12:04
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Those wires form coils, so are long. Every bit of wire has some resistance, and all those bits of wire end to end result in a significant enough resistance to not look like a "short".

These wires shorted across a voltage source is exactly where the stall current of the motor comes from. It is simply the voltage applied to the coils divided by their resistance.

When the motor is running, then another effect is present. The motor actually acts like a generator so that spinning in the forward direction cause a voltage to be generated across the coils. This voltage opposes that applied by the external power source. The current thru the motor is therefore the power voltage minus this reverse EMF produced by the motor acting as a generator, and that result divided by the coil resistance. The faster the motor spins, the less current, because a higher back EMF is subtracted from the driving voltage.

This back EMF effect also limits the top speed of the motor. At some speed, the back EMF generated internally cancels the applied voltage, and nothing is left driving the motor. Of course it wouldn't spin at that speed since nothing is driving it anymore, but it works at a little lower speed if nothing is loading the motor.

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    \$\begingroup\$ +1 Nice and clear (of course). Noting that your explanation extends to brushed motors may be useful. In the spirit of the season, and the regrettable internationality of the site, the (singular) incidence of US spelling will be ignored :-). | Hmm - slip speed - whole new avenue for digression into synchronous motors. No! Stop while ahead :-). \$\endgroup\$ – Russell McMahon Dec 23 '14 at 22:13
  • \$\begingroup\$ Whoever downvoted this: What exactly do you think is wrong, misleading, or badly written? \$\endgroup\$ – Olin Lathrop Dec 24 '14 at 14:13
  • \$\begingroup\$ I didn't downvote, but IMO, the order in which you present the two effects (wire resistance and back-EMF) feels a bit backwards. When I first opened the page, I only saw the first two paragraphs of your answer, and I was ready to downvote it before I scrolled down and saw the rest. \$\endgroup\$ – Ilmari Karonen Dec 24 '14 at 15:33
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A "short" is a low resistance, low inductance connection. A motor coil is a moderate resistance, high inductance connection. The inductance reduces the rate at which the flow of current increases from 0 (at the initial time of connection), and the resistance limits the maximum amount of current that can flow. So even if you have a motor coil directly across a battery, as long as the connection isn't maintained for very long there isn't an opportunity for the amount of current flowing through it to get out of control.

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  • \$\begingroup\$ Thanks. Does the same explanation apply to transformers? \$\endgroup\$ – user1527227 Dec 23 '14 at 21:45
  • \$\begingroup\$ In general, yes. But a transformer is also (usually) presented with an AC voltage, which magnifies the effect the inductance has on the circuit. \$\endgroup\$ – Ignacio Vazquez-Abrams Dec 23 '14 at 21:56
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You mention a brushless machine. These require a form of AC excitation (inverter, controller etc...). Take a single phase brushless DC machine (same argument for single phase BLAC, or polyphase...)

If you were to take a battery and connect them directly to the phases of this brushless machine A few things would happen

  1. instantaneously the machines stator would appear like an open circuit as essentially it is a real-world inductor: R+L (for a 100uH line inductance you could have 0.1R)

  2. The current would build up as per V = L di/dt

  3. Once current is flowing there would be rotor airgap flux generated which in turn would impose a torque on the rotor & it would turn (unless already aligned) & essentially lock the rotor.

  4. Depending on other considerations, a few other points of interest need to be noted.The current limit will be based upon the Equivalent Series resistance of your source, plus the windings resistance. Now the stator acts as a really good heatsink so taking that into consideration.

    4a. If the steady state short-circuit current is a value that the machine can tolerate then it will sit there generating torque at its shaft

    4b. If however it isn't (more than likely for anything over a small battery...) the stator windings will heat up & the insulation will breakdown, shorted turns, propagated shorted turns, hard short.

If an inverter+controller was present, depending on the complexity they may PWM the stator to a fixed current level to control the stall torque within the bounds on the machine+inverters design.

Simply put, Depending on the battery, its cranking capability, and the stall current capability of the machine, you would not put it directly across a brushless machine for exactly the same reasons.

Now if they were connected to a PMDC machine (brushes) the rotor would start to rotate and the current will be limited due to commutation and the rotor speed

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I think he is trying to understand why the coils that are touching don't all short. I believe the answer he is looking for is that the wire has a thin resin coating before it is wound around the stator laminations. It is that resin coating that keeps the current flowing "in line" down the wire. When a brushless motor is pushed past its current rating, it can overheat and burn off some resin. When this happens, the wires are now free to touch and exchange electrons, and the motor is now dead. This coating is clear, so unless you look close or scratch it off, you can't really see that it is there.

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  • \$\begingroup\$ This doesn't really address the question. The enameling on the wire likely was taken as a given, as it is common in motor windings. The concern was about the overall DC resistance of a coil and how the motor operates. The question has also been effectively answered with the correct context, both from the accepted answer and my reading of all the other answers. \$\endgroup\$ – user2943160 Aug 8 '16 at 2:41
  • \$\begingroup\$ this actually was one part of my question. thx. \$\endgroup\$ – user1527227 Sep 11 '16 at 22:40

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