The shaft of an unconnected motor is easy to rotate relative to a motor with shorted terminals. If a resistive load is connected to the terminals, the turning difficulty is somewhere in between.
Why is this? (I'm using a BLDC motor.)
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4\$\begingroup\$ Normally electrical power sources are constant voltage, so a load with a smaller resistance is considered to be a larger load. Could you edit your title, please? \$\endgroup\$– TimWescottCommented May 12, 2019 at 20:32
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\$\begingroup\$ Not in my my experience, shorted terminals is more difficult with a permanent magnet DC brushed or BLDC motor. Be specific about the type of motor you are using. \$\endgroup\$– Neil_UKCommented May 12, 2019 at 20:33
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\$\begingroup\$ @Neil_UK I agree with you. I think that's what I stated in the description. \$\endgroup\$– abcCommented May 12, 2019 at 20:36
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3 Answers
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I have to start with some terminology -- sorry if it's esoteric, but this will bring things into line with how folks talk about this subject.
When you turn a permanent-magnet DC machine*, the armature generates a voltage internally. This is called the "EMF"** of the armature, or the "back EMF" if the machine is running as a motor. This EMF is always generated when the machine turns.
When you run current through a DC machine, it generates a torque. This torque is always generated when the machine turns, regardless of whether it's a motor or a generator.
When you put a resistance on the terminals of a machine and turn its shaft, it generates that EMF. With the resistance connected, this EMF causes a current to flow that's proportional to the EMF divided by the external resistance plus the machine's armature resistance. This current, in turn, generates a torque that resists motion (due to conservation of energy, it must be in a direction to resist motion).
Shorting the machine puts the smallest possible resistance on it -- you can't get lower than 0 without resorting to active circuitry. The back torque in this case is purely a product of the EMF and the armature resistance. Increasing the resistance by putting a resistor on there means less current for the same machine speed, which means less back torque. In the extreme, you have no resistor at all, which means infinite electrical resistance -- this means that the back torque will be from mechanical effects such as friction (and windage, if you're turning it that fast), and possibly mechanical and electromechanical effects as the field magnets work against the iron in the armature.
* I'm calling it a "machine" instead of a "motor" because it can be a motor or a generator, depending on how you use it. But you don't have to change anything internally to change how it's used -- hence, "machine".
** EMF stands for "electromotive force", which is just and older term for "voltage". It seems silly to have two terms, but sometimes it's useful.
answered May 12, 2019 at 20:42
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1\$\begingroup\$ I appreciate the fundamental explanation. I find a lot of information regarding the "whats" of DC motor operation, but the "whys" are harder to come by. \$\endgroup\$– abcCommented May 12, 2019 at 21:14
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\$\begingroup\$ You mention active circuitry--are there examples of motor drives that actively introduce current in response to a back EMF to provide better braking than shorting the terminals can provide? \$\endgroup\$ Commented May 12, 2019 at 23:44
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3\$\begingroup\$ @AndreyAkhmetov yes. In fact, it's possible to build an amplifier whose output impedance is negative and equal in magnitude to a motor's armature resistance. Then for the purposes of motor dynamics, the system comes close to acting like a motor with zero-resistance winding. Speed regulation is much (but not perfectly) improved, including regulating down to speed = 0. I'm not sure if it's been used for motor braking, but it was used for a while in the 1970's to regulate the motor speed of cassette tape drives in sorta-high-end consumer audio equipment. \$\endgroup\$ Commented May 12, 2019 at 23:52
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1\$\begingroup\$ @AndreyAkhmetov If you want a fine level of control, you could do what Tim said, but for a quick-and-dirty method, you could just drive the motor in the opposite direction. (staying mechanically in sync, of course) This also ends up with regenerative braking, so make sure that your upstream circuitry can handle that. \$\endgroup\$– AaronDCommented May 13, 2019 at 4:34
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"applying a resistive load" to a running motor is essentially how an electric brake works. As a first approximation, the torque produced by the motor is proportional to the current, that's turning the motor is harder as the load resistance gets smaller. When you short the terminals, there's only the internal resistance of the motor which limits the current.
answered May 13, 2019 at 11:11
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As I read the accepted answer my brain came up with the following simplification, which I think is loosely accurate (?):
Motors are both dynamos and electromagnets.
Turning a motor invokes its properties as a dynamo.
Because the motor's terminals are shorted together, the generated voltage is applied to the motor coil windings, invoking the motor's properties as an electromagnet on its own axle.
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\$\begingroup\$ Every motor is also a generator. Put a mechanical drag on it, and it draws electrical power. Apply torque (nagative drag), and it delivers electrical power. The first law of thermodynamics is in control. \$\endgroup\$ Commented May 16, 2019 at 21:30
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\$\begingroup\$ Wouldn't the "electromagnetic properties" to which you refer actually assist its turning? \$\endgroup\$ Commented Sep 15, 2021 at 18:21