What is the difference between overload and short-circuit protection for motors? I do not understand the need for SC protection when a motor has overload protection. Shouldn't the overload protection trip under SC conditions (and really fast due to the inverse time characteristics)?
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asked Oct 7, 2014 at 12:50
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2 Answers
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Overload and short-circuit protection protect against two different parameters being exceeded: heat (power integrated over a relatively short time) and current.
A high current can be a an almost instantaneous problem: a wire that can't shed its heat can get too hot and burn through, or set fire to surrounding materials. In a real short-circuit situation the voltage is ~ zero, so the power delivered by the supply is ~ zero.
Overload is a slightly longer-term situation, where the heat in the motor is the problem. A motor has much more heat capacity than a wire, so a peak is not an immediate problem, but over a slightly longer period the heat must still be limited.
answered Oct 7, 2014 at 13:12
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\$\begingroup\$ I'm not convinced that the power delivered to a chafed wire is ~ zero because the voltage (V) is ~ zero. You're correct so far, but the current (I) is ~ infinite. Power (P) = (V) x (I). In practicality, (V) is not zero and (I) is not infinite, both because of the resistance of the wire, which makes the math still work. Ever put a short across 480V? youtube.com/watch?v=Hp1JdVwbN_U \$\endgroup\$– AaronDCommented Oct 7, 2014 at 16:18
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\$\begingroup\$ I don't think the current will be infinite, the power suplly (if it is a suriver) will see to that! My point is that the aim of an overcurrent protection is mostly to prevent local short-term problems that have more to do with local heat than with the overall deliverd power. \$\endgroup\$ Commented Oct 7, 2014 at 16:23
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\$\begingroup\$ @WoutervanOoijen: If a power supply is sized to power dozens of motors simultaneously, short-circuit current might not be infinite, but could easily exceed stall current by an order of magnitude. If under stall conditions a motor driver would normally drop 5% of the supply voltage, and under short-circuit conditions current increases tenfold while the supply voltage hardly budges, heat dissipation in the driver could increase by a factor of 200. Not 200%--20,000%. \$\endgroup\$– supercatCommented Oct 7, 2014 at 23:38
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I'm not sure in what contexts a motor itself would require short-circuit protection, but motor driver circuitry certainly can require short-circuit protection separate from overload protection. Many parts of a motor and its associated control system will heat up and cool down relatively slowly, and may be expected to endure for brief periods of time levels of current which would cause them to heat if applied for too long. It's not uncommon for a motor driver to be sized in such a way that it would overheat if it tried for too long to drive a stalled motor with maximum current. Current will be limited by the motor resistance, however, and most of the power that gets taken from the supply will be dissipated in the motor.
In case of a short circuit, however, not only will the driver generally pass more current than would have been able to pass through the motor resistance (implying that more power gets taken from the supply), but all of the power from the supply will be dissipated as heat in the motor driver. This may heat dissipation in the supply to be 100 times greater than it would be with a stalled motor. If the driver doesn't include short-circuit protection, parts of the driver may melt almost instantly--before the over-temperature circuitry has time to react. Once the driver starts melting, overtemperature-protection circuitry will become irrelevant. I've seen a driver chip for a motor with a 3-amp stall current glow like a flare when the output was shorted (I think it was hooked up to a 50A 24V supply), most likely for exactly that reason.
