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I'm trying to design a circuit to turn on and off an electrical 230V 1A 1-phase motor using a relay. I'd like to design it such that if the user inadvertedly short circuits the output, the relay is immediately opened.

My best idea so far is to use two relays. The first relay is in series with a 30 ohm resistor and with the load. The second relay is in parallell with the resistor.

If there is a short circuit when the first relay is turned on, this is sensed by a current sensor, and the relay is immediately turned off before the resistor overheats (a few ms). The current here will be <10A because of the resistor.

If there is no short circuit, the second relay, which is connected in parallell to the resistor, is switched on. This lets the full current through, and saves the resistor from overheating.

The whole thing is controlled by a microcontroller.

I think that if I only use one relay, and there is a short circuit, the inrush current may be in the 1000 of A, which I think a regular small relay cannot break.

Does this sound reasonable?

If not, how would you build a short-circuit disconnect function for a relay-based 230V motor controller?

It doesn't have to handle short circuits which occur during operation, only before startup.

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  • \$\begingroup\$ How do you short circuit a motor? \$\endgroup\$ – Leon Heller Jun 20 '15 at 16:36
  • \$\begingroup\$ I'm thinking screwdriver across the terminals. Or simply if the phase is miswired to ground or simular. \$\endgroup\$ – avl_sweden Jun 20 '15 at 17:05
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    \$\begingroup\$ 230V mains easily provide 1000A in the few ms before the fuse (s) blow. \$\endgroup\$ – avl_sweden Jun 20 '15 at 18:54
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    \$\begingroup\$ That's exactly what circuit breakers do. If the mains circuit breaker is insufficient then put one in the device. \$\endgroup\$ – Ignacio Vazquez-Abrams Jun 20 '15 at 18:54
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    \$\begingroup\$ Think about it: How many amps do you think the power grid in a large apartment complex can deliver? In the ms before the fuse blows? \$\endgroup\$ – avl_sweden Jun 20 '15 at 18:56
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While the comments have gotten a bit acrimonious, you should be aware that your idea has a number of problems.

I've produced a pair of your circuits, one with a shorted output, and one with a nominal 10 amp load.

schematic

simulate this circuit – Schematic created using CircuitLab

First, consider the shorted condition. Power is applied, RLY2 is activated. This removes the coil drive, the relay opens, the short is reconnected, and the relay chatters until the contacts are burned out.

Now consider the situation of a "normal" 10 amp load. For this to work, RLY1 must operate with a coil voltage of (23/(23 + 30)) * 230, or 99 volts. This might just work (more or less) with a standard 115 volt relay. However, consider that in the case of no load, the coil voltage on RLY1 will be essentially 230 VAC, and this will burn out the coil before very long - it is, after all, only rated for 115 volts.

Finally, you should be aware that relays have what is called pull-in and drop-out ratings. That is, it takes a certain percentage of the nominal voltage (the pull-in value) to activate the relay, but once it is activated it will stay activated until the voltage has dropped below a much lower percentage of nominal (the drop-out value). I've seen pull-in/drop-out values of 85%/15% for real relays. This means that, once the circuit has activated, for the values shown the contacts of both relays can experience currents of ~60 amps without RLY2 opening up.

So the quick answer is, no, your idea won't work. It will oscillate and chatter in the presence of a short, and one or more relays will burn out during normal operation.

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  • \$\begingroup\$ Thank you for a constructive answer! I realize I forgot to mention that the relays are controlled by a microcontroller. It will of course not reapply power without operator intervention once a short circuit is detected. \$\endgroup\$ – avl_sweden Jun 21 '15 at 5:57
  • \$\begingroup\$ Well, no. RLY2 is controlled by voltage across the 30 ohms. That's the point of the circuit. \$\endgroup\$ – WhatRoughBeast Jun 21 '15 at 8:17
  • \$\begingroup\$ WhatRoughBeast: What I mean is the circuit you have drawn isn't like the circuit I was proposing in the question. Although, from the way the question was originally ambiguously worded, your interpretation was the most reasonable. \$\endgroup\$ – avl_sweden Jun 21 '15 at 16:53
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how would you build a short-circuit disconnect function for a relay-based 230V motor controller?

Hmmmm...... Short-circuit overcurrent protection for large AC loads....... I wonder if an existing solution is commercially available in the $5-15 range?

I think that if I only use one relay, and there is a short circuit, the inrush current may be in the 1000 of A, which I think a regular small relay cannot break.

Won't happen. The circuit breaker already installed in the building electrical system will trip long before 1000 amps goes down the wire.

The motor and building wiring are already covered. If you want to protect your controller from accidental shorts, put a fuse in it.

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    \$\begingroup\$ The part about 1000A not flowing is in fact wrong. For a short moment, a very high current will actually flow, before the circuit breaker can react. This is why fuses for mains equipment must be rated to interrupt >1000A, even though there are certainly fuses with a nominal value smaller than that in the current path. \$\endgroup\$ – avl_sweden Jun 21 '15 at 6:03
  • \$\begingroup\$ (At least such an interrupt rating is required if you want the fuse to blow before the other fuses, in a short circuit condition) \$\endgroup\$ – avl_sweden Jun 21 '15 at 16:55

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