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(This is a continuation of Failsafe for relay contact welding.)

I'm trying to wrap my head around something like a "crowbar" circuit, only, instead of tripping on over-voltage, it should trip when a signal is sent (i.e. when a certain wire goes from low to high). I'm interested in avoiding a TRIAC or such, both due to the added complexity of driving it via an opto-isolator, but also so I'm not introducing an additional current path under normal operation. (When an EMR is off, it's off; an EMR design will see zero current unless a fault occurs. A TRIAC would have to be interrupting mains almost continuously.)

In other words, I have something like this (incomplete):

schematic

simulate this circuit – Schematic created using CircuitLab

The problem is, this is intended for a device whose normal load is rated for 20A... but might be < 1A. In order to actually trip in a reasonable time, that means I need to dump at least 40A through the relay in order to ensure the fuse blows in a "reasonable" amount of time. (IIUC, even at 40A, it could take up to a minute.)

Of course, if I just dead-short this, I'll probably reduce the blow time significantly, but the instantaneous current through the relay (and also my mains wiring) will be... "a lot".

There are some reasonably-priced, high-amperage relays out there (e.g. T9VV1K15-12S is rated for 40A, G4A-1A-E is rated for up to 200A inrush). What I'm not sure about is how I would go about limiting the current through the relay... or even if I should. (Obviously, once the fuse goes, the relay won't see any current, though it will remain energized.) Also, if I could find a plug-in relay at a reasonable price (~$5 or less), I'd be okay considering the entire relay as expendable as long as it takes the fuse with it.

...And no, I can't simply slap in something else to interrupt mains. That would be missing the entire point, which is that if this circuit ever goes live, it's because the thing that was supposed to interrupt mains has failed to do so.

Without spending a fortune, how would I go about limiting the current so as to not a) destroy the relay, or b) start a fire? (Or should I not be trying to do so? Or is it better to use something other than an EMR, keeping in mind a) I still have the problems of not frying anything besides the fuse, and b) any leakage current is going straight from AC-L to AC-N for hours on end?)

Matthew
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    You probably have to assume the relay will be destroyed (possibly weld shut) by the fault current. Fault currents vary widely depending on the situation and unless your load current is small it won't be easy to limit the current. This kind of feels like an X-Y problem. Isn't there a better way to ensure safety in your particular case? – Spehro Pefhany Jan 28 '23 at 00:54
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    Is this going to be UL approved? Somehow I think not. – Harper - Reinstate Monica Jan 28 '23 at 00:56
  • @SpehroPefhany, the scenario here is that the device that was *supposed* to break the circuit has failed to do so. You'll note I'm not using a resettable fuse or trusting the breaker here. A plain old fuse *will* fail when subjected to enough current, due to good old *reliable* physics. If you have a better idea for a last-ditch way to break a circuit, that doesn't require human intervention, I'd be happy to hear it! – Matthew Jan 28 '23 at 03:55
  • What are the consequences of the circuit not breaking that you are trying to prevent? – Spehro Pefhany Jan 28 '23 at 04:18
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    I was watching the other question and kept thinking “this is a job for pyrotechnics!”. I wonder if there’s some off the shelf parts you can use to (safely?) break the circuit. – Bryan Jan 28 '23 at 06:47
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    I understand what you are saying but terminology wise, when a fuse blows due to over-current, that does not mean the fuse failed. Failure would be if it doesn't blow, or if it blows, but arcs over and does not prevent current from flowing. The term is "operate" or "blow" is fine too. I was legitimately confused and had to read your comment twice to understand it. – user57037 Jan 28 '23 at 07:50
  • What causes relay contacts to weld? Assuming the contacts are closed, excessive overcurrent will cause the relay contacts to get red hot and weld. That would mean that the wiring would also be damaged as you’d expect the wiring and relay would be sized for the expected load and fused accordingly. An overload to damage the relay would be way in excess of this. Now if you’re switching a load with a high startup current, then the arcing can cause the contacts to get red hot and weld. Choose a relay or contactor to suit the load. – Kartman Jan 28 '23 at 12:29

1 Answers1

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And no, I can't simply slap in something else to interrupt mains.

Why not?

Unlike relays, circuit breakers are designed to withstand huge currents and prevent contact welding.

If you use standard DIN rail circuit breakers, you can get a shunt release. It looks exactly like a circuit breaker, but it's actually an actuator with a little arm on the side that will flip down the lever of the circuit breaker next to it. So when it is powered by your circuit it will trip the breaker. It's latching, so it will stay down until someone resets it.

Breakers can also be linked together so one tripping will trip the whole bunch (they sell accessory linkage bars for this). So if you can't find a shunt release, you can also use a low current breaker to trip a high power one.

You can also add a RCD (residual current detector) circuit breaker, and use the relay to create enough residual current to trip it. As a bonus, you get extra user protection from the RCD.

How to trip a RCD (pic source):

enter image description here

RCD is a common mode current detector based on a transformer. If you make some current flow between one wire on one side of the RCD and the other wire on the other side of the RCD, just like the test button does on schematic above, it will register this as a ground fault and trip. However, unlike a real ground fault, all the other RCDs in the installation on the cables leading to this one will not trip, so it's more civilized.

If you're feeling extra paranoid, you can add a loud alarm that will cause someone to cut the power.

FYI, interrupting a high current circuit safely is a problem for electric car batteries. Some simply use an explosive cartridge to actually cut the wire with a blade.

Your "Relay blowing the fuse" solution has a problem: when it trips it may kill the relay, and no-one will think about replacing it, so if the relay is kaput the protection will not work but no-one will know. It's not possible to test if it works without blowing a fuse, which is annoying. And the test may kill the relay, so even if the test worked and the fuse is replaced, that doesn't mean it still works!

bobflux
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  • Let us [continue this discussion in chat](https://chat.stackexchange.com/rooms/142473/discussion-between-matthew-and-bobflux). – Matthew Jan 30 '23 at 22:50