Short trip circuit question.

[Ian.M]

The current coil you wind to go round the reed switch goes in series with your high current load.  The reed switch's single lead (moving contact) + the NC lead from the other end go in series with the feed to your main relay coil.

Here's one possible way of wiring it:


C1 provides enough charge to pull in the main relay briefly.  You need to choose its value so its big enough to fully pull in the relay for a fraction of a second.

During this time there is NO protection against over-current, so an initial surge can flow to charge reservoir caps etc. in the load.  As C1 only charges when the ON button is released, holding it down wont override the over-current protection.

D1 and D2 prevent load current flowing through the ON button and prevent C1 re-charging from the load circuit.  D3 and R2 handle the back-EMF from the relay coil preventing arcing in the OFF switch or the reed switch when they break the coil circuit.

If you move the reed switch to directly in series with the OFF button, it will provide over-current protection while C1 is discharging, but that is likely to cause the main relay to chatter briefly, which risks it welding closed.

[davelectronic]

I've been trying to visualise it for some time now, the reed C switch is where I eventually thought it might be. To me a while to see that, but I wouldn't have known off the bat where the other in circuit protection components would have gone. Thank you for the drawing much appreciated. I will certainly have a go at this, health is a bit south atm,  but itching to give it a try. Thanks again for the help Ian.

[MagicSmoker]

While I don't disagree with the comments that this circuit isn't really as safe or effective as it could (nay, should) be, it is relatively easy to limit the maximum short circuit current and ensure reliable tripping by inserting a bit of resistance directly after the reservoir capacitor. This will cause the output voltage to collapse when a short occurs and do so at a predictable current level. Use at least a 5W wirewound resistor, even if the continuous power level is well below that, because of the high pulse power rating required. Note that most relays will tolerate quite a bit of overcurrent if they are already closed - a nominal 5A "ice cube" relay should easily withstand 10-20A for the time it takes for the contacts to open while an automotive cube relay rated for 20A or more should easily handle 50A peak.

One caveat: the peak short circuit current should be at least 10x the average current to minimize dissipation in the resistor under normal operating conditions.

[Ian.M]

Actually, the relay needs to be massively over-rated, as it may be called on to make or break the full short-circuit current.  If you are planning to trip at four or five amps, you'd better use a 40A rated automotive relay.   If you need a higher trip current you are into DC rated contactor territory, as relays don't have the arc quenching features or contact separation required.

Adding enough resistance to the circuit to limit the short-circuit current may be difficult - it degrades the PSU regulation and, supposing you decide to limit the worst case 'bolted short' current to 60A, for a 12V PSU, that's 0.2 Ohms with a peak pulse power of 720W.  Even a 7W wirewound resistor will be marginal, so you'd probably need to go up to 10W.

The electromechanical trip I described is a fairly pointless exercise in ingenuity - It might have been 'hot stuff' way back before the transistor but as a practical over-current protection circuit for today's electronics, its seriously lacking.  However, if you keep the trip current down to a few amps it does make an interesting experiment, and the concept of reed switches + current coils to monitor loads is old-fashioned, byt still useful for low-cost applications.

[MagicSmoker]

Actually, the relay needs to be massively over-rated, as it may be called on to make or break the full short-circuit current.

Not necessarily... with the resistance after the reservoir capacitor but before the momentary switch/relay coil/contacts, voltage will collapse if the momentary switch is closed into a shorted load and the relay won't energize. The momentary switch needs to be beefier than expected but, again, the (approximate) short circuit current can be set with the resistor.

If you need a higher trip current you are into DC rated contactor territory, as relays don't have the arc quenching features or contact separation required.

You only need "magnetic blowouts" (aka "arc quenching features") if the load is both high current and inductive in nature, otherwise, you can always put an RC snubber across the relay contacts.

Adding enough resistance to the circuit to limit the short-circuit current may be difficult - it degrades the PSU regulation and, supposing you decide to limit the worst case 'bolted short' current to 60A, for a 12V PSU, that's 0.2 Ohms with a peak pulse power of 720W.  Even a 7W wirewound resistor will be marginal, so you'd probably need to go up to 10W.

Yes, I already addressed this when I said that a 5W wirewound resistor letting through a peak current of 10A-20A should be sufficient to protect a 5A relay as long as the caveat that the short circuit current be at least 10x the max load current is followed. That implies a max load current of 1-2A, and so with a 12V supply a resistor in the range of 0.5-1 Ohm should work well.

Really, this isn't rocket science and nobody is (likely) going to lose an eye if things go pear shaped as long as the OP does not try to use this in between, say, the AC mains or a car starting battery, and/or to supply a >10A load. For a 1-2A load at 12V or so it should be fine, and work at least as good as a fuse.

And yes, there are numerous better approaches to limiting current to a load, but in the Beginners section I like to try to encourage the OP to attempt whatever idea they've come up with as long as it isn't too dangerous or flat-out impossible.