Short Circuit Protection

Hmm, this seems to rely on the contact resistance of the relay (which can’t be relied on to have a consistent value), which also happens to be extremely low, resulting in a 2kA circuit breaker. That is a LOT of current. Try adding a 100mOhm resistor to the relay switch for a 20A breaker.

Ah... but it's not a current protection circuit, it's a short protection circuit! ...Meant to quickly cut-off power to everything in the event of a short circuit. The resistor is only there to provide another method of simulating the short, or really -just close to it. This is really a good application for a diy power supply, or for a small battery powered breadboard, (which is never going to be able to draw anywhere near 2k amps) -but again, its protecting against a short, not an overload. EC doesn't account for mah or Ah or maximum currrent draw of a power source. (This particular design is only a very rough and simple one though, that only took me a few minutes to come up with).

FLASH - BANG ....

Hmm, 2kA is still a lot though. The way I see it, in the time it takes for the relay to trip won’t the 2kA current spike already have stopped because it vaporized whatever it was passing through? Wouldn’t, like, everything in the path of the short melt and explode a vaporize stopping the short before the relay even trips? That’s about 10kW being dissipated through some short wires and small components. Not a large array of fins or long heat resistant wire with lots of surface area like in even a 1kW heater. And if it’s a breadboard, won’t the plastic melt and possibly catch fire? Also, about half or more of those 10kW is being dissipated through the relay contacts. Won’t that weld it shut before it can trip, therefore preventing it from tripping? Also, what are you even doing with a 10kA power supply? Won’t that be the first thing to explode spectacularly like a box of fireworks?

*2kA, not 10kA

Hmm, with a 1mS or so trip time with a 0uOhm short circuit (closed switch) and 10kW dissipation through the relay makes for about 10J of energy dissipated.

That’s enough to heat one gram of copper by 26 degrees celsius.

That’s enough to melt 0.026 grams of copper. I hope the relay contacts weigh more than that.

The relay might be warm after the trip if the small contacts have enough mass to not melt from the heat. I wouldn’t recommend this. Try for 100A, not 2kA. In my opinion, as soon as you let things get into the kA range in case of failure, well, let’s just not let that happen, like with a fuse that has a resistance of 100mOhms. That will keep things from getting past 100A with it’s resistance and then a few mS later the fuse will have vaporized. Or if fuses don’t have that much resistance try putting a 100mOhm resistor in series with the fuse. That would work too. Worst case scenario, the 100mOhm resistor ends up as the fuse immediately popping from the major wattage overload. Also, what if it’s a 200mOhm short? Ever thought about the short having too much resistance? In that case it would still blow the fuse or make the 100mOhm resistor pop like a firecracker. That actually gives me an idea: a battery powered reusable firecracker. Load it up with a 100mOhm resistor, press a button, and it pops.

try a 2-3omh resistor in series with the reset button to not burn the led when you reset.

So you think, in a couple milliseconds, that something, somehow, could draw 2kA through a battery or even a power source with a small voltage of around 5 maybe 10, 12v... and instantly melt this rare, extremely low resistance component, and immediately draw 2000 Amps from this magical battery, as if it would even survive a quarter of that! Even car batteries can't handle that, most are rated for 600-800 CCA (cold cranking amps). This component would definitely have to be a wire for sure, and a large one too, otherwise it probably would melt like a fuse, but fuses are designed to be fast-acting by keeping them thin, but increasing the surface area at the same time, within the same volume, and.... THEIR NOT MADE OF COPPER! Besides, I don't know about you, but I'm never building any experiments that would come even anywhere close to drawing that many amps. And if I were, it would be no accident, and I wouldn't want any protection circuit in the middle of it, and I wouldn't be using a 9v battery for it like most of my other projects. In other words, this little circuit is good enough for what I'm doing, and it doesn't need to be that high in quality anyhow. Maybe I would be lot more concerned about making it much more responsive and fast-acting if it were for something much more sensitive, such as an expensive laser diode (and when I say expensive, I mean one tiny little component smaller than your fingernail, and as expensive as a crapy car. Ok, I guess the ones that are that expensive are a little bigger than your fingernail, but still.). If you really want, you can get one the size of a soldering station from Mouser Electronics for roughly $8,000+ american dollars. But... till I become crazy enough to decide to get something like that for a crazy amount of money, I'll just stick with this simple little protection circuit.

This simple little protection circuit that happens to trip at over 1kA. Try adding a 100mOhm resistor to make it trip closer to 15A-20A instead of 1.5kA-2kA.

Haha, ok

I totally agree with jason. People here see a short by zero Ohm but forget whats already happen to its equipment at 1Ohm or 500mOhm. The definition of a short is not zero Ohm. I would define it as overload which mostlikely happens much sooner at higher resistance.

True, plus zero Ohms would mean infinite current, which is impossible. Isn't such a circuit that detects a short called a crowbar, though?

A crowbar is more what jason was directed, but its used to avoid overvoltage. Incase you detect overvoltage, you fireup a short circuit, you provocate overcurrent which will break a fuse and the overvoltage is gone. The short is mostly made by a thyristor or triac, cuz its much faster then a relay.

Oh ok, thanks