Short trip circuit question.  - Page 1
Forum section: electronics / beginners

author: davelectronic
author_rank: Frequent Contributor
timestamp: June 18, 2017, 09:24:19 pm
content: Just a question about a trip circuit, I'm looking at the circuit in the diagram. What I was looking in to is the final diode. I know there's a single diode voltage drop across it, but my question, or dilemma is to use a stud mount diode, or use a single diode in a bridge rectifier package by using two pins for a single diode. I'm using high current, so a standard diode package of the axial type is not rated high enough. Can you see any draw backs using a single diode from two pins of a bridge rectifier over a stud mount diode. Thanks for reading, any thoughts appreciated.

author: Zero999
author_rank: Super Contributor
timestamp: June 18, 2017, 10:51:29 pm
content: What's the purpose of the diode?The circuit is a latching relay cannot be relied upon for safe short circuit protection. You might find that it works but it's a bad way to implement this.If the transformer has a high enough impedance, then when the output is short circuited, the current will be limited to a safe value, the output voltage will collapse and the relay will turn off because the coil voltage will be too low to hold it closed. The current at which the voltage drops below the relay's holding voltage is unpredictable but could be estimated by looking at the transformer and relay's data sheets.If the transformer has a very low impedance, then when it's short circuited, the diodes might blow up or the relay contacts will weld shut and the circuit will not be disconnected, causing a large current to flow. Hopefully the transformer will contain a built in thermal fuse, otherwise there will be smoke and possibly a fire.The correct way to implement short circuit protection is with a fuse. If it needs to be resettable, then use a circuit breaker or a PTC fuse which will auto-reset when the power is disconnected.

author: davelectronic
author_rank: Frequent Contributor
timestamp: June 19, 2017, 12:29:28 am
content: Yes I think it should work, I thought the diode protects the relay coil from the short circuit. It's something I found, thought it might be of some use in short circuit scenario problems.

author: davelectronic
author_rank: Frequent Contributor
timestamp: June 19, 2017, 12:37:21 am
content: I would bread board it first to see how it behaves, I was looking to see what diode option would be better. Guess I will try stud mount, and isolate a single diode from a bridge, see which works best. There's always a better way, you don't find out unless you try. In my opinion I think it looks viable. I wouldn't use it if experimental use showed it as dangerous.

author: Zero999
author_rank: Super Contributor
timestamp: June 19, 2017, 08:14:28 am
content: How can the diode protect the relay? The diode will probably be the first thing to fail.Please do not use this circuit. Use a proper form of over-current protection to protect you against fire. People can get killed, as a result of poor design. Even if it works on a bread board, do not assume that it will be reliable or trip at a safe current. The current draw probably has to exceed the transformer's rating by several times before the voltage across the relay coil will drop low enough to turn it off.Do you have any idea of what current rating you need? If in doubt, choose a fuse/breaker with a tripping current around 50% higher than the rating of the transformer. If you have large capacitors in the circuit, then pick a slow blow fuse/ time delay breaker.Here's an example of a 1A fuse, resettable fuse (PTC resistor) and a breaker. Even if you have to order parts or it works out more expensive, then it's worth it.https://www.rapidonline.com/siba-70-007-65-1a-1a-20x5mm-antisurge-ceramic-fuse-26-0386http://uk.rs-online.com/web/p/resettable-wire-ended-fuses/6571772/http://uk.farnell.com/potter-brumfield-te-connectivity/w58-xb1a4a-1/circuit-breaker-1a-push-reset/dp/1436201

author: newbrain
author_rank: Super Contributor
timestamp: June 19, 2017, 08:58:18 am
content: I see what you did here...Quite spot-on.

author: Ian.M
author_rank: Super Contributor
timestamp: June 19, 2017, 09:19:23 am
content: The only way it could work as intended would be if you added a load current sensor. e.g a Form C Reed Switch (http://www.mouser.co.uk/ProductDetail/Hamlin-Littelfuse/MDRR-DT-15-20-F/) with its NC contacts in series with the relay coil, and a coil carrying the load current wound round the reed switch. The number of turns of the reed coil sets the trip current as that reed switch needs a nominal 20 Ampere.Turns to operate.

author: davelectronic
author_rank: Frequent Contributor
timestamp: June 19, 2017, 09:50:26 am
content: What ever i use as output protection would be safe, I've used zenner and fuse in some projects, and a simple single thyristor zenner and fuse in other projects, i just thought it interesting to try the circuit out. I wouldn't rely on that as the only output protection. I'm certainly not that wreckless.

author: davelectronic
author_rank: Frequent Contributor
timestamp: June 19, 2017, 09:58:28 am
content: Here is a link where i found this circuit.http://www.homemade-circuits.com/2012/04/add-this-short-circuit-protection-to.html?m=1

author: Shock
author_rank: Super Contributor
timestamp: June 19, 2017, 10:45:04 am
content: If that is aflyback diodeisn't it backwards?

author: davelectronic
author_rank: Frequent Contributor
timestamp: June 19, 2017, 11:12:23 am
content: I don't think its an implementation of a flyback diode, another reference to the same circuit is here.https://modelrail.otenko.com/n-scale-models/simple-short-circuit-protection

author: MagicSmoker
author_rank: Super Contributor
timestamp: June 19, 2017, 11:32:16 am
content: This circuit is a bit of a hack; the intended mode of operation is that a short will - by definition - drag the output voltage of the supply down to near 0V which then opens up the push-to-latch relay circuit. Response time will be in the range of 10-100ms which is about the same as a typical breaker or fuse.

author: davelectronic
author_rank: Frequent Contributor
timestamp: June 19, 2017, 11:43:48 am
content: Quote from: MagicSmoker on June 19, 2017, 11:32:16 amI think your right, whilst it may not be the best devised means of short circuit protection, it looks like it works from what i can see. My original question was on the diode used, but the second reference to this circuit has kind of answered that. I will put it together just to try it. I definitely wouldn't use it on high voltage applications. I always incorporate a fuse in the final output of projects anyway. But I'm interested to see how it responds.

author: MagicSmoker
author_rank: Super Contributor
timestamp: June 19, 2017, 11:54:28 am
content: The diode in series with the output? That can be deleted unless the load is a battery or a PMDC motor operating in generator mode, as either would cause the relay to automatically re-close upon removal of the short (ie - rendering the whole circuit moot).

author: Zero999
author_rank: Super Contributor
timestamp: June 19, 2017, 11:58:38 am
content: Quote from: MagicSmoker on June 19, 2017, 11:32:16 amBut unlike a circuit breaker the tripping current will be unknown, quite possibly above the breaking capacity of the relay and will only work with a dead short.Quote from: davelectronic on June 19, 2017, 11:43:48 amYou don't understand. Voltage doesn't even come into the equation. It's power which starts fires, not voltage. If the current is high enough, less than 1V can cause a fire big enough to kill hundreds of people. Fire is more dangerous than high voltage.QuoteIf you have a fuse then why bother with this dangerous circuit? The fuse will also probably blow before the relay opens so it's pointless.

author: X
author_rank: Regular Contributor
timestamp: June 19, 2017, 12:18:22 pm
content: When the load at the output draws enough current, the coil will activate and the relay contacts will short-circuit your transformer via the bridge rectifier and diode. If the current is high enough, either the diode will fail, or the relay contacts will heat up and possibly weld themselves together right at the point of contact.This is not a good circuit to use. You are better off using a commercially available circuit breaker, but even these take time depending on how far above the rating the current is. Most circuit breaker datasheets specify how long the current needs to be held before it trips, based on % of rated value.If you are after inrush protection, you can get high-current NTC thermistors rated specifically for this purpose.

author: davelectronic
author_rank: Frequent Contributor
timestamp: June 19, 2017, 05:48:20 pm
content: I'm going to try it to see its action, even under a heavy load, it will be monitored at all times. I wouldn't leave an unknown circuit unattended ever. The relay I've ordered is plenty big enough to handle the current involved. As is the choice of diode now, stud type. If it proves a flop I won't use it. I'm aware it's power that creates heat and overload situations which can become dangerous. If it's an unviable circuit I wont use it. As I said I don't think it's a perfect solution, but I'm interested to see the action of it. And if I did use it, I'd test it repetitively first. Sometiemes simple circuits can be affective. I fully understand overload parameters that might accure.

author: davelectronic
author_rank: Frequent Contributor
timestamp: June 19, 2017, 05:56:10 pm
content: Quote from: X on June 19, 2017, 12:18:22 pmIt won't trip under heavy current load conditions, the relay coil voltage has to collapse for it trip, I'm fairly sure the rectifier and transformer are going to be fine. Like I said if it's not 100% safe I won't be using it. And I would never use it at mains voltage potential.

author: Ian.M
author_rank: Super Contributor
timestamp: June 19, 2017, 09:08:29 pm
content: If you intend to use it for short-circuit protection, *DON'T*!. Its a crappy circuit that relies on the transformer having a high enough impedance to survive a brief short-circuit without damage and without tripping the primary supply fuse or breaker. Most transformers, especially higher VA ones have a low enough impedance to trip the primary side if shorted. Also, the drop-out voltage of the relay will be a fraction of the pull-in voltage, and the supply voltage must be greater than the pull-in voltage, so the trip point will be less than half the supply voltage, maybe as low as a third, so it will allow the transformer to be grossly overloaded to the point of burning up.About the only reasonable use for it would be, without the transformer and bridge, on the output of a SMPSU that has a short circuit rating >5 seconds, to protect it against continuous short circuits.It can be used as a loss of supply protection for robotics and machine tools, where it is absolutely safety critical that they should not restart automatically if the supply is lost then restored.Add the form C reed switch current sensor I suggested and it becomes a practical over-current trip circuit. However its better rearranged with a separate pair of diodes (and reservoir capacitor) for the relay coil + supply, so the output diode can be eliminated to reduce the losses.Another way to fix it would be to add a polyfuse immediately after the reservoir cap. When the polyfuse trips, the relay would disconnect the load, allowing the polyfuse to cool off and reset. The circuit could then be reset manually with the button.

author: davelectronic
author_rank: Frequent Contributor
timestamp: June 19, 2017, 09:35:08 pm
content: Quote from: Ian.M on June 19, 2017, 09:08:29 pmIf it behaves anything like you say on a transformer of high VA I certainly wouldn't implement it for that. I do have a number of server smps to try it with. But sure if it's action on a linear transformer was iffy, I wouldn't use it. Thanks for the information.

author: davelectronic
author_rank: Frequent Contributor
timestamp: June 21, 2017, 09:05:24 pm
content: Okay, it's all become clear, bit of practical trying this. It works on power supplys that impedance is effected by a short circuit, think that's how you describe it. But I can clearly see its no good for high current applications. In fact whilst it triped out a 20 watt load on a 2 Amp psu, and also triped out a 50 watt load on a 5 Amp psu (both linear) (lamp loads) it would not trip on a higher current psu. Those it wouldn't trip out on where a server psu (converted) and a big 25 Amp linear psu. I can clearly see its usless in a high current power supply. So just asking, as I have a couple of high current relays... Does anyone know of a circuit that uses a relay in its circuit, where it would be better suited to high current use ? Or is there a way to get the relay coil voltage to collapse in a high current set up ? I doubt it, but thought it worth asking anyway.

author: Ian.M
author_rank: Super Contributor
timestamp: June 21, 2017, 09:49:33 pm
content: That's why you need the Form C reed switch similar to the one I linked earlier.Form A reed switches have a pair of magnetic contacts hermetically sealed in a glass tube with a small gap between them. When you apply a magnetic field, the contacts are attracted to each other and when the field is strong enough to overcome their springyness they touch, making the circuit. Form C reed switches are a little different. They have an additional non-magnetic contact at one end, and the end with the two contacts has stubbier stiffer ones. The moving contact from the other end is longer, and sits in-between the fixed contacts, and is held against the non-magnetic contact by its own springyness.You can operate a reed switch with a permanent magnet or with a coil, typically wound round it. Use lots of turns of fine wire to make a reed relay controlled by its coil voltage or a few turns of heavy wire to make a current controlled relay.A form A (SPST-NO) reed switch is the commonest type, but as you need to interrupt power to the main relay coil when the load current is excessive, you need the form C (SPDT) type so you can use its moving contact and NC contact to open the main relay coil circuit when the load current exceeds the desired threshold.Wind the heavy wire for the current coil round a dowel, round rod or cross-point screwdriver shaft a little thicker than the reed relay body, as the reed switch is glass and rather fragile and its far too easy to snap or crush it if you handle it roughly. When bending reed switch leads, to avoid chipping the glass seal or breaking the glass, *ALWAYS* hold the lead between the desired bend position and the glass body with pliers, not touching the glass, and bend the free end of the lead against the pliers without touching the other leads or the glass body.Reed switches are not precision devices so you'll need to experiment with the number of turns to get the desired trip current. Also the position of the coil around the reed switch has a considerable effect so you can use that to fine trim the trip current. When you have the reed switch and current coil correctly adjusted, lock them together with a little hotglue - you can always use a hot air gun to soften the hotglue if you need to remove or adjust the coil.To protect the reed switch contacts against arcing you should use a RC snubber across the main relay coil 100R and 0.1uF should do nicely unless the main relay is a contactor or other heavy duty brute with massive coil inductance. In that case, use an anti-parallel diode that can carry the main relay coil current + a resistor directly in series with the diode, equal to the DC coil resistance of the relay coil. Don't leave out the resistor - as a relay with an anti-parallel diode directly across its coil is slower to release - not what you want from an over-current trip.The combo of a read switch current sensor and a contactor or heavy duty relay to break the circuit is obviously not going to be as reliable or accurate as a proper DC rated circuit breaker, nor does it offer the anti-surge delay characteristic typical of a thermal or magneto-termal circuit breaker. Another DIYable option, would be an electronic 'fuse' circuit - an analog current sensor feeding a comparator to determine if current is over the desired threshold, triggering a flip-flop to remove the gate drive from a high-side P-MOSFET power switch. Some automotive high side power switch chips have a current sense output so all you need to add is the comparator circuit and the flip-flop.

author: davelectronic
author_rank: Frequent Contributor
timestamp: June 22, 2017, 12:01:55 am
content: Thanks for taking time to go through some ideas Ian. I will order a couple of these reed switches and have a try, experiment with it a bit. The electronic breaker sounds interesting also, but I'm a modest novice, so might take me a while to get my head round that one. I just find it interesting to try and use a relay in some kind of resettable breaker set up. I'll give it a go. Cheers Ian.

author: Ian.M
author_rank: Super Contributor
timestamp: June 22, 2017, 12:27:57 am
content: If you start off with a current sense coil of 30 close spaced turns of magnet wire, the trip point with the reed switch I linked to should be well under 1A, and within the capabilities of a cheap bench supply.A coil former made by supergluing a strip of paper to itself round a well waxed dowel or rod makes it much easier to transfer such a high turn count coil from the rod to the reed switch as you can fasten the magnet wire ends to the former so the coil doesn't unwind.Use a DMM on continuity to indicate whether the reed contacts are open or closed so you dont have to assemble it in the final circuit yet. Once you've got the reed switch figured out, applying it should be fairly simple.

author: davelectronic
author_rank: Frequent Contributor
timestamp: June 22, 2017, 09:59:58 am
content: I've found a couple of these spdt reed switches on ebay, 1 and 2Amp ratings, there are lower rating one's as well. I know the coil of magnet wire closes the switch in the event of a short. What I'm having trouble with is visualising its position in the circit with the relay. Is it in parallel with the + Volts side of the relay coil ? I'm not thinking its directly in series with it. Thanks for your time Ian.

author: Ian.M
author_rank: Super Contributor
timestamp: June 22, 2017, 12:41:07 pm
content: 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.

author: davelectronic
author_rank: Frequent Contributor
timestamp: June 22, 2017, 06:19:32 pm
content: 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.

author: MagicSmoker
author_rank: Super Contributor
timestamp: June 23, 2017, 10:41:01 am
content: 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.

author: Ian.M
author_rank: Super Contributor
timestamp: June 23, 2017, 12:49:23 pm
content: 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.

author: MagicSmoker
author_rank: Super Contributor
timestamp: June 23, 2017, 04:34:22 pm
content: Quote from: Ian.M on June 23, 2017, 12:49:23 pmNot 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.Quote from: Ian.M on June 23, 2017, 12:49:23 pmYou 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.Quote from: Ian.M on June 23, 2017, 12:49:23 pmYes, 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.