Overcurrent and overvoltage - Page 1
Forum section: electronics / beginners

author: AxleD
author_rank: Regular Contributor
timestamp: May 17, 2015, 10:50:25 am
content: Hi All,As i'm going through the process of building a power supply, i was wondering what the overcurrent and overvoltage protection features in an entry level power supply does? Does it protect the supply or the load? Typically how would this work?My guess is that to protect the supply from voltage spikes (transformer problem for example) a crowbar circuit could be used. It is not clear what overcurrent protection does in a supply that can control output current and is fused.Just curious.Thanks

author: tron9000
author_rank: Frequent Contributor
timestamp: May 17, 2015, 03:45:27 pm
content: I would watch Dave's video's on building a power supply on YouTube first. Should help explain a few things

author: Shock
author_rank: Super Contributor
timestamp: May 18, 2015, 03:57:57 am
content: Perhaps you should share what power supply, design and website you are looking at.They should be talking about the power supply features and output protection not the AC-DC circuit. Terminology can be found herehttp://www.bkprecision.com/support/downloads/power-supply-guide.html

author: T3sl4co1l
author_rank: Super Contributor
timestamp: May 18, 2015, 09:45:56 am
content: "Protects the supply" is probably more accurate.Typically the supply has a big fat electrolytic at the output, which obviously isn't going to be very friendly if already charged. Such supplies recommend that you start at 0V (or short the terminals together), connect the load, then set V to max, to start a load in constant current. Which is silly, because an ideal current source should have infinite open circuit voltage, and an inductive rather than capacitive characteristic. (To be fair, a CCS should never be ~open circuit, because that's how it delivers maximum power, just as a CVS delivers maximum power into a ~short circuit.)Of course, even a good CV-CC PSU can't really do a good job of approaching ideal characteristics in both, but the fact is, they're usually optimized for one, the cheap ones quite strongly so (like with the cap..).Tim

author: jeroen79
author_rank: Frequent Contributor
timestamp: May 18, 2015, 03:11:15 pm
content: Quote from: T3sl4co1l on May 18, 2015, 09:45:56 amIt can do both.The PSU's regulator should be designed to allow no more current or voltage than the supply can handle.When you are designing it for a specific load then you can take the load's Vmax/Imax into account as well.When you are designing a general purpose PSU to experiment with then can you add adjustable voltage and current controls that will go from 0 to the PSU's Vmax/Imax.You can then adjust them to the Vmax/Imax of whatever load you are powering.If you want to use it with loads or other PSUs that may 'overpower' the PSU when something goes wrong then something like a crowbar could be used to protect the PSU.

author: Dave
author_rank: Super Contributor
timestamp: May 18, 2015, 04:30:04 pm
content: As far as I am aware, the protection circuits on power supplies are mainly there to protect the DUT. The protection of the supply itself is of secondary importance.A power supply might cost 1000-2000 bucks, but the circuit it is powering might be invaluable.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: May 19, 2015, 09:59:34 am
content: Riddle:I you set an adjustable bench supply to 30V, 20mA and connect a red LED (forward direction), how many connect-disconnect cycles can you expect the LED to survive?This riddle applies from $50 POSs to kilobuck units, so don't fool yourself there, either!Tim

author: Psi
author_rank: Super Contributor
timestamp: May 19, 2015, 10:29:15 am
content: i blew up a 900lumen LED like that.So im going with "once"

author: jeroen79
author_rank: Frequent Contributor
timestamp: May 19, 2015, 10:29:29 am
content: How quick is the regulator at switching from CV to CC?

author: T3sl4co1l
author_rank: Super Contributor
timestamp: May 19, 2015, 10:33:13 am
content: Quote from: jeroen79 on May 19, 2015, 10:29:29 amGood question.Some qualification: are you asking internally, or at the terminals?Which one matters to the riddle?Tim

author: jeroen79
author_rank: Frequent Contributor
timestamp: May 19, 2015, 12:33:07 pm
content: The load will only meet the terminals.And the U in PSU stands for unit so anything inside that affects what goes out of the terminals matters, including a regulator IC and any capacitors that may be across the terminals.

author: AxleD
author_rank: Regular Contributor
timestamp: May 19, 2015, 02:55:00 pm
content: Thank you all.I think i will have a fuse and crowbar type circuit for the supply and I can use some comparators to trip a relay if the output voltage or current is very different from what the user sets.Hopefully that will work.T3sl4co1l, my guess is only once as i assume the CC kicks in way to late make a difference . I'm a noob though, so that's a total guess - have never even tried to run a led on more than a few V.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: May 20, 2015, 01:01:26 am
content: The best take-away is this:If you wish to have a PSU that's as gentle as possible, it must have a modest impedance at AC frequencies (so that it is neither a very good voltage or current source), and a controlled, dependent impedance at DC (so that, after some time, say microseconds to milliseconds, it "makes up its mind" and stabilizes to a constant setting -- voltage or current, as the case may be). With minimal overshoot when crossing between the two conditions -- a part of the "soft AC" requirement.Indeed, putting a capacitor at the output of the power supply might be exactly the wrong thing to do, for many reasons.The best reason not to is because it's simply futile! ...Or should be.Suppose you're testing with some average banana jack cables strewn about the bench. Each is 3' (~1m) long, so you've got very roughly several uH of stray inductance spread about (depends on how "strewn" they are..). You can never possibly hope to have a well-bypassed and stable voltage at the ends of those cables, no matter how much capacitance you put at the beginning!What's worse is: if there's a bypass cap at the PSU end of the cables, it presents a very low impedance, making an effective AC short. So both the +/- cables act in series as one solid inductor, and plugging that into a circuit that has just a few bypass caps (a few 0.1's, say) will ring like one hell of a bell when suddenly connected! If you set the power supply to +/-15V and connect it to an analog circuit, don't be surprised if it explodes from the +/-30V peak transient!Overshoot is lessened if the circuit has electrolytic capacitors, because their ESR tends to be dominant, and comparable to the ESR and reactance of the capacitor inside the PSU too (if applicable). But this brings a second problem: if you have anything electronic in-line with those power cables, the inrush current can easily be in the >100A range, and pop goes -- diodes, transistors, whatever. Shorting capacitors together is an extremely stressful event, and if a semiconductor has to bear that transient, it won't much appreciate it!So the whole point of putting that cap there is to make it a "nice stiff" supply. Well considering you'llneverhave a circuit plugged directly into the binding posts with zero lead length, that's a useless concern! And anyway, a properly designed power supply should be as fast as an electrolytic capacitor, so it should be redundant in the first place!For reference, I have a bench supply, which is more of a high current audio amplifier than a power supply as such. It's got +/-25V rails (with well over 10A DC available), complementary darlington outputs, and over 10kHz bandwidth. It would make a fine subwoofer amplifier, if you have 1 ohm speakers handy. I normally use it as a power supply by wiring a variable DC reference to the input. I normally leave this beast on the floor, and connect to it through a ~6ft (1.8m) twisted pair cable. I measured its short-circuit transient output atover 40A. It took some 10 microseconds to climb up to that current, largely because of the sheer inductance of the cable -- about 2uH. Since V = L * dI/dt, for an applied 20V, it should take 4us to reach 40A -- the additional time being due to resistance in the circuit, and the actual response time of the amplifier. The delay is longer with more cable, and less without.This is the kind of performance you should be shooting for -- any faster and it doesn't matter, because no one will be using little enough cable to matter. Any slower, and you will see the potential consequences of overshoot, or short circuit transients, or stuff like that. This can be more trouble in some designs than others -- a switching supply might be limited to run much slower than this, for a chain of reasons; and probably can't avoid having some amount of capacitance (often a large amount!) on its output, for filtering. In that case, you're kind of stuck. A hybrid (switcher + linear postreg) might be okay, but might also have some unsightly side-effects (like slow recovery following a short, as the switcher recovers), and the challenge is to hide those as gracefully as possible.Tim

author: Asmyldof
author_rank: Regular Contributor
timestamp: May 20, 2015, 01:15:55 am
content: Quote from: T3sl4co1l on May 19, 2015, 09:59:34 amThat will actually depend on the topology. There's mid-range units that use a switching supply with large capacitance and linear post-filtering with insane fold-back response (inherited from chip-testing rigs), that have let me do that trick with a normal few $ for 100 3mm LEDs repeatedly. Because, yes, I intentionally try those things. The old Phillips Lab supply with Thyristor Rectification and 0.68F smoothing capacitance, yes, that red LED will be red very shortly and black in a very different way long after that on the first attempt.And of course you failed to mention what type of LED. In a way a 5W COB can be said to be "a LED" but might have a Vf of 37V, in which case: infinitely many!:-P