Power resistor overload rating vs time - Page 1
Forum section: electronics / beginners

author: fmzambon
author_rank: Contributor
timestamp: February 06, 2022, 02:17:21 pm
content: Hello everybody,just a quick question: if a wirewound power resistor (such as the WHS3 typehere) is rated for a 5s overload of 15W, does it mean that it can also take 7.5W for 10s (or, more generally, the same energy over a longer period of time)?I understand that short, high power pulses can be problematic since there is not enough time to dissipate the generated heat. But, on the flip side, stretching the same energy over a longer period of time should actually give the resistor more time to shed the heat. Am I correct?Cheers,Andrea

author: Siwastaja
author_rank: Super Contributor
timestamp: February 06, 2022, 02:45:17 pm
content: Somewhat more-or-less. There is no generic answer, thermal modelling is difficult. Many datasheets include complete graphs where you can find the answer directly, based on the manufacturer's modeling or testing. If this is not available, the best you can do is to find data for a similar resistor, in same size of package and ratings.

author: Picuino
author_rank: Super Contributor
timestamp: February 06, 2022, 04:35:54 pm
content: Thermal protections to protect motor windings usually follow a curve I·I·t = constant. This means that if you double the current, it should reduce the on time by 4 times.It comes to what you say, doubling the power (P=I·I·R) should cut the on time in half.In short times (such as a few seconds) there is no time for the heat to dissipate, so the heat is stored in the winding, which increases its temperature linearly depending on its thermal capacity. The more energy it stores (E=P·t) the higher the temperature increases and that has a limit before the coil breaks down (abobe 250ºC).Edit:The problem comes when you want it to support an overpower after running at full power or near full power. That cannot be done because it would raise the temperature above the maximum. The average RMS current must be less than the maximum current.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: February 06, 2022, 09:35:59 pm
content: Power ratings tend to scale as 1/t^2 to 1/t^3, depending on geometry.Though at these time scales, it may be enough time that it's gone proportional (~1/t), being limited by total thermal mass, and then by convection cooling over still longer time scales (10s, 100s s?).It's hard to say in general; pulse ratings, when given, are usually relevant over much shorter time scales (related to element-substrate heat transfer, or even within the element itself i.e. skin effect at ~µs scales). Ratings can vary all over the place, even within a given product line; I recall some TE metal-body resistors, I think it was, that give all sorts of power/energy ratings for various resistances, some e.g. 50W parts even outperforming other 100 or 200W parts. (Similar to the jagged lines on that WHS family in fact, but somehow greatly exaggerated.)On the upside, the overload factor you're asking about, is quite modest, and I would guess that your assumption is reasonable. You might do a qualifying test, measuring body temp as indicated (in the middle of the body, and mind to use thermal paste with the thermocouple to get good thermal contact to the device), and confirm that it's within ratings. These things can handle quite a lot of heat (~300°C) and are welded construction, so should be fairly robust to thermal cycling.For proper design assurance, I'd want to either:- Use the most conservative assumption: maximum power rating vs. time is no more than as given. So, three points are given for this family: continuous (rated for t > 5s), momentary overload (5 x rated, square pulse, t <= 5s), and surge (waveform specified, t ~ µs). Which, they don't say whether the surge power is the same as the surge voltage waveform, I would assume so but it doesn't actually say, so that would be something to confirm with the MFG.Since t > 5s is not specified otherwise, the continuous rating would have to be chosen. So, the WHS10 would do for 7.5W, and probably the 7 is close enough too.- Ask MFG for derating at relevant pulse durations. Or any other clarifications, like the surge ambiguity above.Can also shop around to see if anyone uses a 10s overload condition. 5s is pretty traditional so you'll be looking a while, but you may get lucky and find something.- Full qualification testing. Do a representative sampling (across various lot numbers and ages), using however many thousands of temp cycles are required, at the required worst-case condition. Obviously quite an undertaking, which is why basically no one does quals these days.Tim

author: fmzambon
author_rank: Contributor
timestamp: February 07, 2022, 08:32:54 pm
content: Hello,well, I thought I'd get an unqualified "yes" as an answer, but it turns out it's not so simpleI guess more context is in order. The role of the resistor I was talking about is a current limiter in a crowbar circuit. See the attached pics of the relevant parts of the schematic. It's still work in progress, so the positive line coming out of the bridge rectifier would be connected to the "dead end" trace to the left of C28. And the resistor in question is, as you may have guesssed, R41.Its purpose is to limit the initial current through the SCR (2N2329) when the crowbar is first activated. At first I was thinking of using a value of 330R, to limit the maximum current to under 1A even when C28 is charged to a voltage way above what it normally sees (around 190 - 200V with normal mains voltages).The transformer (Hammond 229A120) is a 6VA type used to convert 230V to 120V, and it has a 30% regulation. This means that the full load output voltage is 120VRMS, while the open circuit voltage is:120 * 1.3 = 156VRMSThe transformer is a 6VA unit, so the max rated current is:6 / 120 = 50mARMSThe equivalent resistance of the transformer (imagining it all reflected on the secondary side) is then:(156 - 120) / 0.05 = 720RWith R41 = 330R, the max continuous current through the crowbar is:156 / (720 + 330) = 148mARMSAnd the power dissipation through R41 is:0.148^2 * 330 = 7.3WThe primary-side current is:0.148 * 156 / 230 = 100mARMSThe 32mA fuse (datasheet here) would be operating at 3.13 times its rated current, so it could take up to 10s to blow.Hence my original question of 7.5W for 10s.But then I realized that if I accept a higher initial current through the SCR, I can lower the power dissipation of R41. Setting R41 to, say, 100R, the initial current spike will be higher (still way within the 15A single-pulse limit of the SCR), but the subsequent power dissipation will be lower.With R41 = 100R, the max continuous current becomes:156 / (720 + 100) = 190mARMSAnd the power dissipation through R41 is:0.19^2 * 100 = 3.6WThe nice thing is that the primary-side current will also be higher:0.19 * 156 / 230 = 128mARMSWhich is good because this value reaches 4 times the rated current of the fuse, which in turn guarantees that it will blow in 3s at most, according to the datasheet.Do my calculations look correct?With R41 = 100R, the resistor would need to whitstand the initial discharge of C28 (less than 1.5J even with an unlikely initial voltage of 300V) and then a dissipation of 3.6W for at most 3s.That should be within the capability of the smallerWHS2 type: almost 2J pulse rating and 10W for up to 5s.So, by lowering the value of R41 I should be able to use a physically smaller resistor, stay within its datasheet limits and ensure that the fuse blows sooner. As a bonus, turns out I already have that specific 100R WHS2 part, so I don't have to purchase anything else. Win-win-win-winAndrea

author: T3sl4co1l
author_rank: Super Contributor
timestamp: February 07, 2022, 10:35:33 pm
content: 2N--, how quaint. Spare parts?So, under what condition would this crowbar fire? The transformer will saturate at high line conditions, not delivering much more peak output voltage; and it won't clear the fuse very quickly (they stop short of calling this an "impedance protected" transformer -- so a fuse is the correct option -- but given the shite regulation, it won't draw many times more current than rated, under fault). If it's protecting an output (say a linear regulator), fair enough; but there may be more fruitful consideration of that circuit's robustness protection. Note that crowbarring the input capacitor, can only discharge it rapidly; it can't prevent an output voltage surge event due to device failure. (Typically a pass transistor fails shorted.)Your analysis is quite reasonable.Note that it'll take on the order of RC = 33uF * 100R = 3.3ms to discharge the cap alone. Plus whatever current the transformer is delivering, which, as you show, isn't much more current, so will act to increase the duration only modestly. This isn't a sine pulse (as the SCR's rating is given for), but a decaying exponential, so isn't automatically worse (most of the energy is spent in the first ~3ms, decaying gradually over time), and matching peak currents should be a safe assumption.Indeed, although you can't avoid the capacitor energy, you can minimize the transformer/fuse let-through energy by shunting it as hard as possible. With such a small SCR, not much more is possible, but your average modern part would do some hundred amperes surge quite easily: far more than the transformer can supply, and maybe even more than the capacitor can (but, you might want some resistance or inductance to limit turn-on dI/dt).Which, well, youcanavoid capacitor energy, if the drop across its ESR dominates -- this is true if the SCR has much lower average resistance. (Average, meaning instantaneous Rav = Vf / If during operation. As opposed to incremental or dynamic resistance, r = dV/dI, the slope of the Vf(If) curve. A turned-on SCR looks something like a diode curve, so these are not the same thing!)dI/dt is simple enough: take the rating, or say half of it for safety, and set V = L dI/dt. V being maximum crowbar voltage, and L the series inductor in the circuit. Probably you won't have to worry about damping this, for two reasons: one, cap ESR usually dominates; two, ringing only causes reversal (not healthy for the electrolytic, but, well, neither is fucking shorting it out;if it dies, presumably it's a small price to pay compared to whatever's after the crowbar. And in practice, it'll probably survive like thousands of cycles, not actually a big deal.) The SCR turns off once current decays below zero, which will stop the capacitor right about in the middle of the negative voltage peak. For the damping condition, ESR >= sqrt(L/C). For 170V max supply, dI/dt = 50A/us and C = 33uF, L = 3.4uH (3.3 is common) and ESR >= 0.32Ω. A cap of that value might be around an ohm so like I said, no problem. Likewise we expect peak current around I = V / sqrt(L/C) or 530A, but this is an overestimate not accounting for resistance; likely it's closer to 50-70% of this figure. And the surge takes ~ pi sqrt(LC) ~= 33us to finish, so if you can find an SCR of >300A surge for such short time scales, you're set; or if all you can find is the 8.3 or 10ms half-sine surge figure, then possibly a little less than that would also be acceptable given the much shorter time scale (but, again, not an ideal assumption to make if 100% reliable operation is required -- reliable in this case being that it turns off for repeat cycling. If your idea of "reliable" is, it firesat leastonce -- SCR destruction might not be much of a failure condition after all.Hm, one downside of the series inductance is, the SCR turns off at the bottom; it'll take some ms before the transformer charges the cap back up. The fuse just thinks it's a random turn-on inrush -- nevermind that it was just in normal operation, or under load, right? So a gate drive circuit somewhat more complicated than a mere zener diode and resistor should be used. Something like, a little RC filter as a local supply independent of the crowbarred supply, then dump a small SCR (or PUT, or BJTs arranged in a similar way, or some kind of latch in any case) into the big SCR so that its gate current is maintained until the fuse is cleared. Not much gate current should be needed (for a sensitive gate type, anyway), so this shouldn't take too much capacitance. Like, 10mA for 3s, over a span of say 6 to 2V, is I = C dV/dt or C = 7.5mF.Terribly overkill for such a small transformer, but I haven't done the design exercise in a while, so here you go.As you can see, there might indeed be advantage to using a stiffer transformer (faster clearing the fuse), echoing the improvement you noticed with lower series resistance.Tim

author: fmzambon
author_rank: Contributor
timestamp: February 07, 2022, 11:54:33 pm
content: Hehe,actually I'm trying to use as many "tin can" parts as possible since this is part of a Nixie clock and the circuitry will be visible (under a clear polycarbonate cover). I'm striving for a vintage look, if you will. TO-18 2N2222 and 2N2907, TO-5 2N5416, TO-39 2N3439 and so on. Hence the choice of that particular SCR: more tin cans!The crowbar is on the input of a discrete, high voltage regulator that outputs a fixed 180V and its purpose is to blow the fuse if something goes wrong with the regulator. Since the regulator features foldback current limiting, I couldn't place the crowbar on the output and expect it to blow the fuse, so I had to place it on the input. A 1N4007 diode across the regulator allows the crowbar to also pull down the output rail and discharge its small cap (100n), while a 5W 190V zener on the output serves both to detect an overvoltage condition (in which case the SCR is triggered) and to clamp the output voltage until the SCR does its job. The SCR is also triggered if the error amplifier of the regulator starts dumping too much current to ground, another condition that indicates that something's wrong.I need to re-read your reply tomorrow (it's almost 1AM here and some of my neurons must be asleep already), but I seem to understand that you're suggesting not using R41 at all, instead using an inductor and the cap's ESR to limit the current through a beefier SCR, correct? If that's the case, using a resistor has the advantage that the current would not fall to zero during mains zero-crossings. That should keep the SCR on for however long the fuse takes to blow, without the need for elaborate triggering circuits.One last thing: I chose that transformer PRECISELY because of its 30% regulation. I need a high enough input voltage to the regulator in order to get 180V on the output. Ideally I would have used a 140V transformer, but that's a non-standard output voltage. 120VAC is far more common, but the problem is that 120V RMS, if rectified, only produces around 170V DC, not enough to get 180V on the output. But since I only need a few mA, I solved the problem by using a small transformer with a lousy regulation in lieu of a proper 140V one. At very low mains voltages it may still drop out of regulation, but that's fine, no custom transformers neededAndrea

author: T3sl4co1l
author_rank: Super Contributor
timestamp: February 08, 2022, 12:12:36 am
content: I sure don't remember what I pulled 'em from all those years ago, but I have a few 2N3439 on hand, quite useful indeed for stuff like exactly this...You can think of the R vs. L as limiting extremes of a range of options. Obviously peak current goes up in the latter case, so you'll be limited on how far you can go between those extremes, while using that part.Why bother at all, though? Nixies are just metal plates in gas. A current limiting resistor (on the anode, or drivers) is overkill!Tim

author: fmzambon
author_rank: Contributor
timestamp: February 08, 2022, 08:00:18 am
content: Quote from: T3sl4co1l on February 08, 2022, 12:12:36 amSimply because fuses, resistors and SCRs are cheap and plentiful. Nixies are notAndrea.

author: magic
author_rank: Super Contributor
timestamp: February 08, 2022, 09:11:00 am
content: It seems intuitively fairly obvious that if you can dump 75J of energy into the resistor over 5s then you can spread the same 75J over 10s too.Peak temperature raise will be the same (if the resistor is perfectly insulated thermally from the world) or less, because some of the "early" heat will be dissipated by the time the "late" heat arrives.The rate of heat flow through the resistor and related thermal gradients and thermal expansion mismatch stresses will be lower too.It would be risky to go the other way, 75W for 1seditBTW, I have used a wirewound resistor as heater, continuously far above its rated power. The first thing to give up was the external case.