High voltage, low Current "fuse" - Page 1
Forum section: electronics / projects

author: Amper
author_rank: Frequent Contributor
timestamp: October 07, 2021, 05:42:22 am
content: Hi,for components testing im looking for a way to apply up to 1kV at low currents (<100uA) to a set of 10-100 devices. My Problem is, its for testing life expectancy of the components so some will always fail.To keep fires from happening and to make sure the test continues for the others even when a component has failed i want something to interrupt the power to parts exceeding a threshhold, essentially i need a 100uA 1kV fuse.Is there some sort of industry standard for such tests?Is there soem kind of simple component taht could be used (even if single use)?Or do i have to design some simple measuring circuit that controls the load via some igbt or reed switch?I guess that would turn out to be some shunt-reference-schmitt-driver-switch thing.

author: Ian.M
author_rank: Super Contributor
timestamp: October 07, 2021, 06:09:28 am
content: Assuming the supply has sufficient reserve current capacity, you could use a few hundred Kiloohms flameproof resistor soldered in series with a small diameter tension spring using low temperature solder, then stretched and soldered in circuit. I'd also recommend fibreglass sleeving over the assembly for containment. If the load shorts, the resistor will rapidly heat up, and the solder will melt, releasing the spring and breaking the circuit. Odds are the resistor its self will also fuse, but wont be able to stand-off 1KV, so will arc over and continue to heat up till the spring rapidly opens the gap.

author: PartialDischarge
author_rank: Super Contributor
timestamp: October 07, 2021, 06:43:02 am
content: That fuse does not exists. Not even in low voltage because it's very difficult to make low current fuses with accuracy. The variability of metal thickness during manufacturing just would represent +-300% variation for currents <10mA. Have you seen the price of a 2mA fuse? Adding the 1kV requirement makes things more difficult. In practice a low mA value for a fuse will provide enough protectionQuote from: Amper on October 07, 2021, 05:42:22 amDo not confuse a current limiter with a fuse. A fuse is a safety device, not very accurate but that will stop large currents and arcs. Semiconductors are not for that.That said in my company we are developing kV fuses for low currents (mA range) for internal use but we will eventually probably sell them.

author: magic
author_rank: Super Contributor
timestamp: October 07, 2021, 07:22:23 am
content: I have built a current limited 100V generator using a transformer, capacitive voltage multiplier (~120V output), series resistor and 100V zener. Short circuit current is higher than normal output limit, but still manageable. This could be scaled to higher voltages with beefier resistors.

author: PartialDischarge
author_rank: Super Contributor
timestamp: October 07, 2021, 07:30:17 am
content: Quote from: magic on October 07, 2021, 07:22:23 amUnfortunately the power of a resistor grows with the square of voltage. A kind of foldback limiting with a mosfet would be much easier for 1kV

author: Amper
author_rank: Frequent Contributor
timestamp: October 07, 2021, 07:45:10 am
content: @ Ian.M: Yes, im sure it would work, but it would be a bit bulky, for a set of up to 100 DUT it would require quite a bit of mechanical work. A solution on pcb would be much nicer and easier to explain to my boss.In reality i guess one of those 1/4" fuses with a spring inside is what i would use if i didnt have to stay within official ratings, they would pretty certainly work but i also dont want to be responsible for a fire alarms going off at night.@magic: That would definitely work but i need the 1000V to be applied until the compliance current is reached, so every single supply would have to be regulated. Thats way too complicated for the ammount of channels i need.@MasterTech: I know about the issues with fuses, thats why i used "" in the titleWhy do you think semiconductors are no use? I dont want to limit the current, i need a cutoff when a part fails to interrupt the supply to a single channel.

author: PartialDischarge
author_rank: Super Contributor
timestamp: October 07, 2021, 08:03:02 am
content: Quote from: Amper on October 07, 2021, 07:45:10 amI'm just saying that semic are not fuses and not safety devices. For your application I would personally limit the impedance of the line with a resistor and build a current limit using a cheap 1500V mosfet for each device, assuming this is DC you didn't mention

author: MegaVolt
author_rank: Frequent Contributor
timestamp: October 07, 2021, 08:25:40 am
content: Perhaps you can control the consumption in the primary circuit? Increasing the secondary current will cause a noticeable increase in the primary current. You only need to adjust the threshold.

author: Amper
author_rank: Frequent Contributor
timestamp: October 07, 2021, 09:21:38 am
content: Its not so much a safetydevice, its not actually there to protect people or property but it has to be reliable.To prevent an arc or to limit its effect i can still use a normal fuse rated at higher current or the powersupplies safetyfeatures, in case my channel cutoff fails to open the current will rise to the limit set by the supply and trip the fault.@Megavolt: The issue is that lets say i expect the device to draw up to .1mA and i want to have a setup for up to 100 of them i will have a sum of 10mA. If most parts draw roughly 20uA but one draws .5mA because it failed the sum will still be below the 10mA threshhold. Also it would still not be possible to have the test run for the other parts after one has failed.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: October 07, 2021, 09:25:34 am
content: Guys, nothing wrong with dissipating a fraction of a watt, it's not even 1mA. A plain simple constant current source will do.Seconding the resistor, it's preferable for simplicity, and is basically what a hi-pot tester is all about. One to each DUT and you can check which one(s) fail, independently. Maybe put a neon bulb + resistor across the limiter resistor, so you can see when it's dropping more than 80V or so (corresponding to whatever current).1kV isn't even very much in the grand scheme of things, but it depends. For example, anything touching mains should be tested over 2.5kV, and should also be AC. If it's testing of semiconductors, 1kV is fairly high (though board-level parts are available up to 4kV these days). Beware of static discharge: the average currents may be small, but the peak current delivered from a capacitance, even a current source's drain capacitance, can be extreme, however brief. The only meaningful, wideband, current limiter is a resistor -- having maybe a few pF, the impulse is small, and at such low currents, the voltage drop can still be low. (So, I would recommend for example, using a transistor current source AND a resistor, if you choose to go that way, that is.)Tim

author: Amper
author_rank: Frequent Contributor
timestamp: October 07, 2021, 10:01:57 am
content: But a resistor has voltage dropAssuming my 100uA "good" current and a short current of 1mA (100mA 1kV worst case total) implies i need 1M 1W (or better 2W) resistors. -> .1uA will be 100V drop wich is 10% and unacceptable for the test completely ignoring the effort to have 100 powerresistors housed, cooled and wired. The next issue is, the test has to be interrupted when the device fails or the results will loose signifficance, its a HTRB test, for example a diode or a fet is subjected to high temperature and high voltage to accelerate its degradation. Breakdown voltage will be measured before and after the test and the conditions have to be repeatable. If the device fails partially to a breakdown voltage of 500V and has lets say a Ptot of 100mW i will have 500V over the resistor and .5mA running -> 500mW and i will completely frie the component by the losses so the results will not be useable. (const current source will do the same, even more losses in the device)All in all i need a way of turning off 1kVDC, there is no way around that im afraid.

author: tszaboo
author_rank: Super Contributor
timestamp: October 07, 2021, 10:13:35 am
content: Quote from: MasterTech on October 07, 2021, 06:43:02 amSemiconductor based safety devices are acceptable for explosion safety (if properly designed and rated, and redundant) so these solutions are acceptable for some generic safety device.Quote from: T3sl4co1l on October 07, 2021, 09:25:34 amI agree. Probably I would add a series resistor to limit transient energy, and add a current limiter in series with it to limit power. You want a generic 2 pole circuit design. I would use an LM334 or s TL431 based circuit, with some protection for the ICs themselves.

author: Berni
author_rank: Super Contributor
timestamp: October 07, 2021, 10:56:39 am
content: I have a Fluke 10kV precision power supply that has an odd way of over current protection.The supply current is ran trough the coil terminals of special potted relay. When the output draws too much current it also causes current trough the relay coil. At some point the relay switches over and this sends a signal to shut down the power supply. A bit odd but it solved the problem of high side current sensing. That being said this is not a super fast acting device and the PSU has quite a bit of capacitance. So when the output of this supply arcs over it makes a pretty loud pop as all of that capacitance discharges into the arc. This is very damaging to any sensitive electronics and i would imagine hurts like a bitch if you happen to touch it( luckily i have not found out yet). As a result i keep a big high voltage resistor in series on the output when using the higher voltages, this limits the current to something much more gentle in the event of an arc over.Fuses at these high DC voltages are very difficult. For something like this the only way is some sort of pyrotechnic fuse. You need to have a super thin short piece of wire surrounded by explosive that gets set off by the wire burning out. The resulting explosion blowing the fuse contacts apart in order to get them far enough to extinguish the arc.When high voltage stressing electronic devices like this they will typically fail suddenly and abruptly where they go from being an open circuit to a sudden short. So you would typically want some series resistance on each one to limit the amount of current in such a failure. If you do it correctly it might be possible to use the resistor as a thermal fuse. A spring can be soldered to the end of a large SMD resistor and then put under tension. When the device shorts out the resistor is going to get very hot very quickly, melting the solder, letting go and making the spring snap away from the resistor to disconnect it with a large gap. The resistor prevents the single device from dragging down the power supply voltage too while the spring provides a visual trip indicator.

author: rsjsouza
author_rank: Super Contributor
timestamp: October 07, 2021, 11:11:54 am
content: Quote from: Amper on October 07, 2021, 10:01:57 amPerhaps an isolated amplifier such as the one below, which has an input excursion of 2V, allows you to make a current sensing using very low power dissipation and voltage drop. Calculate a sense setpoint at 500 µA (1 kΩ) or 1 mA (2 kΩ) and use an IGBT or a relay to cut the power.https://www.ti.com/product/AMC1411

author: Amper
author_rank: Frequent Contributor
timestamp: October 07, 2021, 11:59:07 am
content: Oh, in case i will go the route of active components the current sense can be a straight forward shunt in the low side, a resistor from ground to source and a schmitt trigger/ latch opening up when the voltage drop reaches a certain level, kind of like a reverse crowbar. It also has the advantage of the only really endangered part beeing the switch and the switch beeing easily replaced by any voltage rating available as a component.About the speed im actually not really worried so much since im not raising the voltage until there is a sudden breakthrough but the parts will age and slowly reach the threshhold. The failure usually is a thermal runaway that takes at least seconds, if not hours.

author: Zero999
author_rank: Super Contributor
timestamp: October 07, 2021, 12:07:45 pm
content: Just make a current limiting circuit with some transistors.

author: Amper
author_rank: Frequent Contributor
timestamp: October 07, 2021, 12:14:22 pm
content: Did you read the thread?

author: Alex Nikitin
author_rank: Super Contributor
timestamp: October 07, 2021, 12:30:38 pm
content: 1) You need to sense the current. Can you afford, say, 100K resistor in series with the DUT? 10V drop at 100uA? That will limit the maximum current at 10mA to be on the safe side and to use a relay I've linked below.2) There are somenice low power HV reed relays from Pickering, you can use these as 1kV control devices. Not the cheapest way but reliable and clean.3) If you can sense the current at the "cold" end, the circuit can be pretty simple, if you must sense at the "hot" end, you can use an optoisolator.CheersAlex

author: magic
author_rank: Super Contributor
timestamp: October 07, 2021, 01:58:22 pm
content: Maybe like that:1005V positive supply shared by all DUT.100kΩ plus an N-ch MOSFET in series at the bottom of each DUT.All gates at +15V initially.When any source voltage exceeds 10V, drive the corresponding gate to ground.All DUT see 1000V ±0.5% when drawing <100µA.Fault current can't exceed 100~150µA, depending on FET.Faults won't last longer than microseconds.Tolerances can be tuned, the limiting factor is how well you can control/match FET threshold voltages.

author: Zero999
author_rank: Super Contributor
timestamp: October 07, 2021, 04:42:31 pm
content: Quote from: Amper on October 07, 2021, 12:14:22 pmYes. I don't see why a simple current source wouldn't be suitable. It's just a couple of transistors. Nothing complicated.If you want it to disconnect the load, then you need positive feedback, to make a bistable.

author: Amper
author_rank: Frequent Contributor
timestamp: October 07, 2021, 05:05:31 pm
content: Quote from: Amper on October 07, 2021, 10:01:57 am@magic i agree, that will most likely be the outcome. Maybe il throw together one channel on a proto board next week.Having the shunt from source to ground is a bit annoying still, to make matters easier with the reference voltage a shunt voltage of 5V or so would be good but that may be enough to interfere with the gate voltage. On the other hand thats negative feedback so at least it wont oscillate and 15V gate drive should take care of the rest. Just has to be a UGth of > shunt voltage or ill need a negative rail.@Alex Nikitin I really like those relays, you are totally right, they would be the ideal switch... Sadly even thugh budget is not that big of a deal, 100pcs is still a pretty hefty tagIll go with igbts first and then go reed in case there is problems. We have >1600V switches on stock for production all the time, dozends of reels so if one blows up thats no issue as long as the process of swapping is not to much of a pain.

author: magic
author_rank: Super Contributor
timestamp: October 07, 2021, 05:27:51 pm
content: Quote from: Amper on October 07, 2021, 05:05:31 pmYou can use 50kΩ and drive the gates to 10V. The point is that the FETs need to still be low RDS(on) at 100µA but limit current to something reasonable when this is exceeded. Or dunno, divide the shunt in two segments and monitor voltage at the midpointNote that this is just a handwavy sketch of the idea. More "interesting" things may happen in practice: the FET is hard on initially so a brief surge current will flow if a DUT chooses to go completely short circuit all of a sudden and the source may jump a little, some current still flows through parasitic capacitances even when the FET is turning off, FETs don't like too much dV/dt on their drain, etc. etc. Some 10kΩ in series with the drain may be a good idea too.

author: jbb
author_rank: Super Contributor
timestamp: October 07, 2021, 09:33:33 pm
content: It’s ugly, but might some PTC thermistor do the trick? Maybe as a backup ‘don’t catch fire’ method?Would a depletion mode FET make sense? They’re similar to a ‘standard’ MOSFET with gate threshold below zero (for N channel). They can be used to implement current limiting without an auxiliary supply. I like that because you could stack them in series with some other protection method.They’ll still have a bit of an inrush problem from the gate charge, so adding a drain resistor as previously suggested seems like a good idea. Note in the current limiting case you might get significant dissipation (eg 1mA limit * 1kV = 1W). Maybe part of the source resistance could be a PTC thermistor placed close to the FET, which could reduce temperature rise a bit.Ixys IXTA1N170DHV (1700V) with a 22k source resistor might be effective.