High Current Inductive Load Switching Issues - Page 1
Forum section: electronics / projects

author: Pizzashape23
author_rank: Contributor
timestamp: September 18, 2020, 03:18:19 am
content: I recently made a post about a safety circuit for a high current solenoid which was being used to feed a smelting furnace. That post is much more detailed but probably not a necessary read to understand my Issue.After ordering and assembling the trigger circuit for the solenoid, it is a low side mosfet switch using the NVMFS5C604NL from On Semi triggered by a photo interrupter, it worked great for about a day. After a while I noticed the solenoid was staying on for way longer than it should have been, upon closer inspection my mosfet had failed short and thankfully my safety system had worked. As I was switching a really big inductor running upwards of 40A through it I figured it was a voltage spike when the mosfet switched off that had killed it, turns out my flywheel diode had died and probably took the mosfet with it.I replaced the mosfet, and hoping for a quick fix, putting two of the same diodes in series, at the time I figured I might have been on the upper voltage limit of the diodes. The cirucit started working again but only survived half a day this time, the mosfet was dead but according to my multimeter the diodes were still fine.I don't have any tools that can properly trouble shoot a circuit with such high current and potentially very high voltage spikes, so I was wondering if anyone could provide any insight on what went wrong, particularly why the mosfet died while the diodes survived. They are FERD60M45CT's.I'm pretty sure that I just don't have the right diodes, but I don't really know what to look for in a flywheel diode. If anyone could point me in the right direction or recommend a specific part, any and all help would be appreciated.As a side note: The solenoid is running at about 26V, has a resistance of 0.6ohms and an inductance of around 3.3H. The mosfet is getting 12v at the gate.

author: TimNJ
author_rank: Super Contributor
timestamp: September 18, 2020, 04:34:17 am
content: I think you want a diode with fastforwardrecovery time. Most datasheets don't specify forward recovery...just reverse recovery. I think it may be true that a fast reverse recovery implies a fast forward recovery. Forward recovery just means the time it takes to go from a non-conducting to conducting state. This diode needs to "catch" the inductor energy before the voltage builds up across drain-source, as you know.43A + 3.3H means the peak voltage can be very high..depending on di/dt at switch-off. You can simply try to slow down the MOSFET shut-off transition time. Since the voltage spike is a product of di/dt, if you use a higher value gate resistor, the peak voltage should be lower.Also, 43A through 1.3mOhm RDSON is over 2W dissipated in a poor little DFN5x6. Do you have plenty of copper area on your PCB? I've used that package before, but kept it to less than ~0.5W, since there wasn't much available area for PCB heatsinking. Check Note 2 on the datasheet. The thermal resistance from junction to ambient is 39C/W with a 25mm x 25mm 2oz copper area. That's a pretty large area, and even then you'll get 85C rise @ 2.2W

author: exmadscientist
author_rank: Frequent Contributor
timestamp: September 18, 2020, 04:40:22 am
content: Slowing down the FET has the opposite problem: now it's in linear mode dissipating heat for a longer period of time, which heats it up and tends to induce thermal runaway. This is a pretty good way to kill FETs. So this is something that's not trivial to balance well. I think I usually see more people err on the side of fast switching and then deal with the pulse, but this inductor is on the larger side, so one must be careful....Speaking of gate transition times, what's the gate driver here?

author: Pizzashape23
author_rank: Contributor
timestamp: September 18, 2020, 05:05:13 am
content: Quote from: TimNJ on September 18, 2020, 04:34:17 amYeah, I get where you are coming from, but at the gate voltage that it is being driven at the 1.3mohm is much closer to 0.8mohm. I also made sure to include a massive copper area with large thermal vias going to more copper on the other side, i also filled in the vias with solder and made sure to layer it on both sides. This does still have it's limtations which is why I am trying to reduce the on and off times as much as possible, there is no gate resister and the gate to ground resistor is 100ohm.Quote from: exmadscientist on September 18, 2020, 04:40:22 amThe gate driver is a super simple board of my own design, it takes the output from the photo sensor, inverts it using an npn transistor then drives a pnp transistor directly switching the mosfet to the lower voltage supply rail at 12v. I don't have a schematic right now, but if it's needed i could put one together.I really think it's an issue with the snubbing of the voltage spikes, in testing I was able to keep my finger on the mosfet while it was switching multiple times which means that it isn't getting very hot. I tried to keep the entire switching ciruit as simple as possible so that troubleshooting and repair would be fairly easy.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: September 18, 2020, 05:26:50 am
content: What is the switching loop, from GND, to MOSFET, to catch diode, to supply, to GND?There should be a bypass cap right beside the transistor and diode.What does the drive circuit look like?Tim

author: TimNJ
author_rank: Super Contributor
timestamp: September 18, 2020, 05:30:17 am
content: Quote from: exmadscientist on September 18, 2020, 04:40:22 amDepends on how frequently it's being switched? If you're actuating once a second continuously, maybe not a great idea, but once every few minutes, I can't see how it would run away. I'm also not suggesting to slow down the switching time to something ridiculous. But I don't think it needs to switch in (a few) nanoseconds either.

author: exmadscientist
author_rank: Frequent Contributor
timestamp: September 18, 2020, 05:48:20 am
content: Quote from: TimNJ on September 18, 2020, 05:30:17 amSounds about right to me, yeah. The selected MOSFET is pretty robust and might even be able to get away with a gate transition time as low as a millisecond or so.I guess I've been spending too much time with less-forgiving HV superjunction parts lately, where that's a relative eternity!

author: Pizzashape23
author_rank: Contributor
timestamp: September 18, 2020, 09:44:36 am
content: The "driver" is attached, please mind the hand writing, its only a rough sketch.It works in conjunction with the safety circuit mentioned in the other post, in a fail state the relay should prevent the solenoid from being active for longer than would cause damage .Also the solenoid is being activated once or twice a minute, when in use. This means I would much rather correctly spec the snubber circuit than try to reduce the switching time. I was lucky the first time it failed because I was in the shop and I could see that it had failed short, I dont want to risk a thermal runaway while someone else is using it as they probably wont see it fail and try to conti ue to use it.

author: TimNJ
author_rank: Super Contributor
timestamp: September 18, 2020, 03:41:33 pm
content: Your "switching frequency" is about 0.03Hz if you actuate twice a minute. I think the fear of thermal runaway is not 100% relevant in this particular case.What's the approximate rise time of the drain voltage during switch-off, as is? Here's a worst case SOA approach:The peak power dissipation in the device (theoretically) happens when Vds = 0.5*Vdd and Ids = 0.5Idd. This can vary a little based on gate characteristics. But, let's just try it. 0.5*26V = 13V; 0.5*43A = 21.5A. Looking at the SOA curve from the ON Semi datasheet, we see that this roughly aligns with the 1ms curve, meaning it can handle a pulse with the aforementioned power for 1ms or less. In reality, the MOSFET only sees the peak power for part of the switching cycle. Theoretically, the power waveform roughly looks like a half sine wave. You can probably estimate that the real power capability (or max pulse duration) is at least 20% higher.My point is: You don't have to slow the thing down to 1ms if it makes you uncomfortable, but you also don't need the switching time to be 50ns. Somewhere in between may get the peak voltage down to something reasonable. Of course, you can just take care of it with snubbers, flyback diodes, maybe even TVS...but there too you need to understand peak power handling etc.

author: Pizzashape23
author_rank: Contributor
timestamp: September 19, 2020, 01:19:45 am
content: That's a great response and makes sense, but it would also imply that the culprit of the mosfet death is definitely the voltage spikes and also that the snubber i have got in place is working temporarily but still not doing its job properly.If im not mistaken the switching speed is based on how hard the mosfet gate is driven, so adding a gate resistor should slow it down. How would i find the right resistance value?In my case it is much easier to fix the snubber than to add a resistor as the pcb has already been printed, in hindsight i should have added pads for it and just bridged it with wire. Even If i end uphaving to slow it down, I am still curious about how i would have correctly spec'd my snubber and would probably include it anyway just to be safe.Something im still confused about is why after adding another diode in series with the first, the mosfet lasted half as long.

author: julian1
author_rank: Frequent Contributor
timestamp: September 19, 2020, 02:14:55 am
content: > Something im still confused about is why after adding another diode in series with the first, the mosfet lasted half as long.Because the flyback voltage got worse after it was blocked by an additional slow switching diode?Are you able to wire in a jelly-bean schottky 1n5822 as a test. Only 3A but it merely has to shunt the voltage spike for an instant ( "once or twice a minute") . Unless I am misunderstanding something.

author: Pizzashape23
author_rank: Contributor
timestamp: September 19, 2020, 02:35:20 am
content: It makes sense for normal diodes, but the ones I have used are schottky diodes, they should be switching fairly quickly.Sure, if they are in stock at my local altronics i might be able to pick some up today and try it. But I don't think it will make much of a difference, the max reverse voltage is only 40V no where near the voltage spikes.

author: julian1
author_rank: Frequent Contributor
timestamp: September 19, 2020, 02:56:53 am
content: The reverse voltage of 40V is sufficient to stand off the 26V of your supply. To shunt the spike it's forward biased no? And it should shunt/cap to limit the spike to lower than that to protect the FET drain.The FERD60M45CT datasheets states they are a 'Field effect rectifier '. I have no idea what that is.

author: TimNJ
author_rank: Super Contributor
timestamp: September 19, 2020, 03:03:13 am
content: Quote from: Pizzashape23 on September 19, 2020, 01:19:45 amIn general, I would find the value experimentally. Start with a few ohms (1 to 5, maybe) and move up from there until you reach the desired rise/fall time.Before trying to implement any solution, you should really put a scope across drain-source to see what it looks like right now. 10x probe should be good up to ~500V, and given this is a 60V MOSFET which worked for a little while, doesn't seem like it is too much more than 60V...maybe 100V-ish?Can you share a picture of your board? Or a screen-capture of the layout?

author: TimNJ
author_rank: Super Contributor
timestamp: September 19, 2020, 03:09:43 am
content: Quote from: julian1 on September 19, 2020, 02:56:53 amMaybe this is a dual "diode-connected" MOSFET type thing? If so, seems like a bit of an odd choice. Seems that these are targeted at some special application, but I do not know what.

author: Pizzashape23
author_rank: Contributor
timestamp: September 19, 2020, 03:30:20 am
content: Sure, it worked for a little while, but it did have the "schottky diode" across the solenoid, and that's what is confusing. Without the diode the mosfet blew after the first trigger, using one diode it survived about a day and using two diodes about half a day. I also don't think that such a large solenoid with such high current would generate spikes of only 100V.Note that on the arrow.com listing it said it was a schottky diodeThe board is attached below, the 4 terminal connector goes to the sensor while the 2 terminal connector goes to a regulated 12V. The leads go to the solenoid and gnd

author: T3sl4co1l
author_rank: Super Contributor
timestamp: September 19, 2020, 03:40:01 am
content: Quote from: T3sl4co1l on September 18, 2020, 05:26:50 amAh, that answered my question then. It's not. Quite possible the loop length is causing excessive peak Vds.How long is the cable from MOSFET to clamp diode, and to the supply?Tim

author: Pizzashape23
author_rank: Contributor
timestamp: September 19, 2020, 03:50:27 am
content: The length from mosfet to diode is about 30cm and can be cut down, but if loop length does cause issues it might be coming from the metre between the diode and the supply.Could you explain how that would increase Vds?

author: T3sl4co1l
author_rank: Super Contributor
timestamp: September 19, 2020, 04:48:28 am
content: Wiring has inductance too. The whole path from MOSFET to diode to supply (assuming the supply has a big fat capacitor in it) and back gets charged by the load current, then discharged when it turns off. V = L * dI/dt simply generates excess voltage proportional to the rate of current, and if that exceeds the transistor rating, it will avalanche. It might not fail right away, or even after many cycles, but it tends to lead to failure and I recommend against using MOSFETs in repetitive avalanche breakdown.So then, how fast does it turn off? Without a scope waveform, we can still simulate it, or estimate it from component values.Using this datasheet:https://www.onsemi.com/pub/Collateral/NVMFS5C604NL-D.PDFIf we assume Vgs = 12, and the BC327 turns off*, then we have 100 ohms discharging the gate capacitance. Drain is carrying ~40A.*For best results, there should be some resistance B-E to ensure it turns off cleanly. Anything from 60 to 10k ohms would be fine here, I think. Should still work as-is, though.Following this instant, Vgs discharges. Ciss is about 10nF so this goes exponentially with a time constant of 1us.Nothing much happens until Vgs ~ 3V, where Rds(on) increases hyperbolically. Vds starts rising, and Vgs levels off -- the Miller plateau.The Miller plateau takes about 13nC. While Vgs is in the 2-3V range, Ig is 20-30mA, so the plateau will be about 0.5us long. In this time, load current is still 40A (that big fucking inductor isn't going anywhere) and Vds has risen to its peak (whatever that is).As the plateau ends, Id finally drops, and we can calculate dI/dt.g_FS = 180SCiss = 9nF (it's dropped slightly, because Vds is high; the difference is basically Crss)Ig = 20-30mAIg = Ciss d(Vgs)/dt ==> d(Vgs)/dt = 2.2-3.3 V/usdI/dt = d(Vgs)/dt * g_FS = 400-600 A/usVpk = Lloop * dI/dt = well, a lot even if Lloop is a mere 0.1uH. Which it's definitely not (that would be about 20cm of cable for the entire loop), so for sure, it will avalanche.How much avalanche is permissible?Datasheet gives one-shot rating for Ipk = 22A, E = 776mJ (which means it's a pretty darn big die, barely any plastic making up that SMT package, I suspect). There's a curve, Fig.12. Say we pick the hot condition, Ipk = 40A. That gives 0.6ms.If Lloop ~= 0.8uH, and V(br)dss = 70V typical, then Vloop is 70-26 = 44V, so the 0.8uH will discharge from 40A to 0 indt = L * dI / V= 0.7uswhich sounds pretty safe versus 600us.No repetitive figure is given, so it's not obvious how many shots of this it can withstand. Unlimited? Does it cause wear, and if so how much? It's about 1/1000th the maximum, but does that mean it can handle 1000 shots? Maybe it doesn't derate proportionally, maybe it's only ~30 shots? Maybe it's unlimited? Is the plot even for a single event, or what?Possibly, a manufacturer FAE could answer some of these questions; it will be faster to change the circuit and try again, though (at least, for a couple tries). Might still be worth a shot.I also wonder about the path between solenoid GND and the 12V supply: are they common ground (elsewhere)? Isolated? What length of wiring?I think it would be pretty safe to put a SMAJ36A from source (anode) to drain (cathode), and a 1N4745 or such from GND to +12V. This at least prevents hazardous voltages from reaching the transistor itself. I would try that and see. (This clamps voltages locally, without needing to reroute power lines or add a capacitor.)Some filtering on the optointerrupter leads might also be desirable. Preferably this circuit would be placed nearby..?Also, how hot does that thing run in practice? Might be worthwhile soldering a copper fin on it, or a soft thermal pad and heatsink. Shouldn't need much to dissipate a couple watts.Tim

author: Pizzashape23
author_rank: Contributor
timestamp: September 19, 2020, 06:11:33 am
content: Look, you went in to way more detail than i expected, and while I can appreciate it, im going to be honest and say that most of it went over my head.I will answer what i can and say that the 12v supply is derived using a dc/dc converter module directly from the supply of the solenoid at 26 or so volts. The photo sensor is less than 5cm away from the trigger circuit and while the mosfet was working it did not run very hot (Keep in mind i was not able to monitor it constantly). Also i think it is important to know that the supply at the moment is 2 12v lead acid car batteries, I planned to change it after i could see that the system was working reliably.My questions now are;(Before reading these questions, a couple will probably be pretty stupid, keep in mind this is my first time using mosfets and i acknowledge that i am way over my head. However, I am definitely keen to learn.)-Is there anyway to avoid avalanche/failure entirely?-If running at avalanche does cause lasting damage and thus the life span is temporary, would there be any way to extend it or reduce the risk of failure?-When you said that the zener should go from GND to 12V, could it go directly across the input terminals of the board or does it need to be near the mosfet?-What is the zener diode doing?-Why is it that the mosfet can handle being in avalanche for any amount of time?-Im assuming that while in avalanche as long as the breakdown voltage is not exceeded and it's not avalanching for too long it might survive, is this correct?

author: T3sl4co1l
author_rank: Super Contributor
timestamp: September 19, 2020, 08:11:20 am
content: Quote from: Pizzashape23 on September 19, 2020, 06:11:33 amAhh... and I guess the converter is common ground, a three wire module more or less?That sounds alright.QuoteYes! Divert the excess voltage into a clamping device.Technically, TVS diodes also avalanche; they're just more reliable at it. We could still dodge that, if we wanted to -- say by using a clamp diode into a large capacitor (this is called a voltage peak clamping snubber).QuoteThe MOSFET by itself, can dissipate a fair amount of power for a short time, at pretty much any voltage or current it can handle. Avalanche occurs when the voltage is pushed too high (avalanche breakdown), somewhere above 60V. This is a different sort of current flow (there's physics under there), and it cannot be controlled by gate voltage.One way to avoid this type of current flow, is to make the transistor turn on, just enough to hold drain voltage below breakdown.Which can be done with two diodes anti-series, one a zener. Zener points to drain, regular diode points towards gate. This is a common approach in "protected switch" devices (which are a MOSFET with a controller built in, powered by the "gate" terminal itself). See diagram here:http://diodes.com/design/support/technical-articles/self-protecting-mosfets-deliver-improved-reliability-in-the-harsh-environment-of-automotive-applications/I'd still include the S-G zener, to protect against rapid changes, ESD and the like.Another way is, since we know the load is inductive -- just don't turn it off as fast. Evidently, it's about 20 times too fast. If we slow it down 20x, the excess voltage will be manageable. This is easily done by increasing the pull-down to 2kΩ.Downside: now the whole turn-off time (Miller plateau), spent with high drain voltage and full drain current, is 20 times longer, too. Which is 0.5us * 20 = 10us. If it's dissipating up to 50V * 40A = 200W for 10us, well, that's still well within the safe operating area, we can check that on Fig.11.So this seems to be a fine option. Which echoes what some have said earlier in the thread, but this puts a number to it.QuoteThe board is small, at the connector is fine.The trick here is, the inductor doesn't care where its current is going, it's just a current coming down a wire. When the transistor turns off, that current just goes into, whatever it goes into. If nothing's there to catch it, it keeps going into the transistor (as avalanche breakdown). It's the transistor's fault that the voltage shows up -- it's the causative event.Aside:This is actually true to a deep level. If we do the same thing with faster transistors, we can see the event propagate through space (or at least along wires) at the speed of causality itself, i.e., the speed of light. A transistor that switches in 1ns for example, goes from off to on in but 30cm at light speed! You can imagine a spherical shell expanding away from the transistor, encompassing all space that can possibly know about its switching. Think of the shockwave from an explosive, made visible by the refraction of air; except in this case, instead of pressure, it's the...rather abstract idea of being able to tell an event has happened. Though conveniently in this case, at least, we can also measure something more concrete, like the change in voltage.I suspect this is just more confusing than useful..... but the point is it's ultimately very fundamental to the universe itself, and the reason your transistor experiences a voltage peak is, in an overly basic but fundamentally true sense, because of causality at the speed of light. I think that's pretty frikkin cool.I digress.So, er, in any case, to close that causal loop, you need something to "cause" current to flow, soon after the transistor stops doing it. Which means putting the TVS beside the MOSFET, where it can "know about it" immediately. The inductance can't stop flowing current until it "knows" about it, and it's  *waves hands* far away.Likewise, the clamp diode being at some distance, isn't helping as much as it could be. Yes, it's clamping the bulk of the solenoid's charge. Which is, by a huge margin, most of the energy in this system. But there's still some in the wires stringing everything together.QuoteBecause... slimey wimey semiconductor physics?I actually don't know much about this myself. I guess that's no accident, according to this appnote?https://www.infineon.com/dgdl/Infineon-ApplicationNote_Some_key_facts_about_avalanche-AN-v01_00-EN.pdf?fileId=5546d462584d1d4a0158ba0210977cdeSomething about "hot carriers", and I would assume dislocations and other stuff, in a sense causing breakdown of the semiconductor or insulator materials themselves, in a similar way as they would when just utterly blasted (with excess voltage and/or current), but a tiny bit at a time. Eventually I guess enough defects occur in some point to forms a conductive channel, or it aggravates localized heating or something, and then current draw starts increasing, and pretty soon it shorts out.Also, this one from ON Semi seems to agree the avalanche curve is single only,https://www.onsemi.com/pub/Collateral/AN-7515.pdf.pdfso it doesn't apply to repetitive avalanche.(Huh weird, that link is also a .pdf.pdf. I never noticed that before, I wonder how many of their files are named like that...)QuoteNot quite, avalancheisbreakdown. It seems it's more or less the case that a device can withstand a certain amount of avalanche, total. So, they're happy to give big fat single-pulse ratings, or even test them in production (as the appnote above says OptiMOS are?), but a few too many of those, or a few too many hundred or thousand or million or whatever of very brief ones, and...not so much.But yes, if the voltage is kept below breakdown, avalanche does not occur, and operation should be reliable.So, it does look quite likely that this is the problem, and it would be easy enough to verify with an oscilloscope, if one is handy. And the symptom should go away with a TVS installed, or any other change that avoids MOSFET breakdown.TVS diodes do not suffer from the reliability issue, as far as I know. The problem is that MOSFETs are complicated, they have a bunch of stuff in them that....transists. Probably, the same wear mechanism occurs in diodes, but the only insulator that could fail, is around the edges of the diode, and stuff getting blasted around there can't really cause problems, partly by careful design I'm sure, and partly because it's just a thin surface around a big fat lump of semiconductor. (That's another thing that MOSFETs are: surface area. MOSFETs work purely on a surface effect. It's all surface, millions of tiny structures etched into the silicon provide enough surface area to handle huge currents.)Tim

author: Circlotron
author_rank: Super Contributor
timestamp: September 19, 2020, 09:15:36 am
content: Quote from: Pizzashape23 on September 19, 2020, 01:19:45 amWhat are you currently using as a snubber? If you are switching off 40 amps then I'd suggest putting a 1-2uF polypropylene cap directly from source to drain, reasonably close to the mosfet. This will slow things down enough to give the diode a much better chance to catch the voltage despite the stray inductance of wiring etc.

author: Pizzashape23
author_rank: Contributor
timestamp: September 19, 2020, 09:32:02 am
content: This post has cleared things up soooo much, seriously Tim you are a life saver, without help I probably would have gone through my whole tub of mosfets without thinking and still not have found a solution. If only this post had existed before i started it would have saved me a whole lot of confusion starting out with mosfets.Going forward i will probably change the 100ohm resistor to 2k (there is a small resistor pad under the high power one) and definitley include the TVS and zener. (still not 100% on what it's doing but you have more than proven yourself in my mind)Thanks AgainAntonio

author: Pizzashape23
author_rank: Contributor
timestamp: September 19, 2020, 09:36:53 am
content: Quote from: Circlotron on September 19, 2020, 09:15:36 amThe original snubber was a schottky diode anti-parallel to the solenoid, it failed and blew my mosfet, so I tried again placing 2 in series and it worked even worse than the first time. I still don't know why the diodes didn't work but either way Tim has provided a more than adequate solution.

author: TimNJ
author_rank: Super Contributor
timestamp: September 19, 2020, 01:28:24 pm
content: Ah just a correction to some of my thoughts above: Your circuitdoeshave series gate resistance, but only at MOSFET shut-off. The 100R "pulls" the charge out of the gate once the BJT is shut off. Not sure why I couldn't see that before, but just to clarify. Your circuit does not really have any gate turn-on control, but that's usually okay, as there's no inductive load current at that point.I'm not sure if there's a reason to do it differently with a very large inductive load, but I would've went for a totem-pole gate driver (2 transistors, complementary NPN/PNP) so that you don't have to dissipate ~1.5W in the big wirewound resistor while it's on. But, if you're going to bump the resistance up to ~1-2K anyway, then the dissipation is way more reasonable. No need to do anything.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: September 19, 2020, 01:40:05 pm
content: Yeah the beefy resistor is quite out of place, but it's also a tiny fraction of the solenoid's power consumption, so, it's really quite meaningless in terms of overall efficiency. Fun, huh?Speaking of totem poles -- I would go even further, and recommend a 2-switch configuration:https://e2e.ti.com/cfs-file/__key/communityserver-blogs-components-weblogfiles/00-00-00-07-88/2450.Capture-6.JPGIwould... but it is more complicated, and it's not necessary here. So, this is just FYI.Anyway, say you were to use this -- the advantage is, you have two transistors, with respective clamp diodes, all right there on the one board. The transistors switch each end of the solenoid, so when they turn off, one end flybacks up (pushing into +V) and the other end flybacks down (pulling from GND). You need a bypass cap to account for the sudden reversal (going from a +40A load, to a -40A return), but the turn-off is as fast as turn-on (faster, actually) because it's applying the full 26V, in reverse -- rather than shorting it out with a clamp diode (at ~1V). And, all the energy stored in the solenoid is returned to the power supply -- not that that's a problem with a slow cycle like here, but it can be valuable for high speed driver applications (like print heads and injectors).Tim

author: Marco
author_rank: Super Contributor
timestamp: September 19, 2020, 01:42:45 pm
content: Quote from: TimNJ on September 18, 2020, 04:34:17 amDiodes don't have forward recovery time. There's resistance modulation, which will start high but the higher the dI/dt the faster the resistance drops, the forward voltage is self limiting (although it can be still quite high).

author: T3sl4co1l
author_rank: Super Contributor
timestamp: September 19, 2020, 01:47:56 pm
content: Quote from: Marco on September 19, 2020, 01:42:45 pmWell whatever it is, itlooksthe same -- I've (semi-accidentally) applied almost 60V to a diode before -- in about 20ns. It looks like it has series inductance, but what would've been stored reactive energy is lost as heat in the diode instead, it doesn't come back out.Tim

author: Pizzashape23
author_rank: Contributor
timestamp: September 19, 2020, 02:30:54 pm
content: Just curious, but assuming I go the TVS route, could I put a fairly low value resistor in series with it across the mosfet to try and collapse the field faster?

author: TimNJ
author_rank: Super Contributor
timestamp: September 19, 2020, 04:58:38 pm
content: Quote from: T3sl4co1l on September 19, 2020, 01:40:05 pmTrue. I was mostly talking about heating up the MOSFET PCB more than necessary. If this board is jammed in a little, sealed enclosure, 1.5W can make the ambient temperature pretty high within the box, since we were talking about thermal runaway and semiconductor stress, etc.

author: TimNJ
author_rank: Super Contributor
timestamp: September 19, 2020, 05:08:19 pm
content: Quote from: Marco on September 19, 2020, 01:42:45 pmPoor terminology on my end, I suppose. I have seen the term "tfr" stated in some datasheets, just not commonly.After some additional reading, it seems that trr and "tfr" may in fact be in inversely proportional to each other. That is, lower trr may come at the expense of higher "tfr". I need to do some reading on that...and maybe pull out my semiconductor physics textbook for once.I wonder what metrics we can use to gauge "tfr", when it's not stated.

author: Marco
author_rank: Super Contributor
timestamp: September 19, 2020, 05:26:58 pm
content: The problem with using the term forward recovery time is that it gives the impression that it will be like reverse recovery time, just with an open instead of a short.The forward recovery time is just the time till the peak forward voltage, that time is mostly independent of dI/dt ... but it's also not really valuable to know, because the diode is not an open at all. For a given dI/dt there will be peak forward recovery voltage and that's what you're generally going to be interested in.

author: TimNJ
author_rank: Super Contributor
timestamp: September 19, 2020, 05:36:36 pm
content: Quote from: Pizzashape23 on September 19, 2020, 02:30:54 pmIf you put resistor in series with the TVS, the peak drain-source voltage will be higher, which may defeat the purpose of the TVS a little. On the other hand, if your TVS cannot handle the energy, it may be necessary. Higher series resistance = lower average power dissipated in TVS while clamping = higher peak voltage on the drain. A trade off.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: September 20, 2020, 12:36:14 am
content: A 36V TVS will discharge the inductance faster than a clamp diode. Mind, I'm recommending using both -- you'll need a HUGE TVS to handle the charge in the solenoid, better to let it discharge through its internal resistance (which takes some seconds!), using the clamp diode.Tim

author: Pizzashape23
author_rank: Contributor
timestamp: September 20, 2020, 12:50:28 am
content: The clamp diode was one of the issues I had in the first place. It just didn't seem to work. How would I go about correctly choosing the type of diode and then choosing the right specs? Clearly my judgments not that good from the two dead mosfets sitting on my desk.This would then mean I am increasing the gate discharge resistor, placing a zener across 12V and ground, placing the TVS between the Drain and source and using a clamping diode across the solenoid.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: September 20, 2020, 12:57:51 am
content: The clamp diode is fine, as long as it's heatsinked (really, just some thermal mass will do I think) and both sections are in parallel.The problem is it's not beside the transistor with a capacitor, so you need the TVS to deal with the inbetween.Quote from: Pizzashape23 on September 20, 2020, 12:50:28 amYes; for "across/between X and Y" meaning, cathode and anode respectively. They're still regular diodes in the forward direction (anode to cathode), don't want to short out the supplies.Tim

author: Pizzashape23
author_rank: Contributor
timestamp: September 20, 2020, 01:27:36 am
content: Would this be a sufficient TVS Diode?https://au.mouser.com/ProductDetail/Littelfuse/SMDJ36A?qs=zL8PTTxXkF3y4QOmGeQP1A%3D%3D

author: T3sl4co1l
author_rank: Super Contributor
timestamp: September 20, 2020, 02:27:19 am
content: Yes. Even an SMAJ would do, but you aren't going to hurt anything with a bigger one.Tim