Preferred diode type for high current flyback diode? (1uH 120A) - Page 1
Forum section: electronics / projects

author: tinfever
author_rank: Regular Contributor
timestamp: September 17, 2022, 04:33:24 am
content: I'm working on building a very high power 12V-only electronic load that relies on switching in resistors of various sizes to set the load current. The highest current stage for short pulse loads will be pulling 120A through a 0.1 Ohm load resistor. The resistor is rated for 300W but can do 10x power for 5 seconds.Obviously these 300W wire-wound resistors are going to have significant inductance and I'm planning on doing short 100us pulses every 1ms or so, thus I need some sort of flyback diode to help with the inductive spikes. (The MOSFETs switching the load will also have TVS diodes across them to prevent them from going in to avalanche mode.)What would be the best type of diode to use for this flyback diode? Schottky since they will have low forward voltage and be fast? Could I use a unidirectional TVS diode in forward biased mode since I read large TVS diodes have a lot of silicon inside which might handle these pulses better?Since no one rates their Schottky diodes for 100us pulses, and instead they are rated for like 10ms pulses, do you think I could safely exceed the 10ms pulse rating?

author: Someone
author_rank: Super Contributor
timestamp: September 17, 2022, 04:47:30 am
content: Look for more modern diodes designed for switching supplies, they're often giving thermal and power handling data for continuous duty factors at faster rates 20,50,100kHz100us pulse of 0.1 DF is "slow"

author: T3sl4co1l
author_rank: Super Contributor
timestamp: September 17, 2022, 09:24:41 am
content: No need, let the TVS handle it.Or just don't switch the MOSFET so fast, a generous gate resistor plus an R+C between D and G will afford easy control of slew rate. (R+C because a C alone is prone to oscillation; the R also allows controlling the slope / corners / leading edge a bit better.)Also consider using two transistors in parallel (one resistor each) to reduce the peak current, making snubbing that much easier. Especially if you intend to have sharp edges (sub 1µs?).That said, I made this circuit, a current-limiting electronic fuse,https://www.seventransistorlabs.com/Images/LimitingFusePcb2.jpgyou can pretty much read off the output stage as-is: 1/4" QC terminals, a transistor, current sense shunt, and TVS. Under the most aggressive conditions I could muster, I measured this waveform:https://www.seventransistorlabs.com/Images/Wvfm_30V_ExtSw.jpgThat's a 30V edge, with low source resistance, and evidently about 200nH inductance, into the switch, already turned on. Calibration is -20A/div. Current rises steeply, peaking at 120A, then reducing to 40-60A as Vgs is pulled down (initially, Vgs stays high due to Ciss, and it has to discharge through the series gate resistor). The current is then limited by the shunt resistor (acts as source degeneration), Vg, and Vgs(th) which decreases over time as the die rapidly heats up. After about 3µs, it turns off, with a small reversal as capacitance and clamping act on it. During this time, the drain voltage is as you expect: initially low, then pulled up to 30V, then flyback up to 40V or so, then stable at 30V for the rest of time. (I don't have that trace handy, unfortunately.)So, with a TVS placed as close to the switch as possible, it'll be fine. At least at low duty cycle.If you're expecting 1uH max / worst case, and 1kHz max PRF, that's 7.2mJ each pulse, and 7.2W at 1kHz. You'll want a few TVS in series to dissipate that, but SMC package diodes can handle that just fine. Put a thermal pad on top and a heatsink. Probably two or three will handle that?As for clamping diodes, the problem is what to clamp into. Instead of one component, you have three: the catch diode, a capacitor that needs very low ESR, and a shunt resistor to bleed off the extra (or another TVS, redundantly enough). The path is longer, making stray inductance ever more of a concern.For ratings, I would take the 10ms surge value probably, sure. Going shorter probably isn't a good idea because the voltage drop will be much higher -- schottky diodes are already forward-biasing their guard-ring diode at that level, let alone for shorter pulses at even higher currents. It's the PN junction that handles huge peak currents; go for a PN diode in the first place. Any will do, really, since you don't need reverse recovery, and forward recovery is rarely poor. (Avoid 1kV general-purpose parts, really.)TVS as rectifiers, isn't something I've played with, but could be interesting for two reasons:1. The high doping level (that gives the low and controlled breakdown voltage) should also give low recovery. (Compare Vsd, t_rr of MOSFETs -- Vds is simply the nominal Vrrm of the body diode. Low voltage MOSFETs recover faster than high.)2. It should also reduce Vf, at least moderately so. Even a highly-doped PN diode will still be inferior to a schottky of the same ratings, but when you need the surge rating as in a snubber, it should be advantageous.2a. Normal rectifiers lie. For example, I've measured 1N4001s with breakdown over 400V; they really don't care, after all the spec is quite loose, and evidently it's easy enough to meet the Vf specs while offering greatly inflated (if undocumented) Vrrm. Low Vr leads to random failures (failure from transients or mains surge) so it makes sense they would optimize towards high Vr as a way to improve robustness. Anyway, this is all to say: you don't simply get a lower Vf by choosing a lower Vrrm, not among regular diodes. I haven't surveyed high speed diodes (maybe their breakdown voltages track better), but they aren't available in low voltages anyway. So considering TVSs might indeed be a way to explore that (low Vrrm) range today (since at-actual-ratings parts don't exist anymore).Of course, Vf and t_rr are never rated on these parts, so you need to characterize them yourself. Also, capacitances are massive, because of course, which mostly defeats the purpose for snubbers in switching circuits, but needn't be much of a problem for certain pulsed applications -- like this for one.Tim

author: tinfever
author_rank: Regular Contributor
timestamp: September 18, 2022, 01:47:44 am
content: Thanks for the feedbackWhile it hadn't occurred to me to let the TVS handle it or just switch slower, I think having a flyback diode across the load resistor might still be the best option. With the flyback diode, the current can recirculate through the load resistor and most of the energy can be dissipated in that resistor, which is what it's designed for. This has the added benefit that the DUT will see a higher current slew rate since the current recirculating through the load resistor isn't seen by the load.I also think switching slow enough to not exceed a 20V Vds max on a 12V rail during an inductive spike might increase the switching losses in the FETs excessively.I ran some simulations and here's what I'm seeing:(Simulation is with 10V gate drive source with 100R gate resistor. Using an array of four SISA40DN FETs in parallel. Single gate resistor is shared between all FETs. 100us on, 340us pulse period, 1.02ms simulation time)With no TVS across the FET and no flyback diode, the voltage spikes until the MOSFET goes in to avalanche, obviously. The combined MOSFETs dissipate 8.70mJ during the switch-off transition (of which 7.2mJ is from the energy stored in the inductor/load resistor). Average power dissipation per FET is 6.65W x 4 FETs! During "on" steady state, each MOSFET is dissipating 975mW so the power dissipation from the avalanche is substantial, as expected since 7.2mJ from the inductor energy alone at ~3kHz is 21.6W.Adding a 20nF + 10R between drain and gate, the turn off slew rate is reduced to 6A/us, which limits the inductive spike to ~18V. The combined MOSFETs now dissipate 11.56mJ during turn off transition. Average power per FET is now 8.78W. Way too much.Removing the slew rate control on the gate, and adding a SS10PH45 Schottky flyback diode across the load resistor, the combined MOSFETs now dissipate 245uJ during each turn off transition. The best part is that the Schottky only dissipates 1.20mJ during the same transition. Most of the energy previously being dissipated is now "gone" and is dissipating in the load resistor. Average power per FET is now 430mW. Schottky is dissipating 3.5W avg. This alone clamps the Vds to ~14V which would be safe.Adding a SMDJ12A TVS across the FETs, since the cable inductance (not in this simulation) to the load resistor and from the DUT will also be a problem and no flyback diode can fix that: Combined FET switching loss is 243uJ. Schottky loss is 1.14mJ. TVS loss is 267uJ. Average FET power is 425mW. Schottky power is 3.36W. TVS power is 787mW. Vds now clamped to 13.8V.Thinking about it more, if the goal is to dissipate as much current in the load resistor, and as little as possible in the diode, getting a diode with the lowest Vf as possible, like a Schottky, makes the most sense. I'm glad I looked in to this a bit more though since 3.5W is significant in the Schottky. It may require two of them.

author: tinfever
author_rank: Regular Contributor
timestamp: September 18, 2022, 05:26:53 am
content: Here's something I found and thought others might find interesting:When adding the cable inductance to the simulation, this caused most of the inductive spike energy to go through the TVS diode, instead of the flyback Schottky diode. I'm pretty sure this is because the inductance of the supply cables causes the 12V rail to increase, and the GND rail to go negative, thus even with the breakdown voltage of the TVS diode, it's still easier for current to flow through the TVS instead of through the flyback diode to the positive rail. This makes the TVS diode now average 15W...An interesting conundrum. I'll have to think on this some more. Perhaps there is a way to encourage the current to recirculate rather than just breakdown the TVS. A higher breakdown voltage on the TVS might work, but then I run in to the limit of the 20V Vds rating of the MOSFET.I briefly considered adding more MOSFETs to replaced the flyback diode and them driving them with the inverse of the main switching MOSFETs, but that's just too complicated and almost certainly wouldn't be fast enough....or would itEdit: No, that won't work since you'd have to drive the gate higher than the source, and the source would be at 12V-ish.(Cable to load resistor will be three 24" cables assemblies each with six 16AWG wires, three for 12V and three for GND. I measured 0.6uH for a single 24" 16AWG conductor, so with 9 wires in parallel on both the 12V and GND side, that's 66nH)

author: T3sl4co1l
author_rank: Super Contributor
timestamp: September 18, 2022, 01:35:25 pm
content: Again, you need massive bypass caps to use a clamp diode. For an antiparallel diode, this must be as close to the switch as possible, all three rather (switch, diode, cap). Since supply inductance is part of the loop otherwise. The TVS just does everything already, or slow enough switching does.What's wrong with 8W dissipation anyway,,, oh, you've got 3x3mm DFNs? Why? Why make this harder for yourself when a TO-220 or few is a slam dunk, clamp on a small heatsink and be done with it?Your simulation has too many grounds in too many places. Decide on just one. Probably removing the ground on the far right (12V source) is sufficient. Put everything common on the "low" node.Ed:Also, Q5 has a Baker clamp D3, but massive speed-up cap (C2) from a 1ns edge source, and emitter degeneration (R8)? Why not Q6 as well (Baker clamp)? Or why Q5 at all, or the driver at all since you can put in a suitable source (inverted or gate leveled) at each point? (Modeling a level shifter / gate drive at the same time, perhaps?)Also also, you've got two TVSs, which one were you looking at for power, D2 or D1?Let's see, the L2 between Q5 and Q6 emitters (roughly speaking), will have the effect of... positive feedback at high frequencies? So it should tend to "chunk" on fairly sharply as it goes. Turn-off speed better be limited though, there's not many volts you can afford to drop there. I mean, SPICE won't care, and the real thing probably won't be connected that way, but it would be a weird way to go, for sure...Tim

author: tinfever
author_rank: Regular Contributor
timestamp: September 19, 2022, 07:38:52 am
content: QuoteI don't quite understand what cap you are referring to. Do you mean a much larger RC snubber across the MOSFETs? I just simulated that and a 100uF + 0.1R snubber works remarkably well, and limits any Vds overshoot to about 12.3V total with no flyback diode or TVS diodes. However, now that resistor is burning an average of 42W. I can reduce the cap size a bit to reduce resistor losses at the expensive of increased Vds. I guess a few 10W THT resistors wouldn't be completely insane. If there is no way to coax most of the current to just recirculate through a flyback diode across the load resistor, and the only option left is to burn the power somewhere, a resistor would be the cheapest place to do it. Although I wish I could use the 300W resistor I already have!I read up a bit about RCD clamps but I can't quite see how to fit one here.(Any RC snubbers already shown in the schematic were just me trying to get the simulation to run smoothly. Not because of any thoughtful placement.)QuoteI do like to make life harder for myself... Perhaps just switching slower and using bigger FETs with heatsinks would be logical, as suggested. Or maybe I could do something with an aluminum PCB as a heatsink...just because I like to make life harder for myself.The real reason for those specific FETs was that in a big spreadsheet of Rds losses, number of FETs required, and costs, they were the cheapest solution. And I'm a terrible cheapskate. They also have a max gate voltage of +12V/-8V so I'll also have to add a zener to the gate drive circuitry to drop a couple volts and add another zener at the gate to ground just in case... because I like to make life harder on myself. The FETS are probably easily available because they're a PITA to use!QuoteThe odd grounding shown is a side effect of how I'm thinking of laying out the entire system. Diagram attached. Since I'm planning on having a current sense amp on each load resistor stage control PCB, that's going to need a ground reference that isn't the same one bouncing all over the place with 120A being switched. And who can afford optocouplers anyways!The FET enable signal is also going to be 3.3V or 5V, hence the level shifter circuitry before the gate driver. However, the enable signal will be referenced to the GND on the microcontroller, and I thought it made sense to keep the emitter current on Q5 returning on the signal GND, knowing that I'll also get a small amount of current from 12V through R4 as well. R8 was just there to simulate having a not-quite-as-low-impedance path to ground for the 5V enable signal being simulated.Thinking about it more, it might make more sense to tie the emitter on Q5 to the power GND, because it looks like there are some current spikes up to 0.5A going that way with all the switching somehow. Then there'll just be a small current from the enable signal that has to flow back through the the power GND back to wherever I tie everything together and back to the MCU.QuoteI didn't include a Baker clamp on Q6 since I didn't find it to help speed things up at all. No good reason for the sizing of C2 except IIRC, going smaller was slower and going larger didn't have many negative effects. R8 is just my attempt at simulating different grounding paths as mentioned earlier. Yes, this is modeling a level shifter from 5V and gate drive at the same time. I also didn't just want to use just 5V levels to drive the FETs to get that sweet sweet low Rds_onQuoteD2 is the TVS I was looking at for power. In the simulation I ran to get all the numbers I was referencing, I was using a more ideal circuit with an ideal 10V gate signal, and with only one TVS in the position of D2. D1 is just where I was playing around but it's clear that D2 makes D1 pretty much redundant.I also realized my estimate for the cable inductance might be off and it might be closer to 20nH on each side. I should also mention I really have no idea the real inductance of the 300W load resistors until they arrive from Mouser in a few days so I can measure them. 1uH was just a wild-ass-guess. Although I think the principles stay mostly the same, and only the magnitudes change. I did notice that reducing the cable inductance to 20nH per side in simulation caused more current to recirculate through the flyback diode and load resistor, and took the load off any TVS diode, which I think is *really* what I want since then I don't have to dissipate that energy in a different part.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: September 19, 2022, 10:34:53 am
content: Quote from: tinfever on September 19, 2022, 07:38:52 amRight, C1+R10. But not puny values. Like 1n isn't enough compared to Coss anyway, but a proper supply bypass, that's going to be like 1000uF and 20mohm.Using a clamp diode across the resistor means you're assuming it's where all the inductance is. That your local supply it's returning to, is ideal, a supernode with MOSFET source. Because that's the critical loop. But it's not (L1, L2). You need a bypass cap on the supply to make it more nearly ideal.QuoteU5 into a big cap, instead of "high". Then a bleeder resistor from cap to "high". This would be a peak clamp snubber. A rate snubber could also be used but you might as well slow the MOSFET to do the same thing.QuoteSounds like any MOSFET... they'll usually be ran at 5-8V. A 7805 or so before the drive circuit would do fine... can use logic for drive too, no need for BJTs.QuoteOh lord...Well, if you're going to fail, fail hard and often, right?Reminder you don't need isolation (analog isolation sucks anyway). Use differential amps. Keep as solid a common ground as you can (ground plane). Use local sense and diff amps to resolve AC and DC offsets later.Quote1uH will probably be an overestimate, but in the right order of magnitude.Wiring is roughly 1nH/mm. So, 100s of nH is easy enough to see inside a project box.Also why the MOSFET, diode and capacitor must be very nearby. Or MOSFET and TVS. Every mm of distance between them (including their own lead lengths) is stray inductance.Tim

author: tinfever
author_rank: Regular Contributor
timestamp: September 21, 2022, 06:12:29 am
content: QuoteThat makes a lot of sense and works quite well. Since this causes the "high" rail to be greater than 12V as the resistor discharges the cap, I'm seeing current flowing back in to the source (PSU DUT). Should I be concerned about this?QuoteHmm. Perhaps this is why it takes me so long to do everything...Out of curiosity, what part of my plan screams "fail" the most?For any future readers, I measured the following inductances:0.1R 300W Wire Wound (Tubular) Resistor - 0.836uH1R 300W Wire Wound (Tubular) Resistor - 5.382uH0.25R 1000W Wire Wound (Tubular) Resistor - 16.134uH0.1R 100W Wire Wound (Aluminum Housed, Heatsink required) - 55nH

author: T3sl4co1l
author_rank: Super Contributor
timestamp: September 21, 2022, 01:09:03 pm
content: Quote from: tinfever on September 21, 2022, 06:12:29 amIt's always and necessarily less than was drawn in the first place.In a truly extreme case, perhaps it could be a concern? Like, if you had some henries to charge up, some monster solenoid or something, or maybe a superconducting magnet, in which case the clamp capacitor needs to also be truly massive, and the supply would be drawn upwards pretty well (and if it has downprogramming, it might well not handle enough current, or power dissipation. But that's still rather contrived, because the sheer scale of energy storage is obvious from the application.For which, again... the correct answer is to dump it into a load, perhaps a resistor, perhaps a stack of TVS, anything that can dissipate the heat. Or for the superconductor, you might just let it quench (vaporize all the cryo refrigerant).The point that I find most convincing is that:1. The MOSFET is the source of the transient, and this is true from the shortest time scales (read: the transmission line between MOSFET and load), to the longest (as above);2. This is the only node that matters for protective purposes (MOSFET overvoltage causes failure), and3. It's again scale invariant, in terms of energy: whether you need to dump 1uJ (as in a switching snubber) or 1MJ (as above), this is the correct place to do so.And, again, and as you already know, or have discovered from the simulation -- it's not enough to clamp the wirewound resistor, there are other inductances in the loop. It's not enough to clamp the bulk of the inductance (e.g. a relay coil) at its terminals, because there may be wiring inbetween as well. And so on.Clamping a bulk inductive load, may still besufficient for a particular application: say your current level is low enough, or switching rate slow enough, or voltage rating high enough (or avalanche rating high enough, or other considerations related to that), to deal with up to some amount of wiring inductance. But it cannot apply in general, for arbitrary and unknown amounts of wiring, or if someone decides to add extra inductance just because (maybe they thought it was noisy and needed filtering in the worst way possible?..), or if someone forgot the clamp device, or its connection loosened over time (or rusted off or something). Exactly zero cases of which, are left uncovered by the clamped switch arrangement.The main reason you would employ something different, is if the repeat rate is high enough, or energy must be conserved. In that case, your best option is a boost converter motif, for which the switch-diode-capacitor loop must be minimal inductance. The boosted rail can then at least be bled back into the main supply (RCD clamp snubber), which saves on dissipating the extra "+ Vin" that a clamp to GND would otherwise have to dissipate. Or if you want to go to the trouble, a second converter could be used to return that energy back to the original supply entirely, ultimately only costing the converter's efficiency. Or if the energy to be snubbed in the first place, is rather well known (as can be the case for switching converters), there are quasi-resonant / "lossless" snubber networks (using L, C and D) that can return that energy automatically (by the switch's action alone).I understand that the argumentsIfind most convincing, may be rather high level / abstract, so I don't know how well you will appreciate them.Or worse? still: at least in principle, I prefer that people understandwhy, and when, something should be done some way -- not to take a single-case solution as wrote, which will probably lead to misapplication later. But that can also takeyearsto learn, for deeper topics.QuoteYou seem to be aware that your grounds won't all be. But you're indicating making loops of them anyway, and worse still, making loops across low-level circuitry, inviting all manner of ill behavior: from measurement error to oscillation, perhaps destruction I don't know.Maybe I'm jumping to conclusions there, but it felt worth adding the differential suggestion at least. Proper grounding takes much more to explain, unfortunately.QuoteMeasured how? Mind that a random meter at unknown frequency might be reading the resistance more than the inductance (magnitude impedance versus reactance, or parallel vs. series equivalent inductance, or at a mixture of harmonics).Tim

author: tinfever
author_rank: Regular Contributor
timestamp: September 26, 2022, 01:49:54 am
content: I think I understand your arguments for just using beefier FETs with heatsinks and just switching slower. It would be the simplest solution for sure. I'm just not a fan of the cost at this point. And any additional cable inductance directly puts more heat in to the FETs, vs. into some power resistors that are easier/cheaper to scale. Is the cost savings worth the time I've spent on this? No, probably not, but I learned a lot.While it sounds like it would probably be okay for some current briefly to flow back in to the DUT on turn-off, it just doesn't "feel" right to me. It doesn't feel like something a well behaved electronic load would be doing. So I'm going to try to avoid it if I can.I've come up with a circuit that feels just crazy enough to work, assuming the parasitic inductances to the TVS diodes are kept extremely low. Picture attached.I've changed the FETs to 30V SO-8 models to give more voltage overshoot headroom. Then each FET will have a small TVS as close as physically possible to clamp the initial voltage spike. Then each FET will have a diode to a capacitor to ground to store the bulk of the energy. The capacitors for all the FETs will all be tied together, effectively creating a boost converter rail as you alluded to. When the voltage on this boosted rail reaches 15V or so, a BJT will turn on a separate FET which will short the boosted rail to ground through 4x 5R 10W power resistors. A flyback diode on the main load resistor will continue to recirculate a significant amount of the current from the load resistors inductance, reducing the load on everything downstream.This ends up behaving like a big RC snubber, where most of the energy is burnt in the R, but it doesn't force the DUT to sink any current at load turn-off.I think the only parts that need as little parasitic inductance as possible are the TVS, foremost, and then the Diode-Capacitor pair, but a little less so. The load dump resistors and FET are much less critical.As for cost, I've also attached the numbers I put together for this solution vs bigger FETs with heatsinks. I estimate this solution might cost $7.45 per unit vs $13.9 for a solution using FETs and heatsinks. I think there is probably some room for trimming on the FETs and dump resistors, especially if I don't insist on being able to handle an extra 500nH of wire inductance on both the + and - supply connections. The caps are already downsized because getting ceramic caps in this amount of effective capacitance after DC bias loss gets expensive.There is now an issue with inrush currents when the DUT is connected and the load caps are charged, but barring some sort of pre-charge mechanism and a message for the user to not connect the DUT until pre-charge is complete, I don't see a way to avoid that.In simulation with 66nH added wire inductance, I'm seeing a design switching slower with bigger FETs and heatsinks is dissipating about 32.9W combined in the FETs, and 1.38W in the flyback diode, for 34.3W total. With this solution I'm seeing 4.73W combined in the FETs, 2.99W in the flyback diodes, 0.87W in the storage rail diodes, 10.68W in the dump resistor, and 0.13W in the dump FET, for 19.4W total. The difference is being dissipated in the primary load resistor via the flyback diodes.Regarding the ground loops, I'm going to add a footprint for an optocoupler to replace everything around Q5, since this would make the power/DUT side and the sense/control side be completely isolated, assuming I don't have the control MCU connected via non-isolated USB to a computer or anything. Of course, I'm stubborn so I'll probably try it with the ground loops first just to see if it will work at all.QuoteThey were measured with a DE-5000 LCR meter at 100kHz in equivalent series inductance mode. Admittedly I've never paid attention to the series vs parallel equivalent setting before, and was only using the 100kHz setting to get the highest resolution measurement, but I've read up on it a little now and double checked those are the settings and results. That 55nH is pretty weird though.