Killing Mosfets - Page 1
Forum section: electronics / beginners

author: smus
author_rank: Newbie
timestamp: February 19, 2023, 05:57:45 pm
content: HiI'm a novice and I need a little help. I'm trying to build a dual constant current power supply with a 5V input to turn it on and off. But I keep killing my Q1 mosfet. It happens the second I input 5V.A little about my circuit:Batt = 84VQ1 gate voltage = 10,8VI have tried to put a 10k resistor in series with the gate. No Difference.I ran the circuit at 31V and it worked great.Is my 84V just to close to the 100V Vdds?

author: Picuino
author_rank: Super Contributor
timestamp: February 19, 2023, 06:16:57 pm
content: How is the PCB you are using?Can you upload a picture of the tracks?If you have a large parasitic inductance, it is possible that the mosfet is exposed to higher voltages than Vcc.A low resistance is usually placed in series at the gate of the mosfet to achieve low switching times.

author: Picuino
author_rank: Super Contributor
timestamp: February 19, 2023, 06:21:36 pm
content: Make sure that the gate of the mosfet does not exceed 20V.https://www.mouser.es/datasheet/2/268/HV9910C_Data_Sheet_DS20005323-3002607.pdf

author: smus
author_rank: Newbie
timestamp: February 19, 2023, 06:32:01 pm
content: Quote from: Picuino on February 19, 2023, 06:16:57 pmQuote from: Picuino on February 19, 2023, 06:21:36 pmQ2 and Q3 is running great no problem there. Gate voltage on Q1 is 10,8V

author: wasedadoc
author_rank: Super Contributor
timestamp: February 19, 2023, 06:53:53 pm
content: Excessive current? Q1 is switching 84 Volts to C2. How large is the current spike?

author: smus
author_rank: Newbie
timestamp: February 19, 2023, 07:08:01 pm
content: Quote from: wasedadoc on February 19, 2023, 06:53:53 pmWhen running the circuits is using around 0,4A combined. About the current spike I don't know. But would a little 33uf cap be enough to kill a 7,5A mosfet?

author: wasedadoc
author_rank: Super Contributor
timestamp: February 19, 2023, 07:13:43 pm
content: Quote from: smus on February 19, 2023, 07:08:01 pmRemove the capacitor and see.

author: exe
author_rank: Unknown
timestamp: February 19, 2023, 07:15:25 pm
content: How mosfets die? Short circuit between drain and source? It's worth checking drain-source voltage with oscilloscope to make sure there is no inductive kickback or something.I once had a curious case when small mosfets where dying while driving a relay (there was a diode across relay coil). I tried two mosfets in a row from the same reel, both died instantly. Then I replaced it with a similar mosfet from another manufacturer that was spec'd very similarly. It worked flawlessly ever since.

author: Siwastaja
author_rank: Super Contributor
timestamp: February 19, 2023, 07:25:11 pm
content: Quote from: smus on February 19, 2023, 07:08:01 pmVery possibly yes. If the little 33uF capacitor is ideal and has zero ESR, then the current is infinite.Battery load switches are notoriously difficult to get right. I had to recall a design where I blindly followed, in hurry, an appnote of such load switch, which did basically the same, driving a series pass switch with large Rg. IIRC, some 200-300uF caused enough inrush to kill the MOSFET which was somewhat larger than yours.You can test it without the capacitance, or with series resistance added to the capacitor, or you can take a look at the SOA curves for the MOSFET.

author: Siwastaja
author_rank: Super Contributor
timestamp: February 19, 2023, 07:26:07 pm
content: Quote from: exe on February 19, 2023, 07:15:25 pmWhere was the diode located physically - was it at the relay coil, far away from the transistor, or was it next to the transistor? Asking for a friend

author: Picuino
author_rank: Super Contributor
timestamp: February 19, 2023, 08:10:17 pm
content: The PCB design is far from optimized.You may have voltage spikes due to parasitic inductances. You can check this with an oscilloscope by running the circuit at low voltage (30V).Another possibility, if you don't have an oscilloscope, is to place a snubber network between the Q2 drain and Vcc to check with a multimeter how many volts above Vcc you have in Q2 drain. To make the snubber network you simply need a 1N4148, in series with a 100nF capacitor, to which you must place in parallel a 10k resistor.

author: Kim Christensen
author_rank: Super Contributor
timestamp: February 19, 2023, 08:34:50 pm
content: Is there any reason why you're not controlling the PWMD (pin 5) on U2 & U3 via your opto-isolator instead?

author: smus
author_rank: Newbie
timestamp: February 19, 2023, 09:11:01 pm
content: As suggested I removed C2 and the problem went away.Quote from: Picuino on February 19, 2023, 08:10:17 pmAny tips for making a better PCB designQuote from: Kim Christensen on February 19, 2023, 08:34:50 pmStandby power. If I read the datasheet right at standby it draws 0,5ma per driver. It might be the way to go.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: February 19, 2023, 09:23:24 pm
content: It's not spikes and inductance, it's a slow switch. It's driven with ~3k.It could be oscillation, but with high gate drive impedance, that's a stretch. (There doesn't seem to be enough layout area for that to be provided by trace impedance, either.)Most likely it's SOA.Never hurts to link the datasheets:https://ww1.microchip.com/downloads/en/DeviceDoc/20005344A.pdfhttps://www.diodes.com/assets/Datasheets/DMN10H220LK3.pdfThis is a tricky application, because lots of things are moving at once:- Vgs doesn't rise instantly, it ramps up at an (approximate) RC time constant of 2.9k + 670pF (Thevenin of R2 || R3, equivalent capacitance of Qg(tot) / Vgs(on))- Vds doesn't fall instantly, it ramps down at Id / C2 = dV/dt (or (Id - I_L), if the regulators turn on at some point before C2 is mostly fully charged)- Id isn't constant, it's rising (roughly as Vgs^2, once above Vgs(th))- Power dissipation isn't constant, it's rising, peaking, then falling (as Id ramps up, the Vds ramps down)It would be convenient to jam a straight line through most of these so we can use simple calculations (like we often can with SMPS), but there aren't many opportunities to do so; and they're all interdependent so any such approximation has knock-on effects.We might first look at an oversimplified bounding case, instead.Consider a typical charging current like 1A. 33uF charges to 84V in 2.8ms. If controlled at 1A, the MOSFET operating area follows a horizontal trajectory on the SOA plot (Fig.12), starting somewhere just right of the 100us curve, and moving left.In 100us (at 1A), Vds has dropped only 3V, so it's still on the top/right side of the 100us curve. Smoke shortly ensues.Well, that's not a great sign.What about gate drive? If it's not instant, maybe Id is a lot less than that.Well, 2.9k * 670pF = 1.9us, so the gate will come up pretty quickly, regardless of what the drain is doing. Id(pulsed) is 30A, maybe it's capable of more, who knows (not without destruction anyway, it would seem), but if we say it goes up to 30A in a couple microseconds, that'll pull the drain down in a hurry (30A / 33uF ~= 1V/us), but with 84V to go (so, as many us), again, we're not even in left field, we're way way out in the parking lot here.This is a terrible choice for a linear switch as the SOA curves are grouped so closely together at any time scale that would be helpful here (i.e. ~ms): not only is it a small die device (hence the short SOA times at much power, which the fairly small Qg and E_AS (only 1.1mJ!) agree with), it's prone to 2nd breakdown (the curves are slightly steeper above 10V 1A or so), and hardly has any voltage headroom either.We could still consider slowing gate drive (by putting a capacitor in parallel with R3, but mind to also put a resistor in series to the gate -- from the R2-R3-cap node to G, to keep the gate impedance high and resistive -- a capacitance across G-S is sure to cause oscillation!), but eventually we run into the problem that the regulators start up and carry load current, while the voltage is still ramping up, and power dissipation increases dramatically. Even if we run the MOSFET at a steady 18W (assuming water cooling to keep it at Tc = 25°C!), that's an initial 214mA, ramping up (along a constant-power curve, a hyperbola) to 1A at 18V (66V at VCC).A probably better option is to use the regulators' disable function: switch PWMD instead. Now, HV9910 doesn't claim a very impressive shutdown current: ~0.5mA. That's not great for battery use. Perhaps VIN should be gated as well, or instead. There's very little inrush current to the chip (its VDD regulator will charge C3/C4 pretty quickly). Maybe it (they) could even be directly powered by the opto?It may be worth noting that, at the instant of turn-off, the battery cable inductance flies back, and Q1 needs to be able to handle that. I'm guessing load current is fairly modest (under 1A??), and even for fairly long cables (a handful of meters?), 5 or 10uH let's say, so dI/dt must be less than (100V - 84V) / (10uH) = 1.6 A/us. A somewhat larger gate resistance would probably help out here, but since a larger MOSFET is needed anyway, it'll probably have enough Qg or Ciss to deal with that.(This is also, more trivially, solved by moving C2 to the input side. Beware hot-plugging (inrush) which may blow the fuse. Which... perhaps the soft start circuit is less a nice-to-have and more a functional requirement..?)Tim

author: exe
author_rank: Unknown
timestamp: February 19, 2023, 09:56:15 pm
content: Quote from: Siwastaja on February 19, 2023, 07:26:07 pmLook at the bottom of pcb (e.g., D3, ZD3 and Q3 drive K_SER relay):https://github.com/eez-open/modular-psu/blob/master/previous%20designs/bp3c/Images/EEZ_DIB_BP-3C_PCB_top.png

author: T3sl4co1l
author_rank: Super Contributor
timestamp: February 20, 2023, 12:16:54 am
content: Quote from: exe on February 19, 2023, 09:56:15 pmWas there a bypass cap on the coil VIN?

author: exe
author_rank: Unknown
timestamp: February 20, 2023, 07:30:02 am
content: Quote from: T3sl4co1l on February 20, 2023, 12:16:54 amNo, I don't think so, not at the pins (does it need it?). Probably there was a bypass cap at the supplying rail. Here is the schematic:https://github.com/eez-open/modular-psu/blob/master/previous%20designs/bp3c/EEZ%20DIB%20BP3C%20r2B3.pdf.As far as I remember it was failing on activating the coil (I could hear a click, but only once), not when releasing. But can't be sure, it's been two years ago.

author: Siwastaja
author_rank: Super Contributor
timestamp: February 20, 2023, 08:21:46 am
content: Quote from: exe on February 20, 2023, 07:30:02 amThat's your problem. Forget the relay and just think about the switching loop parasitics, from Vcc to GND, through the diode + transistor.The layout regarding transistor-diode loop is not too bad (although it could be better), but if the nearest capacitor is somewhere else, then your switching loop includes the Vcc and GND tracks all the way back to the capacitor. Fundamentally, your relay driver (transistor+diode) it's no different from a logic IC or MCU switching it output pin, and you putting the capacitor right next between its Vcc/GND pins as the datasheet instructs. Attached is a layout which I have never had problems with. Mid layers (full ground planes) are not shown. (I have used bidirectional TVS here which does the same as what you have, but with one component less.)

author: T3sl4co1l
author_rank: Super Contributor
timestamp: February 20, 2023, 09:08:39 am
content: Hmm, that would be a turn-off-only problem. Turn-on would have to be something like inrush (on a coil??), or, I don't know,extremelyslow gate turn-on (even so, R_coil is usually enough you'd have a hard time burning up a SOT-23), or maybe oscillation (driven by the same issue, too slow turn-on). But as you say, it was a while ago, maybe it was different. Also, possible (albeit an outside chance) the logic input was chattering or something, thus burning up multiple hits of avalanche, or incomplete turn-on/off, or etc.I also prefer a TVS/zener, from GND to drain, when clamping inductive loads like this, and a nearby supply bypass isn't available.One downside is the leakage current, which can be surprisingly high for certain purposes; my last example of that was a 5.1V zener on a battery-powered motor and solenoid application. Leakage was a couple tens of uA, I think? At room temperature of course. Compared to the couple uA the MCU was consuming in the power-down state, that's not fantastic! (We later had reason to change to diodes anyway: we moved to PWM drive for both outputs.) Another example: LED lights, which either glow very dimly (but definitely visibly, in low ambient conditions!), or flash slowly (as for some mains powered LEDs).(Which, if anyone's wondering, no it doesn't really do any good to ramp a solenoid. It still pulls in very suddenly, even if you try and release it just as it starts to move. At least, with the clamp diode, the current falling rate isn't fast enough to be able to do that. You can't get the timing right, either, not open loop -- the mechanical movement itself isn't consistent enough. At least on the part I tested. Oh well.)Anyways, this all isn't too relevant to OP's case, don't mind this mini-thread here.Tim

author: exe
author_rank: Unknown
timestamp: February 20, 2023, 09:28:54 am
content: Quote from: Siwastaja on February 20, 2023, 08:21:46 amI don't get it. We are talking about may be 20-30cm of wires (combined +5V and ground path). Do you think this much inductance could kill anything? I don't remember the part number, but those transistors have some 60V Vds and some noticeable Rds(on) (in the region of 1-3 Ohms) that would, I assume, dump the oscillations. I also thought that mosfets are somewhat rugged devices. Perhaps, it's time for a test if I have time and an inductor at hand...PS What I noticed, is that in the BOM there are 2N7002 variants that have vastly different Ids(max), from 115mA for 2N7002L (onsemi) to 380mA for 2N7002K (Diodes). But again, I don't remeber what I had populated on my pcb.

author: Siwastaja
author_rank: Super Contributor
timestamp: February 20, 2023, 09:59:12 am
content: Quote from: exe on February 20, 2023, 09:28:54 amOf course 20-30cm is a colossal problem. Even the very old rule-of-thumb says that bypass capacitors need to be within 1 inch.But is it the cause for the killed MOSFET? Possibly yes, possibly not. Maybe indirectly?I don't see the 20-30cm of wire creating a 60V spike on the MOSFET Vds. But maybe something couples to the MCU or gate.As you are not using a diode free from reverse recovery, accidentally turning the MOSFET back on quickly after turn-off would cause short circuit current to flow, as the diode is still conducting from the stored energy in the coil (Continuous Conduction Mode switching), and slow reverse recovery devices can't stop conducting quickly when the MOSFET turns on, so now you have both devices on, shorting out the supply, and without local bypass, storing quite a lot of energy in the 20-30cm of wiring inductance. Maybe this could even exceed the 60V rating. ½LI^2 is not much if I=100mA, but if it's a short circuit only limited by capacitor ESR and RdsON of the MOSFET, it could be several A, in which case the squared term starts to really hurt.Scoping the gate could reveal something like this happening. Even just a short excess pulse. With a schottky, it would be possible to turn the gate back on at any time (making PWM operation possible, too) without causing significant short circuit currents.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: February 20, 2023, 11:16:29 am
content: Quote from: Siwastaja on February 20, 2023, 09:59:12 amMeh, even if it's a 1N4007 (a shite one at that), I don't see this blowing from a relay, not at any sane (for a relay) frequency anyway. (Maybe above 10s kHz?) Reverse current isn't many times higher than forward, even with a fast change; the diode is conductivitymodulated, not a mechanical contact all-or-nothing.Bringing it back on topic -- whereas OP's example has 33uF, even with a (somewhat) bigger transistor, it's way too much (and also a much higher supply); a 1N4007 taking a good couple microseconds to recover, out of say 12V in this case (or whatever it was?), and drawing an ampere or so peak in the process, is about like a 0.25uF cap.In comparison, if the inductance was 200nH and peak current 2A, that's still only 0.4uJ. You're probably hard pressed to find a device with such a low avalanche rating (if it has one at all, which is the real hitch here I think). It's possible a 2N700xy might die from as little as one hit, but it seems unlikely to me.Heh speaking of 200nH, here's a waveform with a much harder short circuit:-20A/div, through a "hyperfast" current limiting fuse I made (nominally rated 30V 8A, with a response time in the 10s of us at -- I forget how much, 2x? 5x? overcurrent). This test was externally switched, using a whack of STP50N06s in parallel to dump a few 10s mF at 30V across the poor fuse. In the first 1us, current ramps up through the switch loop (~6" twisted pair); in the next 1us, device capacitances discharge and it goes into current limiting (the equivalent circuit is 9V to the gate, applied to a MOSFET with, what was it, 50mohm source resistance?, so it acts as a current limiter passively already, albeit at rather high currents). After the last 1us, the current sense comparator detects it and latches off. Note the upward tick in that final us, probably due to device tempco -- 30V at 50-60A is pretty far into any D2PAK's SOA,but turning off in less than 10us means it's perfectly healthy.I made this fuse to help power things like this, actually, but it's rather shite at it because the comparator is too sensitive and the current limit is too high -- it's not able to start into much load capacitance at all, even if you sit there mashing the "start" button. Let alone if the load has any current draw pulling the capacitance back down between presses!I later made one with a more modest current limit, and a couple ms maximum fault time; it's pretty comfy to use. Such a solution could be used by OP; even just the simple method of a source resistor limiting short-circuit current (while Vg is fixed, say with a zener clamp), combined with a bulkier transistor (say, IRF640?*), should do fine, albeit maybe at some expense to operating efficiency (that source resistor will drop a bit of voltage under normal load conditions; how much depends on the ratio of peak to average current you're willing to run at). (I used a '50N06, or something similar, for its 30V 10A operating range.)*None are rated for DC SOA today, but Fairchild's IRF640A was:https://datasheetspdf.com/pdf/283837/FairchildSemiconductor/IRF640A/1These classic parts don't usually receive a die shrink, so I wouldn't be too leery of using them this way. Beware of new and improved versions; significantly lower Qg, Ciss or E_AS are clues.I even made a current-limiting fuse based on a (proprietary? unique?) switching architecture, that's capable of dropping 30V 20A for up to 150ms at a time (and is bidirectional too, so can be used for AC), while using only two MOSFETs (the same ones as the 30V 8A hyperfast fuse, actually). I wouldn't recommend this approach to OP though; it's a bit of a bother to make the control circuitry. (I'm not aware of such a controller IC on the market today, unfortunately; I did it discrete(!).)Tim

author: exe
author_rank: Unknown
timestamp: February 20, 2023, 11:34:42 am
content: Tbh, without exact reproduction of the experiment it's now more of a guesswork of what happened.Having said that, I just hooked a small signal mosfet (XP231P0201) to drive a 6V relay. No bypassing, no diodes, long wires everywhere. I plugged it incorrectly as I thought it's an N-channel devices (hint: it's not). I'm "driving" the circuit by hand, I just connect wires. Anyway, it died instantly.I connected my oscilloscope. For some reason it doesn't always show spikes (idk why, I set to 100us/div, trigger at 66V, 10X probe). But when it triggers I've got 200-600V p-p. A bit of satisfaction as I don't like said mosfet (too much gate capacitance, 70-80pf according to transistor tester vs ~30pf typical value in the datasheet). I'm also a bit surprised, I thought mosfets can absorb some avalanche breakdown. But, I guess, the mosfet I have is too small for that.PS can I kill my scope probing this way?PPS no spikes when connecting relay, only when disconnecting.PPPS Set my scope to peak-detect mode, now it seems I've got a reliable triggering

author: jmelson
author_rank: Super Contributor
timestamp: February 20, 2023, 05:01:13 pm
content: Quote from: wasedadoc on February 19, 2023, 06:53:53 pmBINGO!This is the first error, but then the second one is that the 22K/3.3K resistor divider is giving a pretty slow rise/fall in the gate voltage of Q1, causing it to linger in the linear range for quite a while. Use a real gate driver chip there that can deliver several amps of gate current, both source and sink, to turn it on and off quickly. With 84 V available, lingering for even a few us in the linear range will cause failure of the MOSFET. C2 appears to be electrolytic, and hit with repeated 84 V square waves, it will blow its top within seconds. You CAN'T do what you have there. Charging/discharging that cap will give a peak current of 84 Amps assuming a 1 Ohm equivalent series resistance, and likely iit is much less, meaning the current will be higher!Jon

author: Picuino
author_rank: Super Contributor
timestamp: February 21, 2023, 03:43:03 pm
content: Quote from: smus on February 19, 2023, 09:11:01 pmSearch internet for: PCB Layout Design Switch Mode Power Supplieshttps://www.analog.com/media/en/technical-documentation/application-notes/an139f.pdfhttps://www.analog.com/media/en/technical-documentation/application-notes/an136f.pdfhttps://www.ti.com/lit/an/snva021c/snva021c.pdfhttps://www.ti.com/lit/ml/slyp762/slyp762.pdf

author: T3sl4co1l
author_rank: Super Contributor
timestamp: February 22, 2023, 01:50:30 am
content: FYI, I don't see anything overtly wrong with the PCB. It's not crazy frequencies or anything, and the currents are low. You can get away with quite a loose layout at these levels. (Consider all the <100W PSUs that are single layer boards! ...Consider all the ~500W PSUs that are, too..!)Quote from: jmelson on February 20, 2023, 05:01:13 pmThis works if the transistor can saturate* into the ~short-circuit current of C2. But that will not happen with the transistor shown, not by a long shot. So linear mode is forced regardless of Vgs(on) and speed, and SOA will be violated. If anything, slower is better -- give more time for the heat to spread out -- but as I said, I doubt that's reasonable with this device. A bigger MOSFET is needed, either way. (Another option is a two-stage turn-on, with two MOSFETs in parallel but one delayed, and the faster one with some series resistance so it can saturate* properly while the resistor does most of the work (with a capital 'W', since we're putting energy into the cap!). (The resistor can be cheaper than a MOSFET of the same energy/power handling; but maybe not cheaper than two plus sequencing logic?)*Voltage saturation (ohmic region), not "current saturation". I'd prefer people move past the archaic " saturation" distinction between BJTs/logic and FETs. But for now, I always add this note to be clear.Tim

author: Siwastaja
author_rank: Super Contributor
timestamp: February 22, 2023, 11:15:56 am
content: Yeah, the amount of energy dissipated in the MOSFET pre-charging the C is always the same - equal to the energy finally stored in the capacitor, 1/2 CU^2, so slower would be better. (This is assuming zero ESR in rest of the circuit. If this is non-zero, then part of this energy is spent elsewhere, and rest in the MOSFET.)1) Increase ESR of the C, if rest of the circuitry can accept this, so that larger part of 1/2 CU^2 is spent in that capacitor, and less in the MOSFET2) Add series resistance to the whole circuit, if that is acceptable (same effect as 1)3) Create a separate pre-charge path with series resistance, to be bypassed with the main MOSFET after low enough VDS is detected (needs control, and may not be possible if the main load is dumb and tries to start pulling power during precharge, which never finishes then.)4) Just use larger MOSFET which can take the precharge energy.

author: Picuino
author_rank: Super Contributor
timestamp: February 22, 2023, 11:17:59 am
content: I have mounted power supplies on a single layer development PCB with no problems.The important thing is not the layers, but the current loops that are formed and that should be as tight as possible.On the other hand, the conductors with sudden voltage changes (mosfet drain and associated connections) should have the smallest possible area to reduce their parasitic capacitance and so that they do not act as an antenna.Another important point highlighted in the documents I have linked is that ceramic capacitors with low ESR should be used in high current loops to limit voltage drops.

author: Picuino
author_rank: Super Contributor
timestamp: February 22, 2023, 12:09:14 pm
content: You can replace R3 with a 15 volt zener diode, switch R2 between the phototransistor collector and Vcc and add a capacitor between the phototransistor collector and ground.With that the transition of Q1 will be faster and you avoid burning Q1.

author: exe
author_rank: Unknown
timestamp: February 22, 2023, 04:48:57 pm
content: Quote from: T3sl4co1l on February 22, 2023, 01:50:30 amAh, so that the slower transistor is effectively shunted by the resistor and thus has to operate over smaller voltage range? That's a nice trick. I think I did something similar in my bridge rectifier when mosfets were shunted by diodes:https://www.eevblog.com/forum/projects/semi-active-bridge-rectifier/. Diodes limited voltage drop across mosfets, which helped with SOA.

author: wasedadoc
author_rank: Super Contributor
timestamp: February 22, 2023, 05:20:36 pm
content: Quote from: exe on February 22, 2023, 04:48:57 pmNo. The intention is that by the time the second MOSFET is turned on C2 has been charged and that second MOSFET:1. Does not have to carry C2 charging current.2. Has very little voltage across it.3. Only needs to handle the vast majority of the operating current of the two dc-dc converters.Prior to that the first MOSFET is handling the C2 charging current limited by the resistor and has very little voltage across it as most of the 84 Volts is dropped across the resistor.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: February 23, 2023, 02:56:05 am
content: Quote from: exe on February 22, 2023, 04:48:57 pmSomething like that. Well, that was extra steps, the body diode is there already.* But if the diode were another MOSFET (with resistance), and forward not reverse current, then exactly that.It's a hell of a lot easier to switch into a capacitor when it's doing 0.84V instead of 84.*More than a few times I've tested a sync rectifier and noticed it getting really hot really fast. Then I turn on the VCC (conveniently supplied from a bench PS for the test) and all is well.Quote from: wasedadoc on February 22, 2023, 05:20:36 pmNot quite -- the capacitor will still be charged further.There is no delay for load current here, so it will be doing whatever it does, probably in a fraction of a millisecond. (This could perhaps be set by the VDD bypass: it has to charge past its UVLO threshold first, and apparently current limit is around 5mA. Sadly they don't specify the VDD pin as a full and proper linear regulator, though it evidently is one. The LD pin could also be used with an RC as a soft start. Which is maybe a nice-to-have feature anyway, say if one finds it useful to replicate the slower turn-on of incandescent, or even slower.)Therefore, the resistor must be small enough to carry full load current (and accounting for the higher current draw at lower Vin because this is a SMPS load!), and at a low enough voltage drop that the low-resistance switch does not exceed its SOA delivering the remaining charge.For example, it could be something like 84V 1A maximum load, and let's say a 10 ohm is used, we have the intersection of a line and a hyperbola. The solution of which is:$$ V = \frac{V_\textrm{in}}{2} \left( 1 + \sqrt{1 - \frac{4 P_o R}{{V_\textrm{in}}^2}} \right) $$In this case, R = 10Ω, Vin = 84V, Po = 84W, so the steady-state across the resistor will be around 11.6V. At 33uF, this is only 2.2mJ, which the original transistor will easily handle; and at low enough voltage that it's clear of 2nd breakdown so you get the full power rating too. (Note that 10V, 3A for example is on the 1ms SOA curve, which is 30W for 1ms or 30mJ. The 2nd breakdown "kink" on this curve is around 16V.)Note that rapid switching doesn't help anything, even at this (much gentler) stage: supposing the transistor is still 0.22 ohms at such currents (doubtful), and the capacitor is equal ESR, that would be 26A peak, if the gate is switched on as fast as possible (which could be within a couple nanoseconds). This peak would be reached in -- let's say minimum 20nH loop inductance, so an L/R time constant of 45ns so it'll be carrying a lot of current by 100-150ns. Now we have an RLC series network with R = 0.44Ω, C = 33uF and L = 20nH; Zo << R so we can ignore L for the remainder, and RC = 14.5us so it will charge in a couple 10s of us. But, the SOA is 20A for only 10us, so we've justblown the transistor from switching it too hard.So, y'all please stop suggesting that the MOSFET can be switched hard enough to solve the problem: it can't. It's a teeny transistor. I don't know how many times I have to reiterate this......Note that 20nH assumes local input bias, which isn't the case; we could have up to ~3uH ESL (i.e., battery cable inductance -- this is a few meters worth of twisted pair) and still have a well-damped system, and the peak current will be only somewhat less (15-20A?). This is still marginal at best, and we can't depend upon undefined cable lengths.A rising gate voltage, on the second parallel MOSFET, of between ohh say 80 and 5000us, should be safely within the SOA.The 10R + 33uF precharge has a time constant of 330us so needs a good ~1ms to approach the 11.6V steady state value. It may well be the case that a fast (~1us?) gate time constant for the first one, and slow (~5ms) for the second, is adequate -- no sequencing logic required. That is, the circuit as already shown, but with another MOSFET added in parallel, and one gets a series drain resistor; both have independent gate divider resistors, one using larger resistance and extra G-S capacitance.This is still kind of a lot of board area compared to OP's original sample so it might not fit, but if the outline can be expanded slightly, it'll do just fine.Oh and the resistor shall have a pulsed capacity of 0.1J or ~350W peak (either way, for t_p ~ 1ms). A small wirewound (maybe even an SMT one) should handle that fine.Tim