Mosfet killed by back EMF - Page 1
Forum section: electronics / beginners

author: generic_username
author_rank: Regular Contributor
timestamp: January 19, 2023, 01:06:22 pm
content: Hi,a couple of years back I worked on a prototype for a heater control.the heater was used in a wind tunnel for another prototype.In total it was 10 heaters with varying from 1 to 3 kW.The heater was moslty ohmic with a strong PTC behaviour, thus a good control was needed.Our supply voltage was 200V DCwe were going for a PWM of 100 HzMosfet (IPW65R019C7) used as low side switchSo the prototype was build an after an initial function test it was tried to run on 200V and the FET was instantly destroyed.Soon the problem was discovered that we had a spike when switching off the Mosfet, screenshot attachedThis was the first time we used a Mosfet for voltage > 60V DC but we build different PWM controls for similar heaters.The old projects never required a flywheel diode so we didn't put it in (thinking the load is only ohmic)But was we had to use long cables 10m 2.5mm² / (32 feet, AWG 14) the inductance was around 30 nH.So we added a flyfwheel diode between drain and source and did some tests.BIG BRAINFART HEREwe added a diode between drain and source in case the internal diode couldn't handle the spikeSeems this did little to nothing as we killed the next FET.next we tried adding a diode in over the load 1 attempt before the cables and 1 attempt right after the cables, both attempts did not solve the problemIn the end we got it working when we added a 1.5KE480 Zener diode between drain and source.So years later I'm still wondering why the flywheel did next to nothing.Any ideas how we could've solved this issue differently?kind regardschanges in italic

author: RoGeorge
author_rank: Super Contributor
timestamp: January 19, 2023, 01:39:23 pm
content: Quote from: generic_username on January 19, 2023, 01:06:22 pmNot sure what you mean by a flywheel diode between DS. The datasheet shows there is an internal diode already between DS:https://www.infineon.com/dgdl/Infineon-IPW65R019C7-DS-v02_01-en.pdfMaybe that's why yet another diode doesn't change much.About the exact cause that was keep destroying the MOSFET, hard to tell from that piece of schematic alone. See the "Application Notes" section from Infineon, they might have the answer that applies to that setuphttps://www.infineon.com/cms/en/product/power/mosfet/n-channel/500v-950v/ipw65r019c7/

author: Brianf
author_rank: Regular Contributor
timestamp: January 19, 2023, 01:44:05 pm
content: What's the cold resistance of the heater bank? Heaters often take a substantial surge current at power on.

author: Siwastaja
author_rank: Super Contributor
timestamp: January 19, 2023, 01:45:34 pm
content: "So we added a flyfwheel diode between drain and source and did some tests. "In parallel with MOSFET? It's in the wrong place. It needs to be over the load. And physically close to the MOSFET so that it bypasses as much inductance as possible.The idea is simple once you get it: when current flows in an inductance, the inductance stores energy, trying to keep the current from changing. You need to close the loop for current by adding diode in parallel with your inductance (heating element plus any wiring running to it). This way, it won't generate more voltage than the Vf of the diode, since diode starts conducting when that happens. Heater will use that stored energy to heat (minus the small diode drop). Current ramps nicely to zero in a linear ramp.

author: generic_username
author_rank: Regular Contributor
timestamp: January 19, 2023, 02:14:30 pm
content: Quote from: Siwastaja on January 19, 2023, 01:45:34 pmyes, I mixed up 2 things we did back then.first was to add another diode between drain and source, in case the internal diode couldn't handle the voltage spike.second we added a diode as close to the load as possible (my bad,had a brainfart and wrote some garbage) with the same result of the FET dyingchanged this in the OP as wellQuote from: Brianf on January 19, 2023, 01:44:05 pmone of the strongest heater had 4.94 ohm at 23°C and 12.35 ohm settledchanges in italic

author: Terry Bites
author_rank: Super Contributor
timestamp: January 19, 2023, 04:25:46 pm
content: Video killed the radio star- so they say.Cable capacitance will create a ll resonant circuit. I'm guessing about 1n.This could create voltages which exceed the FETs ratings.The waveform looks like clipped ringing. It short duration sugests a low Q circuit.A "freewheel" diode needs to go across the load not the FET.

author: Manul
author_rank: Super Contributor
timestamp: January 19, 2023, 04:42:42 pm
content: That is a powerful mosfet. They can eat quite high energies while in avalanche (high voltage DS breakdown), it is not so easy to kill them. With low frequency PWM of 100 Hz I doubt it is a problem. What is a problem is driving these mosfets properly, because gate is very sensitive. Ringing on a gate or some kind of ground bounce can kill them easily. Also gate drive should be reasonably high current when total gate charge is so high, otherwise the mosfet might get cooked during transitions (high switching loss).

author: jmelson
author_rank: Super Contributor
timestamp: January 19, 2023, 05:25:12 pm
content: I don't quite understand your scope trace. it shows a pulsing behavior at about 2 us rate, and about -250 V from the top baseline. Perhaps this was due to bounce on the source terminal causing the transistor to oscillate. When the transistor was on, any resistance or inductance in the source terminal would cause the source (inside the transistor) to rise, and then when shut off the source would drop relative to the gate. If the source voltage drops below the threshold voltage, the transistor might turn on weakly and put it in the linear mode. the gate driver needs to be "stiff" ie. have low output impedance and be VERY near the power transistor, with low inductance connectios to the source terminal. Operation of these power transistors in the linear mode can cause them to fail shorted in microseconds. That business where the trace seems to taper off downward very slowly is likely where the transistor developed a local hot spot and went resistive, then the damage spread across the die.Jon

author: Doctorandus_P
author_rank: Super Contributor
timestamp: January 19, 2023, 05:38:06 pm
content: I also have a suspicion it has something to do with the gate. Either a voltage spike getting above it's 20V limit, or insufficcient drive voltage which causes the FET to dissipate a lot of energy and get too hot. A15V or so TVS diode directly between Gate and Source can help prevening overvoltage, for insufficient gate drive... use a gate driverAnother possibility is the cable inductance. To remedy that, use a diode between the Drain and the power supply close to the FET. The Idea is that when the Fet turns off, the current can circulate a bit though the load and the cables. The fet can withstand quite some Avalache energy though so this seems less likely.

author: Bud
author_rank: Super Contributor
timestamp: January 19, 2023, 07:06:08 pm
content: Quote from: jmelson on January 19, 2023, 05:25:12 pmI do not understand it either. Where is the zero volt level reference ?

author: floobydust
author_rank: Super Contributor
timestamp: January 19, 2023, 07:29:54 pm
content: Heating elements have significant inductance, they are not purely resistive. For some reason EE's don't understand that. Just measure it already. Most are wound coils.Then add on the wiring inductance, and at the high currents involved it's quite a bite. 32' is not 30nH, redo your calculations or measurement, that is out to lunch.You would typically use a back-EMF diode or snubber to tame that. So your diode is an OK solution.Heating elements have significant inrush current, their cold resistance can be many times less than their hot resistance. How long this is true depends on the construction, some use fine others thick nichrome wire which can warm up in msec or seconds, and some glow hot orange, others do not. So sizing the switch (mosfet) exactly to the heater (hot) current is not a good idea, always over design it.

author: Doctorandus_P
author_rank: Super Contributor
timestamp: January 20, 2023, 01:20:48 am
content: Quote from: floobydust on January 19, 2023, 07:29:54 pmYah, however, the used fet is an IPW65R019C7https://www.infineon.com/dgdl/Infineon-IPW65R019C7-DS-v02_01-en.pdf?fileId=db3a30433e78ea82013e791a9f140093... which is rated for 700V and 62A (@110c), some 500A pulsed and that should be plenty for a mere heater @200Vdc and 3kW.

author: floobydust
author_rank: Super Contributor
timestamp: January 20, 2023, 01:52:14 am
content: Quote from: Doctorandus_P on January 20, 2023, 01:20:48 amQuote from: generic_username on January 19, 2023, 01:06:22 pmI'm not sure if this is total or per heater? Assuming the gate drive voltage is proper and 10nF can be driven, the part looks suitable.

author: Siwastaja
author_rank: Super Contributor
timestamp: January 20, 2023, 06:27:48 am
content: Quote from: generic_username on January 19, 2023, 02:14:30 pmStill wrong place. The load (heating element) is only part of the total inductance. Wiring inductance acts in series with it. You did not bypass the wiring inductance.It's all about layout. Place the two switches (transistor + diode) right next to each other. Furthermore, add a capacitor between Vsupply and GND right next to the two switches, so that you also bypass the input wiring inductance. You can get the parasitics down to a few dozen nH, very little energy stored. Then you can slow down gate drive a tad by adding series resistance to gate if you still have voltage peaking.Other failure modes include MOSFET with too much Rds_on, or overvoltage on gate.

author: generic_username
author_rank: Regular Contributor
timestamp: January 20, 2023, 09:06:54 am
content: Thanks for the replies.Regarding the gate driver, I did examine this voltage and couldn't see any issue with the voltage there, it seems I don't have any scope pictures of this measurement so if I can get my hands on a working PCB (project was years ago) I will redo the measurement and share a screen.attached shematic of the gate driver and how the 15V Gate voltage was realised.Quote from: floobydust on January 19, 2023, 07:29:54 pmThe heating element in question is not like traditional heating elements.It is a thin film of printed carbon on a flexible carrier, in this case a thin FR4 witch eched copper serves as the contacting pad and traces to thin "squares" of the carbon.depending on the "paint" used the PTC effect is stronger or weaker as well as the temperature where the effect would start to be significantI attached a poorly drawn curve as I don't have acess to the original cuves (the last part where the PTC effect goes down again can occour when the maximum temperature is exceeded, but the behavior of the elements is not my expertiese).So the heating elements are apart from the copper traces only resistive in nature unlike classical wire wound heating elements.Quote from: floobydust on January 20, 2023, 01:52:14 amyes this is the power per heater, depending on the size of the heater, the heating elements here used to be between 1 and 2.5 W/cm² the power given here is when the elements are settled so they reached their temperature and have higher resistance, however the problem with theQuote from: Manul on January 19, 2023, 04:42:42 pmI actually did run a test where I just turned the FET on without any PMW and then turned it off, still one single turn off event was sufficient to kill the FETI did rerun this and turned the FET on and then turned off the Power supply to see if the FET can still be switched on, as the FET was always conductive dying (meaning always on)One thing I forgot to mention is that the dying of the FET occoured at around 130V, below this no FET died.Quote from: Siwastaja on January 20, 2023, 06:27:48 amAdded a system overview of the PCB.I added the layout, the part where we shorted the connector "J200" in the screenshotI'm not to happy with the layout

author: jmelson
author_rank: Super Contributor
timestamp: January 20, 2023, 05:32:28 pm
content: You need to minimize loop area of Gate-L-1, HV-common and Current-1 on the PC board. I couldn't quite trace these on the PC board layout, but minimal loop area is quite critical to cut down on these inductances. Oh, if you are using a MAGNETIC current sensor, it would be WAY better to put it in the drain lead. Some of those sensors with Hall elements have a pretty big ferrite core to collect the magnetic field. That adds inductance, and you DON'T want that in the source lead. This one seems to not have room for much ferrite, but the datasheet gives NO inductance measurement. But, anyway, routing the source terminal through the sensor adds inductance, and it should not be done there, but in the drain.Jon

author: David Hess
author_rank: Super Contributor
timestamp: January 20, 2023, 07:24:03 pm
content: Quote from: Manul on January 19, 2023, 04:42:42 pmThe gate is a possibility but also easy to protect with a shunt zener.Big MOSFETs are very rugged, however they can still be quickly destroyed by exceeding their dV/dT rating which will activate the parasitic bipolar transistor. TheSiliconix MOSPOWER Applications bookhas good discussions of these issues.

author: xavier60
author_rank: Super Contributor
timestamp: January 21, 2023, 02:48:43 am
content: The 10Ω Gate resistor will be causing unnecessarily fast turn off time for that application.As well as looking at the Dran waveform while running at a safe HT voltage, the Gate turn off plateau can be observed to get an indication of the turn off time. Although the presence of the current sensor might cause some distortion.https://www.jedec.org/standards-documents/dictionary/terms/gate-plateau-voltage-vgsplTo implement a flyback clamp diode, a HT connection needs to be made to the PCB with bypass capacitor to ground. Or add some sort of snubber.

author: MrAl
author_rank: Super Contributor
timestamp: January 21, 2023, 10:15:00 am
content: Quote from: generic_username on January 19, 2023, 01:06:22 pmHello there,This discussion addresses most of the problems associated with the turning on and offof a power switch such as a MOSFET transistor. See the attachment for the variousdiagrams.FIG 1:The circuit drawn with the included power supply filter cap.FIG 2:Showing the various inductances that are also in the circuit.FIG 3:The circuit drawn with the load shown in a slightly different position but that's the same as FIG 2except the inductances are lumped.FIG 4:A boost circuit where the input 200v is converted to an output of 1000v.Noting the position and polarity of the diode, we can see that when thetransistor turns off the diode conducts and supplies current to theload (the load is now in the circuit where it would be for a boost circuit.What this shows is that the back EMF spike appears as a positive voltageat the drain caused by the back EMF of L1 and L2 and also a negative voltageat the source due to L3. This means the transistor sees a very high spikeacross the drain and source. This allows us to determine where the backEMF diode should go, and if it is even enough to protect the transistor.FIG 5:A typical diode placement that may seem to solve the problem, but due tothe analysis of FIG 4 we can see that the diode will not conduct whenthe positive spike appears at the drain, so that diode does not helpfor suppression of the back EMF.The diode will help with another problem that is unrelated to the back EMFhowever. That is, when the internal diode conducts it has to dissipateheat, and that is heat that is inside the MOSFET. By placing a betterdiode outside of the package some of the heat is moved to outside ofthe MOSFET and thus the transistor does not heat up as much. Becausewe can also use a lower forward voltage diode we can also reduce thetotal dissipation of the circuit. That's if the diode ever conductshowever, and how often and for how long it conducts. In a circuitlike this it may never conduct or conduct only for a short time.In a transistor bridge circuit it would be more important.FIG 6:Here we look at placing the diode in a more useful position in the circuit.Placing the diode where D4 is drawn, the problem now is that we have not yetbeen able to bypass the effects of L1, L5, L6, and L7, so that's not thebest place. By placing the diode where D3 is drawn, the problem now isthat although we have bypassed the effects of L5 and L6 we have not yetreduces the effects of L1 and L4, and even L3. To do that we have toadd another small circuit called a snubber.FIG 7:This drawing shows the diode D3 in place because that helps with some ofthe spike energy after some small time period, while the snubber absorbsthe fast rising leading edge of the spike plus it is also connected veryclose to the transistor drain and source as to not introduce significantinductance which would also reduce the effectiveness of that snubber.We should still talk about another important issue but the conclusion sofar is we are now able to protect the transistor against high drain tosource spikes due to D3 and the snubber. The next issue is oscillationin the gate to source circuit due to L3.FIG 8 shows the current flow in the source to drain circuit. To turnon we see I1, and when turning off we see I2. The problem is that L3causes the voltage between the internal gate and source to vary, whichin turn causes the MOSFET to turn on and off repeatedly, and thisappears as ringing at the drain. To mitigate that we can place aresistor in series with the source, and it looks like that has beendone already. The size of the resistor is of concern because we wantto reduce the oscillations, but that resistor also slows the transistorswitching speed down. This means we have to make a tradeoff when choosingthe value of the source resistor.The nice thing is, this resistor also slows down the fall time of thedrain, and the means instead of a faster falling wave it will be moreof a ramp, and a ramp excitation with an inductor means lower backEMF also. That relaxes the requirements on the snubber circuit a bit.With a gate resistor alone however we introduce another problem, and thatis the turn on time of the drain to source. It would not hurt if wecould turn on faster if needed, so it would be nice if we could turnon fast and turn off slower. This brings us to another circuit thatallows us to use an asymmetrical drive to the gate. Another fast diodein parallel with the resistor allows a higher current gate drive forturning on and lower current for turning off. Thus a diode connectedwith the cathode on the gate and anode on the other side of the resistorallows us to obtain this kind of drive circuit without too much effort.In some cases we may want to slow down the turn on also although lessthan we slow down the turn off, so we can use a different valuerresistor in series with the diode.Unfortunately, we are not exactly done yet as there is one more issueto talk about. That is the speed at which all the inductor energydecreases versus the switching frequency. Often times we want toreduce the energy in the load to near zero when the transistor turnsoff, which allows the PWM technique to work correctly.Now since the load also may have inductance, we have to think aboutthat energy too. From a little circuit theory, in order to reducethe energy in an inductor quickly (before the next switching cycle)we have to allow the voltage to go higher while it is discharging.The energy is sometimes looked at in terms of volt seconds, and thatmeans that the higher we allow the voltage to reach, the faster theenergy is dissipated. The diode D3 unfortunately has a low forwardvoltage so it would mean the energy dissipates slower, meaning theswitching frequency can not be too high. By using two diodes inseries, we increase the drain source voltage slightly while allowingthe inductance energy to dissipate faster. Using three diodes evenfaster, etc. This means that if we can allow some extra voltageacross the drain to source we can get the energy of the inductanceto dissipate faster, meaning we can use a faster switching frequency.This may or may not be of concern because some circuits just have toturn on once and not turn off again for several seconds or even hours.That concludes the basic description of how the power switch circuit works.To sum up here is a short list.The back EMF diode placement should be as shown as D3 in FIG 7. The totaldiode or diodes (or other component) voltage should be chosen withrespect to the required switching speed and energy dissipation in theload inductance.A snubber circuit is required when inductances cause a high spikeenergy, and must be placed as close to the transistor drain and sourceas possible with the shortest leads possible. We are talking twoinches if possible.Asymmetrical drive helps to reduce gate to source oscillations whichimproves switching losses. This circuit should be close to the gateand drain terminals. A separate lead from the driver ground terminalto the source terminal should be used to avoid the problems with theextra current through the source from the power supply capacitors.

author: Siwastaja
author_rank: Super Contributor
timestamp: January 21, 2023, 10:34:28 am
content: In above drawing, Fig.6 is helpful: start from there, minimize L3, L4 and L1, and note that L3 and L1 go all the way to the DC link capacitor drawn, so minimize distance to the cap, too, not just the two semiconductors.Fig 7 shows an RCD snubber, but I would primarily go in this order:1) minimize stray inductance L3, L4, L12) Increase gate series resistance to slow down switching (heater is PWM'd at low frequency, so switching losses do not matter)3) Add simple RC snubber over the MOSFET (possibly another over the diode) - R and C in series. (heater is PWM'd at low frequency, so RC snubber losses do not matter)You can very likely stop after 2).