MOSFET inductive kick causing alot of heat
Philldapill
Posts: 1,283
First off, this isn't exactly Parallax product related, but I'm getting desperate and I know there are a few wizards on here. I need some serious help, as this problem is going to cause me to go to a psych ward if it can't be fixed. I mean that literally...
Now, here's the problem in a nutshell. I have a switching circuit that uses some MOSFETs to do the switching. The fets are driven by a top-of-the-line mosfet driver - UCC27322. The driver kicks these fets on hard, and off hard. I have a big fear that this might be my problem... in a not so straight forward way. The problem on the surface, is that my fets get pretty hot. I've determined that they are NOT getting hot due to excessive current or slow transition times. When I have the switcher just remain on 100% duty, I am drawing nearly 40 amps through 6 fets. They get slightly warm, but nothing major.
With the steady state I^2 * R out of the question, I have also determined that slow transition times are also·NOT giving me a headache. In fact, I believe the opposite to be true. I think the fets are turning off so quickly, that even the tiniest inductance in my load(parallel plates dipped in water/electrolyte), is causing a large voltage spike. My fets have an internal 55V Zener diode, which I believe is catching this inductive EMF. This, in turn is causing the excess power dissipation in the fets at higher frequencies.
My solution to this, was to add a schottky diode between the drain of the fets, and the positive supply(a car battery). This seemed to help a little bit, but not enough. My next solution, was to add a large 1000uF capacitor between the source of the fets(GND), and the positive supply, with the diode still in place. This keeps the fets nice and cool at high frequencies, but my capacitor not overheats and goes boom, or the legs melt...
Basically, my lateout goes like this. When the fets turn off, the inductive spike·goes through the diode, and into the capacitor, charging it up a little. The reason for the capacitor is my lack of wiring back to the source - it provides a shorter route for the induced current to flow, slowing the di/dt. As the fet remains off, the capacitor discharges back to the supply, and goes back down to the nominal 12V of the supply. The problem with the capacitor·arises from the large charge/discharge cycle allowing lots of current to flow back and forth through the legs, heating them. In addition, the capacitor is also acting as a buffer for the supply, discharging even lower if the voltage sags when the fets turn on - more discharge current.
My next though, is to add a small inductor between this capacitor and the positive supply. This way, instead of doing a rather large discharge to supplement the supply, the inductor slows down the discharge. I don't have any inductors like this on hand, so I'll wait and see...
With all that explained, and my lack of circuit diagrams(on public computer), I'm hoping someone can pull a bunny out of their hat and think up a solution to all this. Any thoughts?
Now, here's the problem in a nutshell. I have a switching circuit that uses some MOSFETs to do the switching. The fets are driven by a top-of-the-line mosfet driver - UCC27322. The driver kicks these fets on hard, and off hard. I have a big fear that this might be my problem... in a not so straight forward way. The problem on the surface, is that my fets get pretty hot. I've determined that they are NOT getting hot due to excessive current or slow transition times. When I have the switcher just remain on 100% duty, I am drawing nearly 40 amps through 6 fets. They get slightly warm, but nothing major.
With the steady state I^2 * R out of the question, I have also determined that slow transition times are also·NOT giving me a headache. In fact, I believe the opposite to be true. I think the fets are turning off so quickly, that even the tiniest inductance in my load(parallel plates dipped in water/electrolyte), is causing a large voltage spike. My fets have an internal 55V Zener diode, which I believe is catching this inductive EMF. This, in turn is causing the excess power dissipation in the fets at higher frequencies.
My solution to this, was to add a schottky diode between the drain of the fets, and the positive supply(a car battery). This seemed to help a little bit, but not enough. My next solution, was to add a large 1000uF capacitor between the source of the fets(GND), and the positive supply, with the diode still in place. This keeps the fets nice and cool at high frequencies, but my capacitor not overheats and goes boom, or the legs melt...
Basically, my lateout goes like this. When the fets turn off, the inductive spike·goes through the diode, and into the capacitor, charging it up a little. The reason for the capacitor is my lack of wiring back to the source - it provides a shorter route for the induced current to flow, slowing the di/dt. As the fet remains off, the capacitor discharges back to the supply, and goes back down to the nominal 12V of the supply. The problem with the capacitor·arises from the large charge/discharge cycle allowing lots of current to flow back and forth through the legs, heating them. In addition, the capacitor is also acting as a buffer for the supply, discharging even lower if the voltage sags when the fets turn on - more discharge current.
My next though, is to add a small inductor between this capacitor and the positive supply. This way, instead of doing a rather large discharge to supplement the supply, the inductor slows down the discharge. I don't have any inductors like this on hand, so I'll wait and see...
With all that explained, and my lack of circuit diagrams(on public computer), I'm hoping someone can pull a bunny out of their hat and think up a solution to all this. Any thoughts?
Comments
You do mention that the drains are grounded, so perhaps I can assume that the load is between the supply (car battery) and the drain of the FETs. Or perhaps not.
In any case, you should bag the idea of the added inductor, which will only cause ringing. I'd bag the capacitors too, or use much smaller ones. Are you using capacitor types with low equivalent series resistance (ESR)? Apparently not, if they're heating so extremely.
My first working hypothesis would be that your circuit is ringing at turnoff. As you describe your load, it seems to me that it may be rather capacitive, rather than inductive, anyway. Did you connect the kick diodes directly across the load, or did you connect them to the battery terminals? If the battery is some distance away (a couple feet would be "some distance"), that would add wire length, and therefore inductance, in the wiring.
Don't add any more inductance. Get rid of the capacitors. Connect the kick diode directly across the load, with short fat wires. Make sure yo don't have a lot of inductance in the ground circuit, either. If all that doesn't work, perhaps we can come up with other ideas. It's not clear from your prose that the kick diode is across the load at all, but that's where it belongs. Report back.
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· -- Carl, nn5i@arrl.net
Also, contemplate slowing down the gate drive to the FETs. If the sharp turn-off edge is causing you woes, well, don't do that.
There's obviously a balancing act between switching losses and flyback losses, but it's (probably) clear that you're right over at the flyback end at the moment.
For better answers, post better questions - schematics, drawings, photos, scope plots...
Steve
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· -- Carl, nn5i@arrl.net
If it is a just a dc PWM maybe throw a diode in series with the FET then a free-wheel diode across the load to catch the inductive feedback.
I have no idea if i am on the right track at all. Just some ideas.
Thanks guys for the input.
@Carl, I think the schematic will clear some things up. Now, as for the "capacitive load" - I highly doubt it. Granted, there may be SOME capacitance created due to the parallel plates, but I don't think this would cause the inductive reactance and the inductive "kick" upon turn-off. The mosfets are indeed hitting their voltage limit. I can see this on the scope - upon turnoff, the voltage will shoot way up, and instantly level off at the 55V mark - the zener voltage. The voltage stays level, nearly proportional to the duty cycle, then goes back down, oscillates for just a bit, then goes back to 12V.
Now, for an explaination of the circuit. D1 is actually a high speed schottky - not a 1N4004. The diode is only rated for 25V, so I will change this diode out for a better one on the final design. So, what happens on turn-off, SHOULD be that any inductive kick goes through the diode, into the capacitor and charges the cap up slightly. This charge build up should be able to bleed off via L_LOWPASS, an inductor. This inductor is there so that upon turn-on, the capacitor does not instantly discharge, then charge right back up during turn-off. i.e., a low pass filter. The capacitor is simply a buffer for the inductive kick. The R_LINE and R_SUPPLY are there to just show the stray resistances through my crappy connection points(yes, just jury rigged atm).
The Analog_Supply capacitor is to buffer the voltage for the analog circuitry that drives the whole PWM signal for the mosfets and mosfet driver. The inductor is there, again, to act as a low pass filter so that the capacitor doesn't charge/discharge during switching events. This part works very, very well.
The circuit is actually working rather well at lower frequencies. I can pull a solid 30A with virtually NO heat from any component. At higher frequencies, that's another story(20kHz-100kHz).
@Tesla, I don't think a diode would work. My goal here is to reduce any power loss. Even a schottky diode at 0.35V, 30A would be dissipating over 10W. At the high frequencies, the driver is fine. I have 4.3 ohm resistors between the driver and gates.
@kwinn, yes, I agree, there is probably SOME capacitance, but we are talking nF - probably no where near a single uF. I have the inductor, L_LOWPASS,·in there to attenuate any large rushes of current.
@SteveW, the high freq. is a requirement of the device. I'd REALLY rather not slow the transition times for the mosfet gates. My goal is to get rid of any power dissipation - not just get rid of one form, only to give rise to another(I^2R). I'd love to post some scope plots, but I don't have access to a camera atm...
So, anymore suggestions?
Post Edited (Philldapill) : 4/12/2009 1:12:40 AM GMT
Across the FET's D-S --
http://www.onsemi.com/pub_link/Collateral/SMF05C-D.PDF
BTW, the mosfet Part # is STD60N55. These already have a built in Zener diode which acts as a TVS in a way. These Zener's are what is absorbing the inductive kick and causing the mosfets to heat up - if I'm not using the diode/capacitor pair.
@Tesla, oh yeah, I'm pretty sure the caps are rated for at least 3kHz... However, when the frequency goes up, so does the rms current(more switching events = more charge/discharges).
Post Edited (Philldapill) : 4/12/2009 3:09:24 AM GMT
If your switching frequency is in the megahertz range, I wonder if you're suffering from reflected power due to an impedance mismatch. This is a common issue with transmitting antennas, wherein extreme mismatches can destroy final amplifier stages, and for which matching networks exist in profusion. If this is a remotely realistic hypothesis, perhaps you could borrow a reflected power meter from a ham radio buddy to investigate.
Beau? Carl? Am I off base here?
-Phil
Post Edited (Phil Pilgrim (PhiPi)) : 4/12/2009 3:17:32 AM GMT
I'm waving my magic wand, hoping Beau will appear and chime in, as well. I could be wrong, but it seems like Beau has a bit of experience with power electronics.
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· -- Carl, nn5i@arrl.net
Post Edited (Carl Hayes) : 4/12/2009 4:01:09 AM GMT
The diode is there to give the current an alternate path through the capacitor. The charge in the capacitor is then bled off through the inductor acting as a low pass filter. Due to the long wires leading back to the external power source, it seems like a "buffer" capacitor is the only option atm.
(1)· C1 will charge up to the supply voltage and stay there.· Its only discharge path would be through D1, which (being turned the wrong way) will never conduct.· Ever.· So, during operation, the current through D1, Llowposs, and C1 is zero at every instant.· Take'em out, they do nothing.
(2)· That being so, there is no kick diode at all·in your circuit.· Disconnect D1 from the junction of Llowpass and C1 and connect it instead somewhere in the positive supply.· The easiest way to proceed would be to remove C1, and to replace Llowpass by a wire.
(3)· You need, for a kick diode, something with a higher surge current rating than a 1N4004.· A 1N4004 can conduct an occasional 30A surge, but it can't last long with a couple hundred thousand of them per second.
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· -- Carl, nn5i@arrl.net
Post Edited (Carl Hayes) : 4/12/2009 4:23:54 AM GMT
(1) - You are right about C1 - sort of. The capacitor DOES indeed charge to the supply. When the FET switches off(assuming current is flowing through the load), Inductive Kick will force current into the capacitor, through D1, filling it up a bit more. While it is filling up, there is now a higher voltage on the capacitor end of the inductor than on the Supply side of the inductor. This will cause the capacitor to DISCHARGE through the inductor, but at a near DC rate, rather than high-freq. I DO NOT want the capacitor to discharge through D1. D1 is in the correct position. The current flow after the fet turns off, is through the load, into D1, into C1, into GROUND, back to Supply. And I disagree. They do something. If I take them out, the mosfets get very hot, and start melting the solder within 10 seconds. Obviously, they are doing something right with respect to the mosfets. They may be dissipating the energy in the components, rather than in the mosfets, but they are doing something.
(2) D1 IS the kick diode in the circuit. In a sense, it IS connected across the load, via the capacitor and voltage rise from GND to Supply. I CAN replace L_LowPass with a wire, however, there is more current through C1, resulting in C1 getting hotter, faster. If I have the inductor in there, the inductor now gets hot - nothing else does. However, this is due to my inductor having a resistance of about 18 ohms. This is a 15mH inductor and rather large.
(3) As for the diode, as I stated above, it is NOT a 1N4004. It is a schottky rated for 25V, 10A. I will be replacing this diode with a larger one with more current/voltage handling ability. 1N4004's evaporated in seconds, even with 3 in parallel.
Post Edited (Philldapill) : 4/12/2009 5:38:33 AM GMT
I'm trying to understand your circuit... what are the inductor values that you are using. All of the suggestions thus far are things that I would consider, but unless I have something that I can see and play "what if" on a simulator, I'm not sure that I can give a better answer.
Perhaps more detail on what it is your are using this for might help me understand.
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Beau Schwabe
IC Layout Engineer
Parallax, Inc.
I don't understand why exactly, but if I don't have C1 in there, and just connect the diode to the Drain and Supply, the diode does virtually nothing. The mosfets get very hot, and when I look at the Drain-Source Voltage, it still peaks at the 55V internal zener voltage. Again, this is WITHOUT C1 in place. With C1 in place, everything BUT C1 stays cool. Taking L_LowPass out, just makes the capacitor even hotter. My thinking is, L_LowPass is slowing the discharge rate of the capacitor to a near DC value, causing the rms current to be at it's minimum. This, in turn, is causing the lowest I^2 * R in the capacitor.
I know this doesn't make much sense, but all this is what actually happens. It's nuts to me, too, but is starting to make some sense.
-Phil
I did notice in the simulation, that there is an interesting current loop that occurs in a counter-clock-wise rotation the way you have your schematic drawn.
This occurs from R_LINE --> R_LOAD --> I_LOAD --> D1 --> L_LOWPASS back to R_LINE ... C1 is taking 1 Amp pulses each time the MOSFET turns "ON".
I'm not sure if the simulator is being "tricked", since the current going through R_SUPPLY (set at 1 Ohm) measures·497 mA in the simulator, but yet the current going through the current loop that I mention is reading·1.1 A ... of course I have R_LOAD and R_LINE set to 0 Ohms... but still, unless I'm missing something, the current through the loop should be less than or equal to the total current at the input.
Another thing... The junction between I_LOAD and D1 measures at 12.32V with C1 connected. With C1 removed this node increases to 263.97V and the current through the loop drops to about·82 mA.
In place of the LOWPASS_ANALOG inductor I replaced that with a diode with the Anode facing the supply... this seemed to provide a steadier analog supply with less interaction from the current loop.
All said, I don't see why the MOSFET should be stressing, again.. unless I'm missing something.
Edit: I'll have to physically try this circuit....·this might be an effective method for DC Power factor correction to compensate·for·IR drop ... assuming that the simulator isn't getting confused.
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Beau Schwabe
IC Layout Engineer
Parallax, Inc.
Post Edited (Beau Schwabe (Parallax)) : 4/12/2009 7:46:08 AM GMT
They can be, like most anything else, if that's how you design 'em in.· Not good lightning arrestors either.
You are right about this being a type of buck converter. Sort of. Let's say I do exactly like you'll say and put the diode right aross the load. This way, current flows through R_Load -> L_Load -> D1 back to R_Load. However, in that arrangement excess power is being dissipated purely in the load/D1. What my circuit is trying to accomplish, is taking that excess energy, and allowing it to slowly go back into the battery. I'm not too concerned with recharging the battery as much as getting RID of the excess power that would just go to heat. Heat is a big issue in this device since it will be in a small ABS enclosure. Basically, any method that just shunts the inductive kick away and turns it into heat, will not work for this application. The energy must be recycled and NOT heat anything but go back into the battery.
Also, Beau, if I remove the Analog Supply inductor and replace it with a diode, any HV transients that occur will be absorbed by the Analog Supply Capacitor and will not be bled off because of the diode. This was happening in an earlier version of the circuit before I thought of adding that inductor. The inductor works very well, acting as a low, low pass filter, only letting DC through - for the most part. The voltage on the Analog Supply Cap stays at a VERY steady voltage; the voltage of the supply.
However, forget about the "analog supply" part of the circuit, i.e. the cap and inductor. Those parts are not related to the left hand side of the circuit and the heat issue.
Beau, you said in your simulation that without the capacitor/diode/inductor, the voltage on the Drain rose to over 290V? Since the zener is 55V, is it possible to then see how much power is being consumed by the Zener? I know this is hard to understand WHAT the circuit should be doing, but I think you are right when you mentioned "power factor correction". It is meant to smooth the induced current when the MOSFET turns off. If this current is just one large wave, then the I^2R loss goes way up. However, if this current is immediately stored on the capacitor, then SLOWLY allowed to return to the Supply, then the rms current is lower, which gives lower I^2R losses - less heat. I think. I'm not at my home computer, and I lack the simulation software atm.
If you cant put a freewheeling diode in because of losses what about a synchronous switch system or would that be adding to much to something you where trying to keep simple?
Post Edited (Tesla) : 4/12/2009 6:26:55 PM GMT
The classical way to do that is by supplying the load (your wet plates, in this case, rather than a transmitter) with DC (your battery) in series with a modulating voltage (your rectulangular wave, which should have a p-p value equal to 1/2 the supply voltage). When the modulating signal is audio, this is done by putting a transformer secondary in series with the DC supply. The same would work for you, but the transformer design must be suited to your modulating signal.· It'll be a big transformer, because you wil need kilowatts of modulating signal.
However you do it (by switching with FETs or by AM), you're running kilowatts of extremely wideband RF (square wave modulation!), and it's going to be really difficult to prevent it from radiating. Good luck.
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· -- Carl, nn5i@arrl.net
Post Edited (Carl Hayes) : 4/12/2009 6:30:54 PM GMT
http://pdf1.alldatasheet.com/datasheet-pdf/view/158108/STMICROELECTRONICS/STD60N55.html
@Tesla, the zener reverse recovery time is 47ns. Hardly anything to worry about at low frequencies.
@Carl, ironically, the device seems to dissipate slightly LESS power at higher frequencies. I think this is because my current is less due to the parasitic inductance slowing the di/dt during turn on. I must be missing something, because this seems, to me, to be clearly a switching problem. When I scope the Drain-Source voltage, without any diodes/caps/inductors/what have you, there is CLEARLY a clamping effect going on at 55V. This is CLEARLY an inductive response, is it not? If I'm wrong, PLEASE correct me. I am going insane with this. This is one of those "It SHOULD work... but it doesn't?" problems.
BTW, carl, why do you say "2400W"? At my maximum 40A, that's only 480W. At 50% duty cycle, it's 240W. This just doesn't seem like THAT much power to be radiating in the RF band.
I don't know. I mean, you guys have more experience, while I'm just an EE student, so maybe there's more going on than my simple "kirchoffs current law" mind can think up.
I'm sure the zener IS dissipating that huge power for very short periods. The transient looks pretty short - maybe 1-2% of the period, but 2.2kW for a short bit of time, is still a good bit of power for a 1" x 3" area of copper to dissipate.
I still have NO idea why the voltage still rises to 55V WITHOUT a capacitor, and the diode shorted back to +Supply(D1 from Drain to +Supply, i.e. across the load).
Sorry if i didnt get the whole picture of the problem and stated the obvious.
I have a Capacitance meter on my multimeter, but I'm afraid that the load will have a dynamic capacitance while running. I'd love an Inductance tester, but don't have one ATM.
Yea LCR meters are the best for problems like these. But mine only goes to 10khz because i didnt have the cash for a bench top top of the line one that goes to 100khz
For your catch capacitor are you using a PP or PE . 1000uf sounds like overkill to catch a bit of stray inductance. If it is melting and exploding sounds like too high of RMS current for it. You want to run this at 100khz right?