Forward converter question

I built a 2-switch forward converter, whcih worked OK but the IGBT switches (IRG4BC20U) failed after a few months (probably because of inadequate heatsinking). WHen I replaced them I must have indavertently stripped the track on the gate of the upper transistor. This made the gate float. When I powered up, it worked at low output but near the max load the upper transistor got fried and the fuse blew. The lower IGBT was fine. I rplaced the upper one (not realizing the problem), and the exact same thing happened.

Im assuming a floating gate will make the IGBT conduct all the time. This means during the OFF period, the primary, instead of returning the magnetizing current to Supply, will be shorted. But why does this kill the transisotr? And why does this result in the fuse blowing? The lower transistor is still functioning.

Any thoughts?

vkj

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vkj
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you presumed correctly..

And why does it kill it? Lets think about that for a moment.

First of all, you're working it against the other side of the bridge and there for it is conducting a lot more than it normally would. Secondly, the floating gate control signal is most likely partially turning on the IGBT, because it is not at full bias level and causes higher resistance to be present in the IGBT but not infinite, most likely the reason why you don't get an immediate short when the other side is on, because it is not fully on.

This higher resistance relates to wattage, heat dissipation that would otherwise not be there.

You must look at the specs of the IGBT on the subject of watts not so much current. A lost of IBGT's can deliver loads of amps, like Power Fets do.. Calculate the watts and required heat sink to maintain it at that level.

The correct thing to do would be to fix the gate driver circuit.

Jamie

Reply to
Jamie

ches .

Yes lets!!

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No!! first of all find out what a 2 switch forward converter is.

You got something right

Reply to
cbarn24050

Yes, I just happen to know what forward converter is and still, my answer is the same. Guess you have not seen the various types of configurations that are out there..

What a dip shit, its people like you that make's the like we're in now, and it's not pretty.

You mean I got something right that you understand..

It seems you and those like you do have trouble comprehending basic 101.

waste of space..

Jamie.

Reply to
Jamie

But if the upper transistor is only partially conducting, then I wouldnt get the max output, right? The duty cycle is limited to 45% by the controller. Also why does the fuse blow? The lower transistor is fine (and the two are in series). I guess a more basic question arises: What exactly is the purpose of the upper transistor? If the upper transistor is shorted out, then during the off cycle, the primary of the transformer is shorted (by the diode). Why is that bad?

vkj

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Reply to
vkj

I am trying to visualize your circuit by your description and all I can see is an error in your design?

You either have a full forward bridge which uses 4 units or 2 transistors with 2 diodes ?

The 2 transistor 2 diode version does not have themselves in series with each other, the primary coil of your transformer should be between them and the diodes crossing over from each side to the opposite side of the rails.

Please look at this file and see if there is any design there that resembles yours.

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Jamie

Reply to
Jamie

Yes, what I meant was that we have (a) upper transistor, (b) primary (c) lower transistor all in series. The upper transistor gate is open and therefore is always in conduction.

Anyway, after some thought here is my explanation for what is happening: The upper transistor is always conducting. When the lower transistor conducts a magnetizing current, Im builds up in the magnetizing inductance Lm. Im = Vg*Tc/Lm, where Tc is the conducting time (=Ts*D). During the rest of the cycle the lower transistor is OFF. The magnetizing current now flows thro the diode and the upper transistor. im(t) = Im - 1/Lm (Integral(Vm(t)dt)) the limits of integration from Tc to Ts. Vm = approximately -Vdiode. The time to reduce im to zero is therefore Tc*Vg/Vdiode, and since Vg ~ 300V and Vdiode ~ 1V, im will NOT go to zero in Ts-Tc! Therefore im builds up, leading to saturation. SInce the upper transistor carries this current in OFF and ON times while the lower only in the ON time, I presume the former fails first.

vkj

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Reply to
vkj

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The two transistors share the off voltage. This means you can use lower voltage devices than with a single device.

Reply to
cbarn24050

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you are still wrong!!!

figure6 dimwit. notice the series connection.

Reply to
cbarn24050

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Yep youve pretty much got it, all on your own to. As you say the core will go into saturation once the duty cycle gets large enough. The top transistor fails first because it on resistance will be higher than the driven one but carries the same current and so gets hotter quicker

Reply to
cbarn24050

Oh really, how misinformed you have been these years, you must've of been a problem child and your parents kicked you out on your ass.

Not only can't you understand what is going on, you can't even understand what I said. Shad state of affairs we have these days.

I see what your problem is now, you have a case of the dumb asses. It's a disease brought on by the large majority of what is being pushed out of the education system these days. And that isn't much to talk about, like yourself for example.

Did you father abuse you or something? Showed you early in life just how life really is?

Jamie

Reply to
Jamie

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I notice you avoid defending your position. When all else fails resort to personal abuse.

Reply to
cbarn24050

It wouldn't destroy the upper IGBT if it remained fully saturated however, that does not seem to be case here. The transistor is partially on which can generate lots of heat.

I did a fuel injector driver using a switch method like this however, I used a pair P and N channel Power MosFets, instead of employing a gate driver. But the circuit is the same otherwise, like you'd being using in a forward converter. The only reason I did this was to regain as much energy as possible back to the supply rails from the magnetizing collapse, instead of wasting it through a clamping diode to common and it worked out very well.

In my circuit, the same would happen if the upper transistor happen to short. It simply would then cause the magnetizing collapse energy to drain off in the other diode and shorted transistor.

In your case however, you are correct in assuming the upper transistor would be active in the circuit at all times but its the fact that it is only partially on and not saturated is what the factor is. Heat is being generated in the IGBT. If you look at the specs on those IGBT's on wattage and j temperatures. I think you'll agree on that.

At work, we have a couple of drives we need to check the IGBT's once in a while to make sure the fans are operating otherwise, these little babies go up real fast due to heat and these drives run to their max in a not so well ventilated closet. 500 HP AC drives tend to get hot.

P.S.

I don't know the size of your IGBT's but the ones we have are very expensive and we do use thermo protection that does not always work but saves the day on slow heat ups.

We just recently designed a very simple but practical mass air flow detector for irradiation process units and been thinking of employing these same detectors in the ventilation grills of the dog houses that contains these large drives. Air current is very mild and hard to detect using conventional devices.

Jamie

Reply to
Jamie

Youre right. Thanks for your help.

vkj

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vkj

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