Output transistor temperature changes with load characteristics..

[volkantr]

hi there,

I'm designing a class d amplifier using IRF6645 transistor and IRS20124S mosfet driver with full bridge topology. I'm using pwm as modulator. you can see simplified schematics attached.

My problem is about the output transistors. When I connect single 8 ohm resistor as load, the circuit can operate @100W and output transistors' maximum temperature is about 65C degree and steady. When I connect a filter suitable for 8 ohm load and then operate again @100W, the output transistors' temperature continuously increase more than 100C degree in a short time.

I can't understand the reason of this situation. I made 2 seperate simulation by using resistor as a load and filter+resistor as load with exact models of the fet driver and fets in both simulations as you can see in attached pictures. In the simulation that single resistor as load, there is no shootthrough current, but there is shootthrough when I connect filter+resistor as load. Is that possible inductive characteristic of the filter cause shootthrough and heats the transistor more?

or another reason, comes to my mind,(may be not logical) the reactive power heats the transistors. is that possible?

this made me sick. I couldnt find the reason why the transistor heats more and more when I changed the load. May be some of you confront with the same situation and survive it, can give me some idea or another aspect for solution..

this is my first class d design and my questions can seem easy but I'm struggling here..

[Pafi]

Diode recovery charge + reactive load.

[volkantr]

thanks for the quick answer I'll look for "diode recovery charge" title but can you gıve some detailed explanation pls. how can reactive load heats transistors? reactive current return back to the supply so no enegy loss.. Am I wrong?

[volkantr]

I think, you mean the situation high side mosfet's turn on when the body diode still conducting, I searched for some tricks preventing this situation, but the ways I found like using faster (low recovery charge) diodes than body diode of transistor is inefficient. if somebody know better ways of preventing diode and transistor conducting at the same time pls let me know thanks

[volkantr]

is it possible to prevent this shoot through current?

[Denniz]

You shold add them as shoun in attachment.

1N5400 is a good choice, by the way.

Or You can use so-called "copac" power transistors from IR: IRGB4045, for instance.

[Denniz]

Originally Posted by volkantr

is it possible to prevent this shoot through current?

The answer is Yes, You can.

You should generate so-called "deadtime" beforehand.

[volkantr]

Hi denniz..
I tried shunt diode, I used 1N4148 as shunt diode, but it didnt work neither in simulation nor in real board.may be 1N4148 was wrong choice, I preferred it because of its low Trr. I read other topics about body diode problems and I learned that Vf and Qrr important parameters. and I thiink 1n4148 is better than 1N5400 in this way. If you can explain why you advise 1N5400 I'll be glad.

thnx

[Pafi]

Originally Posted by volkantr

thanks for the quick answer I'll look for "diode recovery charge" title but can you gıve some detailed explanation pls. how can reactive load heats transistors? reactive current return back to the supply so no enegy loss.. Am I wrong?

The power on the load/filter can be reactive, but not on the MOSFET! MOSFET is resistive, so any type of current creates real power (loss) on it.

I think, you mean the situation high side mosfet's turn on when the body diode still conducting,...

I'm glad you found some information!

...is it possible to prevent this shoot through current?...

Only a schottky has small enough Vf to carry the majority of the current, but I'm not sure if it's the best solution, because it has a relatively high capacitance. Try it!

The other solution can be the setting of minimal dead time. The channel can carry quite high current below Vf. If you turn on the other MOSFET exactly when the first MOSFET turns off, the diode doesn't have time to build up stored charge. Not an easy task in reality...

An other important factor is switching speed. The faster you want to switch off a diode the higher the Qrr is. You can play with turn on speed of MOSFET.

And finally: sometimes you have to accept that a loss cannot be eliminated. Then increase cooling!

[Pafi]

1N5400 must be a joke, but not a good one. IGBT and dead time also.