Sorry. I meant drain source breakdown voltage.

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It is the voltage between the drain and the source at which the FET begins to conduct with no gate voltage applied. The gate to source voltage is zero and the FET is supposed to be off but you have applied too much voltage (exceeded the drain source breakdown voltage) so avalanche current flows and your FET blows up.

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Thank you.

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Not that simple, most MOSFETs are avalance rated to some degree, for example the IRL540 can take a single shot of 440mJ without dying.

To put this into perspective, you could switch 20A though a 2mH load or 663mA through a huge 2H load with no back EMF diode and it won't die.

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hero the question was breakdown voltage not conduction of 10,20 amps. if it breakdown at 400v and the leakage is .001amps that is 4 watts and because it gets hot at 4w it will beging to heat up more LEAKING MORE current and eventualy blow up.

 

That's a form thermal runaway which is a different phenomena to breakdown voltage.

You've got one hell of a leaky MOSFET, the IRL540 only leaks 25µA at 100V 25° which is just 2.5mW.

Even so if you use a good sized heat sink thermal runaway is unlikely to occur.

It was good idea to bring it up as it is certainly something you should be careful of at higher voltages.

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I do not answer private messages asking for help because no one else can: benefit from advice I may give or correct me if I'm wrong.

Please ask on the open forum if you have a question and I'll be happy to help, if I know the answer.

 

ok here is my question so that you will finaly understand. on a square wave where is the maximun heat dissipation rise time, saturation, or fall time, or off. asuming a transistor switch.
finaly if I LIMIT THE CURRENT DURING BREAKDOWN of any device it can sustain that indefenetly. is true or false?

 

Hi,
In my point of view source breakdown voltage is a voltage when the device starts conducting. Then it will be ended at a particular point.

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martin

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As far as i know...kchriste ans is rite.....