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What will happen if the drain voltage in the below image increases up to 200V? (the Mosfet could be IRF9640 or could be p-ch mosfet in IRF7105) I want to know if the Mosfet temperature goes up and also if the Drain current is almost zero. enter image description here

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3 Answers 3

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It will heat up, and if you indeed manage to get it up to 200V it will probably violently blow. The internal body diode is forward based and causes excessive current to flow. It'll probably be destroyed long before you actually reach 200V.

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As others have said, for positive voltage, you cannot get to 200V without total destruction. But if you meant for V Drain to be -200V, that should be OK for the IRF9640, which is rated to -200V. FET will stay off.

But the IRF7105 is only rated for 25V. So it will fail due to Vds breakdown if you try to go to -200V. Also, the IRF7105 is a dual FET with a P-channel and N-channel in the same package.

It is kind of strange that you say it could be either one when they are so very different in specifications and package.

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This is an insane configuration, and the FET will die at almost any voltage. The body diode is forward biased and will always conduct. Depending on the heat sink, the lethal voltage will probably be in the range of 1 to 5 volts.

If the voltage on the drain is negative, and the gate is positive, the FET will not conduct at all. In theory, the drain current will be extremely small until the breakdown voltage of the FET is reached.

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