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I am building a high power ZVS induction heater. I originally used this famous design. It worked like a charm when I used a 48v 31A power supply. But I want an upgrade. I recently got a few of these FCH072N60F 600v 52A power mosfets. They have a low Rds on, fast switching and low gate charge. All of which are needed for a good ZVS driver.

However, I recently got myself a 96v 30A power supply. I simply replaced the IRFP250N mosfets with the beefier FCH072N60F mosfets. I then realized that at double the voltage, my mosfets would burn out. The 470Ω 2w resistors were for the 48v version.

Could I simply double the resistance of these resistors or is it not that simple?

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  • \$\begingroup\$ Your MOSFET gates would burn out? Do you mean your 12V zener gate clamps would burn out and THEN with that gate voltage protection gone your MOSFET gates would burn out? \$\endgroup\$ – DKNguyen Jul 6 at 0:08
  • \$\begingroup\$ I guess so, yes. \$\endgroup\$ – ElectronicsNoob Jul 6 at 0:55
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(96V - 12V) across 470 ohms is 180mA through your zeners. That is 12V * 180mA = 2.16W dissipated in the zeners. I don't know what zeners you have but that's probably near the zener's limit so you probably do want to increase the resistors.

Your resistors are also dissipating power so they also will heat up more unless you increase resistance. About Four times more due to the voltage doubling if you leave the resistance unchanged.

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  • \$\begingroup\$ 12v 1w zener diodes. 10w resistors. Anyway thanks for telling me. I know what to do now. I will try it out. \$\endgroup\$ – ElectronicsNoob Jul 6 at 1:22
  • \$\begingroup\$ What about 750 ohm 5w resistors? I got a few of them. I don’t know how you got 18mA through the zener though. \$\endgroup\$ – ElectronicsNoob Jul 6 at 1:26
  • \$\begingroup\$ How did you get 18mA through the zeners? \$\endgroup\$ – ElectronicsNoob Jul 6 at 1:29
  • \$\begingroup\$ The resistor drops all the voltage that the zener doesn't. Since you know the resistor voltage drop you can find its current. The resistor current is the same as the zener current. \$\endgroup\$ – DKNguyen Jul 6 at 1:30
  • \$\begingroup\$ I mean what is the formula you used? \$\endgroup\$ – ElectronicsNoob Jul 6 at 1:30

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