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I am designing a BMS using BQ76940 IC. I will be implementing high side MOSFETs for charging and discharging. I read in a theory document that in order to switch ON the MOSFETs in high side, the gate voltage should be higher than the input to the drain of the MOSFET by atleast 10V. The MOSFETs have a maximum gate voltage 20V, the battery would pass approximately 48V through the MOSFETs. So, If I apply 58V at the gate of the MOSFETs, won't it destroy them?

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  • \$\begingroup\$ If you don't need high switching speed, then a typical solution is a large series gate resistor, with a 15 V zener clamping GS. However, this is one situation where use of a P channel FET high side may make life easier, doing away with the need for a super-voltage. \$\endgroup\$
    – Neil_UK
    Jan 11, 2022 at 7:27

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The MOSFETs have a maximum gate voltage 20V

No this is an inaccurate statement. MOSFETs are likely to have a maximum gate-source voltage of +/- 20 volts. There's a big difference between what you said and what I'm saying because, when the gate voltage is driven above the drain voltage, the source voltage follows due to the MOSFET turning on.

Anyway, when there might be a risk of overvoltage between gate and source, a 15 volts or 18 volt zener diode is usually used to protect the MOSFET.

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  • \$\begingroup\$ can you please provide a basic schematic and explanation of where the Zener diode would be placed in a circuit to make this answer more useful? \$\endgroup\$
    – Jay Dee
    Apr 23 at 8:14

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