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In many circuit, like buck converters, MOSFET used in high side for switching, I've came across the term that Gate voltage with respect to ground must be greater than source voltage by +8V.

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For example, Vs = 60V Vg = 68V (should be).

We generally implement a bootstrap circuit to achieve this higher voltage.

But, ideally as datasheet of any general MOSFET says, Gate voltage should be VGS = 20V. If I consider this MOSFET.

So, if VGS should be 20V then how MOSFET will work on Vg = 65V?

NOTE: Consider Gate is triggered at 13.5V PWM according to this datasheet.

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  • \$\begingroup\$ You are misled in your opening paragraph - where did you read this? \$\endgroup\$
    – Andy aka
    May 14 '20 at 9:18
  • \$\begingroup\$ that Gate voltage with respect to ground must be greater than source voltage by +8V. That is nonsense. Realize that the source of the MOSFET isn't connected to ground. Indeed when Vs = 60 V you would need for example Vg = 68 V to turn the NMOS on. And Vg = 60 V to turn it off. \$\endgroup\$ May 14 '20 at 9:26
  • \$\begingroup\$ @Andyaka I read it here \$\endgroup\$ May 14 '20 at 13:29
  • \$\begingroup\$ @Bimpelrekkie yes in high side switches, source is not connected to ground. But, basically considering the basics of MOSFET, The voltage at grain will appear at source when gate is triggered. We can say like as a switch application. I'm considering those basics in school. And wondering where the voltage gets dropped ? If drain is at 60 V, after gate triggering, save voltage should appear at source. Please excuse if this sounds non sense as I'm new to power electronics and gaining knowledge from internet and basic of school education. Request to correct me if I'm wrong. \$\endgroup\$ May 14 '20 at 13:34
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But, ideally as datasheet of any general MOSFET says, Gate voltage should be VGS = 20V. If I consider this MOSFET.`

No. It's the "absolute maximum" which means that higher Vgs can break the MOSFET down. Check the datasheet again and look for the "gate threshold". That's the minimum voltage that turns the MOSFET on. So, according to the datasheet, 4V is sufficient.

So, if VGS should be 20V then how MOSFET will work on Vg = 65V?

Higher drain currents may need higher Vgs. So check the transfer characteristics at p.5 of the datasheet and determine if Vgs=5V is sufficient for your application. For most situations, 10-15V is enough though.

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  • \$\begingroup\$ Can you please refer this link. It was suggested to get higher Vgs to reduce heating of MOSFET \$\endgroup\$ May 14 '20 at 7:25
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    \$\begingroup\$ @HrishikeshDixit It was suggested to get higher Vgs to reduce heating of MOSFET that's because of the deviation of the Rds. Check the graphs at p.5 of the datasheet. Especially the Transfer Characteristics (Graph-7) and Rds Deviation (Graph-6) according to your needs. The heating of the MOSFET comes from the power dissipated by it (i.e. \$P_D=I_{Drms} \cdot R_{DSon}\$). \$\endgroup\$ May 14 '20 at 8:35
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    \$\begingroup\$ @HrishikeshDixit, no! Vgs is not needed to be 68V. Vgs means gate voltage wrt source voltage. If you need the Vgs to be 8V then gate voltage should be 68V wrt ground. \$\endgroup\$ May 14 '20 at 15:53
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    \$\begingroup\$ @HrishikeshDixit 1) Depends on the peak drain current and Rds of the MOSFET, but we can say "nearly 60V". 2) Yes, but wrt supply ground. It'll be still 13.5V wrt MOSFET's source because IC's ground is tied to MOSFET's source. \$\endgroup\$ May 15 '20 at 6:42
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    \$\begingroup\$ @HrishikeshDixit If you are to go for UC3845 then I recommend you to use 1:1 transformer for gate driving (Example). It saves you from designing a bootstrap circuit. Or, you can use another controller having built-in bootstrap circuit. \$\endgroup\$ May 15 '20 at 7:15

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