Using a MOSFET to lower current, temperature concerns SOLVED

I am unclear on MOSFET ratings, I would like to switch a load through a MOSFET, for example 200vdc/3amp. Can the MOSFET gate be lowered to reduce amperage by 1/2, with the goal of reducing current pulled from the source?

If I understand correct, lower gate voltage increases impedance and generates heat - what if the source voltage was also dropped ie to 50v, still keeping with 1.5amp current? would that lower the temperature? A higher dissipation rating should help keep temp lower?

Would this unit fit these needs: IRFP460C Drain to Source Voltage (Vdss) 500V
Current - Continuous Drain (Id) @ 25°C 20A (Tc)
Drive Voltage (Max Rds On, Min Rds On) 10V Vgs(th) (Max) @ Id 4V @ 250µA
Gate Charge (Qg) (Max) @ Vgs 170nC @ 10V Vgs (Max) ±30V
Input Capacitance (Ciss) (Max) @ Vds 6000pF @ 25V
Power Dissipation (Max) 235W (Tc)
Rds On (Max) @ Id, Vgs 240 mOhm @ 10A, 10V Operating Temperature -55°C ~ 150°C (TJ)

If I were moving 200v, 3amp constant, that is 600W, almost triple the dissipation rating, what is the best method to properly lower temps, a heatsink, or active cooling fan is needed? At 100v / 1.5amp, = 150W, so now within the dissipation rating?

by totalimpact
January 12, 2018

If you use a FET to switch 200V 3A then you need a FET which will stand 200V when off and pass 3A when on. Your choice of FET looks adequate. The dissipation rating for the FET relates to the voltage drop across the FET times the FET current. With say 10V on the gate the on resistance is 0.24 ohms so the dissipation is 3^2 x 0.24 or about 2W. At 40C/W this sounds reasonable without a heatsink, but use of a small sink should keep you fireproof. Your method of lowering the gate drive is undefined and is not recommended. Other methods, if necessary, are possible.

by mikerogerswsm
January 12, 2018

Thanks Mike, that answers my question pretty well, not sure how to make your comment an answer.

by totalimpact
January 12, 2018

Copy-paste.

by mikerogerswsm
January 12, 2018

1 Answer

Answer by mikerogerswsm

If you use a FET to switch 200V 3A then you need a FET which will stand 200V when off and pass 3A when on. Your choice of FET looks adequate. The dissipation rating for the FET relates to the voltage drop across the FET times the FET current. With say 10V on the gate the on resistance is 0.24 ohms so the dissipation is 3^2 x 0.24 or about 2W. At 40C/W this sounds reasonable without a heatsink, but use of a small sink should keep you fireproof. Your method of lowering the gate drive is undefined and is not recommended. Other methods, if necessary, are possible.

ACCEPTED +2 votes
by mikerogerswsm
January 12, 2018

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