# Gate Drive Power Loss

Discussion in 'Homework Help' started by AlexMcDuffMiller, Sep 16, 2012.

1. ### AlexMcDuffMiller Thread Starter New Member

Sep 16, 2012
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Hi everyone,

When you turn a MOSFET on/off there is a very small amount of power required to do so. I'm given the amount of charge needed to turn on the MOSFET, Vgg (assume it is a step voltage).

Does anyone know how I would go about calculating how much power is lost trying to turn on the MOSFET?

2. ### crutschow Expert

Mar 14, 2008
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The energy (not power) is simply q*V where q is the total gate charge and V is the charging voltage. It's independent of the risetime of the waveform. The power is energy per unit time or q*V*f where f is the frequency of the drive signal.

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3. ### AlexMcDuffMiller Thread Starter New Member

Sep 16, 2012
5
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Thanks for the help!

Can't believe I didn't see that before.

Power = Energy/Time so Energy = qv and then it happens a lot of times every second so you multiply it by the frequency to get energy/time