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
    13,056
    3,245
    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
    0
    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
     
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