understanding power dissipation

Discussion in 'General Electronics Chat' started by ian123, Sep 19, 2011.

  1. ian123

    Thread Starter Member

    Aug 24, 2011
    hi there

    I have a mosfet h bridge design which works with a passive load of 3 amps @300vdc.

    therefore p=v x i = 300 x 3 = 900 watts

    so the whole circuits power disspation = 900 watts

    how do i work out the power going through the mosfets?

    is it 900 watts or is it P=I² x Rdson = 3² x 0.27 = 2.43 watts

    which means that 2.43 watts is going through the mosfets and the rest in the passive load??
  2. SgtWookie


    Jul 17, 2007
    You have 2.43 Watts being dissipated in a MOSFET. Since you say it's an H-bridge, and if all of the MOSFETs are identical, you will have 2.43 Watts dissipated in one high side and another 2.43 Watts in one low side MOSFET, for a total of 4.86 Watts. Then subtract that from the 900 Watts to get 895.14 Watts power being dissipated in the load.
  3. ian123

    Thread Starter Member

    Aug 24, 2011
    So in order to calculate the heatsink for the mosfets i can get a heatsink 200°c per watt per meter

    therefore a 60mm heatsink would do for the two mosfets and a 12mm lenght for four mosfets?
  4. Adjuster

    Late Member

    Dec 26, 2010
    Unless the period of the PWM is extremely long relative to the switching time of the MOSFETs, there may be significant heating due to switching losses. This is because as the FETs are in the process of turning on and off, there are short intervals when the IDS.VDS product is larger than when fully on.

    Simulation could help you to get an estimate of this, but parasitic effects like track inductances may affect what you get in practice.

    Edit : I suppose some motor drives do work at such low frequencies that switching losses could be made negligible, at least if a good gate driver is used, but best to be sure: better safe than sorry!