Mosfet switching question ?

Discussion in 'General Electronics Chat' started by curry87, Oct 23, 2010.

  1. curry87

    Thread Starter Member

    May 30, 2010
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    I have a irf510 n channel fet which im using to switch the low side of the load of around 4 amps at 12v the problem is the fet is heating up too hot i have around 9v on the gate and the drain to source resistance is .4ohm which is rds(on) of around .45ohm according to the datasheet.Ive tried putting 12v on the gate still the fet is getting too hot,any ideas ?
     
  2. bertus

    Administrator

    Apr 5, 2008
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    Hello,

    Can you post a schematic?
    That way we can see if the is a mistake in the wiring.

    Bertus
     
  3. curry87

    Thread Starter Member

    May 30, 2010
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    Added schematic.
     
  4. arunsjoshi

    New Member

    Oct 22, 2010
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    4 amps flowing through the rds of 0.4 ohms makes 1.6 watts of power. So, the mosfet is bound to heat up. You can try to put the device on an air cooled heat sink of adequate capacity. But better to use another Mosfet with a much lower rdson.
     
  5. curry87

    Thread Starter Member

    May 30, 2010
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    When i say hot i mean almost smoking literally.

    Does the fet drain source resistance change much depending on the load ?

    If the fet is on when drain source is .4ohms and gate is at 12v then what changes does increasing the gate voltage all the way to max gate voltage have ?

    How can you tell when a fet is too hot to be safely used without risking fet failure or worse?
     
  6. bertus

    Administrator

    Apr 5, 2008
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    Hello,

    Next to changing the fet to one with a lower Rdson,
    you can add a resistor at the second transistor to have it shut down better.

    [​IMG]

    Bertus
     
  7. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    Lets do some calculation
    Power dissipate in the MOSFET is equal
    P = Id^2 * rds(on) = 4^2 * 0.4Ω = 16 * 0.4 = 6.4W
    In reality the power dissipation will be larger becaues rds(on) will be 3 times largen then 0.4Ω
    So you need a heatsink.
    Without heatsink you can dissipate a power:
    Pd = (Tj_max - Ta )/Rja = ( 150 - 30 )/ 80 = 1.5W
    So yes you need a heatsink
    Rth < ( (150°C - 30°C ) / 6.4W ) - 4.5C°/W = 14 C°/W

    http://sound.westhost.com/heatsinks.htm
    http://www.irf.com/technical-info/appnotes/an-994.pdf
     
  8. curry87

    Thread Starter Member

    May 30, 2010
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    So without a heat sink how much current can i safely sink though the drain at 12v ?
     
  9. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    Well if we assume Ptot=1W and rds(on) = 1.5Ω then
    Id_max = √ (Ptot/rds ) = 0.8A
     
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