How do Power mosfets absorb energy?

Discussion in 'The Projects Forum' started by newbieateverything, May 4, 2013.

  1. newbieateverything

    Thread Starter Member

    Feb 25, 2013
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    from what i have been reading power Mosfets absorb energy during avalanche operation.
    However the sources i am using are slightly confusing
    Need help breaking it down.
    How does a power mosfet absorb energy?
     
  2. wayneh

    Expert

    Sep 9, 2010
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    Nothing can absorb energy without heating up.
     
  3. newbieateverything

    Thread Starter Member

    Feb 25, 2013
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    i know that
    heatsink can dissipate the heat
    what i would like to know is the operation of a mosfet when its absorbing energy.
     
  4. ScottWang

    Moderator

    Aug 23, 2012
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    When you turn on the bjt, the Vce will be around 0.2V, but when you turn on the Mosfet there is different, whatever it is a N Channel or P channel, because the mosfet has different Impedance inside between DS, so the Vds is not regular as bjt, the Vds depends on the Impedance of Rds and the current flow through DS when the mosfet is a N channel and the current flow through SD when the mosfet is a P channel.

    The Vds will be like this Vds = I * Rds, and the Watts = Vds * I, that's what the mosfet absorb energy and transfering to the heat.

    Different MOSFET has different Rds, so the Vds is different, you can check some of the Rds of mosfets.
     
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  5. newbieateverything

    Thread Starter Member

    Feb 25, 2013
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    Ok say it is an N channel Mosfet the IRF150 specifically
    The current flow from drain to source is between 0.1 and 4.1
    Rds of IRF150 is 0.055ohm

    so could you please help me explain how the mosfet absorbs energy in a circuit if the Vds at t=0 is 0.1 and at t=1ms increases to 4.1
    Then a few microseconds later drops back down to 0.1

    How would the mosfet absorb energy in that situation
     
  6. ScottWang

    Moderator

    Aug 23, 2012
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    According to the given data from you to calculate the following values.

    Vds = (t=0) → 0.1V → (t=1ms) → 4.1V → (x uS) → 0.1V

    (t=0,0.1V) ==> a low Volts on Vgs and a current(0.1V/55mΩ=1.818A) flowing through Rds
    (t=1ms,4.1V) ==> a high Vlots on Vgs and a current(4.1V/55mΩ=74.545A) flowing through Rds
    (t=x uS,0.1V) ==> a low Volts on Vgs and a current(0.1V/55mΩ=1.818A) flowing through Rds

    So there are something wrong about the given data of current and voltages.
     
  7. newbieateverything

    Thread Starter Member

    Feb 25, 2013
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    Sorry i dont understand what is wrong please explain
    can anyone else clarify
     
  8. ScottWang

    Moderator

    Aug 23, 2012
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    See the V and I data that what you given and the values of current that what I calculated.
     
  9. newbieateverything

    Thread Starter Member

    Feb 25, 2013
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    I only gave you current data 0.1A and 4.1A and the RDS on which is 0.055
    you did a few calculations
    which i dont understand
    What i would really like is for you to please explain to me how a mosfet actually absorbs energy
     
  10. #12

    Expert

    Nov 30, 2010
    16,261
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    It's about Watts Law: P = IE The current through the mosfet times the voltage across it creates heat. The mosfet absorbs energy and turns it into heat the same way a resistor does, or even a sausage connected to an AC power line. There is nothing magical about amps times volts in a mosfet.
     
  11. newbieateverything

    Thread Starter Member

    Feb 25, 2013
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    charming!!....:/
    could you please explain what the whole avalanche operation in a mosfet is about
    does that have anything to do with how it absorbs energy?
     
  12. ScottWang

    Moderator

    Aug 23, 2012
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    Please google Ohm's Law.

    Ohm's Law:
    E = I * R or V = I * R
    I = E / R or I = V / R
    R = E / I or R = V / I

    W = E * I or W = V * I

    Ohm's Law Calculator.
     
  13. newbieateverything

    Thread Starter Member

    Feb 25, 2013
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    Jesus people i know Ohm's law that is not what i am asking at all
    Avalanche operation does that have anything to do with the mosfet absorbing energy
    yes or no
    I know the power dissipated is current through the mosfet* voltage across it
    Power dissipated is energy absorbed

    Are you saying the mosfet avalanche operation mode has nothing to do with the Mosfet ability to absorb energy
    if so please just say yes

    based on the attached schematic could you please tell me how much energy the mosfet absorbs during the fault condition
     
    Last edited: May 4, 2013
  14. ScottWang

    Moderator

    Aug 23, 2012
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    What do you mean "fault condition", does the mosfet open or shorted?

    If you want to reduce the heat, the one is to choose some other mosfets has less Rds as bolow, also you need adding the heatsink:

    FDP5800,Nch,80V/60A,4.6mΩ.
    FDP8440,Nch,40V/80A,2.4mΩ.
     
  15. newbieateverything

    Thread Starter Member

    Feb 25, 2013
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    the mosfet is to turn off when the current in the 2.2k ohm increases past 9.09mA
    Its at 9.09mA when the switch in series with the 47ohm resistor is open

    my question is not about how to reduce the heat
    i already attached my mosfet to a heatsink

    i really just wanted to understand the process by which a mosfet absorbs energy

    same way a capacitor can absorb energy by charging
    http://hyperphysics.phy-astr.gsu.edu/hbase/electric/capchg.html
     
  16. joeyd999

    AAC Fanatic!

    Jun 6, 2011
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    I found this App Note.

    Does it help?
     
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  17. newbieateverything

    Thread Starter Member

    Feb 25, 2013
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    I read that already
    Its quite short
    have u had a look at it?
    from what i understand
    avalanche is when max drain to source voltage when mosfet is off has been exceeded.

    what i am unsure of is how a mosfet would for example absorb energy during a fault condition. As per my attached circuit
     
  18. joeyd999

    AAC Fanatic!

    Jun 6, 2011
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    I've no experience with what you're asking. Just trying to give your thread a boost. Sorry.

    As a suggestion, you could consider calling/emailing an application engineer of the manufacturer of the part in question.
     
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  19. kubeek

    AAC Fanatic!

    Sep 20, 2005
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    I still don´t quite see what you´re asking. The energy lost on the mosfet simply gets converted into heat, and it doesn´t matter whether this is through normal operation or avalanche breakdown. The mosfet has some thermal capacitance, so the energy is stored there and the temperature of the mosfet rises. If you ask it to store this energy too often the temperature will rise high, possibly with hot spots in the structure, ultimately causing the mosfet to fail.
     
  20. ScottWang

    Moderator

    Aug 23, 2012
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    You focus on the absorbs energy, but in your circuit that you need is adding a rectifier diode in parallel with the motor, and you have to reverse the polarity, and the negtive is conecting to D of mosfet, the positive connecting to the another pin of motor.

    where do you want to protect when you using the rc integrator circuit? (Vgs?)
     
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