MOSFET driver circuit, when is 5V too low?

Discussion in 'General Electronics Chat' started by Nicholas, Jan 21, 2016.

  1. Nicholas

    Thread Starter AAC Fanatic!

    Mar 24, 2005
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    Hi guys

    Yet another question. I hope you can answer this. The MOSFET is an IRL540.

    I was given the good advice, on this forum, to insert a diode on the mosfet gat, to protect the microprocessor should the
    MOSFET give in or something, and direct 30V back through the gate.

    I have made one of my famous drawings(see below). When the diode and the 10K resistor is in place, my 5V goes to
    4.48V. Will this ensure complete saturation of a logic level MOSFET?

    On a less-important note; if I want to connect an LED + resistor from 5V to gnd (before the diode etc) then my current draw
    goes from nothing, to 10-20mA...on a pin that has a maximum of 30mA. Will this not affect saturation of the MOSFET?

    Thanks for your time!

    Nick
    mosfet (2).jpg
     
  2. Jony130

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    Feb 17, 2009
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    Modern LED's will have a good brightness if the LED current is >1mA. So increase your led resistor value and Iled around 5mA or lower.
    As for the MOSFET we need to know what is the load ?
     
  3. Nicholas

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    Mar 24, 2005
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    Hi, thanks for your answer!!

    The load will be 36-40V

    Thanks,

    Nick
     
  4. Jony130

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    And the load current is??
     
  5. Nicholas

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    Mar 24, 2005
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    I would think about 5 amps!
     
  6. mcgyvr

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    Oct 15, 2009
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  7. Nicholas

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    Mar 24, 2005
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  8. crutschow

    Expert

    Mar 14, 2008
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    Don't use "typical curves" for a design. They don't indicate worst case characteristics.
    However the data sheet does shows that, for a Vgs of 4v, the ON resistance is a maximum of 0.11Ω with a drain current of 14A, so you are fine at 5A.
     
  9. Jony130

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    The power dissipation in the MOSFET is around Ptot = Id^2*Rds(on) = 5A*5A *2*0.077Ω = 3.85 watts. So, the small heatsink will be needed.
     
  10. crutschow

    Expert

    Mar 14, 2008
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    It's only half that value at 1.92W.
    You multiplied by the "2" which is used in that formula as the "squared" indicator.
    But even for that dissipation you should use a small heatsink.
     
  11. Jony130

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    This "2" in the formula I use because, as you know the Rds(on) resistance will rise with junction temperature and this "2" includes this effects (more or less).
    http://www.vishay.com/docs/91300/91300.pdf (Fig 4)
     
  12. crutschow

    Expert

    Mar 14, 2008
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    Okay. You just threw that fudge factor of 2 in there with no explanation so I didn't know.
    But that's reasonably conservative since, according to the data sheet, the ON resistance doubling occurs at a junction temperature of about 140°C and you likely don't want to operate much above that junction temperature for good reliability.
     
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