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P channel power mosfet IRF9540 or similar

Discussion in 'Electronic Basics' started by Pete D, Apr 14, 2007.

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  1. Pete D

    Pete D Guest

    What happens if I put a bigger negative Vgs on this that is specified on
    the datasheet?, I have a convenient -25V to put on the gate but I see
    fromn the datasheet it expects between -5 and -12 ish volts. I am a
    complete noobie to FETs and any advice would be welcome.

    I suspect I have to get a more appropriayte voltage but space is a concern.
     
  2. The International Rectifier datasheet I looked at had +/- 20V as an
    _absolute maximum_ rating, so -25 doesn't seem like too good an idea.
     
  3. Wouldn't that depend upon "the rest of the story"? I guess if he hooks the
    drain to -12V, then it should be ok, but if he has ideas of hooking it to
    +25V from somewhere "convenient", then there's going to be a stiff penalty.
    ;-) I'm really curious why he's wanting to connect it to a static voltage.
     
  4. Pete D

    Pete D Guest

    Right thanks, as I mentioned, i am a real noobie with FETs and hadn't
    spotted that maximum on the datasheet, i can change the voltage
    hopefully simply, I was trying to svae real estate on the PCB, but it
    wont take too many components.

    Thanks for the adice

    pete d
     
  5. He did say he was talking about gate-to-source voltage (Vgs), but
    maybe the actual picture is more complicated.
     
  6. Pete D

    Pete D Guest

    I can see why you think I want to connect it to a static voltage, I
    missed part of the explanation, what I realy menat was mmy PIC would
    switch a transistor to connect g to the -25. I generate a more suitable
    gate output. These FETs have a steeper learning curcve than bipolar.

    thanks both



    pete mc
     
  7. Driving P-channel MOSFETs is not a trivial task, and there are not nearly
    so many drivers available as there are for N-channel. Depending on the
    frequency at which you will be switching, it is very important to drive the
    gate hard and fast to reduce switching losses.

    Linear Technology has a P-channel driver LTC1693-5, but it needs a 5-12 V
    nominal Vcc. They have a high voltage high side driver LTC4440, but it is
    designed for N-channel, and utilizes a boost circuit and level shifters to
    drive the MOSFET at 80 V on the high side. TI has a complementary driver
    UC3714 which can drive bith N-channel and P-channel, but the P-channel
    drive appears to be for an auxiliary switch in a forward converter.

    You can make a three-transistor level shifter to produce -12 V gate drive
    from a 5 V logic level, but it will be inefficient and slow, unless you can
    create a -12 VDC rail.

    Depending on current requirements and speed, it may be better to use an
    optoisolator or a small relay to switch the negative voltage.

    Paul
     
  8. Pete D

    Pete D Guest

    thanks for that but-

    Wow that sounds complicated, am I trying to make it too simple with



    25V
    |-------------------------.
    /+\ | |
    ( ) z '------||-+
    \-/ 10V A | ||->
    | | | ||-+
    | '--------. |
    === | Load
    GND .-.
    | |
    | |
    '-'
    |
    |
    |\ ___ |/
    -| >O----|___|--|
    |/ |>
    |

    GND
    (created by AACircuit v1.28.6 beta 04/19/05 www.tech-chat.de)

    This seemed easy and I nicked the concept off somebody elses circuit,
    did I misunderstand the concept?

    pete d
     
  9. Ahah! With a +25 VDC supply, that changes things immensely. Your circuit
    will work if you add a resistor across the zener to bleed off the gate
    charge and turn off the MOSFET. In fact you can just use two equal value
    resistors (about 1k each) to apply 12 VDC to the gate when the NPN
    transistor is on. The resistors and gate capacitance will cause slow gate
    voltage change and high power dissipation during the transition through
    linear range.

    The simplest way to apply power to the load would be to use an N-channel
    logic level MOSFET and have the load from drain to 25 VDC, if it does not
    need to be grounded.

    You could also use a high-side driver and an N-channel MOSFET. They come in
    small SMT packages and cost about $1, and are almost essential if you are
    switching at high speeds.

    Paul
     
  10. Pete D

    Pete D Guest

    I see, thanks fr the clarification, there are several different voltages
    all referenced to 0V so the load does really need to be grounded, it is
    interesting about the slow switching, on this particular circuit the
    load will only be switched occasionally so it isn't really a problem,
    but another project I am working on will be PWMing a solar panel to
    charge 24V lead acid battery (yes 2 or actually 3 jobs on at the same
    time) so I guess the switching tine will become a factor then that will
    mean going to one of the methods you suggest. the high side driver
    sounds interesting I'll go look for one and study the datasheet see if I
    understand how to use it. FETs are a bit of a mystery to ne, so there's
    lots of learning to do.

    Many thanks again for the advice and for taking the time to give such
    thorough answers.

    pete d
     
  11. jasen

    jasen Guest

    it's mostly right, but

    How is the mosfet that supposed to turn off?

    when the NPN transistor turns on current flows through R1 and pull the
    voltage on the MOSFET base down to 15V, that's enough to turn on the MOSFET
    and led current flow into the load.

    but.

    when the transistor turns off nothing happens, well if you wait long
    enough the MOSFET will catch fire - the problem is there's nothing that can
    pull the gate back up to 25v to turn the mosfet off,

    if you're unlucky the zener will leak a little and the voltage on the gate
    will increase slowly until the mosfet is turned half way on, then half the
    power goes to your load and the other half overheats your mosfet.

    here's one way to fix it.

    Change the zener to a resistor equal to R1. That way the two resistors form a
    voltage divider giving the gate half of 25V. and when the NPN is turned off
    the gate can discharge back to 25V through the top resistor.

    Bye.
    Jasen
     
  12. Pete D

    Pete D Guest

    Thanks, that makes a lot of sense, i have added the resistor but left
    the zener in as it is just possible that if the connectons to the
    circuit are made in the wrong order there could be a lot more volts than
    25. I think from what Paul said that is okay to do.

    Thanks again

    pete d
     
  13. Jamie

    Jamie Guest

    remove the ZEner, and put a resistor there., remove the resistor
    from the collector and run straight to the Gate.
     
  14. Pete D

    Pete D Guest

    Thanks, but unfortunately I have to allow for the possibilty of a much
    higher voltage on the 25V line temporarily and I think that would allow
    the entire possible max volts across the gate source junction, only for
    a very short time, but i don't know hopw the FET would cope with that.
    As usual the devil is in the detail and I only put a small part of the
    circuit up


    pete d
     
  15. Jamie

    Jamie Guest

    ok, Keep the zener along with the Resistor then. but do the rest as i
    suggested.
    if protection is what your worried about, you may want to consider
    using a TVS diode (AC Type) because it's also possible you could get a
    pulse from the common side. We had that problem not to long ago. the
    TVS took care of that. We no longer have FETS dropping out mysteriously.
     
  16. Pete D

    Pete D Guest

    Okay thanks I'll look into TVS diodes, not come accross those before.
     
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