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How to use MOSFETs ?

Discussion in 'Electronic Basics' started by Jan Nielsen, Jun 21, 2007.

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  1. Jan Nielsen

    Jan Nielsen Guest

    Hi,

    I was told on another forum, that the way to switch "large" currents is
    a MOSFET since a regular transistor cant, atleast not from a mcu output.

    Now I have a few of these IRF530 MOSFETs, and are trying to get them
    working.

    I connect ground/common to one of the outer legs, a led (cathode) to the
    other outer leg.
    The anode of the led to VDC+ (4V)
    Then I connect the base to a 1K resistor and then to common/ground.

    So in my understanding, when I apply power the led should be off, well
    its not, no matter where I connect the base with 1K resistor, the led is
    always on.

    I tried searching google for some guides, and found some, but I cant get
    their exsamples to work, the led is either always on or off, no matter
    what the gate/middle led is.

    /Jan
     
  2. Lord Garth

    Lord Garth Guest

    You don't connect ground to "one of the outer legs". Insure you connect
    the Vss leg ground. That is the 'S' lead or source.

    Connect the LED cathode to the Vdd leg. That is the 'D' or drain pin. You
    then need a current limiting resistor in series with the LED anode. The
    value of the resistor depends upon the power supply voltage and the current
    specified for the LED. 10mA is a good round number. Connect the other
    end of the resistor to the positive power source.

    The 'G' or gate of the IRF-530 doesn't require a limiting resistor as the
    device
    is a MOSFET. It is an N channel MOSFET so applying a positive voltage to
    this pin will activate the LED. Grounding the pin will deactivate the LED.

    The device is sensitive to static electricity so you can damage it by
    touching
    the pins or by not observing ESD precautions.
     
  3. The first hit is the data sheet:
    http://kitsrus.com/pdf/irf530.pdf

    Terminal 3 is the source, which goes to ground.

    Jumper that to terminal 1 to turn the gate off, and connect
    your led and current limiting resistor between the positive
    supply and terminal 2 (also the tab), which is the drain.

    This should result in a very dim or off led (just the
    leakage current).

    Turning the mosfet on is accomplished by bringing the gate
    up to a positive voltage with respect to the source.
    Unfortunately, from the Electrical Characteristics section,
    you will find that 10 volts is needed to get the thing
    essentially all the way on. The gate threshold voltage
    (where the channel just begins to turn on) may be anywhere
    between 2 and 4 volts.

    If you want to switch a large current from a logic level
    signal, you need a logic level (low gate threshold) device.
    For instance:
    http://www.fairchildsemi.com/ds/RF/RFP3055LE.pdf
     
  4. ehsjr

    ehsjr Guest

    _____
    | o |
    |_____|
    /_____/|
    | IRF ||
    | 530 ||
    |_____|/
    | | |
    | | +----Source
    | +------Drain
    Gate

    Source to ground, drain to LED, LED to resistor,
    resistor to +

    Ed
     
  5. Jan Nielsen

    Jan Nielsen Guest

    ehsjr skrev:
    Oh, I thought gate was the middle, since the datasheet didnt show a
    clean pinout, atleast not the one I found.

    Thanks, it works.

    /Jan
     
  6. Hello,
    A MOSFET gate looks superficially like a Capacitor. It gets more
    interesting when you begin to drive them at higher frquencies. At the
    same time if is a large device then you will begin to see an appreciable
    charging current pulses from the output of your Micro. In this regard,
    I would install a current limiting resistor, in order not to overload
    the output pin spec.

    In the case of a large MOSFET, the drive power is equal to P = f*C*V^2,
    where f is drive frequency, C is the gate capacitance, and the V is the
    drive voltage. Some people like to drive them with a negative voltage
    (me included) when off, such that any dv/dt seen on the drain-source
    route, does not take the gate positive due to the drain-source
    capacitance, and so turning it on mistakenly.

    There is usually a resistor stated as suitable for the gate drive of the
    FET, you see it in the spec data sheets. This should be taken as the
    minimum resistance to drive the fet with. Also remember if you are
    switching the FET *really* smartly you will see some RF emission due to
    the harmonics of the very fast transtition of voltage due to the
    switching speeds.

    Is that of use?

    Cheers,

    Rob Wilson,
    Robstech Consulting Ltd, UK.
     
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