Help with shorting out Mosfets.

Discussion in 'The Projects Forum' started by hobbyist, Aug 10, 2008.

  1. hobbyist

    Thread Starter Distinguished Member

    Aug 10, 2008
    I'm experimenting with Mosfets and I keep on shorting them out.
    I'm using IRF540 Mosfets, I put a 2 ohm series resistor at the source terminal between ground. The load is a DC. 5 volt 3 amp battery drill geared motor in series with the resistor. I'm using a 18 volt supply, I have a 47k ohm resistor at the gate to ground. I have a variable resistor from gate to positive 18 volts. I have 3 voltmeters set up.
    1. Voltage across Mosfet Vds.
    2. Volt. from gate to ground. Vg. (note. the reason I'm going to ground
    with this reading and not to source
    because I need to use this voltage
    to determine the gate driver output)
    3. volt. across motor. Vm.
    My recorded data:
    // Vg. // Vds. // Vm. // Im. //
    Min Vm // 2v. // 0 // 0 // 0 //
    Max.Vm. // 15v. // 3.2v. // 5.5v. // 2.8A. //

    I'm using less than 9 watts, and MF is rated at 92W.
    I have no diode across the inductive load.
    and no heatsink.
    It runs find as I switch it on and off with the variable resistor.
    but as I let it run longer and experiment with the var. res. the mosfet shorts out. I checked the short with a continuity checker. and it is shorted from drain to source.
    I'm no where near the rated power factor, even if the motor ran full 3A current at Vds of 17v. still the power diss. would be 51W.
    But the Mosfet is not operating at that extreme.
    They just want to burn out running at Vds 3v.
    What am I doing wrong do I need to heatsink, and put the flyback diode in the circuit. I'm at a loss.
    I hope I gave enough information so someone can steer me in the right direction with this.
    Thankyou for your Time.
  2. SgtWookie


    Jul 17, 2007
    You're attempting to use the MOSFET as a variable resistor; ie: in linear mode.
    MOSFETs work best as on/off switches; ie: Vgs = 0 or Vgs=10. If it is somewhere in between 0 and 10v, then it is operating in linear mode, and is dissipating a great deal of power as heat. You are probably attempting to use it without a heat sink, thus the heat has noplace to go, and the MOSFET is burning up internally.

    The proper way to control your drill using a MOSFET is to use PWM, or Pulse Width Modulation. The gate is brought high for a period of time, and then low for a period of time. The ratio of ON time to OFF time works out to be equivalent to limiting the voltage across the motor, only you're switching the power on and off instead of limiting it in a linear fashion.

    You could use a 555 timer in PWM mode to control the gate; however you'll also need a voltage regulator to stay within the limits of the 555's operating voltage, as 18v is a bit too high.

    See the attached. It's an adaptation of Bill Marsden's 555 PWM circuit. Not shown is a 12v regulator IC and appropriate capacitors on the output to supply the 12v for the 555 timer IC.

    There is a heavy duty Schottky diode across the motor to take care of the motor's reverse EMF when the MOSFET turns off; otherwise you would get voltages in excess of what the MOSFET could handle. The small 470pF cap is to "buy time" for the diode to begin conducting, as they don't turn on instantaneously.

    In the simulation, there is a "CMD1 4V" balloon - ignore that; it's required to set a known state for the simulation to run.
    Last edited: Aug 11, 2008
  3. hobbyist

    Thread Starter Distinguished Member

    Aug 10, 2008
    Thankyou for your help.

    I probably need to use a heatsink and other protective configurations.

    My hobby in electronics, is to try to design circuits using all discrete components. I designed and built one PWM, with torque controle for a small 300ma geared motor and it worked real well. I used a IRF610 no heatsink and no diode across motor. and got nice results.
    So when I tried to use the same design for this heavier motor, I ran into mosfets shorting out. IRF540
    Finally when I determined the switch on gate voltage needed I reworked the circuit around that voltage.
    At this point I stopped burning Mosfets. I would switch manually the gate voltage from millivolt region to 15 volt and watch the voltmeter at Vds go from 18v. to 3v. respectively.
    When I hooked up my audio generator to the PWM circuit and adjusted the output I got nice slow motor rotation. But my voltmeter was reading Vds.13v. and after a little bit the motor current began increasing and went full on. As the mosfet was now shorted out.

    So I gather from what you said I should use a heatsink for sure. as well as diode.

    Thankyou again for your time and expertise.
  4. SgtWookie


    Jul 17, 2007

    You're welcome.
    Yes you do, unless you like frying MOSFETs. :eek:

    Good for you - but that was a very small motor.

    If it were being driven by PWM, you would expect an average voltage reading somewhere between 0V and 18V. However, you said that you were using a pot controlling the gate voltage. That's quite different from PWM.

    Was your audio generator putting out a square wave, or something else?
    Your audio generator would not likely source/sink enough current to charge/discharge the gate of the MOSFET in a timely manner. This would cause the MOSFET to run in the linear region, generating a great deal of heat.
    Heatsink, Schottky diode and a very small cap across the diode - somewhere between 270pF and 1nF (1,000pF).
    You also need a gate driver circuit that's capable of putting out a decent amount of current. The circuit I posted should work pretty well for you, and it's using a minimal number of parts.

    You could build an oscillator out of discrete components - but why work that hard?
  5. John Luciani

    Active Member

    Apr 3, 2007
    If you are using a IRF540 in a TO-220 package without a heatsink at 9W you
    are way above the rating of the device. The specification you need to look at
    is thermal resistance.

    The thermal resistance (Rja) is 62DegC/W without a heatsink.

    62DegC/W * 9W = 558DegC

    The junction temperature rise is 558DegC above ambient which exceeds
    the 175DegC maximum specification.

    To determine the temperature rise you need to calculate the junction
    to ambient thermal resistance (Tja) ---

    Tja = Tjc + Tcs + Tsa all the dimensions are DegC/W

    Tjc thermal resistance between the IC junction and IC case. This is
    from the IC datasheet.

    Tcs thermal resistance between the IC case and your heatsink. This
    is in the IC datasheet. Usually you place a thermal
    compound between the IC case and the heat sink to decrease the
    thermal resistance. The decrease occurs because the thermal
    compound fills the air gaps between the two surfaces and conducts
    heat better than the air.

    Tsa thermal resistance between the heatsink and ambient air. This is
    in the heatsink datasheet. Most heatsink datasheets specify Tsa
    with and without airflow. You can determine if you need a fan by
    reviewing this number.

    (* jcl *)
  6. hobbyist

    Thread Starter Distinguished Member

    Aug 10, 2008
    SgtWookie and JCL.
    for taking the time to explain in detail and getting me on the right path in designing power electronics projects.
    I will take all this advise and aquire it into my project.
    Again Thankyou you were a great help.