FET voltage drop

Discussion in 'General Electronics Chat' started by dingeling, Apr 8, 2010.

  1. dingeling

    Thread Starter New Member

    Apr 8, 2010
    1
    0
    Hiya,

    We're constructing a weird light-dimming device. The light-intensity is controlled
    from an IO-board, and transformed to a analog signal with a R-2R ladder. This gives a voltage between 0 and 5 volts. This is then amplified so the range is 0-12 volts (nb: please ignore the fact that the op-amp in the screenshot isn't rail-to-rail, and yields a slightly lower output voltage).

    This is in turn fed to an IRF530N to get enough current to handle a 20w light bulb. The problem is that the output we're getting is barely 8 volts, instead of the wanted 12 volts. Can anybody clarify why that happens?

    Following image is a simulation from pspice showing our schematic
    http://peecee.dk/uploads/042010/Capture_Scematics.png
    And the output-window:
    http://peecee.dk/uploads/042010/Pspice_Simulation.png

    Thanks in advance,
    Adam
     
  2. beenthere

    Retired Moderator

    Apr 20, 2004
    15,815
    282
    You might get better control by placing the lamp in the drain circuit. A FET's conduction is determined bu the Vgs. It must be at least 10 volts above the source voltage to go into full conduction.

    With the load in the source circuit, all the voltage drops over the filament. As soon as the FET conducts, that voltage starts to rise. The upshot is that the source voltage rises with conduction, and there is no way to compensate for the decrease in Vgs. Even with 12 volts applied to the gate, the source voltage seems to have risen to 8 volts, making Vgs = 4. That is just Vth.
     
  3. SgtWookie

    Expert

    Jul 17, 2007
    22,182
    1,728
    Another problem you will face is high power dissipation in the MOSFET, as you are using it in a linear mode.

    Incandescent lamps have very low resistance when the filament is cold. When one first has power applied, current is very high until the filament starts to warm up. When the filament reaches operating temperature, the current will be normal; for a 12v 20W bulb, the current will be I=P/E = 20W/12v = 1.667 Amperes.

    However, when cold, the low resistance may result in anywhere from 3 to 10 times that amount of current. Measure your filament's resistance with an Ohmmeter, and divide 12v by that resistance to determine the current flow.

    A power MOSFET is best used as a switch, rather than a linear current limiter. Instead of controlling the current through the MOSFET in the linear region, consider using PWM techniques. When used as a switch, power dissipation in the MOSFET will be minimized, resulting in greatly improved efficiency.
     
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