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Dimming a 6V bulb by half.

Discussion in 'Electronic Basics' started by Tibur Waltson, Jan 15, 2004.

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  1. My spot light is too bright (500,000CD). I like to dim it down when-
    ever I toggle a switch. It uses a 6 volt lead-battery with a 25W bulb.
    I'm trying to dim this in the most simplest way. How could I build one
    that could dim the light by half? I have resistors, mosfet, etc.
    TIA, Tibur
     
  2. rob

    rob Guest

    You CAN use a mosfet in a 'linear' control scheme , but it is goung to
    dissapate a lot of heat and will require a large heatsink.

    A simple PWM would be much better.You can then vary the bulb
    brightness from off to full on with very little dissapation in the
    FET. You can build a nice simple one with a lm393 a few resistors and
    caps and a mosfet. (logic level mosfet for your 6V app)
    Use half of the 393 as a oscillator to generate a rough sawtooth
    waveform.
    Use the other half as a comparator to compare a reference(from a pot)
    with the
    sawtooth. Output is a PW Modulated square wave which you feed to the
    mosfet.
    Used this design with a IRF1010e mosfet to control 100W 12v bulb.
    Worked like a charm , did not even need a heatsink on the Fet. I made
    one small enough to fit in the handle of my "spotlight" with a pot on
    the bottom of the handle for brightness.
    If you want I'll send you a schematic (email).Can't seem to get to
    a.b.s from here :0(

    Cheers
    Rob
     
  3. Tibur,

    Bulbs are made to do their job at some voltage. 6V in your spot. Lowering
    the voltage will reduce the brightnes but more than proportional. It will
    also lower the current, but less then proportional. I wonder whether you
    will have any usefull light left at 3V. The best way to dim your light is
    looking for a 6V/10W or 6V/15W bulb.

    petrus
     
  4. James

    James Guest

    Why dont you try adding a few diodes in series until you get your
    desired brightness?

    A very simple idea, but it may accomplish what your after.

    James
     
  5. Garrett Mace

    Garrett Mace Guest

    You CAN use a mosfet in a 'linear' control scheme , but it is goung to

    PWM gets some more bonus points in this application, since the spotlight
    will actually use less battery power and last longer when dimmed. Light
    where you're walking, not the whole countryside, unless you need it.
     
  6. I have a square wave PWM. If it actually save more power I will
    use this to drive the mosfet. If I understand correctly the PWM acts
    like turning the switch on or off quickly. Mosfets are happy either on
    or off. But why do my sewing machine speed control transistor
    don't become hot. They don't seem to use PWMs or do they?
    Thanks
    Tibur
     
  7. I have a 4.7V/10W. How can I use this bulb without burning out the filament?
    Thanks
    Tibur
     
  8. Would diodes work on DC bulbs? What
    mechanisms in the diodes make the lights dim?
    Thanks
    Tibur
     
  9. The bulb will not like it so you'll need some extra spares. The power
    savings are also questionable. A cold buld needs much more current then a
    hot one.

    petrus
     
  10. Ian Stirling

    Ian Stirling Guest

    The better way is probably to cut a small hole in the reflector and
    add a small spotlight bulb (with reflector) in it.
    The newer crop of these lights have a small light along with the big one,
    so practically, it may be easier to just buy one.
     
  11. Bob Masta

    Bob Masta Guest

    The bulb should be perfectly happy with PWM, as long as you don't
    make the frequency so low that the filament cools down between
    pulses. That's not likely to be an issue here, since you'd have to
    slow it down so much it became a flasher. So you will indeed get
    excellent power savings, inversely proportional to the duty cycle.


    Bob Masta
    dqatechATdaqartaDOTcom

    D A Q A R T A
    Data AcQuisition And Real-Time Analysis
    www.daqarta.com
     
  12. A diode only passes one half the waveform, and leaves an effective
    70.7% of the voltage, less the forward drop of about .6 volts. 12.6 *
    ..707 = 8.9082 -.6 = 8.3082 volts.

    Some series string tube TV sets were built with a diode in series
    with 84 volts worth of tubes, and ran on 120 volts. if you plugged the
    set into DC, the tubes would either not light, or you would blow the
    filament in at least one tube.

    --
    We now return you to our normally scheduled programming.

    Take a look at this little cutie! ;-)
    http://home.earthlink.net/~mike.terrell/photos.html

    Michael A. Terrell
    Central Florida
     
  13. James

    James Guest

    Each diode would drop the voltage by about .6 of a volt, therefore
    dimming your bulb slightly. Very simple, and effective.

    James
     
  14. James

    James Guest

    The guy did say that this was d.c. so there is no waveform. They are
    used here for dropping voltage slightly.

    James
     
  15. Guest

    | Tibur Waltson wrote:
    |>
    |>
    |> > >My spot light is too bright (500,000CD). I like to dim it down when-
    |> > >ever I toggle a switch. It uses a 6 volt lead-battery with a 25W bulb.
    |> > >I'm trying to dim this in the most simplest way. How could I build one
    |> > >that could dim the light by half? I have resistors, mosfet, etc.
    |> > >TIA, Tibur
    |> > >
    |> > > Why dont you try adding a few diodes in series until you get your
    |> > desired brightness?... but it may accomplish what your after. James
    |>
    |> Would diodes work on DC bulbs? What
    |> mechanisms in the diodes make the lights dim?
    |> Thanks
    |> Tibur
    |
    | A diode only passes one half the waveform, and leaves an effective
    | 70.7% of the voltage, less the forward drop of about .6 volts. 12.6 *
    | .707 = 8.9082 -.6 = 8.3082 volts.

    What kind of waveform do you get from a lead battery (like the OP has)?
     
  16. I cannot get a 12-volt PWM to work on
    a 6-volt power supply. What is wrong?
    Tibur
     
  17. a) Pick a higher PWM frequency if you are worried about that

    b) Surges are not as hard on lightbulbs as many people think.

    Although household incandescent lightbulbs usually burn out from a
    cold start, the famous surge of a cold start does not cause extra wear
    so much as kill lightbulbs that are already so badly aged (and by
    run time rather than starts!) that their hours are numbered.

    details in http://www.misty.com/~don/bulb1.html
    At equal voltage, a cold filament needs more current than a hot one.
    But in a PWM scheme, the "average current" (battery drain) will vary
    directly (but less than proportionately) with the duty cycle. Give a
    lightbulb full voltage for 50% duty cycle (1 millisecond on, 1 millisecond
    off) and the average current (and this is actual power consumption) will
    be about 70% of that with the same voltage applied steadily.
    (Light output will be about 30-33% of full light output from that
    roughly 68-72% of full power, and that means if the lightbulb in question
    is usually being dimmed then you get better energy efficiency by using a
    lower wattage lightbulb than from using one that is normally being
    dimmed.)
    - Don Klipstein ()
     
  18. The effective voltage is the RMS voltage, and in a variable duty cycle
    application of full voltage the RMS voltage is the peak applied voltage
    times the square root of the duty cycle.
    If you want PWM "effective voltage reduction", then the duty cycle
    should be the square of the ratio of desired effective voltage to peak
    applied voltage.

    Examples:

    4.7 volts desired effective voltage from 6 volts: Take the square of
    (4.7/6), and that is .614, or 61.4%

    4.7 volts from 6.3 volts, higher side of lead-acid "6-volts":

    Square of (4.7/6.3) is .556, or 55.6% - a 3/5 duty ratio is probably close
    enough to run a 4.7V bulb from "6V"

    4.7 volts from 12V: Square of (4.7/12) is approx. 15.3% - a 1/6 duty
    cycle should be close enough for "4-cell" flashlight bulbs to be
    powered from 12-12.6V

    The main "gotcha": Slight chance the PWM frequency or one of its
    harmonics will excite a resonance of the filament seriously enough to make
    it vibrate enough to suffer damage. Try projecting an image of it onto a
    wall with a smaller magnifying lens of shorter focal length - if the ends
    appear skinier than the middle region (even considering that the ends may
    not be coiled while the middle is usually coiled), then try a different
    frequency. Try variable frequency to see what is best and worst for
    shaking up the filament. Then again, most lightbulbs last a little longer
    on 60 Hz AC than on DC among the ones where a difference is measurable.

    - Don Klipstein ()
     
  19. Diodes have a voltage drop. Most rectifier diodes have a voltage drop
    close to .8 volt at usual current and usual temperature, maybe .7 volt
    at lower currents or where they have less "heatsinking" than "usual good
    design" for the amount of current being conducted.

    At lower currents .5-.7 volts is common (At microamps, many common
    rectifier diodes at room temperature can drop .35-.4 volt). Schottkey
    diodes are special lower voltage types and many of that type drop near .5
    volt at full current. On the other hand, diodes heavily heatsunk and
    conducting higher currents can drop 1-1.1 volts. Diodes conducting spiky
    current waveforms tend to have a slightly higher voltage drop than ones
    conducting more continuously.

    - Don Klipstein ()
     
  20. Problem solved. I just add an old 9-volt battery in series.
    Thanks all.
    Tibur
     
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