Dimming a white led

Discussion in 'The Projects Forum' started by bigkim100, Oct 21, 2007.

  1. bigkim100

    Thread Starter New Member

    Oct 21, 2007
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    I need to simulate a bright white flash of light, for a movie prop. When a button is pressed I need a white led to flash on , full brightness, then slowly decay in brightness for approximatly 3-5 seconds to fully off.
    Any Ideas?
     
  2. John Luciani

    Active Member

    Apr 3, 2007
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  3. SgtWookie

    Expert

    Jul 17, 2007
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    Actually, that should be fairly simple.

    You need just a few parts:
    1) Your LED(s) (normally, the short lead is the negative supply)
    2) A power source (battery)
    3) A capacitor
    4) A momentary pushbutton-type NO (Normaly Open) switch
    5) A resistor to limit the maximum current through your LED
    6) Interconnecting wire or PCB.

    In the attached schematic/pictorial representation, the switch is normally open. When the switch is momentarily pressed, the capacitor charges nearly instantly to the battery's voltage.
    The resistor limits the amount of current through the LED(s). The power in the capacitor will be discharged through the resistor and LED over time, becoming dimmer as time goes by. How fast this happens depends upon how large the capacitor is, and how much current is being drawn through it.

    Figuring out the value of the resistor will depend upon your LED's rated voltage and current, and the voltage rating of your battery.

    For example, if your LED is rated for 70 mA at 2.6 volts, and you're using a 3.6 volt rechargeable battery borrowed out of a portable telephone:

    3.6v - 2.6v = 1 volt

    You have 1 volt left that you need to limit to 70 mA of current going through it.

    Ohm's Law is: I = E / R, or Current = Voltage / Resistance
    We need to calculate resistance for a known voltage and current, so we change the formula:
    R = E / I
    R = 1v / 70 mA, or 0.07 A
    R = 14.28 (approx)

    So, we need to have at least 14.28 Ohms to limit the current. 15 Ohms is the closest standard value, so we'll try that. Going back to Ohm's Law:
    I = E / R
    I = 1 / 15
    I = 0.0666 Amperes, or 66.6 mA - that'll work.

    Let's check the wattage:
    P = E * I (Power in Watts = Voltage * Current)
    P = 1 * 0.0666 Amperes
    0.0666 Watts
    You could safely use a 1/10 Watt resistor in this case.

    Now for the capacitor sizing - you want the light to fade over a few seconds.
    For that, we'll need to figure the total circuit resistance
    Our resistor is dropping 1 volt; it's 15 Ohms.
    The LED is dropping 2.6v, and has 66.6mA going through it. Back to Ohm's Law:
    R = E / I
    R = 2.6 / 0.0666
    R = 39 Ohms (approx)
    Rtotal = 39 + 15
    Rtotal = 54 Ohms
    So, we need to find a value for the capacitor that will allow it to discharge over a 2-second time span.
    Time = R x C
    Time = 54 x ?
    C = 54 x 2
    108
    But we want 1/5th of that, so around 22,000 uF should work pretty well. (In the first 1/5 of the time constant, the capacitor is 80% discharged. In your case, cap voltage would be down to 0.72, current through the resistor and LED would be 13mA, or very dim.)

    So, whatever your specifications for your LED, plug them in.
     
  4. thingmaker3

    Retired Moderator

    May 16, 2005
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    To the Sargent's design I would add a high-value resistor in parallel with the capacitor. Say a megOhm or two. This will have little effect on how the LED flashes, but will provide a means of discharging the capacitor after the device has been used.
     
  5. Voltboy

    Active Member

    Jan 10, 2007
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    How exactly did you ended up with C=54 x 2.
    I think it's.
    Time = R x C
    2 = 54 x C
    C = 2/54
    C = 0.37037
    C = 37.0370 mF
     
  6. SgtWookie

    Expert

    Jul 17, 2007
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    Actually, you're right - I messed up that part. :rolleyes:

    Decided I'd check it out for myself with what I have onhand; an LED rated 70mA @ 2.6v, a 15 Ohm resistor and a 10,000 uF (10mF) electrolytic rated 6.3v.

    Something I hadn't figured on was what Thingmaker alluded to; the resistance of the LED apparently increases as the voltage across it decreases; visible (albiet dim) light was still being produced by the LED even a minute after the 10mF cap was charged! :eek:

    I added a 680 Ohm resistor across the LED. This resulted in a seemingly linear decay from maximum brightness to dark in just over 3 seconds.

    Changed that resistor to 380 Ohms, and wound up with 2.2 seconds. That seems just about right for the setup I made. Might make it a bit more dramatic for a particularly poignant scene to "drag it out" a bit to over 3 seconds.

    bigkim100, are you still with us?
     
  7. thingmaker3

    Retired Moderator

    May 16, 2005
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    ..........
     
  8. Voltboy

    Active Member

    Jan 10, 2007
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    I was bored and tried this circuit.. but with other values, similar though.
    When the battery is on the LED is very bright. When You disconnect the battery. The LED's brightness go low very fast, and keep up a time in a very low brightness (like enough brightness to see when its on, you see a small light dot inside the LED). I guesses it's because capacitors maintain voltage, and the LED needs current to be bright, so I think this circuit doesn't really works for dimming a LED.
    When I disconnected the battery, the cap was still giving the same voltage, but about 0.01mA, compared to the 20mA from the battery.
     
  9. arthur92710

    Active Member

    Jun 25, 2007
    307
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    well I just tryed a simple circuit that dims an led.
    As Yoda said "the LED needs current to be bright" I said a resistor resists current.
    So i connected a led 9v battery and a pot. when you spin the pot the led is bright then dims and shuts off.
    So, if you don't mind manually dimming the led you can use this.
     
  10. Voltboy

    Active Member

    Jan 10, 2007
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    I think he wants it automatically.
     
  11. SgtWookie

    Expert

    Jul 17, 2007
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    Of course, the connection from the battery to the cap has to be simply momentary.

    As long as the battery is connected to the capacitor via the switch, the LED will be lit @ maximum intensity. But with the LED that I used, with the components I specified (including the after-mentioned 680 or 380 Ohm resistors across the LED) the intensity of light emitted by the LED faded in a somewhat linear fashion from full brightness to extinguished in about 3 or about 2.2 seconds, respectively (680 vs 380 Ohms across the LED)

    If your components don't match what I used, you're going to get different results.
     
  12. arthur92710

    Active Member

    Jun 25, 2007
    307
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    using a 680k resistor in parallel to the cap works well.
     
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