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CREE Q5 LED light

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  • CREE Q5 LED light

    Got some of these LED's.

    http://www.dealextreme.com/details.dx/sku.11621

    I have one that runs on a 4.5 volt wall wart and gives out a good light without getting hot. These will run from 3.7v to 18 volt but at the higher voltages they get hot according to the feedback.

    I'm fed up with these machine lamps on stays as fitted the the Bridgy [ hawk - spit - ding ] no matter how you place them they get in the way of the hight precision quill fed lever [ laugh ] or the job.

    What I want to do is scrap these and fit a couple of these under the ram to shine down and another one or two on magnetic bases and plugs to go where needed, not always the same for different jobs.

    The lamps fitted ar 25 V AC Edison screw, another reason to get rid as the bulbs are expensive here, UK default are bayonet fitting, get hot and don't last long.

    The Bridgy has provision for 12v AC and 24v AC.

    Now the question?
    What resistance would I need to drop the 12v down to say 6v to handle 3 or 4 of these ?

    I know I'll need a bridge rectifier to get DC but it should still be at 12v when rectified.
    .

    Sir John , Earl of Bligeport & Sudspumpwater. MBE [ Motor Bike Engineer ] Nottingham England.




  • #2
    I don't think LEDs are made that take 18V. I think those already some voltage dropper circuitry in there.

    With LEDs it's a question of feeding the required current, controlled. If you can check the current taken by the one you currently use (pun intended), then just choose a resistor to feed the same current from your new voltage, and go from there. I've had no problem running many LEDs in parallel from a well controlled voltage source. Just mess with a resistor in parallel with a pot and tinker till the brightness is about right.

    If you buy LEDs from someone like RS, Maplin or Farnell, they tell you the brightness at certain currents, and the voltage you have to cover before anything happens. You don't get that with these made up units. You just have to experiment.
    Richard - SW London, UK, EU.

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    • #3
      Yes these LED's are self contained with their own on board regulator hence the 3.7 to 18 volts, at 5v they work well and don't get hot and this is what I want to drop down to
      .

      Sir John , Earl of Bligeport & Sudspumpwater. MBE [ Motor Bike Engineer ] Nottingham England.



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      • #4
        I looked at the link and saw that someone calling themselves "junkmeister" posted pictures of Evans "ray of death" flashlight.

        I thought Evan was the "Crapmeister".

        Brian
        OPEN EYES, OPEN EARS, OPEN MIND

        THINK HARDER

        BETTER TO HAVE TOOLS YOU DON'T NEED THAN TO NEED TOOLS YOU DON'T HAVE

        MY NAME IS BRIAN AND I AM A TOOLOHOLIC

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        • #5
          John, I'm guessing you mean you have 12V DC? A simple way would be to use an LM317, which is a 3 terminal adjustable voltage regulator. Depending on the variant, takes up to 27V input, and outputs 1.2 - 37V @ 1.5A.

          $2 at Radio Shack, I'm sure you have some equivalent.
          "Twenty years from now you will be more disappointed by the things that you didn't do than by the ones you did."

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          • #6
            I have a datasheet for these chip LEDs at work, but basically they are something like 3.5V LEDs. They get real hot at full power so the fact that you say they get hot is not all too surprising. The fact that it takes from 3.7-18VDC means there is most likely a current controller in the module, not just the resistor. Is the bulb a constant brightness, or does it vary with input voltage?

            I would try just using a bridge rectifier and a capacitor across it to give some filtering if needed (This is all that is usually in a non regulated "DC" walwart power unit). While we are at it, why not just use a 300-500mA walwart in the 5-12V range? Are you trying to package everything nicely onto the machine or something?

            To answer directly, to drop 6V, assuming the drawn current is 300mA (typical for these type of LED units) you would need a 20 ohm resistor. You would then need a 2 Watt power resistor:

            6V/0.3A = 20 ohm
            V^2/R = 36V/20ohm = 1.8Watts

            Also keep in mind that 12VAC is really 12*sqrt(2) in peak voltage, or about 16.8V. If the description is right you should be OK since it is under 18V.


            Let me know how your experiments go. It might be of help if you could put a current meter on it to see what it draws at different voltages.

            KEJR

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            • #7
              Go to any car boot sale ..there you will find hundreds of power supllies at lots of different voltages ..all mostly 50 pence each .

              BTW Home and Bargain were doing a 75 led array for £4.00..they work well ..equal to a 60 watt lamp

              all the best.markj

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              • #8
                Update,
                I have 12v and 24 volt from an internal transformer, this machine is supplied 440 volt 3 phase only, no neutral so I can't fit an internal wall wart and I don't want miles of cable going to outside sockets.

                Lazlo, good idea about the LM XX05's we have here, feed 12v DC in and get 5v DC out.
                I can make a simple board up with 3 or 4 of these, one per light and a bridge rectifier to turn the Ac to DC.

                I'll look for the bits tomorrow.
                .

                Sir John , Earl of Bligeport & Sudspumpwater. MBE [ Motor Bike Engineer ] Nottingham England.



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                • #9
                  I think the LM7805s will take a watt or two (check the datasheet) but if you use them I'd hook them up to a heatsink of some sort.

                  The voltage regulator isn't doing much more for you than a dropping resistor though (the LED is a constant load), so if you really do need to drop the voltage it might be simpler with the resistors.

                  KEJR

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                  • #10
                    Those are regulated dropins. If you add an lm-XX05 voltage regulator you will simply be buring off the excess volatage in a resistor at a different spot.

                    I'd run it straight from the 12 volt supply ( with filters to depress ripple).

                    Dan
                    At the end of the project, there is a profound difference between spare parts and extra parts.

                    Location: SF East Bay.

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                    • #11
                      Originally posted by danlb
                      Those are regulated dropins. If you add an lm-XX05 voltage regulator you will simply be buring off the excess volatage in a resistor at a different spot.

                      I'd run it straight from the 12 volt supply ( with filters to depress ripple).

                      Dan

                      Agreed, but OP stated he wanted to have the LED module run cooler. Hence the suggestion to use dropping resistors or external regulators.

                      KEJR

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                      • #12
                        A dropping resistor will work fine and moves the heat out of the module. Unless you use a switching regulator that extra power must be dissipated somewhere and outside the module is the best place.

                        The Q5 is about a 3 watt LED so it needs around 850 milliamps to run. Use a 10 ohm resistor in series with the 12 volt supply. The resistor should be wirewound and at least 10 watts. It will get hot enough to burn your fingers so keep that in mind. The resistor can be mounted on a heat sink with a clip to help it run cooler.
                        Free software for calculating bolt circles and similar: Click Here

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                        • #13
                          Wire three in series, then drive the string directly from your rectified 12vac. For the least flicker, put a capacitor on the output of the bridge rectifier, anything over about 470 uf and with a 25 volt rating will do. The on-board regulators will divide the voltage equally as they regulate the current to each led. If you want more light, add another string of three in parallel with the first string.

                          Each module will get about 5 volts or so- no need to add any regulator or dropping resistor. Everything will run cool.
                          I seldom do anything within the scope of logical reason and calculated cost/benefit, etc- I'm following my passion-

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                          • #14
                            This will handle every contingency:

                            http://www.electronics-lab.com/blog/?p=1955

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                            • #15
                              Originally posted by darryl
                              Wire three in series, then drive the string directly from your rectified 12vac. For the least flicker, put a capacitor on the output of the bridge rectifier, anything over about 470 uf and with a 25 volt rating will do. The on-board regulators will divide the voltage equally as they regulate the current to each led. If you want more light, add another string of three in parallel with the first string.

                              Each module will get about 5 volts or so- no need to add any regulator or dropping resistor. Everything will run cool.
                              I thought there was some reason why you couldn't run these in series like that. I'm sure I recall Evan explaining why, unless I'm remembering wrong.
                              Peter - novice home machinist, modern motorcycle enthusiast.

                              Denford Viceroy 280 Synchro (11 x 24)
                              Herbert 0V adapted to R8 by 'Sir John'.
                              Monarch 10EE 1942

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