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converting LED flashlight to mains power?

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  • converting LED flashlight to mains power?

    I have a flashlight with flat LED array of 24 emitters that is powered by 3 1.5volt AA batteries. I would like to mount this on my drill press in the housing that is there for a light. The existing light is too dim.

    Will it work if I just connect a 5volt power supply to the connecters that the batteries connect to on the PCB?

    In the picture the black looking box is the on off switch. The only electronic component on the board is the unit you see just under the black box switch. It has written just above: YC- WL1015

    How to become a millionaire: Start out with 10 million and take up machining as a hobby!

  • #2
    Most of those displays use the driver board to supply the
    correct current to the LED's. The driver board usually will
    take a range of voltage. Might try ebay to see if you can
    find something similar to guide you. If it were mine I would
    give it a try. If you look at three AA batteries fresh they are
    more than 4.5 volts.
    olf20 / Bob

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    • #3
      That resistor between the switch and the LED array is the current limiting device. You should be able to connect a 5 volt supply to where the batteries go. No problem.
      https://www.youtube.com/channel/UCIF...7S66kX1s8rd0qA

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      • #4
        +1 with Joe_B. Get the polarity right and it'll work great.
        Definition: Racecar - a device that turns money into noise.

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        • #5
          you might find that the powersupply doesnt put out enough current.

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          • #6
            I have done just what all of you have recommended. I am using a 5volt dc 1 amp WART. It worked fine for the first hour. As of now only 8 of the 24 are still working. Answers????

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            • #7
              Your wart is most likely unregulated and is only putting out five volts with a heavy load. Lightly loaded, the voltage on these wall warts can get pretty high. Do you have a volt meter?
              https://www.youtube.com/channel/UCIF...7S66kX1s8rd0qA

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              • #8
                Can you post a photo of the other side of the circuit board? How were the three batteries installed? It could be the batteries were installed parallel rather than in series, giving 1.5V and more current, rather than 4.5V.
                Kevin

                More tools than sense.

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                • #9
                  They would have to be in series, 1.5 volts would probably not be enough voltage to forward bias an LED.
                  https://www.youtube.com/channel/UCIF...7S66kX1s8rd0qA

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                  • #10
                    I did a quick lookup - "white" LEDs run from 2.2V to 3.6V. So series it would be. (Infrared LEDs run at 1.5V -- that's what made me think of it -- the TV remote has the batteries installed in parallel.)
                    Kevin

                    More tools than sense.

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                    • #11
                      that component below the switch in the picture is the current limiting resistor, with a value based on the input voltage (4.5V max, 3.6-3.8V typical) and current draw of the LEDs (probably ~20mA per LED, these are probably all in parallel). If you can look up the resistor value, you can find out what current the board is supposed to be regulated at. Then just work out what resistor you'll need for 5.5V and wire that in its place. These things are about as simple as you can get, so you can't really go wrong if you get the right resistor in there.

                      If you want more light that what this provides, high power LEDs are where it's at. I've made a lot of LED lights (just finished one last night ) so if you have any questions, feel free to ask.

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                      • #12
                        I would recommend calculating a new value for that resistor. Base on the picture, the colored stripes appear to be brown gray gold
                        gold which would make that a 1.8 Ohm resistor. Since you are increasing the voltage by 11% that gets you to 2 Ohms. You should
                        also add the internal resistance of the three AA batteries you eliminated from the circuit which puts you at about 2.6 Ohms.
                        2.7 Ohms is a standard value. If the supply is unregulated and you want to allow for line voltage fluctuations of about 10% go to
                        the next higher value which would be 3 Ohms. To determine the power rating required, connect three fresh batteries to the light
                        and measure the voltage across the 1.8 Ohm resistor. Divide that by 1.8, square it, and multiply by 3. That will tell you how many
                        Watts the new resistor will need to handle.
                        If that's too much work, at least add a diode in series with the existing resistor to keep the voltage more in line with the original
                        design.

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                        • #13
                          Unfortunately, you can't just throw a random power source on a flashlight and expect it to work perfectly for a long time. If 8 of the 24 are still working, it means that one of two things have happened....

                          Scenario 1) The wall wart is putting out LESS voltage than the LEDs need. LEDs light up when the voltage reaches a minimum value. That minimum value is slightly different from one LED to the next due to manufacturing variances.

                          Scenario 2) The wall wart is putting out significantly MORE voltage than the LEDs need. LEDs are strange devices. A voltage as little as 10% too high can overpower the LED by 100%. This causes them to be really bright for a short time (minutes to days) before they burn out.

                          To diagnose the situation you have to check the voltages. If you have a volt meter, measure the voltage across the legs of any of the LEDs. If the voltage is above 3.6V chances are that they are burning up. If it's below 3 volts, it might not be enough to light the LED.

                          Small LEDs are usually rated at 1/50 of an amp at the correct voltage. 24 of them will pull 1/2 amp from your wall wart if the voltage is close to correct. If the voltage is as little as 1/10 of a volt higher the current can double, making the draw on the power supply over an amp. Double A battery voltages drop very quickly as they are used, so even if the voltage starts too high it drops to a safe level within 5 minutes.

                          The key to adding a power supply to an LED device is that you need to be able to measure the amps that are being used and change the power levels to keep it in a safe range.

                          I have a small under cabinet lamp from Costco that I run from a USB power supply. It has only 6 LEDs and was originally run by 3 AAA batteries. It works OK because the power supply puts out 4.5 volts and there is a small regulator chip that keeps the amps at the right level. If it used just a resistor I would have altered the resistor to keep it in range.

                          The calculator at http://www.hebeiltd.com.cn/?p=zz.led...tor.calculator can be used to decide on the correct resistor value.

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

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                          • #14
                            Using the method above you can even run it directly from a 110VAC source, just have to change the resistor to keep the current at the required level. you will get a drop in light as the LED's will only run 1/2 of the cycle, which you could solve by putting in a couple of diodes to make a bridge rectifier (of buy one ready made). Just make sure if you do that you have it well insulated as you dont want any "line voltage accidents"

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                            • #15
                              Originally posted by theGallery View Post
                              I have done just what all of you have recommended. I am using a 5volt dc 1 amp WART. It worked fine for the first hour. As of now only 8 of the 24 are still working. Answers????
                              Is this BlackForest on a different account, or a different person?

                              In either case, look at the other side of the board to see if there are any other components. Since a white LED takes around 3.2V to light up, with just a 4.5V battery and no voltage boosting circuitry, they must all be in parallel to work. If they were in 3 parallel strings of 8 LEDs in series, then it would take 25.6V (8x 3.2V) to light them up, or if they were all a single series string, it would take over 75 volts.

                              If all 24 LEDs are in parallel, then all bets are off when you connect a wall wart, since they won't all draw the same amount of current, so some of them will suck down too much and die. Many (most?) of these flashlights are very poorly designed in an attempt to minimize cost. The ones I have at home depend on the internal resistance of the batteries to limit current, so if I were to replace the cheap AA's with good quality ones, there's a chance I'd blow the LEDs.

                              If the wall wart is a regulated one, then you have a better chance of it working, but even then it's touch and go.

                              To add to what Dan said above, LEDs are really current-driven devices. That's why a tiny change in the voltage can have such a huge impact: that little voltage difference caused a big change in current.
                              In another thread yesterday, I showed a diagram for a low-cost current here. It will supply enough current to light an LED over a wide 7-15 volt supply range.
                              Last edited by lwalker; 01-17-2014, 02:13 PM.

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