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OT: LED Headlamp Power Replacement Question

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  • OT: LED Headlamp Power Replacement Question

    I have an LED headlamp in the shop that gets well used. The lamp is 3- AAA powered so I go through quite a few AAA’s over time. In the mail I get a small USB lithium ion pack for mobile phone back-up power from some magazine subscription gimmick. It sits on the shelf for a couple of months until it occurs to me that it might work well in my headlamp being conveniently rechargeable. It does quite well actually except its 5 volts runs the lamp a little too hot as compared to the 4.5 volt of 3 AAA’s. The lamp has 6 LED’s in parallel and draws .5 Amp at 5 volts.

    My knowledge of electrons is a little weak but is this as simple as placing a resistor in the circuit to drop the voltage down a notch or are there other complications?

  • #2
    Your led array draws a fair amount of current and will drop the voltage from the aaa's so they won't burn too hot. This probably is not true of the lithium pack, so yes a single resistor in series is probably a good idea. A wild guess suggests to me about 5 ohms, so the standard value of 4.7 ohms should be close to right. You could instead put a single rectifier in series- that will drop the voltage about the right amount. Something like the 1N400_ series, like a 1N4001 would be fine. The resistor won't care, but the rectifier cares about polarity, so it will only work one way.
    I seldom do anything within the scope of logical reason and calculated cost/benefit, etc- I'm following my passion-


    • #3
      1.5 x 3 = 4.5

      One li-ion cell will run from 4.2 to 3.5 full swing. I have replaced alkaline 3S batteries with a single 18650 or a few in parallel with no issue. Make sure you get one with a protection circuit to prevent over-discharge.


      • #4
        Thank you psomero. Will the resistor need to be rated for at least the 2.5 watts of the circuit .5A x 5V?


        • #5
          The resistor is only dropping part of the voltage, therefore it would only need to be probably 1 watt.
          I seldom do anything within the scope of logical reason and calculated cost/benefit, etc- I'm following my passion-


          • #6
            Thank you all! I went with the diode. It's much smaller than a 1 watt resistor. Lamp is plenty bright and now draws .25A. All runs cool.


            • #7
              5 V - 4.5 V = 0.5 V

              That is the additional Voltage you want to drop.

              0.5 A, you say so you want to drop that 0.5 V when the current is about that much of a bit less. Not knowing the actual current with the three batteries makes us guess at this point, but it is a fairly accurate guess.

              So, good old Ohm's law:

              R = E / I

              R = 0.5 V / 0.5 A

              R = 1 Ohm

              That should be a very good starting point.

              And you are right to ask about the power rating.

              P = E * I

              P = 0.5 V * 0.5 A

              P = 0.25 W

              That is 1/4 Watt, but it is not good practice to have the resistor's power rating exactly match the expected value. So I would step up to a 1/2 W resistor.

              You may not have any 1 Ohm, 1/2 W resistors but you can put two 2 Ohm, 1/4 W ones in parallel to get the right value and power rating. Four 4 Ohm, 1/8 W ones would also do.

              If you install the resistor and find that it drops more or less than 0.5 V, you can adjust the value until you hit the right size. This is because you did not give the original current, with the three batteries. Or you could just measure the current with the batteries and start over with the calculation chain above.
              Paul A.
              Golden Triangle, SE Texas

              And if you look REAL close at an analog signal,
              You will find that it has discrete steps.


              • #8
                Measure the current through the LEDs with the 3 AAAs, then put in a resistor to limit the current to that amount when you are using the Li-Ion pack.


                • #9
                  Alternatively, a silicon diode will dropp about 0.6V at almost any current. So use a single 1A diode to do the job.

                  We did this with a stack of diodes to drop a laptop power supply voltage from one of our laptop supplies to that needed for a foreign visitor's (who'd forgot to bring it with him) machine back in 1998.

                  Then the sod took it home with him, so we were left with an expensive laptop without a supply...
                  Location- Rugby, Warwickshire. UK


                  • #10
                    Say you had an old laptop power supply that produces 24v. How could you split that voltage to give 12v supply? Still rely on resistors? Or is there another way?
                    West Sussex UK


                    • #11
                      I do some gaming and use a controller rather than the keyboard. I have been using rechargeable AA batteries in the controller for years with great success. Simple & cheap.


                      • #12
                        An alternative to driving leds and led arrays is a constant current circuit. You will always need about one volt more from the power supply than your led array wants. This is my preferred way of driving led arrays, as you can maintain a steady and safe brightness as the supply voltage degrades- within limits. The circuit is fairly simple, though we're now beyond what you might make for use in a headlamp. I like the constant current circuit as you can arrange for variable brightness with a fairly linear control voltage, and still have a current cap at some fixed level. I used this method in one of my stage lighting schemes.
                        I seldom do anything within the scope of logical reason and calculated cost/benefit, etc- I'm following my passion-