Make Your Own LED Bulb Replacement for Regular Torchlight

LED torchlight are pretty common these days, but if you happen to have an incandescent filament light bulb based on 100 years old technology, here's your chance to have it updated with LED that last 8000 years! (if incandescent has a human lifespan)

What this instructable does:
Take a regular PR 2 bulb with a P13.5s base (i.e. those most commonly in torchlight), reuse the base and insert 3 x 5mm diameter white LEDs. See image of end result below.

The shortcut to this project is to go into a shop and buy one of these. But hey, what's the fun in that when you can make your own!

Step 1: Removing the Glass Bulb

This task is pretty hazardous, so be very careful, tiny shards of broken glass cut easily and are painful to remove (from my own encounter with a broken bottle), I suggest covering the bulb with a kitchen towel and gently grip with a plier, twist and nudge it out.

The glass bulb is fused to the metal base with a clay-like cement, and should crumble easily. Once the bulb is removed, use a mini screwdriver scrape the metal base clean. The internal surface wall of the base has to be electrically conductive.

What's left is seen in the image below.

Step 2: Desoldering the Remnant of the Glass Bulb From the Base

Having done this, you can now insert the LED legs through this hole, where it will be the positive connection to the battery.

Step 3: Shaping the Legs of LED

Now for some Uri Geller magic tricks!

What you need to do is bend the legs, solder the anodes together (that's the positive polarity leg, the longer one, or see ). A bit of tape would help securing them together while you do the soldering.

Next, fold the cathode legs into a u-shape, about 7mm long, these have to make contact with the internal wall of the metal base.

Once you have done that, what you have is something like the image below.

Insert this into the metal base, through the hole at the end, solder and cut away the protruding excess.

Step 4: An Important Final Note

The electrical characteristic of the LED bulb is completely different from a normal incandescent bulb, if you have chosen to use white LEDs, you would typically need at least 3.3V to turn it on, producing optimum light output ( say, 3 x 1.2V rechargeable nickel metal hydride) , if you are using red LEDs, these may require only 2.2V, and can be used as a direct replacement bulb for a 2 cell torchlight, no problem.

When using a white LED, the easiest method of driving these, is a connection in series, with a shunt resistor. Working on typical values for a 5mm white LED.

For 3 battery cell, between 3.6V to 4.5v, depending whether you're using alkaline or rechargeables, assuming a nominal of 4V, with the white LED operating voltage and current at 3.3V, 30mA respectively, (4 - 3.3)/30e-3 , a 22 ohm resistor will do.

For 4 battery cell, between 4.8V to 6V, assuming a nominal of 5.4V, with the same typical values for a white LED, (5.4 - 3.3)/30e-3, a 68 ohm resistor will do.

A direct placement of the resistor in the metal base could save you more hassle.



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    12 Discussions


    12 years ago

    I don't thin the resistor's function or importance is adequately explained. A lot of people are going to burn out a lot of LEDs just throwing them in there. Also, it's worth emphasizing that this will only work if the battery voltage is higher than the LED's forward voltage. No white LEDs in 2-cell lights, unless you build a boost converter! My favorite circuit for driving white LEDs from a single cell is the "Joule Thief", explained so well by Big Clive.

    4 replies

    Reply 10 years ago on Introduction

    The only thing needed to not burn out the LEDs is a resistor somewhere in the circuit in serial with the LEDs.


    10 years ago on Step 3

    Great thinking. i just use this method and eureka i got my flash light to light up with led


    11 years ago on Step 4

    Hi, Great item! Will try it soon with my underpowered CatEye HL-330H. By the way, based on your calculations you mean of course a series resistor, not a shunt (parallel) resistor. A shunt resistor would be in parallel with the battery and would, rather uselessly, draw power from the battery independently of what is flowing through the LEDs while providing no protection against overcurrents. Whether to put the LEDs themselves in series or parallel is another issue. Since white LEDs require higher voltages you might be tempted to put them in parallel (cathode to cathode, anode to anode), and that may work OK. You may get unequal currents through the LEDs, though, unless either the LEDs are rather well matched or you have a series resistor for each LED - tying each anode to the positive through its own dropping resistor. Series connection guarantees equal currents thru the LEDs and requires only a single current limiting resistor but requires a voltage equal to triple the amount for a single LED, plus half a volt or so - more for long chains - across the dropping resistor. P.S. Do not hook the LEDs (or the battery) up backwards. A careful design might use a series diode (or a largish shunt diode with a fuse at the battery) to avoid this. Remember the anode attaches to the positive supply voltage, the negative to the other. For LEDs in series put the anode end of each diode closest to the positive end of the series chain.


    12 years ago

    I'm running a 2D-cell flashlight with 3 LEDs; white 10000mcd ones from the seller that Dan's-Data talks about here, and it works just fine, because they have builtin voltage regulation:

    Next step, to make a color-changing flashlight.. The tricky part is that the plastic base of the LEDs sticks slightly over the metal lip of the bulb, so it took some fiddling to get everything to contact properly in the flashlight I used, but it works in the end.

    Nice idea -- thanks!


    12 years ago

    Right on! Every once in a while I see an istructable that makes me wonder how I didn't think of it myself. This is one, I can't wait to start making these--THANKS!