How to Turn a Dead Flashlight Into an Uber LED Light




I've had this dead flashlight lying around in my shop for a few months, and I decided that it was time for it to live again!

I've broken this instructable (and the flashlilght) in to a few easy steps:

Step 1: Open up the flashlight,

Step 2: Clean out the dead bulb,

Step 3: Calculate the resistance needed,

Step 4:  Solder the LEDs and the resistor together

Step 5:  Solder the LEDs to the bulbs, and

Step 6:  Reassemble the flashlight.


Step 1: Open Up the Flashlight

The rubber handle was glued in place, but since this flashlight was a few years old, the glue was brittle and the handle peeled right off. After that, it was just a case of prying the light open.

Step 2: Clean Out the Dead Bulb

I wanted to use the bulb socket, so I needed to be rid of the glass bulb. After wrapping some paper towel around the bulb, I crushed the glass with a trusty pair of pliers

Afterwards, I used a power drill to clean out the insulating material out of the base. Remember TO USE SAFETY GLASSES! There will still be bits of glass in the socket. I will use a non conducting epoxy later to replace what I've removed.

Step 3: Calculate the Resistance

Ah, math! Isn't math great! without it we wouldn't have spaceflight,  medical advances, or grilled cheese sandwiches! (2:1 bread to cheese ratio. See? Math!)

In this instance however, we're just using using math to make sure we don't blow up the LEDs (which I've done in the past. I mean, like Pow!)

Part one:

Since my source voltage (VS) is the three AA batteries in series, the total is 4.5 volts. I intend to connect the LEDs in parallel, so I'm going to take the easy route, which is to calculate the resistance for one LED, then using Ohm's law, calculate the resistance for three LED's in parallel. White LEDs require about 3.6 volts to light (known as forward voltage, or VF), and draw 20 miliamps (Ma) of current , known as I


VS= 4.5V,
VF= 3.6V
I= 20 Ma= .02A

What does this all mean? Using the formula R= (VS-VF)/I  we get this:

R= (4.5V-3.6V)/ .02A
R= 45 Ohms

Now then, since the next highest resistor is 54Ohms(Always go up in resistors, never down), we will use this value for the next step, Calculating the equivalent resistance for three LEDs in parallel!

Part Two

This is by far the easiest part. When you have to calculate an equivalent resistance (Req)for three LEDs in parallel,  all you need to do is take your resistance (R) from step one, and divide it by the number of LEDs (n LED for this instructable) you intend to use.


Req= R/n LED
Req= 54 Ohm/ 3
Req= 18 Ohms! Easy peasy!

Step 4: Solder the LEDs and the Resistor Together

Just a quick note about LEDs.

LEDs have a polarity, The long lead is the positive, the short one is negative. If you've already snipped your LED leads, you can also tell which one is the short  (negative)by looking at the base of the "bulb". One side of the base has a flat spot. The negative lead is the same side as the flat spot .

Here you can see that I have soldered the like polarities together, and I wrapped and soldered the 18 Ohm resistor (this is the negative end). The other three leads will be soldered to the bulb socket.

Step 5: Solder the LEDs to the Bulbs

Insert the resistor/ LED assembly into the socket, so that the reisistor lead pokes out of the bottom of the base. Solder the resistor lead and then solder each of the remaining leads to the top of the socket.

After the soldering was done and the socket had cooled, I mixed a bit of Aves Apoxie Sculpt (Oh, that wonderful Aves!) and pressed it into the voids around the resistor. Aves is a non conductive two part epoxie that scale and figure modellers use for sculpting. You could just as easily use epoxy glue, I just had the Aves on hand.

Step 6: Reassemble the Flashlight.

Put the new and improved bulb back in the light, and reassemble. I used Methylene Chloride (AKA liquid glue) to glue the rubber grip back onto the flashlight body, and I put fresh batteries in while I was at it.

To the casual observer, it looks like any other el-cheapo flshlight. The only clue to its true nature is found by looking right at the bulb.

...although some people miss that clue...


So, there's the Uber light, sitting on the workbench, when along comes one of my coworkers. Having seen the light sitting on the counter for weeks before hand, he picks it up and pointing it at his face, he turns it on. (He wasn't aware of my jiggery-pokery)

He lets out a girlish squeal (think nine years old, with pigtails), and laments,"Why is this thing so bright!"

Hopefully you've all found this instructable entertaining and educational. Happy hacking!



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    25 Discussions

    Just as a medical doctor thinks & performs his/her duty within a certain paradigm,

    there are others that do the same thing because that is how they were educated....

    Be it high school, college, or some other educational institution, these people work, think, & live, within a certain paradigm...

    I do not know anything about bat159, and I hope that he/she doesn't think I'm trying to disrespect him/her, but I think that may be the case here. bat159 is so used to thinking within this paradigm, that he/she often forgets that the rest of us do not think that way & hence, do not understand a lot about LED's, or in my case, I do not understand a lot about electronics in general....but thanks to him/her taking the time, I think I understand now about LED's.


    noun\ˈper-ə-ˌdīm, ˈpa-rə- also -ˌdim\

    : a model or pattern for something that may be copied

    : a theory or a group of ideas about how something should be done, made, or thought about


    Reply 5 years ago on Introduction

    You sir are correct! I've been playing with/ working with electricity/electronics for about 40 years now. Sometimes I have to step back and think about an alternate explanation when I'm talking about it to someone who isn't familiar with the little invisible things running around in the wires.


    By the way, are you familiar with the theory that all things electronic run on SMOKE. It says that if you let the smoke out, the electronics quit working.!!

    (Geek humor)


    5 years ago on Introduction

    If you use the same LED's that db used you will not need a resistor, however , the LED's will not be as bright. His LED's are 3.6v and your batteries will only produce 3 volts.

    To calculate the resistance that you would need for another flashlight, first take the number of batteries that it uses (AA,AAA,C, or D cells) and multiply by 1.5. A 2 cell light would be 3 volt a 3 cell light 4.5 volt and so on This is Vs (Voltage @ supply).

    Next check the specification sheet for the LED's, you are looking for the "forward voltage" rating this may be listed either as Vf or Ef (E is the standard electrical symbol for voltage). This is the voltage that will be "used" or "dropped" by the LED's. The voltage of your batteries should be equal to or greater than Vf. If it is, subtract the Vf form your battery voltage. This gives us the voltage that has to be used or "dropped" by the resistor. We'll call this Vr (Voltage @ resistor)

    The second thing that you will look for on the spec sheet is the current drawn by the LED. This will normally be abbreviated as "I", and will be rated as either a whole number ie 20 mA (20 milliamp) or as a decimal value 0.02 A (0.02 Amp). Both mean the same thing 20 thousandths of an amp of current for each LED you use.

    There are two basic ways to wire something electrically series and parallel. Wiring in series Voltage drops add and current stays constant. In parallel wiring, currents add and voltage drops stay constant.

    Now the calculations.

    We are going to assume that the flashlight we are using holds 3 D cell batteries, and the bulb will be replaced by 3 LED's, each of which has a 3.6 Vf and uses 20mA of current. We will have our batteries in series (normal for a flashlight) and our LED's will be wired in parallel.

    Vs (voltage supply) will equal 4.5 volts (1.5V per battery X 3 batteries)

    Vf (Voltage forward) will equal 3.6 volts (per spec sheet)

    This means that Vr (voltage across resistor equals 4.5-3.6 = 0.9. Vr =0.9

    Now figure the total current across the resistor. This will be the sum of the 3 LED's in parallel, or 0.020A x3= 0.060 A. So Ir (current at resistor) equals 0.060A

    To calculate the Ohm value of the resistor we use the formula R=V/I (resistance = voltage /current) so R=0.9/0.060 or 15 Ohms.

    The other thing we need to know to buy the resistor is what Wattage we need, and luckily we have all the numbers we need to calculate it. Wattage or power is abbreviated as "P". We will calculate the wattage of our resistor by using the formula P=I x E or Power equals current times voltage. So, our resistor wattage will be Pr (Power@resistor) = Ir X Er (Curren@resistor times Voltage@resistor, or

    Pr= 0.06 X 0.9 or Pr =0.054W

    Resistors come in regular wattage ratings of 1/8w,1/4w, or 1/2w, and larger.

    Since our wattage is 0.054 the closest size would be a 1/8w resistor (0.125w)

    So we would need a 15 Ohm, 1/8 watt resistor. Unfortunately Radio Shack only lists a 15 Ohm 1/4watt (catalog # 55049354) on their web site for $1.44. As long as the resistance is correct the wattage can be any size greater than you need and it will work fine.

    3 replies

    Reply 5 years ago on Introduction

    So if I have 4 batteries in the flashlight, and I want to convert it to LED, I need to add up the voltage and the watts of the LED's until the volts & watts of the LED's match the volts & watts of the this correct?

    The resistor, I don't even want to touch on til I get this first half understood....


    Reply 5 years ago on Introduction

    Think of the watts as the amount of heat that each device gives off as it uses its share of the volts. With the small amount of power that this circuit uses we can actually ignore the watt factor, I just threw it in for completeness.

    The batteries are your supply and don't have a watt rating since they don't use any power themselves, they just supply it to everything else.
    What we want to do is to get the total voltage that the LED's and the resistor uses to match the voltage supply from the batteries.
    Add up the voltage of the batteries,1.5+1.5+1.5+1.5 = 6 volts
    If you are using 3.6 volt led's subtract the led voltage from the battery voltage 6-3.6= 2.4. The 2.4 is the voltage the resistor has to use.


    Reply 5 years ago on Introduction

    OK, I think I got it now...

    but I'll copy/paste this just to make sure I don't forget....which I'm good at(forgetting things).

    Bat159, TY very much Sir for taking the time to explain all this to my feeble mind...much appreciated.


    5 years ago on Introduction

    This sounds cool as all get out, but you lost me after step #3.

    the first 2 steps I understood completely, but, I'm not an educated man, so it's easy for me to get lost when trying to calculate anything higher than 6th grade math...& sometimes even that gets me lost....

    Please keep in mind, I'm not saying the instructable was badly written, it's the math & electronics that I don't understand. : (


    8 years ago on Step 3

    With the room you have in the case, could a voltage/current regulating circuit (maybe a zener diode based unit) be installed? My understanding is that batteries are useful down to .8 volts per cell. That would allow us to put 4 cells with a start voltage of 6 volts and it would drop to 3.2 volts. That would allow keep LEDs lit from full charge down to useless. I realize you have 3 batteries in this one but the regulator would work for a variety of input voltages.

    1 reply

    Reply 8 years ago on Step 3

    Remember that the regulator is essentially just wasting power: the current I through it is the same as through the LED, so the wasted power is I*(VS-VF) (battery voltage minus the output voltage of the circuit, equal to the LED forward voltage). The same goes for the resistor in the Instructables circuit---essentially it regulates the required current, and so the only way to minimize the wasted power is to select VF to be as close to VF as possible


    8 years ago on Introduction

    Perfect, I have a few dead lights kicking around, was going to replace bulbs anyway! Good project to do with my 11yr old Daughter. How much longer will batteries last over incandescent?

    1 reply
    darph bobomccauleymon

    Reply 8 years ago on Introduction

    Hem, I can't say for sure how much longer it will last, but it will be considerably longer with the LEDs, as they have a very low current draw. I'm glad that you find that this will make a good project for you and your daughter, make sure to document it and post it here afterwards!


    8 years ago on Step 2

    Electrical or other tape around the bulb works better than paper towel. If you heat the bulb base the epoxy/glue will melt and you can free the bulb in one piece in many cases.

    1 reply

    8 years ago on Introduction

    Most excellent work. Good use of stuff on-hand... like cheese & bread.


    8 years ago on Introduction

    NICE PROJECT!!!!!!! I really like the fact that you didn't modify the flashlight itself, only the bulb. You can use this techneque to convert almost any flashlight to LED. This is a perfect project for someone with limited skills (like myself). I've got a couple expensive camping lights at home that I never use because the batteries are always dead. I woud hate to end up destroying an expensive flashlight but have no fear if the only thing i'm destroying is a bulb. Thanks


    8 years ago on Introduction

    Nicely done. I went the other way around when I upgraded my xenon headlamps. It just had the 2 AA batteries so I did a joule-thief as a driver circuit for it to work.

    2 replies