In this Instructable I will show you how to make a super bright, USB powered LED with a compact form factor, which I have lovingly named, "The Plugbulb".
This little bulb can be plugged into any USB jack. Great for turning your portable power bank into a powerful, long lasting flashlight!
Step 1: Ingredients
Let's start with the materials. One Plugbulb requires:
- A USB plug (preferably from a broken cable)
- A 3W LED bulb
- An LED heat sink
- 2 diodes, of the non light-emitting variety (any sort should do) OR a 5ohm, 1/2W resistor
- your favorite plastic bottle cap (here's mine)
- 1/2 packet of Sugru (or similar)
- a tiny, itty-bitty amount of thermal compound
Along with the following tools:
- soldering iron and solder
- hot glue gun
Feel free to scale up your recipe as desired for larger batches of Plugbulb.
Step 2: Tear Apart That USB Connector
Be careful to preserve at least a couple inches of the wires. I found that pliers worked well for peeling away the plastic. It might depend on the type of plastic surrounding your cable. It is also a good idea to use one with the cable coming out of the back of the plug, as opposed to the side.
Step 3: Make LED Circuit, Part One
Here's the technical part. I am going to dive into some theory for those interested in understanding how to design with power LEDs. For those who would rather just get on with the project so you can start blinding your friends with your cool new flashlight, feel free to skip to the next step.
Diodes can be tricky to design with at first because they are non-linear devices. This means that the voltage and current are not linearly proportional like they are in resistors. The first image above, courtesy of https://www.allaboutcircuits.com/textbook/semicon..., shows a typical IV curve, or the relationship between current and voltage, for a diode.
LEDs are special diodes that are designed to emit a certain wavelength of light. The high power LEDs that we will be working with will have a similar curve as the above, except with the exponential slope elongated horizontally (the bend upward is shifted toward a higher voltage). The second image above is a curve I made with data I gathered while investigating the characteristics of the 3W LEDs I used in this project (the same ones I linked to, but I would guess that all 3W white LEDs will look pretty similar).
From my testing, I found that between 200 to 500 mA seems to give the best balance between brightness and power consumption. Beyond 500, the brightness gains are minimal as the current increases. Below 200, the LED is not nearly as bright as it can be. So easy enough. If we want to pass a given amount of current, all we have to do is follow the curve and find the voltage it corresponds to. If I was powering this with an adjustable voltage source, and could dial in that specific voltage, it would indeed be that easy.
The tricky part comes in when you want to power this from a source without the correct voltage. In this project, we want to power the LED from 5 volts. If we connect the LED straight to 5 volts, we would pump far too much current through it and it would burn out in an instant. So how do we limit current?
We have several options. We could use a voltage or current regulator IC, and some might argue this is the best way to go about this task. However, size is a limitation in this project, so we need something smaller. Fortunately, since we are powering this off of a stable, regulated 5 volt source (as USB supplies typically are), we can simply use diodes and/or resistors to hone in on the current/voltage we need.
I will describe how to correctly pick resistors first, even though I chose to use diode method in my build. To size the correct resistor we would take the current we want, let's say 300mA, and the voltage the resistor will see, 5V-VLED, where VLED is the voltage across the LED at 300mA (using our graph) and use ohms law (V/I=R) to calculate. In the graph we can see that at 300mA the LED is dropping about 3.25V. Therefore our resistor will drop 5-3.25=1.75V. Using ohms law, our resistor should be 1.75V/300mA=5.83 ohms.
If you don't have a nice IV curve for your LED, you can always resort to math, however it is not pretty. The last image I attached to this step is the equation for the typical IV curve of a diode. We can combine this equation with ohms law for the resistor (V=IR) and solve for R (if you know the LED's saturation current). We know the I's are equal and the V's have to add to 5. Two equations, two unknowns. But gross... right?
Long story short, a resistor of about 5 ohms will do the trick. You also must take power dissipation into consideration, though. 5ohms at 300mA will dissipate .3^2*5=.45W of heat, so we need a 1/2W resistor. 5ohms is an awkward resistor size, however we can make this with more commonly available resistors in parallel, such as two 10ohm resistors, or four 20ohm resistors. If you do this method, make sure your resistors are 1/4W or, preferably, even larger in terms of acceptable power dissipation, otherwise they might get too hot and become a hazard.
The other option is to use diodes to drop the voltage. A standard diode is said to drop .7 volts, however, this isn't strictly the case. It will drop slightly more at higher currents, and slightly less at lower currents. This means that two diodes in series will drop somewhere around 1.4V. In our circuit, this would leave 3.6V for our LED, which should pass somewhere around 500mA according to our graph. While this is a bit high, it is within the range I was looking for, and adding a third diode in series would drop the voltage too low (~2.9V). Also, when passing this much current through the diodes, it is likely that the voltage drop will be a bit more than .7, thus the system will find an equilibrium at a slightly lower current. Again, this can be solved more precisely with math if you had all of the details of the diodes, but I used an easier approach - an adjustable voltage regulator. I just added two diodes (because this was my guestimate) and slowly inched up the voltage while measuring the current. By the time I got to 5 volts it was pulling somewhere around 400mA. Perfect.
If you are using a different diode and two doesn’t work out, you can add or subtract diodes or even try different diodes with a different voltage drop. Or you can use resistors if you have the right values lying around. I cant think of any reason why one method would be better than the other, but if you can I'd love to learn about it in the comments.
One more side note for those playing with high power LEDs: Distilled water is a great heat sink! While I was testing the limits on these LEDs, I completely submerged them in distilled water. Distilled water is an insulator (well, more like a very, very weak conductor) so it is safe for electronics. DO NOT USE tap water, as the dissolved minerals are what make it conductive. As always, use common sense and be careful, but this can be a helpful trick.
Step 4: Make LED Circuit, Part Two
Now its time to solder together the basic circuit.
Place a dab of thermal compound onto the center of your heat sink, then press your LED onto it. It will help hold the LED in place while you solder it to the heat sink. Now do that. Solder the LED to the heat sink.
Next, solder the LED and the two diodes (or your 5ohm resistor) in series. Remember, diodes are polarized, so make sure they are all facing the same direction, or your light will not turn on. Diodes usually have a silver band indicating the low voltage side. Make sure they each go into the circuit with this band on the side further from your 5V source. The LED is also a diode, meaning it is also directional. Make sure you have this pointing in the right direction too. Usually they have a marking on the tiny leads. If your's don't, use a low voltage source (~2-3V, two AA batteries in series will work) to test. You will not damage the LED by connecting it backwards, it just wont work.
I added some electrical tape to the back of the heat sink, then tucked the diodes behind it. It does not matter which order these components go in within the circuit, as long as they all face the correct direction.
Step 5: Connect the Jack
Now solder the USB jack to the circuit. All you need is the power (red) and the common (black) wires from the USB. You can trim the other ones down (but careful not to short them, so as not to damage whatever device you plug this into). Try to do this with as little excess slack as possible in the wires.
Now use some hot glue to hold it all together.
Step 6: Cut a Hole in the Bottle Cap
Yes, I know it is your favorite, but we have to do this.
We need to make a slit in the back of the bottle cap so the USB plug can slide through. I found that I could use a drill bit to drill two holes next to each other that are the right width, and then use a sawing motion with the drill to connect them, forming a slit. I am sure there are better methods and better tools, and I'd love to learn about them in the comments!
Step 7: Add the Bottle Cap
Now push the jack through the slit you made in the bottlecap and add some more hot glue around the seem to hold it in place.
Step 8: Add the Sugru
Use the Sugru to make a nice seal around the top of the jack, and hide the seem. This stuff also acts as a glue, which will make it more durable.
Step 9: Enjoy!
Behold! The Plugbulb!
These lights pull less power than a smartphone charging, so they should be able to be powered from just about any USB battery pack you have. Great for an emergency light or to bring on a camping trip. With a large battery pack, they will run for tens of hours!