Here is a nice, simple instructable for anyone just starting out with LEDs, soldering and electronics. It uses basic parts, with no mucking about with microcontrollers or timers (as fun as those are!) You can build one in an evening if you have all the parts ready to go.
But what is it? The LED Disco Light in a Jar is exactly what the name implies. Over a dozen RGB LEDs in a mason jar, frantically changing colour in a completely random pattern. It's a neat light effect for your next party, or you can use it to entertain a baby for quite a while!
See the last page for videos.
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Signing UpStep 1: Parts
1 mason jar, jam jar or anything with a suitably wide neck and a metal cap
1 switch
1 4xAA battery holder, in a 2x2 configuration
1 9V battery snap (mates with the battery holder)
24 RGB fast or slow-change LEDs*
12 10 ohm resistors
some solid 22 gauge wire
some heatshrink
perfboard or make your own printed circuit board**
some 25mm aluminum standoffs and matching screws
Glass Frosting Spray (optional)
- These LEDs have only two pins, and when power is applied they automatically cycle between red, green, blue, and combinations thereof. You can find them on eBay from various sellers in Hong Kong and elsewhere.















































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I used a nice cut glass vinegar carafe I got from St. Vincent de Paul's and used slow colour changing LEDs with a 6V power supply.
Thanks so much for posting, I really like my lamp and your instructions made it easy and fun.
If anyone is curious about how it looks in motion, reply and I can post a vid to Youtube.
Looking forward to my next project - question is, does mineral oil and UV reactant paint mix?
Regarding mixing mineral oil and UV paint... Hard to say, so I guess it's up to you! You could also try mixing other UV-reactive liquids, like highlighter marker ink and glow stick fluid. Be sure to get your proportions right, though. From what I've read the concentration can't be too low or high, or it won't work.
Well, 4700mAh AAs would indeed power it for much longer.
Basically, could you do this with one of the cheaper solar lights but just use less LEDs?? (I'd love to be able to do this with a solar light lol)
So, if i had a resistor between the batteries and the power rail to reduce the voltage, most of the current might be hitting one of the LEDs first, even thought all the LEDs light up. And that LED might fail first, because of the current, even though the voltage is ok for it. :/
Because the voltage is the same across all the branches in parallel, but the current adds up for each branch, and could be slightly different for each branch. http://www.kpsec.freeuk.com/voltage.htm
Would it be ok if they were all the same LEDs (and therefore the same voltage drop)? I am using all the same FLEDs. (very cheap from china, good tip there).
No, I don't get it. Please tell me where I'm wrong here:
With a resistor after the battery, the current is reduced to a low level, so it can't damage any LEDs, even though the voltage drop will still be the same for every branch (6V). Because resistors limit current.
Maybe i need a different voltage/current for dummies link than the ones i am reading...
Thanks for the help, this is a cool project, and I don't want an LED to burn out early.
The resistor is there to absorb (or drop) some of the voltage going through each branch, according to V=IR. So, if each LED should be consuming 2.5V (total 5V) at 20mA, then the resistor must drop the remaining 1V. Using the equation I listed, the value of the resistor must be 50 ohms.
Now, note where I wrote "should be." In reality, unless you measure the voltage drop of each LED and group them together, you never really know how much the voltage drop could be. It could range from 2.5V to 3.1V, for instance. So, the resistor could be anywhere from 50 ohms to 1 ohm. We just pick a value that will work with the average voltage drop we're expecting to see from the LEDs.
The trouble comes when you hook up the LEDs in parallel. LEDs are not resistors, they behave differently. What happens if one LED drops 2.5V, while the other drops 3.1? The LED with the highest voltage drop determines what the drop along the entire parallel chain will be. So, the 2.5V LED is actually being fed 3.1V, which is too much for it to handle. It burns bright for a while, then dies.
A resistor for each parallel LED ensures that each LED gets only the voltage it needs, with the rest being sopped up by the resistor.
I'll remember though!
For anyone trying to replicate this: note that the PCB board or perfboard is kinda optional here. You could just solder all the terminals together instead. For example, solder all the + terminals to a common + wire running the circumference of the lid, and gather all the - terminals to a wire at the bottom of the jar, leading back up to the battery.