Multi Colour Runway Light





Introduction: Multi Colour Runway Light

This instructable is a modified version of the various LED Sun Jar and Night Light instructables on this site. This is my first instructable posted and is really just a proof of concept at the moment. Revision plans have already started.

This particular version is switchable between 1, 2 or 3 lights. The parts list, as I made it, is as follows:

1 x Red 10mm clear LED ($0.90 AUD)
1 x Green 10mm clear LED ($2.70 AUD)
1 x Blue 10mm clear LED ($2.60 AUD)
1 x 180 ohm resistor ($0.38 AUD 8pack)
1 x 240 ohm resistor ($0.38 AUD 8pack)
1 x 150 ohm resistor ($0.38 AUD 8pack)
2 x diodes ($0.38 AUD 4pack)
1 x 9v battery ($5.65 AUD)
1 x 9v battery snap ($0.29 AUD)
1 x single pole sealed rotary switch (set to 4 position) ($2.95)
1 x Clear glass jar (I used an old coffee jar)
1 x Empty aluminium can
Broken glass (How much is needed will depend on the size of he jar)
0.12mm wire (To connect everything. USE A LIGHT INSULATION COLOUR)

Circuit diagram below is what I drew up after a brief discussion with the guys at work. (Thanks also to the guys at Jaycar for giving me a crash course in resistors)

Step 1: Assemble the Wiring

Due to unforeseen inebriation, the taking of pictures had been forgotten.

Assemble the components as per the wiring diagram below. Bear in mind that the LEDs will be inserted at different levels. So varying lengths of wire may be needed.

Initially I made all the lengths the same, which was the total height of the jar. so any every LED is the same distance away from the switch when held up. This allows for any LED to be at any level but causes some minor problems with the slack being pushed up into the lid.

Step 2: Getting the Jar Ready


First up cut the aluminium can up so as to give a single flat piece of aluminium. If no aluminium can is available, aluminium foil could be used.

Place to jar on top on the aluminium and trace around the base, then cut out the circle of aluminium. Insert the aluminium disc inside the jar shiny side up.

This should help reflect light from the bottom LED. Not much good it shining right into the ground.

Drill the holes in the lid of the jar to mount the rotary switch in. Bear in mind that 2 holes are needed. one for the shaft and another for a locking pin to stop the switch spinning.

Depending on the type of lid, attaching the wiring harness to the lid may be possible. It really comes down to how much slack is in the wiring harness and what type of lid the jar has.

Step 3: Pack the Jar


First place a small amount of broken glass in the bottom of the jar over the reflector. Insert the first LED on top and use a dowel to hold it in position while packing more glass over the top. The glass should start holding it in position fairly quickly so the dowel can be removed.

Make a gap with the broken glass, how big will depend on the size of the jar, and insert the second LED. Repeat the process with the dowel and glass covering.

Once again leave a gap and inser the last LED. This final one should only require a light covering of glass.

Remember that there will be a battery and the base of the switch that will require room. I packed the glass a small amount at a time checking the space left for the battery and switch in between.

Step 4: Finale

The final product in all its ??glory??


  • We did it. Thanks fo...-LeonN9

    LeonN9 made it!


  • Epilog Challenge 9

    Epilog Challenge 9
  • Sew Warm Contest 2018

    Sew Warm Contest 2018
  • Paper Contest 2018

    Paper Contest 2018

We have a be nice policy.
Please be positive and constructive.




i did this except i used colour changing leds looks really cool.

Can someone explain to me how the calculations for the LED resistors to work?

I'm using...:

Red: 3.1v-3.8v, current is 20-30 mA
Blue: 3.1v-3.8v, current is 20-30 mA
Green: 2.8v-3.2v, current is 20-30 mA

Yes, i will be using a 9-volt akaline battery which has 565 mA-h (mA hour).

As Scromple stated or just use an online LED resistor calculator.

brooklyn:  Okedoke... The math for calculating the resistors needed for a given LED and power source is as follows.

Resistance = (Voltage Supplied - LED Voltage) / LED Current

If you use a search engine and do a look up, I think there are some people that have actually written web pages that where you can just plug in the info and it spits out the resistor needed.

Also if you take a look at the circuit diagram, I put the formula on there.


That should be.....

R = 5.5v / 25mA

R = 6v / 25 mA

Correct? but compared to the water flow theory this would be making the tube smaller for more resistance but with the same flow rate.....

Or is it okay?


Also, when i'm activating 2 LEDs at the same time, should i put a lesser ohm resistor instead of 2 resistors?

Can someone tell me how to make one of these so it plugs into the wall?

You need an AC/DC power adaptor. Remember that LED's cannot run on AC, which is what comes out from your wall sockets. I'm sorry I don't know how resistors work, otherwise I'd tell you how many resistors you'll need.

Same circuit but with mains power instead of a 9v battery. Were you going to go directly into the mains socket in a house or do you have a power pack that will step the voltage down from mains to 12v???? Either way you would need to work out the resistors to use as the one I state above would not be correct. The formula to work out the resistors to use is in the comment for COMMODORE64 and in the instructable above.

Not really sure the number of hours exactly... I generally only use them when I have friends around. I can vouch for at least 6 hours. But that is not constantly on for 6 hours. I know that because of the 3 reasonably powerful LEDs the efficiency is not over the top. If you insert another resistor to drop the voltage that can be drawn from the battery you could make lights like this last quite a while on one 9v battery.