This is the most comprehensive step and requires the most skills (most fun part!)

1. Glue the battery holder in the part that has the holes for the LEDs and the switch (so that it can easily be taken apart afterwards when the battery is empty).

2. Glue the switch in the provided opening (this can sometimes require some filing to place it nicely).

3. Solder the red wire from the battery holder to the first terminal of the switch (it doesn’t matter which connection).

4. Solder the 150 Ohm resistor to the other terminal of the switch.

5. Place the LED’s in the right position in the two openings.

6. Then solder the other end of the resistor to the anode (plus or the long leg of the LED) of the first LED.

7. Go from the cathode (the short leg of the LED) to the anode of the second LED.

8. Go from the cathode of the second LED back to the black connection of the battery holder.

9. Place the battery in the battery holder and test the circuit.

10. Place the steel tubes (I took a piece of a long bolt for this and attached it to the opening) and fix it in the provided holes of the two pieces.

The wiring diagram of the circuit can also be found in this step.



If you like to use other LEDs (green, red, white, yellow, …) you will also have to use a different resistor.
The calculation of this is simple.

You build a circuit in serie, so the voltage is divided over the various components.

The LEDs I use each operate at a voltage of 2.6V. The LEDs consume a current of 25 mA.

9V (battery) — (2 x 2.6V) = 3.8V

3.8V / 0.025 A = 125 Ohm.

The resistor that matched the most was a 150 Ohm.resistor.

Always make sure your LED’s are getting the correct voltage and current to make sure they don’t get overheated.

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