Introduction: A Super Simple UV Flashlight
Step 1: Gather the Materials
1. 9 volt battery
2. A momentary push button switch or other switch of your flavor
3. LEDs of your choice. I used UV considering it doesn’t need a resistor to run off of a 9v battery
4. Resistor (depends on the voltage of the LEDs used, with UV, a resistor isn’t needed (2x4.5v=9v))
5. A case to put it all in
1. A soldering iron
3. Drill bit and drill
4. Electrical tape
5. Wire cutter
Step 2: Drilling for LEDs
Drill the necessary number of holes in the case where you want the LEDs to be.
Step 3: Drilling for Switch
Drill holes for the switch and do a test fit.
Step 4: Solder the LEDs
Solder the LEDs in series taking care that they have the right spacing for the holes that were drilled.
Step 5: Solder the Switch
Solder wires onto the legs of the switch as shown.
Step 6: Install the Switch
Fit the switch into position. The holes that were previously drilled may have to be enlarged slightly for the wires to go through.
Step 7: Solder the Rest
Solder the rest of the connections as shown.
Step 8: Insulate
Add tape where necessary and put it back together. You now have a simple flashlight!
5 years ago
nice. I would think you need resistors unless they are built into the LEDs.
Reply 5 years ago
Resistors are only required if the input voltage is greater than the maximum forward voltage of the LED or string for example if the forward voltage of the LED is 3v and 3 are hooked in series the total forward voltage is 3x3=9V so for 3 hooked in series with an input of 9v the amperage will remain at a nice level but much higher than will result in the death of the LED(s) unless a resistor is used. Just find the forward voltage for the particular LED and divide it by the total maximum input voltage and that will tell you how many you can wire in series without smoking any of them. If this number comes out to be a non-whole number, you'll need to use a resistor or add an extra one but adding an extra LED in series increases the total forward voltage of the string therefore decreasing the brightness for the same voltage.