# IR/UV Light

Discussion in 'LEDs and Optoelectronics' started by davedotwojo, Feb 5, 2014.

1. ### davedotwojo

3
0
Feb 5, 2014
I recently modified an old point & shoot camera to be a full spectrum camera. So, I decided that I would like to make a light for this camera. I picked up almost everything I need from Radio Shack: 8 940nm 1.2V 100mA LEDs, 4 405nm 3.3-4.0V 20mA LEDs, an enclosure, perf board, 9V connectors, and 9V batteries. I still need to get some resistors, but I could assemble most of this project now.

I also picked up a center-off toggle switch, which brings me to 2 questions.
1) How do I wire this? I can find plenty of diagrahms online for controlling the direction of a motor, but none for choosing which set of lights is on.
2) The data on the switch says 3A 125V AC or 1A 250V AC, canI use this with a 9V DC power supply?

I probably should have mentioned this earlier, but I know very little about making electronics. I am, however, no stranger to soldering due to R/C Cars.

The IR lights will be run as a series of 4 pairs of LEDs run in parallel and therefore will require me to use a 21ohm resistor. While the UV lights will be run as a series of 2 pairs of LEDs and will require a 40 ohm resistor.

2. ### duke37

5,364
769
Jan 9, 2011
Do not run LEDs in parallel without individual current control.

5,164
1,081
Dec 18, 2013
Is this what you are after, and as Duke says you need a resistor for each LED.

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4. ### davedotwojo

3
0
Feb 5, 2014
So, if I have this right, I will have to use 4 40ohm resistors for the UV lights and 8 21ohm resistors for the IR lights.

5,164
1,081
Dec 18, 2013
Oh sorry didn't realise the reduction in UV lights. You have to calculate the resistor value for the current you require.
IR = (9V-1.2V)/0.1=78R each. Resistor Power =I^2*R=0.78W
UV =(9V-3.3)/0.02= 285R each or 250R if they run at 4V. Resistor Power =I^2*R= 0.114W