With the LED panels, you shouldn't need any sort of current limiting resistor or special current driver source - just connect to 12V and go. I can't say they are the highest quality, but they should work fine for what you are trying to do, although, I think the using RGB LEDs would work better as the different colors would blend better to form white than these separated panels will.
I think this could be solved very easily using many different methods. The best method for you will depend upon what you know how to do. A simple microcontroller with a couple of external buttons and transistor switches would work flawlessly, but if you don't know how to program, that is not a good route. A few logic gates driving transistors with button inputs would also work fine, but I imagine you would have already done that if you knew how.
If you would be interested in seeing a simple logic circuit, let me know and one can easily be provided.
But what about using a couple of different simple switches. Take a look at this circuit:
Switch 1 is 6 pole rotary switch. It uses rotating knob to connect the main input to one of the 6 outputs. There are many types of rotary switches with many poles (number of contacts). You will need at least a 4 pole switch, but a 6 pole switch may be better. I'll explain why in a minute.
Switch 2 is a triple-pole, double-throw switch. It has three different poles, and has two ways to "throw" them. If the switch is throw in one direction, all three LED box positive leads are connected to 12V, so they all three turn on. If it is thrown in the other direction, each LED positive lead is connected to one of the poles on the rotary switch.
If using a 4 pole rotary switch, pole 4 should be left open - this would be all LEDs off. Pole 1 would go to LED 1, pole 2 goes to LED 2, and pole 3 goes to LED 3. As you turn the knob, a different LED will be powered. There are two reasons why a 6 pole switch may be better. First, there could be an off state between each of three LED off states (1 ON, all OFF, 2 On, all OFF, 3 On, all OFF). Secondly, some rotary switches are not completely off in between states, meaning that as you turn the knob, the two adjacent LEDs will both be on for a moment, and I don't think you are wanting that to happen.
The great thing about using the 6 pole rotary switch in combination with the 3PDT switch like this is that no two LEDs can be on at the same time - it's either none, one, or all of them. The down side to using the rotary switch is that you can just select an LED bank to turn on, you might have to cycle through the colors depending on the current state of the switch.
One way to avoid this is to use a third "main power" switch to disconnect power/ground to the whole thing while you select a specific color with the rotary dial. Since you are working with color sensitive paper, this might be the best way to go.
A different way to do it is to replace the rotary switch with three common single-pole, single-throw switches like this:
With this setup, you can easily choose which LED bank to turn on. The bad thing is that you could accidentally turn on more than one at a time if you aren't careful...
Of course, no matter what you do, the switches need to be at least rated for 12V DC and as much current as the LEDs are expected to draw.
There are other mechanical switches that will select one while automatically deselecting other inputs - like how a lot of old RCA selectors used a spring loaded lever to select between your VCR, Nintendo, DVD player, etc to input into the TV RCA jacks. But again, these switch selectors typically allow more than one input to be select as they change states.
A resistor, R4, is used to pulldown VLOAD when the SPST switch, SW1, is open. This means that when SW1 is closed, VLOAD is high, and when SW1 is open, VLOAD is low.
An NPN digital logic inverter is used to provide the complement of the VLOAD signal. The input to the inverter is VLOAD, and it is powered by the voltage source. The output of the inverter is VLOADN.
D1 and D2 have shared cathodes. Since VLOADN = ~VLOAD, there is always one LED that is on while the other is off:
- If SW1 is closed, the inverter input is high. D1 is on and D2 is off.
- If SW1 is open, R4 pulls the input of the inverter low. D1 is off and D2 is on.
simulate this circuit – Schematic created using CircuitLab
Best Answer
simulate this circuit – Schematic created using CircuitLab
The red led has 1.4V forward voltage drop while the green one has 2.2V. When the red one is on, the green one will not have enough voltage drop and will be very dim.
simulate this circuit
If you want to turn off the green led completely off, you can use a MOSFET switch like this.