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Discussion Starter #1 (Edited)
I recently made an "add-on" pod for my center console with some switches and controls, as can be seen in this photo...
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The four square buttons near the shifter are audio selector buttons that allow me to easily select different audio sources that play through the factory AUX-IN jack. Well, each of those switches has an LED in it. They were originally red, but I wanted the switches to be blue, so I popped off the cap of each switch and simply pulled out the red LED and inserted a blue one -- they are only held in with little clips. As far as I know, there is no resistor inside these switches, so I apply the appropriate resistance externally to the connections on the switch for the LED's.

I've done this before with another modification and had no problems, but on this application, the LED's in individual switches would dim or cut out. Pushing the button or jiggling it would cause it to dim and brighten, making it appear it was some sort of loose connection. Same thing if I took off the cap of the switch and directly jiggled the LED. I even tried a different set of LED's and got the same issue. While it seemed unlikely that I fried the interior wiring of all four switches, that is what I was considering, until I decided to try one more thing...

First, keep in mind that these LED's come on and stay on when the car is switched on -- they do NOT turn on and off depending on whether the switch is on or not (there's a reason for this, but I don't want to clutter this post and make it longer than it already is). So, I originally wired all four LED's in parallel, with one resistor in line to drop the voltage to the entire circuit of all four LED's so that each LED gets about 2.8 volts or so (these are 3 volt LED's, but I don't want them at full brightness). I had a resistor of about 620 ohms in parallel with the circuit. So, what I decided to try was to have each LED get its own resistor, instead of using one resistor for the entire circuit. The LED's are still wired in parallel, but instead of one 620 ohm resistor going to the entire circuit, I now have a 2000 ohm resistor in line with each individual LED (okay, one has a 2400 ohm resistor because I ran out of 2000 ohm ones).

This has worked! At least now for two days, the LED's are rock solid and no longer flicker or cut out like they were doing before. I'm guessing it has something to do with the fact that the LED's are just held in by clips, as I have wired parallel LED's with one resistor before without this problem, but those LED's were soldered into a circuit (you just have to be aware that the required wattage of the resistor you need can be quite a bit greater than you might originally expect when you do this). Maybe only having one resistor in line with the four LED's causes any imperfections in the LED's clip-in connection to be "magnified"? Because the LED's are still mounted exactly as they were before. And, also note that with the other project where I wired up similar switches without any problems, those LED's do each have their own resistor, as well.

I'm just curious as to what may have been going on here.
 

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Discussion Starter #2
I did some experimenting on a test console/breadboard, and I think I have a basic idea of what may be going on. First, I want to point out that the tests I did are very general. As I said above, I ran out of some of the values of resistors I was using, so I substituted other values. Plus, for measuring current, only the 10 amp scale of my multi-meter is currently working (I think there's a blown fuse in it), so I can only measure current with a resolution of 0.01 amps (10 mA), so I cannot measure fine nuances of low-current circuits.

With that said, it appears that when the LED's are in parallel after going through one resistor, the current going through them is lower than when the LED's each have their own resistor -- but again, that could be due to the resistor values I was using for my testing.

I think the important thing is that, when the four parallel LED's are all in series with one resistor, that network of LED's is like a football field, allowing the current to go anywhere it needs to. Electricity follows the path of least resistance, so when there is even a minor defect in the connection at one LED, it encourages the current to mostly go to the other LED's in the parallel network, causing the first one to dim. The one resistor does limit current to the LED's, so based on the resistor value I was using, I don't think the LED's would be in danger of getting damaged, but because the LED's are connected directly in parallel after that one resistor, it enhances the possibility of the circuit being unstable if the LED's and all other connections are not securely soldered in place.

When each LED has its own resistor, then the current for each LED is controlled individually, so that if there is a slight flaw in the connection of one LED, then all of the available current cannot just be drawn directly to the other LED's (because of the resistors at those LED's), which means a more serious flaw in the LED circuit has to occur for the LED to dim or go out.

This is just a theory. Again, I welcome any input from people who are better trained in electronic circuits than I am. I know another possible factor is that an LED is not just like a simple light bulb. The "D" in LED means "diode", so it is a semi-conductor, and may behave differently then expected under certain circuit conditions.
 
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