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LED Lighting problem


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I bought a 24 LED tent light in order to try to replace my RV lights that use an incandescent bulb like an auto tail-light bulb.

It contains 4 batteries in series, so I figured, I could simply use a series resistor to drop the remaining 6 volts and use the light. I measured the amperage and calculated the resistor value. The light was very dim. I reduced the resistance, but incurred a lot of heat from the increased current.

My question is - since these are 6-volt lights, can I simply put 2 in series (and get twice the amount of light)?

My electronics training is from many years ago, and I suspect I'm missing some special characteristics of the LED array that is being used.

Any help would be appreciated.

Jere

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Posted · Report post

There are various types of LEDS, there are the standard ones which need 20ma to give max light, and the voltage across them varies with the colour, make it an average of 1.8v So if you have a 6v battery the voltage difference will be 4.2v divided by 20ma gives 210 ohms (Make that 220)

Thw ultra bright leds draw the same current, but there is a new range of leds that really push out a lot of light and they draw quite a lot more. I saw a PCB with three of these leds that draw 330ma, and they are as bright as a 100w incadescent lamp! Supply voltage was 12v.

One could put two such units in series and if they are 6v each they should work off 12v. If you want to go into the unit and rewire the thing for 12v, just see how they connected them and what was their series resistor. I would think using 24 leds they would have 12 banks of 2 in series, 12 resistors for current limit. In this case you could make 6 banks of 4 leds and two resistors, depending on how they did the wiring inside, you could bank 2 sets together 6 times. (Draw the diagram for yourself first and then start the modification, it will prevent you from making mistakes)

Cheers Trevcharl

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