How do I use christmas LEDs with 12V?

I have a string of 60 LED Christmas lights and would like to use them with 12V in a car for interior lighting. I dont need to have them all together in 1 run but rather set them up as "pods". After removing the plug from 1 end and the socket from the other, I removed the resistors and the extra strands of wire from the string. I am left with the 60 lights on 1 stand connected in series. How many lights in series can I use for 12V without using a resistor? What about polarity?? I don't want to extract them from the sockets as seen in some Ibles

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lemonie7 years ago
That colour 1.5-2V each I reckon. L
junits157 years ago
An inverter to kick the voltage to 120 AC
lemonie7 years ago
What colour(s) are they? L
I guessed they were multiple colors, (hence my answer) but that's a good question.
RYG are going to be low voltage, white/blue higher, you can say "try 1.5V ea. for your cheap coloreds" but "whiter" is a different matter.

I'd rather have them measure a few.
professor2005 (author) 7 years ago
According to the box, 43 sets of 60 lights would equal 210watts.
professor2005 (author) 7 years ago
The lights are amber and actually Halloween lights. I called them Christmas lights to avoid confusion. I'm guessing the longer of the 2 contacts is +??
seandogue7 years ago
Here's what I'm going to suggest, since I don't know the Vf for the LEDs in question and can only guess.

Use the circuit shown in the picture below to measure Vf for a few different colors, by starting with the pot adjusted to full resistance and adjusting it to lower resistance until the LED glows. Measure the voltage between the anode and cathode of the LED, as well as the current draw at that Vf .

Then its just a matter of adding up the values and applying a resistor that covers the difference between the supply voltage and the grouping. Parallel groups to achieve the desired number of LEDs for your application.

For using a 12V source instead of the 5V shown, I'd replace the 1K pot with a 5K or 10K pot so that you don't inadvertently overdrive the LED.

the resistor required in each group is

R = (Vs - (N x Vf) ) / If
oops... , where N is the number of series LEDs in each group