re-wiring yet another christmas light string

rickpaulos

Jan 30, 2026
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Resistor question for re-configuring a crazy christmas string for another use.

Yes Christmas light strings are made in an endless variety. This one has 300 identical white leds, wired in 12 groups in series, each group or column has 25 leds wired in parallel with alternating polarity so only 12/13 in each group are on at any given instant. 29 volt power supply from Golden Power output: 29 volts 6 watts. My KillAWatt meter confirms 6 watts in "steady on" mode which is actually reversing polarity so fast you can't see the LEDs flicker. The string as a separate controller box with 2 wires in, 2 wires out and 8 modes. I want to use 5 of the groups with the existing power supply & controller. I need to add a resistor to replace 7 groups.

I used LED Calculator . net to determine a needed resistor for 5 LEDS in series. 910 ohms, 1/2 watt. As seen below.

But I want to use 25 LEDS (only half will be on at any instant) in each location which LED calculator doesn't appear to accommodate (no surprise there).

I'm guessing that the resistance is the same but the power rating should go up. (by 12x?) so a 6 watt resistor? That seems to much, don't the LED's disperse some of the power? With all 300 leds the draw is 6 watts. If I remove 7/12ths of the LEDs, the resistor should substitute 7/12ths of 6 watts or 3.5 watts.

From LEDCalculator dot net



LEDx1.jpg

Effective load, with 12 LEDS for each of the 5 columns,
LED5x12.jpg
 

Harald Kapp

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I'm guessing that the resistance is the same but the power rating should go up. (
No. If you do this, the LEDs will be very dim.
  • When you connect 12 strings of 5 LEDs each in parallel, the total current is also increased by a factor of 12. Since the voltage drop across the resistor stays the same, the resistance has to drop by said factor to R = 910 Ohm / 12 = 76 Ohm.
  • With 76 Ohm and current being 20 mA per 5-LED substring (as per your calculation), the minimum power rating is P>(12 × 20 mA)² × 76 Ohm = 4.4 W.
  • Connecting LEDs in parallel as per your schematic can (and very probably will) lead to uneven brightness of the LEDs. LEDs do have slightly differing operating voltages. When you connect them in parallel, the LEDs with a slightly lower operating voltage will draw more current and shine brighter. A better scheme would use the single 5 LED + resistor string of your first drawing and connect this 12 times in parallel (meaning 12 resistors of 910 Ohm instead of the single 76 Ohm resistor). Thus the inequality of the LED's pass voltage (or operating voltage) is irrelevant as each string of 5 LEDs has its own resistor.
 
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