I have a LED string with four different lights. The lights are in order red, green, yellow and blue and this pattern is repeated. It has a controller with a button to cycle through fading/blinking modes. All red and yellow lights form a group and all green and blue lights form a group. The lights in a group are always on at the same time and equally bright.


First (left) and second LED:

enter image description here


  • Two wires are going from the controller to the first LED.
  • Every LED is connected with two wires to the previous LED and with two wires to the next LED.
  • There is a wire from the first LED that skips the other lights (I thought all other lights, but I didn't check them all, still have to check) and goes into the last LED, so three wires are going into the last LED.

I expected there to be three wires from the controller to the first LED, one for each of the groups and one common wire that is either + or -. How can two groups of lights be controlled independently with only two wires?

Based on the current answers I drew this diagram, but I'm not sure if it's right. I still have to count the LEDs to see if it is possible with the 31V output specified on the adapter.


  • \$\begingroup\$ Can you double-check. It looks to me as though the wire that skips the first LED is connected to the second one and that one of the other wires skips the second one. \$\endgroup\$ – Transistor Jan 13 '19 at 14:44
  • \$\begingroup\$ @Transistor I don't have LED string here, but I checked like 6 in row at a few places. But maybe the wire only connects at 4 points or something to get 4 parallel circuits. \$\endgroup\$ – Paul Jan 13 '19 at 14:46
  • 2
    \$\begingroup\$ The 31 V spec on the PSU is a clue. How many LEDs in total? \$\endgroup\$ – Transistor Jan 13 '19 at 15:39
  • \$\begingroup\$ Your sketch would be a typical way to do it but there are two wires in and out of the second LED in the photo. \$\endgroup\$ – Transistor Jan 13 '19 at 16:14
  • \$\begingroup\$ @Transistor That is compatible with the sketch if the top wire goes in and out the second LED tape without being connected to anything. \$\endgroup\$ – Paul Jan 13 '19 at 16:19

I expect there is an ordinary diode across the two pins of each LED such that when driven in the forward direction the LED lights and in the reverse direction the diode conducts. This diode is hidden in the black tape under the LED.

If a series string of these LEDs is formed where the one color (eg blue) is connected one way and another color (eg yellow) is connected in the reverse direction then the blue will all illuminate with positive current and yellow with negative current.

Two of these strings can then be assembled with a common return wire resulting in 3 wires in total. (put different colors in the second string).

The controller has to provide one polarity or the other depending upon the combination required.

The AC supply provides a convenient way of providing the two polarities.

Since each LED will be driven with 60Hz, flicker is normally fairly noticeable in the LEDs, especially if you move your head (or the LEDs) when viewing them.

| improve this answer | |
  • \$\begingroup\$ Wouldn't a diode on every LED be too expensive? I think diodes can be avoided with extra wire. Or is this common practice? \$\endgroup\$ – Paul Jan 13 '19 at 16:11
  • \$\begingroup\$ @Paul - it would cost but maybe cheaper than the additional wire and its connections. \$\endgroup\$ – Kevin White Jan 13 '19 at 17:21

Probably something like this:


simulate this circuit – Schematic created using CircuitLab

This arrangement has 2, 3 or 4 wires between the LEds. but is compatible with what's shown in the photo.

| improve this answer | |
  • \$\begingroup\$ That would colocate two LEDs or require more wires between adjacent LEDs. I suspect there is a hidden ordinary diode across the LED that is hidden in the black sleeving in the picture. \$\endgroup\$ – Kevin White Jan 13 '19 at 4:45
  • \$\begingroup\$ note: D7 - 2 - wires - D6 - 3 wires - D5 - 3 wires - D4 that matches the wire counts in the photo. \$\endgroup\$ – Jasen Jan 13 '19 at 5:09
  • 1
    \$\begingroup\$ But you've got four wires between D1 and D2. \$\endgroup\$ – Transistor Jan 13 '19 at 15:40
  • \$\begingroup\$ my guess is that the original poster miscounted. \$\endgroup\$ – Jasen Jan 13 '19 at 20:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.