Christmas lights: 8 groups in the chain, only 6 cables?

Question

How do you control 8 groups of LEDs independently with only 6 cables?

So, today I've put up my Christmas decoration. It's a long chain with several independent groups and a controller with a couple of programs, you know the deal.

At first I was puzzled, in my case: 8 LEDs and a ground => 9 cables.. but there are only 6 cables. I tried following the cables to draw a diagram, but they are twisted so tight it's impossible.

After a while I realized that the LEDs could be wired in both directions: at first only the odd ones light up, change the direction of the current and only the even ones light up. But this implies that an AC current is present when all LEDs are on.

Is AC and differently positioned LEDs the (most plausible/used) solution to my conundrum? Or is there perhaps a very obvious way of cabling, or another trick that I just haven't thought about?

They may be multiplexed, or more specifically, Charlieplexed. — Ron Beyer, Dec 15 '17 at 21:40
6 wires for 8 groups sounds more like traditional multiplexing than charlieplexing to me. — Peter Green, Dec 15 '17 at 21:50
Ah, that is exactly what I was hoping to find/learn. Thanks for the read @RonBeyer and Peter Green. — Frank Vermeulen, Dec 15 '17 at 22:19

score 3 · Answer 1 · answered Dec 15 '17 at 21:52

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I would expect that the lights are in a 2x4 multiplexed grid. As long as the multiplexing is fast enough the human eye won't notice that only half the LEDs are on at any given time.

answered Dec 15 '17 at 21:52

Peter Green

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score 0 · Answer 2 · answered Dec 15 '17 at 23:03

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I suspect wires are shared by bipolar 3 level PWM current as your result indicates for even odd positions. Thus an H bridge design with commutation of each polarity for average current.

Sometimes this is called a Polar Inverter design.

answered Dec 15 '17 at 23:03

Tony Stewart EE75

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Christmas lights: 8 groups in the chain, only 6 cables?

2 Answers2