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I am purchasing 50x - 1 meter / 12 V / 72 LED / Rigid SMD 5730 strips directly from China. I don't know the LED manufacturer, nor the resistor being used. (Yes, this is a gamble.)

The strip is one that could be cut at every 3rd LED. So, it seems that every 3 LEDs is wired in series, and there are 24 groups of 3 wired in parallel.

Each strip is specified at 12 V/18 W. I would like to take 2 strips, and wire them in series with a 24 V 18 W power supply.

(By series, I do not mean chaining them in parallel and calling it series like the billion articles polluting my google searches)

My question is, can I wire two 12 V strips in series using a 24 V power supply? It seems like I can, but am worried I am overlooking something that I don't even know to ask, or I do not have enough information without the LED spec sheet.

example layout

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  • \$\begingroup\$ You've got it right (the pic on the right). Any time two 12v loads are wired in series, it's now a 24v load. But what your sixth sense is telling you is correct - the supply is not big enough. It's only good for one strip. V x A = W \$\endgroup\$
    – user57849
    Commented Nov 9, 2014 at 4:00
  • \$\begingroup\$ Anecdotal, I've been running a set of 12V strips in this exact configuration for years straight now without issue. Of course it's a small setup of 2x 7 segments. \$\endgroup\$
    – Passerby
    Commented Dec 13, 2021 at 18:58

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Well, yes and no. On the one hand, connecting 2 in series ought to work. On the other hand, you will need 24 volts at 36 watts.

Since each string takes 12 volts at 18 watts, its current must be 1.5 amps. Connecting them in series will still take 1.5 amps, but the total power will be 1.5 x 24, or 36.

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  • \$\begingroup\$ 24 volts at 36 watts what? You must mean 24 volts with 1.5 amps \$\endgroup\$
    – Funkyguy
    Commented Jul 25, 2014 at 0:32
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    \$\begingroup\$ As long as power equals volts times amps, I fail to see the difference. \$\endgroup\$ Commented Jul 25, 2014 at 1:36
  • \$\begingroup\$ Yes, sorry, I had a brief moment of duh. Of course adding a second fixture will double the wattage demand on the power supply regardless of voltage. \$\endgroup\$
    – Cody
    Commented Jul 25, 2014 at 16:57
  • \$\begingroup\$ Heh. Let's hear it for brain farts, our constant companions :) \$\endgroup\$ Commented Jul 25, 2014 at 17:00
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This is an old post, but i just wanted to contribute my observation. Both your parallel and series would work, however that is if we assume that none of the LEDs would ever burn out.

If half of the leds on the 24V would have burned out on the left side (see image). enter image description here

The result would be catastrophic because the remainder leds on the burned side would receive an increase in voltage (about 16 V) and on the right side the voltage would drop to about 8V.

In conclusion, as the led burn out the voltage increases on the same parallel side and if your LEDs are rated at 12V it would be just a matter of time until all lights in the same parallel burn out due to high voltage.

That is why it is recommended to always connect led strips in parallel.

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  • \$\begingroup\$ Excellent point! Hopefully people read this and understand the theory vs the reality that LED invariably burn out prematurely. Of course the longer the individual (smallest serial) runs are the worse this can get since most LED strips in series act like old christmas tree string lights, when one goes out, they all go out in that small serial run. (Which would correspond to the each of the rungs of the ladder in the original post...those rungs are each 3 LEDs wired in series...so the more LEDS in those rungs...the more impact each LED burnout will have on the overall misbalancing you discuss) \$\endgroup\$
    – Questor
    Commented Jun 11, 2018 at 14:11
  • \$\begingroup\$ I haven't quite thought this through but this doesn't quite seem right - wouldn't all the LEDs still see the same voltage drop (since the fwd voltage drop is a constant for an LED, within reason), but the current passing through the intact strip would drop (along with brightness) because now the series resistance of the blown strip is higher (having fewer current-limiting resistors in parallel)? The strings on the blown strip can't just decide to carry more current, they're limited by their series resistors. I get the feeling the truth might be somewhere between our two theories, maybe. \$\endgroup\$
    – agittins
    Commented Jul 12, 2021 at 11:35
  • \$\begingroup\$ @agittins you could test it with a few spare leds and resistors. Or a sacrificial led strip. Dime a dozen at this point. \$\endgroup\$
    – Passerby
    Commented Dec 13, 2021 at 18:55
  • \$\begingroup\$ @agittins I think you are missing a point that the supplied voltage is in accordance with the number of LEDs that are supposed to be present. For example, if two LEDs are connected in series and is provided with 6V, if one of them burns but still somehow manages to keep the circuit close, the first led will be feeded with high voltage now. Please correct me if I am wrong. \$\endgroup\$ Commented May 8, 2023 at 23:59
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This is of course correct and a real problem as long as you have only a few LEDs on each side. However, if you connect 2 X 2 meters with 2 X 120 LEDs, a burnt unit (3 LEDs and a resistor) will increase the voltage only about 0,6 Volt on the other side. This will not do any harm especially if you turn down the power supply a little bit (if the power supply can be adjusted). Furthermore you can balance the systen anew by cutting of 3 units as soon as 2-3 units are burnt on the other side. Sorry, no native speaker of English.

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  • \$\begingroup\$ Welcome to EE.SE. This site is not a forum - it is a Question & Answer site. Your "answer" will move up and down with votes and user sorting preference. It is not clear which answer you are referring to. When you have enough reputation you will be able to post comments. \$\endgroup\$
    – Transistor
    Commented Aug 13, 2017 at 7:39

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