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I have 5 stripes of 5m length and 300 RGB LEDs (type 5050) each. One LED consumes about 20 mA at 12 V. Therefore, my whole 25m setup would use about 30 A.

5050 RGB LED stripe

Is it safe to connect the 5 stripes series and powering one end with a 30 A power adapter? I wonder whether the small traces on the PCBs can handle that. If that's not possible, what would be a safe setup?

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  • \$\begingroup\$ You are best off actually measuring the actual current that each 5m length consumes. I highly suspect that your current is significantly less than you are thinking. In strips like this, there are usually 3 (or more) LEDs in series to allow operation from higher voltage at lower current. \$\endgroup\$ Commented Feb 7, 2015 at 19:29
  • \$\begingroup\$ In fact, looking at the picture you show, that is exactly the case. Notice the cut line is between the 3rd & 4th LED from the end. I can't make out the value of the middle resistor but I'm guessing that the 150 Ohm resistors are for the green and blue LEDs. Assuming about 3.2V per led, we get 12V - 9.6V = 2.4V divided by 150 Ohms equals about 16mA per group of 3 LEDs per color. \$\endgroup\$ Commented Feb 7, 2015 at 19:34
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    \$\begingroup\$ @DwayneReid It's 331 Ohm for red and 151 Ohm for blue and green respectively. 16mA per 3 LEDs per color is 48mA per 3 LEDs, which would still give me 25 A for my 1500 LEDs, right? At the moment, I can't measure the actual current, but I'll do the next days. \$\endgroup\$
    – danijar
    Commented Feb 7, 2015 at 20:22
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    \$\begingroup\$ @danijar - A 331 resistor is actually 330 ohms, that is, 33 x 10 ^1. Likewise a 151 is actually 150 ohms - same arithmetic. "16 mA per 3 LEDs" is 16 mA per color, since 3 LEDs (1 red, 1 green, 1 blue) are needed to produce a color. So your 1500 LEDs will produce 500 "color units", and at 16 mA / color will need .016 x 500, or 8 A. \$\endgroup\$ Commented Feb 7, 2015 at 22:13
  • \$\begingroup\$ @WhatRoughBeast Thanks for the clarification. However, there actually are 3 color units every 5 cm segment, thus 1500 in total. So it's 24 A unfortunately. \$\endgroup\$
    – danijar
    Commented Feb 9, 2015 at 21:19

2 Answers 2

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See the comments I left with your question.

Personally, I would power each 5m strip with its' own cable coming back to your controller or power supply. Note that the common lead has to handle the sum of the currents for each of the colors and thus should either be a larger conductor or multiple conductors in parallel.

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  • \$\begingroup\$ I want to mount the stripes around the room near the ceiling, so I'd have to lay cables down from the ceiling every 5 meters which would be a bit annoying. Is there a way to estimate how much stripes I can power in a row? Having only two power supplies would be easier to hide in the room. \$\endgroup\$
    – danijar
    Commented Feb 7, 2015 at 20:27
  • \$\begingroup\$ After reading your answer again, I have a second question. The LEDs have a shared anode, and I plan to mix colors by using a PWM signal for the kathodes of the channels. How can I use multiple power supplies then? I guess that would only work with synchronized PWM signals or did I get it wrong? \$\endgroup\$
    – danijar
    Commented Feb 7, 2015 at 20:31
  • \$\begingroup\$ Use one power supply for each 5m string if you have to. Just be sure to connect the negative lead from each supply to your controller. The controller doesn't know or care if the individual LED strips have separate power supplies so long as they are all working. Do note that the PWM controller has to be rated to handle the full current of ALL of the LED strips. \$\endgroup\$ Commented Feb 7, 2015 at 22:13
  • \$\begingroup\$ What I've done in terms of getting wire to all of the LED strips is to run the wire behind the LED strips. This is usually done when the place is being built and it was easy to have the carpenters provide a space behind the surface where the LED strips mount to hide all of the wires. \$\endgroup\$ Commented Feb 7, 2015 at 22:15
  • \$\begingroup\$ Alright, since I'll install the LEDs all around the room, they'll form a circle. The largest distance from the power supply will be 12.5 m then. Can I power this using a single 8 A power adapter then? \$\endgroup\$
    – danijar
    Commented Feb 8, 2015 at 1:03
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Powering 25 meters of continuous Flexible Circuit Board, from a single power connection, will not work. There are heat and voltage droop issues from that much current over high resistance copper tape.

The higher the current and the longer the led strip, the lower the voltage at the far end will be. Accordingly, it will result in dimmer leds the farther down the cable we go. This is slightly noticeable on a single 5M run, a 25M run will probably not even light up at the end.

You will need to run multiple parallel regular 18 or 16 awg cable to different parts of the led strip.

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  • \$\begingroup\$ One question, would it somehow damage the stripes or would the light just be dimmer? I'll get some awg cables but I'd like to test this first. \$\endgroup\$
    – danijar
    Commented Feb 8, 2015 at 12:18
  • \$\begingroup\$ Too much heat can cause the cable to unglue and the leds and resistors to desolder, or worse. 25A at 12v is 300 watts you are putting through fpc meant for alot less. \$\endgroup\$
    – Passerby
    Commented Feb 8, 2015 at 16:33
  • \$\begingroup\$ Thanks to @Dwayne Reid, I expect the current drawn from the 25 m stripe to be around 8 A. Does that change the situation? \$\endgroup\$
    – danijar
    Commented Feb 8, 2015 at 16:56
  • \$\begingroup\$ That's not right. A typical 300 led per 5M roll RGB strip uses 3 leds per segment per color. That's 100 segments per color. That's ~20mA per segment so 100 *.02A = 2 Amps per color. That's 6A at 12v when full white (72 Watts). Per 5 Meters. So your initial current usage was right. 30 Amps, at full white, taking ideal numbers and no voltage droop into account. Actual numbers will be less due to actual resis tor value, or that the resistors chosen give ~18mA at 12v, or the length of the cable causing voltage droop from the copper resistance. \$\endgroup\$
    – Passerby
    Commented Feb 8, 2015 at 17:15
  • \$\begingroup\$ Ah, he said about 8A per color, so 24A at full white, all three colors on. That's close to my calculations with ideal numbers. \$\endgroup\$
    – Passerby
    Commented Feb 8, 2015 at 17:17

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