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I have 300 LEDs (WS2812B) wired in a series. They were sold as 150 LED/5 meters and rated to power at 9W/m, so 45W/5m strip. I would need 90 Watts to power my entire 300 LED strip. I have a 3-rail power supply that has an output of 5v 40A (200W), would this be adequate to power the LEDs?

Naively, I thought I could power them straight from the PS but got major color change around the halfway mark (expectedly where the second strip is connected). I soldered +V and -V of the second rail to the halfway point and it helped my situation but not entirely and I am worried I am going about it wrong. If anyone could shed some light (haha) on my situation I would greatly appreciate it.

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  • \$\begingroup\$ I don't think the traces on these strips can handle 40A. What's your voltage and the far end? \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 May 24 '19 at 0:53
  • \$\begingroup\$ Thanks! Posted there as well \$\endgroup\$ – user2544427 May 24 '19 at 1:05
  • \$\begingroup\$ Typically, the forward voltage of an LED is between 1.8 and 3.3 volts. so if you have 10 in series you need an 18v to 33 volt supply depending upon how bright you want them ( the brighter the less long they last). Experiment with 1 led and find the voltage you want. \$\endgroup\$ – PhysicsDave May 24 '19 at 2:23
  • \$\begingroup\$ Crossposted from physics.stackexchange.com/q/481965/2451 \$\endgroup\$ – Qmechanic May 24 '19 at 9:10
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    \$\begingroup\$ @PhysicsDave these are "digital" LEDs, they are not in series, they are in parralel and controlled individually... \$\endgroup\$ – BarsMonster May 24 '19 at 10:47
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Afaik you are doing right.

300 leds / (150 leds / 5 m) = 10 m of LEDs. 10 m * 9 W/m = 90 W. So 200 W is more than twice you need.

The reason you get color problems, is that the voltage reduces halfway so much, the LEDs do not have the entire 5 V but less.

By splitting the complete circuit in two, you reduce the voltage reduction, but probably not enough, you might to split it in four segments if possible.

By doing this, you have 4 segments of 50 W, but you also need 90 / 4 = 22.5 W per segment.

(probably there is a way to exactly calculate how many LEDs in one chain you can power with a certain PS but some more variables are needed, which I don't know).

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  • \$\begingroup\$ Thank you for this!! So essentially if I were to stick my voltage to four equal points along the whole strip I would see better power dissipation? I assume I need fairly thick gauge wire running into the power supply? \$\endgroup\$ – user2544427 May 24 '19 at 1:12
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From the user questions online, as I expected, "BUYER BEWARE if you run this strip at max power you'll see about 50% voltage drop from the input to the output which will cause the color temperature to change on one side of the strip."

I expect you would need AWG 16 x5m feed to each end and locate PS near centre.

AWG 16 .... 12.5mΩ/m x5m x 18A = 1.12V drop per wire and feeding from centre reduces at least half this in the middle.

If that user was correct (?) voltage drops 50% on 1 reel at full power then 2.5V/18A/2wire = 70mΩ per 5m power trace

My suggestion: ( if possible)

schematic

simulate this circuit – Schematic created using CircuitLab

40A on FPC traces .not.

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  • \$\begingroup\$ So if I understand you correctly, you're saying run one rail to the beginning and end of one 5m strip (half of the whole lot)? Is this because the ws2812b can run power from either side? Can this be accomplished from a single rail? Maybe I just don't understand you properly. And are you saying 16 AWG is adequate to run this power? Luckily 16 AWG is what I'm using. I appreciate your time put in this, that CircuitLab is sick never seen that site before \$\endgroup\$ – user2544427 May 24 '19 at 1:29
  • \$\begingroup\$ As I show, DC is close the centre two feeds 5m LEDstrip connections. Current sharing with 5m pairs AWG 16 to both ends will reduce the drop by >50% by sharing current at both ends with 3 DC connections for +/- \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 May 24 '19 at 1:55
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You are doing it right and your power estimation is correct. ~50W PSU should do fine, 200W will have plenty of safety margin.

You are suffering from resistance of power and ground tracks on the LED strip, as current at maximum power is quite large (~10A). Your approach (connect power/ground at several points) is also correct.

Optimally, you always need to solder additional wires in the middle, like this. X is where you connect the power to the LED strip:

  1. ===X======X=== is the best (maximum distance to power point is 3). You can extend this approach to have 3-4 connection points.
  2. X======X====== is worse (maximum distance to power point is 6).
  3. X============X is also worse (maximum distance to power point is 6).
  4. X============ is the worst (maximum distance to power point is 12).

As your wires would carry 10A - you need to use quite thick wire to connect power, at least 1-2.5mm².

Few other notes:

  1. Please watch your heat dissipation. My experience shows that 9W/meter would lead to self-distruction of LED strip due to overheat if you don't glue it to some heat dissipator. You need to avoid temperature >85°C.
  2. To ensure more reliable operation you might also need to solder larger electrolytic low-ESR decoupling capacitors (something like 10000uF 6.3V). On the other hand - it could be an overkill. Small decoupling capacitors are already there.
  3. You might not need full power. In my projects full power was always too bright, so I limited it in software to something 1/10 of max.
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  • \$\begingroup\$ Thank you for this comment!! So just to clarify, in scenario 1 you're saying to hook my voltage up at 4 points along the whole strip NOT starting at the beginning? Is this because ws2812b can handle power from either side (or so I was told)? Should I run the ground to each point along with the voltage? Or to each end of the whole strip? \$\endgroup\$ – user2544427 May 24 '19 at 1:04
  • \$\begingroup\$ @user2544427 Yes, if you connect it closer to the middle - it is just more efficient. LED strip can take power at any point. Yes, you need both ground and power at each point. Ground rail resistance is as important as power rail. So you need same number of connections to ground as you will have to power. \$\endgroup\$ – BarsMonster May 24 '19 at 2:54
  • \$\begingroup\$ @user2544427 If you only had 1 ground connection before - than it was the problem. If you will start with 2 connections for power and 2 connections for ground, all in the middle (not edges) - you will see very significant improvement, and maybe 2 connection points will be enough. So do not hurry with 4 point connection. \$\endgroup\$ – BarsMonster May 24 '19 at 2:55
  • \$\begingroup\$ Ground rail resistance is MORE important than power rail. The same current flows through both but the control signals are referenced to ground. \$\endgroup\$ – Finbarr May 24 '19 at 11:35

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