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I need to put an LED strip (ws2812) which uses 18watt per meters at 5V.
I intend to use 1 meter but the power supply is 10 meters away.

I can't use bigger wire than awg20 which I know will cause big voltage drop.
I thought about mounting multiple leds in parallel but it would need a tone of wires which is quite expensive.

Could I use something like a supercapacitor to help with the voltage drop?

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  • \$\begingroup\$ From a brief look at an online resistance chart, your voltage drop will be around 0.5V. Will this actually cause any problems? \$\endgroup\$ Oct 15, 2016 at 16:51
  • \$\begingroup\$ These leds are quite sensible, they need to be feed with 5V, they burn easily. \$\endgroup\$
    – user126774
    Oct 15, 2016 at 16:55
  • \$\begingroup\$ I've run (WS2812B) them at 3.3V with not too much color shift. I can't imagine that 4.5V would do much. \$\endgroup\$ Oct 15, 2016 at 16:57
  • \$\begingroup\$ They tend to burn over 5v and under their behavior is more random, I change colors, make them blink...quite often. \$\endgroup\$
    – user126774
    Oct 15, 2016 at 17:12
  • \$\begingroup\$ I got roughly 10 ohm/1000 feet, which would be something like 1.2 volts at 18 watts/5 volt. \$\endgroup\$
    – pipe
    Oct 15, 2016 at 17:14

6 Answers 6

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Ok so you need a stable voltage without an overly thick wire. You also don't want to put bulky components near the load.

So what are our options.

I wouldn't nessacerally rule out a regulator chip near the load. There are some impressively tiny switched mode converters available nowadays though heatsinking can be an issue.

Another option is "remote sensing". A pair of thin wires is brough back from the load to the power supply. This lets the power supply measure the voltage at the load and compensate.

If we know the resitance of the wires we can just build a power supply that compensates based on he known cable resistance but that risks overvolting the load if the cable is shortened.

Linear techonlogy has a clever soloution to this which they call virtual remote sensing. They use the decoupling capacitors in the load to measure the cable resistance and compensate for it without the need for physical sense wires or known cable lengths. The downside seems to be that it's measurements will cause some output voltage ripple.

http://www.linear.com/docs/30159

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  • \$\begingroup\$ I didn't know that. Remote sensing is a wonderful idea. Thank you sir for this information. \$\endgroup\$ Jul 29, 2017 at 17:34
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Use a PSU with a higher voltage and then use one or more buck regulators near the strip to bring the voltage down to the strip's requirement.

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  • \$\begingroup\$ I thought about a voltage regulator but it is bulkier than what I need. \$\endgroup\$
    – user126774
    Oct 15, 2016 at 16:31
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    \$\begingroup\$ Then just feed a higher voltage in from the power supply to take into account the voltage drop. This could work if your current draw is constant. \$\endgroup\$ Oct 15, 2016 at 16:45
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    \$\begingroup\$ " it is bulkier than what I need" No. It is bulkier than what you want. And, as the Rolling Stones so famously put it, "You can't always get what you want." \$\endgroup\$ Oct 15, 2016 at 18:07
  • \$\begingroup\$ @silverscania the current draw will not be consistent, so a higher voltage will only be workable if it is feedback controlled by a sense wire, or if there is substantial headroom between the desired voltage and the maximum voltage - and in that case brightness may be inconsistent too. \$\endgroup\$ Oct 15, 2016 at 18:23
  • \$\begingroup\$ Point-of-Load (POL) regulation makes all kinds of sense in this application. You do not necessarily need to crank up the primary supply if the POL regulator is a boost type. But there would be a limit on how much power you could transfer based on the primary supply voltage and the wire resistance. 5 V and 0.65 ohms would limit you to a little over 19 W delivered. \$\endgroup\$
    – Dave Tweed
    Oct 15, 2016 at 18:40
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Since you don't want to add cables or components on the LED side, it seems to me that your only option left is to compensate for the voltage drop in the source. Fortunately, your cable will have a relatively constant and fixed resistance that you can measure or calculate, and with this information and ohms law you know how much the loss will be. Remember that you will get the loss in both wires.

Fortunately, there are some ICs that does just this. I know because Linear Technology sometimes brag about them in the dead-tree journals they send me. The technique is simply called cable drop compensation. I'm sure there are other solutions from other suppliers.

Hacking together your own would be an option, and probably a fun exercise.

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  • \$\begingroup\$ Egads. Fancy. I was merely thinking of telling the OP to use sense wires brought back from the remote end (said he can't use 'bigger' wire, but sense wires can be very small.) Hadn't considered this idea, though. \$\endgroup\$
    – jonk
    Oct 15, 2016 at 18:13
  • \$\begingroup\$ This will be very difficult to practically implement as it will require determining the time constant of the at-load capacitance and modeling that. The sense wire is a far better idea. \$\endgroup\$ Oct 15, 2016 at 18:20
  • \$\begingroup\$ Yes, cable compensation is a thing, but it can't be 100% effective. The harder you push it, the less stable it is, ultimately turning into an oscillator. I suspect it would be very hard to get it working well with a highly-dynamic load such as a bunch of PWM LEDs. \$\endgroup\$
    – Dave Tweed
    Oct 15, 2016 at 18:49
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You can power it from your desktop 12V supply HDD port or laptop charger with a 5V step down regulator at the LEDstrip. ~$8.36 +shipping on Amazon

enter image description here It's also waterproof.

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  • AWG20 is 66 mΩ/m for paired resistance or 660 mΩ for 10m
  • The LED strip uses small series R drops from 5V to each colour Vf max for max current and thus max brightness and then PWM for dimming.
    • Thus brightness and power dissipation will drop with supply drop to very dim for BL/GR below 3V and RD below 2V.
    • At 18W, 5V or 3.6A the Req of the Ledstrip assuming it is an ideal 3V zener, the series Req. becomes 2V/3.6A= 556 mΩ.
    • I know the Zener is not ideal and all the 18W LED's would make the Zener have an additional approx. ESR of 1/18W=56 mΩ.
    • So your load power and current drops to about half from doubling the ESR.
    • from 556mΩ+56mΩ = 622mΩ to (622+660)=1282 mΩ with 10m of AWG20
  • if you used 10 pairs of AWG20 this would work with only 10% loss in brightness (neglible)
  • if you raised the supply voltage by V=IR=3.6A*1.28Ω=4.8V above3V =7.8V you would get exactly 18W at full brightness but then if the PWM's are not all synchronous at 10% dim, now the tiny SMT series R's that drop each sub-string may have twice the voltage drop from 5V now at 7.8V and burn up. or possibly never go as dim as originally.

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So your installation will not work and you need AWG 10 cable

But this costs more than the LEDstrip, so Plan B is feed a 10m AC extension cable with remote DC power and try to figure out your next question of send PWM signal 10m long with good signal integrity.

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  • \$\begingroup\$ or you can use a 5V 20W buck regulator from 12V at the LEDStrip \$\endgroup\$ Oct 15, 2016 at 19:30
  • \$\begingroup\$ I need to avoid anything bulkier than components like resistances, capacitors... at the led strip \$\endgroup\$
    – user126774
    Oct 15, 2016 at 22:06
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The solution is not using a voltage regulator but a current regulator.

Nearly all LEDs intended for lighting are rated 350mA (but check the datasheet if you find it) and there are power supplies in the market which do exactly that. When you use such a power supply, the cable cross section doesn't matter any more as the regulator measures the current and adjusts the voltage accordingly. Automatically.

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  • \$\begingroup\$ The LED is specified in the question, and it is a led module, with programmable current. It needs a voltage. \$\endgroup\$
    – pipe
    Oct 15, 2016 at 17:08
  • \$\begingroup\$ Ah, see it now. It's a driver chip plus LED, not an LED. Sorry. \$\endgroup\$
    – Janka
    Oct 15, 2016 at 17:12
  • \$\begingroup\$ It's a 5050 sized led, which run at 20 mA per diode as well. Not a 350 mA or 1W led. And there are LEDS that are above 1W as well. So thus advice is completely inappropriate. \$\endgroup\$
    – Passerby
    Oct 15, 2016 at 21:04
  • \$\begingroup\$ Got it before. Sorry again. \$\endgroup\$
    – Janka
    Oct 15, 2016 at 23:22

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