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I'm currently helping a friend who is renovating a VW camper to fit some LED lighting as she has no electrical experience. She has a 10 metre set of LED fairy lights that are powered via USB (bigclive reviewed something similar, or indeed the same here). She has until now been powering them by plugging them into one of these mounted in the van but would like a more permanent solution. As a result I bought a 12V to 5V DC-DC Converter and a small switch which I wired up to draw power from the vans leisure battery.

I cut off the USB plug a few inches down the cable and connected the two wires directly to the output side of the converter. As soon as they were turned on though only about the first quarter of the LEDs lit up and they trailed off in brightness which would suggest an obvious drop in voltage. The DC-DC converter was also getting warm so I turned the whole lot off before any of the LEDs burnt out.

Assuming that the LEDs were drawing far too much current I dissected the USB plug looking for a resistor but couldn't find one. I'm hoping I haven't done permanent damage as they would be a pain to replace! These lights do have a bundle of 3 twisted wires so going from what bigclive says it would appear they do not require a resistor.

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  • \$\begingroup\$ Why can't you make the panel-mount USB charger into a permanent option? It is a 12V to 5V DC-DC converter. And just because it can be mounted in the dash or in some other visible panel, that doesn't mean that that's where you must mount it. \$\endgroup\$
    – Solomon Slow
    Commented Nov 20, 2017 at 17:41
  • \$\begingroup\$ @jameslarge - Mostly because she wanted a more professional installation using mounted switches for all of the lighting rather than having to plug in a USB lead each time you want to turn the lights on. \$\endgroup\$
    – Dan Workman
    Commented Nov 21, 2017 at 12:42
  • \$\begingroup\$ OK, so why can't you hide the USB power supply, same as you would have hidden the other one, and why can't you switch the 12V line feeding the USB power supply, same as you would have switched the feed to the other one? \$\endgroup\$
    – Solomon Slow
    Commented Nov 21, 2017 at 14:17
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    \$\begingroup\$ Because it's already mounted elsewhere in the van so she can charge her phone etc. I could have ordered exactly the same item and hidden it away with a switch but I felt it seemed an inelegant solution to keep a USB plug when all it should need (to my understanding) was a 5V supply. In hindsight it would have caused less issues but as I've already destroyed the plug I guess its a moot point. \$\endgroup\$
    – Dan Workman
    Commented Nov 21, 2017 at 14:29
  • \$\begingroup\$ The video you linked explains that if you power from one end it will become dimmer as you go further down the thin wires. You want to connect to the third wire to get even illumination. As for the over heated PSU I wonder if you have connected the remote feed wire to the other supply and you are trying to warm up the camper van with the wiring (not good). Check your connections as per the video around the 3 minute mark. \$\endgroup\$
    – KalleMP
    Commented Dec 11, 2023 at 10:21

2 Answers 2

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It is difficult to say what is happening with out measuring the voltage before and after the modifications.

Consider what may happen if the load on the replacement power supply is too low. A 12 to 5 volt DC to DC converter is likely a buck switching power supply. Such supplies may require a minimum load to operate correctly. This is discussed in detail in this stackexchance question.

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  • \$\begingroup\$ Blimey looks like I'm a little out of my depth here. Assuming the power supply I'm using does indeed require a greater load can I not add a resistor to the circuit to achieve this? \$\endgroup\$
    – Dan Workman
    Commented Nov 21, 2017 at 13:03
  • \$\begingroup\$ Yes, I thought of that. But resisted mentioning it in hopes you would first measure the voltages. If you do go this route remember V=IR so that if V was really 5 Volts you should be able to pick out a resistor for a given current. Also, that resistor needs to get rid of W= IV Watts of power. Most calculate this and double the Watts needed for safety. But since we do not actually know the voltage - this may not work - the power supply could even fail and cause harm. \$\endgroup\$
    – st2000
    Commented Nov 21, 2017 at 17:33
  • \$\begingroup\$ Yep, point taken - I'll get my multimeter on it first. Assuming I am getting the correct 5V then something like 200 ohm's should probably do it. \$\endgroup\$
    – Dan Workman
    Commented Nov 21, 2017 at 19:18
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If the dc converter is getting really hot, it sounds like your trying to draw more current than it is rated for. You may want to get a more powerful 5 volt power supply. Alternatively, you could measure the resistance across the string of lights. $$\frac{5V \times R_{lights}}{7V} = R$$ If you connected a resistor in series with the lights that had a value of R, you could simply hook it right up to the 12 volt battery. As long as the resistor could handle the voltage, you would have 5 volts over the string of lights, and 7 volts over the resistor. Should be relatively cheap as well.

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  • \$\begingroup\$ Well this is exactly what is confusing me. Yes I think they are currently (no pun intended) drawing too much but they certainly didn't used to. The USB port I had them plugged into was rated for 1A and the vans fancy leisure battery charger showed a load of 0.1A when they were plugged in. Also there are three wires running down the length of the lights which suggests they are wired in parallel (as per bigclive's video). \$\endgroup\$
    – Dan Workman
    Commented Nov 21, 2017 at 19:11
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    \$\begingroup\$ The lights aren't wired in parallel, that would be way too many wires, but the current of 0.1 amps is normal. The reason is that the DC converter must be overloaded in some way. That's the only reason that a converter should be heating up, if it is being asked to output more power than it is rated for. \$\endgroup\$
    – Sam Spade
    Commented Nov 21, 2017 at 20:43
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    \$\begingroup\$ Also, note that the battery is losing 0.1 amps. This is 0.1A * 12V = 1.2W. The transformer is therefore providing 1.2W/5V = ~.24A. The current output by the battery does not equal the current output by the transformer. \$\endgroup\$
    – Sam Spade
    Commented Nov 21, 2017 at 20:45
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    \$\begingroup\$ Ah of course, didn't factor in the power supply when thinking about the load. As for the wiring of the lights themselves it would appear that the third wire provides the negative to the far end so as not to have the LEDs fade along the length of the run. Not sure how that could affect it but the video describes it better. \$\endgroup\$
    – Dan Workman
    Commented Nov 21, 2017 at 21:14
  • \$\begingroup\$ I'm not in a position to watch the video right now, but I am confident you can figure out which wires are positive and negative. If your transformer is heating up, my experience tells me it's overloaded, and I would try just wiring it up directly to the 12 volt outlet with a resistor. Better yet, get another string of lights and hook them up in series. 5v + 5v = 10v which is close enough to 12 to be quite safe. \$\endgroup\$
    – Sam Spade
    Commented Nov 21, 2017 at 21:18

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