I'm going to be powering some 20 m of LED strips at 24 V. After testing the strips, I can see that this will draw about 6 A.

Of course when inserting power from one side the voltage drop across the strip is large enough to make a significant difference to the brightness of the LEDs towards the end of the strip.

I would like to insert power from both sides of the strip, like shown below where I have modelled to resistance between across the power lines.

My concern in that the two 24 V lines might somehow compete with each other, having some unwanted power loss between them.

My question is whether this is a genuine concern or not? Would it be preferable to cut the 24 V line towards the middle of the strip?


simulate this circuit – Schematic created using CircuitLab

Edit: I think I was a bit ambiguous. To be clear, I will be using a single power supply, inserting power at both ends (not sure if that makes a difference).

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    \$\begingroup\$ The two-dimensional diagram here has, as far as I can see it, four corners. How do these corners correspond to "both ends"? \$\endgroup\$ – hmakholm left over Monica Nov 6 '17 at 11:05
  • \$\begingroup\$ @HenningMakholm: The 0V and 24V lines of such a LED strip run along the length of the strip. No matter where you cut it, you'll have those lines. Thus, you have 2 terminals at either end. \$\endgroup\$ – MSalters Nov 6 '17 at 11:08
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    \$\begingroup\$ @MSalters: But which are those "2 terminals at either end" in the diagram? Is it "two on the top and two on the bottom" or "two on the left and two on the right"? \$\endgroup\$ – hmakholm left over Monica Nov 6 '17 at 11:10
  • \$\begingroup\$ @HenningMakholm see my answer here: electronics.stackexchange.com/questions/307110/… The second diagram illustrates a 12V LED strip that can be cut every 2 inches. For 24V the principal is the same, only LEDs are 6, not 3 in each group. \$\endgroup\$ – Todor Simeonov Nov 6 '17 at 11:44
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    \$\begingroup\$ @Agent_L You can't use thicker wires - LED strips have the conductors baked into their design. They're made to be run in legs of a certain length and no more. It's up to the installer to arrange a suitable topology to feed those legs correctly. \$\endgroup\$ – J... Nov 7 '17 at 0:22

You don't need two power supplies. One is enough. Just connect the positive wire to one side of the strip and the negative to the other. Then you will have equal voltage at each LED group and this will ensure equal brightness of all LEDs.


simulate this circuit – Schematic created using CircuitLab

Using two separate power supplies has some drawbacks:

  • In the middle of the strip voltage will be lower than both ends. LEDs in the middle will shine less than LEDs at both ends.
  • One of the supplies may fail and not be noticed for long time, until someone sees the difference from one end to the other.
  • Two power supplies almost always are more expensive and installing mains supply (230VAC/115VAC) connection leads to additional costs.

Update: When current through the strip is higher that the strip could handle safe, you need to cut the strip to pieces and power each piece separately. However the cut pieces can also be powered as I suggest above to ensure equal voltages and brightness.

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  • \$\begingroup\$ That's a really neat way of getting equal brightness. My concern is that all the current will be going through through that first section on the strip, which at 6A might be problematic. Also the brightness while equal won't be nearly as bright compared to full brightness at 24V \$\endgroup\$ – Makoto Nov 6 '17 at 10:00
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    \$\begingroup\$ @DrFriedParts Yes but the resistance of the extra ground return wire is now an obstacle for ALL LEDs so brightness will be equal. You might want to do some calculations / simulations to confirm that you're wrong. \$\endgroup\$ – Bimpelrekkie Nov 6 '17 at 10:09
  • \$\begingroup\$ @Makoto - about 6A through the strip traces - yeah it might be a problem. Consult manufacturer's datasheet if you have one. However even if you cut the strip to reduce current flow, it is better to power it from both ends for equal voltage and brightness. About the drop: brightness will be a little less, but most power blocks have a potentiometer and you can raise the voltage a bit. \$\endgroup\$ – Todor Simeonov Nov 6 '17 at 10:12
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    \$\begingroup\$ This isn't exactly great - now you've got to run the low leg all the way out to the other end of the strip. Over 20m for standard 18ga wire (for this type of install) you're just adding more length to the overall loop and more overall voltage drop. This is not a recommended installation practice. This question really belongs on Home Improvement and not electronics SE, to be honest - unless you're a lighting installer you probably don't know what standard industry practice is for these types of systems. I'm quite sure this is not one of them. \$\endgroup\$ – J... Nov 6 '17 at 15:18
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    \$\begingroup\$ I didn't see a picture but I agree, if they are equal width traces, then any manufacturing discrepancies between the two would be negligible. I just wanted to point this out because some strips are made with a ground plane on the bottom layer of the board and thinner supply trace on the top layer. \$\endgroup\$ – kjgregory Nov 6 '17 at 18:33

Note: As originally asked, the question posed connecting two supplies in parallel.

Don't do that (please).

As you correctly suggest, the supplies must be in perfect balance for this to work correctly. That means unless the reference voltages, architecture, environmental conditions, manufacturing conditions, and binning are identical, which cannot happen in practice, one supply will carry some fraction of the other supply as a load.

Exactly what happens depends on the details of the power supplies' design. It can range from one supply will effectively just disconnect (in which case it's pointless to add it) to large circular currents which result in additional heat generation (inefficiency, wear, damage), part destruction, fire, or even propulsive damage.

Many power supplies out there are poorly constructed and mislabeled. At best you are relying on the safety mechanisms (fuses, diodes, et al.) in the nominal case to provide regular operation. Crashing your car deliberately because you have air bags is not a good plan.

Connecting the outputs of 2 uncoordinated power supplies together is dangerous and should never be attempted.

Do this instead...

  1. Either you can insert power from the middle of the strip to reduce the apparent change in brightness or...

  2. Just cut the strip in the middle and power each half separately from both ends (1 strip powered from 1 end each).

Updates: Single supply; both ends

OP now suggests that s/he will use a single supply and connect to both ends. If you use a single supply at both ends the question becomes how will you get the single supply to the far end?

There are many ways to do that but let's look at some extremes:

Power supply is at one end

This scenario is exactly the same as before (above), but now we have an external path to carry the current that is not part of the LED strip (the LED strip also has this pathway built-in). Assuming the conductors we use externally are much larger than the ones internally, it will help reduce the losses at the far end you will raise the relative brightness between the near and far ends, at the cost of more cable and installation effort. The middle will still be dimmer and the distribution cable you use to get the current to the far end will need to be very low resistance (large diameter or many strands) to maximize the effect.

Power supply is in the middle

The best thing to do (easiest, cheapest, fastest) is to just center-tap the LED strip. Apply power from the center of the strip instead of the ends. You don't need any additional cabling. The current load in each leg is now half of what it was before and therefore the voltage/brightness changes between the two legs are half of what they were in an end-fed configuration.

Add an additional return wire

Some of the other answers have proposed this solution, but did not discuss the consequences of that additional wire. To understand it, you will need a significant wire for that pathway. A 24AWG wire is approximately 0.08 Ohms/meter. At 6A and 20 meters, you will lose almost 60W of power in this return line and your LED voltage will drop from 24V to less than 15V. Your LED's will be very dim. If you increase the wire to 8AWG, the power loss will fall to about 1.5W and LED supply voltage will be about 23.8V, but the cost of that wire will exceed the cost of the LED's -- suggesting that you could achieve better overall illumination (not to mention ease of installation, lower cost, and higher efficiency -- longer life/lower power bill) with shorter strips (e.g. center-fed).

Managing Brightness

People key to sudden changes in brightness. Standard installations will manage the human eye as best as possible. For a long straight run, for example, keeping balance and symmetry is the key.

enter image description here

This type of topology avoids sudden "sawtooth" brightness patterns or long fades that people tend to notice. The idea is to break up your runs in a sensible way and feed them such that you minimize overall variation and abrupt changes.

For even longer runs, or to further minimize variations, as above, it is acceptable to feed both ends of the strips and continue subdividing as needed. Most LED strip manufacturers will issue installation guidelines about the maximum recommended length of a leg for any given strip type so you should do best to read and follow those guidelines.

enter image description here

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  • \$\begingroup\$ Check my edit: I'm using a single supply connected to both ends \$\endgroup\$ – Makoto Nov 6 '17 at 10:00
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    \$\begingroup\$ Connecting the outputs of 2 uncoordinated power supplies together is dangerous and should never be attempted. That is ONLY TRUE when the supplies do not go into current regulation when their maximum current is flowing. Try it with two lab supplies, set supply A to 24 V, 0.5A and supply B to 24 V, 0.5 A, now load with 24 V 0.7 A. If the currents are shared equally (this is very unlikely) then increase the voltage slightly on one supply. Then that one will go into current limiting. \$\endgroup\$ – Bimpelrekkie Nov 6 '17 at 10:14
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    \$\begingroup\$ Relying on current regulation is dangerous, especially in consumer applications. That's the equivalent to suggesting that sleeping next to an open fire in an apartment is safe because the fire alarm will sound if the walls catch fire. Safety mechanisms are a last resort in case of an off-nominal situation. Many LED strips you find online are sold with cheap China-designed power supplies that lack basic protections, are mislabeled as being more capable supplies than they actually are, run hot, and use sub-standard (often recycled) components. UL wouldn't approve this for sale. \$\endgroup\$ – DrFriedParts Nov 6 '17 at 10:35
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    \$\begingroup\$ @DrFriedParts it would be good if you try to explain why my solution would not work. I have made numerous installations using this approach and they work for years. And I see no theoretical or practical reasons they should not work. I'm not saying I'm 100% right, but when you say someone's wrong, please point out the mistakes. That is why we are all here - to learn something new every day. \$\endgroup\$ – Todor Simeonov Nov 6 '17 at 10:46
  • \$\begingroup\$ @Todor -- Could you help me understand your concern a bit better? First, it's become unclear as to what your solution is exactly at this point. That said, I didn't say any of the proposed ideas wouldn't work. I said some configurations are dangerous. \$\endgroup\$ – DrFriedParts Nov 6 '17 at 11:13

This is less of an EE problem, and more of an installation-and-packaging problem.

You can't put 6 amps on a 2-amp bus

The LED strip is made of numerous segments. Each segment is 6 LEDs in series with a resistor, and in between each segment is a "CUT" line.

Down the entire strip are two PCB traces, which are a positive and negative bus. They serve every segment. These traces have finite ampacity. On 12V monochrome strips, traces are sized to power one 5m segment (2A), or absolutely at most two 5m segments in series, counting on the voltage drop (well shy of 4A). Running 6A through a 12V strip is out of the question. With 24V strips, they might be built the same as 12V strips, or their traces might be thinner, since they only need half the ampacity to do the same thing.

But even if you used the strips within their design limits, this can still result in noticeable dimming. Remember LEDs are not linear - their V-I curve is rather steep, and the I-lum curve is not flat either. They're not linear (quadratic) like incandescents, a little voltage drop goes a lot dimmer. It can be seen even in a single 12V strip, comparing one end to the other. You're running 24V, which will helpby a factor of 2 to 4 depending on how much they shrunk the PCB buses.

Feed from the 1/4 and 3/4 points...

A 20m strip is probably four 5m strips. Place your power supplies between strips 1 and 2 ... and between strips 3 and 4. That way each leg of your power is supplying only one 5m strip. The strips are built for this.

... Or use feeder

If you ever look at trolley-bus wires, they have 2 bare conductors over the street about 2' apart. Then every few poles, they have a jumper over to two much heavier power wires running along the poles -- those are feeders. Here's a photo from Dayton. Note the hanger and cross-wire are "hot" and feed the left (+) wire, insulating the right (-) wire, then jumpered to a feeder (note the double feeders). These jumpers exist every 4-6 poles.

enter image description here

You do the same thing. You choose a feeder cable of sufficient size to avoid voltage drop, and then tie it together everywhere you can. This results in the PCB traces and the feeder being fully paralleled, but there is nothing wrong with paralleling feeder when it is DC. That's what feeder is.

There are unrelated reasons not to parallel AC or parallel circuits with onboard circuit protection.

If voltage drop is bad, multiple power supplies aren't

Two DC power supplies right next to each other might have a problem because they are closely connected. But here, you're talking about needing multiple power supplies because of voltage drop. That electrical "distance" means you're far less likely to have a problem with the supplies fighting. As long as their output voltages are closer to each other than the voltage drop between them, they both must carry their weight. The trolley bus system has many sub-stations feeding 600VDC into the system. They are simple machines, and they can't fight each other because of the distance and having diodes (as part of the rectifier) which prevent backfeeding.

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In theory you could do what you propose however there will be practical issues you need to be aware of.

It is likely that both 24 V supplies will not deliver exactly the same voltage. Let's say the supply V1 delivers 24.00 V but supply V3 delivers 24.05 V.

In that case V3 will take most of the load and V1 will only take less than half of the load. Only when the voltages are exactly the same will they share the load equally.

This isn't a problem as long as you would use two 6 A supplies and not two 3 A supplies.

The LED's brightness will be more equal though, now the LEDs in the middle will burn the least bright.

Look at Todor's answer for a better solution!

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    \$\begingroup\$ You're forgetting that the power supplies are regulated that means they fight to keep the voltage where they want it. No I am not, supplies can only supply current so connect two in parallel and the one set to the lower voltage will not deliver any current until the output voltage drops below the voltage it wants to maintain. I suggest getting two lab supplies and try it out for yourself. Only when using an electronics load (these can act as a supply and a load) would you be right. But these are costly lab instruments, not needed for driving some LEDs. \$\endgroup\$ – Bimpelrekkie Nov 6 '17 at 10:06
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    \$\begingroup\$ @DrFriedParts Obviously you 1) have never tried this and/or 2) if you did you are using an "odd" power supply without any reverse input current protection. Use proper lab supplies and no current will flow. To make current flow the 3 V supply would have to be an active load. \$\endgroup\$ – Bimpelrekkie Nov 6 '17 at 10:30
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    \$\begingroup\$ All but those designed for parallel operation will present a load. Certainly the cheap consumer SMPS contemplated here do Hmm, usually the cheap consumer SMPSs are not designed to be used in parallel. Most will deliver more than their rated current (they do not have a proper current limit). I suggest that you connect such a cheap SMPS and connect a slightly higher voltage than it's rated voltage to its output and see if any current flows. 99 out of 100 have a diodes at the output preventing current flowing in so I don't see how they even can load the output. \$\endgroup\$ – Bimpelrekkie Nov 6 '17 at 10:37
  • \$\begingroup\$ That's definitely not true of most Lab supplies as they almost always have some kind of linear output stage after the AC/DC rectifier. Take the HP/Agilent/Keysight E3631A as an example: gerrysweeney.com/wp-content/plugins/download-attachments/… \$\endgroup\$ – DrFriedParts Nov 6 '17 at 11:05

This is a classic case of a little knowledge being a dangerous thing. Your error is clear from the phrase "resistance across the power lines". If your power lines are kilometres long, then you need to consider that. If not, you don't. It's that simple.

Of course, if you use really thin cable then you might need to consider its resistance. However then your biggest problem is that the cable is now a heater element and will be at risk of melting the insulation and/or starting a fire. If you've paid attention to current ratings on your cables, then the resistance in the cables is neglible and can safely be ignored.

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    \$\begingroup\$ The power lines on an LED strip are really thin, and the resistance is significant, evidenced by the clear drop in brightness. \$\endgroup\$ – Makoto Nov 6 '17 at 12:34
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    \$\begingroup\$ This is just wrong. Voltage drop is a real concern in low-voltage lighting applications. If you think you can ignore voltage drop when doing this type of installation you simply don't know what you're talking about. \$\endgroup\$ – J... Nov 6 '17 at 13:27
  • \$\begingroup\$ @Makoto How significant? At 6A, every 1V drop means you've got 6W of power dissipated in the resistance, and that is absolutely a fire hazard. You most likely want to check how you're wiring them. Maybe you can get away with the thin cable it came with for a single strip, but for multiple strips you'd need to replace that cable with something able to handle the current. \$\endgroup\$ – Graham Nov 6 '17 at 13:53
  • \$\begingroup\$ with 5m, (about 1A), powered like Todor suggested there is a 3.5v drop. so about 3.5 watts. But the current plan is to separate the lines into 5m sections \$\endgroup\$ – Makoto Nov 6 '17 at 13:57
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    \$\begingroup\$ 6W of power dissipation isn't a big deal if spread over 10 metres. The problem is, putting 6A on a PCB trace made for 2A is going to cause a lot more than 6W. @Agent_L you're allowed a lot higher ampacity on the same wires inside a chassis, because distances are short and users can't contact it. For instance 14AWG is not unreasonable for 30A in a chassis. I for one use 20AWG stranded wires to connect LED strips, because 18AWG solid is too stiff and will pull connectors off or tear soldered wires off the PCB pads. Mind you, I typically have at most a 4' section of strip. \$\endgroup\$ – Harper - Reinstate Monica Nov 6 '17 at 22:04

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