I decided to give a try to a LED strip/rope (already fastened above a window to make autumn/winter days a bit brighter and it works fine so far).
The LED strip parameters are 12V / 11.5W / 1m.
Its total length is 3m, so the total power should be 34.5W. That's 2.875A.

I originally ordered a 12V/3A power supply. The reseller contacted me on my cell phone and suggested me a 5A power supply (in the attached picture) as the 3A power supply was borderline. They said it should work but the 5A power supply will be better.
(They didn't try to sell me a more expensive item - actually the 5A power supply was a tiny bit cheaper possibly because of a different manufacturer and missing mount holes).

I wonder why there is such an output power reserve needed?

In power supplie like that, the manufacturer's attitude at rating their units is more important than the exact printed current itself. 3A unit from one manufacturer may well be more capable and built with higher rated components and better heatsinking than 5A unit from another. You cant really know untill you open and evaluate the units, or atleast weight them

Some Ebay units are built quite dangerously - No satisfied clearance distance between primary and secondary tracks on the board, Who knows how much islation between primary and secondary in the transformer, Electrolitic cap of primary positioned so it would touch the secondary heatsink (that is usually connected to the secondary) if it bulges, .... Another reason to open and inspect power supplies before you get to use them

The LED strip would be drawing a bit different current at different voltages. If we assume the provided rating to be correct, and LED forward voltage to be constant, then very approximately, according to "diode = constant voltage source" model :

dI = dV / R

dI = I(real) - I(rated)
dV = V(real) - V(rated)
R = resistors that are in series with the LEDs

But this is important for the LEDs, that usually allready work quite on the edge in such strips. The power supply current rating is not that precise - a properly made one will supply right up to its rating without damage (and few have some more safety margin), bad one will slowly burn up at half of its rated load

And i'd be also concerned with 3A going through the conductors that go along the LED strip, close to the end from which the supply comes. It may be enough to raise the temperature there a bit, so push the LEDs harder than they allready are

These LED stripes have ballasts directly on the strip itself (most frequently just resistors, but constant current regulators are also available for that purpose). That means the stripe need as accurate 12V as possible. The thing is, the resistors drop just about 2.5V, so 0.5V extra drop means 20% less LED current and so less light, therefore the need for the accuracy (standard requirement for LED stripes is 2%).
And that is the main problem with the high current and narrow supply bus traces on the stripes. The problem is not the stripes wopuld get damaged by that, but there will be quite significant voltage drop. And that may lead to brightness nonuniformity along the stripe length.

Therefore generally it is better (if possible) to:

- Split the stripe to multiple sections and feed each one by separate wires. But still keep all wires the same length (or better to say to length corresponding to the stripe section length - the key is to have the same voltage drop). But if you use thick enough conductors, their length differences would not matter anymore (again, it is all just about the voltage drop: Once you stay below 0.1V difference, it is OK; the 0.1V corresponds to about 4% in current, so brightness difference)

- Supply the stripe section from the opposite ends (positive connection on one end, negative on the other). The wire to the far end then runs parallel with the stripe. With this each LED will have same supply bus length to the supply (on one end it is none for positive and the full on negative, on a middle one it is half and half,...), so the voltage drop seen by each of the LED segment will be the same.
With long stripe sections this is quite important, because the conductors on the stripe have very low cross section, so build up their resistance pretty fast.


Supplying the whole string from just one side and even chaining multiple stripes behind each other (how it is set up with many commercial kits) is completely wrong, it may work only with really short and low power stripes, not with 3m 36W one.

 With your setup (assume you need a single 3m long illumination stgrip) I would at least cut the stripe to 1.5m halves, connect one supply terminal to the middle of the stripe and the other terminal split among two wires, each going to one end of the stripe and connected to the strip there.


Generally if the stripes use the constant current regulators instead of simple resistors (diode-like looking device instead of the resistor), they are then quite immune towards these voltage drops (nearly 1V means no significant difference in brightness), but these stripes are very rare and I would guess they will be more expensive.


Regarding the supply: The 5A would be indeed better, but only if that means really a higher power design (higher efficiency, better cooling,...).
What worries me is, the 54A is cheaper than the 3A version. That may be just some resaler's margin difference, but it may as well be, the 3A would be well made and the 5A being the barely sized piece of ****. In that case the 3A would be obviously better choice...

Moreover on the unit displayed on your photo is a unit rated "For IT use only". The LED lighting is not an IT equipment. Problematic could be really the lifetime when heavily loaded all the time (as with the LED's), the output voltage accuracy and mainly the frequent switching of the input (the "IT" supply needs larger input capacitor to cover supply dips, but the lighting type needs rather lower capacitance to limit the inrush stress associated with primary switching).

The "IT equipment" those 3A power supplies come with are LCD monitors, they draw fairly constant power so that can be considered as "heavily loaded all the time". I think this is some regulatroy requirement to write this on the power supplies as it is written on many of them from different manufacturers. I guess not really related to their characteristics

The decent ones have a NTC in series with the input, this gotta suffice if you let them cool down before switching back on

The "IT equipment" is related to an assumption it will run on an outlet, where is sitting a high computer supply as well (and so providing quite strong overvoltage suppression). Plus the emission RF limits are different (more relaxed; relies on the fact the computer will be radiating as well). And the hold off time specification as well.
At the same time it expects free surrounding (for cooling). So in tight lighting installation it may overheat more than with the expected IT installation.
But it does not expect frequent input switching - so it may get damaged by that.

On the other hand the lighting does not have to provide any hold off time, the fact the lights dim down during mains undervoltage is of no problem at all (unlike the reset of a computer or so).

The fact the higher power unit is cheaper than the supposedly lower power one is really strange, I would really suspect the higher power one would be more aggressively "cost optimized"...

I've just checked that. There is 12.16V at the input (measured at the ends of the black-red wire) and 11.14V at the very end of the stripe. That's a 1V drop at 3m of the strip length.
Actually, the brightness decrease is barely visible - I haven't noticed anything before and now I rather guess it (I can "see" it because it should be there. )

My strip is this. You can click the image to check the details. (I know I could order such a strip for a half price directly from China but I prefer local shopping because of the warranty.)
Another reseller (GM Electronics, IIRC) suggests the maximum length for similar kinds of strips up to 5m.
My original power supply choice (the ordered one) was this.

Btw. I find a bit strange another suggestion. You shouldn't put the switch at the 12V side but only at the 230V side (before the power supply). You could destroy the LED stripe otherwise. Why? Is there such an overvoltage on the idle power supply terminals that could damage the LEDs?

IT equipment includes laptops and networking devices (switches, accesspoints and such). I seen the "IT equipment" notice on some power supplies for them too. But they must be expected to be installed alone - laptops are laptops, and switches and accesspoints are often installed away from everything else. So the "relies on presence of other devices" is eliminated

Normal use of laptops includes plugging in and out up to several times every day. So the "dont plug in and out" is eliminated as well

What i think is either :

 - They mean that the EMI levels from the device itself are approved for an IT device but not for other appliances, so the power supply is ok when it is part of the IT stuff and not OK when part of other appliance

 - They mean "dont short circuit the unit by starting motors with big starting current and long starting duration with it" (allthough there should be no damage, the power supplies go into slow pulsing mode when the output is shorted)



What they say about damaging the LEDs is, they are expecting an overvoltage to be present on the output of the power supply after it was powered up without load

In switching power supplies, the power supply volage is regulated all the time by measurement from the secondary, communicated to the control chip in the primary through an optocoupler. There is no overvoltage possible

In linear unregulated (transformer, rectifier bridge and capacitor) power supplies, The capacitor theoretically charges to the peak line voltage * transformer coefficient, so may be quite a bit higher than rated voltage under load (can be as high as 2x, partially because the manufacturers sometimes do the transformer itself with a bit higher output than rated too, to prevent undervoltage under full load). Some 17V..19V on an idle "12V" adapter is sorta normal. Then you connect the LEDs, and they get the inrush at several times their normal current : Instead of 12V supply and 2.5V drop on the resistor, there is 19V supply and 9.5V drop - so 4x the current..

Some LEDs are rated for high current pulses (for permanent work in this condition !), some not, in some the rating is not 4x but less. If we take for exmple 4700uF cap (that s quite big, usually the cap is smaller) * 7V overvoltage / 1A load = about 33 mSec. The LEDs are rated for pulses in the order of 0.1 mSec. But 4x overcurrent is not that much either, i dont know whether a non repetitive 4x can make any damage in 33mSec..

The elevated voltage when unloaded is related to the method used for the voltage regulation - how the secondary voltage is sensed.

Either it is sensed directly and then the error signal is send to the primary side by an optocoupler. With this the output voltage remains exact even when completely unloaded. The drawback is, it needs that optocoupler, which uses to be a bit of trouble maker (LED loosing output, so the coupler loosing CTR, too high dark leakage mainly at elevated temperatures, more likely failures leading to an overvoltage even with high load).
Because of the frequent no load operation with very strict internal consumption limits, this method is the most common one for the IT equipment.
As the output is usually regulated again inside of the equipment, the eventual overvoltages due to optocoupler failures could be swallowed for a short time (sufficient for the fault detection in the mains supply controller to detect the fault and shut down). But for the LED's it could be killing (in case of optocoupler failure, the voltage may double for about a second or so, sometimes with restart attempts yielding about 1:50 duty ratio)

Other method is to use the main flyback transformer to carry the output voltage information back to the primary. This eliminates all the problems related to the optocoupler, generally is an order of magnitude more reliable, but the drawback is, it is able to extract the voltage information only when transferring power. When unloaded, the power demand is very low, so the regulator looses track on the exact secondary voltage, yielding the secondary voltage to increase. However usually very minimal load (few mA for an "indication" LED and/or the secondary electronic switch controller consumption) is sufficient to keep the voltage in check. This method is usually usedf for the lighting (either for the main voltage regulation for these voltage fed stripes, or for the over/undervoltage protection for the current fed LED modules).
Depend on the required accuracy and minimum load power for regulation, the second one uses to be very simple, so it is popular with the cheap cheepeese "phone chargers". The LED there has the main function to provide that minimum load (the very simple regulation requires few percent of the max load) for the simple voltage regulation to work, so it is really not recommended to disconnect it (it uses to be quite obtrusive).
The more advanced schemes used within the related IC's for the more accurate regulation suffices with order of magnitude lower preload, so even with the 3A rated supply a 3mA preload would be sufficient to keep the output voltage in check.
Such preload is not used within the LED supply itself, because it is a component, which consumes extra power.
The thing is, either the LED's are connected permanently to the supply output (so it never works unloaded), or the secondary side control is usually some more advanced electronic one, which require some idle current to supply the controller. And that supply current uses to be sufficient to act as the required preload to keep the voltage in check.

Where the secondary switching is really prohibited is, when the LED current is dictated by the ballast converter: When the LED's are disconnected, the ballast charges it's output filter to the overvoltage protection threshold (and the auto-restart keeps the voltage there), so when you connect the LED's back, very huge current spike will flow into the LED's, possibly damaging them. In this case there are no resistors limiting the current, so it may be order of magnitude higher than the rating. And that is really way too much.


And when looking at the supply designs: I would really trust way more the white one.
At first really designed for lighting, so switched on the primary.
Second most likely no optocoupler (so limited to primary, or electronic secondary switching)
And 3'rd you may open the case without damaging it, so not that difficult to repair.
And 3'rd these use to be way more full-load efficient than the "IT" ones, so generate less heat (I guess that would be the cause for the higher cost). The thing is with lighting, the reliability requirements are really higher than for the notebook or monitors or so, so even when we would be talking about a cheepeese category, they will still be better than the "IT" of the same "cheepeese" rank.


For feeding the LED's: 1V is about 30% current difference, it is already quite significant. The "5m" limit is really what the traces are able to support without damage, but the voltage drop there would be excessive.
You may try to feed the LED's by each supply wire from the opposite end of the stripe and then measure the voltage the LED circuits really see. It is really very huge difference - from a 1V difference you will get something about 0.2V or so (about 11.5V everywhere). Even when the stripe is already quite long, so there is the 0.5V drop, it will be at least seen by all the LED's as the same.