There are a few questions on here that talk about LED power supplies and how to calculate their ratings but none address the issue that my 5meter long 5v addressable LED strip of 18watts power consumption per meter draws a ridicilous amount or current. Current = (18watts*5m)/5v

Current = 18amps

How do I configure a power supply to supply that sort of current? 5v DC supplies on AliExpress go up to 4A. which supplies only 20watts of power when 90watts are required. The problem with 5v supplies as well is that it would require a current of 18A to deliver 90watts! Sockets are limited to 10A I think, which means getting a 100w 5v power supply drawing 18A makes no sense because it can't draw its required current through the wall socket.

I understand that higher voltage supplies can supply more power at the same current. 12v supplies also come in higher current variants, for example there are 12v 12a supplies capable of delivering 120W without drawing a ridiculous amount of current from the socket.

My question now is whether I can cheat the 10A socket limit by using a 12V power supply (rather than 5v to match the LED strip) at 10A and then use a DC to DC step down converter to change 12V/10A to 5V/24A. That way I get the 5v voltage required by the LED strip as well as enough current (18A) to power the strip at full brightness without drawing more than 10A from the wall socket.

Same power output but different current supplied. Is that how it works? I have no idea if I'm using the power formula correctly there and if power at different voltage levels can be treated that way.

  • \$\begingroup\$ Sometimes you just need a cage power supply. Nothing wrong with that as long as you're careful. \$\endgroup\$ Mar 26, 2017 at 8:06
  • \$\begingroup\$ With that 20A caged supply..Is 20A the output and the unit itself draws less than that? Does it convert the 240V/10A input automatically to produce 5V 20A output? \$\endgroup\$
    – danielbker
    Mar 26, 2017 at 8:11
  • \$\begingroup\$ Did you look at the page? "3.15A/115VAC 1.5A/230VAC " \$\endgroup\$ Mar 26, 2017 at 8:13
  • \$\begingroup\$ I did see that part just wasn't sure if it means what I think it means. I thought in order to create 20A output it needs 20A input. Everything I went through in the question is done automatically in one of those power supplies which I had no idea. I was going to reinvent the wheel with a higher voltage supply and DCDC converter not realising that a caged supply does that already! that makes it so much easier. \$\endgroup\$
    – danielbker
    Mar 26, 2017 at 8:19
  • \$\begingroup\$ A linear supply needs slightly more current in than out. A switched supply trades some of the input voltage for output current, so it has reduced input current requirements. \$\endgroup\$ Mar 26, 2017 at 8:20

2 Answers 2


If some steps aren't taken, then your imagination is right. The input current has to at least equal the output current. However, there are lots of ways to take some extra steps.

For example, you can use a transformer where its primary is designed for the AC mains voltage and where it produces an output voltage much closer to the desired voltage. A transformer will dissipate a small amount of parasitic power (it will leach some power in order to do its job), but it is reasonably efficient and can yield a lower-voltage, higher-current arrangement that changes what you imagined is required. You might use a \$6.3\:\textrm{V}_\textrm{AC}\$ secondary power transformer, for example, in order to get a voltage that is closer to the desired \$5\:\textrm{V}\$ and then add some circuitry to finish the job. If you use this method, then the current load on the AC mains supply will be quite a bit less than the current required by your LED strip.

Or, you might use what's called a "switcher," where it is designed to operate directly from the AC mains supply. These will use inductors and capacitors in a somewhat more complex arrangement than the above-mentioned transformer method and they will usually be smaller and weigh less and probably be cheaper, too (as AC mains transformers tend to be heavy and expensive, by comparison.) These also can be very efficient in terms of transforming power.

A transformer-based method starts out with reasonable efficiency with the transformer itself, but often needs some significant control "overhead" voltage and for \$5\:\textrm{V}\$ systems this might mean you get efficiencies that are perhaps as good as 60%. Switchers can usually achieve better, at the expense of some complexity but lower costs because of avoiding an expensive and heavy transformer. Perhaps 75-80% for an efficiency figure, but just be aware that this value can vary widely depending upon the specific design and how it relates to your specific needs.

You can boil all this down to a simple efficiency figure, such as 75% or 80% (or 60%, let's say, for the transformer method), and a power consideration. In your case, you need \$90\:\textrm{W}\$ of delivered power. Just divide the efficiency figure into your power requirement to figure out the AC mains power. Divide that by the AC mains RMS voltage and you get the required current.

Let's say the efficiency is 70%. Then the power required from the AC mains will be about \$\frac{90\:\textrm{W}}{70\%}\approx 130\:\textrm{W}\$. If the AC mains voltage is \$120\:\textrm{V}_\textrm{RMS}\$ then the current will be \$\approx 1.1\:\textrm{A}_\textrm{RMS}\$. If the AC mains voltage is \$220\:\textrm{V}_\textrm{RMS}\$ then the current will be \$\approx 0.6\:\textrm{A}_\textrm{RMS}\$.

So you can see that this is a lot smaller than you were otherwise expecting.

As others have already mentioned, you probably don't need a high precision \$5\:\textrm{V}\$ supply. But you probably do need a power supply that can hold \$5\pm 0.2\:\textrm{V}\$ at the current rating into the load that you need, which is about \$20\:\textrm{A}\$. So it would be a good idea to make sure that the power supply is rated for at least that much (more is usually better) current and that it is rated to hold the voltage within a reasonable margin around your specified voltage over variations of the AC mains supply voltage and over some range of your load's current requirements, too. This starts to get more technical, now.

A thorough design would take into account various risks, including risks of fire. An LED strip that is dissipating \$90\:\textrm{W}\$ over \$5\:\textrm{m}\$ length isn't likely, by itself, to start a fire. But a power supply that is connected to the mains and which may be, itself, dissipating \$40-50\:\textrm{W}\$ into a small volume buried inside a wall might actually cause a fire. So please also consider other aspects of your work. The power supply itself needs to have some safety measures built into its design and it needs to be arranged in a location where its dissipation can be reasonably accommodated. We often just assume you'd do that. But I've seen cases where people have placed SSRs with \$100\:\textrm{W}\$ of dissipating heat sinks attached and just buried them into their thin walls as though that was a safe thing to do. So be careful.

  • \$\begingroup\$ Thank you, very detailed answer. Would this supply work? It uses PWM for modulation. There is no efficiency rating.bit.ly/2o66n6i \$\endgroup\$
    – danielbker
    Mar 26, 2017 at 10:48
  • \$\begingroup\$ @danielbker It may cut things a little close. I'd rather see you use a power supply designed to deliver at least 150 W. But frankly, it's really more about the company and the quality of their design. Some companies will provide an absolutely bullet-proof 100 W design, with all the right things done, and it will be better to use that over another design specified for 150 W (but where they've used parts operating right at the ragged edge.) Some actually seal their electrolytics so that leakage, if it happens, doesn't destroy the rest of the board -- leaving it repairable. But yes, like that. \$\endgroup\$
    – jonk
    Mar 26, 2017 at 19:56

You are getting confused in thinking that the load current of the power supply directly translates to the input current to the power supply. This is not the case.

Instead the input power has a direct relation to the output power. The two power numbers are not the same because some energy is consumed in the supply such that the efficiency is not 100%. So consider that input power is line RMS voltage times current. Divide that by efficiency to arrive at the maximum you can expect out in Watts.


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