Another LED power question?

I bought some LED modules with the intention of using them to replace florescent tubes in some fixtures. But I am now getting a bit confused about the power requirements because the numbers don't seem to add up.
These are like the ones I bought.

http://www.ebay.com/itm/221546290709?_trksid=p2059210.m2749.l2649&ssPageName=STRK%3AMEBIDX%3AIT

In the details description it says that each led is .25 watts and with 5 LED's per unit that makes the power requirements for each one at 1.25 watts. There are 20 units in a strip or module so that would mean that each module would require 25 watts.  If the LED's are ,25 watts then that makes sense since there are 100 LED's per module. 

Now the thing is when I put power to them they are requiring a lots less.  With only a little AC adapter with an out put of 12 v and 2 amps or 24 watts I am able to power SIX of the modules. The power adapter does get warm after a few minutes but if I drop a module and go with 5 it is just fine. So if I go according to what they are actually using the power draw appears to be no more than 5 watts per module. Did they miscalculate? Or am I wrong in my simple math? I tried several different adapters and get the same results. A simple wall wart of 900MA can power 3 but overheats at 4.  How is it that they are figuring these at 25 watts when real life usage is saying its more like 5 watts? Its the 5 watt figure that would be more reasonable to me anyway. 
I am guessing that it will take 8 modules to replace one dual tube 80 watt florescent.  If they are 25 watts each I am loosing on the power consumption because that would be 200 watts as opposed to 80. At 5 watts I am doing a lot better coming in at about 40 so that gives me an almost 50% power savings. 

If I am correct in my estimates then a simple 30 or 40 watt power inverter like this would work perfectly.

http://www.ebay.com/itm/AC-DC-12V-Transformer-Power-Supply-Adapter-LED-Driver-IP67-Waterproof-10W-200W/271569039906?_trksid=p5713.c100284.m3505&_trkparms=aid%3D555012%26algo%3DPW.MBE%26ao%3D4%26asc%3D29396%26meid%3D624e74e14a1848afb36d865d8590fc01%26pid%3D100284%26rk%3D3%26rkt%3D18%26mehot%3Dpp%26sd%3D221682389057

Any ideas as to why there is such a discrepancy in these numbers?


Picture of Another LED power question?
AdapterC.jpg
LabelC.jpg
sort by: active | newest | oldest
Wired_Mist2 years ago

I know exactly what your talking about !!

I`ve been working with 12V Led strips for years and I Always Noticed the same thing!

The Engineer in me wants to say 1.25W is the Potential Wattage, That is then reduced by the resistor. (Like a crafty way of Altering the Stats of the Unit)

But Still, I've always believed it was because of how the strip is wired... all the groups of Led`s are wired in series, But each group of leds is wired in Parallel

If you drill holes into a garden hose every few inches you create a small water spout. (Think of each of these holes as a group of leds being powered off the 12V Rail, the hose) At first, most of them near the start will spit out water quickly. Towards the end of the hose it won`t spit quite so high.

I know the Voltage should remain constant when it`s wired is Parallel. Perhaps it`s the tracks them selves that are causing this effect?

**Fun Experiment**

I was going to do this test myself, untill I noticed my MM blew it`s 10A fuse >.<

Take a voltage reading from both ends of the strip and the overall current for the Entire Unit. THEN Repeat the process with a smaller group or 2-3 Modules !

Really hope you Post what you find !

Vyger (author)  Wired_Mist2 years ago

After reading through the answers and going back and taking more measurements its still a puzzle.

I measured the actual amps and still find myself scratching my head.

I also got out my AT computer power supply with a 10 am rating on the 12 volt rail just to eliminate the possibility of the smaller adapter being overloaded.

A single unit, string, all by itself draws 1.2 amps. About 15 watts which is half what the specs say. Hook another one up, leads both going to the PS and it jumps to 2.4 amps which makes sense. Now switch it to being on the end of the other one and the power draw is much less but the lights still look as bright. Add a second set just like it, so 4 strings and the amp draw is still not that much. OK add some jumper wires connecting from the PS to the leads of the second strings, effectively making them all connected individually to the PS. The amp draw should be 4.8 but its not, its way less. So I went all out and hooked up 8 strings connected to each other and with jumper wires to the PS and the total power draw was 4.9 amps with all the lights burning brightly. It should be pulling at least 8 amps but not even close.

I guess I am going to write this up into an instructable. Use better wire, better connections, not just twisted together and a step by step power readout. A better set up than just laying all over the floor with the cat playing with the wires.

I was going to do one anyway for the LED strips as there appears to be a lot of interest in them so included in the how to I will put this head scratcher and maybe someone can figure it out. Or maybe its not worth bothering with as long as it works. But from my initial experiments it looks like anything more than a 5 amp power supply will be able to power 8 strips with no trouble. Very strange.

I`d love to see an I`ble !

Would be nice if someone could set it in stone eh? At least they draw Less current rather then More :) When I get my MM back up I`ll try it myself too.

Be sure to give me a poke when you post it :P

Probably what is going on is that the LEDs are wired in series in the modules and the current flows thru several LEDs & they share that current, like water falling over a series of waterwheels. There is no more water (current) used but many times the work can be done (light produced). So if you have 3 LEDs in series you can get 3 times the light with the same current. HOPE THIS HELPS!!

Forgive me if the following is obvious:

The current specification printed on the back of a AC-to-DC adapter is a maximum value. The amount of current actually supplied by the adapter depends on what load it is connected to. For example if no load is connected the adapter, then the current is zero.

The way to observe how much current is actually flowing to the load is to put an ammeter (or a multimeter, set up in current measuring mode) in series with the load.

The current flowing through the load (in amperes), multiplied by the voltage across the load (in volts), equals the power dissipated by the load (in watts).

And that's how you measure electric power, for DC loads. Essentially, measure two numbers, then multiply them together. The math gets easier considering the voltage should be approximately 12 volts, independent of how much current is flowing. You know, just measure the current, then multiply by 12 V.

If you happen to have an AC power measuring gizmo, like the Kill-a-Watt(r) brand, or equivalent, then you can just plug that into the AC side, and read the numbers on it, and it will tell you how many watts are being consumed on the AC side. Then that number is an upper limit for the power consumed on the DC side (assuming the efficiency of the AC-to-DC adapter is <1)

If it turns out your LED modules do indeed draw less power than advertised, then it means the e-monger who sold them to you got it wrong. But from there I cannot say whether this error was due to stupidity or malice. For something like LED light fixtures, there might be an incentive to overestimate, erm lie, about the power dissipation. Although getting it wrong by a factor of 5 seems really careless, or really malicious, depending on which you think it is.

-max-2 years ago

I think to answer your question directly, I think there was a lot of margin in the ratings given. Either that, or the voltage you are powering them with is sagging, or lower than expected. Take a multimeter, measure the voltage before, and after connecting them. If the output is regulated, then the voltage should remain constant, however, if it is the old cluckin heavy Xformer, the output impedance may not be that low, esp. If it is a small Xformer, and the voltage will sag a lot relative to the current drawn, due to either the internal resistance of the thin wire on the primary/secondary or the core saturating and the coupling coefficent between the 2 coils.

iceng2 years ago

121 SMD = 120 ohms per each of five circular multi-LEDs with six leads or three LEDs in series with a 120 ohm resistor, assuming 3.3v/LED makes 12-9.9 => 2.1v for the resistor Ir = 2.1 / 120 =17.5 ma.... LED Power = 9.9 x 17.5 = .175 W.... Strip Power = .0175ma x 5 x 12v yields 1.05 watts

The o.25W each and 1.25W per strip may refer to Luminous Flux ie light intensity.

You seem to have found less strip current then my assumed 3.3v/LED...

---------------------------------------------------------------------------------------

Who knows what China diffuses into their dies ?

That's why the specs are so vague, we get LEDs meant to go in series at 97V

wired in parallel with a single killer resistor.

The Chines engineers are trained to always respond to a FAX which they do ... BUT only one question and out of context ... so you can never establish a discourse leading to a solution.

In a month's time everything will be different anyway.. :-/

CLEDV2.jpg
-max-2 years ago

Well just because something is rated for X amount of power, does not mean it will draw X amount of power.

powering lots of LEDs to their maximum is a bit difficult in some cases, because of the complicated relationship between voltage and current. Look up the IV curves of LEDs to see what I am talking about.

To make things really simple, we assume that LEDs should act like low impedance (short-circuit) loads when the voltage across them is higher than the forward voltage, and draw next to nothing (open circuit) when said voltage is lower than the forward voltage, and that if there is a finite current going through it, the voltage dropped across the diode is exactly the forward voltage. In most cases, thats fine. It makes the math easy.

To see how easy, assume 12V is our power supply, that there is some finite resistance we need to know to power the LEDs with a certain amount of current must flow, and at ANY finite amount of current, the voltage across the LED remains the same. Thus, it can be thought of like a voltage source connected backwards, so you literally just do ohms law, and subtract away the drop voltage of the LEDs from the power supply, and do ohms law like normal! V=IR, ( (Vpower_supply -- Vforward_voltage_drop = I*R).

verence2 years ago

Just guessing here...

The LED strips are designed for 12V. That's where they work best. A bit more voltage, they will be brighter but die earlier, a lot more and they will be very bright for a very short time. A bit less voltage and they will be a bit darker (but live longer). A lot less voltage and they will merely glow a bit.

With 5 or six modules, your power supply is overloaded. It will probably not deliver 12V any more but a bit less. Still enough to let the LEDs light up (just not as bright as nominal). Did you measure the voltage?

Vyger (author)  verence2 years ago

With no load the power adapter is putting out right at 12 volts. I have it hooked to a dimmer on/off module and after it runs through that it drops to 10 volts. reading at the end of the last module is 9.5 so it is dropping but very little. With out the dimmer in place, just powering direct from the adapter, the lights are brighter, the voltages stay the same as before, 9.5 at the end of the string. The adapter stays cool to the touch as do the lights but they are brighter.

I am thinking that the manufacturer made a mistake and that each unit is .25 watts and not each LED That would make a strip of 20 units to be 5 watts which is right were I am getting results. But why all the companies selling them would carry over the mistake is puzzling. Nobody noticed?