LED strips: 46% resistive losses?

I recently bought these LED strips. They are supposed to run on 12V/72W and have 300 x the SMD 5050 LEDs on it. Schematic:

simulate this circuit – Schematic created using CircuitLab The voltage values on the schematic have been measured with a multimeter. "560" is written on the resistors.

As you can see, the resistors account for 5.6V - that makes 46% of the voltage (or power) consumed by the strip!! Is that normal?

• They're not built to be efficient, they're built to be cheap. Commented Oct 29, 2014 at 13:24
• Inexpensive Chinese-made products solving everyday engineering problems in the cheapest way possible? Yes, that's normal. Commented Oct 29, 2014 at 13:51
• Not sure your schematic is accurate. 5050 leds are 3 diode packages. You should sacrifice a segment to figure out the layout. Commented Oct 29, 2014 at 14:04
• Also these strips were designed with automotive 12V in mind, which is between 10 to 14V typically. So just get a lower voltage supply... Commented Oct 29, 2014 at 14:56
• @Michael the LEDs are efficient, the circuit on the other hand is not... Commented Oct 29, 2014 at 15:48

This is normal enough. This is done because

• People are being lazy, cutting corners and cost and reusing strip backing which is also used for white LEDs.

• Because they can

• You are 1000's of km away, they do not expect repeat business, they do not care what you think.

These strips allow 3 LEDs in series to be connected across the supply. when white or blue LEDs are used Vf (forward or operating voltage) per LED is typically in the 3.0 to 3.5 Volt range or 9 to 10.5V typical. When operated on 12 Volts they thus dissipate 9V/12v tp 10.5V / 12V or 75% to 85%+ in the LEDs. when Red LEDs are used Vf is about 2.0 - 2.2vper LED or 6V to 6.6V total resulting in the result that you see. If they cared they could make a design that uses 4 LEDs in series (8V - 8.8V) or just possibly 5 in series (10V-11V) although the latter has too little head-room voltage dropped acrtoss the resistors.

If you care enough about the excess dissipation and if the resistors are accessible you could sort out one series resistor per 3 series LEDs and operate then from around 10V or short out two resistors and run then from about 8V. This is quite a lot of work and unlikelt to be attractive except where power use if of vital importance.

Interest only: If dissipation at 12V is 72 Watt then dissipation when modfied to run on 8V would be abpt 48 Watt - or 24 Watt less. If run "24/7 for a whole year (8765 hours) and if electricity costs 25 cents per unit then after one year of continuous use you would have saved about $50 in power costs!. [ 8765 hours x 24 W/*1000 W/kW) x$0.25 = $52.59 Adjust figure for actual hours/day and unit cost of energy. If you ran these continually for 8765 hours it's likely that their light output would be much reduced. The reasons for those are the same as those at the top of this answer. Added: As Olin says, they will be 56 Ohm resistors. I = V/R = 2.8/56 = 50 mA. As far as I can tell from the image the LEDs use 3 die in parallel (each brought out on two pins) and then a resistor connects to the next LED in a set of 3. Each die is probably rated at 20 mA max so 60 mA for 3 - so running at 50 mA is slightly conservative which is good. This is a very unusual design as only one series resistor is strictly necessary per set of 3 LEDs. Thy may have done this so each resistor dissipates 140 mW instead of having one only but needing it to dissipate 420 mW. This MAY be OK but it is likely that th ADDED: Efficiency, seller integrity, ... Note: I might buy these if I needed them if the price was right. They probably represent typical offerings. The following is re what they claim - not why you shouldn't buy them :-) Claims: • In the ad they claim A++ energy efficiency. • In the ad they claim "Save power more than 90% by ordinary bulbs". Reality: Use 5 lumen/Watt for now for "ordinary bulbs". To achieve their claimed 90%+ less they'd need 50 l/W LEDs at 100% of input energy used by LEDs. Factor in the fact that only about 53% of the input energy is used by the LEDs and you'd need 50/0.53 = 94 l/W. A thin clear polycarbonate cover gives 10% light loss. So their waterproof covering made of ??? is liable to cause 20%+ loss 20% loss means LEDs need to be 94/0.8 = 117 l/W. (Or 104 l/W at 10% loss). Apparently identical LEDs on Alibaba advertise "15 l/W (24 l/W available"). This may allow for the cover but not for the resistors. Even if it did allow for the resistor loss as well the effective lumen's Watt in the Alibaba case mentioned would fall FAR short of the levels needed to meet the advertised claims. This is not untypical for this class of product. Even allowing for filtering losses, assumptions of l/W of incandescent bulbs etc (1) There is no way they meet their 90%+ claim (2) The A++ claim would be actionable legally in my country (NZ) by a government department at no cost to the complainant. The 50,000 hour claim is garbage. Ask me how I know! :-) The price is an utter ripoff, but blaim the reseller for that. I bought Chinese made 5m reels of LEDS in India retail in 3 reel lots for about$US6/reel (no remote control). Remote adds minimally. Astoundingly good.
Out of China they cost more (than Chinese made LEDs bought in India) but not vastly so. BUT if you are buying in 1's on ebay and that's the best value available and it suits you then buy them. Just don't believe anything they say :-).

The 300 LEDs in the heading and 150 in the body is (probably) due to a 12V strip heading being used for the 24v description. The 72W does not match your calcs = probably about 90W if Volt figures correct.

LED Z (efficiency - usually in l/W (lumen/Watt)) depends on colour, % of rated output run at, temperature, MAKER, ... & more. Use white as reference as these are common. Nowadays at full power a Z of 90-110 l/W is goodish. Best reasonably available tend to be 120 to maybe 140. Run LEDs below full power and Z rises maybe 10% to 20% at very low I. Run cold it is maybe 10% up. Very very best top bin LEDs at say 30% power at 25C (BIG heatsink+) are around 200 lumen/Watt. Lab samples now exceed 300 l/W.

For low power high efficiency LEDs I use small NSPWR70CSS-K1 that have been out for ?4? years now. 125 l/W at 70 mA. 165 l/W at 30 mA. Superb.

• What you called lazy id call effecient and resourceful and pound-wise. Commented Oct 29, 2014 at 14:58
• @Passerby Yes- many people would call a fraction of normal efficiency "efficient" if it meant they could thereby make a sale, call taking advantage of people "resourceful", and making increased profits at someone else's expense "pound wise" - but I try not to deal with such if I can help it, try not to recommend such to others, tend to highlight such anti-social behaviour when I come across it and generally deplore the impact that such shonky behaviour has on the ability of people to provide a good product at a fair price. Commented Oct 29, 2014 at 15:12
• +1 for shonky - oxforddictionaries.com/us/definition/american_english/… - had to look that one up (O; Commented Oct 29, 2014 at 15:51
• Effecient because they didn't waste time overengineering something simple. Resourceful because they reused existing material instead of letting it go to waste. Pound wise because time is money and wasting it recreating the wheel is a waste of money. then again you sound like a mechanic that gets mad that another mechanic provides a free or 5 dollar code check when you want to charge a person 50 dollars to plug a cable in and read off the screen. Who's the antisocial one? Commented Oct 29, 2014 at 20:12
• @Passerby Are you a lawyer? If I thought that you genuinely thought that any of your 3 arguments had other than trolling merit I'd be somewhat concerned. Number three is closest to reality except that you'd need to swap what each of the exemplars was offering to get somewhat close to my point. As it is your example grossly misrepresents and in fact inverts what I was saying. ie my objection was to fraudulent behaviour, lack of design and people fitting the wrong wheel and overcharging for it. But, hopefully for you, you know that and are just 'having fun'. Commented Oct 30, 2014 at 6:51

First "560" on a resistor means 56 Ω, not literally 560 Ω. This is essentially a floating point format. The last digit is the exponent of 10 to apply to the integer value formed by the previous digits. "560" therefore means 56 x 100 = 56.

Yes, this LED array is quite inefficient. If that's a primary concern, don't use it. It seems the design was optimized for cost. I expect it was cheap, so you got what you paid for.

• Uses same "PCB" as 3 x white LEDs per unit. Commented Oct 29, 2014 at 13:38
• To be fair, the strip is ineffecient due to the specific led color. Other leds with a higher Vf have 80% or higher effeciency. Commented Oct 29, 2014 at 14:48
• @Passerby You sound like an apologist for Asian (and other) shonky business and electronic "design" practices. "To be fair" is saying what I have said - that they have taken a strip designed for white or blue and used it in an inappopriate application because they can, because they don't care and because there are no consequences. Commented Oct 29, 2014 at 15:07
• @russell and you sound bitter, if not a bit racist and xenophobic. Commented Oct 30, 2014 at 11:45
• @Passerby Better than sounding like a lawyer :-). You may as well add "sound like a Nazi", invoke Godwin, and be done. To add some perspective: I'd hope I'd sound like someone who'd visited China 15 times in about 6 years and spent ~= 6 months there (about 1/2 that running around with a camera looking at everything and enjoying the country immensely). I love the country, and I have and had no problems with the overwhelming majority of the people there. I now have some good friends there and continue to do business there. AND/BUT: if people do not appreciate that what I said reflects the ... Commented Oct 31, 2014 at 2:04

560 on an SMD resistor is typically 56Ω - that's a five, a six, and zero noughts.

So if you have 12V, three LEDs, and 112Ω of resistance, and assuming 2.1V voltage drop across an LED, the sums work out as:

$V_{FT} = 2.1*3 = 6.3V$

$V_{RES} = 12 - 6.3 = 5.7V$

$I = V/R = 5.7 / 112 = 51mA$

Blue LEDs have a higher forward voltage, normally - say 3.2V. Using the same components that would work out as:

$V_{FT} = 3.2*3 = 9.6V$

$V_{RES} = 12 - 9.6 = 2.4V$

$I = V/R = 2.4 / 112 = 21mA$

Blue LEDs are visually brighter than red ones, so less current through blue LEDs is acceptable (though not necessarily "nice").

By using the same components throughout they keep costs down. Three blue LEDs is the limit for a 12V supply, so you are stuck with 3 LEDs on the red version (same PCB), and keeping the same resistors, while not "ideal", is fairly normal. Yes, it means the resistors drop more voltage on the red one than the blue one, and the current is higher, but that's a small price to pay for having very cheap LED strips.

• Rgb 5050 strips have different values for red resistor than they do the green and blue resistors Commented Oct 29, 2014 at 14:33
• Also these can come in 24v versions with more leds per segment. Commented Oct 29, 2014 at 14:36
• It's a shame though that they are not CHEAP LED strips. Such is life. Commented Oct 31, 2014 at 12:53

Solder in bypasses on the first resistor on each series, and replace the second resistors with a 20R, and make sure you use a switching power supply and keep the voltage under 7V - this should dramatically increase efficiency. You can pick up a reel of surface mount 20R resistors, and set if you create a custom soldering head to "stamp" them in it should be fairly efficient to get the job done.

Another way I have done it is to buy bulk flat LED lights and wire them all up myself.

Ideally each the LED's should be wired in groups of each colour with a set resistor for each colour, or even better, each colour is run on its own complete circuit (allows for colour mixing).