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I've always found circuits containing LEDs hard to understand, please bear with me. I know most people find it easy, but I'm confused by them so some of my assumptions might not be correct, please correct me if that's the case.

So onto the question: Since LEDs are, after all, diodes, they essentially act as conductors with forward voltage, right? Which is why we need a pull-down resistor to regulate the current that flows through the circuit.

For example, let's say we have an LED with a Vf of 2 V and an operating current of 20 mA. (I think those numbers are reasonable right? Again, if not, please let me know.) And our power supply is a constant 4V. This means we need the resistor to draw 20 mA at 2 V, so it would be a 100 Ω resistor, with 40 mW going through it. That's a tiny power usage, but half of the power supplied is wasted through heat. So in this case, isn't the best case efficiency 50%? Which isn't really efficient in terms of DC power supplies, I would have thought.

So when people refer to LED's high efficiency, are they referring to the fact that the LEDs themselves convert the power they use into light efficiently, or is it considered efficient even after considering the 50% max wall plug efficiency?

Or is it just that I've given an example that happens to be a horrible circuit design that would never be found in production applications?

Travis
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kumowoon1025
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    *we need a pull-down resistor*? No. we don't. We need a *series* resistor. High efficiency means that the ratio of the output power to the input power is high. Output of LED is light. Input is electrical power. That's it. – Eugene Sh. Jan 25 '17 at 15:21
  • Perhaps the efficiency-problem is with the overvoltage/overcurrent rather than the LED technology itself? – user95482301 Jan 25 '17 at 15:22
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    They refer to the incandescent bulb as a reference. Even 5% efficiency is really a big deal, if the alternative you are replacing things with is 1% efficient. – PlasmaHH Jan 25 '17 at 15:24
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    Look up the wikipedia article https://en.wikipedia.org/wiki/Luminous_efficacy. Also, forget about resistors and look up a constant current power supply. No resistor but a current feedback loop instead. – winny Jan 25 '17 at 15:28
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    The LED itself is a more efficient converter of electrical energy into light than an incandescent bulb. Commercial LED 'bulbs' don't use simple resistors to power their LEDs, they use switch-mode power supplies which don't convert all of the 'dropped voltage' into heat. – brhans Jan 25 '17 at 15:29
  • @PlasmaHH: From one point of view, tungsten-filament lamps are horribly inefficient, and even fluorescent lights don't seem too great. On the other hand, I can't think of anything prior to the 21st Century which was particularly efficient at converting *any* sort of energy to light. – supercat Jan 25 '17 at 15:33
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    And some cheap-ass LED bulbs (yes, from China) that do not use a proper switched-mode converter use a capacitive dropper circuit which drops the voltage in a more sneaky way but also more efficiently than using only a resistor. These still provide more lumens per Watt than an oldfashioned lightbulb. – Bimpelrekkie Jan 25 '17 at 15:34
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    @supercat: Then you have rarely been outside in the night. Everything is yellow there, because sodium lamps are about as efficient as LEDs. – PlasmaHH Jan 25 '17 at 15:36
  • 20mA@2V = 40 mW – Andy aka Jan 25 '17 at 15:37
  • Just for reference 2 V and 20 mA are reasonable. The lowest you'll normally see is ~1.6 V for red LEDs, yellow and green are normally a little higher but not much. Blue, white and "true" green are more, normally just over 3 V. For small LEDs 20 mA is normally the design maximum with 10 mA being a more normal working level. High power ones will take more current. – Andrew Jan 25 '17 at 15:39
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    @PlasmaHH: I meant to make two points and sorta munged them: (1) other means of lighting like lamps and candles are even less efficient; (2) 21st LEDs were the first technology that would be suitable for use many of the places incandescent lamps were used. – supercat Jan 25 '17 at 17:03
  • Tangentially, there is another efficiency: they have a longer life than incandescent bulbs so there is less cost involved in replacing them. E.g. I repaired a VCR once by replacing a failed internal bulb used as part of a position sensor. – Andrew Morton Jan 25 '17 at 17:37
  • @EugeneSh. Err... to be honest I'm not sure what a pull-down resistor is precisely lol. I just thought from learning about transistors that they were just a way to bridge a floating voltage point to ground (or 0VDC). I understand the resistor being wired in series part, but could you explain what exactly a pulldown resistor refers to? – kumowoon1025 Jan 26 '17 at 09:58
  • @Andyaka Well thats embarrassing haha thanks for catching that. I think what happened was I did I^2R but used 2 as the resistance instead of the voltage. – kumowoon1025 Jan 26 '17 at 10:03
  • @winny How does that work with semiconductor type components? Wouldn't the effective resistance pretty much flip flop every time the voltage adjusted above/below the Vf to keep the current constant and make the LED flicker? Or do they not behave exactly the same as regular diodes? – kumowoon1025 Jan 26 '17 at 10:08
  • So thanks everyone, I think I understand. Efficiency refers to efficiency of just the LED, from anode to cathode / there are more efficient power supplies / it's efficient in the context of optoelectronics is what I got. Btw, when responding to multiple comments, is it better to do it in separate ones for each response, or merge them into one comment? – kumowoon1025 Jan 26 '17 at 10:16
  • Nothing flip-flops unless you design it that way. Just like you have a closed loop controlling the output voltage of a power supply by modulating the PWM duty-cycle you can do the same for a current loop. Moreover, the Vf varies with current which makes it even easier with a soft/bad/slow control loop but there is no inherent problem in regulating current around a "tight" Vf of a diode/LED. – winny Jan 26 '17 at 10:23
  • The efficiency refers to the driver too. You can have it as efficient as you can afford. You will find manu commercial ones in deep 80 % territory. – winny Jan 26 '17 at 15:57
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    Re, "...a way to bridge a floating voltage point to ground..." You might want to learn some basic theory before you stat throwing around words and phrases that you don't understand. The first couple of chapters of this book would be a good read: https://en.wikipedia.org/wiki/The_Art_of_Electronics – Solomon Slow Jan 26 '17 at 17:19
  • @jameslarge sorry... I thought that was the term to use with an open circuit with a resistance thats not linear? could you help me say what I'm trying to in a way thats accurate? Also that book's a little expensive... Is there a site where I can get a preview of it or something? – kumowoon1025 Jan 28 '17 at 00:44
  • All I can say is, after twenty years of tinkering with electronics, reading books about it, and even taking university-level courses; that book really opened my eyes. It's kind of an odd-ball. It's not meant for serious electrical engineering students, nor is it a hobbyist's cook book. It's meant for scientists who need to design their own simple circuits for monitoring & controlling experiments, who need to really understand how things work, but who don't want to waste time learning any _more_ than what they need to know. It's clear, it's concise, and the authors really "get it". – Solomon Slow Jan 28 '17 at 21:33

3 Answers3

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You seem to be getting confused between the efficiency of the LED and the efficiency of the circuit to drive the LED.

In terms of light output per unit of energy used by the LED they are an efficient way to generate light. In absolute terms they aren't great, they are around 10%[1] efficient in that respect however that is still far better than the ~1-2% of a conventional incandescent bulb.

But what of that power wasted in the resistor. A series resistor is the simplest way to drive an LED, it is far from the only way to do so.

Even sticking to a resistor what if we put 20 of your 2V LEDs in series and supply it with 45V? Now you are using 45*0.02 = 900mW of which 800mW is going into the LEDs and only 100mW (11%) is being used by the series resistor.

But we can make it even more efficient, the reason for the resistor is that the LEDs needs a constant current and most electronics are designed to supply a constant voltage. The easiest way to convert from one to the other (assuming a constant load) is to throw in a series resistor.

You can get constant current power supplies. If you use one of those to drive your LED then the resistor can be eliminated and you can get an efficiency of well over 90% of your total system power going into the LEDs.

For a home project or a simple indicator on a signal a resistor is a lot cheaper and simpler but if you are driving a lot of LEDs then the logical choice is to pay a bit more, have a slightly more complex circuit and use a dedicated constant current LED driver IC.

  1. As noted in comments, 10% is a good ballpark for current household lighting and probably also about correct for cheap commodity LEDs using older processes. Newer single colour parts can achieve significantly higher levels of efficiency.
Andrew
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    Somehow I thought modern, non-phosphored (single color) LED's were around 25 to 35% efficient. Can you add a link for that 10% power efficiency estimate? Great answer btw! – uhoh Jan 26 '17 at 10:11
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    The 10% number was a rough approximation from memory rather than a hard number. A quick look here https://en.wikipedia.org/wiki/Luminous_efficacy#Lighting_efficiency would indicate that for household lighting 10% is a reasonable number for the final product. Which would imply that a single colour high efficiency LED could well be in the ~30% region if you look only at the LED and not the supporting electronics. I'll update the answer. – Andrew Jan 26 '17 at 10:52
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    Top bin of best on-market white phosphor LEDS achieves >50% of energy input leaving as light. Above 200 lumen/Watt output. – Russell McMahon Feb 01 '17 at 14:22
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    @RussellMcMahon Do you have a source for that? Wikipedia would indicate that 100% efficiency is around 680 lm/W meaning that 200 lm/W is around 30% not > 50%. It is however well over 50% of the theoretical maximum efficiency. While wikipedia is hardly a definitive source it's better than nothing. – Andrew Feb 01 '17 at 14:35
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    @Andrew - see my answer. Basically - lumen rating is human eye rtesponse weighted. 683 l/w max is ONLY at 555 nm. – Russell McMahon Feb 02 '17 at 06:54
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The efficiency of a LED refers to how efficient the LED is. This has nothing to do with how efficient or not the driving circuit is.

In many cases, the overall circuit efficiency of LEDs is not much of a issue. If the LED is just being used as a indicator, it is low power in the first place. A typical green LED drops 2.1 V and is plenty bright enough for indicator use at 20 mA. That's 42 mW of power going into the LED. Even if a additional 50 mW is lost in the circuit driving the LED, the total power consumption is still inconsequential in many cases.

In some low power applications, 100 mW can be a major amount of power. In such cases, more care will be taken in the circuit other than the cheap and simple series resistor to some handy supply. Various tricks include using a higher efficiency LED and running it at lower current, using a supply that is only a little above the LED voltage, adjusting the user interface so that blinking or otherwise keeping the LED off part of the time is acceptable, and a high-efficiency constant current power supply to drive the LED.

Efficiency also matters in high power applications, such as lighting. In such cases, more effort and production cost is put into the electronics to minimize additional power dissipated outside the LED. Often the main reason for maximizing efficiency isn't so much not wasting the power as not having to deal with the heat caused by the wasted power.

Olin Lathrop
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    Also, a lot of lighting applications are having the efficiency of LEDs compared to the efficiency of incandescent and CFL bulbs. LEDs could still be said to be high-efficiency even if they weren't particularly, if they were nonetheless more efficient than the alternatives. Which I believe is the case by quite a large margin (particularly so in the case of incandescents. – KRyan Jan 25 '17 at 16:39
  • They aren't very much more efficient than CFL bulbs. However, the light quality is much better. IMO too much focus is on the efficiency of light production and too little on the quality (spectrum, CRI) of the light. The effect of poor light quality is to cause home-owners to opt for quantity where quality is lacking. I search out LED bulbs with a CRI >= 95, which are almost as good as halogen-incandescent. These LEDs are slightly less "efficient" than ones rated CRI >= 80, but I find that the much better light quality means I'm happy with less watts of lighting. – nigel222 Jan 26 '17 at 12:28
  • BTW, on the CRI front, high CRI led lights also cost quite a lot more than cheapo CRI>=80 lights. IMO they are worth every penny. – nigel222 Jan 26 '17 at 12:31
  • The efficiency of a lighting LED is its system efficiency, usually expressed in Lumens per Watt. It doesn't matter whether the waste heat is coming from the driver or the LED chip or the CRI-correction filter in front of it. – nigel222 Jan 26 '17 at 14:59
  • Data point only: Top bin of best on-market white phosphor LEDS achieves >50% of energy input leaving as light. Above 200 lumen/Watt output. – Russell McMahon Feb 01 '17 at 14:23
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The question re how efficient actual LEDs are is a good one, but the answer is more complex than may be expected. Illumination capability is usually expressed in "lumens".

LED efficiency is usually expressed in terms of either

  • light energy output or

  • illumination capability

per unit of energy input.

For a given lumen output, efficiency is usually expressed in lumens per Watt (l/W) or in light energy output per Watt W/W). The first figure is more useful in practical illumination applicatios, but the second is more meaningful in terms or energy conversion efficiency.

If lumens and light energy had a fixed relationship then efficiency determination would be simple. However, What a given lumen figure represents in terms of "light energy" varies with the spectral composition of the light.

Lumens are expressed in terms of the theoretical response curve of the human eye. The same amount of light energy will produce a different number of lumens as light wavelength or mix of wavelengths varies. As a consequence, the wavelength or wavelengths of the source plays an important part in the lumens produced per energy input.

At the short wavelength end of the visible spectrum (not quite UV) eye sensitivity is extremely low, so lumens/Watt are low - so much so that it is usual to quote the output of deep blue and "Royal Blue" sources in terms of mW/W (light energy per electrical energy). This is highly useful as an LED family which includes phosphor-less and phosphor based LEDs allows some comparisons to be made. For example, the "top flux bin" of the Cree Royal Blue XT-E LED when operated at Vf= 2.85V and If = 350 mA produces 613 mW typical (600, 613, 625 mW min/typ/max) at a wavelength of 465 Nm.
That equals an electrical to light conversion efficiency of 60.2% / 61.5% / 62.7% min/typ/max.
See page 19 of Cree XT-E datasheet top right of table - XTEARY-00-0000- 000000Q01

The top white phosphor version of the same LED produce 180 lumen at 25C at 2.77V, 350 mA = 970 mW DC in or 186 lumens/Watt.

IF the light energy from the Royal Blue & White LEDS was the same then the white LED would have a 100% l/W figure of 186/61.5% = 302 l/W at 100% efficiency. However, light outputs are not identical (quite) as in the white LED a portion of the LED die's blue light is used directly and the remainder excites the phosphor(s) with some loss in light/light conversion efficiency.

As has been noted, Wikipedia (correctly) states that the maximum theoretical l/W figure is 683 l/W.
How can this be reconciled to the claim that 100% white LED efficincy is ~= 300 l/W - and the fact that various manufacturer's are now making white LEDs with efficiencies > 300 l/W?

The answer lies in the useful but arcane (or arcane but useful) fact that the lumen rating is related to eye response. Maximum eye sensitivity occurs at a wavelength of 555 nm. The maximum possible efficiency in l/W is achievable with a monochromatic source at 555 nm. ANY other source, monochromatic or multi wavelength, will have a lower 100% theoretical possible l/W figure.

The "ideal" white light source is a black body radiator at 5800k with its spectrum truncated to the 400-700 nm range and has a max efficiency of 251 l/W !!!!

By making various adjustments to maintain "white" light while altering the % of various wavelengths increasing white efficiencies can be achieved. A 2800k black body truncated asymmetrically to achieve a CRI of 95 has a 370 l/W max theoretical efficiency.

But wait - there's more, but, later maybe.
I'll come back and add sources and more detail, but the above shows that the answer is harder than the question, and demonstrates that in true energy out per energy in terms the top modern LEDs achieve energy conversion efficiencies of > 50%.

More anon - light fades - rootop job beckons ...

References WIP

https://en.wikipedia.org/wiki/Luminous_efficacy

Analysis on the Luminous Efficiency of Phosphor-Conversion White Light-Emitting Diode

https://en.wikipedia.org/wiki/Light-emitting_diode#Efficiency_and_operational_parameters

http://www.hi-led.eu/wp-content/themes/hiled/pdf/led_energy_efficiency.pdf

http://www.philips.com/consumerfiles/newscenter/main/design/resources/pdf/Inside-Innovation-Backgrounder-Lumens-per-Watt.pdf

2014 http://www.forbes.com/sites/peterdetwiler/2014/03/27/leds-will-get-even-more-efficient-cree-passes-300-lumens-per-watt/#258823b870b4

http://www.cree.com/News-and-Events/Cree-News/Press-Releases/2014/March/300LPW-LED-barrier

Useful:

http://boards.straightdope.com/sdmb/showthread.php?t=719499

Russell McMahon
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