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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 2V and an operating current of 20mA. (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 20mA@2V, so it would be a 100Ohm resistor, with 0.8mW 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?

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8  
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. 16 hours ago
10  
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 16 hours ago
3  
Look up the wikipedia article 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 16 hours ago
6  
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 16 hours ago
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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. – FakeMoustache 16 hours ago
up vote 17 down vote

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% efficient in that respect however that is still far better than the ~1% 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.

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up vote 16 down vote

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.

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5  
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 15 hours ago

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