Sound good, but you also want to be thinking about efficiency, too.
Most linear LED circuits have three components: the power source, a series current limiting device (CLD), and the load (i.e. LEDs). The CLD is usually a resistor or a LM317 configured as a current regulator. The efficiency of the circuit is primarily determined by the power dissipated by the CLD. The less power that is dissipated by the CLD, the more efficient the entire circuit is.
The power dissipated by the CLD is I[sub]CLD[/sub]*V[sub]CLD[/sub], where I[sub]CLD[/sub] is the current through the CLD (which is equal to the current through the LED), and V[sub]CLD[/sub] is the voltage across the CLD. To minimize the power dissipated by the CLD, you need to make I[sub]CLD[/sub] small, V[sub]CLD[/sub] small, or I[sub]CLD[/sub] and V[sub]CLD[/sub] small.
Well, you can’t do much about I[sub]CLD[/sub]. After all, that’s the current through the LED, and it is specified right up front. But you do have control over V[sub]CLD[/sub]. The lower you can make V[sub]CLD[/sub], the better.
The easiest way to make V[sub]CLD[/sub] small is to make the voltage of the power source (V[sub]PS[/sub]) just slightly higher than the nominal voltage across the LED (V[sub]LED[/sub]). And then use a resistor as your CLD. (If the voltage across the CLD is less than 3 V, you can’t use an LM317 as a CLD. If it is higher than 3 V, you can use a LM317 as the CLD, and it is the preferred method.)
Let’s look at some examples:
Let’s say your LED has a maximum operating current of 20 mA, and a nominal voltage drop of around 3.4 V when operated at max current. To make the circuit efficient, you make the voltage of the power source V[sub]PS[/sub] = 4 V. This means you’ll use a 30 Ω resistor as a CLD. Using these values, 85% of the power will go to the LED, and 15% will go to the CLD. Not bad.
But instead of using a power source of 4 V, let’s say you use a power source of 6 V (V[sub]PS[/sub] = 6 V). This means you’ll use a 130 Ω resistor as a CLD. Using these values, 57% of the power will go to the LED, and 43% will go to the CLD. This is not nearly as efficient as using a 4 V power source.
It obviously gets a lot worse when you make the power source voltage even higher. Say you use a power source of 9 V (V[sub]PS[/sub] = 9 V). This means you’ll use a 280 Ω resistor as a CLD. Using these values, only 38% of the power will go to the LED, and 62% will go to the CLD. This is sucks. (Also note that you can use a LM317 instead of a resistor. While the current regulation will be a lot better, the efficiency will not be any better.)
Now having said all of that, minimizing the voltage drop across the CLD may not be the most important thing on earth:
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While minimizing the voltage drop across the resistor improves efficiency, it decreases its ability to properly regulate current if V[sub]PS[/sub] is not well regulated.
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For most applications, maximum efficiency is only important for battery operated devices. And besides, if efficiency were very critical, the designer would opt for a PWM circuit to drive the LED.