So bearing in mind the aim of my original question and all the excellent informative replies, it sounds as if all one really needs to do for a quick and dirty solution is connect the LED to the intended voltage with some high value series resistance and gradually reduce this till the LED is bright enough to be visible.
Hi,
That's a sure fire way to do it every time
As others have pointed out, you can calculate a rough estimate for the resistor based on the power supply voltage, but there are times when this might not work very well and that is when the power supply voltage is comparable to the LED voltage. The best bet then is to assume the LED voltage is the lowest possible.
For example, if we take the lowest LED voltage to be 1.5v and the highest to be 3.6v, we can do two calculations knowing these two voltages (and set current of 10ma):
R1=(Vs-1.5)/0.010
and
R2=(Vs-3.6)/0.010
So lets say the power supply voltage Vs is 20v, we get two values:
R1=(20-1.5)/0.010=1850 ohms
R2=(20-3.6)/0.010=1640 ohms
We can see that the value of the resistor will always be between 1640 and 1850 ohms. If we start with the higher value, we should be ok, then we can check the current and decrease if needed.
But lets say the voltage is only 10v, we get two values again:
R1=(10-1.5)/0.010=850 ohms
R2=(10-3.6)/0.010=640 ohms
and now we see that it would be a good idea to start with 850 ohms.
But what if the voltage is only 5v, we get two again:
R1=(5-1.5)/0.010=350 ohms
R2=(5-3.6)/0.010=140 ohms
Here we see that we should start with 350 ohms, because if we start with 140 ohms we'll get a current that is too high if we happen to have an LED with a forward voltage near 1.5v:
I=(5-1.5)/140=0.025 amps.
but even that wont blow out the LED if it is rated for 20ma.
When the voltage gets REALLY close to the LED voltage that's a different story as with 4v:
R1=(4-1.5)/0.010=250 ohms
R2=(4-3.6)/0.010=40 ohms
From this we can see that a good starting value is (Vs-1.5)/I where Vs is the power supply voltage and I is the desired current.
Also worth mentioning is in a battery application we might want to calculate the current with a new battery and with a battery that is run down a bit so we know how much brightness we will loose as the battery runs down over time.