On October 10, 2017 at 08:19, buzz said...
Parallel is not a viable plan unless the the two emitters (diodes) have absolutely identical characteristics (there is virtually no chance of this fortunate accident occurring in the field).
EXACTLY, except since LEDs are made in batches, with all in the batch having values very close to one another and packaging happening in batches, it's quite possible for them to turn on at the same voltage. The problem is that you can't plan on it, so a method should be used that solves the problem before you even see it.
Since they are not identical, one emitter will turn on first and cap the voltage below the other emitter's turn-on voltage -- exactly duplicating your results.
Yes, EXACTLY.
The solution is to wire the emitters in series. Don't carry this too far, two or three will usually work, sometimes four. The output of each emitter is lower, but this is typically not an issue.
This will probably work if the IR system uses a 12V supply, but I've seen them using 5V supplies. Two LEDs in series off of five volts gives you limited output current. Three probably simply won't work.
So it's good that there's another solution, which is exactly how IR distribution blocks are made: wire them in parallel, but add a resistor of, say, 100 ohms in series with each LED. (In some power amplifier circuitry where transistors are more or less in parallel, such resistors are called sharing resistors.)
Think of what happens when you have two LEDs in parallel that don't start to conduct at PRECISELY the same voltage, and you apply an IR emitter voltage that increases from zero.
A crucial point is that an "IR emitter voltage" is a signal with resistance in series with it.
At some voltage, one of the LEDs turns on. A further increase in the applied voltage will only make the LED conduct more current. The voltage across the LED won't change* and the other LED won't turn on because that voltage isn't high enough to make that one start to conduct.
If you have a resistor in series with each LED, once the first one turns on, the voltage at that LED won't get any higher, but the voltage applied to its series resistor will get higher... this voltage is also applied to the second LED through its series resistor, and when that voltage gets high enough, the second LED will also turn on. The fact that the two LEDs don't turn on and off at exactly the same voltage won't matter because the added resistors have forced the voltage and current to be shared between the two LEDs. This is why those resistors are called sharing resistors.
*In the real world, this is not exactly true, but it's okay to assume it's true as the real voltage change is minimal.
