Sure.
The fastest way to blow the
LED is with too much
current. I looked up the linked spec sheet and it listed the maximum as 30ma. It also indicated the forward
drop voltage was 3.3 volts. We know that's not enough
voltage to light 3 in series, but would be sufficient to light two. A 9 volt battery will produce a lot more than 30ma, so we need to throttle, or
ballast it a
bit. Since the LEDs will be dropping 6.6 volts, we need to
drop 2.4 volts while limiting the
current to below 30ma. Using
Ohm's law, we calculate 2400mv / 24ma and we come up with 100 ohms. Since the
current was in thousandths of an amp, the
voltage needs to be changed to thousandths of a volt, thus 2.4 volts = 2400mv. So basically, a 100
ohm resistor will pass 24ma if it has 2.4 volts across it. I could have used an odd resistance value to get it to 30ma like this - 2400/30 and come up with 80 ohms, but that would operate the
LED at exactly 30ma which is its maximum value, but there would be no
safety margin. Conversely, we see a 120
ohm resistor would run them at 20ma, which is safer yet, but as the value goes up, the
LED will be
dimmer. So, the + side of the battery goes to the 100
ohm resistor, then to + on the first
LED, the - on the first
LED goes to the + on the second
LED, the - on the second
LED goes back to the - side of the battery. (Unless you are from the vacuum tube error, but that's a who other story
)
