I want to wire 10 of these LEDs into a circuit with a 9-V battery. The site states that the forward voltage is 3.2-3.8 V at 20 mA current. Should I assume 20 mA is the max amount of current that can flow throw this circuit?
Using V = RI, I calculate that I need a resistor that is 450 Ω (9/0.02). Is this correct? Secondly, if I want to have 10 LEDs here do I need a higher voltage (38 V)?
This is the first LED circuit that I am making, so I'm not familiar with most of the basics.
Answer
If you put all these LEDs in series you'll indeed need 38V, plus a bit for the series resistor. But you can make a circuit with several branches. Put 2 LEDs in series and you'll need 7.6V. So the remaining 9V - 7.6V = 1.4V is the voltage over your resistor. If you want 20mA through your circuit You divide this 1.4V / 20mA = 70 ohm, so a standard 68 ohm will do nicely.
Now that's for 2 LEDs, you could go for more if you had a higher voltage. the way to calculate the resistor is the same. If you want 10 LEDs from 9V, in theory you could put the circuit 5 times in parallel. That would be a total current of 100mA though (5 x 20mA), and that's a bit much for a 9V battery.
edit
It's been suggested that you could replace the 5 resistors by a single one, and branch from beneath this. In an ideal world that would be true; the resistor value would then be 1.4V / 100mA = 14 ohm. But this isn't an ideal world, and there may be small differences in LED voltages. In that case the branch with the lowest voltage will draw most of the current (100mA!) while the other LEDs will hardly light at all.
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