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edit - much better way:
Rework your circuit so you only have a single resistor R, so its battery in series with R then LED in parallel with each other but in series with R.
Now to provide your LEDs with 3.6V and 20mA. Since they are in parallel they will all be at the same voltage. And since LEDs are all identical, we can assume the current will split itself equally amongst. So 20mA each for 6 LED is 120mA in total. So now we need to find R such that when there is 2.4V across it (since the LEDs have a voltage drop of 3.6V), it delivers 120mA to the LEDs:
2.4/120mA = 20 Ohms.
Power = I^2R = (120mA)^2 * 20 = 0.288 - you might get away with a 1/4W. If you used 2 x 40Ohm in parallel you definitly can as the power in each resistor is only (60mA)^2*40 = 0.144W.
Oh, you won't actually find a 40Ohm resistor. The closest value is 39Ohms.
Your existing calculations are however, fine.
Cheers,
Steve
Last edited by freespace; 19-05-2007 at 02:31 AM.
Reason: Better idea.
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