Which one of these three solutions is the best for the best light intensity?
http://img329.imageshack.us/img329/5950/solucijenw2.jpg
More info:
source voltage - 12V
diode forward voltage - 3.7V
diode forward current - 20mA
number of LEDs - 23 (blue, 470nm)
Provided that the current through the LEDs is the same, all three
solutions will give the same light intensity.
The first one wastes most power - 2/3 of the power is dissipated in
the series resistor, and only 1/3 in the LEDs.
The third solution probably won't give a particularly equal
distribution of current through the LEDs - three nominally 3.7V LEDs
in series gives a total nominal voltage drop of 11.1V.
Since I don't know which blue LED you have in mind, I don't know the
tolerance on the nominally 3.7V forward voltage drop or the
temperature coefficient of the voltage drop, but I do know that
current through the different strings of three LEDs won't be
particulary close to 20mA, and the total light output will probably be
lower than you will get from the second and third solutions.
Hope this helps.
