the first light i need to repair is my 18v Black and Decker
5 LEDs in series is 3.6v each so i only need to limit the current correct? the formula i found is confusing unless it means that I dont need a resistor since the voltage is equal to the requirement of the 5 LED chain..?
does the formula discussed here
apply ony to configurations that have excess supply voltage? those woule be for using less LEDs probably to consume less space or battery..?
would using less LEDs and more resistance save battery life?
"newsdude" wrote in news: snipped-for-privacy@g10g2000cwb.googlegroups.com:
LED's (and laser diodes, too) are highly dependent on current for their safe output level, so you MUST limit that current. If the voltage is steady, you can get away with a resistor, as in this case. Therefore you must keep some headroom, or that resistance will be too small to control easily, and one moment you might have darkness due to battery undervoltage, the next, a burned out resistor if the voltage rises across its tiny resistance burning it out with the current pushed through it.
The simple answer in this case is use four LED's in series with one resistor, as the remaining 3 volts will give you the headroom you'll need.
Measure your battery on no-load one hour after standing by from a fresh charge, use that as the voltage for the standard formula, Rx = (Vin-Vf)/If where Rx is your resistance needed, Vf is total forward voltage of the LED's in series, and If is the forward current wanted.
If you want more light, make sure each Vf in parallel has its own series resistor, or the lowest Vf will steal current from the other(s) by reducing the available voltage to below the higher Vf for all other parallel diode(s). If you're making one series network, ignore this paragraph.
As to less LED and more resistance, don't, it saves nothing, it just wastes energy in heat dissipated by the resistance. :)
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