D
**DONOTDELETE**
Guest
For a white LS with four AA bats.
Thanks
Thanks
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Jonathan
Enlightened
You can use a very simple dropping resistor circuit. This won't be regulated, meaning that the current (and brightness) will drop as the battery voltage drops, and the efficiency when the battery is fresh will be poor.
The circuit is just the battery in series with a resistor, (optional) a switch, and the LED. You determine the LED voltage drop from a datasheet (about 3.3V for a white Luxeon). Subtract this from your design battery voltage (6V for alkaline AAs, 5V for NiMH AAs), giving the voltage that the resistor has to drop.
The voltage drop in a resistor is just the resistance in ohms times the current in amps. You want to run the Luxeon somewhere between 0.2 and 0.4A, depending upon how good your heat sinking is and how bright you want the light to be. Say you want to run at 0.2A and you are using an alkaline pack. The resistor needs to drop 2.7V, so 2.4V/0.2A = about 14 ohms. You would need a 13 or 15 ohm resistor.
The power dissipated in the resistor is current times voltage, or 0.2A* 2.7V for this example, or about 1/2 watt. Use at least a 1 watt resistor.
-Jon
The circuit is just the battery in series with a resistor, (optional) a switch, and the LED. You determine the LED voltage drop from a datasheet (about 3.3V for a white Luxeon). Subtract this from your design battery voltage (6V for alkaline AAs, 5V for NiMH AAs), giving the voltage that the resistor has to drop.
The voltage drop in a resistor is just the resistance in ohms times the current in amps. You want to run the Luxeon somewhere between 0.2 and 0.4A, depending upon how good your heat sinking is and how bright you want the light to be. Say you want to run at 0.2A and you are using an alkaline pack. The resistor needs to drop 2.7V, so 2.4V/0.2A = about 14 ohms. You would need a 13 or 15 ohm resistor.
The power dissipated in the resistor is current times voltage, or 0.2A* 2.7V for this example, or about 1/2 watt. Use at least a 1 watt resistor.
-Jon
D
**DONOTDELETE**
Guest
thanks
, for the heat sink should I use a small peice of 3/16 aluminum or could I use some .020 guage aluminum sheet metal? Or would copper work better?
JoeyL
Enlightened
I would suggest using 3 AA's if possible. That way you have less voltage to drop, the resistor is a lower value, and less heat is dissipated. I've generally used about 5 ohms which gives a current of about 270 ma. When it gets dim, I short the resistors for about another hour of brighter light. This involves fooling around with the innards of the light, however.
Happy tinkering!
Happy tinkering!
