##

Its really hard to get diodes to track in parallel without ballast resistors. If you have 12V around, go series. No sense wasting energy making heat in resistors.

If you only have 5V, or only have one Lasorb, you need to go parallel, and you need to make sure that you use a pair of ballast resistors, and make sure one diode is not hogging much more current then the other. The voltage across the series resistor is proportional to current, so you can at least see whats going on. 1-2 ohms should be a start.

Your really concerned with the "headroom" available to the driver, if a diode has a 2.4 volt drop, and you have two in series, that is 4.8V volts, leaving only .2 volts with a 5V supply, which is not enough headroom in most cases.

I'm not familiar with the LOCs forward voltage, so that was a made up example.

Each ohm of series resistance burns up 1 volt of headroom, so at 250 mA, 1 ohm in series burns 250 mV of headroom.

Since P=IE, or P = I squared R, you can calculate how many watts of resistor capacity you need.

Example, 250 mA diode current, 1 ohm ballast, (.250 x .250) * 1 = .0625 watts * 1.5 safety factor = 0.9 watts, so choose a 1 watt resistor.

Take the number you calculate and multiply by 1.5, so if you need a 1 watt resistor in the math, multiply that by 1.5, and the nearest standard resistor is a two watt, so that is what you buy or build. By build, wattage in parallel adds, so two one watt resistors in parallel is a two watt resistor.

Steve

Last edited by mixedgas; 09-29-2010 at 16:53.

**Qui habet Christos, habet Vitam!**

I should have rented the space under my name for advertising.

When I still could have...