Re: Brighter LED options?

Windell Oskay

So that LED can handle up to 350 mA, running at about 3.5 V.  If you’re running from a car, that probably means 12 V power. To run the Larson Scanner PCB from that you would need to use a voltage regulator, perhaps a 7805 to reduce the 12 V to 5 V.

For the LEDs, you will need 9 transistors, each wired up like those in the first illustration (“Transistor switch”) here:
Each transistor requires there to be two resistors: R1 (the “base resistor”) that goes between your microcontroller output and the base of the transistor, and R2 (the “load resistor”) that goes between the 12 V power supply and the LED, attached to the collector of the transistor. 

 A good, cheap, and tough transistor that you could use would be the TIP31– about $5 for 10 of them. 

 With the current that you’ll be using, that transistor can give a gain factor out of about 100. So, to get 350 mA of collector current through the transistor, you will need about 35 mA out of the microcontroller outputs. If you’re driving the Larson Scanner at 5 V, and each of its outputs goes to the base resistor, then you’ll get about 4.5 V out of the microcontroller outputs. The base-emitter connection of the transistor takes about 7 V, leaving about 3.8 V. To get 35 mA at 3.8 V, you need the base resistor to be about 100 ohms.

Now for the load resistor. With 12 V in, 3.5 V taken by the LED, and only about 0.1 V taken by the TIP31 collector-emitter voltage, that leaves about 8.4 V. To drop 8.4 V at 350 mA, you’ll need a resistor of about 24 ohms. No problem there. However, its power dissipation is a concern: P = I * V, or 350 mA * 8.4 V = 2.94 W of power to dissipate. I’d recommend using at least a 5 W resistor for each of these.