George

Its not that hard, just a bit of understanding, planning and maths.

You need to know the specs of the LED's, there voltage and current ratings and what configuration your going to use.

lets take a simple example, you want a high beam indicator.

You have a "Blue" LED with specs of 4 volts and max current rating of 500milimamps.

You are going to feed it of a 12 volt power source. thats 3 times the voltage you need, so how do you get just 4 volts?

You have to use the rest of the power somewhere else, in a dropping resistor.

Using the Ohm's law formula V = I x R where V is volts, I is current in amps and R is resitence and the specs of the LED we can work out what size resistor we need.

See

http://en.wikipedia.org/wiki/Ohm's_law for more info

For this simple cct the resistor is connected with the LED in a line from the +ve to the -ve (this is called a series cct), the maximum rate of current in the circuit is defined/controlled by the max amperage of the LED.

So for V we need 12 - 4 = 8.

We know the max amps as 500 milliamps

So we know 2 of the 3 values and transposing the formula to R = V divided by I, plugging the values in to the forumal we get

R = 8 / 500mA

R = 8 / .0005 = 16000

So the resistor you need is a 16000 ohms or 16Kohms.

Now to expand on this cct if we want to add another LED we can do it either in series or in parrallel.

The difference between a series and parrellel cct is that in series they are linked one AFTER another and in Parrallel they are linked one BESIDE another.

Series circuits will have a voltage drop per LED, a parrallel cct will share the same voltage drop.

A series circuit will have the +ve side of a LED feeding the -ve side of the next LED (one AFTER another) if one LED fails the whole cct is broken

A parrallel cct will have the -ve & +ve sides feed from the same point. (one BESIDE the other) if one LED failes the rest keep working

Series cct with 2 LEDS

Voltage for LEDS 4V X 2 = 8 votage for droping resistor = 12 - 8 = 4

R = 4 / 500mA

R = 4 / .0005 = 8000

Series cct with 3 LEDS

Voltage for LEDS 4V X 3 = 12 votage for droping resistor = 12 - 12 = 0

R = 0 / 500mA

R = 0 / .0005 = 0

So you could run 3 LEDs without any resistors

For the Parrallel LEDs cct

You will still need a droping resistor and only the Max current has changed

So for V we need 12 - 4 = 8.

We know the max amps is now 500 milliamps * 2 = 1000 mA

R = 8 / 1000mA

R = 8 / .001 = 8000

Note that the car voltage is a nominal 12 volts and could go up to 14.7 V so this would need to be accounted for.

In the situation where you run all LEDs and no resister an extra lead could be added to allow for the 14.7 volts and protect the LEDs from over voltage.

This is a basic description and more things will need to be taken into consideration but this gives a general outline of how it works.

Warren