LED's resistors and batteries?

I have lit a few models in the past couple of years but I'm always relying on the guy in Maplins to sort out what I need.
Can anyone tell me if there is a site that I can visit that explains in simpleton terms, what size LED's need what size resistors and to what power batteries lol.

Thanks in advance

First off get the datasheet for the LED. On the datasheet find Vf and If, the forward voltage and forward current. Next decide on a battery voltage. For this example we will use a 9V PP3. Subtract the LED forward voltage from the 9V. Then divide the answer by the current you want. The luminosity is directly proportional to the current. Do not exceed the forward current on the datasheet, you will burnout the LED. A good place to start is 5mA, 0.005A.

Here is a fully worked example;

If = 15mA
Vf = 2V
Battery = 9V
Required current = 5mA

9V - 2V = 7V
7V ÷ 0.005A = 1400 Ohms or 1.4k

If you are using one resistor for each LED then you can use 1/4 watt resistors. I will post up later how to run strings of LED's and how to calculate the required resistor power rating.

This might be helpful.

Ralph

Just saying, driving a led with a 9v battery is the wrong way to go since you have to dissipate the unnecessary high voltage through a high value resistor and heat.
If you're not using a linear regulator, use batteries with voltage close to the Vf of the led and use a low value resistor, but even with a regulator you're only generating heat via resistance.

Use small LiPo cells for leds instead. They cost peanuts and comes with overcharge/overcurrent protection.

That aside, calculating the resistor value is as simple as Wizzzbang writes - or use any of the hundreds of online calculators. Picked this on random.

http://led.linear1.org/1led.wiz

Example: 3.3v, led with 2v Vf and forward current 20 mA gives you a 68 ohm resistor 1/4 w

If you're charlieplexing or building a series, or parallel, led layout you have to calculate a bit differently, but this is probably outside your scope right now.