Electronics help...

Im going to use This LED (http://www.radioshack.com/product.asp?catalog%5Fname=CTLG&category%5Fname=CTLG%5F011%5F006%5F002%5F000&product%5Fid=276%2D316) for the lighting and want to use either a 9 volt battery or some other source for power. What is the best set up someone has used before and can explain what type of resistor I would need.

Im going to use This LED (http://www.radioshack.com/product.asp?catalog%5Fname=CTLG&category%5Fname=CTLG%5F011%5F006%5F002%5F000&product%5Fid=276%2D316) for the lighting and want to use either a 9 volt battery or some other source for power. What is the best set up someone has used before and can explain what type of resistor I would need.

Well I'll try and give you a quick electronics 101 course.

First thing is to study the specifications included on the web page you supplied. If you look closely there are three important items which are

1. If mA MAX 20 or in other words the current it draws is 20 milli Amps.

2. The Vf 3.7V typ or in other words it has voltage drop of 3.7 volts.

3. The Vf 4.5V Max or in other words the maximum voltage it can stand is 4.5 volts. Exceed this and you risk destroying it.

Knowing the above info we can now calculate what size resistor is needed to operate the LED on 9 volts. To do this we use a formula called Ohm's Law which in it's normal form is V=IR where V equals volts, I equals current in Amps and R equals resistance in Ohms.

Consequently we know the supply voltage is 9 volts and we know that the typical voltage drop of the LED is 3.7 volts. Therefore the amount of voltage that has to be dropped across the resistor is 9 minus 3.7 which is 5.3 volts. Knowing that the current is 20mA or 0.020 Amps we can now add them to Ohm's law as follows

5.3 = 0.020 X R or

R = 5.3 / 0.020 which becomes

R = 265 Ohms

You won't be able to buy a 265 Ohm resistor however you will be able to buy a 270 ohm resistor which is close enough. Place it in series with one of the legs of the LED (remember that the LED is polarity conscious) and you should be fine. You can calculate also the smallest resistor that you could use by utilising the maximum voltage spec from the LED. This may give a very slighly brighter light but will shorten the life of the LED. It is up to you.

I hope this helps.

either that or if your gonna be running 2 which i assume you are! then why not run them in series! i.e connected together. that way they will draw 9 volts together as in series you add up the voltage as you go along.

2x4.5v = 9v! no need for a resistor

What do you mean..in a series?
Like one after another?

What do you mean..in a series?
Like one after another?

The term "in series" in electrical jargon is as follows

Connect the positive terminal of the battery to the positive leg of the first LED. Connect the negative leg of the first LED to the positive leg of the second LED. Connect the negative leg of the second LED to the negative terminal of the battery. Hope this helps