# The complete guide to using LEDs

6 Steps

## Step 5: Update: Resistors.

Update:
Due to popular community request, I have added a section about resistors and ohm's law.
First, resistors. Resistors are small electronic devices that limit current and/or voltage. (hence the name resistor) In "power sorce" I put a big, fat X over the 9v battery. That was wrong-ish. You can use a 9v IF and only if you have a resistor. "But what resistor should I use?" is a qestion I can hear through you computers. You have to use "Ohm's Law". Ohm's Law is a mathematical formula used to find resistence, voltage, or amperage. (it can also determen wattage, but that is a little complicated.) It states: R=V/A where "R" is resistance, "V" is voltage, and "A" is amerage. Using that, you will be able to find the proper resistor to use with you LED. (You have to do the math on your own, that way you learn)

Remove these ads by Signing Up
saurabh_1603 says: Dec 3, 2011. 6:30 AM
can u tell how to light led using AC.
Do i have to use diode.
i used 100k resistance for a 230v AC supply, red led worked properly but white led stopped working after 5 seconds and was damaged.Can u tell me why.PLZ
karlpinturr says: Dec 24, 2010. 12:04 PM
Right..., so, taking your "R=V/A" to the circular chart, the resistance can be found by dividing I (volts) by E (amps), dividing P (watts) by the square of I (amps), OR dividing the square of E (volts) by P (watts)..?

And the same principal (ie, the central value can be found by ANY of the 3 methods shown) applies to the other segments?

So, if I know the voltage (E) of my power supply and the amps (I) of my LED/LED's, I divide the first by the second to find the necessary resistance..? And I assume that if there's no exact match available, I'd be best using the nearest HIGHER resistance?

But, just a minute. LED's are measured in MILLIamps - does that mean I have to divide/multiply the result of my calculations by 1,000? And how would I know which?
techturtle2 (author) in reply to karlpinturrDec 28, 2010. 1:27 PM
yes.
if you had a 9v power supply and a 30mA LED, you would divide as follows:
9/0.003=R=3000=3kOHMs