Help with a simple LED circuit

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LEDnewb
Posts: 1
Joined: Fri Aug 15, 2014 6:23 am

Help with a simple LED circuit

Post by LEDnewb » Fri Aug 15, 2014 6:40 am

Hi there everyone, I'm new to the forum and electronics and hopefully with your help i will learn loads and get my LED circuit working.

I am currently starting work on a costume based on star lord from the new guardians of the galaxy film, but i thought it best to try and get my head around the eyes first. If anyone has seen the film they will know his helmet has two rings of LEDs in the eyes and i would like to recreate that. I have brought myself some 3mm LEDs which apparently run at 1.7v - 2v and at 20mA, and i have resistors and battery packs. I have already discovered that i can only run a few LEDs from a battery so what i would like someone to help me figure out is how to run lots of them.

I would like to make TWO rings of LEDs with each ring being made up of 12 LEDs, what i cannot work out is what power source i will need to run those and which resistor to use. I think i wont be able to mount the power pack in the helmet so size hopefully wont be a problem, but the smaller in size i can get it the better. I feel like there is some real simple solution to this that i cannot see, whether it be running lower numbers in series then linking each set up in a series of small circuits. (i can imagine some of you are cringing at my terminology, sorry i really am a newbie at this)

Anyway any help would be massively appreciated, if i can get this figured out then i can defiantly build the rest of the helmet

Pauldf
Posts: 170
Joined: Mon Jun 10, 2013 7:42 pm

Re: Help with a simple LED circuit

Post by Pauldf » Fri Aug 15, 2014 3:48 pm

The easiest power source will probably be a normal PP3 9v battery. If the LEDs are identical you may run them in series of banks of 4, then parallel the 6 banks up to the power supply.
For a resistor for four 2v LEDs (8v) it is simply going to drop 1v at 20mA, hence as R=V/I, 1/0.02= 50 ohms or there about. For wattage I^2xR therefore 0.02 x 0.02 x 50= 0.02w so a quarter watt resistor is going to be plenty.

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