Led Sign - Noob Question

[Mogler]

Ok. So im somewhat new to electronics and Im trying to make a LED Sign out of 34 Blue 3.3v leds (i think there 20Ma). Im also using a 9v battery for this project. Iv been thing to figure out the whole resistor in ohms thing because i keep finding different posts that say different things about Ma or Voltage resistance. So what im trying to figure out is what kind of resistor im going to need for these 34 leds and if a 9v battery can even power all of these leds or if im going to need 2 9volt's.

Schematics - *not very good at drawing them lol.

-Way 1?


- Or Way 2?


Correct me if im wrong with this circuit but i just need to know what one will work and what resistor i will need for it.

[FOOKz™]

Here is a drawing of Two Circuits that you can Tackle your problem:



To Calculate your resistor value you need just use Ohm's law and a bit of algebra:

We know The supply voltage you have is 9 Volts
We know The LEDs require 3.3 Volts.

But we Need to know the Voltage drop across the LED if it doesn't use a resistor.
So to find this you just subtract:

Supply - vLED = Surplus Voltage
9v - 3.3v = 5.7v

The Voltage Dropped across the LED Never Changes because LEDs are Non-Linear Components. Resistors are Linear Devices and drop a proportional percentage of voltage upon the supply.

OK now we have to calculate the percentage ratio of 3.3volts to 5.7volts.
(smaller number on top)
3.3v / 5.7v = 0.579 (57.9%)


The resistor must drop 57.9% of the voltage.
So just multiply the percentage to the supply...
0.579 * 9.0v = 5.21v

============================================

Time to use Ohm's Law:
V = I * R

Plug the numbers in the equation:
R = V / I
Resistance = Voltage To be Dropped / Current Through the Resistor
Resistance = 5.21 / .02A
Resistance = 260.5 Ohms

One Last Step. I promise.   :

Finding Power Consumption:
Ohm's Law Power = I * V
Each LED Consumes 20mA of current.
You want to use 34 LEDs in paralell.
0.02 = 20mA
So. 34 * 0.02A = 0.68A
Current total = 680mA

Voltage Supply = 9v
Plug it in the equation:
P = I * V
P = .68 * 9
Power = 6.12 Watts

That's One Huge Resistor!!!!!

Normally Resistors come in 1Watt 1/2Watt and 1/4Watt Packages.

You must put 1 Resistor for each LED individually since the Power is so great; if you don't then you run the risk of smoking the Resistor possibly starting a fire.

I would Round this value down a bit no more than 50 Ohms.
I  recommend 1/4 Watt 220 Ohm Resistors for your circuit. 1 Resistor for each LED.

=========================================

** Remember, current isn't dropped across a resistor its dropped through a resistor. When resistors are put in series together they use the same current but drop different voltages, when resistors are paralleled they use different current and drop the same voltage.(assuming they are all the same value)

======================================

Extra Info:

Resistors are Linear Components since their voltage drops' change as the voltage changes; resistors drop a percentage of the total supply.

The LED will drop 3.3volts in a 9volt circuit and it will drop 3.3volts in a 12volt circuit; a resistor will drop a proportional percentage of the voltage in that circuit.

[Mogler]

THANKS! So much man it helped alot. haha i finally understand it  !

[FOOKz™]

Hey No problem, anytime. What exactly are you using this for? I'm curious!

EDIT: You should stick around here more we need more people handy with electronics. If you understand math like that for the first time then you're going to be good at designing circuits after a bit of practice.

Hey it really pays off... I have a real good job right now at a company called Cumberland Electronics as an on-call Electronics Technician and i make $15.67/hr but still i'm still in high school. Quite a bit of people rely too much on PICs and that's hard to get new people's attention to build something on their own. I like to make the old school circuits because the parts are always available.

[Mogler]

Im making a LED Sign for my father sense his bday is coming up and i don't got any money for stuff. and i don't know what to get him so i thought he would like it haha. also iv been trying to figure out the whole ohms law thing because im hooking up 100leds to my desktop for a sound reactive mod. i already have my case light kit (2 florescent tubes) hooked up to one and the front leds on the case but they already had resistors on them so it was easy.

Edit:
Yea ill stick around iv been checking out this site for awhile just looking at posts and stuff. when i get into collage i want to get a degree in electrical engineering so im hoping learning this stuff now will help me in the future lol. and its fun :P i enjoy making project and stuff like this. ha

[FOOKz™]

Lol I have a similar setup with a few White High-Intensity Surface Mount LEDs that made up my Desktop Light. It Uses an old Router 6 Volt Power supply 5 LEDs and 5 1/2 Watt 12Ohm Resistors to power the LEDs Behind my computer screen to make it look really cool in the dark.

Its so bright it doesn't make a difference.  =D

In the Light:


In the Dark:


Behind it:
         

[Mogler]

Thats sweet im eventually going to get to that. Im putting like another 30+ 3mm Leds in my computer case. and im putting some around my stereo and desk area. Ill send u some pics when its done. waiting on leds from china lol.

My desktop right now.
Sound Sensitive Light Tube Mod For the PC

[pigoesfar]

LOL, just for the purpose of finding one, I found the resistor you would need for the parallel setup.  6.5W and 270 Ohms!  IDK how hot it will get but here it is.
http://search.digikey.com/scripts/DkSearch/dksus.dll?Detail&name=CWD-270RCT-ND

[FOOKz™]

Work ethic FTW

[pigoesfar]

Digikey is the best place to find like any part.  It has so many things.  But it is not the best for learning about parts or ordering like one.  You have to know what you are looking for.

BTW +1 for that detailed explanation.  That just saved that guy like 3hrs in internet research.

[FOOKz™]

But it was like 1 hour of typing that whole thing and trying to remember it off the top of my head.