Music LED Light Box

Step 5 Preparing the circuit

Since I have no experience in electronics whatsoever, this was the toughest part for me to find out.
Other instructables had no good explaining on this subject.

So for everyone who's new to this electronics stuff and have no idea what they're doing, here is some explaining for the circuit.

1. How many LED's to use?
The average LED needs 3v to work. If you are going to put the LED's in series (like I did) you need to calculate how many LED's you can use with your adapter. The formula you can use is adapter output voltage / LED voltage = Total LED's

So if you use a 12v adapter with the 3v LED's its: 12 / 3 = 4 LED's

You can also put LED's parallel in the circuit. But I'm not getting into this, just so that this instructable stays simple and easy to understand for everyone. Let's just focus now on the LED's in series.
If you want to experience with more LED's, you can always calculate your LED's and resistors here.

2. The real voltage on a adapter.
Before I went building this circuit, I thought it would be useful to measure the real voltage on the adapter. The sticker on the adapter (photo 1) says the output 12v. But once I hooked it up on my multimeter, it shows that the actual output is around the 18v (photo 2).

So that means I can calculate the LED's again: 18 / 3 = 6 LED's.

Since I'm going to make the LED's in series I can use 6 LED's in my circuit. 

3. The 3.5 audio jack plug
Which wire is what? That's what I was asking and trying to find out.
As you can see on photo 3, the plug itself has 3 metal parts, and 3 wires in the black protective layer.
On the photo I explained which wire is what.


Now with this information you can go to the next step, building the circuit.

Comments

[yaly]

the adaptor says 12v 200ma that means if it is loaded with 200ma it gives out 12v and unloaded is about 19v so see how much current per led for 3v and calculate using this formula:(N)=(unloaded)/(load per led 3v).
I think it is right can anyone correct me if I'm wrong.

[beehard44]

Unregulated wall warts use transformers and follow what you said. That's why a regulated source would be better

[Kruegem]

Are Skull Candy ear phones 3.5 jacks?

[Moofish]

I think that any ear/headphones that fit into an iPod or an mp3 player are 3.5 jacks.

[Saturn V]

3.5 ear phone jacks can plug into basically anything audio. If they can plug into a standard stereo system/computer/iPod/MP3/DSi/PSP, than they are 3.5 mm jacks.


Short answer: Yes. 

[Bluedove2897]

It would usually be 3.5 if it's not USB

[Bluedove2897]

I'm pretty sure!

[gabrielguy]

so 20 leds would be 60 volts!?D:

[Anthrix]

in one line yes in bridge no :D

[Master T]

So lets say i wanted to make a box for left and right... i would just split the ground wire from the audio jack plug and do everything exactly the same just using the left wire for one box and the right wire for the other box right?

[roofy99]

 When I stripped my headphone cable it didn't look anything like the ones in your photo. I used an old pair of apple ipod headphones. They had 3-4 seperate strands but they weren't like wires like yours are. Am I using the wrong type of headphones, or what should i do to fix it?

Thank You

[TyBoyTwist]

same thing! have u figured it out?

[beehard44]

melt the enamel with a small flame 'till they are gold colored then solder as normal

[vocaxil]

my 3.5 audio jack plug juz hv 2 wire..which should i solder....on each wire it have blue and red

[Salsa766]

I have a 12 AC wall adaptor but I only have the jack and not the thing that goes on top of it. Is it ok to just solder wires to the positive and negative terminals like how you hooked it up to the multimeter?

[squirrelMLM]

When I stripped open my sony headphones, there were two copper wires (like your ground wire) and a green wire and a red wire. What do I have to do differently? Also, another question. If I plug this into my iPod Touch, it won't play the music while the LED Light Box is glowing, will it? Great instructable BTW!

[BlackHatCracker]

Love this project. I did this as a light organ. Some of the "elders" might know what  I am saying when I use that word, from back in the '70's.. lol.. But I used x-mas lights, 3 different strings, and 3 scr's, and a  terminal for the speaker wire, a switch, and a small transformer to step down the voltage. After I got the one done I had to build a second one for the right channel... lol.. Loved it! Now I am going to have to do this again, but with LEDs. Great paper trail, details, keep up the great work!

[yellow thumb]

when you have 18v coming out of the power supply, does this mean you would have to use six or could you use less than six. i tried something and shorted everything out. and, if you cant use less than six, how could you put a ressistor in?

[13r1an]

A typical LED has a 2.1V drop across it.  Applying more voltage may burn it out.  Another important note is that an LED is still just a diode, which means current flows freely in one direction.  Therefore, doing a string of them with no resistor is basically a short circuit.  It's good practice to *always* use a resistor in series, to dissipate the current in heat form.  In this case, that means you'll need to break it into 2 or 3 strings of 2 or 3 LEDs per string with appropriate resistor.

By using 6 LEDs in series though, you kinda over corrected yourself so the circuit isn't dangerous, per se, but it isn't exactly something that would please the electrical gurus out there, if you know what I mean.

[kadris3]

red, yellow, and orange(amber) have a forward voltage of 2.1 vdc at 20 ma. white, blue, green and UV have a forward drop of 3.4 vdc at 20 ma.
w 3.4v leds- 3 in series and 180 ohms across 12vdc works well.
for 2.1v LEDs use 5 in series and 180 ohms across 12vdc.
for mix and match use ohms law. 13.8 (supply) or 12.7 battery minus drop of the led string divided by .02(ma)= value in ohms. r,g,b,= 2.1+3.4+3.4=8.9.
12.7v -8.9v= 3.8v divided by .02 =190 ohms. 200 or 220 is next highest fixed resistor.  good luck with your projects.

[manouttatime]

At the risk of sounding dumb, how exactly do you find the appropriate resistor?  I'm going to try a dual circuit, Left and Right version of this so how do I find the resistance needed on each of them?

[13r1an]

IR = V is your best friend.  Ohm's law.

First step, find out the stats on ur LED.  That should be available from the supplier.  You're looking for the current capacity, specifically.  Or, if you're feeling lazy, just assume it's 20mA, which is pretty standard for a LED.  Then, figure out your voltage supply.  In this particular instructable (and 99% of cars), your supply is 12V.  So, plug in your knowns into oh-me-oh-my's-law so you get
(0.020A)*(?R?)=(12V), use some complicated differential equations or calculus algorithms to solve (no, really just divide 12 by .02) to get R=600ohm.

What you just solved is that if you are going to put a single LED rated at 20mA in a circuit with a 12V supply, you must put a 600ohm resistor in series in order to prevent blowing out the LED.  You have a bit of a fudgefactor here, so dont' worry if you only have a 500-700ohm resistor handy.  Just be aware that if you go too low on resistance (or if you supply voltage fluctuates considerably higher, although unlikely), that you run the possibility of blowing your LED.  Generally, it's a better idea to just accept a *slightly* dimmer LED (with a larger resistor value) than risk it popping and then going through the hassle to replace it.

Your other option, if your feeling TOTALLY lazy, is google "LED resistance calculator" or something of that nature.  There are a ton of sites out there with little website calculators for that purpose.  But, IMHO, I think that takes the joy out of building things if you just cheat every step and don't understand the concepts behind why what you're building is working.

[manouttatime]

Since I am running 2 branches of leds from the power supply, do I use a resistor on the branch, or on the main power lead?

I'm actually planning a switch in mine so could I possibly run a resistor from the  switch to the power supply?

I believe that the total amperage will be 180mA (6 red 20-30mA LEDs) since it is 20-30, should I assume it will draw the max while figuring resistance? If I use 180mA this will need 66Ω of resistance in this case, will a 100Ω resistor work, or would I be better buying some new 75Ω resistors?

And actually, for 6 LEDs I would be drawing 13.2V from a 12V supply at the max for the LEDs. Will this be a problem if both banks max out at once?

Thanks

[danystatic]

The resistor should be place on each branch, consider it a best practice. I wouldn´t recommend placing it on the switch or anywhere else in order to have a well designed circuit.

100ohm vs 75ohm??  Consider it like this....
live 100 boring years
or
live 75 happy lightened up years  ??

And I understood your 4th question, I have to add that it´s always a problem messing with electronics when we´ve got little experience, but making a few Ohm´s excercises will get you on shape. 
a. If both banks(you mean branch?) max out at once??

[manouttatime]

Nevermind that last question.

Thanks for the info about the resistors. I understand Ohm's Law equations.I'm just wondering, do you still use 12V when you have 2 branches from the power supply or would it be split through the branches, and do you just figure the amperage draw for that specific branch or the total draw?

If it's total draw at 12V, I need about 75Ω, if it's branch draw at 12V, I need 130Ω I have some 100Ω (They're cheap so I don't care if I need new ones, just need to know what's needed)

[Jerb]

 13r1an has done an excellent and well thought out explanation of what's going on here. this page needs to be updated to reflect the use of a current limiting resistor or other means of controlling the current to our LED's so readers dont burn out their LED's or burn themselves on hot wiring.

cliff notes:

a current limiting resistor should always be used to dissapate heat and stop the circuit from shorting out. 
1/8 watt resistors will be fine for most string of leds 3-4 long

the power supply in this case is cheap so there will be voltage drift. that is to say that as you add a load to this power supply it will become more efficient and the V will drop. if you put your multimeter leads across it it may read higher than when you attach a circuit to it and measure the V again