Step 5: Power

Picture of Power
Power Supply
Your power supply can be from pretty much anything. An old PSU from a computer, a leftover hard drive power brick; I prefer small and compact power supplies such as those in cordless phones. Laptop power supplies are an amazing power source, they are super cheap and available, usually can supply 3-5 amps (which means a max of about 230 LEDs), and are regulated [which are normally super expensive] at 12v DC. No matter what the plug on the end is, you can bet there are two wires in there: Positive and Negative.

I obtained three 9 volt power supplies at my local goodwill for $3. eBay also has a plentiful selection of power supplies, though it will take some searching. The supply needs 350mA or more to power 18 LEDs. mA determines the maximum amount of LEDs. 99% of 5mm LEDs use 20mA each, so just multiply the number of LEDs by 0.020A (18 LEDs * 0.020A = 360mA, which is technically overloading it, but it still works). 

The speaker wire will carry the electricity from the power supply wires to the LED wires. Decide right now which of your two speaker wires will be your positive, and which will be your negative.

As a general rule, the red, yellow, white, or lined wire is the positive, while the solid (usually black) is negative. After the next step, you will have two LEDs and a resistor soldered together. Just try touching them both to the speaker wire, whichever lights up for you is your correct solution.

Quick Disconnects
This step is optional. It makes it easy to plug and unplug the light bar from the power supply. If you don't want to use Quick Disconnects and want the power supply always attached, just twist the wires of the speaker wire and the power supply together and solder them. Wrap them with electrical tape to keep anything from shorting out.

Quick Disconnects have a wire going into them, and usually you just crush it with pliers (Called crimping) and the wire stays in place. An optional purchase to skip this awl & solder method is a pair of crimping pliers. Regular pliers were incapable of doing the job for me, so I pierced and soldered them into place. Knowing from experience now, I strongly recommend just buying crimping pliers than this next awl + solder method. If you don't feel like spending money, then go for this method, which actually is a stronger connection than crimping alone.

Because its not possible to stick solder into the Quick Disconnect (its covered in plastic) and melt it and the wire together, you should coat the end of your wire with solder. Stick the wire into the quick disconnect. Now hit the Awl on top of the Quick Disconnect, this will pierce the casing and squish the metal and wire together. Follow it up with sticking the tip of your soldering iron into the hole to melt the solder. You should have a very solid connection between the wire and Quick Disconnect. Finally, wrap any exposed areas up with electrical tape.
did u mean dat 2500 mAH is capacity of each battery?
if so then totally it will be 5000 mah but just as the capacity is lessened,it shud still remain 2500 mAH,right?
QuackMasterDan (author)  mayur.phadte4 years ago
I was correct in my statement, and yes, your statement is also correct. Adding two AA, 1.5v, 2,500mAh batteries in parallel will keep the voltage at 1.5v but add their capacities to 5,000mAh. Capacity effectively is mAh. mAh stands for milliamp hours, as a way of describing drain over time. Capacity can also refer to current, which is simply amps, A.

I threw together a quick picture in photoshop to illustrate how it works.
But I was taught in kindergarten that Series=voltage divider in ratio of resistances
Parallel =current divider in ratio of resistances
wat about that?
I am not pointing out anything wrong but trying to learn something.
not arguing but trying to turn this into an informative debate.
Sorry If I hurt you in any way
QuackMasterDan (author)  mayur.phadte4 years ago
Your are correct, but note that those statements apply to the wiring of resistors together. Demand from the LEDs (in terms of volts and amps) remains the same, no matter how much power you provide them, they will take as much as they can use. The LEDs conduct through one another in series, so yes, with each LED in the chain a bit of power is lost through light being emitted. This becomes a problem when you go beyond three LEDs, where the fourth can be noticeably lower than the first. Within just two LEDs, the loss of current from LED to LED is low enough to not be noticeable.

As far as series being a voltage divider, and parallel being a current divider, yes, that is accurate in both terms of supplying power from a power source (generally that refers to batteries), and the wiring of resistors. It works both ways, not simply reducing as in division. If you add power sources (batteries) in series, the voltage *multiplies* (or you could say, divides by a ratio below 1:1) If you add power sources in parallel, the capacity for amperage *adds* while voltage remains the same. This is very useful when trying to get extended battery life from your power sources. Here is an example.

Let's say you have a Luxeon LED Star that runs at 3.0v and 700mA. You have two AA batteries, each which puts out 1.5v and has a capacity of 2,500mAh (milliamp hours). If you were to add the two batteries in series, you would double the voltage at the cost of halving the capacity. So you would get 3.0v but now only 1,250mAh. If you have four AA batteries, you can have the the benefits of both, at the cost of greater physical size. If you have any more questions, feel free to ask.
I dnt really agree with 18*20=360ma thing
With 12 V DC u connect 4 LEDs in series,right?
so they share the same 20mA
So technically every 4 LEDs share 20mA.
SO u could run about 18*4=76 LEDs on that thing.
Correct me if I am wrong
QuackMasterDan (author)  mayur.phadte4 years ago
Power drain does not work like that. For one, the LEDs are in series of two, not four, which makes a large difference on the amount of stress put on the resistor. Second, the resistor chosen with the ledcalc calculator has been calculated to allow two LEDs in series to receive 20mA each, so in reality, the resistors are allowing 40mA through, in which each LED consumes 1/2 of the power.

The hard limit for how many LEDs are allowed is calculated in watts. To calculate watts, it's simply amps times volts. So if your power supply pumps out 350mA at 12v, you have 0.35A * 12v = a 4.2W power supply.

Let's count the drain for one LED, a green 5mm running at 3.2v and 20mA. So 0.02A * 3.2v = 0.064W. Now, let's divide the drain of an LED into the capacity of the power supply, which is 4.2W / 0.064W = 65 LEDs. 65 is the ideal limit, in reality, due to heat and resistance and other inefficiencies, power supplies need a bit of buffer room, so let's just cut off a few LEDs, and say the realistic limit is 60 LEDs.

I hope that makes sense, if you have any more questions, feel free to ask.
brigs1644 years ago

i was wondering if a 12v 1amp dc plug would work if im hooking up 5 bars together
QuackMasterDan (author)  brigs1644 years ago
To measure maximum power possible by your power supply, you need to measure everything in Watts. A Watt, is simply volts * amps.

So if you have 18 LEDs on your light bar, with each LED running at 20mA (milliamps) and 3.2 volts.
18 LEDs * 20mA = 360 mA.

Then let's turn that into Watts
3.2v * 0.36A = 4.32 W.

For five light bars
4.32 W * 5 Light Bars = 21.6 W
Realistically, we want some buffer power since real electrical drain isn't perfect, let's just add 5 Watts to be safe. Thus, 26.6 Watts is our power drain.

Now for what your power supply can put out,
12v * 1 Amp = 12 Watts.

Sadly, that power supply cannot provide enough juice to five light bars. It will realistically be able to light three at full brightness, and at three, it will become extremely hot (like, burn your finger hot, which is bad). So the safe amount for that power supply is two light bars.

If you want a cheap power supply that can handle a ton of light bars, go find a laptop power supply, They are usually in the range of 50-75 Watts, and can be found for a couple dollars at a Goodwill, or bought on eBay for $13 including s/h.

Good luck.
samsonation5 years ago
 I was wondering is it possible to make these with batteries to make these more portable should the need arise?

QuackMasterDan (author)  samsonation5 years ago
 Definitely. If you're going to use battery power, try to adjust your source voltage more towards the lower end like 4-6 volts. You could achieve 6 volts with 4 AA or AAA batteries. For a connector, I would suggest purchasing some 9 volt battery connectors (one octagon and one circle plugs with wires). Just solder one plug to your battery pack, and the other to the light bar. Your light bar can be as long as you choose.

Measurement of capacity is in milliamp hours mAH. A standard one-time use AA battery holds 2,500 mAH. So if a single LED runs at 20mA, you could run one LED for 2,500/20= 125 hours. That's very idealistic though, since the LED would start to get pretty dim since battery voltage drops off with capacity. If you want longer capacity battery packs, consider wiring multiple packs in parallel, or getting bigger batteries (like C or D). If you're going to use it a lot, consider rechargeable, eBay has great deals on rechargeables compared to brick and mortar stores.
I'm curious as to what QD's are needed. Do specifically need male or female? Also does the gauge range matter? Would 14-16 gauge QD's not work?
I no longer recommend using this method of connecting power actually, as quick disconnects are more fragile and more of a hassle than DC cylindrical plug connectors. Just search 2.1mm x 5.5mm x 9.5mm DC plug on google, these things are a real pain to get a hold of for a good price, despite them being in practically every electronic device. I suggest eBay for the easiest to find deals, or RadioShack for a low stock and overpriced supply.

Anyways back to QDs, you will need two male plugs and two female to make one connection. The males go on the power supply, the females on the light bar (or the other way, it's your choice really). The gauge range is more for how much power you are expecting, a 14-16 gauge plug is for much larger power amounts than are involved here, I would recommend a 18-20 or 20-22.
j03tv5 years ago
I'm curious.
In the first pic you note that 18x20mA does not equal 360mA, it in fact equals 144mA.
From Ohm's Law, your spec's of using a 150Ohm resistor in series with two LED's in series, your current would be 3v / 150Ohm = 20mA.
From that : 18(LED's) x 20mA = 360mA (Current total), but since 2 LED's are sharing one branch of current, u divide 360 by 2 and that would leave u with 180mATotal Current. Correct? If not, how do you get 144mA? What values(Voltage,Current or Resistance) are you calculating or formula are you using to get 144mA?

Thanks :)
QuackMasterDan (author)  j03tv5 years ago
I'll be honest, I didn't calculate the measurements myself, I cheated and had ledcalc.com do it for me.

I put into the GURU box
Source: 9v, Voltage Drop: 3.3v, Desired Current: 20ma, # LEDs: 18. The GURU popped out 144mA for the total amount of current being used, and thats what I based my measurements on. I'm pretty sure the GURU takes into account real life inefficiencies, for instance each LED actually ends up drawing 16mA of power rather than the theoretical 20mA, which we use for planning out measurements anyways.

Thanks for your checking the numbers, am I incorrect (or more precisely, is the GURU)?
Lisa Fla7 years ago
I still have the power supply units from just about every piece of electronics I've ever bought. (Yes, I'm a world-class pack rat.) I'm thrilled to see there might be some use for them. My irrational fear of electricity has prevented me from even considering a project like this, but your instructable did a fine job of making me feel a little more comfortable about the prospect. So even if I never make this, Thank You!