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Step 5Power
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.
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.
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if so then totally it will be 5000 mah but just as the capacity is lessened,it shud still remain 2500 mAH,right?
I threw together a quick picture in photoshop to illustrate how it works.
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
:P
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.
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
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.
i was wondering if a 12v 1amp dc plug would work if im hooking up 5 bars together
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.
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.
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.
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 :)
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)?
That calculator is accurate, but I don't usually trust those things and use a calculator and just use Ohm's, Kirchoff's and Watt's Law for raw numbers.
GURU says: a 120Ohm resistor would = 20mA, but your using a 150Ohm which is the closest resistor, so its actually 16mA.
So I see now how you get 144mA, 18x16mA = 288mA. But 2 LED's in series makes it divide by 2 so its 144mA.
Overall 16mA is good current for a LED's life span.
I try to use a lower amperage as well. If its 20mA, i usually give a LED 1-2mA of head room, so mine would be 18-19mA.
The LED calculators all basically use Ohm's law and Kirchoff's Voltage Law(KVL), depending on what informationg they give.
Using a normal calculator you can do this quickly and get the same thing.
Basically all you have to do is subtract how much Voltage you plan to use on an LED from your Supply voltage. i.e. 9v - 3.3v x 2(how many LEDs) = 6.6v in voltage drop. To calculate resistance needed you just use the remaining voltage (2.4v) and divide by mA, 2.4v / 0.02A = 120Ohm. and theres your resistor. All values are same on LED calculators.
BTW iacchus you can use 1/4 Watt resistors for these LED's.
I am currently taking Elec 101, so this is basic electronics fundamentals.
I recommend taking this course in a school wherever you live if you plan on doing more electronic stuff.
Like i said earlier, you will learn the Laws, Ohm's, Kirchoff's, and Watt's to help you build the proper circuits and for technicians skills.
I love electronics, its so fun. :)
I stumbled upon your instructable from trying to find a material to mount the LED's on.
I just created an LED light bar with 20x bright white LED's.
I used a 2f x 1in piece of woold, cut a line thru the middle for wires and drilled holes for the LED wires and LED placement in 1 inch placements.
I then calaculated my own values from standard USB, 5V @500mA.
So using that formula earlier ive got 1 100Ohm resistor in series with a 3V Vf LED in parallel with the rest to make 20.
Then I just drowned that sucker(contacts/wires) like the movie "Open Water" with hot glue
The results are incredible. I mean its a lot of light for low power and I just plug it into my 4port standalone USB hub.
I just wanted a simple but cool looking light display for my shelf to show off my Hot Toys Collectibles :D
I'm going to work on a instructable for hardwiring the lighting features of some Hot Toys figures pretty soon. They use those AG3 and AG9 tiny hearing aid/watch batteries that dont last long,. So then you'll be able to just plug into a USB or DC adapter for displaying while lit, all day for 10 years if you wanted. :D
I'll have to find a solution or more to figure out how to hide wires, with little to no damage to the figure, the figures run roughly $150 - $200 a piece so I really dont even want a finger print on em but we'll see.
If you have any other questions about how to calculate any other values or whatever, feel free to message me. Just don't ask about memory controllers, PIC's and all that stuff. I know a bit of programming but not on how to program actual hardware, ill get an arduino for that.
Sorry its such a long post lol
Cheers!
www.instructables.com/id/ESTRWXVFYTCLJ5J/