100 LEDs in Parallel

Thread Starter

vthokie11

Joined Jun 9, 2010
30
I am working on a project and I need to wire (100) 2V, 30mA LEDs in parallel...I also need to be able to plug into a 120V AC wall outlet. I guess my questions are as follows:

1. How can I convert 120 V AC to 12 V DC?

2. In terms of resistors, etc... how can I make this happen?

Any help will be greatly appreciated!
 

Bychon

Joined Mar 12, 2010
469
The best way is to buy some sort of 12 volt power supply. You'll need at least 3 amps to do 100 times .03 amps. Trying to plug a resistor into a wall socket and power some LEDS is just not done.
 

SgtWookie

Joined Jul 17, 2007
22,210
You need to either build or buy a power supply to convert mains power to low voltage DC.
If you purchase a supply, it will have a transformer inside it to provide isolation from mains power.
If you build a supply, you must use a transformer to provide isolation.

It is usually cheaper to buy a pre-made supply than to build your own.

We don't know where you live, so it will be difficult to suggest suppliers. Please edit your Profile by clicking on the "User CP" link near the top of the page, and then click the "Edit Your Details" link on the new page that comes up.
Put your general location in the space provided, about 2/3 of the way down the page. Then click the "Save Changes" button at the bottom; otherwise it'll throw them away.

Instead of wiring all of the LEDs in parallel, it is better to wire them in strings of multiple LEDs. For example, you say 2v @ 30mA (you should post a link to the datasheet of your LEDs so we can make sure)

If you have a 12v supply, then you could have 5 LEDs in series
2v x 5=10v
You'd have 2v left over.
Then, 2v/30mA=66.7 Ohms. Looking at a chart of standard resistance values:
http://www.logwell.com/tech/components/resistor_values.html (bookmark that page)
at the green columns (E24 values) we find that 66.7 Ohms is not a standard value, but 68.0 Ohms IS a standard value.

2v/68 Ohms = 29.4mA - that's just about right.
Then you need to calculate the power dissipation requirement for the resistor.
P=EI, or Power in Watts = Voltage * Amperes
P = 2v * 29.4mA = 58.8mW; we double that for reliability's sake and get 117.6mW. You could use a 1/8 Watt or higher rated resistor.

So, that's 1 string of 5 LEDs.
You say you want 100 LEDs.
100/5=20 strings total.
20 strings, each drawing 30mA will require 600mA or 0.6 Amperes current. You will need a 7.2W or higher supply.
 

Wendy

Joined Mar 24, 2008
22,134
Wall warts come in many flavors, both AC and DC. You want DC, around 24VDC, regulated if you can get it. It simplifies the math enormously. Different color LEDs drop different amounts of voltage, different lots can vary too, you need to find a data sheet for the LEDs, baring that measure the voltage of 10 or so LEDs with approximately 10ma (current isn't that important) going through them and average the results. If you need talked through how to do this it just ask.

The link I gave, focus on chapter one and two (mostly chapter one).

Any wall wart between 10 to 24VDC will work, but you will need at least 3A to 1A, the more voltage the less amps.
 

Thread Starter

vthokie11

Joined Jun 9, 2010
30
So a wall wart that is listed as having an input of 120 V AC and an output of 12 V DC--500 mA won't work for this project?

Also does this circuit schematic look appropriate? (using 68 Ohm, 1/8 W resistors)

 

Wendy

Joined Mar 24, 2008
22,134
LEDs, 555s, Flashers, and Light Chasers

You need to read chapters one and two. There is spec called Vf, it is how much voltage the LED drops. It is close to a constant. If your battery voltage (or wall wart) does not exceed the total voltages of these LEDs then it won't work.

You didn't mention color. As I said, different colors drop different voltage. White LEDs tend to average around 3.6V (but it does vary). With 6 LEDs in a chain your total Vf adds up to 3.6 X 6, or 21.6VDC. For the circuit you have above it needs a 24VDC power supply.

For a 12VDC power supply and white LEDs it would be 3 to a chain. Understand though, this is a generalization. Specific LEDs can vary as much as ±0.3VDC within a batch, and depending on when the LED was made it can be a lot more. Modern red LEDs usually average around 2.5VDC, older generations of red LEDs can go as low as 1.5VDC.

One note about images, it is better to use the local resource of this site. You can download images to the post or to an album.
 

SgtWookie

Joined Jul 17, 2007
22,210
"Wall wart" type plug supplies are generally not regulated. Their voltage out will be within about 10% of specifications with the load stated. If your load is less, the output voltage will be higher.

There are "desktop supplies" that are regulated switching supplies. Marlin P. Jones & Assoc. sells quite a few different types of power supplies. Their site is here: http://www.mpja.com/

Regulated "desktop" supplies are on this page:
http://www.mpja.com/products.asp?dept=475&main=1

This one would be great for what you want to do:
http://www.mpja.com/prodinfo.asp?number=16790+PS
15v, 800mA, $2.49.
You'll also need a $1.25 power cord: http://www.mpja.com/prodinfo.asp?number=15447+CB
 

Wendy

Joined Mar 24, 2008
22,134
All depends on where you shop. Tanners selection of wall warts have about ½ regulated, with a small smattering of AC output wall warts. There are no standards concerning these devices, but most of them have the specs printed on the case.

If you do wind up with an unregulated model, it isn't a big deal. It will still work, you just have to pay attention a bit more to what you're doing. If you're going to buy one mail order, or if there is a selection, use the highest voltage you can get along with current.

This was mentioned earlier, but the LEDs will use around 7.2 watts. This means you need a power supply with about double that, maybe more. Watts is calculated by voltage X current. So if you have a 24VDC power supply 7.2 watts is 0.3 amps, and since you need double figure 0.6 amps (or 600ma). A 1A would work fine in this case.

Even with the simplest electronics you can't get around the math.
 

Potato Pudding

Joined Jun 11, 2010
688
6 LEDs x 12 strings so reducing to 72 LED's. Fewer LEDs will work to fit into 12 volts at 500 mA but you need to drop back to 5 LED's in a string. With 15 x 5 you get 75 LEDs and 450mA.

Adding the sixth 2V LED means that your currents will drop very low.
You would have so little current because the LEDS will drop 12V and starve on what current gets through the resistor.

You might make it work by carefully matching your LEDs in each string.

LEDs in an underpowered series can have a tendency for only some of them to "turn on."

If they aren't lighting up having them in the circuit isn't working.

Then again, if your specs are not exact and the LEDs are actually 1.75v each then 6 LEDs should work, and if the LEDS are actually 2.6 volts then 4 LEDs in each 12volt series is all you would be able to use.

If these are a bulk bag LEDs be prepared for possible off-specs, 2nd or even 3rd graded for being out of tolerance.
 

SgtWookie

Joined Jul 17, 2007
22,210
In a given batch of LEDs, you'll find that roughly 75% of them fall very close to the "typical" Vf at the specified current. You'll get a very few that are right at min Vf and another very few that are at max Vf.

With a large batch of LEDs, running them in long strings helps to minimize the difference in Vf's - the Monte Carlo random distribution will tend to make all of the strings pretty even in voltage. However, the fewer LEDs you have in a series string, the more likely that you will have poorly matched Vf's.

With 5 or more LEDs in a row, chances are good that you'll have an even distribution.

However, if you really want to be certain, you will test each LED for forward voltage at the rated current, and match them up in strings so they all have the same voltage drop.
 

Thread Starter

vthokie11

Joined Jun 9, 2010
30
Here is the data sheet for the LEDs and I will be ordering them in bulk so do you guys suggest that I measure them prior to hooking them up to be sure I get them as close as possible? I really appreciate all the help and I'm sure I will have several more questions.

Also I determined that I'll only need 72 LEDs.
 

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Wendy

Joined Mar 24, 2008
22,134
Individually measuring all of them isn't needed, but it could help. At 2.0Vf (that is a really low voltage, these might be older parts). I would measure one just to make sure the numbers aren't totally bogus.

From the link I keep recommending...



You could use something like this setup to measure the Vf of an LED. It would put 21ma through your LED setup.
 

Thread Starter

vthokie11

Joined Jun 9, 2010
30
The link is very informative, it is just that I want to make absolutely certain that my circuit will be correct before wiring it up...that is why I keep asking project specific questions
 

Wendy

Joined Mar 24, 2008
22,134
First the parts, then the schematic. We need your power supply specs (regulated or not). I strongly suggest you show the part you are interested in, and we'll tell you if it will work or not (wall wart).
 

Thread Starter

vthokie11

Joined Jun 9, 2010
30
I'm leaning towards the regulated power supply that SgtWookie suggested above, I'm going to use the LEDs that I provided the data sheet for, I would like to use 22 AWG wire, and I have yet to determine the resistors that I will need.
 

Wendy

Joined Mar 24, 2008
22,134
Resistor values depend on power supply and the LEDs Vf. We can give a value for the resistor then.

Which specific power supply are you interested in, I think Wookie suggested two.
 

SgtWookie

Joined Jul 17, 2007
22,210
I don't know what your eventual application will be, but note that the LEDs you are interested in have a VERY narrow viewing angle. They will appear extremely bright viewed head-on, but if you're more than 6° off from the center line, they will appear very dim.
 

Audioguru

Joined Dec 20, 2007
11,249
Of course the cheap LEDs are old. They are not very bright but are focussed in a narrow angle (16 degrees) to make them appear bright (but only on-axis). They might not be seen a little to one side.
 
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