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  #1  
Old 04-11-2011, 08:57 PM
spinnerrulz94 spinnerrulz94 is offline
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Question Wiring multiple LED's to one 12V DC power source

So to start, I know almost nothing about LEDs, and neither does anyone else doing this project. Since I'm the only one who knows how to use a soldering gun, I'm the one assigned to do it.

Right now, I have 3 different types of LED's. Aqua, Pink, and Green. I have a "high performance" 110AC to 12DC converter.

I have about 30 each of the LEDs and I would like to wire them all into this one power source.

First, do I connect them in parallel or series?

Second, Do I need a bread board? Or can I just solderer them all to one wire?

Third, besides LEDS and resistors, are there any other supplies I need for the circuit?

I really don't know how I would go about doing this, or if it is even possible? Was hoping for some help on here.

Thanks for your help in this matter!
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Old 04-11-2011, 10:55 PM
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Bill_Marsden Bill_Marsden is offline
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LEDs, 555s, Flashers, and Light Chasers

Chapter 1 and Chapter 2 (first half) are tutorials on LEDs.

The main thing you need is a stable power supply. 9V batteries shift their voltages radically, dropping quickly to 7V (and the are still considered fresh at this point). You just have to design around them.

Another very recent thread was asking pretty much the same question, so my reply looks very similar to both.
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Old 04-12-2011, 05:55 AM
spinnerrulz94 spinnerrulz94 is offline
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Default Still kinda lost...

Well to start, I have a stable power supply I will be using. It's taking a standard house current and converting it into 12V of DC power.

I read the article you mentioned, but I guess I just dont understand. It's not possible to link that many to a source? I can't parallel link them, so how do I connect them all to a source? Or will I need more than one source?

Also, there is going to be quite a bit of wiring that is connecting the LEDs to the power supply. Does the long wire affect anything?
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Old 04-12-2011, 07:06 AM
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Bill_Marsden Bill_Marsden is offline
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The article has schematics that describe exactly what you are asking. Displaying the illustrations, what is it you don't understand?



Are you having trouble understanding the schematics? Serious question, no sarcasm intended.

Some electronics cares about long wires. LEDs are not one of them.
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Old 04-12-2011, 05:55 PM
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All the info is in those articles. Try to think of it like this. Each LED is going to "use up" some voltage. That's called the forward voltage drop or Vf. So, for example if Vf on your LED was 3.0V, you could connect 4 in series to your 12V supply (3x4=12). It's not that simple though. You should have at least one ballast resistor (Google calculating voltage drop across a resistor, once you understand that the rest should fall into place).

So if you had 3 LED's in series with your resistor, then to use 30 LED's you'd have 10 strings of 3. Each string is in parallel to the next string, the LED's in a single string are in series.
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Old 04-12-2011, 10:18 PM
spinnerrulz94 spinnerrulz94 is offline
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Default That makes a little more sense, but...

So for the sake of simplicity here... It I wanted to string 30 red leds, I could do so by stringing three in paralell to a single wire that has the power running to it? I of course would need a resistor in there. Then just wire ten "strands" of three to the wire?

Now I'm running into another question. I have a 3 amp push-button switch that I want to use to be able to turn the LEDs on and off. Am I going to run into a problem with the switch, or do I just wire it in and have no problems doing so?

Now, an LED has two legs, one positive and one negative. To wire in paralell, I solder the positive legs end to end? Then, what do I do with the negative legs? How do I ground the LEDs?
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Old 04-12-2011, 10:38 PM
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Well, you need to get some info about the LED's you are using. I used 3.0V for Vf as an example. Your LED's are probably different. Look at the example schematic that Bill Marsden posted. The picture on the left has only one LED and one resistor. The schematic on the right has two strings of LED's in parallel. One string starts with R1 at the top and one string has R2 at the top. The arrow looking symbol at the bottom of each string represents ground. So everywhere you see that means you connect that wire to the ground, or negative, of your supply.

Notice how he uses a much smaller value resistor when he's got the three LED's in series?

Technically LED's don't have positive and negative. They have Anodes and Cathodes. The Anode is usually the longer of the two leads and you connect that to the positive supply (if you want them to light up) and the shorter Cathode to the ground of your supply.

So start by finding the info (datasheet) for your LED's, then you can figure out how many and what resistor goes in each string. You want to learn the Vf and the current at that voltage.

I expect that you're 3A switch will be sufficient but without knowing what LED's you're using there's no way to be sure.

You should read that article that Bill posted. You shouldn't solder the legs together. Each string needs it's own ballast resistor. If you just use one resistor then some LED's will pull more current than the others and burn out sooner.
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Old 04-12-2011, 11:01 PM
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Well, first you need to know what your LED's typical Vf (forward voltage) at their rated current.

A typical red LED might have a Vf of 2v at 20mA current. However, you really need to find out what your particular LED's specifications are; normally shown in a datasheet.

Then you can calculate how many LEDs you can operate in a series string. Basically, calculate that as:
Max_LEDs_in_series = Integer((Vsupply-1v)/LED_Vf)
So, if Vsupply = 12v and your LED Vf is 2v, then:
Max_LEDs_in_series = Integer((12-1v)/2v)
Max_LEDs_in_series = Integer(11/2)
Max_LEDs_in_series = Integer(5.5)
Max_LEDs_in_series = 5 (you round it down)

Then you need to calculate what value current limiting resistor to use in the series string.
Rlimit >= (Vsupply - (LED_Vf *LED_count))/Desired_Current
Rlimit >= (12v - (2v * 5))/20mA
Rlimit >= (12v - 10v)/20mA
Rlimit >= 2v/0.02A
Rlimit >= 100 Ohms.
100 Ohms is a standard value of resistance.
A standard table of resistance is here:
http://www.logwell.com/tech/componen...or_values.html
Bookmark that page. Use the green E24 columns.
If there were not a standard resistance value that was the same as the result, you would need to go to the next higher value.

Then you need to calculate the wattage requirement for the resistor.
Power in Watts = Voltage * Current.
We know there will be a 2v drop across the resistor, and 20mA current through it. So,
P= 2v*20mA = 2*0.02 = 0.04 Watts, or 40mW. We double that for reliability, so 80mW. You can use a 1/10 Watt rated resistor or higher (1/8W, 1/4W, etc).

That is for one series string.

You can have multiple of these series strings in parallel.
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  #9  
Old 04-13-2011, 12:54 AM
spinnerrulz94 spinnerrulz94 is offline
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Default So are these calculations correct?

White LEDís:
-Intensity: 690mcd
- Forward Voltage: 3.6V typical, 4.0V Max @20mA
-Max Forward Current: 30mA
Aqua LEDís:
-Intensity: 4000mcd
-Forward Voltage: 3.5V @ 20mA
-Max Forward Current: 30mA

ML=X((12-1v)/3.6)
11/3.6 = 3.05
Max number of WHITE LEDís = 3

ML=X((12-1v)/3.5)
11/3.5 = 3.14
Max number of AQUA LEDís = 3
WHITE LEDíS:
Rlimit >= (12v-(3.6v*3))/20mA
(12v-10.8v)/20Ma
1.2v/20mA
1.2v/.02A
Rlimit>=60 Ohms (So I would Need a 100 ohm resistor for the white LEDS?)
AQUA LEDíS:
Rlimit>=(12v-(3.5v*3))/20mA
(12v-10.5v)/20mA
1.5v/20mA
1.5v/.02A
Rlimit>=75 Ohms (So once again, I would need a 100 ohm resistor for the aqua LEDS?)
WHITE LEDíS:
Watts = Voltage*Current
Pw=3.6V*20mA
Pw=3.6V*.02A
Pw=.072W
.072W=72mW
(So I would need a 144mW Resistor for white)
(That means I need a 100 ohm resistor rated at 144mW?)
AQUA LEDíS:
Pw=3.5V*20mA
Pw=3.5V*.02A
Pw=.07W
.07W=70mW
(So I would need a 140mW Resistor for Aqua)
(That means I need a 100 ohm resistor rated at 140mW?)

So what fraction of a Watt is 140 and 144mW?
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Old 04-13-2011, 01:23 AM
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Quote:
WHITE LED’S:
Rlimit >= (12v-(3.6v*3))/20mA
(12v-10.8v)/20Ma
1.2v/20mA
1.2v/.02A
Rlimit>=60 Ohms (So I would Need a 100 ohm resistor for the white LEDS?)
Well, 100 Ohms would reduce your current too much.
1.2v / 100 Ohms = 12mA.
Remember that decade table of resistance values I posted the link to?
http://www.logwell.com/tech/componen...or_values.html
If you go on there and look at the green columns, you'll notice over on the right that there is 560 and 620. If you divide those values by 10, you get 56 and 62.
56 is less than 60, so that would result in too much current. 62 Ohms is the next larger value, and it will work fine.

1.2v / 62 Ohms = 19.35mA; plenty close enough. 96.7% of 20mA.

Quote:
AQUA LED’S:
Rlimit>=(12v-(3.5v*3))/20mA
(12v-10.5v)/20mA
1.5v/20mA
1.5v/.02A
Rlimit>=75 Ohms (So once again, I would need a 100 ohm resistor for the aqua LEDS?)
Nope - go back to that table of standard resistance values; you'll see 750 in the green column on the right. 750/10 = 75 Ohms, so that is a standard E24 value.

About the wattage values:
1 Watt = 1.0W or 1000mW
1/2 Watt = 0.5W or 500mW
1/4 Watt = 0.25W or 250mW
1/8 Watt = 0.125W or 125mW
1/10 Watt = 0.1W or 100mW

Doubling the actual wattage requirement is easiest to remember. However, you can use 1.6 times the result, which gives you a 60% over the required rating, which is normally plenty good enough.

Now, you kind of messed up on the power calculations; you used the LED Vf instead of (Vsupply - the total Vf's of the LEDs)
For the White LEDs, you would calculate:
Pw=1.2V*20mA*1.6
Pw=1.2V*.02A*1.6
Pw= 38.4mW; 1/10 Watt or greater would be fine.

You also need to re-calculate the aqua LED resistor power requirement.
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