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question about wiring leds and the resistor(s) involved!

post #1 of 13
Thread Starter 
I am in the process of separating the LEDs from an array of led fans to a rocker switch control hub.

Would the resistor included in a 2 LED lighting array ( http://www.koolance.com/water-cooling/product_info.php?product_id=903 )
which is wired:
Code:
+ ------ resistor ----- LED ----- LED ----- g

be suitable for x amount of LEDs?

Or would additional resistors be required for the job?

What would be the optimal way to wire it ? Res to multiple led or just keep adding leds in the above diagram?


Thanks in advance biggrin.gif !

edit:

thought of this possible wiring scheme:
Code:
+ --- (resistor)---LED---LED---[----------- -
               L---------LED----LED--]
               L---------LED----LED--]
                        (...)

Edited by CooooooooL - 1/7/12 at 12:47am
post #2 of 13
Im thinking maybe like this

+ve rail ---+---+---+- ... -+
| | | |
R1 R2 R3 Rn
| | | |
led1 led2led3 ledn
| | | |
g g g g


i think this parallel config would be best IMO
sorry about the crudness of the drawing. Im on phone lol smile.gif
well that didnt come out like i wanted, basically you have a +ve rail and where the + sign is the junction leading to a resistor to the LED to the ground
Edited by stubass - 1/7/12 at 11:48pm
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post #3 of 13
One resistor per LED is best for parallel, but usually you can get away with using 3 LED's in series per resistor.
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post #4 of 13
You can easily wire multiple LEDs in series/parallel and often design the array such that you need 0-1 resistors for the whole array; but you can't just use any resistor, you have to compute the resistor size based on the voltage drop and current across it.

A few notes to start out with:
  • You need to know the typ forward voltage (Vf) and forward current (If) ratings for the LEDs (listed on their datasheets)
  • Voltages add in series, currents add in parallel\
  • All LEDs in the same series string must have the same current draw (voltages can be different)
  • all parallel strings must have the same combined voltage drop (currents can be different)
  • You will need to know Ohm's Law and the Power relation

So to start out, the maximum number of LEDs you can have in series, based on the supply voltage (Vs), is calculated as follows:
Nmax = rounddown( Vs / Vf )
Let's assume Vf = 2.1 V, If = 20 mA = 0.02 A, and Vs = 12 V

Nmax = rounddown( 12 V / 2.1 V ) = rounddown ( 5.71) = 5
note: if for example you were using 3 V LEDs then you could have 4 in series without any current limiting resistor.

Let's say you want 12 of these LEDs with a single resistor
That means you would have 3 series strings ( m = 3 ) each with 4 LEDs ( n = 4 ).

So the voltage drop across the resistor will be:
Vr = Vs - Vled = Vs - n * Vf = 12V - 4 * 2.1 V = 12 V - 8.4 V = 3.6V

And the Current across the resistor:
Ir = m * If = 3 * 0.02 A = 0.06 A = 60 mA

Now use Ohm's law ( V = I * R ) to calculate the necessary resistance
R = Vr / Ir = 3.6 V / 0.06 A = 60 Ohm

Select the actual resistance by rounding up to the nearest standard resistance value; in this case 62 Ohm
Now the power relation ( P = V * I ) is applied in the resistor form ( P = I^2 * R )
The actual power dissipated by the resistor will be:
P = Ir^2 * R = 0.06 A ^ 2 * 62 Ohm = 0.2232 W

So we will use a 62 Ohm 1/4 W resistor (can also use 1/2 W or higher, but never lower than the actual power dissipated)

Quote:
Originally Posted by Furball Zen View Post

One resistor per LED is best for parallel, but usually you can get away with using 3 LED's in series per resistor.
There is no best way of designing an LED array. The only advantages that a resistor per LED brings is that you can very easily deal with LEDs with different forward voltage and/or current; and if all the LEDs are the same then all the resistors are the same. Big disadvantages to this are that you need a lot more resistors and have to deal with tidying up a lot more wire.
Edited by radodrill - 1/8/12 at 9:11am
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post #5 of 13
From a 'n00bs' standpoint, yes. I wish you would stop coming in after my posts and spouting all your knowledge in one post. I know that stuff, but for someone who obviously doesnt even own a soldering iron, you cant answer like that and expect them to not get discouraged. (I dont have anything against you personally, so dont go thinking that, every post doesnt need to show off your EE)
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post #6 of 13
Quote:
Originally Posted by Furball Zen View Post

From a 'n00bs' standpoint, yes. I wish you would stop coming in after my posts and spouting all your knowledge in one post. I know that stuff, but for someone who obviously doesnt even own a soldering iron, you cant answer like that and expect them to not get discouraged. (I dont have anything against you personally, so dont go thinking that, every post doesnt need to show off your EE)

I don't really see the problem. He is giving the OP a detailed explanation of how to properly calculate what resistors to get based on any given setup / # of LEDs, rather than an ambiguous, vague answer.. Well done, Rado.
post #7 of 13
Thread Starter 
Thanks for the help everyone.

And although I may not own the soldering iron that I will be using, I think it's always best to know as much as possible about something before voiding any warranties!


Now I was about to get into the whole calculation process, but couldn't find the technical specs for the built-in LEDs on Yate Loon D12SH-12 (F) from ppc...

I was wondering if anyone had that information handy... Also going to email ppc, see if they can tell me / point me in the right direction!
post #8 of 13
Quote:
Originally Posted by CooooooooL View Post

Now I was about to get into the whole calculation process, but couldn't find the technical specs for the built-in LEDs on Yate Loon D12SH-12 (F) from ppc...

I was wondering if anyone had that information handy... Also going to email ppc, see if they can tell me / point me in the right direction!

For LEDs; 1.8-2.1V, 3-3.2V, and 5V are common voltages and they are normally 20mA or 30mA.

LEDs are pretty inexpensive that it's often easier to just buy new ones as the reseller will list the specs and/or have a link to the datasheet. I don't buy anything at Radioshack anymore since they're way overpriced; Electronics warehouses such as Jameco, Mouser, Newark, AllElectronics, and Digi-Key are all reputable sellers with good prices. Just a comparison of prices; Radioshack sells a packet of 5 resistors (rated at 1/4W) for $1, while the electronics warehouse have a packet of 100 for $1 and even with flat $7 shipping it's more economical if you happen to need a lot of resistors or other stuff as well. BTW, electronics warehouses generally sell LEDs for <$0.20 each

If you want to get an idea of the specs of the LEDs in the fan, you can measure the voltage across the LED and then trace through the circuitry to determine the series/parallel strings of LEDs and resistors. Knowing the number of LEDs and the voltage drop across the LEDs in a series string, you can calculate the voltage across the resistor in that string. You can find the resistance of the resistors from the resistor band code and from the voltage drop and resistance compute the current. This will give you a pretty close estimate of the actual specs of the LED.
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post #9 of 13
Thread Starter 
Well, I tried using the formulae above to try and figure out what the best resistors for my circuit should be and can't seem to get an answer that makes sense...

2ymy3ic.jpg

Feel free to give it a shot, I'm going to give it another go in the morning!
post #10 of 13
Quote:
Originally Posted by CooooooooL View Post

Well, I tried using the formulae above to try and figure out what the best resistors for my circuit should be and can't seem to get an answer that makes sense...
2ymy3ic.jpg
Feel free to give it a shot, I'm going to give it another go in the morning!

Voltage drop across the resistor:
Vr = 12 V - 2 * (2.0 V) = 8 V -- Note, voltages add in series, parallel strings have the same voltage

Current across the resistor:
Ir = 2 * (20 mA) = 40 mA = 0.04 A -- Currents add in parallel, series elements have the same current

Design resistance:
R = Vr / Ir = 8 V / 0.04 A = 200 Ohm

200 Ohm is a standard resistor size, so the actual resistance:
R = 200 Ohm


Now the power dissipated by the resistor:
P = Ir ^ 2 * R = (0.04 A) ^ 2 * 200 Ohm = 0.32 W
Standard resistor power ratings are 1/8, 1/4, and 1/2 W (there are bigger power resistors starting at 1W as well)

So you'd need a 200 Ohm 1/2 W resistor.


If you could put all the LEDs on each fan in series (rather than series + parallel), you could use a resistor with a lower power rating (200 Ohm 1/8 W to be exact).
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