Help with LED Picture. Complete newbie!

[enjunear]

(Hopefully) attached is a picture of the LED specs. I also have 12v Resistors apparently.

**broken link removed**

If the picture doesnt work here is the link to the item, i had to order some more. Scroll down to the section with a red heading titled "Round LED" for the specs.

If you could tell me whether the power supply i posted earlier is suitable, or point me in the right direction to one i can get in the uk, that would be awesome
Regards

The way that website states things, it appears they are giving you one resistor to get with each LED, so you can run a single LED/resistor pair from 12V and not damage the LED. While this will work, it's highly inefficient.

Here's my recommendation:
For the blue LEDs...
Each blue LED requires 3.2 volts. If you put three 3 blue LEDs in series, that's 3.2V * 3 = 9.6V. Then the resistor will have to handle 12V - 9.6V = 2.4 volts. Now select a bias current for the LED string (If = 25 mA, I'd start with 15mA). Since resistors follow Ohm's Law (V = I*R), then you can calculate the needed resistance to be 2.4V / 0.015 A = 160 ohms. So, each string of blue LEDs would wire up like this:
(12V supply) --- LED --- LED --- LED --- resistor --- (ground)

The resistor will dissipate (Power = V*I) 2.4V * 0.015A = 36 mW, so a standard 1/4 watt carbon resistor will be just fine (0.25W).

For the yellow LEDs... a similar treatment.
Each yellow LED needs 2.2V, so you can put 5 in series. That's 2.2V*5 = 11V. The resistor will need to drop 1V at 15mA, which means it should be 1V / 0.015 = 67 ohms (I think the nearest standard value is 68 ohms).

You can also go with four LEDs on a sting, so that would be:
2.2V*4 = 8.8V
resistor drops 12V - 8.8V = 3.2V
3.2V / 0.015A = 213 ohms, so I'd go with a 220 ohm as the nearest standard value.
I'd got this way, so you have some more flexibility in your resistor range (not quite so close to zero).




So, if you have 134 LEDs, half blue, half yellow... that's 67 blue LEDs, 67 yellow LEDs.

67 blue LEDs / 3 LEDs per string = 22.33 strings (round down to whole number.... 22)
22 full strings of blue LEDs @ 15 mA per string.... 0.015A * 22 = 0.33 amps

67 yellow LEDs / 4 LEDs per string = 16.75 (round to 16)
16 full strings of yellow LEDs @ 15 mA per string.... 0.015A * 16 = 0.24 amps

To light up both sets of lights, sum their current needs... 0.33A + 0.24A = 0.57 amps. For that, I'd cheat-up and go with at least a 600 mA capable power transformer.

You should be able to find a 12V, 1000 mA (1 amp) wall-wart transformer for pretty cheap in most Radio Shack-like stores.
**broken link removed**

The PC power supply will be waaaay overkill, since it can supply 14x more current than you'll need. A simple wall-wart transformer will be MUCH easier to find a home for, compared to a bulky PC power supply with a rats nest of wires coming out of it.

 

[een54]

edit: one last question, how do i select the current to set it to 15mA? :S





Thanks a bunch! I've learned a lot. I have a brand new 12v 1amp dc transformer i got with a new router that i could use, I assume this would be ok.

Thanks for all the help and i will give it a go as soon as i get time.

Will post pics when im done

 

[shabazmondal]

since u are a complete newbie...it would be of great help to you if u have someone with you who has a little bit of knowledge in just the basic things.
Then we will be able to help you very effectively and quickly..!!

 

[enjunear]

edit: one last question, how do i select the current to set it to 15mA? :S





Thanks a bunch! I've learned a lot. I have a brand new 12v 1amp dc transformer i got with a new router that i could use, I assume this would be ok.

Thanks for all the help and i will give it a go as soon as i get time.

Will post pics when im done

The 12VDC 1A "wall-wart" transformer should be just fine.

The current is "set" by the voltage/current relationship of the circuit elements. LEDs operate on a very steep voltage vs. current (V/I) curve near their operating point. The voltage across the diode (LED = Light Emitting Diode) changes very little (0.1-0.2V) for a relatively large change in the forward current (5-25 mA, for example). The numbers aren't exact, but give you an idea of how "stiff" the LED bias voltage is, relative to it's forward current. So, we can say that over a large range of currents, an LED will stay very close to it's nominal voltage.

Using this bit of information, I started the calculations using 3 blue LEDs in series (the voltage drops add together... 3.2V+3.2V+3.2V = 9.6V). Since your power source is ~12V, you have to drop 12V-9.6V = 2.4V across a resistor. HERE is where you "set" the current. Remember Ohms law.... V=I*R? You know the voltage drop across the resistor (2.4V), you can choose the current to be anything you like (25mA is the max recommended by the vendor, but I'd go less than that to avoid burning out retinas when people walk by... thus the 15mA value). Once you know V and I, you solve for R with a little algebra.

For 15mA, V=I*R.... 2.4V = 0.015*R, so R = 160 ohms.
If you want to use 10 mA... 2.4V = 0.010*R, so R = 240 ohms.
If you want to use 20 mA, then 2.4V / 0.020 A = 120 ohms

This should make sense, intuitively. If you increase the resistance in the circuit, less current will flow through it. Conversely, if you reduce the resistance in the circuit, more current will flow. Think of current like water in a pipe; if you reduce the size of the pipe (more resistance) the flow rate of the water (current) will decrease. If you increase the size of the pipe (less resistance), more water can flow (more current). In an electrical circuit, instead of water in a pipe... it's electrons in a wire.

So, by using different resistors in series with your LEDs, you can control the current in that single circuit. Try making a string of 3 blue LEDs and a particular resistor and see how bright they are (look straight into them to see the max brightness). Don't forget, you'll be multiplying that by about 40x if you are using all 130+ LEDs that you bought!

---------- Post added at 18:45 ---------- Previous post was at 18:42 ----------

since u are a complete newbie...it would be of great help to you if u have someone with you who has a little bit of knowledge in just the basic things.
Then we will be able to help you very effectively and quickly..!!

Agreed. If you have someone that is savvy in basic electronics (and has decent teaching skills), they'd be a great resource to start you off on the fundamentals like voltage, current and resistance. However, if you don't have anyone, folks on here are pretty good about breaking things down into bite-sized pieces.