Running multiple LEDs in the same box??

[upthepunxxx]

How would one go about running say 3 leds at once instead of one? Is it just a matter of running them together, like connecting all the anodes together and all the cathodes together and continue business as usual. Like 1 big LED that now would have 1 andode and 1 cathode made up of all 3? Or do I need to use to higher resistance resistor and somehow more power?


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[mattlee0037]

I can tell you yes you can run them like that and use the same power, not sure about the resistor though... My physics knowledges says the current is split 3 ways so maye a lower resistor? Not sure though

[RobA]

The best and recommended way to run LED's in parallel is to use a resistor per LED. So, each LED has a series resistor and then the resistor LED pairs are run in parallel. The primary reason for this is that it has better failure characteristics.

[Vallhagen]

The best and recommended way to run LED's in parallel is to use a resistor per LED. So, each LED has a series resistor and then the resistor LED pairs are run in parallel. The primary reason for this is that it has better failure characteristics.

This. Or run them in series, then you only need one CLR.

If you want to read up on the physics around this case, i found this discussion pretty good: http://electronics.stac
kexchange.com/questions/22291/why-exactly-cant-a-single-resistor-be-used-for-many-parallel-leds
Cheers

[mah62]

I recently built a UV box which uses a whole network of 3 LEDs and one CLR. This is what I learnt from the web:

The first thing to do is to have a look at the spec sheet for the LED to see what the forward voltage is. Better still, knock up a little LED/ resistor circuit on a bread board using say, a 2k2 resistor to be on the safe side. Then, using your DMM, measure the voltage across each LED. Typically for a diffused red LED it's about 1.8v/2v. So, the voltage drop across 3 in series would be around 6v. the voltage drop across the CLR would be therefore be equal to the supply voltage minus 6v. I.e. 3v using a 9v supply. Another thing to know is that LEDs will require around 20 milliamps to fire them up.

So, using ohms law and the above info we can do this little sum:

R = V/I

R = 3/0.02

R =  150 ohms

I'm not sure if a 9v supply could work using 3 superbrights as the forward is usually a little over 3v. Experiment. My UV box is powered by 12v so not a problem.

Hope this helps, Mark

[m-Kresol]

I recently built a UV box which uses a whole network of 3 LEDs and one CLR. This is what I learnt from the web:

The first thing to do is to have a look at the spec sheet for the LED to see what the forward voltage is. Better still, knock up a little LED/ resistor circuit on a bread board using say, a 2k2 resistor to be on the safe side. Then, using your DMM, measure the voltage across each LED. Typically for a diffused red LED it's about 1.8v/2v. So, the voltage drop across 3 in series would be around 6v. the voltage drop across the CLR would be therefore be equal to the supply voltage minus 6v. I.e. 3v using a 9v supply. Another thing to know is that LEDs will require around 20 milliamps to fire them up.

So, using ohms law and the above info we can do this little sum:

R = V/I

R = 3/0.02

R =  150 ohms

I'm not sure if a 9v supply could work using 3 superbrights as the forward is usually a little over 3v. Experiment. My UV box is powered by 12v so not a problem.

Hope this helps, Mark

that is neatly explained! As of the problem described at the end. In that case it would be possible in parallel rather than in series. or have to pairs of LEDs in series, which are parallel to another pair for example.

[upthepunxxx]

This place is the best! I knew I could get the help I needed!!