# Infrared LED's project - Help



## Gomo (Apr 29, 2016)

Hello there,

I'm trying to set-up a security camera at my out of town-property which isn't connected to the power nor telephone grid (no electricity/internet). I've bought a decent camera which supports saving video on a SD/microSD card and I'll be using a car battery to power it. This camera has weak IR LED's -> bad range at night, and that's why I bought these LED's http://www.dx.com/p/5w-940nm-infrared-ir-led-emitter-silver-152456#.VyNyMnpg3Os (3x LED's per camera)
. The problem is, I don't know how to connect them .. Should I connect them in parallel or series? And also what resistor should I use? I'm not too good with electronics, and I'd really appreciate if guy could help me out!

Thanks in advance!


----------



## scout24 (Apr 29, 2016)

A one-year belated welcome, Gomo. I moved tour thread here, but left you a re-direct for a few days so you can find it...  Good luck!


----------



## CoveAxe (Apr 29, 2016)

You should connect them in series with a resistor. Simply add up the forward voltage of all the LEDs in series, subtract that off the supply voltage (12V in this case), and then divide that by the forward current. That's your resistance.

You didn't specify what kind of range you were looking for, but you may need to get LEDs with optics on them so the beam is more focused.


----------



## Gomo (Apr 29, 2016)

I tried using this website http://ledcalc.com/#calc to calculate the resistance, but .. after you enter following values: 12V input voltage, 1.8V voltage drop, 2A for current and 3 LED's in series, results kind of worry me.





Am I doing something wrong? Power dissipated by the resistor seems a bit high and power dissipated by LED a little bit low (shouldn't it output 5W?) .. also, 3,9 Ohm resistor seems like nothing (not worth using at all?)


----------



## CoveAxe (May 2, 2016)

For this kind of circuit with that kind of power, this looks fine to me.



> Power dissipated by the resistor seems a bit high




Because you have 2A going through it. Even a 1 Ohm resistor is going to be putting out 2W with 2A going through it.



> power dissipated by LED a little bit low (shouldn't it output 5W?)




Because the calculated value is 3.3 Ohms, and it's bumping that up to 3.9 Ohms for some reason. Using 3.3 Ohms would give you 5W at the LED.




> 3,9 Ohm resistor seems like nothing (not worth using at all?)




It's absolutely critical you keep the resistor if you don't want your circuit to destroy itself after you turn it on. The resistor is what regulates the current.


All this being said, this is a terribly inefficient circuit. It'll work just fine, but it's going to drain the battery faster since you're losing a lot of power in the resistor. If that's a concern, you should be using a constant current power supply of some kind on the LEDs.


----------



## Gomo (May 3, 2016)

Just tested it with a 3.9 Ohm 11 Watt resistor, it works but gets really hot.. How do I make it more efficient? :/ Would running this in parallel help? Or should I add more LED's in series & different resistor?


----------



## CoveAxe (May 3, 2016)

Doing this with resistors is necessarily going to be inefficient. Resistors do nothing but turn electricity into heat.

You can put up to 6 LEDs in series and get the resistor power down to about 2W, but again, you're just losing quite a bit of power to the resistor.

What you need to use is a constant current power supply. Something like this.


----------



## Gomo (May 4, 2016)

Thanks for the info. Btw I've bought http://www.ebay.de/itm/171994769379?_trksid=p2060353.m2749.l2649&ssPageName=STRK:MEBIDX:IT before.. Should I be using this instead? It has a voltage regulator and current is limited at 3A max if I'm not wrong.


----------



## CoveAxe (May 4, 2016)

Yes, that would be much better. Regulators still lose a bit to heat, but not nearly as much as the resistors. I can't read German so I don't know if it will fit your application perfectly so you'll just have to play with it.


----------

