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# How to setup an LED hunting light? Answered

I'm trying to copy something I've seen online...without buying it. So I dont have a light to look at in order to copy it.

I'm doing this for my own use, not to sell, so i'm not trying to take business from anyone.

Online they have hunting lights that are green LED's that hang below a feeder. These are used for wild hog hunting since they usually come out to feed at night. They cannot see the green light, but green is easier to see at night to the human eye.

I'm wanting to power it off of the 12v battery in the feeder since I have it charging with a solar panel.

The LED's i found online for fairly cheap are super-bright 5mm green LED's. It says they give off more than 8000 mcd (light rating?). They require 3.2v (3.4v max) at 20 mA.

I've looked at some of the posts on this and wanted to know if my math is right on how i can set these up.

I figure i would wire 3 in a series, which would require 9.6v. Using Ohms law with a current of .020 Amps, I get (12-9.6)/.020=120 Ohms

So, 3 LED's in series with 120Ohms worth of resistors and my 12v battery will power it fine. The part i'm not sure about is how the current works with this. The battery is 12v 7Ah. Will that fry them or is the current the amount of juice it has?

If i wanted to make a light with say, 30 LED's, can I make these 3 LED units and wire multiple units together in parallel? so each unit would power off of the 12 volt and the parrallel units? Hopefully this makes sense.

Will this work? Is this the best way?

Any suggestions or comments would be appreciated. Thanks.

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Perfect! Awesome homework! Remember, many 12v lead acid batteries charge to 14.5 volts and are dead at 11 volts. You might want to increase the resistor a few notches just to be on the safe side for when the battery is fully charged.

So you have it figured out - each possible string of 3 leds requires one resistor, and you can have multiple in parallel like that.

The string will draw your calculated total voltage (near or at battery voltage) at 20mA. a 7Ah battery is 7000mA hours...
7000/20 = 350 hours. That thing will run on a 7Ah battery for 14 days continuously.

I drew up what i think the circuit would be. I drew 2 different ones. Can ya'll help with what would be the better option?

now that you've done all the homework, check out the wizard and you'll understand exactly what it's doing.

http://led.linear1.org/led.wiz

Whatever has the lowest amount of power wasted in the resistor (smaller resistor) is better :)

The discharge curve is such that it won't maintain 14.5 for very long - and the leds are going to be reasonably happy on 40mA, at night, for 10 hours or so.

Yes you have got the math right. all you need to make sure is that the battery has enough draw to power them all.

You have it dead right. Don't worry, the battery won't fry the LEDs with the maths as you have it !

Parallel multiple units and you can add 30 leds with no problem - though you will draw 10X the current that one chain of three will..

Steve

What would adding more resistance to to performance once the battery discharges to a lower voltage? The light would only be used for a few hours at a time, so the fact that the battery will power it for days doesnt matter. I want the best performance (brightest light) possible out of all of these little guys.

Thanks.

Get the datasheet for the LEDs, and add a circuit to EACH three LEDs like this: You can use the LM317 EMP or MDT packages - thats a Surface mount device, but any old 317 will work.

Use the data sheet to find the MAXIMUM running current of the LEDs.

I;m going to guess at 30mA, and so using my circuit, make the resistor 39 Ohms, and it will push exactly 31mA through each chain of three LEDs, under all input condiions and at all temperatures.

Fancier approaches are possible, but not worth it yet - as Frollard points out, you'll get 350 hours out of these !

Steve