# 100W LED, How to run of batteries?

1. Sep 5, 2010

### DATAcrusher

Okay, to cut a long story short, I got hold of a:
16~18V, 30W, White LED, 2100 milliamps.
And a
32~36V, 100W, White LED, 3500 milliamps.

I got some 6V lead batteries, and I would like to use them to make a crazy flashlight.

I tried searching, I came upon quite a lot of solutions, mostly with resistors, but NONE WITH THIS HIGH POWER.
Which is why I desperately need your help.

2. Sep 5, 2010

### dlgoff

Welcome to PF DATAcrusher.

Those are some fairly large assemblies. I'm curious. How were you able to get hold of them?

If you are going to use lead-acid batteries, you will need some way to limit the current. What have you learned so far about current limiting resistors?

3. Sep 5, 2010

### vk6kro

I wonder if these already have current limiting built into them?

I found this site with some amazing LED stuff and a lot of it is direct replacement products for standard lamps.
They claim efficiencies of 80 lumens per watt (incandescents are about 15) and 50000 hour lifetime. That is about 5.7 years continuous use.

There are even LED floodlights.

Last edited by a moderator: May 4, 2017
4. Sep 5, 2010

### Snoogans

You will also want to consider heat management with that much power or you'll kill the leds very quickly. They may be efficient relative to other light sources, but most of that 100W is still dissipated as heat.

5. Sep 5, 2010

### DATAcrusher

dlgoff-
1.
They were found on a Website.

2.
I know I have to limit the current:
If I have a "load" that's a LED, and luckily this does not require a voltage drop, Then I need to make use of U = I * R;
U = 16
I = 2.1
therefore, R = 7.6 ohm
however, doesn't this stress the limiting resistor far too much?

And thank you ^^ I like it here :P

vk6kro-
Believe me... They do NOT. I see no driver anywhere.

Snoogans-
I'm using a fat copper wire and a heat sink from a PC to cool it down.

Thanks everyone! =D

6. Sep 6, 2010

### vk6kro

I would try for some genuine information first, but you could take a brute-force approach to get it going.
I'd prefer to use a variable voltage power supply with current limiting, but I guess you don't have one of those.

Get 4 of your 6 volt batteries and put them in series. This will give you 24 volts.

Now get a pair of 5.6 ohm 10 watt resistors and put them in parallel with each other. This gives 2.8 ohms.

Now put the combined resistors in series with the positive lead of the battery combination and the positive connection of the LED.
Connect the most negative battery connection of the batteries to the negative lead of the LED.

There is some risk with this. You could try it first with 2 then 3 batteries.

You don't calculate the resistor sizes using the voltage across the LED. It is the difference between the battery and the LED voltages that you have to use.
so, 24 minus 18 = 6 volts. Divide this by 2.1 amps to give 2.8 ohms.

Last edited: Sep 6, 2010
7. Sep 6, 2010

### DATAcrusher

so you limit the 'I' in the entire circuit, and b/c the potential going over the resistor is 8 volts, we know that the diode will get the rest of the potential through it, and maximum 2.1 Amps.

Right? D:
IF IT'S RIGHT I'LL GIVE YOU A CUPCAKE!

Last edited: Sep 6, 2010
8. Sep 6, 2010

### vk6kro

Yes, that is right.

You understand that the battery, the LED and the resistor are all in series and you have to get the polarity right on the LED first time.

+ to +.......- to - except the + side has a resistor in it.

Can you see individual LEDs in this thing or is it just one big LED?

9. Sep 6, 2010

### DATAcrusher

Of cause, I know how to solder stuff together, I was just very concerned about the high energy consumption in this circuit.

The 30W got 30 diodes, the 100W got 100.. But they are close packed together.

One thing I don't "get"- Why resistor between the [+] [R] [+Diode-] [-]

10. Sep 6, 2010

### vk6kro

The resistor limits the current that can flow from the battery to the LEDs.

The voltage across the LEDs will be fairly constant with whatever current you feed into it, so there is a predictable voltage across the series resistor. This means the current can be controlled just by choosing a different resistor.

It is really important that you get more information if you can. This is a high powered device and it could just blow up if we don't get it right.

There is a problem that I can see. If the LEDs don't have inbuilt series resistors, they may not all light up. There are probably 6 parallel branches each with 5 LEDs in them, in series.
Each LED would be getting 350 mA like that.

Here is an alternative procedure. It is a lot safer.
Get a 22 ohm 10 watt resistor. Put it in series with the LEDs.
Put two 6 V lead acid batteries in series.
Try the LED across this.
Then start adding 1.5 volt flashlight batteries in series with the other batteries. You might have to tape them together.
So, you would be trying 12 V, 13.5 V, 15 V , 16.5 V.

If you can, measure the voltage across the LED when it first starts to light up, if it does.

If it does, look at the LEDs to see if they are all lighting up equally brightly.

11. Sep 6, 2010

### DATAcrusher

But that would mean each AA battery would get about 2.1 amps through them. Due to their inner resistance, then they would heat up. Likely.

However, I will try it out, but I'll wait until I get my multimeter- then I can make some diagrams, and check for all the important stuff you mentioned.

Once again. Thank you.

12. Sep 6, 2010

### vk6kro

No, that is the function of the 22 ohm resistor.
Even if it was put directly across 16.5 volts, the maximum current would be 750 mA which the flashlight batteries should be capable of supplying for a short time.
With the LEDs in series, the current should be less than this.

The flashlight batteries are to give small steps of 1.5 volts so we can see what the LED needs to work properly.
This is not the final circuit, of course.

Apart from correct voltages, the LED heatsinking will have to be very good or the LED could easily oveheat and suffer a meltdown. I would leave the 100 watt device for later. Just use the 30 watt one.

13. Sep 6, 2010