# Best battery configuration for powering led?

I have an simple 20mA 3v led that i will power with batteries. I want the led to last as long as possible ant size of batteripack is not an big issue.

I have played with some AA 1.5v 2500mA batteries. 6x solded together in series. I thought that would give me 12v and 15.000mA, but this gives me just 12v and 2.500mA.

But how can i configirate AA batteries to light up a led like this longest possible?

And how do i solder them together? I have seen an youtube video that helped me to solder 6 together in series, after i cooked my other batteries with a mistake.

It's a lot more difficult than i first thought to solder together some batteries.

Soldering batteries is not the best idea even though some people have done it. Batteries contacts are large metal areas and can conduct a lot of heat so you need a high power soldering iron. This will pump a lot of heat into the battery which they do not like.

For battery packs spot welding is usually used to attach the contacts to the battery poles. This is less stressful for the batteries.

Regarding the grouping of your batteries. Try to stay close to the voltage of your LED otherwise you will need to convert the voltage down. 2 x 1.5 would give you 3V. You can connect them in parallel to increase the time (mAh) you get out of them.
Please note, when you group batteries they should all be the same size, age and brand. When they are different you will get current flowing between them because of voltage differences which will reduce your effective run time of the battery pack.

Thanks

But say i want to encrease the current of the batteries. If i want 3v and for example 20.000mA. How do i build and connect AA batteries together? If it's possible... ?

Using a typical AA alkaline battery just as an example. The voltage is let's say 1.5 Volts at 1500 mAH 1.5 AH so that means the battery should be able to deliver 1.5 Volts @ 1.5 Amps for 1.0 hour. If I wire 6 of them in series I get 6 * 1.5 = 9 Volts. The AH rating doesn't change and will be no higher than the lowest in my series circuit. That means I now have 9 Volts 1.5 AH (Amp Hours). Placing the same six batteries in parallel I get 1.5 Volts but 6 times my AH so 1.5 Volts at 9.0 AH. That's how series and parallel play out. Don't always take the AH rating as gospel as the load has a lot to do with it.

So if you have a 20 mA 3 V LED and you take 3 1.5 Volt AA batteries with a 1500 mAH rating you get 4.5 Volts 1500 mAH. Then you have 4.5 - 3 = 1.5 / .020 = 75 Ohms so you want about a 75 Ohm resistor in series with your LED. At 20 mA you take 1.5 AH / ,020 A = 75 hours give or take.

There is more to all of it but in a nutshell it comes out about that way. You can also make series / parallel configurations. Three in series, three more in series ( 6 total) and place three and three in parallel and you now have about 150 hours on your 20 mA LED.

Ron

I have a lot of AA batteries. 100pcs brand new, same brand etc. My question is how i for example could solder together 8 batteries and get 20.000mA and 3v.

I need to se an picture of how i connect theese together to get that power.

I understand that we have to mix paralell and series, but i just want to know how to get it together. Lets assume each battery has 2500mA capacity.

You don't have to run the LED at it's full 20mA rating. If you haven't done so already try 5 or 10mA, or just experiment. And if you get a "high brightness" LED you may get enough light at even lower current.

On the other hand the LED will get dimmer the more the batter discharges so you may need to start-out with a some extra brightness.

OK, let's try and make this simple. A LED is a current device not a voltage device. If you find a good quality LED (not the 100 for a dollar type they have a data sheet. The data sheet will call out a Forward Voltage (in your case 3.0 volts) and a Forward Current (in your case 20 mA (0.020 Amp). Since the LED is a current device we make sure we limit LED current and in this case to 20 mA. I told you three AA batteries in series gives you 4.5 Volts. The formula for a series LED resistor is Vsupply 4.5 Volts - Vfwd LED 3.0 Volts = 1.5 Volts / .020 Amp = 75 Ohms. I showed you that using a 75 Ohm (82 Ohm would be common) with 4.5 Volts and 1500 mAH would last about 75 hours. Now if you want more then take three more batteries in series and parallel them with the first three so now you have 150 hours and you can string the things to infinity and last forever.

You need to understand batteries in series and parallel combinations and how a LED actually works. Too much for a simple single post. Sorry I can't make a picture, schematic yes, picture no.

Ron

You need to understand batteries in series and parallel combinations and how a LED actually works. Too much for a simple single post. Sorry I can't make a picture, schematic yes, picture no.

Yes, i know all the other you say. I just want to know how to connect several AA batteries together to get 3v and so much current 8 batteries gives.

I know how to glue batteries together and solder them. I have done that with 6 AA batteries in series and it worked. I just want to know how i connect AA batteries together who gives med 3V (two batteries in series) and more current.

There are two way you can do this.

1. Solder two AA in series and then connect multiple of these in parallel.

2. You can solder multiple AA in parallel and then solder two of these packs in series.

I suspect (without thinking about all possibilities for hours) case 1 is better in your case. It will probably have a lower loss of energy due to slight differences in batteries. Case 2 could be more effective in other use cases.

You can probably find research papers on cell balancing, because this is a topic for lots of high power battery applications e.g. electric cars, e-bikes, UAV, drones, ... .

Thanks.

Yes, i understand that. But it does not help me if i can't see an actual drawing for how.

This is how i did it with six batteries:

I wan't to to the same with 8...

You soldering skills must be better than mine. Last time I tried that, whatever's connected to the inside of the battery pip and base became dis-connected; presumably I held the iron on too long.

For any battery configuration I would add some switching regulator to keep the LED current constant. Otherwise the LED will be too bright (= consuming too much current) when the batteries are fresh and will be too dim (= not useful) when the batteries still have some power left.

Hi,
Does the LED have to be continuously ON, or can it flash.

If flash is okay, try and track down an IC LM3909.
It is designed to run an LED off single cell for long periods of time.

LM3909 LED

For continues illumination.

joule thief

Tom...

Make this easy and stop heating batteries to solder them. Go to Amazon and buy some common 3 AA Battery Holders for about \$1.00 USD each.

Then tie all the red wires together, then tie all the black wires together and that places the groups of 3 in series with each in parallel. That affords you 4.5 Volts and I already explained the resistor and why it is used. On a scale of 1 to 10 this level is a minus 3. Forget pictures and start learning how to read schematics and some basic fundamentals and formulas.

Ron

This forum need a “like” button.

I would be saving the AA cells for another use and sorting out a suitable battery to drive the project.

If you insist on batteries the attached is as close as I will come up with as far as a picture. You can learn schematics and how to read them or find another hobby. You can place as many three battery series circuits in parallel as you wish.

Ron