Get a 12V COB LED to run for 2-3 hours of of 2 AA or 2 AAA batteries.

[Richmrf]

This is what I should have done from the start...
Can someone walk me through what I can do to get a 24SMD COB LED to run off of battery power?
Similar to a COB flashlight. I just need it to be as compact and light weight as possible and only need 2+ hours of run time.

I have another post where I was using 4 coin batteries but the drop off was too great. Maybe I should have started this way since it is starting to sound like it can’t be done with coin/button batteries.

 

[ronsimpson]

Need more information.
Is this the LED?

or this?

 

[t1d]

A "Jewel Thief Circuit" can help you get the very last drop of energy out of a battery and run a LED on battery voltages lower than its required rating. This is not a joke. Google it. HTH.

 

[Pommie]

According to this data sheet an alkaline AA battery discharged at 500mA will last about 3 hours. Two will give you 3V at 500mA for 3 hours. So available power is 1.5W. How many watts does your LED need?

Mike.

 

[Diver300]

A "Jewel Thief Circuit" can help you get the very last drop of energy out of a battery and run a LED on battery voltages lower than its required rating. This is not a joke. Google it. HTH.

It's spelt "Joule Thief Circuit". If you're going to Google it, that may help, although Google makes a good guess at what you want if you include "circuit"

 

[Diver300]

A lot of 12 V LED bulbs for cars have resistors in parallel with them, to make them take current at all voltages. This is to make the blown bulb detection work, as some cars try to detect the lamps with a low current when turned off. If you are running on batteries, it's a good idea to remove resistors like that.

 

[t1d]

It's spelt "Joule Thief Circuit". If you're going to Google it, that may help, although Google makes a good guess at what you want if you include "circuit"

Yes, its is spelled "Joule!" Great catch! Much appreciated!

 

[rjenkinsgb]

The step-up converter idea I mentioned in your other thread should work fine.

The example I linked to should be suitable, as long as you use a minimum of three dry cells - or one rechargeable lithium cell.
**broken link removed**

Dry cells that are nominally 1.5V start at that, but the voltage progressively drops as they discharge, down to around 1V per cell before they are completely exhausted.

Two dry cells will drop below the minimum 2.5V input for that board, well before they are dead. Using three keeps the voltage reasonable.

Or, a single 18650 rechargeable lithium cell may be simpler than multiple dry cells - and it's definitely cheaper in the long term!
The voltage range on one of those is from 4.2V fully charged down to 3.5V or a bit less when dead.

Get cells with internal protection; that avoids damage due to over-discharge if they are left connected too long.

 

[Richmrf]

Need more information.
Is this the LED?
View attachment 125432

or this?
View attachment 125433

 

[Richmrf]

I posted a picture of the LED’s I would like to use.
I bought 6 different types that would work. They are all similar to the ones in the picture I posted. They all range from 9 SMD to 24 SMD. I tried to use 4 CR 2032’s but the amount of light fell off so fast. I made a little chart of the different LED’s and how much they dropped off over a few hours.
I will also post a picture of a similar light running off of 3 AA batteries. If they can make this flashlight work, I figure I should be able to make one with 2 AAs that last at least 2-3 hours. I just don’t know what components they use to get the right voltage. Resistors or converters maybe?‍

 

[Richmrf]

According to this data sheet an alkaline AA battery discharged at 500mA will last about 3 hours. Two will give you 3V at 500mA for 3 hours. So available power is 1.5W. How many watts does your LED need?

Mike.

I’m not sure. I only know that they are 12V.

 

[Richmrf]

The description says 12V * 0.13A

 

[Diver300]

The description says 12V * 0.13A

That is 1.56 W. You might manage to get enough power from an AA battery, according to Mike's figures.

You have to allow for the boost converter losses. It won't be 100% efficient, so you might well need 1.8 - 2 W.

If you find a boost converter that gives out a constant current, instead of a constant voltage, and you set it to 0.13 A, and you short out the resistor on the COB light, the voltage needed will drop to around 9 V and you will need less power. The resistor is there to limit the current if the supply is a constant voltage. It's not needed if the supply is constant current, which is unusual, but power supplies like that are available.