Most efficient supply of power for 12 LEDs

[luckyvictor]

I want to make something with LED decorated for my girlfriend's birthday. However I am just a newbie and would like to get some advises please.

I will need to light up 12 LED, which has typical forward voltage around 3.5V, so what is the minimum number of battery I need please? It is possible to reduce it to maximum 3 1.5V AA/AAA battery (I can try to built an easy circuit if required) and still give a good brightness please?

Thank you for reading my question

 

[IanP]

Re: Power for 12 LED

3 batteries will give you roughly 4.5V ..
Connect all LEDs with individual resistors ≈68Ω (power rating is not very important, you can use any small resistor you can get) in parallel as shown on the attached drawing ..

IanP

 

[luckyvictor]

Re: Power for 12 LED

really thank you for your advise, but I just wondering it is efficient, and also what is power rating exactly? because I came across this term quite a few times and I did a research for what I wanna do.

For resistors, I saw 1/2 watts and 1/4 watts resistors, is this power rating?

what is the difference for these two resistors if they have the same resistance?

is it one of the spec that I can find on the datasheet of any component?

thanks

 

[KoZu]

Re: Power for 12 LED

The power rating shows how much electricity can run through it without it bursting in flames. I'll assume that you're using small 5 or 3 mm LEDs in which case you'd be fine with 1/4W resistors. The bigger the power rating, the bigger the resistor's package and price tag.

As for efficiency.. Running the LEDs parallel from 3 batteries is as efficient as it gets. And other number of batteries would require some regulation which, at these voltages, is never more efficient than using just a resistor.

You didn't specify the forward current of the LEDs. When you know the current, you can calculate the exact value resistors needed for maximum brightness. Also keep in mind that for battery-powered circuits like this it is always a good idea to add a power switch.