battery for a led bulb

[attchoum]

for now, i only want to build a backup light. so i dont whant to us big battery pack.

i made some test whit 3 aa alkaline battery, it's enouf for me right now and it's really compact.

I think i will use aluminium pipe to make a good case.

on luxeon web site i order a good reflector perfect for the led i use, cheap and easy, i love that.

thx for your help everyone.
etienne

 

[senjaya_chang]

You can add the mAhr ONLY if you have them in PARALLEL. Since they are in series, the three 26Ahr D cells are a 4.5V, 26Ahr battery, not 78Ahr.
The other alternative is to go to 4 cells in series. Although this wastes more power, the current will be more constant over the life of the batteries. Since NiMH batteries have a flatter discharge curve, using 4 NiMH batteries will also give a more constant current, and therefore more constant light output.

Yes, you are right, my apologize of course. I do it in a rush so forgot the principal of series and paralllel.

i'm using 2 resistor, 2 diode, and 2 transistor for my dc-dc converter.
seem's like you have more simple circuit, would you please send me the circuit chart?
Thanks

 

[Charlie99]

There are a variety of simple constant current circuits (or more accurately "current limiting" circuits). These linear circuits are kind of like smart resistors that change value as the battery voltage changes. They are simpler than DC-DC converters with coils/inductors, but are not as efficient. The big advantage over the simple resistor is that they have nearly constant current (and therefore LED light output) over a wide range of battery voltage.

I don't have an easy way to draw circuits, so I googled looking for it. Didn't find the exact one, but look at http://www.radio-electronics.com/in...urrent_limiter/transistor_current_limiter.php

Remove the stuff at the bottom of the circuit, leaving just the two resistors and 2 transistors at the top of the circuit. Once the current going through the circuit gets up to the point that the voltage across the current sense resistor gets up to 0.7V or so, it turns on the "current sense transistor". That transistor steals / bypasses the base current for the main transistor above it, thereby limiting the current.

The LED load can either be on the emitter of the main transistor (in which case the base current also flows through the load), or you can put the load in series with the collector of the main transistor (better current regulation, but you lose the 5 to 10% of current that you send down through the base).

If you don't want to use 2 transistors, replace the "current limit" transistor with two diodes. If your battery voltage is close to the LED voltage, then put the LED in series with the collector of the transistor.
If you have lots of excess voltage, then put the LED off to the right of the junction of current sense resistor and the current limit transistor emitter.

 

[badgers]

To make a current source you need an LM317 from radio shack. This is called a transconductance amplifier. You can use that term if you google (http://en.wikipedia.org/wiki/Transconductance)

1.2V/1.5amps = .8 ohms

High lumen LED's have a lot of heat output. You may want to start with a 1 ohm resistor to see how your heat sinking is working. Typically the specs are quoted at 25C
http://users.pandora.be/davshomepage/current-source.htm
I have some of a similar LED, they are one generation older then the one you bought. But the concept is the same.
The constant current source will provide the same brightness with fresh batteries as with 66% charged batteries.
the back side of a Luxeon emitter is electrically connected to the "plus side" of the LED.
You will be heatsinking 3.42V * 1.5 amps = 5.13 Watts. As the junction temp goes up the lumens go down(while still using the same amount of power...
You will also be rejecting the "whole" system eventually. Lets say you have 4.8V out of the batteries, you really are going to eventually use up 4.8V*1.5Amp the LM317 Heat sink will reject most of the heat but the 1-0.8 ohm resistor will need to be rated for high wattage 1.2 Watts. Most resistors are quarter watt, potentiometers can be rated around a watt. You will need to get power resistors.


keep us up to date with your project. I am mixing red-orange and white leds in a dive light experiment. (3 white and 1 red)
Red light is missing from the spectrum at depth, so I want to add that back to give ballance to what I see. I know blue "penetrates" farther into the water. We will see how it works.

 

[badgers]

here is something
http://www.national.com/pf/LM/LM3405.html

I am going to check mouser for pricing and sourcing.

 

[mddolson]

I'm building a similar project.

As a power source I'd suggest 2 Nimh C cells at 450 mAh capapcity
= 3 hr run time.
While Lithium Ion are denser and don't self discharge,
they are not a cheap date,
nor are they big enough (yet) to be a serious power source.

The luxeon K2 is not a particularly efficient LED.
at 1.5 amp and 3.4 v it's dissipating over 5 watts, and outputing about 1/2 watt of light. You need to conduct 4.5 watts of heat from the LED.
The junction temperature (light emiiting zone) will be about 130°C.
So you need to conduct the heat away and keep the casing below 70 or you'll cook the LED.

I'd suggest the led casing be mounted on a copper plate that extends out to the outer housing (aluminum) which will act as a large heat sink.

Mike D

hi everybody

I just buy a strong led for a light project. This kind of led http://cgi.ebay.ca/ws/eBayISAPI.dll...MEWA:IT&viewitem=&item=110082006135&rd=1&rd=1

I need your opinion on what type of batterie should i use.

This led work on a 4v source. I was thinking about 3v lithium batterie, but i can find how llong i will be abel to turn on my light whit those batterie.
Any help are welcome.

thx
etienne

 

[badgers]

The luxeon K2 is not a particularly efficient LED.
at 1.5 amp and 3.4 v it's dissipating over 5 watts, and outputing about 1/2 watt of light. You need to conduct 4.5 watts of heat from the LED.
The junction temperature (light emiiting zone) will be about 130°C.
So you need to conduct the heat away and keep the casing below 70 or you'll cook the LED.

Mike D

I am not sure that the luxeon K2 is any worse then other white LED's
An incandescent runs 17-20 lumens per watt.

the K2 cat get to about 31 lumens per watt. the trick is that you have to look at losses in the led driver as well.
The reason I posted that LM3405 LED driver is that it can get 90% driver efficiency with a 2-3 led stack.
A compact fluorescent lamp can 70 lumens per watt and some premium lamps can get 80 lumens per watt(including electronic ballast losses.)

A T-5 linear fluorescent can get 95 lumens per watt.

Technically it is called efficacy for lumens per watt, not efficiency.
You are coorect, very correct that the temperature(thermal management) is critical because lumens falls off as it gets hot.