# Boosting voltage of 2AA battery to power Luxeon LED

Discussion in 'Electronic Basics' started by Daniel Towner, Dec 5, 2003.

1. ### Daniel TownerGuest

Hi all,

I'm making myself a head-torch using one of the new Luxeon LED's. At
the moment I power it from a 9V cell, with a series resistor, but this
obviously wastes a lot of power in the resistor. I've seen that some
companies are manufacturing similar Luxeon torches which are powered
from 2 AA batteries, using a voltage booster. How would I go about
building such a circuit? Are there any components I can buy which
would make the circuit simpler (e.g., IC's)?

FYI, Luxeon LED's have a forward voltage of 3.42V and draw 350mA.

thanks,

dan.

2. ### JeBGuest

would 3 batteries make life a lot easier?

3. ### Indian SummerGuest

Was thinking of making one myself. With just a couple of components, I've
tried to copy the circuitry similar to the Luxeon torches but gave up and
used all kinds of available batteries from 3V to 1.5V watch Lithium
batteries.
Indi

4. ### Daniel TownerGuest

That will be my last resort. The argument I've seen against 3
batteries + resistor is that such a setup is not as efficient.
Boosting the voltage of 2 batteries is (so I'm told) more efficient
than using more batteries and dissipating power through a resistor.
Also, since you can only buy batteries in powers of 2, 3 is a strange
number to use (though plenty of manufacturers make products which use
3 batteries, presumably because that is cheaper than making a voltage
booster).

dan.

5. ### John PopelishGuest

Voltage boost regulators are not generally as efficient as voltage
buck regulators. Use whatever number of cells provides a little more
voltage than needed, and build an efficient buck switching current
regulator for highest efficiency.

6. ### Ian StirlingGuest

This is true.
Does it matter though?
If you are using rechargables, I would consider 3 NiMH AA cells.
This is a better voltage match.
These start at about 1.48V, when they are just off charge, and
have voltage around 1.2V per cell for the majority of the cycle.
The right way is to do a switch-mode power converter.
This can have efficiancies over 90%.

The easy way is probably to take three NiMH AA cells, one power transistor
and a variable resistor, and a resistor of a tenth of the value of the
variable one.
Take the NPN power transistor, connect the base to the variable resistors
wiper, and one leg to the resistor.

Connect the resistor to the positive supply.
Connect the emitter of the transistor to ground.
The collector goes to the negative of the LED, and the positive of the LED
is connected to the supply.

The variable resistor is chosen so that it's about a third of
(Vsupply(max) - 0.6) * Transistor Hfe * .35A.

This is bad practice in general, as it's very sensitive to transistor
Hfe, which can vary greatly.
However, it provides a not-too-bad current source over the range of interest.

Set the pot (with fresh batteries) so that it draws 350ma.
It'll go down to around 3.4V (I forget the internal resistance of the luxeon
star) or a bit lower.

Anyone got a transistor suggestion?