I have recently purchased 10, 3W LED Prolight Stars (90L High Power LED’s) with a Forward Voltage of 3.3V~3.5V and a forward Current of 700mA, I have a lot of PP3 9 volt batteries and wish to use them up in a new home made bedside light, obviously if I connect directly to the battery they will burn out, I have been looking for a voltage regulator circuit and I think the LM317 or LM338 voltage regulators would fit the bill, would anyone be able to tell me if this is a good idea or is there a better way of getting 3.3volts out of a 9volt battery while maintaining battery life?

 

PP3 batteries are simply not a practical source of power for those high-power LEDs.

A typical alkaline PP3 battery might be rated for 500mAh; but that is at a 20-hour discharge rate, or 25mA. The faster you discharge them, the more power is dissipated inside the battery itself, due to internal resistance. At a 700mA rate of discharge, the battery would be dead within minutes, and very hot as well.

This is a very common mistake for new experimenters to make, so don't feel alone.

 

3.3V-3.5V Current of 700mA

Let's use 3.5V and 550mA as practical numbers, and consider that:

5.5V = 9V - 3.5V

10 ohms = 5.5V/0.55A

~3W = 5.5V * 0.55A

...that's a 10 ohm, 5W, ballast resistor for each LED. So, using Sarge's number of 500mAh, if you ran these as seperate circuits you would get about 1/2 hour or so of reasonable use per LED & battery before it became too dim (ignoring the effect of high current drain on the battery). Placing many batteries in parallel for one LED may provide a number of hours of reasonable use.

I should think that having to purchase/replace numerous batteries will end up being pretty expensive, once you run out of the batteries you do have.

 

It's worse than that.

9v batteries have an internal resistance somewhere in the vicinity of 4 to 6 Ohms, even when new. You can get a couple of Amperes out of them for very brief periods before they will overheat.

As they heat up, their internal resistance increases, so yet more power is dissipated internally. You wind up wasting much of the battery power in heating up the battery.

Under such high load conditions, you'd be lucky to get 150mAh out of a 500mAh battery.