More AA battery salvage

[cb92]

My AA battery/joule thief lantern is working fine, now I'm trying a regular flashlight. No joule thief this time, the bulb is incandescent, and I'm really cramped for space. I do have a series/parallel setup, hope to accomplish that with a simple wire swap.
My problem is the AA battery receptacle is putting out over five volts from each receptacle. I'm afraid I'll blow the 3 volt bulb. Is there anything I can do to protect the bulb that doesn't draw too much from the batteries? Will this remedy keep the bulb bright as the batteries run down? When the batteries drop to a low voltage, I plan to switch to series.

[ratdude747]

zener diode?

[cb92]

Thanks, Ratdude, this looks good. I'll do the math to determine what resistor is necessary. I'll see if I can find one in my junk pile, but it's just as simple to get some from Radio Shaft.
This will be a good exercise for me, I can learn much better by doing than just reading text books.

[mariushm]

You could use one or several of these in parallel:

http://search.digikey.com/us/en/prod...FTO-ND/1098462

They generate 3v @ 250mA from anything up to 13v and have only about .625v drop so as long as you give it at least 3.6v it will generate stable output.

Besides the 50 cents piece you only need a couple of 1uF capacitors, see the datasheet.

or this... http://search.digikey.com/us/en/prod...FDB-ND/1635997

a bit harder to solder but can manage up to 1a and has only 0.25v dropout

[cb92]

The flashlight bulb measures 1.8 ohms. The input voltage will be 5 to 10 volts. This makes for about 1.6 amps. The second voltage regulator has more power.
My text book shows a zener diode in parallel and a resistor in series with the load for this simple circuit. Does this mean the zener consumes all the current over it's avalanche level? Does the resistor in series with the load (which will be about the same resistance as the load, apparently) also consume current? It seems this setup would draw the batteries down.
I like the idea of a regulator chopping the current to protect the bulb. This chopping should consume less power, right?
I'll check my files for schematics using these power regulators. Maybe a data sheet may have a typical schematic.

[mariushm]

You said you have a 3v bulb and you were afraid of burning it, that's why I gave you links to those regulators. There are lots of designs, with fixed voltage output as I gave you, or with adjustable output.

Digikey has a ton of those, various voltages and maximum amps they're capable - just pay attention to the dropout voltage they have. Ideally, you'd want to use a device with as low as possible dropout voltage.

The datasheets do have schematics but they're basically 3 pin devices, in, ground and out - can't be simpler than that. You just add two capacitors at each end.

These regulators are better than just diodes and resistors but not by much - their efficiency is about 75-80% - it depends on what you choose and what you feed into them.

[cb92]

I went ahead and tried the setup with no electronics. The battery pack is putting out about 5 volts, the bulb did not blow. It seems these bulbs will take extra voltage. I'll see how long it lasts. These used AA batteries should not stay at 5 too long. I did wire the system for series/parallel, so we will see just how much time I can get out of batteries that my cameras and mice rejected.