Battery Drainer Circuit

Discussion in 'Power Electronics' started by Dustin Smith, Dec 9, 2012.

1. Dustin Smith

52
0
Jun 27, 2012
Hi. I'm no electronics wiz, not even an apprentice of much. But I created a circuit to finish draining off my old AA batteries that don't have enough juice to power my other electronic devices. Perhaps I'll call it, Battery Squeezing Flashlight. ha Anyway, the point is, I wonder if there is a way to control the voltage coming out of the batteries automatically. I tried to read up on voltage regulators in Wiki, but I'm confused.

I'm going to try to load a picture of my circuit. Right now, i'm using a potentiometer and a volt meter to adjust my voltage to 3.6 V (or just under) before flipping my switch on. It's not a real handy way of doing it. I may install a volt meter into the circuit for convenience, or maybe some type of componet that will light up at abot 3.5 or 3.6 V so i know when I've got the correct voltage. But I'm hoping someone can tell me if there is an auto way to regulate the voltage to 3.6V.

By the way, in case you didn't figure it out, I got a 4 pack of dead (or at least too dead to do anything else) batteries for my cells. I guess this circuit will drain them down to about 0.9V each before the lights start dimming. (As long as I keep adjusting the resistance when my lights dim too much.)

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2. Harald KappModeratorModerator

12,457
2,985
Nov 17, 2011
With your circuit you're wasting a lot of energy in the resistor / potentiometer.
For light loads (e.g. an LED) look up "joule thief".

3. Dustin Smith

52
0
Jun 27, 2012
Cool, thanks for the information. I always wondered what a joule thief was. I watched this video. It was very easy to follow and understand.

However, it still has a resister in it, in fact, the resister value in that video example is 820 ohms, whereas i run my potentiometer anywhere from 4-5 ohms up to around 50 ohms. I'd have to run it about 150ohms on new batteries, which my potentiometer will do, but this circuit is for draining some of the remaining juice out of old batteries. So I think I'll stick with a potentiometer for now.

But I'd like to see if a linear voltage regulator would be better. However, I read that they should be at least 2 volts above the voltage you are shooting for. So if I got 4 batteries in series with a total of around 4.8 V and I want them down to 3.6V, I guess that wouldn't work very well.

My main point in trying to fix this circuit is my fear of turning the potentiometer the wrong direction on accident and blowing out my LEDs. Which I haven't done so far, but I've only been using it for a couple days. It's bound to happen sooner or later. I'll make that goof, I'm sure of it.

4. BobK

7,682
1,689
Jan 5, 2010
Read more about the Joule Thief. It drains the battery down to a few tenths of a volt while still lighting the LED far more efficiently than your circuit, without a need to keep adjusting it. The resistor in the Joule Thief is used for a different purpose than in your circuit.

Bob

5. Dustin Smith

52
0
Jun 27, 2012
Well, I guess I'll just have to build one and run some tests. Like I said, I'm no wiz! lol

6. Dustin Smith

52
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Jun 27, 2012
Ok, I built a joule thief circuit like the one in the youtube video I mentioned. And I built another circuit with a resistor in it. Here are the test results.

1 battery 1.33 V in joule thief circuit started at 28.5 miliamps and slowly raised to 31.33 after a ten minute test. It appeared to be stable between 31.28 - 31.38.

4 batteries in series totaling 4.15 V with a 30 ohm resistor started at 7.8 mA and dropped to 2.13mA and still dropping after a ten minute test.

I checked the light output by setting the LED being tested at a range of 1 foot and measuring the inner and outer most lighted circle. Both produced light within 1mm difference in the inner circle, and 5mm difference on the outside. I'd call that negligible.

So my conclusion is as follows.

A joule thief only requires one battery. This will save on weight and size of building my device. It does not appear to be dangerous to the LED for overpowering and blowing it out. I do question weather it would reduce the lifespan of the LED. It uses 95% more current than a single resistor.

Using a resistor, you need 3 to 4 batteries, and cannot drain them down as far as the joule thief could, unless you continued to hook them in series, thus increasing the size and weight of the device. However, it does use 95% less current than the joule thief.

So I think I'll stick with my resistor idea for now, and use the joule thief to finish off those batteries when 4 in a series isn't enough to keep my device going. Thanks very much for your useful advice. I've learned a lot.