Saving Your 'half-dead' Batteries





Introduction: Saving Your 'half-dead' Batteries

Don't throw away your 'half-dead' batteries yet! Do you know you can connect the 'half-dead' batteries together in series to provide a greater voltage? It would be as easy as saving all your pennies together to gain a larger sum. This instructable will show you how to make use of all the batteries with very low voltage.

Step 1: Prepare the Components

1. Half-dead batteries
2. Potentiometer/Variable resistor (I used 1k ohms and linear one in this case. It is better to use the one with a rotatory switch but I don't have one so I just used what I have got)
3. Toggle switch (optional, can be any switch)
4. Connecting wires (black and red)
5. Soldering iron and wire
6. Wire cutting tool
7. Multimeter
8. Helping hand (optional but it is always good to have one)
9. Insulating tape
10. LED for testing the circuit

Step 2: Connecting the Batteries in Series

1. Use blu tack to hold the batteries in place. (or other things to hold them in place)
2. Strip out the wires that just reaches the positive and negative terminal of both batteries.
3. Solder them together in series. (Repeat this procedure for all your batteries)
4. Remember to leave out the positive terminal of the 'starting' battery and the negative terminal of the 'end' battery.
5. Check the voltage to see if all the connections are right.
6. Wrap the batteries together with insulating tape.
7. With all my 10 batteries connected in series, I have got 10.84V

Step 3: Connect the Potentiometer and the Switch

1. Solder one end of the potentiometer with the positive terminal of the batteries pack and the middle pin to the switch.
2. Solder another wire on the switch and this is not connected to anything yet.
3. Solder a wire on the negative terminal of the batteries pack.

Step 4: Testing the Circuit

For a 3V LED, I need a 392 ohms resistor provided that the Voltage supply is 10.84V and the current of the LED is 0.02A. Therefore I tuned the potentiometer to near 400 ohms. (Notice that the resistance has to be more than what it is calculated or your LED would be burnt out.

Here is the maths: (10.84-3) / 0.02 = 392 ohms using the ohms law

NOTE: For different electronic components, it would have different voltage and current so make sure you work out what resistance you need and tune it before you flick the switch.

Finally the circuit works and I can re-use my half-dead batteries now!!

P.S. Please give any ideas that you have and feel free to make comments.



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    This is ok, but remember as Alkalines get deader, the internal resistance increases.  Every time you have to "top up" the "pile", more and more power gets wasted as heat inside the batteries themselves, only killing them faster.

    "There's a big difference between mostly dead and all dead."

    Or you could just use a joule thief and then only need to use one battery to power the led.

    Is that some kneadable eraser that I see in the pictures? Those things rock!

    no,blue tack

    This is interesting but may I suggest using 7805 (or 7803 if you can find it) to lower the voltage, because, as you keep on draining the batteries the voltage will get lower and you'll have to be constantly adjusting the resistor. Also beware that some of the batteries will become completely empty before the others and those empty batteries are likely to leak acid or even explode. Oh, and get cheap battery-holder ;-). It makes it much easier to replace the completely empty cells.

    just FYI,there is no LM7803 chip out there,it does not even come up on the obsolete section of National.

    Thanks for the suggestion. I think you are right. A voltage regulator would help to keep the voltage constant.

    quick question about "potentiometer", is it like a resistor that you can tune to what ever you need?

    Potentiometer is same as variable resistor. You can tune it to whatever value (depends on your resistor value). The one that I used here is a 1k ohms potentiometer with linear property.