Old batteries (I mean AA, AAA -sized 1.5v ones) usually have some residual energy, which can be used, for example, to charge mobile phones/pads. I have seen some schematics on the Net, but all of them try to connect old batteries in series without checking for really bad ones.

I was thinking about building smarter power source using old batteries, but my knowledge is not enough to choose the component for switching batteries on and off.

The idea is simple: Switch the battery out of the chain when it consumes more energy than produces. If switches were mechanical, I could have used something like the following:

enter image description here

When non-rechargable battery B1 is low on power, S1 and S2 simulatneously switch to path 1-3 (in mechanical terms it is DPDT-switch). Threshold could be residual voltage. I do not know what is optimal value, but I guess when it drops below 0.6v (without voltage), we can declare the battery dead (it can switch back at some higher voltage, say, 0.7v. These values are arbitrary, optimal old-battery energy-harvester may need them lower.).

The question is, what can be a switching component for this, which will not require a lot of energy for maintaining the state? I can use some large capacitor and voltage regulator to get the stable voltage, so switching time can be larger. However, switching S1 and S2 should be synchronous.

I have an extension of this idea to have several series of batteries, connected in parallel, but it will complicate the switching logic and may be require MCU to control it.

UPDATE: closest (but too costly) component, which in my opinion can serve the purpose is dual coil latching relay, one per battery. E.g. G6SK-2F-H. What can be more economically feasible solid-state counterpart?

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    \$\begingroup\$ Have you heard of a "Joule Thief"? \$\endgroup\$ – Ignacio Vazquez-Abrams Aug 26 '13 at 16:51
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    \$\begingroup\$ You could just connect all the batteries in parallel, each with its own diode. This way, at any instant, the strongest cell determines the output voltage while lower voltage cells are effectively decoupled - until one of them becomes the strongest and starts powering the circuit. Schottky or (simulated) "ideal" diodes can minimize the additional losses involved in this scheme. \$\endgroup\$ – JimmyB Aug 27 '13 at 10:13
  • \$\begingroup\$ @HannoBinder I think, I still need to obtain higher voltage using serial connection, but of course there could be banks of parallel batteries connected in series. \$\endgroup\$ – Roman Susi Aug 27 '13 at 14:07
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    \$\begingroup\$ @RomanSusi Dumb question: is there actually a point in designing such a circuit that consumes the last 5% of batteries' energy? The cost to design and build such a circuit I would think would severely outweigh the cost of the amount of energy you'd save. I would think you'd be much better off getting a mains switching transformer and obtaining your energy at ~$0.1 per kilowatt hour. It's certainly an interesting design problem, but I can't think of any actual application. \$\endgroup\$ – horta Jun 17 '14 at 20:36
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    \$\begingroup\$ @horta I usually carry in my pocket a small LED flashlight that uses a boost converter to drive a white LED from a single AA cell. I feed it "dead" AA batteries removed from other devices and get to find my keyhole in the dark and lost objects in movie theaters. The light output is pretty uniform until the battery is just plain too far gone, then it fails abruptly. I'd probably just buy new batteries for it and carry it regardless, but it is nice that it lives on the cast-offs from radios and such. \$\endgroup\$ – RBerteig Jun 18 '14 at 0:13

The Joule Thief mentioned in another answer is an excellent way to harvest the dregs of battery power. However, if surface mount components are not a problem, one can do slightly better.

Look for ultra low power energy harvesting boost converters from Linear Technologies, Texas Instruments and perhaps other manufacturers.

For instance, the TI BQ25504 can harvest energy down to 80 mV supply, as long as an initial 330 mV is available for starting the process. Typical depleted primary cell batteries will provide a higher voltage under no-load than under load, so getting that initial 330 mV at start-up is not difficult.

The standard application circuit from the datasheet is thus:


This is be pretty effective in sucking residual energy out of your depleted batteries down to nearly the last drop.

For something simpler but with less challenging requirements, the SparkFun LiPower boost converter, which uses the TPS61200 boost converter IC, can be easily modified to work down to 0.5 Volt supply power: The original designer of the LiPower board has posted a blog on how to do this modification.

The key advantage these solutions have over the joule thief is the high switching frequency used, courtesy the highly integrated design, and thus the much smaller inductor required. At the end of the day, they are all conceptually similar approaches, with a quantitative difference in execution.

  • \$\begingroup\$ I really like this answer, thanks! I guess, I can connect N batteries in parallel as suggested by Hanno Binder in his comment. I will try this design when I master QFN soldering. Just one question, is the use of PCB essential for operation of the circuit (due to possibly high frequences), or can I try it first in some prototype setting? \$\endgroup\$ – Roman Susi Aug 29 '13 at 16:59
  • \$\begingroup\$ @RomanSusi batteries in parallel result in unequal discharging, therefore poor results. Also, I am not sure any of those high frequency things would work without proper circuits, that too with minimum length connections on the low power / low voltage side. \$\endgroup\$ – Anindo Ghosh Aug 29 '13 at 19:42
  • \$\begingroup\$ Thanks! I like this way, even though I have not decided yet which chip to use. \$\endgroup\$ – Roman Susi Sep 11 '13 at 17:37
  • \$\begingroup\$ Interestingly, someone implemented the idea commercially: pcworld.com/article/2928997/… \$\endgroup\$ – Roman Susi Jun 4 '15 at 3:45
  • \$\begingroup\$ @RomanSusi I'm pretty impressed with the mechanical size / concept of their implementation. \$\endgroup\$ – Anindo Ghosh Jun 5 '15 at 4:14

Let me make a life a little easier with a simplification of your hard to read circuit: -

enter image description here

The simplification is getting rid of 50% of the changeover switches.

What can you use for these switches? I would consider using N-channel MOSFETs; two per switch element. You will need a supervisory circuit that of course needs to be very lower power but I don't see this as a difficulty a hindrance to efficiency.

  • \$\begingroup\$ Thanks for the simplification. MOSFETs are fine. Half of the question is how to sense the battery's open-circuit voltage is below some threshold? Can I use the battery itself for the switch? Can those be JFETs (I have 2N6027)? How to make switching synchronous? \$\endgroup\$ – Roman Susi Aug 26 '13 at 18:11
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    \$\begingroup\$ @RomanSusi your question appeared to only ask about the "switching component" but, my opinion on the supervisory circuit is that it is based on a low power analogue comparator circuit. JFETs could be a possibility but this will take some careful designing whatever FET you use. It's not a trivial design is my gut-feeling. My simplification means you don't have S1 AND S2 therefore your original concept (as per your question) of "synchronous" is irrelevant. Synchronous switching is not a problem in the individual C/O configured MOSFETs. \$\endgroup\$ – Andy aka Aug 26 '13 at 18:21
  • \$\begingroup\$ have I undestood right, that if a battery is to "self-include" / "self-exclude" itself needs much more careful design than if I have some externally-powered (for example) comparator-based supervisor circuit, which will control MOSFETs? Anyway, it seems that I can already try some experiments. (At some time I even though of some solenoid pushing bad battery out upon complete depletion ;-) \$\endgroup\$ – Roman Susi Aug 26 '13 at 18:30
  • \$\begingroup\$ @RomanSusi it's tricky driving the FETs switching the batteries at the higher voltage end of the stack. I'm not sure what voltage it could go to when all are inseries (despite them being nearly knackered batteries). At the lower end it's easier but at the high end you've got to take care not to damage the gate of the FETs when your turn-off voltage drops to ground - you might find the gate-source max voltage is excceded. \$\endgroup\$ – Andy aka Aug 26 '13 at 20:30
  • \$\begingroup\$ This is true if supervising circuit has "global" ground. I was thinking of making it in chunks, if possible and feasible, even one per battery. Then FETs will always deal with voltage of no more than 1-4 times batteries' voltage. In any case, I was not considering this question as solely about choosing switching element, but in connection of how to sense residual voltage of each battery to control switching. \$\endgroup\$ – Roman Susi Aug 27 '13 at 3:35

As Ignacio points out 'the joule thief' is well known for sucking the final remnants of power from otherwise dead cells.

enter image description here

  • \$\begingroup\$ Thank you. This one is interesting device and I will definitely try to build one, but I need to connect dozens of batteries (and I imagine it may be hard to synchronize joule thiefs) to achieve 5v voltage. However, en.wikipedia.org/wiki/Joule_thief article mentions something about depletion mode JFETs used to similar end (and thermoelectric power sources). \$\endgroup\$ – Roman Susi Aug 26 '13 at 17:18
  • \$\begingroup\$ Actually, I have tried this. Helthier batteries produce nice 40-70kHz Vpp about 3.4v, but batteries below 0.3v fail to ignite oscillation at all. Maybe, I have too thick (~AWG 23, 0.6mm) wires in the coil or not enough turns (just 15 on each side of transformer). \$\endgroup\$ – Roman Susi Aug 26 '13 at 18:18
  • \$\begingroup\$ @RomanSusi you might be interested in this - overunity.com/13175/… \$\endgroup\$ – JIm Dearden Aug 26 '13 at 18:35
  • \$\begingroup\$ Yes, this is interesting. Waiting for parts to try. That 2SK170 seems to be obsolete (a datasheet watermarked it as not for new designs), but still orderable. \$\endgroup\$ – Roman Susi Aug 27 '13 at 6:03
  • \$\begingroup\$ The schematics is not coherent with the physical layout: in the former the resistor is connected between the base and a winding, whereas in the latter it is between the positive terminal of the cell and the winding. Besides the incoherence, do they work the same, or is one of the two actually wrong? \$\endgroup\$ – Lorenzo Donati supports Monica Jul 6 '14 at 11:53

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