I'm working on a low-low power system, which is powered by a little solar panel and a Li-Po (Lithium-ion Polymer) battery, which can be \$20\$ or \$ 50\:\mathrm{ mAh}\$. The capacity by now is not important, since it's only a buffer to store the energy given by the panel, that should guarantee autonomy.

Today we were discussing the fact that Li-Po batteries have a leakage, that somewhere (I don't know the source) is indicated to be the 20% of the capacity in one month. So at first glance, seems that the smallest battery, as long as it can store the necessary energy to survive when there is no light.

But I'd like to measure how much is this leakage; the idea was to feed the battery with a small current (around \$ 1\:\mathrm{ \mu A}\$) and check if the battery voltage increases or decreases over a long time. But I'm not very aware of the chemical processes happening inside, and I think that this technique may involve other factors. So there is a better way than charging the battery and waiting to see it discharge?


For reference, the battery we're currently using are Fullriver toy series 301213HS10C and 501417HS10C (same page). As can you seen, the manufacturer doesn't give these (and many other) specifications, so I want to measure it by myself.

The battery is charged by a LTC4071 shunt battery charger which has a really small (550 nA) operating current and which we have measured having less than 10 nA of leakage when the battery voltage is higher than the panel one.

And for me \$20 \:\mathrm{ \mu A} \$ are critical, since the panel in the worst case we consider (about \$50-100 \:\mathrm{Lx}\$, wich is in a room with no direct light on the panel) deliver \$10-20 \:\mathrm{ \mu A }\$, and the module consumes about \$10 \:\mathrm{\mu A} \$, taking the duty cycle into account.

  • \$\begingroup\$ I can report my study of silver coin battery loaded with 6.5 digit voltmeter (few days experiment). Certainly the battery at such resolution acts as thermometer, barometer, infra microphone, detector of everything. But what the worst discovery is that voltmeter itself has tiny electro force spilling back into battery in pA-nA range, so experiment failed. Wait it was input of voltage follower opamp spilling back. \$\endgroup\$
    – user924
    May 31, 2012 at 1:37
  • \$\begingroup\$ @RocketSurgeon: thanks: I suspected that there was something wrong in keeping the voltmeter attached, but I don't know the entity of the problem and the workarounds :) \$\endgroup\$
    – clabacchio
    May 31, 2012 at 6:53
  • \$\begingroup\$ The solution will involve some ideal diode and rare kind of FET amp. But part were not available for me at the time. So it is not a dead end problem, it is possible to solve. The real problem is trusting noisy voltage data over long interval of time trying to factor out things expected and unexpected. Pure science sort of. Oh. May be to run 2..3 identical batteries for placebo blind method. Say one is loaded one is not. And subtract and compare. Sort of synchronous detection method to factor out temperature variations and tidal forces of Moon \$\endgroup\$
    – user924
    May 31, 2012 at 12:21

1 Answer 1


The discharge rate dependes of several factors and will vary from manufacturer to manufacturer.

A good Li-Poly typically has 2-5% self-discharge rate per month. If it has a PCM (protection circuit module), that module will draw some current, expect less than 1uA for such a tiny cell. Parts of lesser quality may draw 30uA or more.

Self-discharge increases with temperature. I believe it also varies with cell construction (cylinder, prismatic etc.) Quality varies a lot from manufacturer to manufacturer so you have to ask for the exact specs.


I don't have much experience working with the current levels you mention. Wouldn't the ground current in the charger consume most of the 10-20uA? There is a lot of interest in energy harvesting and someone else can probably give a better answer. Slow-charging the battery is OK provided you cut off at 4.2V. My guess is that 10-20uA is not enough to offset self-discharge and other losses, but I'm frequently wrong.

  • \$\begingroup\$ I don't know exactly whay you mean with "ground current in the charger", but we are using a LTC4071 shunt charger which has a 550nA operating current, so it shouldn't be a problem. And if you're right (20uA is not enough) we may be in troubles :) \$\endgroup\$
    – clabacchio
    Feb 23, 2012 at 11:48

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