Working on a long-service device powered by Li-SOCL2 primary battery. I've gathered safe cutoff voltage for this battery chemistry is 2V, so due to that and other reasons I'm aiming for 2.3V cutoff. Probably via support circuitry of my own.

Don't have much Li battery experience. So how low of a post-cutoff current drain is considered safe? The device is basically just a low-q voltage regulator and a microcontroller. Say the micro monitors battery voltage (it has an internal voltage reference & comparator) and when cutoff time comes it shuts down all processes and puts itself into eternal sleep mode. The ongoing circuit current draw afterwards should be well under 2 uA.

Obviously no concern for the undervoltage damaging rechargeability. But is it "dangerous" (long-term overheating, pressure bursting, space-time ripping) to continue discharging it at such low current? My very sketchy math seems to say that's a lot less than the inherent self-discharge of the battery. If that even has relevance.

I've reached out to several battery manufacturers but no replies.


1 Answer 1


The 2V cut-off voltage is just an arbitrary default used by manufacturers to determine the cell capacity. The assumption is that attached circuitry will not be able to operate with less than 2V, which means that the battery is considered to be drained at that point. If your circuitry can operate with less voltage than that, you can extract just a tiny bit more energy from the cell.

As you already mentioned, a depleted primary battery is going to self-discharge beyond the cut-off voltage anyway. It would be pretty bad if discharged batteries suddenly became dangerous after a few weeks or months of storage. That's luckily not the case.

If you're really unlucky, the cell might leak, but that's always a risk regardless of state of charge.

In the particular case of the LiSOCl2 chemistry, the mechanism by which such a cell gets depleted is that it uses up the lithium metal contained within it. Once that's gone, the cell can no longer produce an output voltage and essentially turns into an inert stainless steel container holding a solution of lithium salts and sulfur oxides in thionyl chloride. No further chemical reactions can take place (unless you attempt to charge it, which you should not). This rules out both overheating and overpressure.

You can discharge that battery as low as you want.

  • \$\begingroup\$ Thanks for the reassuring news @Jonathan S. I was (and am still) confused by the barrage of info / disinfo floating around. Which as you noted above didn't seem to make much sense, particularly in this case. But there were enough seemingly-wild claims to make me start wondering if even a tiny excess draw over the self-discharge level would somehow lead to an extinction-level catastrophe. Good to know my device won't lead to that. Well, at least not for this reason. \$\endgroup\$
    – Drone601
    Commented Jan 31 at 21:53

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