first of all: I know that a completely discharged battery is damaged permanently. I know it experiences irreversible chemical changes. I know it might blow up if you try to charge it. But that's not at all relevant to my question:

Is it safe to discharge Li-Ion to 0V and short it permanently afterwards?

Point of this question is: I do recycle lot of old batteries (Smartphones, Laptops, ...) and of course I also have to dispose some of them from time to time. But before I dispose them, I have to store them... And I personally want to store them in the most safe state possible.

Storing them "discharged" to ~3V with the device they were used in still preserves lots of energy in the battery. If I store it with ~3V and, let's say, accidentally drop a huge nail on it, it might still go bang.

Discharging it to 0V and shorting it out, I feel a lot safe - at least I think so. This ensures there is no electrical energy in it left and thus no short circuit current might get it to explode. However, I know pretty much nothing about the chemical changes in the cell and what actually might happen if the battery discharged to 0V (and shorted) might get damaged by a nail or if it even might blow up by its own.

So, is it safe to discharge a Li-Ion battery to 0V and short it? Is it even better than disposing it with 3V charge left (if you have the know-how to discharge it properly of course)?


  • 2
    \$\begingroup\$ I do not have any source so I'll just leave a comment, but a controlled complete discharge is necessary for disposal, as far as I know, so it should definitely be safe. \$\endgroup\$ Oct 21, 2017 at 13:44
  • \$\begingroup\$ I don't have any source either, but the cells don't short until they've gotten a somewhat negative charge. Draining them to 0V supposedly does not destroy them, you'd have to keep forcing current through the cell. \$\endgroup\$
    – Oskar Skog
    Oct 21, 2017 at 14:18
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    \$\begingroup\$ If a battery is s/c, the partially charged cells will drive the fully discharged cells in reverse. This is bad and may cause leakage, bursting, and, in theory, explosion. cells may be stored s/c. In spite of this, it may be best to store the batteries with a high-impedance discharge strap, and simply accept that storing large quantities of Li-Ion is not risk free and should be done with care and attention. Cap and insulate batteries before shipping: do NOT ship with a discharge strap. \$\endgroup\$
    – david
    Oct 27, 2017 at 3:13
  • 1
    \$\begingroup\$ If you discharge each cell independently, and make sure the current is not excessive, it is safe. At least I have never heard anything to the contrary. I have certainly done it before and didn't have any problem. \$\endgroup\$
    – user57037
    Nov 8, 2017 at 0:06
  • \$\begingroup\$ Companies that make Lithium-Ion battery charger ICs say that discharging to a voltage less than about 3V causes some Lithium ions to convert into molten Lithium metal that shorts the battery. Then charging causes a high amount of heat and a fire or explosion. \$\endgroup\$
    – Audioguru
    Feb 16, 2023 at 23:14

5 Answers 5


Storing them "discharged" to ~3V with the device they were used in still preserves lots of energy in the battery.

At 3V a Li-ion battery has almost no capacity left.

In this graph two 800mAh batteries were discharged at various rates. At 0.1A there was virtually no capacity left at 3.0V. Even at 1A they were 99% discharged.

enter image description here

  • 3
    \$\begingroup\$ This, though correct, doesn't answer the question regarding the safety of a 0 volt battery. \$\endgroup\$
    – Sam Spade
    Nov 11, 2017 at 11:02
  • 2
    \$\begingroup\$ The OP was worried that a battery discharged to 3V or below might still hold enough charge to 'blow up'. This graph shows that even at 3V the amount of charge left is negligible (if the battery is discharged at a reasonable rate). I have proved this experimentally by hammering a chisel through a large number of Lipo batteries that were discharged to 3V per cell and below (got a nasty surprise when I chiselled through a battery I had forgotten to discharge - won't make that mistake again!). \$\endgroup\$ Nov 11, 2017 at 20:44
  • \$\begingroup\$ I'm not going to ask why you chiseled through a battery, but I agree and have upvoted accordingly. \$\endgroup\$
    – Sam Spade
    Nov 12, 2017 at 9:45
  • \$\begingroup\$ then how are we able to discharge the phone to almost zero, and then recharge it, and the battery still works? \$\endgroup\$
    – vikrant
    Jun 25, 2020 at 17:07
  • 1
    \$\begingroup\$ @vikrant When your phone shows the battery percentage as 0%, the battery actually still has a lot of charge in it, but the phone is designed not to ever use that charge. \$\endgroup\$ May 4, 2021 at 1:27

The problem with zero volts

It is safely impossible to drop an ideal battery to zero volts. A battery cannot go down to zero volts because of the internal chemistry. In a standard use, you cannot drop the voltage below 2 volts, even if you wired the terminals together. Batteries will vary between 3.8 and 2.4 volts per cell. As voltage drops, internal resistance rises. The higher the internal resistance, the lower the current over the short circuit. I'm not personally sure what the lowest possible safe voltage for a Li-Ion cell is, but as the voltage approaches that lower limit, current will drop to almost zero. See the end of this post for a more detailed proof of this.

NOTE: The above is true for a perfect battery in a perfect world. In reality, you will badly damage the battery quickly after shorting it. At this point, internal resistance, current, and the energy difference between the half cells will all cease to matter.

enter image description here

(I know this graph is an alkaline, I couldn't find a diagram for a Li-ion, I assure you it looks the same)

A safe battery is a dead battery and a dead battery and a dead battery is approximately 2 volts.

If you have dropped the voltage to zero, I can tell you that you have done more than neutralize the cells, you have fundamentally altered the structure of the battery. Li-Ions are sensitive and finicky. I couldn't guess at what exactly is happening inside of a 0V battery, but I can prove to you that it can never get there (see end) and the fact that it has indicates that your battery is now in an unsafe state.

I do like what the other answer said: at 2 volts, the internal energy is ~0. This is true, and a good way to think about it.

What safety measures can I take?

As for storage, I understand wanting to store them safely. If you have concerns, there are 2 things you can protect against: fumes and fire.

To protect against fumes, store either in a well-ventilated area, or a sealed container. A lock-n-lock works well.

To protect against fire, a cinder block with a piece of tile or a paving stone on the top and bottom works well.

Regarding electrical energy, I can tell you that, unless you are talking about a battery for something absolutely massive, the electrical energy in the battery is a relatively small danger. It is the volatile nature of the chemicals which should be your biggest concern.

In summation, shorting batteries is never a good idea. Lithium-Ion batteries were designed to be stored at 2-4 volts. Use them as they were designed to be used.

Why can't I drop it to zero volts?

EDIT: Okay, so a lot of confusion seems to be coming from the battery indicators that all of us 21st-century folks are familiar with. These battery indicators are not voltage indicators and, while measuring voltage is a part of the process, determining the percent of charge remaining in your battery is not straightforward. If you used a laptop 10 years ago, you might remember the flaky battery indicators. They were flaky because calculating remaining battery life is absolutely not straightforward, but I will explain it:

Step 1: First, we need to look at the amount of current the phone is presently using and the voltage at the battery terminals.

Step 2: We can use this information to determine the internal resistance of the battery. When the internal resistance reaches infinity, the battery is completely depleted. Note that this never happens. It would take an infinite amount of time. Still, your phone knows what it's minimum voltage and current requirements in order to function. Now, if we imagine internal resistance is a resistor inline with the battery, it becomes obvious that, as that resistor's value grows there will come a point where \$ V_{battery} - R_{internal resistance} I_{minimum current} < V_{minimum voltage to run phone} \$.

Step 3: Now, we know that the internal resistance has an upper limit, and we know what the internal resistance is right now, but figuring out what percentage of battery life we are at still takes some work. The problem is that internal resistance doesn't increase linearly with consumed power and the curve that it follows is dependant on the specific battery you are using, no two batteries are exactly alike. Thus, your phone constantly tracks the battery's internal resistance and power output and uses that to constantly adapt it's battery life predictions.

Want proof? Have you ever had your phone stay on 1% battery for like 20 minutes? Or sometimes die at 2%? There you go. It's not an exact science, it's an estimate. And it's definitely way more complicated than just measuring voltage.

The scientific reason you can't drop it to zero volts

A battery is composed of two half cells. One half-cell contains dissolved and solid reactant A, the other dissolved and solid reactant B. A transfer of electrons from reactant A to reactant B will cause A to dissolve and bond with a salt, and cause B to be detached from a salt and solidify. For any given chemical reaction, there is a set amount of energy associated.

The hydrogen half cell has a potential of 0 volts, a lithium half cell has a potential of -3.04 volts, a sodium half cell has -2.71 volts. see here for more.

The reason we see the voltage decrease as the battery discharges is that the availability of chemicals in the half-cell decreases, which means that electrons will have more difficulty getting from wherever they are in one half-cell to where they need to be in the other half cell. Imagining we had two half cells each the size of a pop can and one atom of dissolved reactant A in one and one atom of solid reactant B in the other, you can imagine you aren't going to get a hell of a lot of voltage, most of the energy of the reaction will be expended just getting the electrons to the right place.

This rarity of reactants as the battery discharges means that the electrons must do more work to get from one cell to the other. This manifests as an increase in internal resistance and a decrease in CURRENT at the expense of maintaining the nominal voltage. I suppose I could begrudgingly admit that after billions of years of being connected, it is possible that you could get to zero volts when every single atom of A has been used, but the internal resistance at that point would be trivially huge, the current trivially small. Sufficient to say, after only a few minutes or hours, you would have a nominal voltage of ~2 volts.

I feel the need to clarify that I am aware this does not fit empirical data (i.e. that the voltage can be dropped to zero by wiring the cells together). I understand that. The battery ceases to behave this way because it has been badly damaged.

Still not convinced...

Okay, you have this scheme to slowly sap power away. You can't, or rather, you already have. Once it reaches a certain lower limit (close to 2 volts) you can no longer draw significant current from the battery. There are only ppm concentrations of reactants left and there is not enough of them to produce a significant current. Measure the resistance of a Li-Ion battery while drawing from it at a constant current. I searched for the graph online, all I found was Alkaline batteries, but the graph is the same for a Li-Ion. As you draw more and more, the internal resistance will reach a vertical asymptote, growing to infinity.

What actually happens after that? What happens when you try to draw more power from a battery than it can actually supply? I don't know. There are too many variables to accurately predict the reactions, breaches, etc. that could potentially take place. All I can tell you is that there is a limited amount of current in a battery, but that current will always come out at a constant voltage.

The idea of power always coming at a constant voltage seems to be troubling to you, so I ask you to think about it like this: 2 9 volt batteries have MORE voltage than a car battery. Furthermore, you can hook up 100 car batteries in parallel and still only get 12 volts.

This is because the cell-voltage is a function of the reaction: the two chemicals that are in the cell. If you made a car battery cell the size of a grain cylo, it would be 2 volts, because the reaction is two volts. If you made a car battery the size of a dime, it would be 2 volts because the reaction is two volts. Because a given electron will release a given amount of energy as it moves from point A to point B.

That said, how many electrons it is capable of pushing out at once is a function of size and a function of capacity. As the battery becomes 'dead,' it will be able to push fewer and fewer electrons as it runs out of reactant. In a billion years, it will have zero reactants left, but the reaction that isn't happening will still be a ~3-volt reaction.

This is epitomized in the unit of the Volt, which is: $$\frac{N m}{C}$$

The \$Nm\$ is a unit of work. The \$C\$ is a certain number of electrons. A volt is, therefore, a unit of work per electron-a quantity representing how much work a given electron can do. Whether there are 3 electrons or \$1 \times 10^{12}\$ electrons, they do the same amount of work each. A battery therefore only decreases in voltage because it will gain a certain internal resistance and electrons will need to do more work to get to the terminals of the battery. The reaction still releases the same energy on a per-electron basis.

I understand that this concept is difficult to understand and that there is a strong tendency to think of battery voltage as being related to the magnitude of a battery, and to how "full" it is in terms of a percentage value. Nonetheless, it is not an accurate reflection of how batteries operate, and for them to operate otherwise contradicts the very fundamentals of electrochemistry.

If at this point, you are still unconvinced, I must advise you take a course on electrochemistry, the Wikipedia page is very helpful, and I am sure there is an endless supply of YouTube tutorials on the subject matter.

But I tried this and didn't have a problem!

Cool. But the question is not "can it be done safely?" Sure, maybe there is some way to get a Li-ion to be at zero volts without emitting fumes (which you may not, for the record, be able to detect until you get sick from them). The question is not whether or not it is physically possible to do this without any explosions, the question is regarding safety. Although you may be able to do this, and although it may be safe under some circumstances, it isn't any safer than simply leaving them at 2 volts and, I would argue, that there are more risks involved.

Ultimately, it's up to you, but I can think of many reasons it is unsafe to discharge batteries this way, and see no benefit to doing so.

Please upvote or mark correct if you found this answer to be helpful

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    \$\begingroup\$ If you leave a resistor connected across the battery terminals for many days, I can guarantee you that the voltage will asymptotically approach zero volts. Either that, or you will win the Nobel prize for inventing a source of infinite energy. \$\endgroup\$
    – user57037
    Nov 8, 2017 at 0:08
  • 1
    \$\begingroup\$ Why do you think that Li-ion is much safer with 2V than with 0V left? \$\endgroup\$ Nov 8, 2017 at 6:24
  • 2
    \$\begingroup\$ Suppose I discharge a cell by sequentially connecting lower and lower resistors. In no case do I allow large currents to flow (always less than 1C). Eventually, I get down to a 1 mOhm resistor, which we can consider to be a short circuit. I leave the short on the cell for two or three days. Do you believe that current is still flowing in that short after two or three days? Do you believe that the voltage will rebound to 2V when I remove the short? Do you have any references to back this idea up? \$\endgroup\$
    – user57037
    Nov 10, 2017 at 20:26
  • 3
    \$\begingroup\$ The third sentence in your answer says "in a standard use, you cannot drop the voltage below 2 volts, even if you wired the terminals together." I think this warrants a downvote as being absurd and counter-factual. I don't think I am an asshole, but most assholes don't, in my experience. \$\endgroup\$
    – user57037
    Nov 19, 2017 at 22:34
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    \$\begingroup\$ @vikrant I have updated my answer. Check the Why Can't I drop it to zero volts header. \$\endgroup\$
    – Sam Spade
    Jun 26, 2020 at 1:11

Almost every Li-ion battery has copper as anode current collector. When copper is exposed to high anode voltage due to high discharge, the copper dissolute in to the electrolyte provoking internal electrical resistance rise. If the discharge is really deep (i.e. you leave the battery short circuited for two days or so) your battery will became just an useless electrical resistance (seen from terminals). But however you will always see some recover of the open circuit voltage. Once the battery is deeply discharged I would apply a permanent short circuit as safest condition.


Sorry to reply three years after your question, but if anybody read this page, there some simples rules to absolutely follow, related to your sentence : "of course I also have to dispose some of them from time to time. But before I dispose them, I have to store them... And I personally want to store them in the most safe state possible."

First rule to always follow : the safest state of a lithium battery IS between 40% and 60% so never store it higher than that, or below that.

Disposing is not storing it. Storage must be done in the safest state, and it 40-60%, certainly not 0V or even 3V

I hope everyone reading this thread will follow this rule in the future ;-)

  • 1
    \$\begingroup\$ Well, I guess you have to differentiate between storing for later use and storing for later disposal. I clearly meant storing for later disposal and I don't think storing damaged cells at 60% charge is a good idea because if they develop a short, they have a lot of energy to dump. So I guess you missed the point here. \$\endgroup\$
    – xsrf
    May 4, 2021 at 7:35
  • \$\begingroup\$ The OP want to store the batteries, but anyway, I don't see how they would have more chance to develop anything at 60% since it's their most stable state. If the battery is removed from its device and stored, it can only developp any problem by itslef, and generally, any problem resulting from something that happened while in the device would show in the next hour/days at max. Being out of this range is going out of the most stable state, and giving the possibility to generate a problem later. And better storing the batteries in a fireproof container, ideally several small ones. \$\endgroup\$
    – Gigi
    May 5, 2021 at 12:32

Short circuited battery has zero volts between terminals „by definition“ , because there are no two terminals anymore, but only one.

The guy is talking about fundametal things. The real world is what a measurement tells you. For a battery with infinite internal resistance with ANY open curcuit voltage WILL show you zero voltage under ANY small load. A standard voltmeter eats microamps to milliamps if power. So, the battery voltages drops to zero at the moment of connection of a voltmeter to terminals.

Other thing the guy talking about fundamental things forgets is that real battery can be represented by rather complex circuit of ideal elements, like voltage source, current source, resistor, inductor and capacitor. Therefore, dead battery can show you almost any voltage between -4 to 4V after a misuse and at a certain conditions.

  • 2
    \$\begingroup\$ Who is "the guy"? All of the posts have the author's username below them. \$\endgroup\$
    – Transistor
    Jan 11, 2021 at 0:16

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