BU-802c: How Low can a Battery be Discharged?

Discover what causes short runtimes

Not all battery energy can or should be used on discharge; some reserve is almost always left behind on purpose after the equipment cuts off. There are several reasons for this.

Most mobile phones, laptops and other portable devices turn off when the lithium-ion battery reaches 3.00V/cell on discharge. At this point the battery has about 5 percent capacity left. Manufacturers choose this voltage threshold to preserve some energy for housekeeping, as well as to reduce battery stress and allow for some self-discharge if the battery is not immediately recharged. This grace period in empty state can last several months until self-discharge lowers the voltage of Li-ion to about 2.50V/cell, at which point the protection circuit opens and most packs become unserviceable with a regular charger. ( See BU-808a: How to awaken Sleeping Li-ion )

Power tools and medical devices drawing high current tend to push the battery voltage to an early cut-off prematurely. This is especially apparent at cold temperatures and in cells with high internal resistance. These batteries may still have ample capacity left after the cutoff; discharging them with a battery analyzer at a moderate load will often give a residual capacity of 30 percent. Figure 1 illustrates the cut-off voltage graphically.
 

Illustration of equipment with high cut-off voltage

Figure 1: Illustration of equipment with high cut-off voltage.

Portable devices do not utilize all available battery power and leave some energy behind.

Courtesy of Cadex 

 

To prevent triggering premature cutoff at a high load or cold temperature, some device manufacturers may lower the end-of-discharge voltage. Li-ion in a power tool may discharge the battery to 2.70V/cell instead of 3.00C/cell; Li-phosphate may go to 2.45V/cell instead of 2.70V/cell, lead acid to 1.40V/cell instead of the customary 1.75V/cell, and NiCd/NiMH to 0.90V/cell instead of 1.00V/cell. ( See BU501: Basics About Discharging )

Industrial applications aim to attain maximum service life rather than optimize runtime, as it is done with consumer products. This also applies to the electric powertrain; batteries in a hybrid cars and electric vehicle electric vehicles are seldom fully discharged or charged; most operate between 30 and 80 percent state-of-charge when new. This is the most effective working bandwidth; it also delivers the longest service life. A deep discharge to empty followed a full charge would cause undue stress for the Li-ion. Similarly, satellites use only the mid-band of a battery called the “sweet zone.” Figure 2 illustrates the “sweet zone” of a battery.
 

Figure 2: Sweet zone of a Lithium-ion battery to extend life.

Operating Li-ion in the “sweet zone” prolongs battery life because a partial cycle is less stressful than a full cycle. As the capacity fades with use, the battery management system (BMS) may engage the full working range of the battery.

Courtesy of Cadex 


Elevated internal resistance makes alkaline and other primary batteries unsuitable for high load applications. The resistance rises further as the cell depletes. This causes an early cutoff with the device drawing some current, and much energy is left behind. Primary batteries have high capacities and perform well when new, but they soon lose power like a deflating balloon. ( See BU-106: Primary Batteries )

Last updated 2016-03-07


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Comments

On December 16, 2011 at 5:26pm
Ed Vrana wrote:

On my 2.2 amp-hr r/c helicopter Li-Po battery, I land at about 3.8 v/cell and charge to about 4.2 v/cell.  Over this delta V, my charger says it “put in” about 1.47 amp-hrs to get back to 4.2.  If it could, at what voltage would the pack give up all 2.2 amp-hrs?

On January 28, 2012 at 11:17am
David wrote:

You would probably see your charger put back in about 2.2 A-h if your pack were discharged down to the 3.2-3.3 V/cell range. Keep in mind that many ESCs have a soft cut-off (which can usually be changed or turned off) that will protect you from over- discharging the Li-Po battery.

On September 18, 2012 at 6:36pm
Adam Stouffer wrote:

Is it wise to rely on the battery’s internal shutdown circuitry (2.75V) or is it better to use a dedicated circuit with a slightly higher threshold?

On January 26, 2016 at 7:38am
Daniel wrote:

Hi,

I have a 48v lithium phosphate battery pack for a scooter. One day the scooter cut-off and I thought it might be low charge. So i went home and plugged it in and it wouldn’t charge, i checked the voltage only to realise it was at 17.3v which made me think that the BMS system was shorted and needed replacing. Im not really qualified and didn’t really have the time to replace the BMS, so i just left the pack sitting there since october 2015. Came back today to check the voltage and for some strange reason the voltage was back up to 52.3v. So i charged it up to full, put it in my scooter and it ran for about 700 metres and it cut off again. My question is why would this be occuring perhaps something is loose or should i be running a fuse like the my lead acid pack has?

Thanks

Ps. Sorry for the long question/story

On February 13, 2016 at 8:40am
Marion wrote:

I have a mobility scooter with two 100ah 12v batteries. The charger is marked as AC 3.8A input and DC 24V 8A output. It says charging voltage DC 28.8V and floating voltage DC 27.6 V It also says 50/60 Hz. I am using it in the UK/Europe, so 240 v input.  The batteries should travel 31 miles (based on Maximum driving distance is based on an ambient temperature of 20°C, a 100kg driver and a brand new fully charged battery by a constant driving speed at 6 km/h with 70% battery power discharged.) Hope I have put down all that is needed.

I’ve trawled the internet but can’t find an answer to my question and so hope you could help me. I am not very knowledgeable about electricity as you can probably tell from my question. Can you help me work out how much KWh (is that the correct term?) it would cost to charge the battery overnight if I have used it lightly, such as going a mile, how much it would cost if I had travelled say10 miles and how much if I travel 30 miles. Can you show me how to work it out?

I am presuming there wouldn’t be a vast difference between the smaller ones. I know it will be a very small amount but I need the figures to pay back the cost of the electricity. Once I have the figures I can use the unit amount from my electricity bill to work out the costs.

Thank you in advance for your help.

On March 14, 2016 at 9:15pm
Gordon Cooper wrote:

A few thoughts on your scooter battery. One way of estimating would be to look at capacity p
of the batteries. When new and at full spec each has 12 volts and 100 amp hours. So each can store 12 X 100 watt hours, or 1.2 KWh. Lets guess the charger is 80% efficient, (I would expect it nearer 95% as the remainder heats up the charger and can be felt as heat). So to fully charge the battery will require 2 (batteries) X 1.2 (capacity) X 1.25 (efficiency loss) or 3 KWh. That is quite a cost in the UK! But of course that can travel 31 miles, so one could again guess that if you only travelled 3.1 miles or a tenth of the range the re-charge would only cost a tenth.

You can get electricity meters that can measure this usage, as I have ignored the small amount used during trickle or float charge,

On June 14, 2016 at 12:34am
Linus wrote:

My tablet pc has a very short battery life. I tried to calibrate it several times by charging and discharging it in full cycles. Didn’t help. Then I opened the cover and measured the voltage of the “dead” battery with an external multimeter and discovered that Li-Po battery was at 3.7V, when it was declared dead by my tablet. The tablet won’t turn on, displaying low battery icon. To my knowledge, 3.7V is about 50% for Li-Po, the real cutoff voltage has to be around 3V. o it traps around 50% of useful battery charge. What can I do? Is there any way to reset the battery counter / internal voltmeter?

On December 22, 2016 at 10:32am
Salvador Barqueros Provencio wrote:

Linus, as far as I can tell, I see nothing abnormal in what you relate. Damaged cells usually deliver 0K no-load voltages, which plummet under nearly any significant amount of load (unduly high internal resistance). That’s the reason your tablet PC’s circuitry deems it “dead”: it is dead my fellow.
Cheers. (-: