I have a dead Li-ion battery from a Wifi device. Since I didn't use it for a couple of months, Its dead now. I have read somewhere that boost charging is used to charge and bring the battery back to its original capacity. How much volts and current should I supply to bring it back to life?
Electronic – How to boost charge a dead 3.7v li-ion battery
batteriesbattery-charginglithium ion
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Assessing full charge is the easy part.
Method (a) A fully charged Lithium Ion single cell battery will have an open circuit voltage of about 4.2 Volt*. (4.1 to 4.2 OK. 4.0 not quite there. 4.3 - a bit high.) Some cameras use two cells - double the expected voltages. Laptops and other larger devices use 3 or more cells. The voltage should be a multiple of the above voltage. [*There are variants that allow higher voltages. Unless you are CERTAIN that this includes your one, assume that it doesn't. Getting it wrong can be 'upsetting'.
(ie N x (4.1 to 4.2V))Method (b) Use a good quality charger (eg one supplied by camera manufacturer or one of known quality) which has a "charging light.
Place "charged battery on charger". Depending on how long since it was last charged the charge light should either flash or perhaps remain on for a minute or two and then go off.
Remove battery from charger. Wait 10 seconds. Place battery back on charger. Charge light should flash very briefly and go out.
Assessing capacity is harder, but not hard.
(a) you can get some indication, for nominally equal batteries, from the weight. A significant part of the weight in a LiIon battery is actively involved components whether electrically or mechanically (separators, conductors, electrolyte & (of course) Lithium metal. Two batteries of the same nominal capacity should have similar weights. I'd guesstimate that a 10% difference may be due to happenstance and construction, but beyond that I'd be suspicious. In larger & heavier batteries this test will work better than for very small batteries.
For interest, for AA NimH cells this is an excellent indicator. Modern high capacity AA's which claim 2500 mAh + capacity should be in the high twenty gram range - say 26 grams plus with some just over 30 grams. Anything under 20 grams is a complete dud and anything 25 grams or below is suspect.
(b) For any sort of accuracy you need to discharge the battery to an "end point" and measure capacity. No other method reasonably available to you is available. There are other methods such as measuring the change in voltage over a given time under a given load and trying to assess where you are on the discharge curve. This is difficult to get right and needs experience and a degree of luck. Measuring discharge time is "easier".
Best is a constant current load, which can be made very easily with eg an LM317 and one resistor, but I'll assume for now that you don't want to do that. Ask if interested.
A discharge resistor that takes at least one hour to discharge should be used. You could use a motor or lamp or camera or ... but a resistor has some advantages.
R minimum ~= (Cells_in_battery x 4000) / mAh
eg if you have a 1 cell battery (Voc=~4.2V) of 1500 mAh capacity then
- R = cells x 4000 / mAh = 1 x 4000/1500 = 2.666 ohm ~= 3 ohm or 3.3 ohm (std value)
Use the next largest resistor than the value calculated.
Up to Several times larger is OK BUT it will take proportionally longer.
Resistor power rating: Resistor power = V^2/R = (4 x number of cells)/R
eg for the above single cell and 3 ohm resistor the minimum wattage rating is
- 4 x 1 / 3 = 1.333 Watt.
Use a 2 Watt or greater resistor.
Method:
I'll describe this briefly as I don't know your experience level. This may be easy to follow or hard. If hard, ask more questions.
- Attach temporary wires to battery terminals. Two paper clips bent at end resting on terminal is flat and accessible and held with weight or tape. Wires inserted into connector id not openly accessible. Some batteries will not provide power until you give them secret handshakes. but most will.
Battery with accessible terminals.
Below: Harder to access terminals. Two dress making pins or two wires can work here BUT DO NOT SHORT TOGETHER !!! IF YOU ARE NOT COMFORTABLE DOING THIS DON'T DO IT.
- Monitor battery voltage throughout. Multimeter connected to battery wires and set to appropriate range.
http://t2.gstatic.com/images?q=tbn:ANd9GcR4lcHSRViGF_kk58tbzmBWf9G11VxLY3J45qj0lW-_spRMZIiDNg
- Connect resistor to battery leads. Start a timer. Monitor voltage. Stop at 3.2V per cell. DO NOT DISCHARGE BELOW 3 VOLTS PER CELL. STOPPING AT 3.2V IS A "GOOD IDEA". A LiIon battery may be damaged badly by very deep discharge. Set a timer. DO NO leave this running and walk away.
Below: Typical lithium Ion 1 cell 'battery' discharge curve.
Best method is to do this with genuine and clone batteries and compare times.
- Method (c) Easiest :-).
Use a camera. Set to video or timed photos. Note start and end frame times. Compare.
Major advantages are
"set and forget
no playing with battery connections
self timing.
UPDATE - January 1st 2013 - Happy New Year.
I've just been asked offlist by somebody about the LM317 circuit I mentioned for constant current discharge. Here is an example. I copied this from the very useful and relevant webpage on LED driving - here and they in turn copied it from an LM317 data sheet.
The off list query said
- You mentioned a way by using LM317 to determine battery capacity. I need to check a lithium ion battery with about 1700mAh capacity.
What do you recommend to me to measure this kind of battery capacity in a reasonable time like 3-4 hours.
A 1700 mAh battery would be discharged in 3 hours by 1700/3 =~ 570 mA and in 4 hours by 1700/4 ~= 425 mA. So using about 500 mA and seeing how long it takes will give a measure of battery capacity.
The current of the3 load in the circuit above is
Iout = Vref/R1 so
R1 = Vref/Iout
For an LM317 Vref = 1.25V so for 500 mA
R1 = V/I = 1.25V / 0.5A = 2.5 Ohm.
Power in R1 = I^2 R = 0.5^2 x 2.5 or about 0.7 Watt.
A 1 Watt resistor would probably survive this - a 2 Watt or 5 Watt would be better.
The LM317 will dissipate V_LM317 x I = (Vbattery - Vref) x I = (4.2-1.25) x 0.5 =~ 1.5 Watt. So a heatsink or piece of Aluminum or other thermally conductive material on the LM317 will be "a good idea". I use 4.2 V for the battery voltage. It will drop as the battery discharges.
Note that in many cases a 1700 mAh LiIon battery can be safely discharged at up to 1C rate - = 1700 mA in this case. Safer is C/2 = 850 mA. Actual max allowed rate should be set by the manufacturer. Use Imax = C/2 if no data available. This will usually be safe but "caveat emptor" / "YMMV" ... . If using a higher rate the power dissipation in the resistor and LM317 will be higher and changes will be needed. Some LM317 will handle 1A max. Some will handle 1.5A. (Some smaller pkgs < 1A) . See data sheet. The LM350 is a big brother version of the LM317 that works at several amps.
The battery endpoint voltage should be the endpoint Voltage that you will use in your system. As per my comments above, this MUST NOT BE below 3.0V to prevent battery damage, and higher is safer. You need either to keep a close eye on this if stopping discharge manually OR set up an automatic cutoff system. How you do this and how you time the discharge period is up to you.
Added 2023 (from old comment of mine):
Some batteries have internal electronics that talk to the camera and report capacity. The capacity claim may be accurate OR the battery may be fully charged BUT the clone battery may not have emulated the original protocol accurately enough.
An often reasonably continuous load on a camera battery is to put it into USB mode. I have several cameras which do not shut down if there is no activity in this mode but simply drain the battery. This has the advantage of not over-discharging the battery.
You still need to manually time it.
Edited 2017 - changed recommended long life storage voltage and added comments on fast charging using some recent systems. RM.
What YOU do as regards several of these questions depends largely on what YOU are trying to achieve or test.
Discharge to cutoff is fully discharged (to whatever remaining % that voltage represents). That's the easy one :-)
Percent dropoff of current in tail sets final % of max possible charged reached. There was a superb table given here within last week or so. Can supply later if you don't find it.
Real Men™ plateau at 4.2V and tail down to 10% or even 5% of the constant current rate. This gets the battery full and knocks the stuffing out of it.
Others terminate the current tail at say 25% of cc value.
Optimum lifetime for ongoing usage is at about the end of the constant current phase. That makes it very easy to locate - charge at specified current until desired max voltage is reached, then charge at constant voltage as desired. Here "desired" is to stop immediately. This is the point at which batteries tend to give significantly longer whole of life mAh of storage without grossly reducing mAh capacity per cycle. This is liable to be the point where older "fast chargers" tell you they have finished. Actual % total claimed varies but probably 70% - 80% range.
Newer USB input fast chargers use the term differently. In the case of USB the maximum available charge current at 5V is 5A so that the battery MAY be able to be charged at ~= 6A for the CC part of the cycle using an efficient buck converter to drop voltage and raise current.
[For a buck converter: Vout x Iout = Vin x Iin x efficiency_of_conversion]
Some systems such as QuaqlComms Quick Charge system allow the use of higher charger voltages (9, 12, 20) with specifically designed equipment, so battery charging can be faster for a given voltage provided that the battery specification allows this.
Maximum charge rates for LiIon and LiPo batteries are usually C/1 = 1A per Ah of battery capacity.
At 5V, 5A a USB charger can charge a 6000 mAh 1 cell LiPO battery at max rate - so eg a 10,000 mAh single cell battery used in some larger tablets can not be charged at the allowed 10A ! rate.
For long life storage where actual stored capacity is unimportant, LiIon and LiPo cells should be stored at about 3.7V.
___________________
Using cells without protection adds to the rich tapestry of life. As long as you don't mind the occasional scorch mark on the tapestry that's fine. Note that part of the protection is a one time high capacity fuse under the cap for when things get out of control. Undervoltage discharge destroys. Charging from below a certain voltage at full rate can get fun, I'm told. Charging at reduced rate can bring cell up, I'm told. Below another second level they say don't even think about it. I've had very poor success in trying to get LiIon to misbehave. I have a box of unprotected cells that are very uncooperative about venting with lame etc. Strange. Sony and Apple and even HP seem to be much better at it :-).
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Best Answer
You can't "boost charge" a Li-Ion battery. When you say it is "dead", I assume it doesn't deliver any voltage/current. There could be two scenarios of what has happened.
The battery has a built-in protection circuit. When the (internal) cell voltage drops below certain threshold (like 2.5-2.9V), the circutrty disables the battery output.
The cell gets drained to actual zero due to self-dicharge parasitics. Then it is formally "dead".
There is a third scenario when the cell went dead after a catastrophic overdischarge (short), which destroyed internal conductors. I assume this is not the case here.
In first two cases there is a chance for recovery. You should apply a very modest "precharge" current (say, 100 mA). If the cell is just overdischarged and is shut down due to protection, it will quickly gain some voltage above the cut-off thershold, and the protection circuit will disengage. Then you charge the battery using normal means of your device.
If the cell has no protection and is dead by electrochemical depletion, there is still a chance that it will come back to life. However, studies show that the process of recovery is unpredictable, and in no case the cell can be restored to its original capacity. The result can be anywhere. See also some relevant discussion here.