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How do I calculate the approximated time for the Charging and Discharging of the battery? Is there any equation available for the purpose? If yes, then please provide me.

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Discharge time is basically the Ah or mAh rating divided by the current.

So for a 2200mAh battery with a load that draws 300mA you have:

\$\frac{2.2}{0.3} = 7.3 hours\$*

The charge time depends on the battery chemistry and the charge current.

For NiMh, for example, this would typically be 10% of the Ah rating for 10 hours.

Other chemistries, such as Li-Ion, will be different.

*2200mAh is the same as 2.2Ah. 300mA is the same as 0.3A

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    \$\begingroup\$ I would like to point out that some batteries, and certainly all circuits, will not work down to 0 Volts at the supply, so your circuit will stop working a long time before the battery is fully drained \$\endgroup\$ – chwi Aug 4 '15 at 18:24
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    \$\begingroup\$ This answer is about 4 years old now - but it's incorrect for LiIon batteries, which is what he is asking about. (Advised after this answer). See my answer for detail - but, LiIon can typically be charged at the C/1 rate until Vbat = 4.2V/cell. That takes typically 45 minutes to about 75% capacity and then about 2 hours at reducing rate for the balance . \$\endgroup\$ – Russell McMahon Nov 18 '15 at 15:39
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Charging of battery: Example: Take 100 AH battery. If the applied Current is 10 Amperes, then it would be 100Ah/10A= 10 hrs approximately. It is an usual calculation.

Discharging: Example: Battery AH X Battery Volt / Applied load. Say, 100 AH X 12V/ 100 Watts = 12 hrs (with 40% loss at the max = 12 x 40 /100 = 4.8 hrs) For sure, the backup will lasts up to 4.8 hrs.

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    \$\begingroup\$ The charge formula above assumes a 100% efficiency charge, so it's not ideal, but it is a good, simple way to get a rough idea of charge time. For a more accurate estimation, you can assume 80% efficiency for NiCd and NiMh batteries and 90% efficiency for LiIon/LiPo batteries. Then, the formula becomes capacity / (efficiency * chargeRate) or, to use the same values from above (assuming lithium chemistry), 100Ah / (0.9 * 10A) = 11.11 hours \$\endgroup\$ – KOGI Aug 21 '13 at 15:58
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Discharge rates are well enough covered here.

LiIon / LiPo have almost 100 current charge efficiency but energy charge efficiency depends on charge rate. H=Higher charge rates have lower energy efficiencies as resistive losses increase towards the end of charging.

Below LiIon and LiPo are interchangeable in this context.

The main reason to adding an answer to a 3+ year old question is to note that:

LiIon / LiPo should not be charged at above manufacturers spec. This is usually C/1, sometimes C/2 and very occasionally 2C. Usually C/1 is safe.

LiIon's are charged at CC = constant current = <= max allowed current from 'empty' until charge voltage reaches 4.2V. They are then charged at CV = constant voltage = 4.2V and the current falls under battery chemistry control.

Charge endpoint is reached when I_charge in CV mode falls to some preset % of Imax - typically 25%. Higher % termination current = longer cycle life, lower charge time and slightly less capacity for the following discharge cycle.

When charged from "empty" at C/1 a LiIon cell achieves about 70% - 80% of full charge in 0.6 to 0.7 hours ~= 40 to 50 minutes.

The CV stage typically takes 1.5 to 2 hours (depending on termination current% and other factors) so total charge time is about 40m +1.5 hours to 50 minutes +2 hours or typically 2+ to 3 hours overall. But, a very useful % of total charge is reached in 1 hour.

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Peukert's Law gives you the capacity of the battery in terms of the discharge rate. Lower the discharge rate higher the capacity. As the discharge rate ( Load) increases the battery capacity decereases.

This is to say if you dischage in low current the battery will give you more capacity or longer discharge . For charging calculate the Ah discharged plus 20% of the Ah discharged if its a gel battery. The result is the total Ah you will feed in to fully recharge.

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    \$\begingroup\$ How can you advise to charge with 10 A if you have absolutely no specs of the battery in question? \$\endgroup\$ – stevenvh Oct 11 '12 at 10:08
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In the ideal/theoretical case, the time would be t = capacity/current. If the capacity is given in amp-hours and current in amps, time will be in hours (charging or discharging). For example, 100 Ah battery delivering 1A, would last 100 hours. Or if delivering 100A, it would last 1 hour. In other words, you can have "any time" as long as when you multiply it by the current, you get 100 (the battery capacity).

However, in the real/practical world, you have to take into consideration the heat generated in each process, the efficiency, the type of battery, the operating range, and other variables. This is where "rules of thumb" come in. If you want a the battery to last a "long" time and no overheating, then the charging or discharging current must be kept at not more than 1/10 of the rated capacity. You also need to keep in mind that a battery is not supposed to be "fully" discharged. Typically, a battery is considered "discharged" when it looses 1/3 of its capacity, therefore it only needs 1/3 of its capacity to be fully charged (range of operation). With these constraints and the above values, one gets only one answer, t = 33Ah/10A = 3.3hr.

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Rules of thumb given in other answers are often good enough but if you can find the datasheet of the battery it's best to check the relevant graph. As an example here's the datasheet of a low cost 12V battery. In the datasheet you'll find this graph:

Graph of voltage versus time for various currents

Let's say that this is a battery with 7Ahr capacity and that you want to draw 14A. You'll have to observe the 2C curve (2C means to discharge at 7Ahr*2/h=14A). You'll note that this battery will drop to 9.5V-10V after about 15mins. Of-course this is only true for a fresh from the shelf battery kept at 25 deg.Celsius. Temperature, age and usage negatively affect the performance.

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  • \$\begingroup\$ Hi, I couldn't find the original source for that image you included. Can you please edit the answer to add a link to the original web page or PDF file etc., as required by this site rule? Thanks. \$\endgroup\$ – SamGibson Apr 15 at 16:17
  • \$\begingroup\$ Fixed to help you (I'm not at all sure that the rules you linked to concern this case) \$\endgroup\$ – ndemou Apr 15 at 16:32
  • \$\begingroup\$ Thanks and yes, that rule does apply to all answers - it says: "When you find a useful resource that can help answer a question (from another site or in an answer on Electrical Engineering Stack Exchange) make sure you do all of the following [...]" You used something in an answer which came from another site (i.e. not your original work). If you don't believe me, you can flag your own answer, select "in need of moderator attention" and ask them to remove your reference link, if it isn't required - they won't remove it. Or you can ask in the moderator chat room. Thanks again. \$\endgroup\$ – SamGibson Apr 15 at 17:27
  • \$\begingroup\$ 90% of what I post in SE comes from an external resource: other sites, other answers, books & periodicals I've read etc. If I had to give credit to even the tiniest insignificant source I would be contributing 1% of what I do. I'm all about giving credit to others but not without limits. This image is a tiny part of just another free datasheet. Any similar chart would serve just as well because it's used purely as an example. I stand by my first position that a link is not required but it surely doesn't hurt and hey! I've been wrong before :-) \$\endgroup\$ – ndemou Apr 16 at 11:30
  • \$\begingroup\$ Just to be clear, whenever you reuse material from elsewhere (text and images you haven't created yourself), you need to provide the attribution(s). Thank you for providing the link. \$\endgroup\$ – Dave Tweed Apr 19 at 4:20

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