BU-501a: Calculating Battery Runtime

Know about hidden battery losses when estimating the energy reserve.

If the battery were a perfect power source and behaved linearly, charge and discharge times could be calculated according to in-and-out flowing currents. “What is put in should be available as output in the same amount” goes the argument, and “a one-hour charge at 5A should deliver a one-hour discharge at 5A, or a 5-hour discharge at 1A.” This is not possible because of intrinsic losses. The output is always less than what has been put in, and the losses escalate with increasing load. High discharge currents make the battery less efficient.

Peukert Law

The Peukert Law expresses the efficiency factor of a battery on discharge. W. Peukert, a German scientist (1897), was aware of this and devised a formula to calculate the losses in numbers. They apply mostly to lead acid and help estimate the runtime under different discharge loads.

The Peukert Law takes into account the internal resistance and recovery rate of a battery. A value close to one (1) indicates a well-performing battery with good efficiency and minimal loss; a higher number reflects a less efficient battery. Peukert’s law is exponential and the readings for lead acid are between 1.3 and 1.5. Figure 1 illustrates the available capacity as a function of ampere drawn with different Peukert ratings.

Available capacity of a lead acid battery at Peukert numbers of 1.08–1.50


Figure 1: Available capacity of a lead acid battery at Peukert numbers
of 1.08–1.50

A value close to
1 has the smallest losses; higher numbers deliver lower capacities.

Source: von Wentzel (2008)


The lead acid battery prefers intermittent loads to a continuous heavy discharge. The rest periods allow the battery to recompose the chemical reaction and prevent exhaustion. This is why lead acid performs well in a starter application with brief 300A cranking loads and plenty of time to recharge in between. All batteries require recovery, and most other systems provide a faster electrochemical reaction than lead acid. (See BU-501: Basics About Discharging)

Ragone Plot

Lithium and nickel-based batteries are more commonly evaluated by the Ragone plot. Named after David V. Ragone, the Ragone plot looks at the battery’s capacity in watt-hours (Wh) and the discharge power in watt (W). The big advantage of the Ragone plot over Peukert is the ability to read the runtime in minutes and hours presented on the diagonal lines.

Figure 2 illustrates the Ragone plot of four lithium-ion systems in 18650 cells. The horizontal axis displays energy in watt-hours (Wh) and the vertical axis is power in watts (W). The diagonal lines across the field reveal the length of time the battery cells can deliver energy at given loading conditions. The scale is logarithmic to allow a wide selection of battery sizes. The battery chemistries featured in the chart include lithium-iron phosphate (LFP), lithium-manganese oxide (LMO), and nickel manganese cobalt (NMC). (See BU-205: Types of Lithium-ion)

18650 Ragone Plot

Figure 2: Ragone plot reflects Li-ion 18650 cells.  Four Li-ion systems are compared for discharge power and energy as a function of time. Courtesy of Exponent
Legend: The A123 APR18650M1 is a lithium iron phosphate (LiFePO4) Power Cell rated at 1,100mAh, delivering a continuous discharge current of 30A. The Sony US18650VT and Sanyo UR18650W are manganese–based Li-ion Power Cells of 1500mAh each delivering a continuous discharge of 20A. The Sanyo UR18650F is a 2,600mAh Energy Cell for a moderate 5A.discharge. This cell provides the highest discharge energy but has the lowest discharge power.

The Sanyo UR18650F [4] in Figure 2 has the highest specific energy and can power a laptop or e-bike for many hours at a moderate load. The Sanyo UR18650W [3], in comparison, has a lower specific energy but can supply a current of 20A. The A123 [1] has the lowest specific energy but offers the highest power capability by delivering 30A of continuous current. Specific energy defines the battery capacity in weight (Wh/kg); energy density is given in volume (Wh/l).

The Ragone plot helps choosing the best Li-ion system to satisfy maximum discharge power and optimal discharge energy as a function of discharge time. If an application calls for very high discharge current, the 3.3 minute diagonal line on the chart points to the A123 (Battery 1); it can deliver up to 40 Watts of power for 3.3 minutes. The Sanyo F (Battery 4) is slightly lower and delivers about 36 Watts. Focusing on discharge time and following the 33 minute discharge line further down, Battery 1 (A123) only delivers 5.8 Watts for 33 minutes before the energy is depleted whereas the higher capacity Battery 4 (Sanyo F) can provide roughly 17 Watts for the same time; its limitation is lower power.

A design engineer should note that the Ragone snapshot taken by the battery manufacturers represents a new cell, a condition that is temporary. When calculating power and energy needs, engineers must take into account the battery fade caused by cycling and aging. Battery operated systems must still function with a battery that will eventually drop to 70 or 80 percent. A further consideration is low temperature as a battery momentary loses power when cold. The Ragone plot does not show these decreased performances.  

The design engineer should further develop a battery that is durable and does not get stressed during regular use. Stretching load and capacity boundaries to the limit shortens battery life. If repetitive high discharge currents are needed, the pack should be made larger and with the correct choice of cells. An analogy is a truck that is equipped with a large and durable diesel engine instead of a souped-up engine intended for a sports car. 

The Ragone plot can also calculate the power requirements of capacitors, flywheels, flow batteries and fuel cells. A conflict develops with the internal combustion engine and fuel cell that draw fuel from a tank as re-fuelling cheats the system. Similar plots are also used to find the optimal loading ratio of renewable power sources, such as solar cells and wind turbines.


Presentation by Quinn Horn, Ph.D., P.E. Exponent, Inc. Medical Device & Manufacturing (MD&M) West, Anaheim, CA, 15 February 2012

Last updated 2015-06-19

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On May 7, 2011 at 9:03pm

battery full lod discharging chart

On May 25, 2011 at 12:17pm
Samuel Maher Youssef wrote:

please i want to install solar system,and i will use flood deep cycle battery,
i will use four battery 12 volt DC, can i use two battery each other 24 volt and what advantage to use 4 battery.
thanks a lot.

On June 6, 2011 at 10:06pm
Susanta Kumar Sahu wrote:

Dear Sir/Madam,
I am Susanta Kumar Sahu to inform you that i have no knowledge about battery discharge (EX: 180AH battery, connecting with load that depends on customer how much it give back up like this ).Can you tell BACK UP concept of all bettery?

Thank You.
Susanta Kumar Sahu

On June 9, 2011 at 11:42pm
Gaurav wrote:

Is there any empirical formula to calculate battery DEPTH of Discharge for a given lifetime?

On June 20, 2011 at 4:39am
CBT wrote:

Can somebody tell me how to model peukart law im modelling please?
Thanks alot

On August 27, 2011 at 4:47am
Ian wrote:

Peukart (title and sidebar menu) or Peukert (article)?

On November 4, 2011 at 3:27am
Pier wrote:

1) Can Peukert’s constant be used across all battery chemistry’s ?

2) Is there any equation where we can determine the remaining time in a battery upon  
  knowing the discharge voltage during operation?

3) What exactly do you mean by efficiency of the battery ?

4) What is the effect of soldering on a Li-ion cell for paralleling by our self ?

On November 17, 2011 at 9:26pm
gary pione wrote:

can anyone answer question number one from pier ? can peukerts equation be used for lithium type batteries?

On November 17, 2011 at 9:33pm
Ian wrote:

You may have missed this, gary: “The Peukert Law of a battery is exponentialand the readings for lead acid are between 1.3 and 1.4. Nickel-based batteries have low numbers and lithium-ion is even better.” Missing space in there. Different chemistries have different values.

Soldering directly onto a Li-ion cell is not recommended. They can explode.

On November 17, 2011 at 9:55pm
gary pione wrote:

thanks ian. i saw that on the second read thru. i have never been a good student

On November 25, 2011 at 9:53am

Dear Sir,
I live in Pakistan and i am working in lead acid battery company.
My question is, How can i calculate Peukert’s number by theoretical or practical and each battery (size wise) has a different peukert number.


On January 8, 2012 at 9:17pm
Terence Cheong wrote:

Please infrom me how to calculate the time for discharging a traction battery with 560Ah & 420

On February 8, 2012 at 4:30am
Komal wrote:

how to calculate the charging and discharging time of Li-ion battery with specification of 100 Ah, 12 V.and applying 2 A constant current.?????

On February 23, 2012 at 4:41am
Sheldon Patnett wrote:

Yes Peukert’s Equation can be used for Lithium batteries. Their exponents are closer to 1.0 than lead acid Batteries.

On February 23, 2012 at 9:42pm
Pier wrote:

How exactly do you define one cycle in a Li-Ion cell ? I find cycle count in Lenovo thinkpads but not sure as to how they compute it !!!

On May 8, 2012 at 6:13am
Syed wrote:

How to calculate the charging and dischargin time of a Li- ion polymer battery of capacity 1230mAh, 3.7V applying 250mA constant current?
Thank You

On May 8, 2012 at 6:19am
Syed wrote:

How do we calculate the no. of hours a battery can provide on continuous discharge of 250mA of capacity 1320mAh?
Also providing the voltage the battery can provide after every hour of discharge of 250mA would be good.
Nominal voltage of the battery is 3.7V.
max operating range is 2.75V to 4.2V.
max continuous discharge current is 1000mA
internal resistance is 150milli ohm

On July 24, 2012 at 1:20am
Muhammad Ahmed wrote:

My Dear friends;
I designed a solar system in which appliance load is 500 Watt and appliance watt per day is 1500.
So I used 150watt 2 solar panel, 24volt and 12 A Solar charge controller, 2 batteries of 125Ah each.
Normally 7 hours sunlight.

My problem is that I want to calculate in how much time period batteries will again recharge.

Please help me in this problem


On August 30, 2012 at 10:28am
Johnny wrote:

my cell battry is bulging , it’s a li-ion, and i was wondering how i make it safe(er) to dispose of, can you help with that? the phone says “use authentic battery” so i’m afraid it might explode,

On October 18, 2012 at 9:36pm
Zahid wrote:

im using 12v 7.4Ah SLA battery x 3 pcs.. the load 19w 12v dc.. By calculation :-

(12V x 7.2Ah )/ 19W = 4.5 hour x 3 pcs battery in parallel connection can i get 13 hour ??

but by actual run i got 5-7 hour only. i’m i using wrong battery or my calculation wrong..

On October 18, 2012 at 9:45pm
Ian wrote:

7Ah is the capacity at a rate that flattens the battery in 20 hours. Your load is much higher than that so you need to take the Peukert effect into consideration. SLA batteries have quite high Peukert effect. The Peukert calculation is much more complicated than what you have done, but you could use the top graph instead to make it easier.

On February 11, 2013 at 3:48am
Suresh Patil wrote:

If inverter efficiency is around 80%, current from 12 volts for 19W load is approx. 2 amps.
When 3 batteries are in parallel, You have a 7.2 x 3 i.e.21.6AH (@ 20 Hr. rate.)
Using Peukert equation calculation, you should get 8Hr. 30Min. approx. This makes the battery fully discharged. For better life, batteries are not discharged beyond 80%. So inverters’ cut off volts are adjusted to higher value. Probably, the inverter efficiency may be even less than 80%. If you find exact current from the batteries, Peukert formula will give you quite accurate duration. There is no much complication.

On February 23, 2013 at 10:25pm
ashish wrote:

If possible, kindly share the info on making an
SMPS charger of 4000 Watts and above

On March 6, 2013 at 12:41pm
Slocket wrote:

Zahid wrote:

im using 12v 7.4Ah SLA battery x 3 pcs.. the load 19w 12v dc.. By calculation :-

(12V x 7.2Ah )/ 19W = 4.5 hour x 3 pcs battery in parallel connection can i get 13 hour ??

but by actual run i got 5-7 hour only. i’m i using wrong battery or my calculation wrong..
The Peukert effect. Your 7ah battery is too small. The rate in the above chart for 120Ah compared to you 7Ah*3=21Ah battery. Lead acid is 1.3 to 1.4. Your is 19w /12v = 1.58 Amp draw.  120/21 for battery in above chart *1.58 = proportional equivalent~ 9.0 amp so look the 9 amp and it gives capacity reduction of about 50% in the Peukert chart so you will get 20 hours run time down to 10 hours just from that, and even worst as your batteries age. SLA rating is only good for long slow discharge 20AH is 1 Amp for 20 Hours. Take 10x times (10Amps) that out in 2 Hours you get only roughly half to 1/3 that amount if your lucky. That is why my project ebike will have NiMH battery instead of SLA. The higher cost 5x is justified in lifetime use and performance. LiPo4 is still too much expense and funny picky charging/discharge/ quality.

On March 6, 2013 at 12:50pm
Slocket wrote:

Sorry correction,  your example only 13 hours use * 50% = 6.5 hours as you discovered, I was using 20 hour run for 10 hours use.

SLA rating is only for low current draw. Peukert charts are very much relevant. NiMH is twice that of SLA in rating for real world application using any kind of Medium to Heavy use draw. Medium use is >10% current rating of battery capacity. Low is less than 1<%.

Your example of three 7aH batery for 21 Ah array pulling anything more than 2-3 Amps is going to cut your real amp hour capacity to 50% for SLA. You really need to buy 10x more battery or get NiMH—though NiMH self discharge over time quickly over several days in a solar array.

On October 31, 2013 at 12:33pm
LALIT wrote:


On November 15, 2013 at 12:38am
vishal wrote:

Dear sir
I have One 3.7V,600mAh Li-ion battery that battery have DoD is 27% but i want to DoD 70% above that. So what is Procedure for this any circuit have built or other..Please Suggest me.
Give me process for improve battery DoD.

On January 12, 2014 at 11:44am
Anand patel wrote:

i want to know about the normal c20 and c10 rated battery and how to calculate, how much kWh(unit) can be obtained form particular battery like C10 26Ah dc12V battery how much kWh can be obtained at 12V and for C20 Rating Battery.

On March 26, 2014 at 3:01am
okugbesan ibrahim wrote:

how can i calculate 50%volt shutdown on a 24vdc/1400amps inverter batterry.
NB: Inverter capacity is 30kva

On April 2, 2014 at 3:55am
Dj wrote:

i Have 24 numbers of battery, each having 2V,600Ah connected in series with only 400 Watt load that means 48V of my system So can you help me how to calculate the back up time of this system.

On May 31, 2014 at 1:46pm
Bob Sandor wrote:

What is the correct formula to determine how many hours I can draw .028 amps from a 12 volt 700 amp hour lead acid battery before the battery voltage drops to 11 volts?

On June 26, 2014 at 10:13am
Thomas Soares wrote:

A lead acid battery can be transformed to a non acid one. Remove the acid, clean several times with distiled hater and then fill with a solution of KAl(SO4)2 Potassium Alum. This kind of conversion will result in a new kind of energy source: 1) Less voltage; 2) Allows 100% discharge (even shot circuit) with no damage; 3) Fast recharge, but with pulsed DC; 4)Non toxic.

On August 17, 2014 at 5:33am
Solid gold wrote:

A battery of 12v 40ah and 12v 80 ah in parallel , which battery would discharge first?

On September 20, 2014 at 5:55am
santosh wrote:

hello sir, i have 12v 150ah lead acid tubular battery, solar panel giving 4.5A/ hr and avg 6hrs is sunshine available. Load drawing around 20A/ day. battery charging through mppt controller of 30A rated. then, how much time required to fully charge my battery.

On October 11, 2014 at 4:40pm
klaus wrote:

what is 400 cca to amp / hr how to conbert as i have 4 x 12 volt lithium polymer batteries here in my home for a project but need to know the amp/hr on my batteries can you help?

On January 15, 2015 at 2:25am
Baragas wrote:

hi, I need some clarification of UPS battery run time calculation. The detail as follow:

UPS Size : 10kVA
Battery Voltage : 12V
Battery Ah Rating: 100
Battery Qty: 32
Inverter Eff.: 93.5%

I need to know what is the actual battery run time during the failure of main source? does the calculation load shall take on the total load of equipment or the ups rating 10kVA?

Kindly please advise how to calculate and what is the best formula to used.


On February 3, 2015 at 7:56am

hi, I need some clarification of UPS battery run time calculation. The detail as follow:
UPS Size : 10kVA
Battery Voltage : 12V
Battery Ah Rating: 100
Battery Qty: 32
Inverter Eff.: 93.5%
I need to know what is the actual battery run time during the failure of main source? does the calculation load shall take on the total load of equipment or the ups rating 10kVA?
Kindly please advise how to calculate and what is the best formula to used.

On February 28, 2015 at 10:07am
Olga Palizo wrote:

How do I calculate to get the total lead in battery? I am doing the TIER II report. MSDS indicates 60% but how do I get the lbs? The battery is used on stand up riders Model # 118-D125-15 weight of battery is 2485

On April 3, 2015 at 1:05am
NC Malviya wrote:

Sir, I like to know, Inverter of 850 KVA with load 700W (app) installed with 150 Ah battery. If battery is fully charged, how much time it will last in normal time.

On April 17, 2015 at 10:14pm
Thomas wrote:

Dear Sir,
I want to size the battery and its charger for a prime rated generator. I had tried it but not sure whether this correct or not. Kindly advice:

Generator - 1250kVA prime
Starting system data are as follows:
Voltage for the battery = 24V
Battery Charging Alternator - 24V-30Ah
Starting Motor Capacity - 24V -7.5x2kW
Maximum Allowable Resistance of Cranking Circuit - 1.5 m Ω

Recommended Minimum Battery Capacity
At 5゜C (41゜F) and above - 300Ah
Below 5゜C (41゜F) through - 5゜C (23゜F) 600 Ah

I had done the sizing in the following manner as per reference from Battery supplier webpage:

Calculation Steps
1. Determine the current that the starter draws for the entire starting cycle.  Here in our situation:

The starter draws 16,000 amps rolling current, worst case.
(During cranking, current drawn = I=V/R = 24V / (1.5 x 10 -3) = 24*1000/1.5 = 16,000Amps

2. The maximum cranking time per start attempt is 15 seconds, which equals 0.0042 of an hour (this was given in the website. dont know how did the supplier come about this value)

3. The maximum number of start attempts will be 5 (Same situation, this is being given in the webpage.)
Note: My requirement is the battery must run for 24hours and charge within 21hours)

4. Ampere-hours (AH) drawn by the starter is = (16,000) x(0.0042) x(5) = 336 AH

5. It is require to charge the battery for 21hours.

6. The Recharge inefficiency constant for Lead acid battery = 1.4

7. Charger type required =

Total AH drawn by starter x Recharge inefficiency constant / Desired recharge hours

8. Answer = 336 x 1.4 / 21 = 22.4 Amp charger is required.

Thanks & best regards