BU-503: How to Calculate Battery Runtime

Know about hidden battery losses when estimating the energy reserve.

If the battery was a perfect power source and behaved linearly, charge and discharge times could be calculated according to in-and-out flowing currents, also known as coulombic efficiency. What is put in should be available as output in the same amount; a 1-hour charge at 5A should deliver a 1-hour discharge at 5A, or a 5-hour discharge at 1A. This is not possible because of intrinsic losses and the coulombic efficiency is always less than 100 percent. The losses escalate with increasing load, as high discharge currents make the battery less efficient. (See also BU-402: What is C-rate?)

Peukert Law

The Peukert Law expresses the efficiency factor of a battery on discharge. W. Peukert, a German scientist (1855–1932), was aware that the available capacity of a battery decreases with increasing discharge rate and he devised a formula to calculate the losses in numbers. The law is applied 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; the readings for lead acid are between 1.3 and 1.5 and increase with age. Temperature also affects the readings. Figure 1 illustrates the available capacity as a function of amperes drawn with different Peukert ratings.

As example, a 120Ah lead acid battery being discharged at 15A should last 8 hours (120Ah divided by 15A). Inefficiency caused by the Peukert effect reduces the discharge time. To calculate the actual discharge duration, divide the time with the Peukert exponent that in our example is 1.3. Dividing the discharge time by 1.3 reduces the duration from 8h to 6.15h.

Battery Runtime
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. Peukert values change with battery type age and temperature:

AGM:      1.05–1.15
Gel:         1.10–1.25
Flooded: 1.20–1.60

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 have a faster electrochemical reaction than lead acid. (See Basics About Discharging.)

Ragone Plot

Lithium- and nickel-based batteries are 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 discharge power in watts (W). The big advantage of the Ragone plot over the Peukert Law is the ability to read the runtime in minutes and hours presented on the diagonal lines on the Ragone graph.

Figure 2 illustrates the Ragone plot of four lithium-ion systems using 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.)


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. Not all curves are fully drawn out.

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 1,500mAh each, delivering a continuous discharge of 20A. The Sanyo UR18650F is a 2,600mAh Energy Cell for a moderate 5Adischarge. This cell provides the highest discharge energy but has the lowest discharge power.
Source: Exponent


The Sanyo UR18650F [4] Energy Cell has the highest specific energy and can run a laptop or e-bike for many hours at a moderate load. The Sanyo UR18650W [3] Power Cell, in comparison, has a lower specific energy but can supply a current of 20A. The A123 [1] in LFP 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 in the selection of the optimal Li-ion system to satisfy discharge power while retaining the required runtime. If an application calls for a 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. By 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. 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 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 capacity. A further consideration is low temperature as a battery momentarily loses power when cold. The Ragone plot does not take these decreased performance conditions into account.  

The design engineer should further develop a battery pack 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 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 or the fuel cell that draws fuel from a tank, as on-board re-fueling 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 2018-10-20

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Comments (11)

On November 4, 2015 at 7:28am
Ross Wilson wrote:

Hi—After discharging a 127V Lithium ion battery pack to 0V and disconnecting from the discharging unit the voltage begins to creep back up when voltage measured across the battery terminals—the battery unit is completely disconnected from all other circuitry—- is this an inherent characteristic with Lithium Ion cells? And if so how to you fully discharge them and prevent the voltage from re-establishing itself? Any ideas would be greatly appreciated.

On December 26, 2015 at 10:19pm
Adam wrote:

The only way to hold a lithium battery at 0v safely is submerged in salt water and it should only be done when the battery has become unsafe to handle. This battery would qualify. Also, once submerged in the salt water the battery should be taken to to a recycling center that will take lithium batteries.

On January 29, 2017 at 10:38am
khayrul bashar wrote:

can you tell me somebody what is the meaning of life cycle & how many cycle charge/discharge is called life cycle.

On April 10, 2017 at 8:55am
Mike Williams wrote:

Ok, so after completing the discharge of battery, and taken to recycling, what is done with batteries.. Are they discarded or recycled…???

On August 25, 2017 at 6:51am
Ed Pitcher wrote:

My wife and I live off the grid. We are thinking of changing our battery storage from lead acid to LiFePO4 batteries. We have an average winter discharge of 30 amps per hour. How many hours would a 1600 amp hour system last before needing recharge at a 50% discharge level. The Ragone Plot was beyond my pay grade to extrapolate to a larger system. Thank you in advance for any help.

On May 29, 2018 at 9:02am
Masendi Patrick wrote:

Mr. Ed Pitcher from my experience in sizing considering DoD of 50% that leaves you with 800 Ah and dividing that by 30A, you obtain 26.7 hrs ie complete one day plus additional 2.7 hrs into the second day

On August 10, 2018 at 9:12am
Mariano Perez wrote:

Using in complementary explanations in PostGraduate Works for Chemistry Degree Students

On October 9, 2018 at 4:10am
Fred Sterling wrote:

I am in the process of working out what capacity of battery I would require for an application where the draw is 0.5A and the run time is 504hours (21days)

If I were to use a 300ah battery at 85% DoD the capacity is 255ah.
The daily draw would be 12A (0.5 x 24) and 12 x 21 = 252Ah making the battery suitable, however, Using the Peukert law, and taking into consideration the general decay of a lead acid battery at a rate of 1.3.

Doing 12 x 1.3 = 15.6A (As the daily draw) Would I then need a 328Ah battery at 85% DoD, so a 385Ah battery in total?

Could anyone shed some light on this??



On October 19, 2018 at 12:38am
Alan Saunders wrote:

N.B. The Peukert factor (k) is an exponent not a multiplicand! The above example (15A x 1.3 = 19.5A) is nonsense!

On October 19, 2018 at 11:45pm
Alan Saunders wrote:


Batteries are commonly rated at the 20 hour rate, e.g. a 100Ah battery will be discharged at 5A to 10.5V in 20 hours. N.B. It is Peukert’s EXPONENT not multiplicand! i.e. Raised to the power of. k: C = t x I ^k. You first have to calculate the capacity at 1A then recalculate for the desired current. From wikipedia t = H(C/(IH))^k

You want to draw 0.5A for 504 hours to 85% DofD.

Using your figures, a calculator from planetcalc.com suggests 136 Ah.

Do you really know the Peukert exponent for your battery? Better to aim for 50% DofD if you want the battery to last more than a few tens of cycles. Batterystuff.com suggests 202 Ah.

On November 30, 2018 at 4:28am
samphors wrote:

hi guy i have load p=4436w i need backup time 50minit how can i use UPS sizing and battery ? who can explain me ??