Electronics
How To Calculate Battery Life
Battery life in hours is the stored energy divided by the power the load draws. Amp hours give the charge, volts multiply that into watt hours, and the load in watts sets how fast it drains.
Quick Answer
runtime hours = amp hours x volts / watts
- Ah
- The capacity in amp hours
- V
- The pack voltage in volts
- W
- The load in watts
- DoD
- The usable depth of discharge
Multiply the capacity in amp hours by the pack voltage to get watt hours, then divide by the load in watts. A 100 amp hour 12 volt pack holds 1200 watt hours, so a 600 watt load runs for two hours. Drawing only half the charge cuts that to one and a half hours.
What Is Battery Life?
Amp hours are charge, not energy. A 100 amp hour pack could deliver one ampere for a hundred hours or ten amperes for ten, and the product of current and time is what stays constant in the simple model.
Multiplying by the voltage converts charge into energy. A hundred amp hours at twelve volts is 1200 watt hours, which is the number that can be compared with any other battery regardless of voltage.
Runtime follows directly from dividing energy by power. A 600 watt load draws 1200 watt hours in exactly two hours, before any correction is applied.
Real batteries drop their voltage as they empty, so a constant power load draws more and more current over time. That is why the practical runtime is shorter than the simple division suggests.
Depth of discharge is the share of the capacity you are willing to use. Most lead acid batteries are limited to about half to avoid permanent damage, which halves the runtime compared with the nameplate figure.
Peukert's law captures the rate penalty. Drawing heavily reduces the usable capacity, so the same pack runs for less than half as long at twice the current, and the exponent for lead acid is near 1.3.
Lithium packs tolerate deep discharge better and have a Peukert exponent closer to one, so their practical runtime is much nearer the ideal division.
Cold weather cuts the available capacity sharply. A lead acid pack can lose a third of its usable charge at freezing, which is why vehicles struggle to start in winter.
Inverter losses come before the battery is even asked for the energy. Converting twelve volts to mains voltage is around ninety percent efficient, so ten percent of the stored energy becomes heat in the inverter.
Sizing a system means working backwards. Decide how many hours a load must run, multiply by the watts to get the energy needed, then divide by the voltage and by the depth of discharge you are willing to accept.
Cycle life and depth of discharge interact. A lead acid pack cycled to eighty percent typically lasts a few hundred cycles, while the same pack cycled to fifty percent lasts several times as long.
Battery capacity is quoted at a reference temperature, usually twenty five degrees. Both heat and cold change the delivered capacity, and the direction of the change is not the same for every chemistry.
Connecting cells in series raises the voltage and connecting them in parallel raises the capacity. The watt hours come out the same either way, which is why a twelve volt 100 amp hour pack and a twenty four volt 50 amp hour pack hold identical energy.
Formula
runtime hours = amp hours x volts / watts
Formula page →The ideal runtime before depth of discharge and rate effects.
| Symbol | Meaning | Unit | Notes |
|---|---|---|---|
| Ah | Capacity | amp hours | The nameplate capacity in amp hours. |
| V | Voltage | volts | The nominal pack voltage. |
| W | Load | watts | The power the load draws. |
| t | Runtime | hours | The hours the pack can supply the load. |
t = 20 x (capacity / (20 x current))^1.3
Formula page →The runtime once the penalty for drawing at a high rate is applied.
| Symbol | Meaning | Unit | Notes |
|---|---|---|---|
| C | Rated capacity | amp hours | The capacity at the twenty hour rate. |
| I | Current | amps | The current the load actually draws. |
| t | Runtime | hours | The corrected runtime in hours. |
How To Calculate Battery Life
- 1
Convert capacity into watt hours
Multiply the amp hour rating by the nominal voltage. A 100 amp hour pack at twelve volts stores 1200 watt hours.
- 2
Divide by the load in watts
1200 watt hours divided by a 600 watt load is two hours, which is the ideal runtime over the full capacity.
- 3
Apply the depth of discharge
At eighty percent usable capacity the same pack delivers one and a half hours, because twenty percent of the charge is held back.
- 4
Check the discharge rate
Six hundred watts at twelve volts is fifty amperes, which is half the pack's rating. Lead acid loses usable capacity at that rate, which the Peukert figure reflects.
- 5
Add inverter losses if any
If the load runs from an inverter, divide the runtime by the inverter efficiency. At ninety percent efficiency the two hour figure becomes one and a half hours even at full discharge.
Examples
Example 1: A 100 amp hour 12 volt pack on a 600 watt load
- Capacity
- 100
- Voltage
- 12
- Load
- 600
- Depth of discharge
- 80%
| Step | Calculation | Result |
|---|---|---|
| Stored energy | 100 x 12 | 1200 |
| Load current | 600 / 12 | 50 |
| Ideal runtime | 1200 / 600 | 2 |
| Usable runtime at 80 percent | 100 x 0.8 / 50 | 1.6 |
| Peukert corrected runtime | 20 x (100 / (20 x 50))^1.3 | 1.002374 |
Result: The pack stores 1200 watt hours and runs the 600 watt load for two hours at full discharge. At eighty percent depth of discharge the answer is 1.6 hours, and the Peukert correction for drawing 50 amperes brings it down to 1.002374 hours.
Example 2: A 50 amp hour phone battery drawing 5 watts
- Capacity
- 50
- Voltage
- 3.7
- Load
- 5
- Depth of discharge
- 90%
| Step | Calculation | Result |
|---|---|---|
| Stored energy | 50 x 3.7 | 185 |
| Load current | 5 / 3.7 | 1.351351 |
| Ideal runtime | 185 / 5 | 37 |
| Usable runtime at 90 percent | 50 x 0.9 / 1.351351 | 33.3 |
| Peukert corrected runtime | 20 x (50 / (20 x 1.351351))^1.3 | 44.499306 |
Result: A small lithium pack holding 185 watt hours delivers 5 watts for 37 hours in the ideal case, and 33.3 hours if you reserve ten percent of the charge. Because the current is well under the twenty hour rate the Peukert figure rises to 44.499306 hours.
Example 3: A 150 amp hour 24 volt bank sized for half discharge
- Capacity
- 150
- Voltage
- 24
- Load
- 480
- Depth of discharge
- 50%
| Step | Calculation | Result |
|---|---|---|
| Stored energy | 150 x 24 | 3600 |
| Load current | 480 / 24 | 20 |
| Ideal runtime | 3600 / 480 | 7.5 |
| Usable runtime at 50 percent | 150 x 0.5 / 20 | 3.75 |
| Peukert corrected runtime | 20 x (150 / (20 x 20))^1.3 | 5.588183 |
Result: The bank stores 3600 watt hours, enough for 7.5 hours at 480 watts over the full capacity. Holding half the charge in reserve gives 3.75 hours, and the Peukert rate penalty brings the full capacity figure down to 5.588183 hours.
Calculator
Runtime in hours
2
- Runtime at your depth of discharge
- 1.6
- Stored energy in watt hours
- 1,200
- Load current in amps
- 50
- Peukert corrected runtime
- 1.0024
Values update as you type. This calculator covers the single scenario its formula assumes — see Common Mistakes for what it leaves out.
Prefer a full-width tool? Open the Battery Life calculator page.
Common Mistakes
Reading amp hours as watt hours
Amp hours are charge at a particular voltage. A 100 amp hour pack is only 3600 watt hours at thirty six volts and 1200 watt hours at twelve volts.
Ignoring the depth of discharge limit
Lead acid batteries should not be emptied. Using the full nameplate capacity overstates the runtime by roughly a factor of two and shortens the pack's life.
Forgetting the discharge rate penalty
Capacity is quoted at a gentle rate, usually twenty hours. Drawing at half that rate removes a noticeable share of the usable capacity, which Peukert's law describes.
Dividing watt hours by amps
Watt hours are energy and amps are current, so the units do not cancel. Either divide watt hours by watts or divide amp hours by amps.
Leaving out inverter losses
Converting the pack voltage to mains voltage costs around ten percent of the energy. A two hour runtime becomes about one hour and forty eight minutes through an inverter.
Using the nominal voltage after the pack sags
The voltage falls as the pack empties, so a constant power load draws more current late in the discharge. Runtime shortens faster than the simple division suggests.
FAQ
How do I convert amp hours to watt hours?
Multiply the amp hour rating by the nominal pack voltage. A 100 amp hour pack at twelve volts is 1200 watt hours.
How many amp hours do I need for a load?
Multiply the load in watts by the hours required, divide by the pack voltage, and then divide again by the depth of discharge you plan to use.
What is Peukert's law?
It is the observation that a battery delivers less total charge at higher currents. For lead acid the effect is strong enough to matter whenever the current is more than a few times the twenty hour rate.
Why does my battery last less than the calculation?
Age, cold, inverter losses and fast discharge all reduce the real figure. The simple division is an upper bound, not a promise.
What depth of discharge should I use?
About fifty percent for lead acid on a regular cycle, and up to eighty or ninety percent for lithium, which tolerates deep discharge without damage.
Does a bigger battery last proportionally longer?
Almost. Doubling the amp hours doubles the energy and the runtime, but the higher capacity also lets you draw at a gentler rate, so the real gain is slightly larger.
References
- [1]Wikipedia, Ampere hour — https://en.wikipedia.org/wiki/Ampere_hour
- [2]Wikipedia, Peukert's law — https://en.wikipedia.org/wiki/Peukert%27s_law
- [3]Wikipedia, Deep cycle battery — https://en.wikipedia.org/wiki/Deep_cycle_battery