Battery Life & Backup Time Calculator
Device runtime, inverter backup, cell packs and charge time — formulas shown.
How it was calculated
Details
Results are estimates from the formulas shown, not a professional design or certification. Have a qualified engineer verify anything safety-critical (structures, electrical installations, gas or pressure systems).
About the Battery Life & Backup Time Calculator
How long a battery lasts is its usable capacity divided by what the load draws. This calculator does that four ways: for devices rated in mAh — with one average current, or a duty cycle of active and sleep states — for inverters and UPS systems rated in Ah and volts, for a pack you build from cells in series and parallel, and for charging back up.
Inverter backup can include Peukert’s law, which accounts for lead-acid batteries delivering less than their rated Ah when discharged faster than the rating’s hours. You can enter the Peukert exponent, or let the calculator work it out from two capacity ratings on the datasheet (for example C20 and C10). Every result shows its formula with your numbers in it.
How to use it
- Choose a mode: Device (mAh and current), Inverter / UPS (battery Ah, volts and load watts), Battery pack (cells in series and parallel) or Charge time.
- For a device, enter the capacity and either the average current or the duty cycle — each state’s current and how long it lasts per cycle, such as 25 mA for 2 s, then 15 µA for 58 s. Add more states for radios, sensors or displays.
- For an inverter, enter one battery’s Ah and voltage, how many there are, the load in watts, the inverter efficiency and how deeply you let the batteries discharge. Tick Peukert for lead-acid batteries and enter k and the rating hours (C20 → 20).
- For a pack, pick the cell chemistry (or enter its voltages), the cell capacity, S and P and the cell’s discharge limit to get the pack voltage, Ah, Wh and maximum current.
- Read the result and How it was calculated. Copy result copies the figures and the working.
Examples
2,000 mAh; 25 mA for 2 s, 15 µA for 58 s
Average 0.848 mA → 2,359 h ≈ 98.3 days
12 V 150 Ah, 300 W load, 85% inverter, 80% depth of discharge
1,800 Wh × 0.8 ÷ 352.9 W = 4.08 h (4 h 5 min)
Battery current 29.4 A
t = 20 × (150 ÷ (29.4 × 20))^1.25 × 0.8 = 2.9 h — it delivers only about 107 Ah at this rate
3.6 V 2,500 mAh cells, 20 A each
46.8 V nominal (54.6 V full), 10 Ah, 468 Wh, 80 A maximum
5,000 mAh from 20% to 100% at 2 A, 99% efficient
4,000 mAh ÷ (2 A × 0.99) = 2.02 h, longer in practice as the charger tapers near full
The formulas
- Average current of a duty cycle: I_avg = Σ(I × t) ÷ Σt
- Device runtime: t = C × usable share ÷ I_avg; self-discharge is added as a current, C × rate ÷ 730.5 h for a monthly rate
- Inverter backup: t = n × V × Ah × DoD ÷ (P_load ÷ η_inverter)
- Peukert’s law: t = H × (C ÷ (I × H))^k, where C is the capacity rated at H hours and I the current through each battery; below the rated current (C ÷ H) the rated capacity is used instead, t = C ÷ I
- Peukert exponent from two ratings: k = ln(t₁ ÷ t₂) ÷ ln(I₂ ÷ I₁), with I = C ÷ t for each rating
- Pack: V = S × V_cell, Ah = P × Ah_cell, Wh = V × Ah, I_max = P × I_cell
- Charge time: t = C × (to − from) ÷ (I × η_charge)
Peukert’s law and lead-acid batteries
A lead-acid battery rated 150 Ah at the 20-hour rate (C20) delivers 150 Ah when discharged over 20 hours, at 7.5 A. Drawn faster, it delivers less: Peukert’s law (W. Peukert, 1897) models this with the exponent k. For lead-acid batteries k is typically between 1.1 and 1.3 — about 1.05–1.15 for AGM, 1.1–1.25 for gel and 1.2–1.6 for flooded batteries — and it rises as the battery ages. Lithium batteries are much closer to 1, so leave Peukert off for them.
If the datasheet lists capacities at two rates, the calculator can find k: a battery with C20 = 100 Ah and C5 = 85 Ah has k = ln(20 ÷ 5) ÷ ln(17 ÷ 5) = 1.13. In an inverter bank the current through each battery is the DC power divided by the total voltage of all the batteries — the same whether they are wired in series or in parallel.
Below the rated current the law would credit the battery with more than its rated capacity, and without limit as the load gets lighter — about 167 Ah from a 12 V 150 Ah (C20) battery running a 50 W load through an 85% inverter, with k = 1.25. Real batteries gain only part of that, so for loads below the rated current the calculator uses the rated capacity, which keeps the backup estimate on the safe side.
Charge efficiency
Charging puts back a little more charge than the battery gives out. Battery University (BU-808c) puts the coulombic efficiency of lithium-ion above 99% at moderate currents and cool temperatures, lead-acid at about 90%, and nickel-based batteries (NiMH) at up to 90% with fast charging but about 70% with a slow charge. Chargers also slow down near full: lithium-ion switches to constant voltage and the current tapers, so the end of the charge takes longer than the constant-current estimate here.
Sources
- W. Peukert (1897), Elektrotechnische Zeitschrift 18, p. 287 — t = H × (C ÷ (I × H))^k
- IEC 61960-3:2017, Secondary lithium cells and batteries for portable applications — rated capacity and rated energy
- IEC 60896-11:2002, Stationary lead-acid batteries — Vented types — rated capacity at a stated discharge time
- Battery University, BU-808c Coulombic and Energy Efficiency with the Battery
Limitations
- Rated capacity is measured under the maker’s test conditions. Cold, ageing, pulse loads and a device that shuts down above the cell’s end voltage all reduce it — use the usable-capacity and depth-of-discharge fields to allow for them.
- Peukert’s law is for constant-current discharge at a constant temperature. Below the rated current it would predict more than the rated capacity, which real batteries gain only partly, so the rated capacity is used there; at very high currents batteries deliver less than it predicts.
- Self-discharge is modelled as a steady current, which is close enough for a few percent per month but not for batteries that lose most of their charge in storage.
- Inverter backup ignores the inverter’s own no-load consumption beyond its efficiency figure. At very light loads that standby draw can dominate.
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Frequently asked questions
How long will a 150 Ah battery last on a 300 W load?
A 12 V 150 Ah battery stores 1,800 Wh. With an 85% efficient inverter it supplies 300 ÷ 0.85 = 353 W, so using 80% of it lasts 1,800 × 0.8 ÷ 353 = 4.1 hours. With Peukert’s law (k = 1.25, C20) it is closer to 2.9 hours, because 29 A is about four times the 20-hour rate.
How do I calculate battery life from mAh?
Divide the capacity by the current: a 2,000 mAh battery at 20 mA lasts 2,000 ÷ 20 = 100 hours. If the device sleeps most of the time, use its average current: 25 mA for 2 s plus 15 µA for 58 s every minute averages 0.848 mA, so the same battery lasts about 98 days.
What is a typical Peukert exponent?
For lead-acid batteries k is typically between 1.1 and 1.3 — lower for AGM and gel, higher for flooded batteries and old ones. Lithium-ion and LiFePO4 are close to 1. The best source is two capacity ratings from the battery’s own datasheet, which the calculator turns into k.
Do two batteries in series last longer than in parallel?
No — the stored energy is the same: 2 × 12 V × 150 Ah = 3,600 Wh either way. Series doubles the voltage (for a 24 V inverter) and parallel doubles the Ah at 12 V. Use the arrangement your inverter needs.
How do I size a battery pack from 18650 cells?
Put cells in series for voltage and in parallel for capacity and current. 13 cells of 3.6 V in series give 46.8 V nominal (54.6 V when full); four such strings in parallel with 2.5 Ah cells give 10 Ah and 468 Wh. The pack’s maximum current is four times one cell’s rated continuous current.
Why does charging take longer than capacity ÷ current?
Some charge is lost as heat (about 1% for lithium-ion, around 10% for lead-acid), and chargers reduce the current near full to avoid overcharging. Treat capacity ÷ current as the minimum time.