On this page
How to use the battery calculator
- Pick a tab: Runtime, Pack design, Backup and off-grid (an inverter battery or a solar bank) or Charge time.
- Type your numbers. The result updates as you type, and Show the working lists each formula with your numbers in it.
- Pick the battery type. It sets the depth of discharge, Peukert exponent and charge efficiency, and you can change each one.
- Copy link shares the exact problem, for example
/battery-calculator/?tab=runtime&cap=150&v=12&load=300.
| Link parameter | Meaning |
|---|---|
tab | runtime, pack, bank or charge |
cap, capu, v, chem, dod | Capacity, its unit (Ah or Wh), voltage, type (flooded, agm, gel, lfp, nmc), depth of discharge in % |
loadu, load, lt, eff, pk, k, hr | Load as W, A or list; its value; ac or dc; inverter efficiency %; Peukert on (1), exponent and rated hours |
cell, tv, tc, tcu | Pack: cell preset (18650, 21700, lfp, pb), target voltage, target capacity and its unit |
mode, daily, days, lw, hrs, sv, bv, bah | Bank: offgrid or inverter, Wh a day, days of autonomy, load W, backup hours, system voltage, each battery's V and Ah |
Ah vs Wh, and how runtime is worked out
An amp-hour (Ah) counts charge: 1 Ah is one amp for one hour. A watt-hour (Wh) counts energy, and energy is what your appliances use. The bridge is the voltage: Wh = Ah × V. A 100 Ah battery at 12 V holds 1,200 Wh, while a 100 Ah LiFePO4 battery at 12.8 V holds 1,280 Wh. Comparing batteries by Ah alone only works at the same voltage.
Runtime is usable energy divided by the power taken from the battery:
t = (Ah × V × DoD) ÷ (load W ÷ inverter efficiency)
DoD is the depth of discharge you allow, as a fraction. An inverter wastes some energy as heat, so for an AC load the battery supplies the load divided by the efficiency (85% means 0.85). A DC load skips that step. Given the battery current instead, runtime is usable Ah divided by amps.
The C-rate is the current as a fraction of the capacity: 10 A from 100 Ah is 0.1C. The higher it is, the more the simple formula overstates lead-acid runtime.
Why depth of discharge matters
Depth of discharge is how much of the rated capacity you take out before recharging. Lead-acid batteries wear out much faster when they are regularly drained deep, so 50% is the usual planning figure for them. Lithium iron phosphate (LiFePO4) handles deep cycling better, and 80% is a common planning figure. These are the defaults; if your battery's datasheet allows more, type it in.
| Battery type | Default depth of discharge | Default Peukert k | Default charge efficiency |
|---|---|---|---|
| Lead-acid, flooded (incl. tubular) | 50% | 1.25 | 85% |
| Lead-acid, AGM | 50% | 1.1 | 90% |
| Lead-acid, gel | 50% | 1.15 | 90% |
| LiFePO4 | 80% | not applied | 95% |
| Lithium NMC | 80% | not applied | 95% |
The tool shows the runtime to your chosen depth and to empty. Shop formulas quote the second, which is why a "5 hour" inverter battery often gives far less.
Peukert's law: big loads empty lead-acid faster
A lead-acid battery's Ah rating is measured at a slow rate, usually over 20 hours (the C20 rating). Draw current faster and it delivers fewer Ah. Peukert's law models this:
t = H × (C ÷ (I × H))k
Here H is the rated hours, C the rated Ah, I the current and k the Peukert exponent. Published k values run from 1.05 to 1.15 for AGM, 1.1 to 1.25 for gel and 1.2 to 1.6 for flooded batteries. With k = 1 there is no rate effect. Lithium batteries are much less affected, so the tool offers the correction for lead-acid only.
The standard textbook case: a 100 Ah battery rated at the 20-hour rate, with k = 1.2, discharged at 10 A. The simple formula says 10 hours. Peukert gives 20 × (100 ÷ 200)1.2 = 8.706 h, so the battery delivers about 87.06 Ah, not 100. At 25 A the gap grows: 2.899 h instead of 4, or 72.48 Ah.
Series vs parallel: designing a pack
Cells in series add their voltages and keep the capacity. Cells in parallel add their capacities and keep the voltage. A pack label like 13S4P means 13 groups in series, each group holding 4 cells in parallel. The pack tab finds the series count from the target voltage (nearest whole number) and the parallel count from the target capacity (always rounded up).
| Layout | Cells | Voltage | Capacity | Energy |
|---|---|---|---|---|
| 4S2P of 3.6 V, 3,400 mAh cells | 8 | 14.4 V | 6.8 Ah | 97.92 Wh |
| 13S4P of 3.6 V, 3,000 mAh cells | 52 | 46.8 V | 12 Ah | 561.6 Wh |
| 16S2P of 3.2 V, 100 Ah LiFePO4 cells | 32 | 51.2 V | 200 Ah | 10,240 Wh |
| 6S1P of 2 V, 200 Ah lead-acid cells | 6 | 12 V | 200 Ah | 2,400 Wh |
The full-charge and cut-off voltages set your charger and battery management system (BMS). The presets use 2.4 V and 1.75 V per lead-acid cell, so six cells charge to 14.4 V and cut off at 10.5 V. Type your cell's datasheet values to replace them.
Worked examples
1. An inverter battery during load-shedding
A 150 Ah, 12 V tubular battery runs 300 W of fans, lights and a router through an 85% inverter. It holds 150 × 12 = 1,800 Wh, of which 50%, 900 Wh, is usable. The battery supplies 300 ÷ 0.85 = 352.9 W, or 29.41 A. Runtime: 900 ÷ 352.9 = 2.55 h, or 2 h 33 min (5 h 6 min to empty).
That current is 3.92 times the 20-hour rate of 7.5 A. With Peukert correction at k = 1.25 the battery acts like 106.6 Ah, not 150, and the runtime to 50% drops to 1 h 49 min.
2. A 48 V e-bike pack from 18650 cells
Target: 48 V and 12 Ah from 3.6 V, 3,000 mAh cells. Series count: 48 ÷ 3.6 = 13.33, rounded to 13. Parallel count: 12 ÷ 3 = 4. The result is 13S4P: 52 cells, 46.8 V nominal, 54.6 V when full at 4.2 V per cell, 32.5 V at a 2.5 V cut-off, and 561.6 Wh. A 14S pack would be 50.4 V nominal and 58.8 V full, which needs a different charger.
3. An off-grid bank for 3 kWh a day
A cabin uses 3,000 Wh a day through a 90% inverter and needs 2 days of autonomy from LiFePO4 at 80% depth of discharge, on 48 V. Daily energy from the battery: 3,000 ÷ 0.9 = 3,333 Wh; for two days, 6,667 Wh usable; rated, 6,667 ÷ 0.8 = 8,333 Wh, or 173.6 Ah at 48 V.
Four 12.8 V, 100 Ah batteries in series make 51.2 V and 5,120 Wh per string. 8,333 ÷ 5,120 = 1.628, so 2 strings: 8 batteries, 10.24 kWh, 2.46 days. Lead-acid at 50% would need 13,333 Wh: eight 12 V, 200 Ah batteries.
4. Which battery for a 500 W load for 3 hours?
The shop formula, 500 × 3 ÷ 12 = 125 Ah, ignores the inverter and drains the battery flat. With an 85% inverter the battery supplies 1,765 Wh: 147.1 Ah drained to empty, or 294.1 Ah at 50% depth of discharge. That is two 150 Ah batteries in parallel, giving 3 h 4 min to half charge.
Reading the result and common mistakes
- Mixing Ah at different voltages. 100 Ah at 24 V is twice the energy of 100 Ah at 12 V. Compare in Wh.
- Forgetting the inverter. An AC load of 300 W takes about 353 W from the battery at 85% efficiency.
- Planning on 100% depth of discharge. It gives the biggest number and the shortest battery life for lead-acid.
- Ignoring the rate. If the load is well above the 20-hour current, turn on Peukert for lead-acid.
Charge time is Ah to restore ÷ (charge efficiency × charger amps). A 100 Ah LiFePO4 battery from 20% to 100% at 20 A and 95% needs 80 ÷ 0.95 = 84.21 Ah: 4 h 13 min. Lead-acid slows down above about 80% charge, so a full charge takes longer than this estimate.
Capacity falls as a battery ages and in the cold. Size a bank with some margin, and check the battery and inverter datasheets for the limits that apply to yours.
Embed the Battery Calculator on your website
Teachers, bloggers and course sites can add this tool to any web page for free. Paste the code below into an HTML or embed block (WordPress, Google Sites, Canvas, Moodle, Blogger, Wix). It loads the same tool you see above, without our menus or ads. Please keep the credit line under the frame. Sizes, a live preview and step-by-step guides for each platform are on the calculator widget page.
Frequently asked questions
How long will a 150 Ah battery last?
Running 300 W through an 85% inverter, a 150 Ah, 12 V lead-acid battery lasts 2 h 33 min to 50% depth of discharge and 5 h 6 min to empty. With Peukert correction at k = 1.25, the 50% figure drops to 1 h 49 min.
How do I convert Ah to Wh?
Multiply the Ah by the battery voltage. 100 Ah at 12 V is 1,200 Wh, and 100 Ah at 12.8 V is 1,280 Wh. To go back, divide Wh by the voltage.
What depth of discharge should I use?
The defaults are 50% for lead-acid, which wears out faster when drained deep, and 80% for lithium. If your battery's datasheet allows more, type that value in.
What does 13S4P mean?
13 groups of cells in series, with 4 cells in parallel in each group, so 52 cells in total. With 3.6 V, 3,000 mAh cells that is 46.8 V nominal, 54.6 V when full and 12 Ah.
How long does a 10 A charger take to charge a 100 Ah battery?
From empty, 10 hours with no losses, or 11 h 46 min at the 85% charge efficiency used for flooded lead-acid, plus the slow final stage of a lead-acid charge. From 50% it takes half as long.
Why does my battery run out sooner than calculated?
Usually a higher real load than assumed, the Peukert effect at high current, an aged battery, cold weather or the inverter's cut-off voltage. Measure the real load and turn on Peukert correction for lead-acid.