Understanding Battery Runtime and Discharge
Battery runtime—how long a battery supplies power before depletion—depends on the amp-hour (Ah) rating and the load current drawn from it. Dividing the battery's Ah capacity by the load current in amps yields the discharge time in hours.
For example, a 20 Ah battery can supply:
- 20 A for 1 hour
- 10 A for 2 hours
- 5 A for 4 hours
- 1 A for 20 hours
This inverse relationship means lower currents extend runtime proportionally. However, this linear model assumes constant voltage and ignores chemistry-specific effects that occur in real systems.
Capacity Formulas for Lithium-Ion and Lead-Acid
Lithium-ion batteries discharge at a relatively constant voltage and follow a simpler capacity model. Lead-acid batteries, by contrast, exhibit voltage sag during discharge, requiring a correction factor based on discharge duration.
Lithium-ion battery capacity:
B_Li = (100 × I_L × t) ÷ (100 − Q)
Lead-acid battery capacity:
B_Pb = (100 × I_L × t) ÷ [(100 − Q) × (0.02t + 0.6)]
B_Li— Remaining capacity of lithium-ion battery in amp-hoursB_Pb— Remaining capacity of lead-acid battery in amp-hoursI_L— Load current drawn from the battery in ampst— Duration for which power is supplied to the load in hoursQ— Percentage of charge that should remain in the battery after discharge (reserve charge)
Lithium-Ion vs. Lead-Acid Battery Chemistry
Lithium-ion cells maintain near-constant terminal voltage throughout the discharge cycle, with voltage dropping sharply only near complete depletion. This stability makes capacity calculations straightforward and predictable.
Lead-acid batteries exhibit voltage sag—progressive voltage drop as internal resistance increases during discharge. The factor (0.02t + 0.6) in the lead-acid formula corrects for this effect, accounting for longer discharge durations causing greater voltage depression. Consequently, lead-acid batteries require larger nominal capacities than lithium-ion systems to deliver equivalent energy over the same time period.
Reserve charge requirements also differ: lead-acid batteries rarely drop below 50% state-of-charge (SOC) in practice to avoid sulfation, whereas lithium-ion systems tolerate discharge to 5–10% SOC.
Critical Considerations When Sizing Batteries
Real-world battery performance diverges from nameplate ratings due to several practical factors.
- Temperature effects on capacity — Battery capacity declines in cold conditions as chemical reaction rates slow. At 0°C, a lithium-ion battery may deliver only 50% of rated capacity. Lead-acid batteries suffer even greater losses. Always derate capacity estimates by 20–30% if operation below 15°C is expected.
- Peukert effect and discharge rate — Higher discharge currents reduce effective capacity disproportionately, particularly in lead-acid batteries. A battery rated at 100 Ah may deliver only 80 Ah when discharged at high current. The formulas here assume moderate discharge rates typical of continuous loads.
- State-of-charge reserve margins — Specifying 20% reserve charge improves battery lifespan and reliability. Discharging to zero causes irreversible damage in both chemistries. For critical applications, set reserve charge to 30–50% and recalculate capacity requirements accordingly.
- Voltage rating mismatches — System voltage (12 V, 24 V, 48 V) affects calculations via application load (watts = voltage × current). Verify voltage labeling on the battery case; nominal voltage differs from open-circuit voltage when idle.
Practical Application: Inverter Sizing Example
To size a battery bank for an off-grid inverter, multiply the inverter's power rating by the required backup duration, then divide by the system voltage:
Battery capacity (Ah) = (Power × Duration) ÷ Voltage
A 3000 W inverter powering a house for 4 hours on 48 V DC requires: (3000 × 4) ÷ 48 = 250 Ah nominal capacity. After applying the reserve charge (e.g., 30% minimum SOC), the actual usable capacity becomes 175 Ah, so you would install 250–300 Ah of storage.
For variable loads—such as an air conditioning unit running at full power for 2 hours then half power for 3 hours—calculate the average load current and substitute it into the capacity formula.