Home Inverter and Battery Backup Sizing (Lights, Fans, Router, TV)
Where the grid drops out for hours at a time, many homes run a few essential circuits from an inverter and a battery bank: lights, fans, the internet router, a television and phone chargers. The two questions are how big the inverter must be and how long the batteries will last.
The calculator below is preset for a 600 W essential load at a power factor of 0.8, two 12 V 200 Ah lead-acid batteries in series (24 V), 50 % usable depth of discharge, 90 % inverter efficiency and a 4-hour backup target. Replace the load with your own total; motors such as fridges and pumps need extra headroom for starting current.
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For estimation and education. Not a substitute for a licensed engineer, the applicable codes (IEC 60364, NEC/NFPA 70, BS 7671, local regulations) or manufacturer data. Read the disclaimer.
Result
UPS and inverter sizing — Engineerly. Estimate for education only; verify against applicable codes and manufacturer data.
- Required UPS rating (with margin)
- 937.5VA
- Backup runtime
- 216min
The battery bank falls short of your 240 min target; about 222.2 Ah at 24 V is needed.
- Load real power
- 600W
- Load apparent power
- 750VA
- Required UPS rating (with margin)
- 750W
- Next common catalogue size
- 1,000 VA
- Battery bank
- 24 V, 200 Ah (4.8 kWh)
- Battery current at the DC bus
- 27.78A
- C-rate
- 0.139C
- Backup runtime
- 3.6h
- Capacity needed for 240 min
- 222.22 Ah@ 24 V
- Parallel strings of 200 Ah needed
- 2
Working, with your numbers
- 1. Load apparent powerS = P / PFS = 600 W / 0.8= 750 VA
- 2. Required UPS rating with margin (VA)S_UPS = S × (1 + m)S_UPS = 750 VA × (1 + 25 / 100)= 937.5 VA
- 3. Required UPS rating with margin (W)P_UPS = P × (1 + m)P_UPS = 600 W × (1 + 25 / 100)= 750 W
- 4. Bank voltage (series adds volts)V = V_unit × N_sV = 12 V × 2= 24 V
- 5. Bank capacity (parallel adds Ah)C = Ah_unit × N_pC = 200 Ah × 1= 200 Ah
- 6. Battery currentI = P_load / (V × η)I = 600 W / (24 V × 0.9)= 27.778 A
- 7. Runtime (ideal capacity)t = C × V × DoD × η / Pt = 200 Ah × 24 V × 0.5 × 0.9 / 600 W= 3.6 h
- 8. Required bank capacity for the target runtimeC = P × t / (V × DoD × η)C = 600 W × 4 h / (24 V × 0.5 × 0.9)= 222.22 Ah
- 9. Parallel strings neededN_p = ⌈C / Ah_unit⌉N_p = ⌈222.222 / 200⌉= 2
Assumptions
- Design margin 25 % on the load, a common practice for growth and ageing, not a requirement of a standard.
- Constant power load for the whole discharge; battery voltage taken as the nominal bank voltage.
- Usable depth of discharge 50 % of rated capacity.
- Inverter efficiency 90 %, constant over the load range.
- All batteries in the bank are identical, healthy and fully charged at the start of the outage.
Check before you rely on this
- The UPS maker's runtime chart or table (runtime against load for the actual battery model) overrides this estimate. Runtime also falls as the batteries age and at low temperature.
- A UPS has both a VA and a W limit. Check that the load stays within both ratings on the datasheet.
- Laser printers, heaters, motors and compressors draw high starting or surge currents; many UPS makers advise against connecting them.
- Lead-acid capacity falls at high discharge rates, which is normal in UPS service. Turn on the Peukert option for a more realistic figure, or use the battery maker's constant-power discharge table.
- Estimation and education only. Not a substitute for a licensed engineer, the applicable code (IEC 60364, NEC/NFPA 70, BS 7671 or local regulations) or manufacturer data.
Notes for this case
- Runtime here is energy-based. Lead-acid capacity falls at high discharge rates; enable the Peukert option for a more conservative figure.
- Recharge matters as much as discharge: the charger or grid must refill the bank between cuts, otherwise each day starts lower.
- Check the inverter's surge rating against the largest motor it must start.
Open the full calculator page with the formula, worked example and references
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How these formulas were checked: methodology and testing.