When designing an off-grid or backup power system, guessing your inverter and battery requirements leads to either tripped breakers or massive overspending. The core engineering challenge is translating an AC load profile into DC source requirements while accounting for conversion losses and battery chemistry limits. This DC to AC conversion calculator framework provides the exact mathematical model to size your battery bank capacity (in Amp-hours) based on your AC load, inverter efficiency, and DC bus voltage.
The Master DC to AC Conversion Formula
To find the required battery capacity, we must calculate the total DC energy required to sustain an AC load over a specific time, adjusted for the inverter's inefficiency and the battery's safe discharge limits. The master equation is:
| Symbol | Definition | Standard Unit | Typical Real-World Value |
|---|---|---|---|
| CAh | Required Battery Capacity | Amp-hours (Ah) | 100 Ah to 400 Ah |
| PAC | Total Continuous AC Load Power | Watts (W) | 500 W to 4000 W |
| t | Required Runtime | Hours (h) | 2 h to 24 h |
| VDC | Nominal DC Battery Bank Voltage | Volts (V) | 12V, 24V, or 48V |
| η | Inverter Efficiency (DC to AC) | Decimal (0.0 - 1.0) | 0.85 (HF) to 0.93 (LF) |
| DoD | Maximum Depth of Discharge | Decimal (0.0 - 1.0) | 0.50 (Lead-Acid) to 0.90 (LiFePO4) |
Assumptions, Limits, and Unit Traps
This formula applies to steady-state AC loads operating at unity power factor (PF = 1.0), such as resistive heaters, incandescent lighting, and modern switch-mode power supplies. If your load is highly inductive (like an uncorrected AC motor), you must multiply PAC by the Power Factor to find true Watts before running the calculation.
- Percentage vs. Decimal: Entering '90' instead of '0.90' for efficiency or DoD will inflate your required battery size by 10,000%. Always use decimals.
- VA vs. Watts: Inverter nameplates often list Volt-Amps (VA). A 3000VA inverter with a 0.8 PF only delivers 2400W. Always use real power (Watts) for PAC.
- Ignoring Surge: This formula sizes for continuous runtime capacity. It does not calculate the instantaneous voltage sag during motor startup surges. Always verify your chosen inverter's peak surge rating separately.
What does a realistic answer magnitude look like? A 12V system powering a 1000W microwave (PAC) will draw roughly 100 Amps of DC current continuously. A 48V system powering that same 1000W load will draw only about 25 Amps. If your calculated DC current exceeds 150A, your wire gauge and terminal lugs become a severe fire hazard, dictating a shift to a higher DC voltage architecture.
Worked Examples with Unit Tracking
Problem 1: 12V Off-Grid Cabin Fridge
Scenario: You need to run an 80W DC-to-AC powered refrigerator for 24 hours. You are using a 12V LiFePO4 battery bank (safely discharged to 80% DoD) and a high-frequency inverter with 90% efficiency.
- Identify variables: PAC = 80 W, t = 24 h, VDC = 12 V, η = 0.90, DoD = 0.80
- Substitute into formula:
CAh = (80 W × 24 h) / (12 V × 0.90 × 0.80) - Calculate numerator (Total AC Energy):
80 W × 24 h = 1920 Wh - Calculate denominator (Effective DC Voltage factor):
12 V × 0.90 × 0.80 = 8.64 V - Divide and track units:
1920 Wh / 8.64 V = 222.22 Ah
Result: You need a minimum 12V 222 Ah LiFePO4 battery bank. In practice, you would parallel two 12V 100Ah batteries (yielding 200Ah, slightly under) or step up to a single 12V 300Ah server-rack battery to cover the margin.
Problem 2: 48V Workshop Welder and Tools
Scenario: A mobile workshop runs a 2400W continuous AC load (lighting, compressors, and a small inverter welder) for 2 hours. The system uses a 48V LiFePO4 bank (90% DoD limit) and a low-frequency transformer-based inverter (93% efficiency).
- Identify variables: PAC = 2400 W, t = 2 h, VDC = 48 V, η = 0.93, DoD = 0.90
- Substitute into formula:
CAh = (2400 W × 2 h) / (48 V × 0.93 × 0.90) - Calculate numerator:
2400 W × 2 h = 4800 Wh - Calculate denominator:
48 V × 0.93 × 0.90 = 40.176 V - Divide and track units:
4800 Wh / 40.176 V = 119.47 Ah
Result: You need a 48V battery bank with at least 119.47 Ah capacity. A standard 48V 100Ah server rack battery (like the EG4 48V100) provides 5.12 kWh, which falls slightly short of the 5.73 kWh required here. You must install two in parallel for 200Ah total.
Rearranged Forms for Missing Variables
Field conditions often require solving for a different parameter. Use these algebraic rearrangements when your knowns change:
- Solve for Maximum AC Load (PAC):
PAC = (CAh × VDC × η × DoD) / t - Solve for Required DC Voltage (VDC):
VDC = (PAC × t) / (CAh × η × DoD) - Solve for Inverter Efficiency (η):
η = (PAC × t) / (CAh × VDC × DoD) - Solve for Safe Depth of Discharge (DoD):
DoD = (PAC × t) / (CAh × VDC × η) - Solve for Maximum Runtime (t):
t = (CAh × VDC × η × DoD) / PAC
Decision Path: Topology and Component Selection
Calculating CAh is only half the battle. The raw DC current draw (IDC = PAC / (VDC × η)) dictates your physical wiring, breaker sizing, and inverter topology. Use the decision matrix below to terminate your design process with exact component selections.
| Calculated DC Current (IDC) | Required DC Voltage | Wire Gauge & Breaker | Concrete Inverter Pick |
|---|---|---|---|
| < 50 Amps (e.g., < 500W on 12V) |
12V | 6 AWG THHN, 60A ANL Fuse | Samlex PST-60S-12A (600W Pure Sine) |
| 50A to 150 Amps (e.g., 1000W - 1800W on 12V) |
12V or 24V | 2/0 AWG Welding Cable, 150A Class T Breaker | Victron MultiPlus 12/3000 (3000VA, Low Frequency) |
| 150A to 250 Amps (e.g., 3000W on 12V or 24V) |
24V Mandatory | 4/0 AWG Copper, 250A Class T Breaker | Victron MultiPlus-II 24/3000 (Split-phase capable) |
| > 250 Amps (e.g., > 3000W continuous) |
48V Mandatory | 2/0 AWG (due to lower current), 125A DC Breaker | Victron Quattro 48/5000 or Growatt SPF 5000ES |
