Battery Capacity Calculator

Amp hours, watt hours, bank sizing and charge time with step-by-step solutions
Convert a 100 Ah battery at 12.8 V to watt hours
Size a 12 V battery to run a 250 W load for 4 hours at 50% depth of discharge
How long to charge a 100 Ah battery from 50% with a 10 A charger?
How many amp hours is 1.2 kWh at 24 V?

Amp Hours and Watt Hours

Amp hours measure charge, watt hours measure energy, and voltage is what connects them:

E=C×VE = C \times V

  • EE — energy, watt hours (Wh); divide by 10001000 for kWh
  • CC — capacity, amp hours (Ah)
  • VV — nominal pack voltage, volts (V)

Rearranged, C=E/VC = E/V.

Why the distinction matters: a 100100 Ah pack at 12.812.8 V (12801280 Wh) stores twice the energy of a 100100 Ah pack at 6.46.4 V. Comparing batteries of different voltages by amp hours alone is meaningless — always compare watt hours.

The assumption people forget: nominal voltage is a label, not the terminal voltage. A 12 V lead-acid battery sits near 12.712.7 V full and 11.811.8 V empty; a 12 V LiFePO₄ pack (four cells) is nominally 12.812.8 V. Manufacturers rate Ah at a specified discharge rate and temperature, so a rating is only valid under those conditions.

Sizing a Bank and Estimating Charge Time

To size a battery for a load, work backwards through every loss:

Crequired=P×tV×ηinv×DoDC_{\text{required}} = \frac{P \times t}{V \times \eta_{\text{inv}} \times \text{DoD}}

  • PP — load power, watts (W); tt — run time, hours (h)
  • ηinv\eta_{\text{inv}} — inverter efficiency, typically 0.850.850.950.95 for an AC load; omit it for a DC load
  • DoD\text{DoD} — usable depth of discharge as a fraction: about 0.50.5 for lead-acid, 0.80.80.90.9 for LiFePO₄

Charge time:

t=CneededIcharge×ηchgt = \frac{C_{\text{needed}}}{I_{\text{charge}} \times \eta_{\text{chg}}}

in hours, with II in amperes and charging efficiency ηchg\eta_{\text{chg}} around 0.850.85 for lead-acid, 0.950.95+ for lithium.

The assumption people forget: this is a first-pass estimate. Lead-acid capacity falls at high discharge rates (Peukert's law), cold cuts usable capacity sharply, and the constant-voltage absorption stage at the end of a charge is far slower than the constant-current stage. Follow the cell manufacturer's charge profile.

Common Mistakes to Avoid

  • Adding amp hours across different voltages — convert everything to watt hours first, then add.
  • Planning to use 100% of the rated capacity — deep-cycling a lead-acid battery to empty destroys it in tens of cycles. Design to the usable depth of discharge.
  • Ignoring inverter losses — an AC load draws 101015%15\% more from the battery than its nameplate wattage.
  • Treating charge time as capacity divided by current — efficiency losses and the tapering absorption stage make the real time longer.
  • Confusing Ah with A — amperes are a rate, amp hours are a quantity.
  • Mixing series and parallel wiring — series adds voltage at constant Ah; parallel adds Ah at constant voltage.

示例题目

Step 1: E=C×VE = C \times V
Step 2: E=(100 Ah)(12.8 V)E = (100\ \text{Ah})(12.8\ \text{V})
Step 3: E=1280 Wh=1.28 kWhE = 1280\ \text{Wh} = 1.28\ \text{kWh}
Answer: E=1280E = 1280 Wh (1.281.28 kWh)

Step 1: Energy at the load: E=(250 W)(4.0 h)=1000 WhE = (250\ \text{W})(4.0\ \text{h}) = 1000\ \text{Wh}
Step 2: Energy drawn from the battery: 1000 Wh÷0.90=1111 Wh1000\ \text{Wh} \div 0.90 = 1111\ \text{Wh}
Step 3: Rated energy needed at 50%50\% DoD: 1111 Wh÷0.50=2222 Wh1111\ \text{Wh} \div 0.50 = 2222\ \text{Wh}
Step 4: Convert to amp hours: C=2222 Wh÷12 V=185 AhC = 2222\ \text{Wh} \div 12\ \text{V} = 185\ \text{Ah}
Step 5: Round up to the next standard size: 200200 Ah
Answer: About 185185 Ah required, so specify a 200200 Ah bank

Step 1: Charge to replace: Cneeded=(100 Ah)(0.50)=50 AhC_{\text{needed}} = (100\ \text{Ah})(0.50) = 50\ \text{Ah}
Step 2: Effective charging current: (10 A)(0.85)=8.5 A(10\ \text{A})(0.85) = 8.5\ \text{A}
Step 3: t=50 Ah÷8.5 A=5.9 ht = 50\ \text{Ah} \div 8.5\ \text{A} = 5.9\ \text{h}
Step 4: For lead-acid, add time for the absorption stage, which tapers the current well below 1010 A near the top
Answer: t5.9t \approx 5.9 hours of bulk charging

常见问题

Multiply the amp-hour rating by the nominal pack voltage: Wh = Ah × V. A 100 Ah battery at 12.8 V holds 1280 Wh, or 1.28 kWh. Going the other way, Ah = Wh ÷ V.

Depth of discharge is limited by chemistry and cycle life. Lead-acid batteries are normally designed to about 50% DoD, while LiFePO₄ tolerates 80–90%, so a lead-acid bank needs roughly twice the nameplate capacity for the same usable energy.

Divide the charge you need to replace, in amp hours, by the charger current multiplied by the charging efficiency (about 0.85 for lead-acid). The result is the bulk-charge time; the final absorption stage tapers the current and adds extra time on top.

Series wiring adds the voltages while the amp-hour rating stays the same; parallel wiring adds the amp hours at the same voltage. Either way the total watt hours are the sum, which is why watt hours are the honest way to compare packs.

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