Battery Life & Runtime Calculator

Runtime from capacity, depth of discharge and load, with step-by-step solutions

How long will a 12 V 100 Ah LiFePO4 battery run a 300 W load at 90% depth of discharge?

How long will a 100 Ah lead-acid battery last at a 5 A draw with 50% depth of discharge?

Find the UPS runtime for two 12 V 9 Ah batteries in series on a 200 W load

What capacity in Ah gives 8 hours at a 4 A draw with 80% usable capacity?

The Battery Runtime Equation

Runtime is usable energy divided by the rate at which the load takes it. Working in watt-hours:

$t = \\frac{V_{\\text{nom}} \\times C \\times \\mathrm{DoD} \\times \\eta}{P}$

Symbols and units:

$t$ — runtime, hours (h)
$V_{\\text{nom}}$ — nominal battery voltage, volts (V)
$C$ — rated capacity, amp-hours (Ah), so $V_{\\text{nom}}C$ is the stored energy in watt-hours (Wh)
$\\mathrm{DoD}$ — usable depth of discharge, a fraction: about $0.5$ for lead-acid, $0.8$ – $0.9$ for LiFePO₄
$\\eta$ — conversion efficiency of the inverter or converter, typically $0.85$ – $0.95$ ; set $\\eta = 1$ for a DC load wired straight to the battery
$P$ — load power, watts (W)

For a DC load quoted in amperes the same idea simplifies to $t = C \\times \\mathrm{DoD} / I$ , with $I$ in amperes.

The operating assumption: a steady load, a healthy battery at room temperature, and a discharge rate close to the one the capacity was rated at.

Why the Nameplate Hours Never Arrive

Three effects shorten real runtime, and all three are physical rather than pessimism.

Rate dependence. Capacity is quoted at a stated rate — lead-acid is usually rated at the 20-hour rate, so a "100 Ah" battery means $5$ A for $20$ h. Pull $50$ A and you will get noticeably fewer amp-hours out, an effect described by Peukert's relation.

Temperature. Lead-acid loses roughly a fifth of its capacity near $0$ °C. Lithium chemistries hold capacity better in the cold but usually refuse to charge below freezing.

Ageing. End of life is conventionally taken as $80\\%$ of the original capacity, so a battery near the end of its cycle life delivers about four fifths of the runtime you calculated on day one.

Treat a calculated runtime as a planning figure, not a specification. Backup systems for life-safety, medical or telecom service are sized to the applicable standard — for example IEEE 485 for lead-acid sizing — and verified by an actual discharge test, never by arithmetic alone.

Common Mistakes to Avoid

Using the full rated capacity — discharging lead-acid past about $50\\%$ shortens its life sharply, so only half the nameplate is usable in practice.
Forgetting the inverter — an AC load through an $85\\%$ inverter costs you $15\\%$ of the stored energy before the load sees any of it.
Mixing Ah and Wh — amp-hours only become energy once multiplied by the voltage. A $12$ V $100$ Ah and a $48$ V $100$ Ah battery differ fourfold in stored energy.
Adding capacity for a series string — two $12$ V $9$ Ah batteries in series give $24$ V $9$ Ah, not $18$ Ah. Only a parallel connection adds amp-hours.
Ignoring standby draw — an inverter's own idle consumption can dominate on a small load.
Assuming the load is constant — a fridge or a pump cycles, so use its duty-cycle average, not its running watts.

示例题目

Step 1: Stored energy:

E = V_{\\text{nom}} C = (12\\ \\text{V})(100\\ \\text{Ah}) = 1200\\ \\text{Wh}

Step 2: Usable energy:

1200\\ \\text{Wh} \\times 0.90 = 1080\\ \\text{Wh}

Step 3: Delivered after inverter losses:

1080\\ \\text{Wh} \\times 0.90 = 972\\ \\text{Wh}

Step 4:

t = 972\\ \\text{Wh} \\div 300\\ \\text{W} = 3.24\\ \\text{h}

Step 5:

0.24\\ \\text{h} \\times 60 = 14\\ \\text{min}

Answer:

t \\approx 3.24

h, about

3

14

min

Step 1: Usable capacity:

C_{\\text{usable}} = (100\\ \\text{Ah})(0.50) = 50\\ \\text{Ah}

Step 2: The load is DC and wired directly, so no inverter efficiency applies

Step 3:

t = C_{\\text{usable}}/I = 50\\ \\text{Ah} \\div 5\\ \\text{A} = 10\\ \\text{h}

Step 4: The draw of

5

A matches the

20

-hour rating point, so no rate correction is needed

Step 5: At

50

A instead, Peukert losses would make the usable capacity noticeably less than

50

Answer:

t = 10

h at

5

Step 1: Series string: voltage adds, capacity does not —

24

9

Step 2: Stored energy:

(24\\ \\text{V})(9\\ \\text{Ah}) = 216\\ \\text{Wh}

Step 3: Usable:

216\\ \\text{Wh} \\times 0.80 = 172.8\\ \\text{Wh}

Step 4: After the inverter:

172.8\\ \\text{Wh} \\times 0.85 = 146.9\\ \\text{Wh}

Step 5:

t = 146.9\\ \\text{Wh} \\div 200\\ \\text{W} = 0.734\\ \\text{h}

Step 6:

0.734\\ \\text{h} \\times 60 = 44\\ \\text{min}

Answer:

t \\approx 0.73

h, about

44

min

常见问题

Convert the battery to watt-hours by multiplying nominal voltage by amp-hours, cut it down by the usable depth of discharge and the inverter efficiency, then divide by the load in watts. The result is hours. For a DC load quoted in amps, runtime is simply usable amp-hours divided by amps.

About 50% for flooded or AGM lead-acid, and 80-90% for LiFePO4, because deep cycling shortens lead-acid life sharply. Use the figure in the battery datasheet for the cycle life you are aiming at rather than a generic number.

Capacity is rated at a slow discharge rate, so a heavy load extracts fewer amp-hours; cold reduces capacity further; and an aged battery holds about 80% of its original capacity at end of life. Inverter standby draw and cycling loads add to the gap.

Not directly. A series string raises voltage while the amp-hour capacity stays that of one battery, so stored watt-hours do rise in proportion to the voltage. Connecting batteries in parallel is what adds amp-hours at the same voltage.

Battery Life & Runtime Calculator

Runtime from capacity, depth of discharge and load, with step-by-step solutions

The Battery Runtime Equation

Why the Nameplate Hours Never Arrive

Common Mistakes to Avoid

示例题目

常见问题

How do I calculate how long a battery will last?

What depth of discharge should I use?

Why does my battery last less time than the calculation says?

Do batteries in series give more runtime?

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