Battery Capacity Tester Calculator

Measured mAh/Wh, state of health and C-rate from a resistor discharge test.

What it does

Estimates a battery’s real capacity from a simple test: discharge it through a known resistor, note the average voltage and how long it lasts until the cut-off voltage.

Where it's used

Checking used 18650 cells, comparing brands, and judging how much life an old battery has left.

How to use it

Enter the rated capacity and nominal voltage from the label.
Enter the load resistor and the average voltage you measured during the test.
Enter how long it took to reach the cut-off voltage.
Optional: enter your device’s current to estimate its runtime.

The formula explained

I: V_avg / R_load — discharge current
Capacity: I × t, in mAh
Wh: V_avg × I × t — energy delivered
SoH: state of health = measured ÷ rated capacity
C-rate: I ÷ rated capacity (1C empties the battery in an hour)

Worked example

A 3 000 mAh Li-ion cell discharged through 8.2 Ω for 5 h 30 min.

Inputs:

Rated capacity: 3000 mAh
Nominal voltage: 3.7 V
Load resistor: 8.2 Ω
Average voltage during test: 3.65 V
Discharge time — hours: 5 h
Discharge time — minutes: 30 min
Your device load (runtime estimate): 300m A

The calculator gives:

Measured capacity: 2,448 mAh
State of health: 81.6 %
Discharge current: 445.1 mA

Tips and common mistakes

Stop at the cut-off voltage (≈3.0 V for Li-ion, 1.0 V for NiMH) — going lower damages the cell.
The load resistor gets hot; use a power resistor rated well above V²/R.
Capacity is lower at high C-rates and in the cold, so test near the current you will really use.

Questions people ask

How is the capacity worked out?

The average current is the average voltage divided by the load resistor, and capacity is current times time. In the example, 3.65 V / 8.2 Ω ≈ 445 mA for 5.5 hours gives about 2 448 mAh.

What does state of health mean?

Measured capacity as a share of the rated capacity: 2 448 / 3 000 ≈ 82 %. Many devices treat a Li-ion cell below about 80 % as worn out.

Why use the average voltage during the test?

With a plain resistor as the load, the current falls as the battery voltage falls. Reading the voltage every few minutes and averaging gives the average current. An electronic constant-current load avoids this and gives a more exact result.

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