How many hours of operation will a battery capacity (Ah) give in my motor driver application?
Time = usable Ah / average current. Half of a 100 Ah lead-acid battery is usable; in an application drawing 10 A that comes to roughly 5 hours of running.
The rough calculation is a single line: time (hours) = usable Ah / average current. In a lead-acid battery roughly half of the nominal capacity is taken as usable; a 100 Ah battery gives 50 Ah, and in an application drawing an average of 10 A that comes to about 5 hours. What decides is not the peak current but the average current weighted by the duty cycle.
Do the calculation in three steps
- Find the average current. If the motor does not run continuously, use the duty-weighted average rather than the peak current. Example: a motor drawing 15 A that runs for 10 seconds of every minute averages 15 x (10/60) = 2.5 A.
- Determine the usable capacity. In lead-acid batteries, typically half of the nominal capacity is taken as usable in order to preserve life. 100 Ah -> about 50 Ah. The reason for this limit is that depth of discharge sets the cycle life directly.
- Divide. Time = usable Ah / average A.
Factors that spoil the calculation
- Temperature: Cold weather lowers the capacity; on a winter morning the same battery does less work.
- Age: An ageing battery cannot deliver its nominal value, and the gap between its label and reality widens every year.
- The Peukert effect: The energy drawn at high current is more than expected; taking a lot of current in a short time reduces the total work the battery can do.
- Standby currents: The receiver, the control circuit and the indicator draw current all the time; left out of the calculation, they add up over a night.
- Line loss: The energy spent on the cable comes out of the battery as well; a long run in a thin cross-section is a quiet consumer that stays outside the calculation.
How much current the battery can deliver is a separate matter
The Ah calculation gives you the time, not the instantaneous current. The PT500's 30 A continuous and 40 A maximum (15 s) and the KS250's 15 A continuous and 45 A peak (5 s) figures require the battery and the cables to meet that surge without the voltage collapsing. Choose the cable cross-section not for the average current but for the peak current; otherwise the battery stays sound while the voltage drops at the driver supply terminal and the motor starts weakly.
Where the calculation does not apply
The half rule belongs to lead-acid chemistry. In lithium-based batteries the usable proportion is higher, and in LiFePO4 batteries the manufacturer's datasheet sets the limit; do not carry the lead-acid proportion over without looking at the BMS setting of your own battery. Starter batteries do not enter this calculation either: they are designed for short bursts of high current, they are not deep cycle batteries, and they wear out quickly when they are discharged continuously. The cost of taking the house battery below half is paid straight out of the cycle life as well.
What to do
Watch your duty cycle for an hour and count exactly how many seconds the motor runs, measure the current inside that cycle and derive the average from there; calculations that start from the catalogue figure do not hold in the field. Leave a reasonable margin on top of the result, because the four factors above all work in the same direction, against you. If the result comes out marginal, check the cable cross-section, the quality of the connections and the standby currents before enlarging the battery; the gain is usually there, and it costs far less than a new battery.
The full topic: DC Motor Driver Selection Guide