How many hours of operation will a battery capacity (Ah) give in my motor driver application?
Rough calculation: time (hours) = usable Ah / average current. In a 100 Ah lead-acid battery, typically 50% is taken as usable, that is 50 Ah; in an application drawing 10 A on average this comes to about 5 hours. What is critical 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 not the peak current but the duty-weighted average. 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.
- Divide. Time = usable Ah / average A.
Factors that spoil the calculation: cold weather lowers the capacity, an ageing battery cannot deliver its nominal value, the energy drawn at high current is more than expected because of the Peukert effect, and standby currents (receiver, control circuit) add to the total. For this reason leave a reasonable margin on top of the result.
The battery's ability to deliver instantaneous current also matters. 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 handle this surge without a voltage collapse. Choose the cable cross-section according to this peak current.
Full topic: DC Motor Driver Selection Guide