# Electronic – Why is battery C rating not a function of battery topology/voltage

battery-charging

In a battery, there is a 'C' rating, which appears to be a function of its capacity. a 5 Ah battery charged or discharged at 1C charges or discharges at 5A.

However C capability is generally spoken of with respect to battery chemistry only.

If I have four cells, each 3 V, 0.8 Ah, then I can make a battery with three different topologies:

1. 1S 4P: 3 V
2. 2S 2P: 6 V
3. 4S 1P: 12 V

Each variant has a 3.2 Ah capacity*, and 1C in each case suggests charging or discharging at 3.2 A. However in 1), each cell will experience 0.8 A of charge or discharge. In 2) each cell will see 1.6 A, and 3) each cell will see the full 3.2 A. Battery 3 at 3.2 A will charge 4x faster than battery 1, corresponding to four times the power input, totally changing how the chemistry might be expected to behave. C rating sure surely then include the topology, or else it would be a nonsensical metric. What is the reality?

*Incorrect. Amp hour rating varies as inverse of voltage from topology. Amp hour rating alone is insufficient to characterise energy capacity, voltage is required also leading to a common Watt hour capacity. Series topology is cancelled and the Amp hour rating of the battery is equal to the Amp hour rating of the smallest parallel arrangement of cells in the battery.

#### Best Answer

You are miscalculating, or neglecting to calculate, the effect on the capacity of the entire pack of connecting cells in parallel. When you connect two cells in parallel, the current going into or out of them splits. Assuming that the split is equal, then you get:

1. 1S 4P: 3 V
• 3 V 3.2 Ah (9.6 Wh) 1C charge = 3.2 A. Split evenly over four cells, 3.2 A = 0.8 A/cell
2. 2S 2P: 6 V
• 6 V 1.6 Ah (9.6 Wh) 1C charge = 1.6 A. Split evenly over two strings of two cells 1.6 A = 0.8 A/cell
3. 4S 1P: 12 V
• 12 V 0.8 Ah (9.6 Wh) 1C charge = 0.8 A. There's just one string, so 0.8 A = 0.8 A/cell

(As a practical note, you can't trust that a parallel arrangement will split the current evenly -- just be aware).