How is the cable cross-section chosen in a 12V and 24V system?
In low-voltage systems the cross-section is most often determined by voltage drop rather than current-carrying capacity. As a rough guide, on short 12V lines 1.3 mm2 carries approximately 10 A, 3.3 mm2 approximately 20 A, 8.4 mm2 approximately 40 A. On long lines the calculation is mandatory and for copper the resistivity is taken as approximately 0.0175 ohm.mm2/m.
At 12V a 3% voltage drop means only 0.36 V. Such a small budget determines the cross-section on long lines long before the current-carrying capacity does. The common convention is a limit of 3% on the branch and 5% for the supply and branch together.
Rough guide (short 12V lines):
- 16 AWG ≈ 1.3 mm² -> approximately 10 A
- 12 AWG ≈ 3.3 mm² -> approximately 20 A
- 8 AWG ≈ 8.4 mm² -> approximately 40 A
- 4 AWG ≈ 21 mm² -> approximately 80 A
Do the calculation on long lines: ΔV = 2 x L x I x ρ / A (the factor of 2 is for the outgoing and returning run). Example: for 12V, 20 A, 5 m and a 3% (0.36 V) target, 4.68 mm² is required; when the next standard size of 5.26 mm² is chosen, the real drop becomes 0.321 V (2.67%).
Take temperature into account as well: in the motor or engine-bay region, at 90-105 °C the resistance of copper increases by 27-33% relative to nominal, so the voltage drop increases too.
Choose the cross-section according to the driver's peak current: the KS250 can draw 45 A peak (5 s) and the PT500 40 A maximum (15 s). The cross-section must also be capable of carrying the fuse you fit.
The full topic: DC Motor Driver Selection Guide