How is voltage drop calculated, and what is the 3% rule?
Voltage drop is calculated by multiplying twice the distance (out and back) by the current and the copper resistivity and dividing by the cross-section; for copper the resistivity is taken as approximately 0.0175 ohm.mm2/m. The common convention is to leave a limit of 3% on the branch and 5% for the supply and branch together. In a 12V system 3% means a budget of only 0.36 V, and in a 24V system 0.72 V.
The formula is: ΔV = 2 x L x I x ρ / A. Example calculation: 12V, 20 A load, 5 metres distance and a 3% (0.36 V) target. When you solve the formula for the cross-section you get 4.68 mm². When the next standard size of 5.26 mm² (10 AWG) is chosen, the real drop is 0.321 V, that is 2.67%, and stays below the target.
Why such a tight limit? Because in a 12V system every volt lost is a large proportion. When the voltage drops at the motor supply terminal:
- The motor speed and torque drop, and it draws more current for the same job.
- More current creates more drop; the problem feeds itself.
- At the moment of start-up the instantaneous drop is far larger; the start-up current is typically 5-8 times the rated value.
- The supply voltage of the control electronics can fall below the critical threshold and cause the module to behave unexpectedly.
Things to include in the calculation: all connection and terminal resistances on the line, the fuse holder resistance and the temperature effect. At 90-105 °C the resistance of copper increases by 27-33% relative to nominal; on lines passing through a hot region, do the calculation with this corrected resistance.
The full topic: DC Motor Driver Selection Guide