Square-Wave RMS Current

Square-wave RMS current

In power electronics many components carry no steady current at all, but pulses — a switching transistor, a transformer winding, a rectifier diode. To know the real heating that such a pulsed current causes, the average value will not do: what matters is the RMS value (root mean square), since the heat dissipated by the Joule effect is proportional to the square of the current (P = I²R), and the RMS is exactly the equivalent DC current that would produce the same heating. For a pulsed square wave (the current equals I_p during a fraction D of the period and zero for the rest of it), the RMS value is I_rms = I_p × √D, where I_p is the peak current and D the duty cycle. The square root of the duty cycle is the key: a 10 A peak current that flows only 25% of the time (D = 0.25) has an RMS of 10 × √0.25 = 5 A — and it heats like a 5 A DC current, neither like 10 A nor like the 2.5 A average. This calculation is essential to size parts correctly: wire gauge and PCB trace width (which heat according to the RMS), the current rating of transistors and diodes, the copper losses in transformers and inductors, and capacitors, which carry an RMS ripple current limit that must never be exceeded on pain of overheating. Underestimating the RMS — taking the average value by mistake — leads to undersized components that fail from heat. Enter the peak current and the duty cycle.