2 Introduction to Harmonics and Mitigation2.1 Harmonics and Mitigation2.1.1 Linear LoadsOn a sinusoidal AC supply, a purely resistive load (forexample an incandescent light bulb) draws a sinusoidalcurrent in phase with the supply voltage.The power dissipated by the load is:P = U × IFor reactive loads (such as an induction motor), the currentis no longer in phase with the voltage. Instead, the currentlags the voltage creating a lagging power factor with avalue less than 1. In the case of capacitive loads, thecurrent is ahead of the voltage, creating a leading powerfactor with a value less than 1.Illustration 2.1 Current Creating a True Power FactorIn this case, the AC power has 3 components:• Real power, (P).• Reactive power, (Q).• Apparent power, (S).The apparent power is:S = U × I(where S=[kVA], P=[kW] and Q=[kVAR]).In the case of a perfectly sinusoidal waveform, P, Q, and Scan be expressed as vectors that form a triangle:S2 = P2 + Q2PSQφ130BB538.10Illustration 2.2 Sinusoidal WaveformThe displacement angle between current and voltage is φ.The displacement power factor is the ratio between theactive power (P) and apparent power (S):DPF = PS = cos(ϕ)2.1.2 Non-linear LoadsNon-linear loads (such as diode rectifiers) draw a non-sinusoidal current. Illustration 2.3 shows the current drawnby a 6-pulse rectifier on a 3-phase supply.A non-sinusoidal waveform can be decomposed in a sumof sinusoidal waveforms with periods equal to integermultiples of the fundamental waveform.f (t) = ∑ah × sin hω1tSee Illustration 2.3.Introduction to Harmonics a... Design GuideMG80C602 Danfoss A/S © 05/2019 All rights reserved. 92 2