Question 2.12: Harmonic Distortion for a CFL. A harmonic analysis of the cu......

Harmonic Distortion for a CFL. A harmonic analysis of the current drawn by a CFL yields the following data. Find the rms value of current and the total harmonic distortion (THD).

Harmonicsrms Current (A)10.1530.1250.0870.0390.02\begin{matrix} \hline \text{Harmonics} & \text{rms Current (A)} \\ \hline 1 & 0.15 \\ 3 & 0.12 \\ 5 & 0.08 \\ 7 & 0.03 \\ 9 & 0.02 \\ \hline \end{matrix}
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From (2.94) the rms value of current is

Irms =  2 [ I122 + I222 + I322 +    ] = I12 + I22 +  I32 I_{rms} \ = \ \sqrt{ \ 2 \ \left[ \ \frac{{I_{1}}^{2}}{2} \ + \ \frac{{I_{2}}^{2}}{2} \ + \ \frac{{I_{3}}^{2}}{2} \ + \ \cdot \ \cdot \ \cdot \ \right] } \ = \ \sqrt{{I_{1}}^{2} \ + \ {I_{2}}^{2} \ +   {I_{3}}^{2}…}  (2.94)
Irms = [(0.15)2 + (0.12)2 + (0.08)2 + (0.03)2 + (0.02)2]1/2 = 0.211 AI_{rms} \ = \ \left[\left(0.15\right) ^{2} \ + \ \left(0.12\right) ^{2} \ + \ \left(0.08\right) ^{2} \ + \ \left(0.03\right) ^{2} \ + \ \left(0.02\right) ^{2} \right]^{{1}/{2}} \ = \ 0.211 \ A

From (2.95) the total harmonic distortion is

THD = I22 + I32 + I42 +   I1THD \ = \ \frac{\sqrt{{I_{2}}^{2} \ + \ {I_{3}}^{2} \ + \ {I_{4}}^{2} \ + \ \cdot \ \cdot \ \cdot}}{I_{1}} (2.95)

THD =  (0.12)2 + (0.08)2 + (0.03)2 + (0.02)20.15 = 0.99 = 99%THD \ = \ \frac{\sqrt{\ \left(0.12\right) ^{2} \ + \ \left(0.08\right) ^{2} \ + \ \left(0.03\right) ^{2} \ + \ \left(0.02\right) ^{2}} }{0.15} \ = \ 0.99 \ = \ 99\%

so the total rms current in the harmonics is almost exactly the same as the rms current in the fundamental.

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