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               76 Power systems engineering ± fundamental concepts

































                      Fig. 2.48 Third-harmonic oscillation of the star-point potential.


                      in each phase. If the same load is reconnected in wye, the equivalent circuit does not
                      change, but with a three-wire connection the zero-sequence triplen-harmonic currents
                      are prevented from flowing, and therefore the load cannot operate normally. On the
                      other hand, the triplen-harmonic voltage sources may still be active, and in this case,
                      in order to eliminate the triplen-harmonic voltage components from the line±line
                      voltages, the potential of the start point will oscillate, as shown in Figure 2.48.

                      2.12.5   AC line harmonics of three-phase rectifier

                      As an example of a harmonic-generating load, Figure 2.49 shows the circuit of a
                      three-phase rectifier supplying a DC load. If the DC current has negligible ripple
                      (i.e. the DC load has enough inductance to keep the current essentially constant
                      through a 120 period), and if the commutation is perfect (i.e. the current passes

                      from one SCR to the next at the instant when the voltage on the incoming phase
                      exceeds the voltage on the outgoing phase), then the AC line current waveform

                      is a 120 squarewave, Figure 2.50. By Fourier's theorem, using 1/4-cycle sym-
                      metry,
                                           2     Z  p=2           4      hp
                                       I h ˆ    4    I cos (hy)dy ˆ  sin                (2:69)
                                           2p     0               hp     3
                                      p              p        p
                      If h ˆ 1, sin p/3 ˆ  3/2 and I 1 /I ˆ 43/2p ˆ 23/p ˆ 1:103. If h ˆ 3, sin 3p/3 ˆ 0
                                                    p                                    p
                      and I 3 /I ˆ 0. If h ˆ 5, sin 5p/3 ˆ  3/2 and jI 5 jˆ I 1 /5. If h ˆ 7, sin 7p/3 ˆ  3/2
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