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(A)400(V)4002000-200t(s)2000-200t(s)-4000 0.020.04Fig. 30: Line current with filter-4000 0.020.04Fig. 32: Simple voltage with filter(A)4002000-200t(s)-4000 0.020.04Fig. 31: Neutral current with filter4.6 Filtering devicesc Place a passive filter tuned to the third orderharmonic close to the non-linear loads(see fig. 33).Note that this solution requires relatively bulkycomponents, given the low tuning frequency.c Use an active compensator placed close to thenon-linear loads (see fig. 34).Note that this type of device has the capacity tocompensate a harmonic current in the neutralwhose magnitude is three times that of thephase current.Example: Harmonic current per phase 30 ANeutral harmonic current 90 Ac Hybrid filter (see fig. 35): association of anactive compensator which will eliminate the thirdharmonics and a passive filter which willeliminate the dominant harmonics (5 and 7 forexample)iririsitirisitinFig. 34: Active filterActivefilterLoadLoadisitLoadActivefilterinFig. 33: Third order passive filterinFig. 35: Hybrid filterCahier Technique Schneider Electric no. 202 / p.14
Appendix: Calculating Fourier coefficients for arectangular currentConsider the simplified schematic (see fig. 36)for a controlled rectifier, supplying an ideal load,and the current in each of the power supplyphases (see fig. 37).I lineI dcThis gives:⎛ π⎞⎛ π⎞cos ⎜5n = cos n n 6⎟ ⎜ π − 6⎟⎝ ⎠ ⎝ ⎠⎛ π⎞= cos ( n π)cos ⎜n ⎝ 6⎟⎠⎛ π⎞sin n π sin ⎜n ⎝ 6⎟⎠+ ( )n π= ( −1 ) cos n 6And therefore:Fig. 36: Controlled rectifier supplying a load whichdraws a perfectly smooth currentb n =2 I dc ⎡n cos ⎛n π⎞n ⎛ π⎞⎤⎜ ⎟ − ( −1 ) cos nπ⎢⎜ ⎟⎣ ⎝ 6 ⎠⎝ 6 ⎠ ⎥⎦(A)100500-50π/6 5π/6 π-1000 0.005 0.01 0.015Fig. 37: Power supply currentI dcI lineThis function can be expressed in the form of aFourier series:∞nn = 1( ) = ( ) + ( )I t ∑ a cos nωt b sin nωtSince the function is odd, all the coefficients a nare zero.The coefficients b n can be calculated using theequation:2 πb n = ∫ I ( t ) sin ( nωt ) d ω tπ 0bnbnbn2 Idcπn5π6π sin nωt d ω t65πcos nωt6π= ∫ ( )2 Idcπ n[ ]= − ( )2 I ⎡dcn cos ⎛n π⎞=⎜ cos 5n⎝ 6⎟⎠− ⎛ π⎞⎤⎢⎜⎝ 6⎟ ⎥π ⎣⎢⎠ ⎦⎥6t(s)b n =2 I dc ⎡n cos ⎛n π⎞n+1 ⎛ π⎞⎤⎜ ⎟ − ( −1 ) cos nπ⎢⎜ ⎟⎣ ⎝ 6 ⎠⎝ 6 ⎠ ⎥⎦If n is even: b = n 0If n is odd: b n =4 I dcn cos ⎛πn π⎞⎝ 6 ⎠If n π / 6 is an odd multiple of π / 2 , thenb = n 0π( )In other words, for n 6= 2k + 1 2Hence: n = 3 (2k + 1)Put in different terms, if n is an odd multiple of 3,the terms b n are zero.The only non-zero terms are therefore of the form:b n =2 3 I dcnπ( −1) mwhere n = 6m ± 1, m = 0, 1, 2, ...In particular, we get:b 1 =2 3 I dcπThe rms value of the fundamental is therefore:I6 Iπdc1 =The rms value of the non-zero harmonics isequal to:II= nn1πCahier Technique Schneider Electric no. 202 / p.15
Page 1 and 2: Collection Technique ..............
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Page 13 and 14: 3 The third harmonic in transformer
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Page 20: Schneider ElectricDirection Scienti

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