Harmonics Control for Cascaded Multilevel Inverter Applying Waveform Resultant Theory

Transcript

1 28 6 Vol.28 No.6 Feb.25, Proceedings of the CSEE 28 Chin.Soc.for Elec.Eng (28) TM 464 A Harmonics Control for Cascaded Multilevel Inverter Appling Waveform Resultant Theor LIU Qing-feng, WANG Hua-min, LENG Zhao-xia, LIU Ding (School of Automation and Information Engineering, Xi an Universit of Technolog, Xi an 748, Shaanxi Province, China) ABSTRACT: In this paper, the means of harmonics control b selecting switching angles are analzed for improving qualit of output voltage in cascaded multilevel inverter. There is considerable difficult in solving the multi-variable high-order equation sstem when the method of solving equations is applied to eliminate specified harmonics, a novel method of eliminating specified harmonics based on waveform resultant theor is presented in this paper, so the switching angles can be easil obtained b simple trigonometr formula deduction. The exemplifications based on calculation values give all possible solutions of switching angles and show that the different choice of switching angles can also adjust the fundamental-frequenc voltage while eliminating the specific harmonics of cascade multilevel inverter output voltage. Simulation and experiment results verif validit of research work KEY WORDS: cascaded multilevel inverter; waveform resultant theor; total harmonic distortion; harmonics control; switching angle H [-6] (total harmonic distortion, THD, η THD ) [7] [8-] (sinusoidal pulse width modulation SPWM) SPWM PWM [-3] [-3] PWM H

2 [4-5] (N=4 H ) MOSFET 4 Fig. Structure of a cascaded multilevel inverter U π π π π 2 Fig. 2 Output voltage waveform 2 N 4U 4U u= ( cosn+ cosn2 + + n=,3,5 4U cos nn )sin nωt= U sinωt+ U sin3ωt+ U sin5ωt+ () 3 5 H U U 3 U 5 () 4U (cos+ cos2 + + cos N ) = U π 4U (cos3+ cos cos3 N ) = 3 π (2) 4U (cosn+ cosn2 + + cos nn ) = nπ n (2) cos U [-2] 2 H H H H 3 cos3 + k cos3 = k,( k = N, N 2, 2) 3 3 cos ( k + k)cos ( k k) = = f ( ) (3) k k ( π /2) f (3) 3 (2K + ) ( k + k) = π, K = 3 (2K + ) ( k k) = π, K =, 3 4 H 5 cos5k 3+ cos5 k 2+ cos5k + cos5k = ( k= N, N 4, 4) 5 5 cos ( k 3+ k 2)cos ( k 3 k 2) cos ( k + k)cos ( k k) = (4) (3) 4 2 3( k + k) / 2 =π/ 2 (4) 5 π 5 cos ( k 3 + k )cos ( k 3 k ) = f 5 π 5 cos ( k 3+ k ± )cos ( k 3 k ) = = f ( ) (5) k 2 k 3

3 H (3) 8 4 (5) 7 H H H π/2 [4] H ( 3 ) 3(a) (d) 3 =πu /4 η THD [6] η THD3 3 η THD999 3 /( ), 2/( ) (a), ( ) /( ), 2/( ) (b) ( 2 ) (c), ( 2 ) η THD999 η THD999 η THD (d) ( 2 ) 3 Fig. 3 Possible solutions of switching angles and η THD 4(a)~(j) 3 5 /( ),2 /( ),3 /( ),4 /( ) /( ),2 /( ),3 /( ),4 /( ) (a),,, ( ) η THD (b) ( ) (c),,, ( 2 )

4 72 28 Fig. 4 /( ),2 /( ),3 /( ),4 /( ) /( ),2 /( ),3 /( ), 4 /( ) /( ),2 /( ),3 /( ), 4 /( ) η THD (d) ( 2 ) (e),,, ( 3 ) η THD (f) ( 3 ) (g),,, ( 4 ) η THD (h) ( 4 ) (i),,, ( 5 ) η THD (j) ( 5 ) 4 Possible solutions of switching angles and THD 4 ~5 3 3 = =6 2 5 khz Fourier Fourier 6 3 4(a) = =6 =36 = khz Fourier Fourier FFT t/s f/mhz 5 (N=2) Fig. 5 Output voltage(n=2) t/s f/mhz 6 (N=4) Fig. 6 Output voltage(n=4) U(.8V/ ) U(4V/ ) t(5µs/ ) 7 (N=2) Fig. 7 Experiment waveform(n=2)

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