23 5 2018 10 Vol. 23 No. 5 JOURNAL OF HARBIN UNIVERSITY OF SCIENCE AND TECHNOLOGY Oct. 2018 150080 αβ0 MT0 ABC DOI 10. 15938 /j. jhst. 2018. 05. 009 TM351 A 1007-2683 2018 05-0046- 08 Indcton Motor Hybrd Coordnate Transformaton Model and Its Applcaton LIANG Yan-png WAN Yn-long WANG Chen-gang School of Electrcal and Electronc Engneerng Harbn Unversty of Scence and Technology Harbn 150080 Chna Abstract For the strong coplng problem between the rotor crrent and the ar gap flx n the expresson of the electromagnetc torqe expresson of the ndcton motor the hybrd coordnate transformaton s adopted by the αβ0 transformaton and the MT0 transformaton and the electromagnetc torqe expresson of the ndcton motor s decopled. The hybrd coordnate transformaton s dedced and the mathematcal model of hybrd coordnate transformaton of ndcton motor s gven. The smlaton model of the ndcton motor s respectvely establshed nder the hybrd coordnate system and the ABC coordnate system. The mathematcal model of the hybrd coordnate transformaton n the ndcton motor s verfed by comparng the smlaton reslts. In the applcaton of the hybrd coordnate transformaton the ndrect vector control smlaton system based on flx lnkage open-loop and speed closed loop s taken as an example. The smlaton reslts show that the stator crrent of the ndcton motor s decopled nto the stator crrent component generatng the magnetc feld and the stator crrent component generatng torqe by the hybrd coordnate transformaton so the valdty of the hybrd coordnate transformaton s verfed. Keywords electromagnetc torqe hybrd coordnate transformaton decoplng 2016-12 - 05 E2015061. 1963 1984. 1989 E-mal 1129959523@ qq. com.
5 47 0 1 1. 1 1 T e = K 1 Φ m I 2 cosθ 2 1 2-6 K 1 Φ m I 2 7-9 10-11 cosθ 2 1 T 12 e Φ / 13 m I 4 2 Φ m Φ m 14 I 2 Φ m I 2 15 dq0 T e = K 2 ΦI a 2 16 K 2 Φ PI I /F I a dq 17 dq0 Φ I a Φ I a Φ 18 I a 19 20 1. 2 αβ0 MT0 αβ0 MT0 αβ0 1 ABC 1
48 23 1 ABC αβ cosθ snθ 0 N 3 A B C C 2s/2r = - snθ cosθ 0 N 2 α β A 0 0 1 α 1 N 2 α = N 3 A - ( N 3 B - N 3 C ) cos60 = N 3 ( A - 1 2 B - 1 2 C ) 3 N 2 β = N 3 B sn60 - N 3 C sn60 = N 3 槡 3 2 B - 槡 3 ( 2 ) 4 C 3 ~ 4 [ α ] = N 1-1 - 1 2 3 2 A N B 5 2 β 2 0 槡 3 - 槡 3 2 2 C 1. 3 0 = K N 3 ( N A + B + C ) 6 2 5 ~ 6 3 N 3 2 = N 2 槡 3 K = 1 A B C A 7 槡 2 a b c a A 5 ~ 7 φ ω 1 α A β = C 3s/2s B 8 0 C 1-1 - 1 2 2 2 C 3s/2s = 0 槡 3 - 槡 3 槡 3 2 2 1 1 1 槡 2 槡 2 槡 2 MT0 2 αβ MT α β m t ω 1 F s θ M α 3 2 A = R s A + pψ A [ m cosθ snθ ] = [ t - snθ cosθ] [ α ] 9 B = R s B + pψ B β C = R s C + pψ C 11 m a = R r a + pψ a α t = C 2s/2r β b = R r b + pψ b 10 c = R r c + pψ c 0 0
5 49 A A ψ A A A A αβ0 MT0 M A α θ s θ r ( ) A A = 槡 2 R s R r p 3 smcosθ s - st snθ s + 1 槡 2 s0 A = 槡 2 3 smcosθ s - st snθ s + 1 槡 2 s0 ( ) 16 ψ A L AA L AB L AC L Aa L Ab L Ac A ψ ψ B L BA L BB L BC L Ba L Bb L Bc A = 槡 2 3 ψ smcosθ s - ψ st snθ s + 1 B 槡 2 s0 ( ) ψ C L CA L CB L CC L Ca L Cb L Cc C 16 A = R s A + pψ A = ψ a L aa L ab L ac L aa L ab L ac a sm = R s sm + pψ sm - ω 1 ψ st } ψ b L ba L bb L bc L ba L bb L bc b st = R s rt + pψ st + ω 1 ψ sm 17 ψc L ca L cb L cc L ca L cb Lcc c s0 = R s s0 + pψ s0 12 L AA A sm = R s sm + pψ sm - ω 1 ψ st L AB B A st = R s rt + pψ st + ω 1 ψ sm 18 rm = R r rm + pψ rm - ω s ψ rt T e = n p L ms [ ( A a + B b + C c ) snφ + rt = R r rt + pψ rt + ω s ψ rm ( A b + B c + C a ) sn( φ + 2π / 3 ) + ( A c + B a + C b ) sn( φ - 2π / 3) ] 13 ω s n p L ms 12 ψ [ s ψ ] r I [ ] [ s ] = L ss L sr L ψ s = [ ψ A ψ B ψ ] T C ψ r = [ ψ a ψ b ψ ] T c C s3s/2r = C 3s/2s C 2s/2r = I s = [ A B ] T C I r = [ a b ] T c cosθ s cos( θ s - 2π / 3) cos( θ s + 2π / 3) 2 - snθ 槡 s - sn( θ s - 2π / 3) - sn( θ s + 2π / 3) L AA L AB L AC L Aa L Ab L Ac 3 L ss = L BA L BB L BC L 1 / 槡 2 1 / 槡 2 1 / 槡 2 sr = L Ba L Bb L Bc L CA L CB LCC L Ca L Cb LCc 14 L aa L ab L ac L aa L ab L ac C - 1 s2r /3s = C T s3s/2r L rs = L ba L bb L bc L rr = L ba L bb L bc L ca L cb LcC L ca L cb Lcc C r3s/2r = C 3s/2s C 2s/2r = cosθ r cos( θ r - 2π / 3) cos( θ r + 2π / 3) 2 - snθ 槡 r - sn( θ r - 2π / 3) - sn( θ r + 2π / 3) ψ sm sm 3 ψ st 1 / 槡 2 1 / 槡 2 1 / 槡 2 st ψ s0 15 = C s3s/2rl ss C - 1 s2r /3s C s3s/2r L sr C - 1 r2r /3s C - 1 r2r /3s = C T ψ r3s/2r rm [ C r3s/2r L rs C - 1 s2r /3s C r3s/2r L rr C ] s0 = - 1 r2r /3s rm ψ rt rt ψr0 r0 L rs rr I r 19
50 23 L s 0 0 L m 0 0 0 L s 0 0 L m 0 0 0 L s0 0 0 0 L m 0 0 L r 0 0 0 L m 0 0 L r 0 0 0 0 0 0 L r0 sm st s0 rm rt r0 20 L s = 1. 5L ms + L ls L ls L s0 = L ls L m = 1. 5L ms L r = 1. 5L ms + L lr L lr L r0 = L lr C - 1 s2r /3s C - 1 r2r /3s φ = θ s - θ r 13 22 T e = n p L m L r ( st rm - sm ) rm = ( ψ rm - L m sm ) /L r rt = ( ψ rt - L m st ) /L } r rt 21 22 MT0 M ψ r ψ rm = ψ r ψ rt = 0 23 rm = rt = 0 22 ~ 24 18 T r = L r /R r L m ψ r = T r p + 1 sm 24 25 25 ψ r 200 N m sm sm ψ r 22 ~ 23 21 T e = n p L m L r st ψ r 26 26 ψ r T e st 4 6 6 2 30 A 75 A 2. 1 7 Smlnk 0 200 N m 4 Smlnk 4 5 5 5
5 51 7 8 1 500 r /mn 1 420 r /mn 0. 1 s 10 ABC 8 2. 2 11 ABC Smlnk 9 12 12 ABC 0 200 N m 9 ABC Smlnk 9 ABC 10 10 200 N m 13 ABC 1 500 r /mn 1 420 r /mn 11 ABC 0. 2 s 30 A 2. 1 2. 2 75 A ABC 13
52 23 2. 3 Smlnk 14 IGBT 3s /2s 2s /2r PI 16 sm ψ r 14 14 1. 15 148 200 N m sm ψ r st T e 17 st T e 15 15 3 s 1 s 16 sm sm ψ r T r 17 3 s st 0 T e 0 200 N m sm 3 st T e ψ r
5 53 1 J. 2017 34 3 253-257. 2 11 J - 16. 12 J. 2014 19 5 113-119. 14. J. 2011 4 70-72. Matlab 15. d-q J. 2016 35 10 62-65. 1. M. 2004 4-6. 2. J. 2015 43 12 2179-2182. 3 HUANG Xan HUANG Y. Modelng and Smlaton of Vector Control and Varable-freqency Reglatng Speed System of Asynchronos Motor Based on Matlab C / / 2011 Internatonal Conference on Consmer Electroncs Commncatons and Networks CECNet XanNng 2011 2334-2337. 4. STATCOM Performance Drect torqe Control Method wth PWM Approach of J. 2014 18 1 19-25. PMSMs C / / 2014 IEEE Internatonal Conference on Indstral 5. Technology ICIT Bsan 2014 61-66. J. 2016 20 10 3-30. 19. J. 6. 2016 44 3 316-319. J. 2015 19 6 35-40. 20. 7. J. 2016 33 3 300-304. J. 2016 50 12 1-7. 8. J. 2015 30 10 83-89. 9. 10. J. 2016 3 92... 2017 21 4 8. J. 2002 36 6 569-571. 13. 16. J. 2016 20 9 73-79. 17. J. 2012 16 6 13-18. 18 YUN Chang Kwak AHN Jn-Woo LEE D. Dong-Hee An Hgh