Bao cao DTCS_biebn tan 2 bac và 3 bac npc

Mn: in t cng sut nng cao

GVHD: PGS.TS. Nguyn Vn Nh

MC LCPHN A: GII THIU.......................................................................................................2 1 Tng quan v b bin tn....................................................................................................2 2 B chnh lu........................................................................................................................2 3 B nghch lu......................................................................................................................3 PHN B: L THUYT.......................................................................................................9 1 B bin tn 2 bc.................................................................................................................9 2 B bin tn 3 bc NPC......................................................................................................20 PHN C: KT QU M PHNG...................................................................................35 1 B bin tn 2 bc...............................................................................................................35 2 B bin tn 3 bc NPC......................................................................................................45 PHN D: KT LUN........................................................................................................53

HVTH: Nguyn V S

1



PHN A: GII THIU1. Tng quan v b bin tn B bin tn dng chuyn i in p hoc dng in xoay chiu u vo t mt tn s ny thnh in p hoc dng in c mt tn s khc u ra. B bin tn thng c s dng iu khin vn tc ng c xoay chiu theo phng php iu khin tn s, theo tn s ca li ngun s thay i thnh tn s bin thin. Ngoi vic thay i tn s cn c s thay i tng s pha. T ngun li mt pha, vi s gip ca b bin tn ta c th mc vo ti ng c ba pha. B bin tn cn c s dng rng ri trong k thut nhit in. B bin tn trong trng hp ny cung cp nng lng cho l cm ng. B bin tn c chia ra lm 2 loi: + Bin tn gin tip: trong mch c cha khu trung gian mt chiu. Cu to ca b bin tn gin tip gm c b chnh lu vi chc nng chnh lu in p xoay chiu vi tn s c nh ng vo v b nghch lu thc hin vic chuyn i in p (hoc dng in) chnh lu sang dng p hoc dng xoay chiu ng ra. Bng cu trc nh trn, ta c th iu khin tn s ra mt cch c lp khng ph thuc tn s vo + Bin tn trc tip (cn c gi l cycloconvertor): trong mch khng c khu trung gian mt chiu. B bin tn trc tip-Cycloconverter, to nn in p xoay chiu ng ra vi tr hiu dng v tn s iu khin c. Ngun in p xoay chiu vi tn s v bin khng i cung cp nng lng cho b bin tn ny. B bin tn trc tip dng iu khin truyn ng ng c in xoay chiu. Theo qu trnh chuyn mch, b bin tn trc tip c phn bit lm hai loi: b bin tn c qu trnh chuyn mch ph thuc v b bin tn c qu trnh chuyn mch cng bc. Trong phm vi bi bo co ny chng ta ch xt trng hp b bin tn gin tip vi hai b phn chnh l b chnh lu v b nghch lu. B chnh lu B chnh lu c s dng i in p (dng in) xoay chiu mt pha hoc ba pha thnh in p (dng in) mt chiu. y, chng ta ch xt n trng hp b chnh lu ba pha. B chnh lu ba pha c chia thnh hai loi: chnh lu tia v chnh lu cu. 1.1 B chnh lu tia ba pha Gi s cho ngun ba pha l tng: V1 = Vm cos( )2 ) 3 4 V3 = Vm cos( ) 3 V2 = Vm cos(

2.

Khi dng ti lin tc, in p ti ch ph thuc vo in p ngun v c ln tr trung bnh:Vd = 1 2 3

6

+

2 3

Vm cos d =

6

3 3 Vm 2

HVTH: Nguyn V S

2



a. B chnh lu tia ba pha. b. Tn hiu ng vo v ng ra. Hnh 1. B chnh lu tia ba pha. 1.2 B chnh lu cu ba pha

a. B chnh lu cu ba pha. b. Dng tn hiu ng v ng ra. Hnh 2. B chnh lu cu ba pha. A 3 3 2 Vd = 1 = 2 2VLL cos( + )d = VLL (V) 6 6 3

Vi VLL l in p dy: VLL =

Vm 3 2

2. B nghch lu B nghch lu c nhim v chuyn i nng lng t ngun in mt chiu khng i sang dng nng lng in xoay chiu cung cp cho ti xoay chiu. i lng c iu khin ng ra l in p hoc dng in, do ngi ta thng chia b nghch lu ra lm hai loi: b nghch lu p v b nghch lu dng. + B nghch lu p: ngun mt chiu cung cp cho b nghch lu l ngun in p. + B nghch lu dng: ngun in p cung cp cho b nghch lu l ngun dng in. Cc b nghch lu to thnh b phn ch yu trong cu to ca b bin tn. ng dng quan trng v tng i rng ri ca chng nhm vo lnh vc truyn ng in ng cHVTH: Nguyn V S 3



+ ++

+ +

xoay chiu vi chnh xc cao. Trong lnh vc tn s cao, b nghch lu c dng trong cc thit b l cm ng trung tn, thit b hn trung tn. B nghch lu cn c dng lm ngun in xoay chiu cho nhu cu gia nh, lm ngun in lin tc UPS, iu khin chiu sng, b nghch lu cn c ng dng vo lnh vc b nhuyn cng sut phn khng. Cc ti xoay chiu thng mang tnh cm khng (v d ng c khng ng b, l cm ng), dng in qua cc linh kin khng th ngt bng qu trnh chuyn mch t nhin. Do , mch b nghch lu thng cha linh kin t kch ngt c th iu khin qu trnh ngt dng in. 2.1 B nghch lu p B nghch lu p c rt nhiu loi cng nh nhiu phng php iu khin khc nhau. Theo s pha in p u ra: nghch lu p 1 pha, 3 pha, Theo s cp gi tr in p gia u pha ti n mt im in th chun trn mch c: hai bc (two-level), a bc (Multi_level t 3 bc tr ln). Theo cu hnh ca b nghch lu: dng cascade (cascade inverter), dng nghch lu cha diode kp NPC (Neutral Point Clamped Multilevel Inverter), ... Theo phng php iu khin: Phng php iu rng. Phng php iu bin. Phng php iu ch rng xung (PWM) Phng php iu ch rng xung ci bin (Modified PWM). Phng php iu ch vector khng gian (SVPWM Carrier Based PWM). Trong bi bo co ny, ta ch xt phng php iu ch rng xung (PWM) cho b nghch lu ba pha hai bc v ba bc. 2.1 .1 B nghch ba pha hai bc

Hnh 3. B nghch lu ba pha hai bc. B nghch lu hai bc cha hai kho bn dn trn mi nhnh pha ti c gi chung lHVTH: Nguyn V S 4



nghch lu p hai bc ( two-level VSI ). Chng c ng dng rng ri trong phm vi cng sut va v nh. Khi nim hai bc xut pht t qu trnh in p gia u mt pha ti n mt im in th chun trn mch thay i gia hai bc gi tr khc nhau. B nghch lu p hai bc c nhc im l to in p cung cp cho cun dy ng c vi dc (dv/dt ) kh ln v gy ra mt s vn kh khn bi tn ti trng thi khc zero ca tng in th t cc pha n tm ngun DC (hin tng common-mode voltage). B nghch lu p a bc c pht trin gii quyt cc vn gy ra nu trn ca b nghch lu p hai bc v thng c s dng cho cc ng dng in p cao v cng sut ln. 2.1.2 B nghch lu ba pha ba bc Cc u im ca b nghch lu p a bc: Cng sut ca b nghch lu p tng ln. i vi ti cng sut ln, in p cung cp cho ti c th t gi tr tng i ln. in p t ln linh kin b gim xung nn cng sut tn hao do qu trnh ng ngt ca linh kin cng gim theo. Vi cng tn s ng ngt, cc thnh phn sng hi bc cao ca in p ra gim nh hn so vi trng hp b nghch lu p hai bc. Cu hnh b nghch lu p a bc Theo cu hnh ca b nghch lu p a bc ta c 2 dng: dng cascade (cascade inverter), dng nghch lu cha diode kp NPC (Neutral Point Clamped Multilevel Inverter),

a. B nghch lu ba bc NPC. b. B nghch lu ba bc cascade. Hnh 4. B nghch lu a bc. 2.1.3 Cc phng php iu khin b nghch lu a bc Da vo cc k thut iu khin ng ngt linh kin trong b nghch lu ngi ta thng chia thnh cc phng php nh iu bin, iu ch rng xung (PWM), iu ch vect khng gian (SVM), Phng php iu khin theo bin Phng php ny c gi tt l phng php iu bin. Trong phng php iu bin i hi in p ngun dc phi iu khin c. ln in p ra c iu khin bng cchHVTH: Nguyn V S 5



iu khin ngun in p DC. Chng hn s dng b chnh lu c iu khin hoc kt hp b chnh lu khng iu khin v b bin i in p DC. B nghch lu p thc hin chc nng iu khin tn s in p ra. Cc cng tc trong cp cng tc cng pha ti c kch ng vi thi gian bng nhau v bng mt na chu k p ra. Mch iu khin kch ng cc cng tc trong b nghch lu p v th n gin. B nghch lu p ba pha iu khin theo bin cn c gi l b nghch lu p 6 bc ( six-step voltage inverter). Tn s p c bn bng tn s ng ngt linh kin. Cc thnh phn sng hi bi ba v bc chn khng xut hin trn p dy cung cp cho ti. Cn li cc sng hi bc (6k 1), k=1,2,3. cn kh b bng cc bin php lc sng hi. Phng php iu ch rng xung sin

Hnh 5. S iu khin b nghch lu p dng phng php SPWM. Gin kch ng cc cng tc da trn vic so snh hai tn hiu c bn: - Sng mang up (carrier signal) tn s cao. - Sng iu khin ur (reference signal) hoc sng iu ch (modulating signal) dng sin. V d ur>up th cng tc l c kch ng, khi ur< up th cng tc chn c kch ng.

Hnh 6. Gin xung kch ca b nghch lu phng php SPWM - Sng mang up c dng tam gic , tn s up cng cao th lng sng hi bc cao b kh cng nhiu. - Sng iu khin ur mang thng tin v ln tr hiu dng v tn s sng hi c bn ca in p ng ra.HVTH: Nguyn V S 6


GVHD: PGS.TS. Nguyn Vn Nhmf = ma = f sng f i imang khin khin mang

Gi mf l ch s iu ch tn s: Gi ma l t s iu ch bin :

Vm i i Vm sng

Nu ma 1 (bin sng sin nh hn bin sng mang) th quan h gia bin thnh phn c bn ca p ra v p iu khin l tuyn tnh. Bin p pha hi c bn ca b nghch lu 3 pha l:Vt (1) m = ma . V 2

Nu ma>1 (bin tn hiu iu ch ln hn bin sng mang) th bin hi c bn in p ra tng khng tuyn tnh theo bin ma. Phng php SPWM t c ch s iu ch bin ln nht trong vng tuyn tnh khi bin sng mang bng vi bin sng iu ch. Ta c: V V(1) m mSPWM _ max = = 2 = 0.785 V(1) m _ sixsteps 2 V 4

Hnh 7. Thi gian xung kch S1 v ch s iu ch bin ma. Phng php iu ch rng xung ci bin (Modified_ SPWM).

HVTH: Nguyn V S

7



Hnh 8.Thi gian sng iu khin v ch s iu ch bin . Phng php iu ch rng xung sin (SPWM) ch thc hin iu khin tuyn tnh vi phm vi ch s iu ch l: 0 m 0.785 . Khi bin sng hi c bn in p ra:0 Vt (1) m

Phng php Modified SPWM c ch s m ln hn: 0 m 0.907 .mSPWM _ max V = 3 = = 0.907 2V 2 3

V 2

.

Nguyn l thc hin: Gin kch ng linh kin da vo kt qu so snh tn hiu iu khin v sng mang tn s cao. Sng iu khin (ur1, ur2, ur3 ) c to thnh bng cch cng thm phn tn hiu sin vi mt thnh phn sng hi bi ba (thnh phn th th khng). Khi tng ln sng iu khin t ch s iu ch m >0.907, quan h iu khin tr nn phi tuyn.

PHN B: L THUYTHVTH: Nguyn V S 8



B BIN TN 2 BC: Thng s b bin tn: Ng vo chnh lu cu 3 pha diode. Diode l tng. Ngun ac 3f 380V, 50Hz. T lc dc chn kh ln lm in p t phng. B nghch lu 3 pha, linh kin IGBT l tng. Ti 3f di xng gm RL mc ni tip mi pha; R=5 , L=20mH. 1. Xc nh in p DC trn t lc.: Gi Um = 380 V l in p dy th cp my bin p. in p DC trn t lc (vi b chnh lu dng Diode l tng) c tnh theo cng thc sau:Vd = U m .3 2

2. Bin in p hi c bn cc i c th cung cp cho ti ch tuyn tnh0 m 0.785U (1) m = ma . 2

=

380 .3 2

= 513 .44 (V ) 513 (V ).

3. in p pha ti cc i ch qu iu ch. ch qu iu ch th bin hi c bn in p ra tng khng tuyn tnh theo bin ma l ch s iu ch. Vi ma trong b nghch lu ba pha hai bc:ma =

Vd = 0.785

2 513 = 256 .5(V ) 3.14

Cc thnh phn sng hi bc cao c gim n cc tiu, v gi tr in p ti cc i ch qu iu ch c tnh khi cho ma bng 1. Suy ra:U (1) m = ma . 2

V(1) m 2 Vd

K thut iu rng xung (PWM), s dng sng mang tam gic tn s 5khz, bin sng mang trong phm vi (0,1). Kho st qu trnh qu xc lp dng in 3 pha trong cc trng hp sau: K thut sin. Bin hi c bn p pha ti bng 160V, tn s 40hz.Vt1 = 160 cos( 80 t ) (V ) 2 ) (V ) 3 4 Vt 3 = 160 cos( 80 t ) (V ) 3 V0 max =Vd max( Vt1 ,Vt 2 ,Vt 3 ) = 513 160 = 353 (V ) Vt 2 = 160 cos( 80 t

Vd =1

2 513 = 326 .75 (V ) 3.14

V0 min = min( Vt 1 ,Vt 2 ,Vt 3 ) =160 (V ) V0 min V0 V0 max

Chn in p common mode V0 = 250 (V )V10 = Vt1 +V0 =160 + 250 = 410 (V )

V20 = Vt 2 + V0 = 80 + 250 = 170 (V ) V30 = Vt 3 + V0 = 80 + 250 = 170 (V )

Sng mang trong phm vi [0;1] nn in p iu khin:Vk 1 = V10 410 = = 0.799 (V ) Vd 513

HVTH: Nguyn V S

9

Mn: in t cng sut nng caoVk 2 = Vk 3 V20 170 = = 0.331 (V ) Vd 513 V 170 = 30 = = 0.331 (V ) Vd 513 XL = arctg R


Tng tr ti: Z = R 2 + X L 2 = 52 + (2 .40.20.103 ) 2 = 7.088() Gc lch pha: = arctg 2..40 .20 .10 3 = 45 .14 0 5

Z = 7.088 .14 0 ( 45 ) Dng in 3 pha qua ti : V V 160 cos(80t ) it1 = t1 = t1 = = 22.6 cos( 80t 45.14 0 )( A) 0 Z Z 7.08845.14 2 160 cos( 80t ) Vt 2 3 = 22 .6 cos( 80t 165 .14 0 )( A) it 2 = = 0

7.088 45.14 4 160 cos( 80t ) Vt 3 3 = 22.6 cos( 80t 285 .14 0 )( A) it 3 = = Z 7.088 45.14 0 Z

K thut sng mang vi hm common mode trung bnh (medium common mode), tn s ra 30hz; ch s iu ch m = 0.4; m = 0.866; m = 1. a. m = 0.4. Ch s iu ch: U V 0.4 m = (1) m = 0.4 U (1) m = 0.4 d = 513 = 118.476 (V ) Vd 3 1.732 3 Vy bin in p pha ti l 118.476(V). Gi tr in p cc pha ti l: Vt1 = 118.476 cos( 60 t ) =118.476 cos( ) = 118.476 (V ) (vi = 60t = 00 )2 ) (V ) = 59 .238 (V ) 3 4 Vt 3 = 118 .476 cos( ) (V ) = 59 .238 (V ) 3 Vt 2 =118.476 cos( V0 max =Vd m Vt1 ,Vt 2 ,Vt 3 ) = 513 118 .476 = 394 .524 (V ) ax( V0 min = m Vt1 ,Vt 2 ,Vt 3 ) = 59 .238 (V ) in(

Tm in p common mode:

in p common mode trung bnh:V0 =

(V0 max + V0 min ) 394 .524 + 59 .238 = = 226 .881(V ) 2 22 ) + 226 .881 = 167 .643 (V ) 3 4 +V0 = 118 .476 cos( ) + 226 .881 = 167 .643 (V ) 3

V10 =Vt1 +V0 =118 .476 + 226 .881 = 345 .357 (V )

V20 = Vt 2 +V0 = 118 .476 cos( V30 = Vt 31

HVTH: Nguyn V S

10

Mn: in t cng sut nng cao in p iu khin: V 345 .357 Vk 1 = 10 = = 0.673 (V ) Vd 513 V 167 .643 Vk 2 = 20 = = 0.327 (V ) Vd 513 V 167 .643 Vk 3 = 30 = = 0.327 (V ) Vd 513


Tr khng cun dy: Tng tr ti: Z = R 2 + X L 2 = 52 + ( 2 .30.20.10 3 ) 2 = 6.26( ) Gc lch pha: = arctg 2. .30 .20 .10 3 XL = 37 0 = arctg 5 R

Z = 6.26 0 ( 37 ) Dng in 3 pha qua ti: V 118.476 cos( 60t ) it1 = t1 = = 18.92 cos( 60t 37 0 )( A) 0 Z 6.2637 2 118.476 cos( 60t ) Vt 2 3 = 18 .92 cos( 60t 157 0 ) ( A) it 2 = = 0

Z

6.2637

V it 3 = t 3 = Z

118.476 cos( 60t 6.26 37 0

4 ) 3 = 18 .92 cos( 60t 277 0 ) ( A)

b. m=0.866 + Ch s iu ch :m=

U (1) m V 513 = 0.866 U (1) m = 0.866 d = 0.866 = 256 .5 (V ). Vd 3 1.732 3

Vy bin in p pha ti l 256.5 (V). + Gi tr in p cc pha ti l :Vt1 = 256 .5 cos( ) (V )

2 ) (V ) 3 4 Vt 3 = 256 .5 cos( ) (V ) 3 Khi = 00 ta c: Vt1 = 256 .5 cos( 0) (V ) = 256 .5 (V ) 2 Vt 2 = 256 .5 cos( 0 ) (V ) = 128 .25 (V ) 3 4 Vt 3 = 256 .5 cos( 0 ) (V ) = 128 .25 (V ) 3 Vt 2 = 256 .5 cos( V0 max =Vd max( Vt1 ,Vt 2 ,Vt 3 ) = 513 256 .5 = 256 .5 (V ) V0 min = m Vt1 ,Vt 2 ,Vt 3 ) =128 .25 (V ) in(

(vi = 60t )

+ in p common mode:

V0 =

(V0 max + V0 min ) 256 .5 + 128 .25 = = 192 .375 (V ) 2 2 11

HVTH: Nguyn V S



V10 =Vt1 +V0 = 256 .5 +192 .375 = 448 .875 (V ) V20 =Vt 2 +V0 = 128 .25 +192 .375 = 64 .125 (V ) V30 =Vt 31 +V0 = 128 .25 +192 .375 = 64 .125 (V )

in p iu khin: V 448 .875 Vk 1 = 10 = = 0.875 (V ) Vd 513 V 64 .125 Vk 2 = 20 = = 0.125 (V ) Vd 513 V 64 .125 Vk 3 = 30 = = 0.125 (V ) Vd 513

Dng in 3 pha qua ti: V 256.5 cos(60t ) it1 = t1 = = 41cos(60t 37 0 )( A) Z 6.2637 0 2 256 .5 cos( 60t ) V 3 = 41 cos( 60t 157 0 )( A) it 2 = t 2 = 0Z 6.26 37 it 3 = Vt 3 = Z 256 .5 cos( 60t 7.088 45.14 4 ) 3 = 41 cos( 60t 285 .14 0 ) ( A) 0

c. m=1 Ch s iu ch :m=

U (1) m V 513 = 1 U (1) m = d = = 296 .19 (V ). Vd 3 1.732 3

Vy bin in p pha ti l 296.19 (V). Gi tr in p cc pha ti l :Vt 2 = 296 .19 cos(

Vt1 = 296 .19 cos( ) (V ) = 296 .19 (V )

2 ) (V ) = 148 .095 (V ) 3 4 Vt 3 = 296 .19 cos( ) (V ) = 148 .095 (V ) 3

(vi = 00 )


V0 max =Vd max( Vt1 ,Vt 2 ,Vt 3 ) = 513 296 .19 = 216 .81 (V ) V0 min = m Vt1 , Vt 2 , Vt 3 ) =148 .095 (V ) in(


(V0 max + V0 min ) 216 .81 + 148 .095 = = 182 .45 (V ) 2 2

V10 =Vt1 +V0 = 296 .19 +182 .45 = 478 .64 (V ) V20 =Vt 2 +V0 = 148 .095 +182 .45 = 34 .355 (V ) V30 =Vt 31 +V0 = 148 .095 +182 .45 = 34 .355 (V )

in p iu khin: V 478 .64 Vk 1 = 10 = = 0.933 (V ) Vd 513 HVTH: Nguyn V S 12

Mn: in t cng sut nng caoVk 2 = Vk 3 V20 34 .355 = = 0.67 (V ) Vd 513 V 34 .355 = 30 = = 0.067 (V ) Vd 513


Dng in 3 pha qua ti: V 296.19 cos(60t ) it1 = t1 = = 47.31cos(60t 37 0 )( A) 0 Z 6.2637 2 296 .19 cos( 60t ) Vt 2 3 = 47 .31 cos( 60t 157 0 )( A) it 2 = = 0Z 6.2637 V it 3 = t 3 = Z 296 .19 cos( 60t 6.2637 0 4 ) 3 = 47.31 cos( 60t 277 0 ) ( A)

K thut sng mang vi hm common mode nh nht (minimum common mode), tn s p ra 30Hz; Ch s iu ch m=0.4; m=0.866; m=1. Thc hin phn tch Fourier sng hi v nh gi THD p ti. a. m = 0.4. Ch s iu ch: U V 0.4 m = (1) m = 0.4 U (1) m = 0.4 d = 513 = 118.476 (V ) Vd 3 1.732 3 Vy bin in p pha ti l 118.476 (V). Gi tr in p cc pha ti l:Vt1 =118.476 cos( 60 t ) =118.476 cos( )

2 ) (V ) 3 4 Vt 3 = 118 .476 cos( ) (V ) 3 Ti = 00 , ta c: Vt1 =118.476 (V ) Vt 2 = -59 .238 (V ) Vt 3 = -59 .238 (V ) Vt 2 = 118.476 cos( V0 max =Vd m Vt1 ,Vt 2 ,Vt 3 ) = 513 118 .476 = 394 .524 (V ) ax( V0 min = m Vt1 ,Vt 2 ,Vt 3 ) = 59 .238 (V ) in( V0 =V0 min = 59 .238 (V )

Tnh V0max v V0min:

in p common mode nh nht:V10 =Vt1 +V0 =118 .476 cos( ) + 59 .238 (V )

2 ) + 59 .238 (V ) 3 4 V30 = Vt 31 +V0 = 118 .476 cos( ) + 59 .238 (V ) 3 in p iu khin ( = 00 ): V20 = Vt 2 +V0 = 118 .476 cos(

HVTH: Nguyn V S

13

Mn: in t cng sut nng caoVk 1 =


Vk 2

Vk 3

V10 118 .476 cos( ) + 59 .238 = = 0.346 (V ) Vd 513 2 118 .476 cos( ) + 59 .238 V20 3 = = = 0 (V ) Vd 513 2 118 .476 cos( ) + 59 .238 V30 3 = = = 0 (V ) Vd 5132 ) + 59 .238 3 = = 0 (V ) 513 2 118 .476 cos( 0 ) + 59 .238 V30 3 = = = 0 (V ) Vd 513 118 .476 cos( 0 118 .476 cos( 0) + 59 .238 = 0.346 (V ) 513

Ti = 00 , ta c: Vk 1 =Vk 2 Vk 3

Dng in 3 pha qua ti: V 118.476 cos(60t ) it1 = t1 = = 18.92 cos(60t 37 0 )( A) 0 Z 6.2637 2 118.476 cos( 60t ) Vt 2 3 = 18 .92 cos( 60t 157 0 ) ( A) it 2 = = 0Z 6.26 37 it 3 = Vt 3 = Z 118.476 cos( 60t 6.26 37 0 4 ) 3 = 18 .92 cos( 60t 277 0 ) ( A)

b. m=0.866 + Ch s iu ch :m=

U (1) m V 513 = 0.866 U (1) m = 0.866 d = 0.866 = 256 .5 (V ). Vd 3 1.732 3

Vy bin in p pha ti l 256.5 (V). + Gi tr in p cc pha ti l :

Vt1 = 256 .5 cos( ) (V ) (vi = 60t ) 2 Vt 2 = 256 .5 cos( ) (V ) 3 4 Vt 3 = 256 .5 cos( ) (V ) 3 Khi = 00 ta c: Vt1 = 256 .5 cos( 0) (V ) = 256 .5 (V )

2 ) (V ) = 128 .25 (V ) 3 4 Vt 3 = 256 .5 cos( 0 ) (V ) = 128 .25 (V ) 3 Vt 2 = 256 .5 cos( 0

+ in p common mode:HVTH: Nguyn V S 14

Mn: in t cng sut nng caoV0 min = m Vt1 ,Vt 2 ,Vt 3 ) =128 .25 (V ) in(


V0 max =Vd max( Vt1 ,Vt 2 ,Vt 3 ) = 513 256 .5 = 256 .5 (V )

Chn V0 = V0min=128.25 (V), ta c:V10 =Vt1 +V0 = 256 .5 +128 .25 = 384 .75 (V ) V20 =Vt 2 +V0 = 128 .25 +128 .25 = 0 (V ) V30 =Vt 31 +V0 = 128 .25 +128 .25 = 0 (V )

in p iu khin: V 384 .75 Vk 1 = 10 = = 0.75 (V ) Vd 513 V 0 Vk 2 = 20 = = 0 (V ) Vd 513 V 0 Vk 3 = 30 = = 0(V ) Vd 513

Dng in 3 pha qua ti: V 256.5 cos(60t ) it1 = t1 = = 41 cos(60t 37 0 )( A) 0 Z 6.2637 2 256 .5 cos( 60t ) Vt 2 3 = 41 cos( 60t 157 0 )( A) it 2 = = 0Z 6.26 37 it 3 = Vt 3 = Z 256 .5 cos( 60t 6.2637 0 4 ) 3 = 41 cos( 60t 277 0 ) ( A)


U (1) m V 513 = 1 U (1) m = d = = 296 .19 (V ). Vd 3 1.732 3


Vt1 = 296 .19 cos( ) (V ) = 296 .19 (V )

2 ) (V ) = 148 .095 (V ) 3 4 Vt 3 = 296 .19 cos( ) (V ) = 148 .095 (V ) 3 V0 max =Vd max( Vt1 ,Vt 2 ,Vt 3 ) = 513 296 .19 = 216 .81 (V ) V0 min = m Vt1 , Vt 2 , Vt 3 ) =148 .095 (V ) in(

(vi = 00 )

Chn V0 = V0min = 148.095 (V), ta c:

V10 =Vt1 +V0 = 296 .19 +148 .095 = 444 .285 (V ) V20 = Vt 2 +V0 = 148 .095 +148 .095 = 0 (V ) V30 =Vt 31 +V0 = 148 .095 +148 .095 = 0(V )

in p iu khin:

HVTH: Nguyn V S

15

Mn: in t cng sut nng caoVk 1 = Vk 2 Vk 3 V10 444 .285 = = 0.866 (V ) Vd 513 V 0 = 20 = = 0 (V ) Vd 513 V 0 = 30 = = 0(V ) Vd 513


Dng in 3 pha qua ti: V 296.19 cos(60t ) it1 = t1 = = 47.31cos(60t 37 0 )( A) 0 Z 6.2637 2 296 .19 cos( 60t ) Vt 2 3 = 47 .31 cos( 60t 157 0 )( A) it 2 = = 0Z 6.2637 V it 3 = t 3 = Z 296 .19 cos( 60t 6.2637 0 4 ) 3 = 47.31 cos( 60t 277 0 ) ( A)

K thut sng mang vi hm offset cc tr ln nht (V0max), tn s p ra 30Hz; Ch s iu ch m=0.4; m=0.866; m=1. Thc hin phn tch Fourier sng hi v nh gi THD p ti. a. m = 0.4. Ch s iu ch: U V 0.4 m = (1) m = 0.4 U (1) m = 0.4 d = 513 = 118.476 (V ) Vd 3 1.732 3 Vy bin in p pha ti l 118.476 (V). Gi tr in p cc pha ti l:Vt1 =118.476 cos( 60 t ) =118.476 cos( )

2 ) (V ) 3 4 Vt 3 = 118 .476 cos( ) (V ) 3 Ti = 00 , ta c: Vt1 =118.476 (V ) Vt 2 = -59 .238 (V ) Vt 3 = -59 .238 (V ) Vt 2 = 118.476 cos( V0 max =Vd m Vt1 ,Vt 2 ,Vt 3 ) = 513 118 .476 = 394 .524 (V ) ax( V0 min = m Vt1 ,Vt 2 ,Vt 3 ) = 59 .238 (V ) in( V0 =V0 max = 394 .524 (V ) , ta c: V10 = Vt1 +V0 =118 .476 + 394 .524 = 513 (V ) V20 = Vt 2 +V0 = 59 .238 + 394 .524 = 335 .285 (V ) V30 = Vt 31 +V0 = 59 .238 + 394 .524 = 335 .285 (V )

Tnh V0max v V0min:

Chn

in p iu khin ( = 00 ): HVTH: Nguyn V S 16

Mn: in t cng sut nng caoVk 1 = Vk 2 Vk 3 V10 513 = = 1(V ) Vd 513 V 335 .285 = 20 = = 0.654 (V ) Vd 513 V 335 .285 = 30 = = 0.654 (V ) Vd 513


Dng in 3 pha qua ti: V 118.476 cos(60t ) it1 = t1 = = 18.92 cos(60t 37 0 )( A) 0 Z 6.2637 2 118.476 cos( 60t ) Vt 2 3 = 18 .92 cos( 60t 157 0 ) ( A) it 2 = = 0Z 6.26 37 V it 3 = t 3 = Z 118.476 cos( 60t 6.26 37 0 4 ) 3 = 18 .92 cos( 60t 277 0 ) ( A)

b. m=0.866 + Ch s iu ch :m= U (1) m V 513 = 0.866 U (1) m = 0.866 d = 0.866 = 256 .5 (V ). Vd 3 1.732 3

Vy bin in p pha ti l 256.5 (V). + Gi tr in p cc pha ti l :Vt1 = 256 .5 cos( ) (V )

2 ) (V ) 3 4 Vt 3 = 256 .5 cos( ) (V ) 3 Khi = 00 ta c: Vt1 = 256 .5 cos( 0) (V ) = 256 .5 (V ) 2 Vt 2 = 256 .5 cos( 0 ) (V ) = 128 .25 (V ) 3 4 Vt 3 = 256 .5 cos( 0 ) (V ) = 128 .25 (V ) 3 Vt 2 = 256 .5 cos( V0 max =Vd max( Vt1 ,Vt 2 ,Vt 3 ) = 513 256 .5 = 256 .5 (V ) V0 min = m Vt1 ,Vt 2 ,Vt 3 ) =128 .25 (V ) in(

(vi = 60t )

+ in p common mode:

Chn V0 = V0max=256.5 (V), ta c:V10 =Vt1 +V0 = 256 .5 + 256 .5 = 513 (V ) V20 =Vt 2 +V0 = 128 .25 + 256 .5 =128 .25 (V ) V30 =Vt 31 +V0 = 128 .25 + 256 .5 =128 .25 (V )

in p iu khin: V 513 Vk 1 = 10 = = 1(V ) Vd 513 HVTH: Nguyn V S 17

Mn: in t cng sut nng caoVk 2 = Vk 3 V20 128 .25 = = 0.25 (V ) Vd 513 V 128 .25 = 30 = = 0.25 (V ) Vd 513


Dng in 3 pha qua ti: V 256.5 cos( 60t ) it1 = t1 = = 41 cos( 60t 37 0 )( A) 0 Z 6.2637 2 256 .5 cos( 60t ) Vt 2 3 = 41 cos( 60t 157 0 )( A) it 2 = = 0Z 6.26 37 V it 3 = t 3 = Z 256 .5 cos( 60t 6.2637 0 4 ) 3 = 41 cos( 60t 277 0 ) ( A)


U (1) m V 513 = 1 U (1) m = d = = 296 .19 (V ). Vd 3 1.732 3


Vt1 = 296 .19 cos( ) (V ) = 296 .19 (V )

2 ) (V ) = 148 .095 (V ) 3 4 Vt 3 = 296 .19 cos( ) (V ) = 148 .095 (V ) 3 V0 max =Vd max( Vt1 ,Vt 2 ,Vt 3 ) = 513 296 .19 = 216 .81 (V ) V0 min = m Vt1 , Vt 2 , Vt 3 ) =148 .095 (V ) in(

(vi = 00 )

Chn V0 = V0max = 216.81 (V), ta c:

V10 = Vt1 +V0 = 296 .19 + 216 .81 = 513 (V ) V20 = Vt 2 +V0 = 148 .095 + 216 .81 = 68 .715 (V ) V30 = Vt 31 +V0 = 148 .095 + 216 .81 = 68 .715 (V )

in p iu khin: V 513 Vk 1 = 10 = = 1(V ) Vd 513 V 68 .715 Vk 2 = 20 = = 0.134 (V ) Vd 513 V 68 .715 Vk 3 = 30 = = 0.134 (V ) Vd 513

Dng in 3 pha qua ti: V 296.19 cos( 60t ) it1 = t1 = = 47.31 cos( 60t 37 0 )( A) 0 Z 6.2637

HVTH: Nguyn V S

18

Mn: in t cng sut nng cao V it 2 = t 2 = Z V it 3 = t 3 = Z 296 .19 cos( 60t 6.2637 0 296 .19 cos( 60t 6.2637 0

GVHD: PGS.TS. Nguyn Vn Nh 2 ) 3 = 47 .31 cos( 60t 157 0 )( A) 4 ) 3 = 47.31 cos( 60t 277 0 ) ( A)

2.

B BIN TN 3 BC NPC Thng s b bin tn: in p 2 t ngun dc bng nhau v bngVd = 240 V mi 2

ngun. B nghch lu 3 pha, linh kin IGBT l tng. Ti 3f di xng gm RL mc ni tip mi pha; R=5 , L=20mH. K thut iu rng xung (PWM), s dng sng mang tam gic, tn s 5kHz, bin sng mang trong phm vi (0,1) v (1,2). Xc nh bin hi c bn ln nht ca in p pha ti trong phm vi tuyn tnh v vng qu iu ch. a. Trong phm vi tuyn tnh 0 m 0.785 :V =m

b. Trong vng qu iu ch:V6step = 2Vd =

2Vd 2.480 = 0.785 = 240 (V ) 3.14 2.480 = 306 (V ) 3.14

K thut sin. Bin hi c bn p pha ti bng 160V, tn s 40Hz. Ta c:Vt1 = 160 cos( 80 t ) (V ) Vt 2 = 160 cos( 80 t 2 ) (V ) 3 4 Vt 3 = 160 cos( 80 t ) (V ) 3Vt1 =160 cos( 0) (V ) =160 (V )

* Ti 80 t = 00:

2 ) (V ) = 80 (V ) 3 4 Vt 3 =160 cos( ) (V ) = 80 (V ) 3 V0 max =Vd max( Vt1 ,Vt 2 ,Vt 3 ) = 480 160 = 320 (V ) Vt 2 = 160 cos(

V0 min = min( Vt1 , Vt 2 , Vt 3 ) = 80 (V )

Chn

V0 min V0 V0 max in p common mode V0 = 240 (V ) V10 = Vt1 +V0 =160 + 240 = 400 (V ) V20 = Vt 2 +V0 = 80 + 240 =160 (V ) V30 =Vt 31 +V0 = 80 + 240 =160 (V )

in p iu khin: HVTH: Nguyn V S 19

Mn: in t cng sut nng caoVk 1 = Vk 2 Vk 3 2V10 2 400 = = 1.667 (V ) Vd 480 2V 2 160 = 20 = = 0.667 (V ) Vd 480 2V 2 160 = 30 = = 0.667 (V ) Vd 480


Tng tr ti: Z = R 2 + X L 2 = 52 + (2 .40.20.103 ) 2 = 7.088() 2..40 .20 .10 3 X = 45 .14 0 Gc lch pha: = arctg RL = arctg 5 = 7.088 .14 0 ( Z 45 ) Dng in 3 pha qua ti : V 160 cos(80t ) it1 = t1 = = 22.6 cos( 80t 45.14 0 )( A) Z 7.088 45.14 0 2 160 cos( 80t ) V 3 = 22 .6 cos( 80t 165 .14 0 )( A) it 2 = t 2 = 0

7.088 45.14 4 160 cos( 80t ) Vt 3 3 = 22.6 cos( 80t 285 .14 0 )( A) it 3 = = Z 7.088 45.14 0 Z

K thut sng mang vi hm common mode trung bnh (medium common mode), tn s ra 30Hz; Ch s iu ch m=0.2;m=0.6; m=0.866; m=1. a. m = 0.2. Ch s iu ch: U V 0 .2 m = (1) m = 0.2 U (1) m = 0.2 d = 480 = 55.43 (V ) Vd 3 1.732 3 Vy bin in p pha ti l 55.43(V). Gi tr in p cc pha ti l: Vt1 = 55.43 cos( 60 ) =55 .43 cos( ) = 55.43 (V ) t (vi = 60t = 00 )2 ) (V ) = 27 .715 (V ) 3 4 Vt 3 = 55 .43 cos( ) (V ) = 27 .715 (V ) 3 V0 max =Vd max( Vt1 ,Vt 2 ,Vt 3 ) = 480 55 .43 = 424 .57 (V ) V0 min = m Vt1 ,Vt 2 ,Vt 3 ) = 27 .715 (V ) in( Vt 2 = 55.43 cos(


(V0 max + V0 min ) 424 .57 + 27 .715 = = 226 .143 (V ) 2 2

V10 =Vt1 +V0 = 55 .43 + 226 .143 = 281 .573 (V ) V20 =Vt 2 +V0 = 27 .715 + 226 .143 =198 .428 (V ) V30 =Vt 31 +V0 = 27 .715 + 226 .143 =198 .428 (V )

HVTH: Nguyn V S

20

Mn: in t cng sut nng cao in p iu khin: 2V 2 281 .573 Vk 1 = 10 = = 1.173 (V ) Vd 480 V 2 198 .428 Vk 2 = 20 = = 0.827 (V ) Vd 480 V 2 198 .428 Vk 3 = 30 = = 0.827 (V ) Vd 480


Tng tr ti: Z = R 2 + X L = 52 + (2 .30.20.103 ) 2 = 6.26()2

Gc lch pha: = arctg

Z = 6.26 0 ( 37 )

2. .30 .20 .10 3 XL = 37 0 = arctg R 5

Dng in 3 pha qua ti: V 55.43 cos( 60t ) it1 = t1 = = 8.85 cos(60t 45.14 0 )( A) 0 Z 6.2637 2 55 .43 cos( 60t ) Vt 2 3 = 8.85 cos( 60t 165 .14 0 )( A) it 2 = = 0Z 6.26 37 V it 3 = t 3 = Z 55 .43 cos( 60t 6.26 37 0 4 ) 3 = 8.85 cos( 60t 285 .14 0 )( A)

b. m = 0.6. Ch s iu ch: U V 0.6 m = (1) m = 0.6 U (1) m = 0.6 d = 480 = 166.282 (V ) Vd 3 1.732 3 Vy bin in p pha ti l 166.282(V). Gi tr in p cc pha ti l:Vt1 =166.282 cos( 60 ) =166 .282 cos( ) (V ) t

Ti

2 ) (V ) 3 4 Vt 3 = 166 .282 cos( ) (V ) 3 = 00 ta c: Vt1 =166.282 cos( ) =166 .282 (V ) 2 Vt 2 = 166 .282 cos( ) = 83 .141 (V ) 3 4 Vt 3 =166 .282 cos( ) (V ) = 83 .141 (V ) 3 Vt 2 = 166 .282 cos( V0 max =Vd max( Vt1 ,Vt 2 ,Vt 3 ) = 480 166 .282 = 313 .718 (V ) V0 min = m Vt1 ,Vt 2 ,Vt 3 ) = 83 .141 (V ) in(

in p common mode:

in p common mode trung bnh:HVTH: Nguyn V S 21

Mn: in t cng sut nng cao V0 =


(V0 max + V0 min ) 313 .718 + 83 .141 = = 198 .43(V ) 2 2

V10 =Vt1 +V0 =166 .282 +198 .43 = 364 .712 (V ) V20 =Vt 2 +V0 = 83 .141 +198 .43 =115 .289 (V ) V30 =Vt 31 +V0 = 83 .141 +198 .43 =115 .289 (V )

in p iu khin: 2V 2 364 .712 Vk 1 = 10 = = 1.52 (V ) Vd 480 V 2 115 .289 Vk 2 = 20 = = 0.48 (V ) Vd 480 V 2 115 .289 Vk 3 = 30 = = 0.48 (V ) Vd 480

Dng in 3 pha qua ti: V 166.282 cos(60t ) it1 = t1 = = 26.56 cos(60t 37 0 )( A) 0 Z 6.2637 2 166 .282 cos( 60t ) Vt 2 3 = 26.56 cos( 60t 157 0 )( A) it 2 = = 0Z 6.2637 it 3 = Vt 3 = Z 166 .282 cos( 60t 6.2637 0 4 ) 3 = 26.56 cos( 60t 277 0 )( A)

c. m=0.866 + Ch s iu ch :m=

U (1) m V 480 = 0.866 U (1) m = 0.866 d = 0.866 = 240 (V ). Vd 3 1.732 3

Vy bin in p pha ti l 240 (V). + Gi tr in p cc pha ti l :Vt1 = 240 cos( ) (V )

2 ) (V ) 3 4 Vt 3 = 240 cos( ) (V ) 3 Khi = 00 ta c: Vt1 = 240 cos( 0) (V ) = 240 (V ) 2 Vt 2 = 240 cos( 0 ) (V ) = 120 (V ) 3 4 Vt 3 = 240 cos( 0 ) (V ) = 120 (V ) 3 Vt 2 = 240 cos(

(vi = 60t )


V0 max =Vd max( Vt1 ,Vt 2 ,Vt 3 ) = 480 240 = 240 (V ) V0 min = m Vt1 ,Vt 2 ,Vt 3 ) =120 (V ) in(

in p common mode trung bnh:

HVTH: Nguyn V S

22

Mn: in t cng sut nng cao V0 = (V0 max + V0 min ) 240 + 120 = = 180 (V ) 2 2


V10 = Vt1 +V0 = 240 +180 = 420 (V ) V20 = Vt 2 +V0 = 120 +180 = 60 (V ) V30 = Vt 31 +V0 = 120 +180 = 60 (V )

in p iu khin: 2V 2 420 Vk 1 = 10 = = 1.75 (V ) Vd 480 2V 2 60 Vk 2 = 20 = = 0.25 (V ) Vd 480 2V 2 60 Vk 3 = 30 = = 0.25 (V ) Vd 480

Dng in 3 pha qua ti: V 240 cos(60t ) it1 = t1 = = 38.33 cos(60t 37 0 )( A) 0 Z 6.2637 2 240 cos( 60t ) Vt 2 3 = 38 .33 cos( 60t 157 0 )( A) it 2 = = 0Z 6.26 37 it 3 = Vt 3 = Z 240 cos( 60t 6.2637 0 4 ) 3 = 38.33 cos( 60t 277 0 )( A)


U (1) m V 480 = 1 U (1) m = d = = 277 .136 (V ). Vd 3 1.732 3

Vy bin in p pha ti l 277.136 (V). Gi tr in p cc pha ti l :Vt1 = 277 .136 cos( ) (V )

Khi

2 ) (V ) 3 4 Vt 3 = 277 .136 cos( ) (V ) 3 = 00 ta c: Vt1 = 277 .136 cos( 0) (V ) = 277 .136 (V ) 2 Vt 2 = 277 .136 cos( 0 ) = 138 .568 (V ) 3 4 Vt 3 = 277 .136 cos( ) (V ) = 138 .568 3 Vt 2 = 277 .136 cos( V0 max =Vd max( Vt1 ,Vt 2 ,Vt 3 ) = 480 277 .136 = 202 .864 (V ) V0 min = m Vt1 ,Vt 2 ,Vt 3 ) =138 .568 (V ) in(


in p common mode trung bnh:HVTH: Nguyn V S 23

Mn: in t cng sut nng cao V0 =


(V0 max + V0 min ) 202 .864 + 138 .568 = = 179 .716 (V ) 2 2

V10 = Vt1 +V0 = 277 .136 +179 .716 = 456 .852 (V ) V20 =Vt 2 +V0 = 138 .568 +179 .716 = 41 .148 (V ) V30 =Vt 31 +V0 = 138 .586 +179 .716 = 41 .148 (V )

in p iu khin: 2V 2 456 .852 Vk 1 = 10 = = 1.9(V ) Vd 480 2V 2 41 .148 Vk 2 = 20 = = 0.171 (V ) Vd 480 2V 2 41 .148 Vk 3 = 30 = = 0.171(V ) Vd 480

Dng in 3 pha qua ti: V 277.136 cos(60t ) it1 = t1 = = 44.27 cos(60t 37 0 )( A) 0 Z 6.2637 2 277 .136 cos( 60t ) Vt 2 3 = 44.27 cos( 60t 157 0 )( A) it 2 = = 0Z 6.2637 it 3 = Vt 3 = Z 277 .136 cos( 60t 6.2637 0 4 ) 3 = 44.27 cos( 60t 277 0 )( A)

K thut sng mang vi hm common mode nh nht (minimum common mode) , tn s p ra 30Hz; Ch s iu ch m=0.2;m=0.6; m=0.866; m=1. Thc hin phn tch Fourier sng hi v nh gi THD p ti. a. m = 0.2. Ch s iu ch: U V 0 .2 m = (1) m = 0.2 U (1) m = 0.2 d = 480 = 55.43 (V ) Vd 3 1.732 3 Vy bin in p pha ti l 55.43(V). Gi tr in p cc pha ti l: Vt1 = 55.43 cos( 60 ) =55 .43 cos( ) = 55.43 (V ) t (vi = 60t = 00 )2 ) (V ) = 27 .715 (V ) 3 4 Vt 3 = 55 .43 cos( ) (V ) = 27 .715 (V ) 3 V0 max =Vd max( Vt1 ,Vt 2 ,Vt 3 ) = 480 55 .43 = 424 .57 (V ) V0 min = m Vt1 ,Vt 2 ,Vt 3 ) = 27 .715 (V ) in( Vt 2 = 55.43 cos( V0 =V0 min = 27 .715 (V ) V10 = Vt1 +V0 = 55 .43 + 27 .715 = 83 .145 (V ) V20 = Vt 2 +V0 = 27 .715 + 27 .715 = 0 (V ) V30 = Vt 31 +V0 = 27 .715 + 27 .715 = 0(V )

in p common mode nh nht:

HVTH: Nguyn V S

24

Mn: in t cng sut nng cao in p iu khin: 2V 2 83 .145 Vk 1 = 10 = = 0.346 (V ) Vd 480 V 2 0 Vk 2 = 20 = = 0 (V ) Vd 480 V 2 0 Vk 3 = 30 = = 0(V ) Vd 480


Dng in 3 pha qua ti: V 55.43 cos(60t ) it1 = t1 = = 8.85 cos(60t 37 0 )( A) 0 Z 6.2637 2 55 .43 cos( 60t ) V 3 = 8.85 cos( 60t 157 0 )( A) it 2 = t 2 = 0Z 6.26 37 it 3 = Vt 3 = Z 55.43 cos( 60t 6.2637 0 4 ) 3 = 8.85 cos( 60t 277 0 )( A)


Ti

2 ) (V ) 3 4 Vt 3 = 166 .282 cos( ) (V ) 3 = 00 ta c: Vt1 =166.282 cos( ) =166 .282 (V ) 2 Vt 2 = 166 .282 cos( ) = 83 .141 (V ) 3 4 Vt 3 =166 .282 cos( ) (V ) = 83 .141 (V ) 3 Vt 2 = 166 .282 cos( V0 max =Vd max( Vt1 ,Vt 2 ,Vt 3 ) = 480 166 .282 = 313 .718 (V ) V0 min = m Vt1 ,Vt 2 ,Vt 3 ) = 83 .141 (V ) in(



V0 =V0 min = 83 .141 (V ) V10 =Vt1 +V0 =166 .282 + 83 .141 = 249 .423 (V ) V20 =Vt 2 +V0 = 83 .141 + 83 .141 = 0 (V ) V30 =Vt 31 +V0 = 83 .141 + 83 .141 = 0(V )

in p iu khin: HVTH: Nguyn V S 25

Mn: in t cng sut nng caoVk 1 = Vk 2 Vk 3 2V10 2 249 .423 = = 1.039 (V ) Vd 480 V 2 0 = 20 = = 0 (V ) Vd 480 V 2 0 = 30 = = 0(V ) Vd 480


Dng in 3 pha qua ti: V 166.282 cos( 60t ) it1 = t1 = = 26.56 cos( 60t 37 0 )( A) 0 Z 6.2637 2 166 .282 cos( 60t ) Vt 2 3 = 25.56 cos( 60t 157 0 )( A) it 2 = = 0Z 6.2637V it 3 = t 3 = Z 166 .282 cos( 60t 6.2637 0 4 ) 3 = 25.56 cos( 60t 277 0 )( A)

c. m=0.866 + Ch s iu ch :m=

U (1) m V 480 = 0.866 U (1) m = 0.866 d = 0.866 = 240 (V ). Vd 3 1.732 3


2 Vt 2 = 240 cos( ) (V ) 3 4 Vt 3 = 240 cos( ) (V ) 3 Khi = 00 ta c: Vt1 = 240 cos( 0) (V ) = 240 (V ) 2 Vt 2 = 240 cos( 0 ) (V ) = 120 (V ) 3 4 Vt 3 = 240 cos( 0 ) (V ) = 120 (V ) 3

(vi = 60t )




V0 =V0 min =120 (V ) V10 = Vt1 +V0 = 240 +120 = 360 (V ) V20 = Vt 2 +V0 = 120 +120 = 0 (V ) V30 =Vt 31 +V0 = 120 +120 = 0 (V )

in p iu khin: 2V 2 360 Vk 1 = 10 = = 1.5 (V ) Vd 480 HVTH: Nguyn V S 26

Mn: in t cng sut nng caoVk 2 = Vk 3 2V20 2 0 = = 0 (V ) Vd 480 2V 20 = 30 = = 0(V ) Vd 480


Dng in 3 pha qua ti: V 240 cos(60t ) it1 = t1 = = 38.33 cos(60t 37 0 )( A) 0 Z 6.2637 2 240 cos( 60t ) Vt 2 3 = 38 .33 cos( 60t 157 0 )( A) it 2 = = 0Z 6.26 37 V it 3 = t 3 = Z 240 cos( 60t 6.2637 0 4 ) 3 = 38.33 cos( 60t 277 0 )( A)


U (1) m V 480 = 1 U (1) m = d = = 277 .136 (V ). Vd 3 1.732 3


Khi




V0 =V0 min =138 .568 (V ) V10 =Vt1 +V0 = 277 .136 +138 .568 = 415 .704 (V ) V20 = Vt 2 +V0 = 138 .568 +138 .568 = 0 (V ) V30 =Vt 31 +V0 = 138 .586 +138 .568 = 0(V )

in p iu khin: 2V 2 415 .704 Vk 1 = 10 = = 1.732 (V ) Vd 480

HVTH: Nguyn V S

27

Mn: in t cng sut nng caoVk 2 = Vk 3 2V20 2 0 = = 0 (V ) Vd 480 2V 20 = 30 = = 0(V ) Vd 480


Dng in 3 pha qua ti: V 277.136 cos(60t ) it1 = t1 = = 44.27 cos(60t 37 0 )( A) 0 Z 6.2637 2 277 .136 cos( 60t ) Vt 2 3 = 44.27 cos( 60t 157 0 )( A) it 2 = = 0Z 6.2637 V it 3 = t 3 = Z 277 .136 cos( 60t 6.2637 0 4 ) 3 = 344 .27 cos( 60t 277 0 )( A)

K thut sng mang vi hm offset cc tr ln nht (v0MAX)) , tn s p ra 30Hz; Ch s iu ch m=0.2;m=0.6; m=0.866; m=1. a. m = 0.2. Ch s iu ch: U V 0 .2 m = (1) m = 0.2 U (1) m = 0.2 d = 480 = 55.43 (V ) Vd 3 1.732 3 Vy bin in p pha ti l 55.43(V). Gi tr in p cc pha ti l: Vt1 = 55.43 cos( 60 ) =55 .43 cos( ) = 55.43 (V ) t (vi = 60t = 00 )2 ) (V ) = 27 .715 (V ) 3 4 Vt 3 = 55 .43 cos( ) (V ) = 27 .715 (V ) 3 V0 max =Vd max( Vt1 ,Vt 2 ,Vt 3 ) = 480 55 .43 = 424 .57 (V ) V0 min = m Vt1 ,Vt 2 ,Vt 3 ) = 27 .715 (V ) in( Vt 2 = 55.43 cos( V0 =V0 max = 424 .57 (V ) V10 = Vt1 +V0 = 55 .43 + 424 .57 = 480 (V ) V20 =Vt 2 +V0 = 27 .715 + 424 .57 = 396 .855 (V ) V30 =Vt 31 +V0 = 27 .715 + 424 .57 = 396 .855 (V )

in p common mode ln nht:

in p iu khin: 2V 2 480 Vk 1 = 10 = = 2(V ) Vd 480 V 2 396 .855 Vk 2 = 20 = = 1.654 (V ) Vd 480 V 2 396 .855 Vk 3 = 30 = = 1.654 (V ) Vd 480

Dng in 3 pha qua ti:HVTH: Nguyn V S 28



Vt1 55.43 cos(60t ) = = 8.85 cos(60t 37 0 )( A) 0 Z 6.2637 2 55 .43 cos( 60t ) Vt 2 3 = 8.85 cos( 60t 157 0 )( A) it 2 = = 0 it1 =Z 6.26 37 V it 3 = t 3 = Z 55.43 cos( 60t 6.2637 0 4 ) 3 = 8.85 cos( 60t 277 0 )( A)


Ti

2 ) (V ) 3 4 Vt 3 = 166 .282 cos( ) (V ) 3 = 00 ta c: Vt1 =166.282 cos( ) =166 .282 (V ) 2 Vt 2 =166 .282 cos( ) = 83 .141 (V ) 3 4 Vt 3 = 166 .282 cos( ) (V ) = 83 .141 (V ) 3 Vt 2 = 166 .282 cos( V0 max =Vd max( Vt1 ,Vt 2 ,Vt 3 ) = 480 166 .282 = 313 .718 (V ) V0 min = m Vt1 ,Vt 2 ,Vt 3 ) = 83 .141 (V ) in(



V0 =V0 max = 313 .718 (V ) V10 = Vt1 +V0 =166 .282 + 313 .718 = 480 (V ) V20 =Vt 2 +V0 = 83 .141 + 313 .718 = 230 .577 (V ) V30 =Vt 31 +V0 = 83 .141 + 313 .718 = 230 .577 (V )

in p iu khin: 2V 2 480 Vk 1 = 10 = = 2(V ) Vd 480 V 2 230 .577 Vk 2 = 20 = = 0.961 (V ) Vd 480 V 2 230 .577 Vk 3 = 30 = = 0.961(V ) Vd 480

Dng in 3 pha qua ti:

HVTH: Nguyn V S

29



Vt1 166.282 cos(60t ) = = 26.56 cos(60t 37 0 )( A) 0 Z 6.2637 2 166 .282 cos( 60t ) Vt 2 3 = 26 .56 cos( 60t 157 0 )( A) it 2 = = 0 it1 =Z 6.26 37

V it 3 = t 3 = Z

166 .282 cos( 60t 6.2637 0

4 ) 3 = 26.56 cos( 60t 277 0 )( A)

c. m=0.866 + Ch s iu ch:m=

U (1) m V 480 = 0.866 U (1) m = 0.866 d = 0.866 = 240 (V ). Vd 3 1.732 3


2 ) (V ) 3 4 Vt 3 = 240 cos( ) (V ) 3 Khi = 00 ta c: Vt1 = 240 cos( 0) (V ) = 240 (V ) 2 Vt 2 = 240 cos( 0 ) (V ) = 120 (V ) 3 4 Vt 3 = 240 cos( 0 ) (V ) = 120 (V ) 3 Vt 2 = 240 cos(

(vi = 60t )




V0 =V0 max =120 (V ) V10 = Vt1 +V0 = 240 + 240 = 480 (V ) V20 = Vt 2 +V0 = 120 + 240 =120 (V ) V30 = Vt 31 +V0 = 120 + 240 =120 (V )

in p iu khin: 2V 2 480 Vk 1 = 10 = = 2 (V ) Vd 480 2V 2 120 Vk 2 = 20 = = 0.5 (V ) Vd 480 2V 2 120 Vk 3 = 30 = = 0.5(V ) Vd 480

Dng in 3 pha qua ti: V 240 cos(60t ) it1 = t1 = = 38.33 cos(60t 37 0 )( A) 0 Z 6.2637HVTH: Nguyn V S 30



V it 2 = t 2 = Z V it 3 = t 3 = Z

240 cos( 60t 6.2637 0 240 cos( 60t 6.2637 0

2 ) 3 = 38 .33 cos( 60t 157 0 )( A) 4 ) 3 = 38.33 cos( 60t 277 0 )( A)


U (1) m V 480 = 1 U (1) m = d = = 277 .136 (V ). Vd 3 1.732 3


Khi




V0 =V0 max = 202 .864 (V ) V10 = Vt1 +V0 = 277 .136 + 202 .864 = 480 (V ) V20 = Vt 2 +V0 = 138 .568 + 202 .864 = 64 .296 (V ) V30 =Vt 31 +V0 = 138 .586 + 202 .864 = 64 .296 (V )

in p iu khin:Vk 1 = Vk 2 Vk 3

2V10 2 480 = = 2(V ) Vd 480 2V 2 64 .296 = 20 = = 0.268 (V ) Vd 480 2V 2 64 .296 = 30 = = 0.268 (V ) Vd 480

Dng in 3 pha qua ti: V 277.136 cos(60t ) it1 = t1 = = 44.27 cos(60t 37 0 )( A) 0 Z 6.2637

HVTH: Nguyn V S

31



V it 2 = t 2 = Z V it 3 = t 3 = Z

277 .136 cos( 60t 6.2637 0 277 .136 cos( 60t 6.2637 0

2 ) 3 = 44.27 cos( 60t 157 0 )( A) 4 ) 3 = 44.27 cos( 60t 277 0 )( A)

PHN C: KT QU M PHNG1.B BIN TN 2 BC K thut sin. Bin hi c bn p pha ti bng 160V, tn s 40Hz.

Hnh 9. in p ng ra ba pha.

Hnh 10. in p iu khin tn s 40Hz v sng mang tam gic [0,1] tn s 5kHz.

HVTH: Nguyn V S

32



Hnh 11. in p pha ti ng ra b bin tn.

Hnh 12. Dng in trn ti ng ra b bin tn vi m = 0.4

Hnh 13. Dng in trn ti ng ra b bin tn vi m = 0.866HVTH: Nguyn V S 33



Hnh 14. Dng in trn ti ng ra b bin tn vi m = 1

Hnh 15. in p gia cc pha ng ra b bin tn. K thut sng mang vi hm common mode trung bnh (medium common mode), tn s ra 30Hz.

Hnh 16. in p common mode trung bnh vi m = 0.4.

HVTH: Nguyn V S

34



Hnh 17. in p iu khin vi m = 0.4.

Hnh 18. in p pha ti ng ra b bin tn vi m = 0.4.

HVTH: Nguyn V S

35



Hnh 19. Dng in trn ti ba pha ng ra, m = 0.4.

Hnh 20. Dng in trn ti ba pha ng ra khi m = 0.866.

Hnh 21. Dng in trn ti ba pha ng ra khi m = 1. K thut sng mang vi hm common mode nh nht (minimum common mode) , tn s p ra 30Hz.

HVTH: Nguyn V S

36



Hnh 22. in p common mode nh nht vi m = 0.866.

Hnh 23. in p iu khin (vi m = 0.866).

HVTH: Nguyn V S

37



Hnh 24. in p pha ti ng ra b bin tn (m=0.866).

Hnh 25. Dng in ng ra trn ti (m = 0.4).

Hnh 26. Dng in ng ra trn ti (vi m = 0.866)

HVTH: Nguyn V S

38



Hnh 27. Dng in ng ra trn ti (vi m = 1) K thut sng mang vi hm offset cc tr ln nht (v0MAX)) , tn s p ra 30Hz.

Hnh 28. in p common mode cc tr ln nht (m = 0.866).

Hnh 29. in p iu khin v sng mang (m = 0.866).

HVTH: Nguyn V S

39



Hnh 30. in p pha ti ng ra b bin tn (m = 0.866).

Hnh 31. Dng in ng ra trn ti (vi m = 0.4).

Hnh 32. Dng in ng ra trn ti (vi m = 0.866)HVTH: Nguyn V S 40



Hnh 33. Dng in ng ra trn ti (vi m = 1) Kho st qu trnh qu cho trng hp qu iu ch, tn s ng ra bng 50Hz vi hm common mode trung bnh (m = 1.7)

Hnh 34. in p iu khin (m = 1.7)

Hnh 35. p pha ti ng ra b bin tn (m = 1.7).

HVTH: Nguyn V S

41



Hnh 36. Dng in ba pha trn ti ng ra (m = 1.7).

Hnh 37. Phn tch Fourier v ch s THD in p TH qu iu ch Kho st qu trnh qu cho trng hp qu iu ch, tn s ng ra bng 50Hz vi k thut sin vi U (1) m = 326 .75 (V ) c tnh phn l thuyt

HVTH: Nguyn V S

42



Hnh 38. in p iu khin

Hnh 39. Phn tch Fourier v ch s THD in p TH qu iu ch 2.B BIN TN 3 BC NPC K thut sin. Bin hi c bn p pha ti bng 160V, tn s 40Hz.

Hnh 40. in p iu khin.

HVTH: Nguyn V S

43



Hnh 41. in p ba pha trn ti.

Hnh 42. Dng in ba pha trn ti. K thut sng mang vi hm common mode trung bnh (medium common mode), tn s ra 30Hz.

Hnh 43. in p iu khin (m = 0.866).

HVTH: Nguyn V S

44



Hnh 44. Dng in v in p pha a (m = 0.866).

Hnh 45. Dng in ng ra trn ti ba pha (m = 0.866). K thut sng mang vi hm common mode nh nht (minimum common mode) , tn s p ra 30Hz.

Hnh 46. in p iu khin (m = 1).

HVTH: Nguyn V S

45



Hnh 47. in p v dng in trn pha a (m = 1).

Hnh 48. Dng in ng ra trn ti 3 pha (m = 1). K thut sng mang vi hm offset cc tr ln nht (v0MAX)) , tn s p ra 30Hz.

HVTH: Nguyn V S

46



Hnh 49. in p iu khin (m = 1).

Hnh 50. Dng in v in p pha a (m = 1). Thc hin m phng vi vng qu iu ch vi hm common mode trung bnh

Hnh 51. in p iu khin (m = 1.7).

HVTH: Nguyn V S

47



Hnh 52. in p ng ra ca bin tn (m = 1.7)

Hnh 53. Dng in ng ra ca bin tn (m = 1.7)

Hnh 54. Phn tch Fourier v ch s THD ca dng in ng ra.HVTH: Nguyn V S 48



Hnh 55. Phn tch Fourier v ch s THD ca in p ng ra. Thc hin m phng vi vng qu iu ch vi V6step =306 (V ) c tnh phn l thuyt

Hnh 56. in p iu khin

HVTH: Nguyn V S

49



Hnh 57. Phn tch Fourier v ch s THD ca in p ng ra

HVTH: Nguyn V S

50



PHN D: KT LUNSau khi tnh ton l thuyt v m phng trn Matlab b bin tn 2 bc v b bin tn 3 bc NPC c th rt ra mt s kt lun nh sau: - in p ng ra ca b bin tn 3 bc c dc t hn in p ng ra ca b bin tn hai bc. Do gip cho ti hot ng tt hn, nht l i vi cc ng c in.

Hnh 58. in p ng ra ca b bin tn 2 bc (m = 0.866).

Hnh 59. in p ng ra ca b bin tn 3 bc (m = 0.866). - Khi ch s iu ch 0 m 1 th tn hiu ng ra ca b bin tn thay i tuyn tnh vi m. Nhng khi m >1 th tn hiu ng ra khng thay i tuyn tnh theo m na m c gi tr bin cc i, nu m ln hn 1 rt nhiu th tn hiu p ng ra c dng xung vung.

HVTH: Nguyn V S

51



Hnh 60. in p ng ra ca b bin tn 2 bc (m = 2).

-

Hnh 61. in p ng ra ca b bin tn 3 bc (m = 2). Ch s m cng tng th dng in ng ra tng tuyn tnh theo m.

HVTH: Nguyn V S

52



Hnh 62. Dng in trn ti ba pha (m = 0.4)

Hnh 63. Dng in trn ti ba pha (m = 0.866)

Hnh 64. Dng in trn ti ba pha (m = 1)

HVTH: Nguyn V S

53

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