08_chuong 5 Nghich Luu Va Bien Tan

Chng 5 : Nghch lu v bin tn

Chng 5

NGHCH LU V BIN TNB 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. Trong trng hp u, b nghch lu c gi l b nghch lu p v trng hp sau l b nghch lu dng Ngun mt chiu cung cp cho b nghch lu p c tnh cht ngun in p v ngun cho b nghch lu dng c tnh ngun dng in. Cc b nghch lu tng ng c gi l b nghch lu p ngun p v b nghch lu dng ngun dng hoc gi tt l b nghch lu p v b nghch lu dng Trong trng hp ngun in u vo v i lng ng ra khng ging nhau, v d b nghch lu cung cp dng in xoay chiu t ngun in p mt chiu, ta gi chng l b nghch lu iu khin dng in t ngun in p hoc b nghch lu dng ngun p 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 c 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 Trong cc trng hp c bit nh mch ti cng hng, ti mang tnh cht dung khng (ng c ng b kch t d), dng in qua cc linh kin c th b ngt do qu trnh chuyn mch t nhin ph thuc vo in p ngun hoc ph thuc vo in p mch ti. Khi , linh kin bn dn c th chn l thyristor (SCR) Nghch lu l b chuyn i in th DC thnh AC tun hon vi tn s mong mun khc tn s in khu vc (th d nh 400Hz trong hng khng) nhng c dng khng sin. Mun c dng hnh sin ta c th dng cc k thut

124


khc nhau thc hin bin i thnh dng sin. B phn quan trng l b cng tc in t (static contact) cng sut ln thng s dng nh: Transistor, Mosfet, SCR, IGBT cng sut ln Phn loi c nhiu cch phn loi khc nhau nh Theo hnh dng sng ra: hnh sin, hnh vung Theo cch hot ng: ngun th VSI (voltage source Inverter), ngun dng CSI (Current source Inverter), bin i rng xung PWM (pulse width modulated Inverter)

Hnh 5.1 Theo cu hnh: ni tip, song song, bn cu i in, cu i in Theo cu hnh v hot ng: ng dy giao hon (line commutated), t giao hon (self commutated) Thng thng ta c th xt theo cu hnh b i in ng dng b nghch lu c rt nhiu ng dng trong thc t nh iu khin tc ng c cm ng ng b L nung cm ng B ngun cp in UPS (Uninterrupible power supplies) Truyn ti in th cao bng in DC (High voltage DC transmission)

5.1 B nghch lu mt pha 5.1.1 B i in c bn Mch c dng b i in bn cu v cch hot ng hnh 5.2

125


S1 + E + R S2 +

Tt 1 2 3 4

S1 + +

S2 + +

V0 +E 0 -E 0

Hnh 5.2 Khi ch cho mch hot ng 1 v 3, ta c dng sng ra c dng sng vung nh hnh 5.3

Hnh 5.3a Khi cho hot ng c 1, 2 , 3, 4 theo trnh t nh hnh 5.3 ta c dng sng ra c dng ncVo +E 1 1 T -E 2 3 3 4 1 1 2 3 3 t

Hnh 5.3b Vi chu k T v tn s f = 5.1.2 B nghch lu bn cua. Ti in tr thun R+ E TAI G1 SCR1 D1

1 T

+E

Vo1 1 T 2 3 3 4 1 1 2 3 3 t

+ E G2 SCR2

D2

-E

Hnh 5.4126


Trong trng hp sng ra l sng nc nh hnh 5.4 v ti in tr thun R, ta c in th trung bnh ng ra t ON t ON 2 t 0 Edt = 2 E T = 2 Ed ; d = T T in th hiu dng ng ra V AV =ON

(5.1)

VRMS =

t ON 2 tON 2 0 E dt = E 2 T = 2d E T

(5.2)

Dng ti trung bnhI AV = V AV R

(5.3)

Dng trung bnh trong cng tc

I AVsw =

I AV 22 VRMS E2 = 2d R R

(5.4)

Cng sut hp th trung bnhPOAV = (5.5)

b.

Ti cm khng R, L

Trng hp ny dng in qua ti thay i theo hm m ca thi gian trong khi in th vn l hnh vung hnh 5.5Vo+ E R L G1 SCR1 D1

T/2+ E G2 SCR2 D2

t D2 S2

D1 i1

S1

t

+E

Vo SCR1 TON SCR1 t i2 t

-E

SCR2

SCR2

Hnh 5.5

127


Ta c L di + Ri = E dtt t E T 1 e I 01e ; 0 t R 2

(5.6)

Gii ta c nghim ton th i (t ) =i (t ) =

(

)

(5.7)(5.8)

(t T ) 2 E 1 + e R

(t T ) 2 T ; t T + I 01e 2

Vi thi hng =

L ; I(0+) = I01 R

Tnh c ti t = T/2

iT

( )

T T E E 1 e 2 = I 01 = 1 e 2 I 01e 2 I 01 = T 2 R R 1 + e 2

(

)

( (

T

) )

(5.9)

Thay vo ta c

t E E 1 e 2 t i (t ) = 1 e e T R R 1 + e 2

(

) ((

T

) )

(5.10)

Do i xng, ta c thi khong T/2 < t 0, lc ny SCR1 cn dn, in th 2 anod bng nhau; nhng v u t C pha anod 2 b gim th t ngt nn cnh C pha anod 1 cng phi gim cng 1 lng in th, do lm anod 1 c in th m hn mass dn ti SCR1 ngng dn. Khi SCR1 ngng, in th ti u anod 1 tng ln bng E, kt qu l t C np in theo chiu ngc li t pha anod 1 sang anod 2. V chu k mi li tip din Do s thay i dng in trong cun s cp nn s sinh ra dng cm ng cun s cp to nn in th thay i vL ti mc vo cun th cp. in th ny c dng gn hnh vung nu xem l l tng T C c gi l t giao hon hay t chuyn mch, c nhim v lm cho 1 SCR dn v SCR kia ngng v sau lm cho SCR ngng thnh dn v SCR dn thnh ngng. T C phi c tnh sao cho c thi gian lm SCR1 ngng theo mun. T C c tnh theo cng thc sau

C=

It off 2E

=

t off R

t off 0,693R

(5.17)

vi: toff thi gian ngng ca SCR; I dng in ngay trc khi SCR giao hon trnh s tng qu v m bo in th ra c dng hnh vung ta s dng 2 diod D1 v D2 v hi tip bng cun cm L. Cun cm L to nn in th dng catod cc SCR nhm bo m SCR c ngng nhanh khi chuyn trng thi nh hnh 5.12ZL T1

C1 C2 + E xung dk1 L0 C3 xung dk2 SCR1 D4 SCR2 D4

T2

T2

Hnh 5.12 Hoc mch s dng MOSFET cng sut nh hnh 5.13

132


ZL T1

+ E Q1 G1 G2 Q2

Hnh 5.135.1.5 K thut iu khin in th b i in

Trong hu ht ng dng ca b i in i hi cn c s iu khin in th ng ra. C nhiu cch thc hin s iu chnh in th AC ng ra, nhng thng c sp thnh 3 loi sau iu khin in th DC cp vo b i in iu khin in th AC ng ra b i in iu khin in th trong b i ina.

iu khin in th DC cp in

Vi 1 mu giao hon, in th ra b i in t l vi in th vo. Do , s thay i in th DC cp in l cch n gin nht iu khin in th ra. Nu ngun cp in DC, k s dng 1 mch chopper lm thay i in th DC. Ta cng c th s dng ngun cp in DC t mch chnh lu c iu khin l tt hnb.

iu khin in th AC t b i in

Trong phng php ny, s dng mch iu th AC gia b i in v ti iu khin in th AC ca b i in v do iu khin in th ng rac.

iu khin in th trong b i in

Phng php iu bin rng xung PWM l phng php thng dng iu khin in th trong b i in

Hnh 5.14133


Trong phng php ny in th ra l sng iu bin rng xung v in th c iu khin bi thi hn cc xung in th ra Cc phng php PWM thng c xp thnh 3 nhm sau iu bin rng n xung iu bin rng a xung iu bin rng xung dng sng sin PWM n xung Trong cch ny, dng sng in th ra gm 1 xung n trong mi bn k. Vi tn s cho sn (f = 1/T), rng xung tw c th thay i iu khin in th ra Dng sng in th ra ca b i in n pha (xem cu i in) khng c iu bin trong cng tc S1 v S4 dn trong 1 bn k v S2 v S3 dn trong bn k tip theo cho in th ra cc i+EVA S1 VB S4 Vo = VA -VB S3 S4 S3 S2 S1 S2

+Et

VA S1 VB S4 S3 S4 S3 S2 S1 S2

t

+E

+Et

t

Vo = VA -VB

+ET/2 T S2 S3 S1 S4 S2 S3

+Et T/2 T t

-E

S1

S4

-E S4

S1,4 S1,3 S2,3 S2,4 S1,4 S1,3 S2,3

Hnh 5.15 iu khin in th c hon thnh bng cch thay i pha ca S3 v S4 theo S1 v S2. Hnh 5.15 cho thy dng sng in th ra khi thi khong dn ca S3 v S4 sm pha bi 1 gc = 90 0 . in th ra c c bng cch cng 2 in th sng vung chung c dch pha vi nhau. in th ra gm 1 chui xung vi rng xung (180 0 ) = 90 0

134


in th ra c th c iu chnh tuyn tnh t tr cc i n 0 hoc bng cch lm sm pha hoc bng chm pha s khi dn ca 1 cp cng tc so vi cp cng tc khc

Hnh 5.16 Hnh 5.16 m t dng sng iu bin ca mch bn cu i in PWM a xung in th ra c th c giao hon on/off nhanh nhiu ln trong sut mi bn k to nn chui xung c bin khng i Hnh 5.16 biu din 1 PWM a xung l tng. Dng sng in th ra gm m xung trong mi bn k ca in th ra i hi. Nu f l tn s in th ra ca b i in th tn s xung fp cho bi

f p = 2 fmDo xung trong mt chu k: 2m = fp/f+EVo VA

(5.18)

m=21 fp

+E

m=5

0 -E +E -EVo

T m=3

2

t

-E +EVB

m=5

2

t

1 fp

t

t

-E

Hnh 5.16

135


Hnh 5.16 din t dng sng in th ra vi m = 2. rng xung phi nh hn 2 . Hnh 5.16 vi m = 3, r rng l t w < 3 . Tng qut, rng xung: t w m

Hnh 5.17 Hnh 5.17 m t dng sng iu bin ca mch cu i in iu bin xung hnh sin (SPWM) Trong SPWM in th ra c iu khin bng cch lm thay i chu k on/off sao cho chu k ( rng xung) di nht ti nh ca dng sng hnh 5.18+EVo t

-E

T

2

Hnh 5.18 Hnh 5.19 din t thi gian giao hon c xc nh, trong vca(t) l sng sin lm iu bin tham chiu c bin cc i Vm v tn s fm phi bng vi tn s in th ra mong mun ca b i in. Mt sng mang tam gic tn s cao c bin vst(t) v tn s fc c so snh vi sng sin tham chiu im giao hon c xc nh l giao im ca sng sin v sng tam gic. rng xung tw c xc nh bi thi gian trong vst(t) < vca(t) trong bn k dng ca vca(t) v vst(t) > vca(t) trong bn k m ca vR(t) Hai thng s iu khin iu ha in th ra l t s bm N (chopping ratio) v ch s iu bin M

136


N=M=

fc fm

(5.19) (5.20)

V R Vm = ; 0 M 1 Vc Vc

Hnh 5.19 T ta c th xc nh rng ca cc xung v do tr s hiu dng ca in th ra ca b i in. M thng dng iu chnh bng cch lm thay i bin ca sng tham chiu trong khi bin sng mang khng i. Tn s ng ra b i in thay i bng cch thay i tn s sng tham chiu5.1.6 B i in to sng sin

Dng sng ng ra b i in thng c dng sng vung, mun c dng sng sin ta phi dng mt trong cc cch sau Ti cng hng LC Mch lc RLC iu bin rng xung PWM (pulse width modulation)a.

Ti cng hng

Trong b i in song song, ta mc thm t C vo ti cm khng R, L to thnh mch cng hng LC dng b i in song song hoc ni tip hnh 5.20

137


+ G SCR1

R R L + E E C + L CL C/2 R _ + E R C/2 L C L

Hnh 5.20

Mch to ra dng sng sin c tn s bng tn s cng hng1 f0 = 2 1 RL 1 LC 2 L 2 LC2

(5.21)

Thng c1 R2 2L L >> 2 R =2 LC 4L C LC1 1 1 = + C C 2 C 2

(5.22) (5.23)

b.

Mch lc OTT

Trong mch i in song song dng SCR trc hnh 5.11, ta mc mch lc OTT xen gia cun s cp v ti Z hnh 5.22R L2 C2 L3

C1

L1

T1

C1 C2 + E xung dk1 L0 C3 xung dk2 SCR1 D4 SCR2 D4

T1

T2

Hnh 5.22

Mch lc c tnh nh sau138


Tn s

D = 2f DTng trZD ZL 31 6Z D D

(5.24)

( )(F ) ;1 3Z D D

(5.25)

Tr s cc thnh phn c tnh theoC1 = C2 =

(F )

(5.27) (5.28)

L1 =

9Z D ( H ) ; L2 = Z D ( H ) 2 D D

Vipf = cos ; R L =E 2 pf P02

( ) ;

XL =

RL 1 pf pf

2

( )

(5.29)

Th d

Thit k mch lc Ott cho b i in 400Hz c: E0 = 120V; P0 = 360W; f0 = 400Hz; pf = 0,7; E = 28VDC Tnh c Tr khng tiRL =

(120)2 (0,7 )2360

= 20 ; X L =

20 2 1 (0,7 ) = 20 0,7

Z L = 20 2 + 20 2 = 28,30 ; Z L = cos 1 (0,7 ) =

4

Tng tr mch lc thit kZD 28,3 2

Tn s thit k: D = 2(3,14 )(400 ) = 2500rd / s

Tr s cc thnh phn mch lc:C1 = L1 =1 1 = 4,5F ; C 2 = = 9 F 6(15)(2500 ) 3(15)(2500 )

9(15) 15 = 27 mH ; L2 = = 6mH 2(2500 ) 2500

139


Tng tr vo mch lc:Z in = 15(5,5) 16 0 = 80 j 23 Rin = 80; Z in = 23 Z in = 83

in th vo mch lc:E sq = P 2 2 (3,14)(83) 360 = 195V Z in 0 = Rin 80 4 4

Thit k b i in T s bin th n = 195/28 = 7 Cng sut vo (gi s hiu sut 85%): Pi = P0 (100/85) = 360 (100/85) = 424W Dng trung bnh trong SCR:I tb ( SCR ) P0 Z in 2 ERin = 360(83) = 6,8 A 2(28)(80)

in th nh thun qua SCR:Vpk(SCR) < 2,5E = 2,5(28) = 70Vdv dt = 200 V smax

Vi SCR C141A c toff = 10 s ;

Chn tc = 12 s v Ipk(SCR) = 14A. Tnh c: L =6 Et c 6(28)(12) = = 45H I pk ( SCR ) 3,14(14)

3,44(790) dv 3,44 E 2 = = = 4,3V s dt LI pk ( SCR ) 45.10 6 (14)C= 3t c I pk ( SCR ) 8E = 3 12.10 6 (14 ) di = 0,75F ; 8(3,14 )(28) dt

(

)

=t =0

2 E 2(28) = = 1,25 A s L 45

Phng php ny ch s dng trong trng hp cng sut thp

5.2 Nghch lu ba pha 5.2.1 B nghch lu p su tia ti mc hnh saoB i in 3 pha gm 3 b i in 1 pha theo hnh 5.23

140


D1 S1 + E S4 D4 S6 S3

D3 S5

D5

D6

S2

D2

R N

R

R

Hnh 5.23 Mch gm 6 van cng sut v 6 diod dp kt hp. Cc bc dn ngng tun hon theo cch sp xp tun t to dng sng ra mong mun. Tc cc van xc nh tn s ra ca b i in. C nhiu cch hot ng kh hu nhng c 2 cch c bn hon thnh 1 chu k vi 6 bc giao hon: loi dn 1200 v loi dn 1800. Cc bc cng sut l: SCR, MOSFET, Transistor Cng sut, IGBT

a.

Loi dn 1200 (ti R)

Do cch b tr linh kin cc bc s dn trong 1200 v mi cp bc s dn lch nhau 600. Ta c th tm tt trong hnh 5.24iG1 i0 G2 iG3 iG4 iG5 iG6 6,1 1,2 2,3 3,4 4,5 5,6 6,1 1,2 2,3 3,40

60

0

120

0

180

0

240

0

300

0

360

0

420

0

480

0

540

0

600

0

t t t t t t

Hnh 5.24 Ta thy trong mi thi khong (600) ch c 2 bc cng dn, nn theo hnh 5.24 ta c in th pha bng

141


v an =

R E E= R+R 2

(5.30)

Vi Khi bc s l dn Vi > 0 cho: +E/2 Khi bc s chn dn Vi < 0 cho: -E/2 Khi khng c bc no dn cho: V0 = 0E

Hnh 5.25 Ta c kt qu cc in th pha v in th dy T.kh (0) van vbn vcn vab vbc Vca 0-60 +E/2 -E/2 0 +E -E/2 -E/2 60-120 120-180 180-240 240-300 300-360 360-420 +E/2 0 -E/2 +E/2 +E/2 -E 0 +E/2 -E/2 -E/2 +E -E/2 -E/2 +E/2 0 -E +E/2 +E/2 -E/2 0 +E/2 -E/2 -E/2 +E 0 -E/2 +E/2 +E -E +E/2 -E/2 -E/2 0 +E -E/2 -E/2

142


Dng sng in th+E/2 -E/2 vAN 0 60 vBN0 0

120

0

180

0

240

0

300

0

360

0

420

0

480

0

540

0

600

0

t t

vCN

t

vAB +E +E/2 -E/2 -E

t

vBC

tvCA

t

6,1

1,2

2,3

3,4

4,5

5,6

6,1

1,2

2,3

3,4

Hnh 5.26 T cc kt qu trn tnh c Cng sut cho biP0

(E 2 ) + (E 2 ) =2

2

R

R

=

E2 2R

(5.31)

in th hiu dng phaVl n ( RMS ) = Van ( RMS ) = 1

3

0

1 2 3 E 1 E2 E2 E + dt + dt = 3 2 4 3 4 3 2

2

2

143


Vl n ( RMS ) = Van ( RMS ) =

2E 2 E = 12 6

(5.32)

in th hiu dng ng dy

Vl l ( RMS ) = Vab ( RMS ) = 3Vl n ( RMS ) =

E 2

(5.33)

Dng hiu dng qua van

isw ( RMS ) =

E 2 3R

(5.34)

Dng ra hiu dngI 0 ( RMS ) = 2 I sw ( RMS ) (5.35) (5.36)

in th nghch cc i ca vanVSWRM = E

b.

Loi dn 1800 (ti R)

Trng hp ny ta c cch dn nh sau, mi bc lch pha nhau 600 theo th t hnh 5.27iG1 i0 G2 iG3 iG4 iG5 iG6 5,6,1 61,2 1,2,3 2,3,4 3,4,5 4,5,6 5,6,1 6,1,2 1,2,3 2,3,40

60

0

120

0

180

0

240

0

300

0

360

0

420

0

480

0

540

0

600

0

t t t t t t

Hnh 5.27 Theo trn ta thy trong mi thi khong c 3 bc cng dn, do theo H.2.28 in th pha bng

144


v an =

R R R+ 2

E=

2 E 3

(5.37)

Hnh 5.37 Nhn xt Khi bc l dn Vi > 0 Khi bc chn dn Vi < 0 Khi pha c 2 bc cng dn cho +E/3 Khi pha ch c 1 bc dn cho 2E/3 Ta c bng tr s in th T.kh (0) van vbn vcn vab vbc Vca 0-60 +E/3 -2E/3 +E/3 +E -E 0 60-120 120-180 180-240 240-300 300-360 360-420 +2E/3 -E/3 -E/3 +E 0 -E +E/3 +E/3 -2E/3 0 +E -E -E/3 +2E/3 -E/3 -E +E 0 -2E/3 +E/3 +E/3 -E 0 +E -E/3 -E/3 +2E/3 0 -E +E +E/3 -E/3 +E/3 +E -E 0

Dng sng in th pha v ng dy c v hnh 5.38

145


vAN +2E/3 +E/3 00 -E/3 -2E/3 600 vBN 1200 1800 2400 3000 3600 4200 4800 5400 6000

t

tvCN

t

+E -E

vAB

tvBC

tvCA

t6,1,2 1,2,3 2,3,4 3,4,5 4,5,6 5,6,1 6,1,2 1,2,3 2,3,4

5,6,1

Hnh 5.38 Tnh c Cng sut cp cho ti:2 2 0

(E ) (E ) (2E 3 ) P = 3 + 3 +R R R

2

=

2E 2 3R

(5.38)

in th hiu dng pha:

Vl _ n ( RMS )

2 2 2 2 E 1 3 E 3 2E = dt + 2 dt 0 dt + 3 3 3 3 3

146


Vl _ n ( RMS ) =

2 E 3 2 E 3 (5.39)

Van = Vbn = Vcn =

in th hiu dng ng dy Vl l ( RMS ) = 2 2 E 3= E 3 32 3E

Vab = Vbc = Vca =

(5.40)

Vi t s PWM ca Vl n ( RMS ) = Vl l ( RMS ) = 3 E 3 2 E 3 E 3R (5.41) (5.42)

Dng hiu dng qua bc giao hon Vsw( RMS ) = (5.43)

Dng hiu dng ng ra

I 0 ( RMS ) = 2 I sw ( RMS ) Dng DC voIi =

(5.44)

3 2I 0

cos

(5.45)

Vi PWM:Ii =

3 2I 0

cos

(5.46)

trong l gc pha ca ti Cng sut vo cp bi ngun Pi = EIi Dng sng ca b i in PWM hnh 5.39 (5.47)

147


Hnh 5.39c.

Loi dn 1800 (Ti cm khng R, L)

Do cc bc c gc kch lch nhau 600, cc bc dn trong 1800 v ti cm khng ta c dng sng in th tng t nh trng hp trc v dng in nh hnh 5.40

Hnh 5.40 Trong thi khong t t1 n t2 Ta c phng trnh dng ti i(t)

148


L

di E + Ri = dt 3

(5.48)

chn t1 = 0, ti t = 0 :i(0) = I1, nghim tng qut ca phng trnhi (t ) =t t E 1 e + I 1 e 3R

(5.49)

ti t = t2 = T/6: i(T/6) = I2.T T E T i = I 2 = 1 e 6 + I 1e 6 3R 6

(5.50)

Trong thi khong t t2 n t3: Ta cL di 2E + Ri = 3 dtt t 2E 1 e + I 2 e 3R

chn t2 = 0; ti t = 0: i(0) = I2, nghim tng qut ca phng trnhi(t ) =

(5.51)

ti t = t3 = T/6: i(T/6) = I3T T 2E T i = I 3 = 1 e 6 + I 2 e 6 3R 6

(5.52)

thay I2 5.50 vo 5.52 ta c I3 ==

T T T 2E E T 1 e 6 + 1 e 6 + I 1e 6 e 6 3R 3R

(

)

(

)

T T T E 2 e 6 e 3 + I 1e 3 3R

(5.53)

Thi khong t3 n t4: Ta c phng trnh dng ti i(t)L E di + Ri = 3 dt

(5.54)

chn t3 = 0: i(0) = I3, nghim tng qut ca phng trnhi(t ) =t E 1 e 3R

+ I e t 3

(5.55)

ti t = t4 = T/6: i(T/6) = 0T T E T i = I 4 = 1 e 6 + I 3 e 6 3R 6

(5.56)

thay 5.53 vo 5.56 ta c

149

Chng 5 : Nghch lu v bin tn_T T T E 1 + e 6 e 3 e 2 3R

I4 =

+ I e T 2 1

(5.57)

Do dng sng i xng, ta c: I4 = -I1, thay vo 5.57 v sp xp liI1 = E 1 3R 1 + e T 2_T T T 1 + e 6 e 3 e 2

(5.58)

Thay (9) vo (2) tnh c c I2, thay I2 vo (4) tnh I3 Do tnh i xng ta cn c I6 = -I3; I4 = -I1; I5 = -I2; I7 = I1 Ch : C th tnh cc tr s in th v dng in theo phng php phn gii Fourier nh trnh by chng 25.2.2 B nghch lu p su tia ti mc hnh tam gicD1 S1 + E S4 D4 S6 D6 S2 D2 S3 D3 S5 D5

A

B

C

AR RR

a.

Dn 1200 (ti in R)

Hnh 5.41

Mch c mc theo hnh 5.41. Tng t nh trng hp ti mc ch Y, ta c trong mi thi khong ch c 2 bc cng dn, mi bc c kch lch nhau 600 theo th tAE 2

BE 2

E 2

C

Hnh 5.42150


Khi S1 v S6 dn ta c: vab = +E, nhng do c tnh i xng nn ta c vac v vcb ch bng +E/2 hay vca = vcb = -E/2 Khi S1 v S2 dn ta c: vac = +E, nhng L lun tng t vi cc trng hp cn li ta c cc kt qu tm tt trong bng sau T.kh (0) S.dn vab vbc vca 0-60 6,1 +E -E/2 -E/2 60-120 1,2 +E/2 +E/2 -E 120-180 180-240 240-300 300-360 360-420 2,3 -E/2 +E -E/2 3,4 -E +E/2 +E/2 4,5 -E/2 -E/2 +E/2 5,6 +E/2 -E +E/2 6,1 +E -E/2 -E/2

vab = vbc = E/2 hay, vab = +E/2, vbc =+E/2 v vca = -E

Ta c dng sng Hnh 5.43vAB +E +E/2 00 -E/2 -E 600 vBC 1200 1800 2400 3000 3600 4200 4800 5400 6000

t

tvCA

t

6,1

1,2

2,3

3,4

4,5

5,6

6,1

1,2

2,3

3,4

Hnh 5.43 Tnh ton nh trng hp ti dng saob.

Dn 1800 (ti R) Tng t trong mi thi khong (600) ta c 3 bc cng dn

151


Khi S5, S6, S1 cng dn hnh 5.44 ta c vab = +E, vbc = -E, vca = 0A +E 0 +E C

B

Hnh 5.44 Khi S6, S1, S2 cng dn hnh 5.55 ta c vab = +E, vac = +E, vbc = 0 hay: vab = +E, vbc = 0, vca = -EA +E +E 0 C

B

Hnh 5.45 L lun tng t vi cc trng hp cn li ta c cc kt qu tm tt trong bng sau T.kh (0) S dn va vb vc vab vbc vca 0-60 5,6,1 +E 0 +E +E -E 0 60-120 6,1,2 +E 0 0 +E 0 -E 120-180 180-240 240-300 300-360 360-420 1,2,3 +E +E 0 0 +E -E 2,3,4 0 +E 0 -E +E 0 3,4,5 0 +E +E -E 0 +E 4,5,6 0 0 +E 0 -E +E 5,6,1 +E 0 +E +E -E 0

152


+E

Dng sng in th va, vb, vc, vab, vbc, vca c trnh by hnh 5.46vA 600 vB 1200 1800 2400 3000 3600 4200 4800 5400 6000

00

tt

vC

tt

+E -E

vAB

vBC

tvCA

t6,1,2 1,2,3 2,3,4 3,4,5 4,5,6 5,6,1 6,1,2 1,2,3 2,3,4

5,6,1

Hnh 5.46 Vi vab = va; vbc = vb vc; vca = vc va iu bin PWM trong b nghch lu ba pha nh hnh 5.47

Hnh 5.47

153


5.2.3 Ti cm khng R, L, trng hp dn 1800 mc tam gic

Ta c kt qu nh ti R v in th, cn dng in cm ng c biu din nh hnh 5.48 vi in th ng dy vab

Hnh 5.48 V cc dng sng l ging nhau v ch lch nhau 1200, nn ta ch cn tnh dng iab v suy ra cc kt qu ibc, ica Thi khong t1 n t3, ta cL di + Ri = E dt

(5.59)

iu kin ban u cho: t = t1 ta c I = I1, nghim ca phng trnhi(t ) =t t E L 1 e + I 1 e ; = R R

(5.60)

ti t3, t = T/3: i(T/3) = I3I3 =T T E 1 e 3 + I 1e 3 R

(5.61)

Thi khong t3 v t4 ta cL di + Ri = 0 dtt

(5.62)

iu kin ban u t = t3 = 0: i = I3i(t ) = I 3 e

(5.63)

Thay 6.61 vo 5.63 ta c154

Chng 5 : Nghch lu v bin tnT T E T i(t ) = 1 e 3 + I 1e 3 e 3 R

(5.64)

ti t = t4 =T/6 ta c i(t4) = I4:I 4 = I 3eT6

T T E T = 1 e 3 + I 1e 3 e 6 R

(5.65)

Khi iu kin tun hon ti lp, tha dng sng i xng phi c: I4 = -I1 Thay 5.66 vo 5.67 v gii, ta cE e 6 e I1 = R 1 + e T 2T T 2

(5.66)

(5.67)

Cng thc 5.67 cho ta tr s I1 ti t1 l mt tr s ca 4 thi im ti dng sng khng lin tc, I4 l tr ti t4 bng I1 nhng ngc du. Hai tr s khng lin tc khc l I3 v I6 ti t3 v t6. Tr I3 c c bng cch thay I1 t 5.67 vo I3 5.61 v sp xp liE 1 e 3 I3 = R 1 + e T 2T

(5.68)

Ti t6 dng I6 c cng bin nh I3 nhng khc du I6 = -I3

5.3 B bin tn

B bin tn l mch c nhim v bin i in th AC u vo c tn s f1 thnh in th AC ng ra tn s f2 c th iu chnh c C 2 loi bin tn: Bin tn trc tip v bin tn gin tip Bin tn trc tip: l loi bin i t in th AC (f1) c sn thnh in th AC (f2) v d b bin tn trc tip n gin nh hnh 5.47A1 T1 A2f = 10 Hz

t

U1

T2

IZ

Z

UL

f = 16,66 Hz

t

Hnh 5.47155


Bin tn gin tip: bin i t in th AC chnh lu thnh in th DC (nh mch chnh lu) hoc t ngun in DC cho sn, ri sau bin i in DC thnh in AC c tn s mong mun hnh 5.48 bao gm b bin tn gin tip ngun dng v ngun p I f1 id-I ud-I L C id ud-I L id-II ud-II

_Chnh lu I

_

II f2

_f2

Nghch lu p II ud-II

f1

_Chnh lu

Nghch lu dng Hnh 5.48

Cc b bin tn thng c s dng trong vic iu khin ng c AC (ng c ng b, khng ng b) trong sn xut, trong cng nghip, v trong cc lnh vc k thut khc5.3.1 Bin tn trc tip mt pha

S khi b bin tn trc tip

Bin tn trc tip l thit b bin i trc tip ngun xoay chiu c tn s f1 sang ngun xoay chiu c tn s frNguo n Ta n so co nh Bo bie n ta n tr c tie p Nguo n Ta n so Bie n o i K ie u khie n ie n a p /ta n so

Hnh 5.49 Nguyn l hot ng ca b bin tn trc tip B bin tn trc tip gm hai nhm chuyn mch ni song song ngc (s nguyn l ca b bin tn trc tip c trnh by nh hnh v sau). Cho xung m ln lt hai nhm chnh lu trn ta s nhn c dng in xoay chiu chy qua ti. mi pha u ra (a, b, c) c cp in bi hai nhm Thyristor. Nhm T to ra dng in chy thun v nhm N to ra dng chy ngc. Mi nhm l mt b chnh lu (hoc nghch lu ph thuc) ba pha. hn ch dng k sinh

156


chy qua hai Thyristor ca nhm T v nhm N ang dn, ngi ta dng cc cun khng K1 v K6 f1

S nguyn l b bin tn trc tip dng Thyristor~V1

T1

N4 T3

N6

T5

N2

K1

K4

K3

K6

K5

K2

a

b

c

fr Vr Bien oi K

Hnh 5.50 Khi iu khin theo nhm th mi nhm c m trong na chu k in p u ra. Xt s lm vic pha a theo th sauVa(V) t(s)

T1 Tr

T2

Va(V)

Hnh 5.51 Trong khong thi gian t1: nhm T1 m, cn trong khong t2 th nhm N4 m. Cc Thyristor trong cng mt nhm chuyn mch cho nhau nh in p li (chuyn mch t nhin). Mi Thyristor m 1/3 chu k ca in p li. Thay i s Thyristor m trong mi nhm ta s thay i c thi gian ca chu k in p u ra T2=t1+t2 do thay i c tn s u ra ca bin tn. T th ta tm c mi quan h gia tn s li v tn s ra

157


fr T m (5.69) = 1 = f1 T2 2n + m 2 Trong : m: s pha u vo ca b bin tn (m=3). n: s nh hnh sin (tc s Thyristor m mi nhm) trong mt na chu k ca in p ra. Theo cng thc trn ta thy tn s u ra lun ln nh hn tn s li v n l s nguyn nn tn s ra c iu chnh nhy cp. in p ra Vr c thay i bng cch thay i gc chm ca cc Thyristor V s nh hnh sin n trng hp ny ging nh hnh 19.3 nn tn s u ra ca hai trng hp nh nhau, nhng in p hnh b c gi tr nh hn. to ra in p ba pha u ra ta iu khin cc nhm Thyristor m theo th t T1-N2-T3-N4-T5-N6-T1 mi nhm cho m 1/3 chu k ca in p ra. Nu in p ra c lc phng hon ton th bng cch iu khin nh trn ta c th in p ra ba pha nh trn hnh 19.4 (h thng in p ba pha u ra b bin tn trc tip) VA(v)T1 2/3 N4 T3 N6 N6 T5 N2 N2 T1

wt(rad)

Vb

wt(rad)

Vc wt(rad)

Hnh 5.52 c th iu chnh tinh tn s ra v to c in p ra c dng gn hnh sin hn, ta p dng phng php iu khin gc m Thyristor cn thit cho cc Thyristor mi pha ca in p u ra v kt qu ta c th in p ra mt pha u ra nh hnh v 5.53 thnh phn sng iu ha bc nht (theo tn s wr ca in p ny l ng t)

158


2 0

a)

Sng sin c bn

wrt(rad)

b)wt

Hnh 5.53 a) Quan h =f(t) b) th in p ra ca b bin tn trc tip khi iu khin gc theo qui lut hnh sin Nhn xt Hiu sut cao v tn tht nng lng khng ng k, khng cn dng t chuyn mch. Hnh 5.53 Ch cho tn s fr

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