08_chuong 5 nghich luu va bien tan
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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
Chng 5 : Nghch lu v bin tn
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
Chng 5 : Nghch lu v bin tn
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
Chng 5 : Nghch lu v bin tn
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
Chng 5 : Nghch lu v bin tn
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
Chng 5 : Nghch lu v bin tn
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
Chng 5 : Nghch lu v bin tn
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
Chng 5 : Nghch lu v bin tn
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
Chng 5 : Nghch lu v bin tn
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
Chng 5 : Nghch lu v bin tn
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
Chng 5 : Nghch lu v bin tn
+ 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
Chng 5 : Nghch lu v bin tn
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
Chng 5 : Nghch lu v bin tn
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
Chng 5 : Nghch lu v bin tn
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
Chng 5 : Nghch lu v bin tn
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
Chng 5 : Nghch lu v bin tn
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
Chng 5 : Nghch lu v bin tn
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
Chng 5 : Nghch lu v bin tn
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
Chng 5 : Nghch lu v bin tn
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
Chng 5 : Nghch lu v bin tn
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
Chng 5 : Nghch lu v bin tn
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
Chng 5 : Nghch lu v bin tn
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
Chng 5 : Nghch lu v bin tn
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
Chng 5 : Nghch lu v bin tn
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
Chng 5 : Nghch lu v bin tn
+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
Chng 5 : Nghch lu v bin tn
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
Chng 5 : Nghch lu v bin tn
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
Chng 5 : Nghch lu v bin tn
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
Chng 5 : Nghch lu v bin tn
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
Chng 5 : Nghch lu v bin tn
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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