Post on 11-Nov-2015
description
Bi tp ln Thit k h thng in t cng sut
2
Bi tp 9: Thit k cu trc iu khin cho nghch lu c lp ngun p ba pha c
cng sut
5kVA lm vic ch chnh lu tch cc
Tham s s mch lc nghch lu ngun p ba pha
Lng t in p mt chiu trn t Cdc Udc = 700V Tham s in cm pha li Ls = 2mH Tn s pht xung mch nghch lu 5kHz
380V / 50Hz
S mch lc nghch lu c lp ngun p mt pha lm vic ch chnh lu tch
cc
Yu cu:
1. Tnh chn mch lc
2. Thit k khu iu ch rng xung theo nguyn l iu ch vector khng gian.
3. Thit k mch vng iu khin dng in theo nguyn l iu khin vector
dng in is (la chn iu khin theo cu trc PR trn h ta tnh hoc PI
trn h ta quay dq) .
4. Tng hp mch vng in p Udc
5. M phng bng phn mm Matlab kim chng kt qu thit k.
Bi tp ln Thit k h thng in t cng sut
2
TI: Thit k cu trc iu khin cho nghch lu c lp ngun p ba
pha c cng sut 5KV lm vic ch chnh lu tch cc
I. Tnh chn mch lc
1. Tnh ton chn van IGBT
a) in p ln nht t ln van UVmax
Xt ti thi im van V1 m,van V4 dn:
Cc C ca V1 ni vi cc dng ca ngun Vg
Van V4 thng ni cc E ca V1 vi cc m ca ngun Vg
UVmax = Vg = 2.3
2.q . Udm =
2.3
2.0,75 . 380 = 1074,8 V
Trong q l h s bin iu ( 0 < q < 0,866)
b) Dng in trung bnh qua van
Vi S = 3U.I => I = S
U =
5000
3.380 = 4,386 (A)
Trong ch nh mc c:
Zdm = Udm
Idm=
380
4,386 = 86,64
XL = L =314. 0,002 = 0,628
= arcsin 0,628
86,64 = 024
Rtdm = z.cos = 86,64 . 0,9999 = 86,63
Biu thc dng in trung bnh qua van ng lc trong mt chu k in p ra:
IV = 1
2. Im sin( + ) d
0
Rt gn ta c :
IV = Im
2. (1 + cos )
Dng trung bnh qua van l :
IV = Im
2. (1 + cos ) =
4,386
2. (1 + cos 024) = 1,4 A
Dng in v in p nh mc ca van cn chn l :
UVdm = ku . UVmax =1,4. 1074,8 = 1504,72 (V) 1505 V
IVdm = kI . IVmax = 5. 1,4 = 7A
Bi tp ln Thit k h thng in t cng sut
2
T 2 thng s trn ta chn IGBT loi n BSM200GA170DLC vi cc
thng s chnh sau:
Ucemax = 1700 V
IC = 200 A
2. Tnh chn iot
a) in p ngc ln nht t ln van UDngmax
Xt thi im van V1 , D4 kha v V4 , D1 thng:
Cc anode ca D4 c ni vi cc m ca Vg .Cc Kathode ca D4 do van D1
thng nn ni vi cc dng ca Vg. Nh vy in p ngc ln nht t ln van
UDngmax = Vg
b) Dng in trung bnh i qua Diode ID
Biu thc dng in trung bnh i qua diode trong mt chu k in p ra l:
ID = 1
2 Im
0sin( )d
Dng trung bnh qua van l :
ID = Im
2. (1 - cos ) = ID =
4,386
2. (1 - cos 024) = 0,00007 A
Dng in v in p nh mc ca Diode cn chn l:
UDdm = ku . UDngmax = 1,1. 1074,8 = 1182,28 V
IDdm = kI . IDngmax = 5. 0,00007= 0,00035 A
T hai thng s trn ta chn Diode BYX38 c cc thng s c bn sau:
I = 6 A
Ungmax = 1200 V
3. Ga tr t in
Cmax = 1,3 . Lt
Rt2 = 1,3 .
0,002
69,3122 = 0,54.10-6 F= 0,54 F
Chn t C = 0,6 F
Bi tp ln Thit k h thng in t cng sut
2
II. Thit k khu iu ch rng xung theo nguyn l iu ch vecto khng
gian
Sau y l s nguyn l ca b bin tn s dng 6 kha transitor cng sut :
Hnh 2.1. S nguyn l b nghch lu 3 pha
i vi phng php iu rng xung vector khng gian, b nghch lu c xem nh l mt khi duy nht vi 8 trng thi ng ngt ring bit t 0 n
Hnh 2.2. Trng thi ng ngt cc kha bn nghch lu
Bi tp ln Thit k h thng in t cng sut
2
Bng tm tt :
Vector
in p
Trng thi ca kha in p pha in p dy
Q1 Q3 Q5 Uan Ubn Ucn Uab Ubc Uca U0 0 0 0 0 0 0 0 0 0 U1 1 0 0 2/3 -1/3 -1/3 1 0 -1 U2 1 1 0 1/3 1/3 -2/3 0 1 -1 U3 0 1 0 -1/3 2/3 -1/3 -1 1 0 U4 0 1 1 -2/3 1/3 1/3 -1 0 1 U5 0 0 1 -1/3 -1/3 2/3 0 -1 1 U6 1 0 1 1/3 -2/3 1/3 1 -1 0 U7 1 1 1 0 0 0 0 0 0
Ghi ch: ln in p phi nhn vi VDC
i vi ngun p ba pha cn bng, ta lun c phng trnh sau:
ua (t ) + ub (t ) + uc
(t ) = 0
(2.12)
V bt k ba hm s no tha mn phng trnh trn u c th chuyn
sang h ta 2 chiu vung gc. Ta c th biu din phng trnh trn
di dng 3 vector gm: [ua 0 0]T trng vi trc x, vector [0 ub 0]T lch mt
gc 120o v vector [0
0 uc]T lch mt gc 240o so vi trc x nh hnh sau y.
Hnh 2.3. Biu din vector khng gian trong h ta x-y
Bi tp ln Thit k h thng in t cng sut
2
T ta xy dng c phng trnh ca vector khng gian trong h ta phc nh sau
( ) +
Ta xt trng hp b nghch lu trng thi u U1 :
Ta c: Ua= 2/3 Vdc ; Ub=Uc= -1/3 Vdc Xt trn h ta : trong Us = U1 = K * (Ua + Ub + Uc) ; K=2/3 l h s
bin hnh
Hnh 2.4. Vector in p U1 trn ta
Bi tp ln Thit k h thng in t cng sut
2
+ Tng t nh vy vi cc vector U2-> U6 , ta c gin sau:
Hnh 2.5. Vector in p V1(U1)->V6(U6) trn gin
+ Ngoi ra , chng ta cn 2 trng hp c bit l vector U0 =U7= 0
Hnh 2.6. U7 & U0
tng ca vic iu ch vector khng gian l to nn s dch chuyn
lin tc ca vector khng gian tng ng ca vector in p b nghch lu
trn qu o ng trn, tng t nh trng hp ca vector khng gian ca
i lng 3 pha hnh sin to c. Vi s dch chuyn ca u n ca vector
khng gian trn qu o trn cc sng hi bc cao c loi b v bin
p ra tr nn tuyn tnh. Vector tng ng y chnh l vector trung bnh
trong thi gian mt chu k ly mu Ts ca qu trnh iu khin b nghch lu
p
Bi tp ln Thit k h thng in t cng sut
2
Hnh 2.7. Vector Vs trn h trc
Hnh 2.8. in p 3 pha ng ra trong min thi gian tng ng Hnh 2.11
9
Bi tp ln Thit k h thng in t cng sut
Hnh 2.9. Vs sector 1
Xt gc 1 phn 6 ca hnh lc gic c xc nh bng nh ca 3 vector U1.U2,U0-7
Gi s trong thi gian Ts, ta cho tc dng vecto U1 trong khong thi gian TA, vecto U2
trong khong thi gian TB, U0-7 trong khong thi gian T-TA-TB.
Vector tng ng c tnh bng vector trung bnh ca chui tc ng lin tip
Trong in p pha Vs=Us in p pha ng ra ca Ua, Ub, Uc
Chiu phng trnh vecto Vs v dng cng thc Ts, t s m:
10
Bi tp ln Thit k h thng in t cng sut
Bi tp ln Thit k h thng in t cng sut
Page 11
Nh vy vector trung bnh ( Us) c iu khin theo qu o ng trn.
Chiu quay c th thun hay nghch theo chiu kim ng h. ng trn ni
tip hnh lc gic l qu o ca vector ko gian ln nht m phng php
iu ch vector khng gian ca b nghch lu p hai bc c th t c trong
phm vi iu khin tuyn tnh. Bn knh ng trn ny chnh bng bin
thnh phn c bn in p (pha) ti, vi Udc=700V
Bi tp ln Thit k h thng in t cng sut
Page 12
III.Thit k mch vng iu khin dng in theo nguyn l iu khin vecto
dng in is
Thit k biu chnh dng in trn h ta tnh
Do lng t dng in trn h ta tnh c dng hnh sin vi tn s bng s (tn
s c bn dng in hnh sin), ta s s dng cu trc iu chnh kiu cng hng (PR) c
tn s cng hng 0 =s gii quyt vn ny.
Gc(s) = Kp + Kis
s2+ 02
Phng thc thit k biu chnh ny c thit k trn min tn s, trn c s la
chn bng thng (bandwidth) cho hm truyn h thng kn . Thng thng, bng thng
c la chn trong khong 10 ln tn s c bn v 1/10 tn s pht xung vo mch
nghch lu m bo h thng c p ng ng hc nhanh v n nh.
Hm truyn kn mch vng dng in
GPR(s) = i (s)
i(s) =
Kps2+ Kis+ Kp0
2
Ls3+ (Kp+ R)s2+ (Ki+ 02 L )s+ Kp0
2+ 02R
|GPR(j)| = (Ki )
2+ KP2 (0
2 2)2
[Ki+L ( 022)]2 2 + ( Kp+ R )2 ( 0
2 2)2
GPR (jw) = arctan [ Ki
Ki( 02 2)
] - arctan [ [ L (0
2 2)+ Ki]
( Kp+ R )( 02 2 )
]
Cho Ki = 0,ta c:
|GPR(j)| = Kp
[(L )]2 + ( Kp+ R )2
Bi tp ln Thit k h thng in t cng sut
Page 13
c h s suy gim bin l -3dB ( hay |GPR(jbw)|= 1
2 ), vi bng thng bw=
5000,ta c:
Kp = R + (L bw)2 + 2R2
= 86,63 + (0,002 .2. 5000)2 + 2.86,632
= 224,3
a thnh phn tch phn vo biu thc bin
Ki = bw
2 02
bw . [(R + Kp)2 + 2. (Lbw)2 2. Kp
2 - Lbw] =
(314002 3142)
31400[( 86,63 + 224,3)2 + 2. (0,002.2. 5000)2 2. 224,32 0,002.5000]
= 328,27
Thay vo ta c:
Gc(s) =224,3 + 328,27s
s2+ 3142
* s u
PR
PR SVM NLNA
s i
L u
abc
s i
s i
s i
* s u
s u
* s i
* s i
s i
s i
C dng in
Bi tp ln Thit k h thng in t cng sut
Page 14
IV. Tng hp mch vng in p
1. Tnh ton T C-rc
T lc C phi ln m bo p mch ip p u ra v s st p khi ng
ti trong phm vi cho chp. p mch gm 2 thnh phn : do in tr ni tip t C
gy ra, v do t lc cc xung in p t u ra b bin i.
p mch :
CC rV V V
(Error! No
text of
specified
style in
document..1)
Hnh . p mch in p ra
Thnh phn p mch do t C :
Bi tp ln Thit k h thng in t cng sut
Page 15
1 1
2 2 2
L SC
Q I TV
C C
(Error! No
text of
specified
style in
document..2)
Suy ra gi tr t C p ng c yu cu :
.
8.
L S
C
I TC
V
(Error! No
text of
specified
style in
document..3)
i vi cc mch ngun DC-DC, p mch khng nn vt qu 1% Vo.
2. Mch vng iu chnh in p
Hnh. Cu trc iu khin mch vng in p b bin i Buck
Hm truyn in p gia in p u ra v h s iu ch di dng :
2( ) 0
( ) 1 . .( )
( )1
.
in
o cvd vdo
u s
o o o
u s r C sG s G
d s s s
Q
(Error! No
text of
specified
style in
document..4)
Trong : vdoG = Vin = 700V
o =2*500 Q0=RC
L = 1,5
Bi tp ln Thit k h thng in t cng sut
Page 16
D=U0/UIn= 380/700=0.54
GVD(s) = (1+ 0,6s)/(1+s/(1,5*2*5000)+(s/1,5)2)
Bi tp ln Thit k h thng in t cng sut
Page 17
V. M phng
Hnh 5.1. in p dy u ra nghch lu ba pha
Bi tp ln Thit k h thng in t cng sut
Page 18
Hnh 5.2.Mch iu ch rng xung theo nguyn l iu khin vecto
Hnh 5.3. Sng sin ,
Bi tp ln Thit k h thng in t cng sut
Page 19
Hnh 5.3. Mch vng iu khin dng in
Hnh 5.4 . Dng dng in
Bi tp ln Thit k h thng in t cng sut
Page 20
Nhn xt : Dng phn hi Is bm gi tr t
Hnh 5.5. Khu so snh xung rng ca
6
PWM6
5
PWM5
4
PWM4
3
PWM3
2
PWM2
1
PWM1
Sign2
Sign1
Sign
Satu5
Satu4
Satu3
Satu2
Satu1
Satu
RC
-1
Gain2
-1
Gain1
-1
Gain
3
dc
2
db
1
da
6
PWM6
5
PWM5
4
PWM4
3
PWM3
2
PWM2
1
PWM1d0
d1
d2
sector
da
db
dc
tinh da,db,dc
anpha
beta
Udc
d0
d1
d2
sector
Tinh d0,d1,d2
sector
Scope
da
db
dc
PWM1
PWM2
PWM3
PWM4
PWM5
PWM6
PWM
3
Udc
2
beta
1
anpha
Bi tp ln Thit k h thng in t cng sut
Page 21
Hnh 5.6. Khu iu ch vecto khng gian
Ub
Uc
Ua
4
sector
3
d2
2
d1
1
d0
-K-
cong
Switch2
Switch1
Switch
1
2
3
4
5
6
Multiport
Switch
Matrix
Concatenate2
Matrix
Multiply
Matrix
-0.5
Gain
f(u)
FcnDivide
0
C3
1
C2
2
C1
4
C
3/2 sqrt(3)/2
0 -sqrt(3)
A6
-3/2 -sqrt(3)/2
3/2 -sqrt(3)/2
A5
0 -sqrt(3)
-3/2 sqrt(3)/2
A4
0 sqrt(3)
-3/2 -sqrt(3)/2
A3
-3/2 sqrt(3)/2
3/2 sqrt(3)/2
A2
3/2 -sqrt(3)/2
0 sqrt(3)
A1
1-D T(u)
1-D Lookup
Table
3
Udc
2
beta
1
anpha
Bi tp ln Thit k h thng in t cng sut
Page 22
Hnh 5.7. Khi tnh cc secto
]
3
dc
2
db
1
da
1
2
3
4
5
6
Multiport
Switch
Matrix
Concatenate
Matrix
Multiply
Matrix
[A6]
Goto4
[A5]
Goto3
[A4]
Goto2
[A3]
Goto1
[A2]
Goto
[A6]
F5
[A5]
F4
[A4]
F3
[A2]
F2[A3]
F1
0.5 0 0
0.5 1 1
0.5 1 0
A6
0.5 1 0
0.5 1 1
0.5 0 0
A5
0.5 1 1
0.5 1 0
0.5 0 0
A4
0.5 1 1
0.5 0 0
0.5 1 0
A3
0.5 1 0
0.5 0 0
0.5 1 1
A2
0.5 0 0
0.5 1 0
0.5 1 1
A1
4
sector
3
d2
2
d1
1
d0
Bi tp ln Thit k h thng in t cng sut
Page 23
Hnh 5.8. Khi tnh da, db, dc