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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.


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


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


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


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


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


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


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


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



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


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


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


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 :


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


Page 16

D=U0/UIn= 380/700=0.54

GVD(s) = (1+ 0,6s)/(1+s/(1,5*2*5000)+(s/1,5)2)


Page 17

V. M phng

Hnh 5.1. in p dy u ra nghch lu ba pha


Page 18

Hnh 5.2.Mch iu ch rng xung theo nguyn l iu khin vecto

Hnh 5.3. Sng sin ,


Page 19

Hnh 5.3. Mch vng iu khin dng in

Hnh 5.4 . Dng dng in


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


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


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


Page 23

Hnh 5.8. Khi tnh da, db, dc

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