Bai Giang Htdkso All Hk2nh1011 v92 01

7/21/2019 Bai Giang Htdkso All Hk2nh1011 v92 01

1/106

I HC QUC GIA TP. HCH MINHTRNG I HC BCH KHOA

KHOA IN IN T

Bi ging:

HTHNG IU KHIN S

IU KHIN CC MY IN

BBIN TN

Bin son: ThS. Trn Cng Binh

TP. HCH MINH, THNG 1 NM 2010


2/106

(HTKS-KCM) TB

09/12/2009 3

CHNG TRNH MN HCHTHNG IU KHIN S- IU KHIN CC MY IN

Chng 1: Bnghch lu ba pha v Vector khng gian (5T) Vector khng gian. Bnghch lu ba pha.

Chng 2: Hqui chiu quay (2T) Hqui chiu quay. Chuyn i htoabc dq.

Chng 3: M hnh CKB 3 pha ( ), (dq) (8T) Stng ng ca ng cv mt sk hiu. M hnh ng ctrong HTstator ().

M hnh ng ctrong HTtthng rotor (r).Chng 4: iu khin nh hng tthng (FOC) CKB (6T)

iu khin PID iu khin tip dng. iu khin tip p. M phng ca FOC.

(21 tit)

Chng 5: Mt sphng php c lng tthng rotor (6T) Tmv ia, ibhi tip.

Tusv ia, ibhi tip. Tv ia, ibhi tip. c lng vtr (gc) vector r. c lng (r) trong HTdq. c lng tthng rotor dng khu quan st (observer) p ng m phng FOC.

Chng 6: Cc phng php iu khin dng (6T) iu khin dng trong HQC (): vng trvso snh. iu khin dng trong HQC (dq).

Chng 7: Mt sphng php c lngtc ng c (3T) c lng vn tc vng h(2 pp). c lng vn tc vng kn (c hi tip). iu khin khng dng cm bin (sensorless).

Chng 8: Bbin tn (9T) Cu trc mt hthng iu khin ng c. Cm bin o lng Mt su im khi sdng biu khin tc ng c Hthng iu khin sng ckhng ng bba pha Bbin tn

(24 tit)(45 tit)


3/106

(HTKS-KCM) TB

Chng 1: Vector khng gian v Bnghch lu ba pha I.1

Chng 1: VECTOR KHNG GIAN VBNGHCH LU BA PHA

I. Vector khng gian

I.1. Biu din vector khng gian cho cc i lng ba phang c khng ng b (CKB) ba pha c ba (hay bi s ca ba) cun dy

stator btr trong khng gian nhhnh vsau:

Hnh 1.1: Su dy v in p stator ca CKB ba pha.

(Ba trc ca ba cun dy lch nhau mt gc 1200trong khng gian)

Ba in p cp cho ba u dy ca ng c t li ba pha hay tbnghch lu,bin tn; ba in p ny tha mn phng trnh:

usa(t) + usb(t) + usc(t) = 0 (1.1)

Trong :

(1.2a)

(1.2b)

(1.2c)

Vi s= 2fs; fsl tn sca mch stator; |us| l bin ca in p pha, c ththay i.(in p pha l cc sthc)

A

B

C

N

rotor

stator

Pha A

Pha B

Pha Cusc

usa

usb

usa(t) = |us| cos(st)usb(t) = |us| cos(st120

0)

usc(t) = |us| cos(st + 1200)


4/106

(HTKS-KCM) TB


Vector khng gian ca in p stator c nh ngha nhsau:

[ ])t(u)t(u)t(u3

2)t(u scsbsas

rrrr++= (1.3)

[ ]000 240jsc120jsb0jsas e)t(ue)t(ue)t(u3

2)t(u ++=

r

(mt phngba chiu vi 3 vector n v)[ ]00 240120 )()()(

3

2)(

j

sc

j

sbsas etuetututu ++=r

(1.4)

(tng tnhvector trong mt phng phc hai chiu vi 2 vector n v)

[ ])t(u.a)t(u.a)t(u3

2)t(u sc

2

sbsas ++=r

vi0120j

ea=

[ ] 0eeeaa1 000 240j120j0j2 =++=++

V d1.1: Chng minh?a) ( ) ( ) ( )[ ]tsinjtcosutueu)t(u sssss

tj

sss +===

r (1.6)

b) [ ]

+= csbscsbsass uujuuuu

2

3

2

35,05,0

3

2& (1.5)

Hnh 1.2: Vector khng gian in p stator trong hta .

Theo hnh v trn, in p ca tng pha chnh l hnh chiu ca vector in pstator su

rln trc ca cun dy tng ng. i vi cc i lng khc ca ng c: dng

in stator, dng rotor, t thng stator v t thng rotor u c thxy dng cc vectorkhng gian tng ng nhi vi in p stator trn.

I.2. Hta cnh stator

Vector khng gian in p stator l mt vector c modul xc nh (|us|) quay trnmt phng phc vi tc gc sv to vi trc thc (trng vi cun dy pha A)mt gcst. t tn cho trc thc l v trc o l , vector khng gian (in p stator) c thc m tthng qua hai gi trthc (us) v o (us) l hai thnh phn ca vector. Hta

ny l hta stator cnh, gi tt l hta .

Re

Im

A

B

C

o0je

o120je

o240je

s

ur

usas

usc

usc


5/106

(HTKS-KCM) TB


Hnh 1.3: Vector khng gian in p stator sur

v cc in p pha.

Bng cch tnh hnh chiu cc thnh phn ca vector khng gian in p stator( ) ss u,u ln trc pha A, B (trn hnh 1.3), c thxc nh cc thnh phn theo phng

php hnh hc:

(1.7a)

(1.7b)

suy ra(1.8a)

(1.8b)

Theo phng trnh (1.1), v da trn hnh 1.3 th chcn xc nh hai trong sba in ppha stator l c thtnh c vector su

r.

Hay tphng trnh (1.5)

[ ]

+= csbscsbsass u

2

3u

2

3ju5,0u5,0u

3

2u (1.9)

c thxc nh ma trn chuyn i abc theo phng php i s:

=

cs

bs

as

s

s

s

s

u

u

u

2

3

2

30

2

1

2

11

3

2

u

u

(1.10)

V d1.2: Chng minh ma trn chuyn i htoabc?

0

j

sur

usa=us

ususc

usb Cun dypha A

Cun dypha B

Cun dypha C

us= usa

us= ( )sbsa u2u3

1+

usa= us

usb = ss u

2

3u

2

1+


6/106

(HTKS-KCM) TB


=

s

s

s

s

cs

bs

as

u

u

2

3

2

1

2

3

2

1

01

u

u

u

(1.11)

V d1.3: Chng minh:

Bng cch tng tnhi vi vector khng gian in p stator, cc vector khnggian dng in stator, dng in rotor, t thng stator v t thng rotor u c thc

biu din trong hta stator cnh (hta )nhsau:

(1.12a)

(1.12b)

(1.12c)

(1.12d)

(1.12e)

II. Bnghch lu ba pha

II.1. Bnghch lu ba pha

sur

= us+ jus

sir

= is+ jis

rir

= ir+ jir

+= sss jr

+= rrr jr


7/106

(HTKS-KCM) TB


Bin tn ng vo 1 pha, nu tlc nh, in p DC trung bnh l:

(vi Vi l in p pha).

Bin tn ng vo 3 pha, nu tlc nh, in p DC trung bnh l:

(vi Vi l in p pha).

Nu t lc ln (hay khi khng ti), in p DC sc lc phng. Trin pDC trung bnh ca:

_ Bin tn ng vo 1 pha : phaseU2

_ Bin tn ng vo 3 pha : lineU2 .


8/106

(HTKS-KCM) TB


Hnh 1.4: Sbnghch lu ba pha cn bng gm 6 kho S1S6.

V d1.4: Chng minh cc phng trnh tnh in p pha?

a) ( )CnBnAnNn UUU3

1U ++=

b) CnBnAnAN U3

1U

3

1U

3

2U =

Phng php tnh mch in:V d1.5: Tnh in p cc pha trng thi S1, S3, S6 ON v S2, S4, S5 OFF?

Hnh 1.5: Trng thi cc kho S1, S3, S6 ON, v S2, S4, S5 OFF (trng thi 110).

II.2. Vector khng gian in pn v(Udc)

A B

C

Udc

n

N

UAN UBN

UCN

AB C

Udc

S4

S3

S6

S5

S2

S1

S7

R

n n

motor

N


9/106

(HTKS-KCM) TB


Va Vb Vc usa usb usc uab ubc uca U Deg usk S1 S3 S5 UAN UBN UCN UAB UBC UCA us

us

0 0 0 0 0 0 0 0 0 0 U0 U000

1 1 0 0 2/3 -1/3 -1/3 1 0 -1 U1 0o

2 1 1 0 1/3 1/3 -2/3 0 1 -1 U2 60o

3 0 1 0 -1/3 2/3 -1/3 -1 1 0 U3 120

o

4 0 1 1 -2/3 1/3 1/3 -1 0 1 U4 180o

5 0 0 1 -1/3 -1/3 2/3 0 -1 1 U5 240o

6 1 0 1 1/3 -2/3 1/3 1 -1 0 U6 300o

7 1 1 1 0 0 0 0 0 0 U7 U111

Bng 1.1: Cc in p thnh phn tng ng vi 8 trng thi ca bnghch lu.V d1.6: Tnh cc in p thnh phn us v us tng ng vi 8 trng thi trong

bng 1.1?

iu chvector khng gian in p sdng bnghch lu ba pha

V d1.7: Xt bnghch lu trng thi 110:Khi cc in p pha usa=1/3Udc, usb= 1/3Udc, usc=-2/3Udc.Phng php i s: theo phng trnh (1.4):

[ ]

+=++=

0000 240j

dc

120j

dcdc

240j

sc

120j

sbsa1_phase eU3

2eU

3

1U

3

1

3

2e)t(ue)t(u)t(u

3

2ur

( )[ ] 0000000 60jdc180j240jdc240jdc240j240j120jdc1_phase eU3

2eeU

3

2eU

3

2e3ee1

3

U

3

2u ===++= r

,

Hay [ ]

+=++= dc

2

dcdcsc

2

sbsa1_phase U.a3

2U.a

3

1U

3

1

3

2)t(u.a)t(u.a)t(u

3

2ur

vi0120jea= , 0aa1 2 =++

( )[ ] 00 60jdc240jdc2dc22dc1_phase eU3

2eU

3

2aU

3

2a3aa1

3

U

3

2u ===++=r

Phng php hnh hc: c hnh v


10/106

(HTKS-KCM) TB


Hnh 1.6: Vector khng gian in p stator surng vi trng thi (110).

trng thi (110), vector khng gian in p stator pha 1_phaseur

c ln bng

2/3Udc v c gc pha l 60o.

V d1.8: Tm ( ln v gc ca) vector khng gian in p stator )t(usr

ngvi trng thi (101)? (Gii theo phng php i snhtrn hay theo phng phphnh hc)

Xt tng t cho cc trang thi cn li, rt ra c cng thc tng qut

3)1k(j

dck eU3

2U

= vi k = 1, 2, 3, 4, 5, 6.

Hnh 1.7: 8 vector khng gian in p stator tng ng vi 8 trng thi.

3)1k(j

dck eU3

2U

= k = 1, 2, 3, 4, 5, 6. U0v U7l vector 0.

A

sur

B

C

scur

Udc

saur

sbur

scsbsa uuu rrr

++

U2(110)

U1(100)

U2(110)U3(010)

U6(101)U5(001)

U4(011)

CCW

CW

U0(000)

U7(111)


11/106

(HTKS-KCM) TB


Cc trng hp xt trn l vector khng gian in p phastator.

Hnh 1.8: Cc vector khng gian in p phastator.

3)1k(j

dck_phase eU

3

2U

= k = 1, 2, 3, 4, 5, 6

Bng cch iu khin chuyn i trng thi ng ct cc kha ca bnghch lu ddng iu khin vector khng gian in p quay thun nghch, nhanh chm. Khi dngin p ng ra bnghch lu c dng 6 bc (six step).

Hnh 1.9: Cc in p thnh phn tng ng vi 6 trng thi.

V d1.9: Chng minh00j

dc0_phase eU

3

2u =

Xt bnghch lu trng thi 100:Khi cc in p pha usa=2/3Udc, usb= 1/3Udc, usc=-1/3Udc.

Phng php i s:theo phng trnh (1.3): [ ])t(u)t(u)t(u3

2)t(u scsbsas

rrrr++=

hay phng trnh (1.4):

[ ]

=++=

0000 240j

dc

120j

dcdc

240j

sc

120j

sbsa0_phase eU3

1eU

3

1U

3

2

3

2e)t(ue)t(u)t(u

3

2ur

( )[ ] 000 0jdcdc240j120jdc0_phase eU3

2U

3

2ee13

3

U

3

2u ==++=r

,

Up1

Up2Up3

Up6Up5

Up4Up0

Up7

Trc usaa

b

c


12/106

(HTKS-KCM) TB


Hay [ ]

=++= dc

2

dcdcsc

2

sbsa0_phase U.a3

1U.a

3

1U

3

2

3

2)t(u.a)t(u.a)t(u

3

2ur

vi0120jea= , ( ) 0aa1 2 =++

( )[ ]

00j

dcdc

2dc

0_phase eU3

2

U3

2

aa133

U

3

2

u ==++=

r

Phng php hnh hc:c hnh v

Hnh 1.10: Vector khng gian in p stator surng vi trng thi (100).

trng thi (100), vector khng gian in p pha stator 0_phaseur c ln bng

2/3Udc v c gc pha trng vi trc pha A.

Trong mt strng hp, cn xt vector khng gian in p dyca stator.

[ ])t(u)t(u)t(u3

2u cabcabline

rrrr++=

hay [ ]00 240jca120jbcabline e)t(ue)t(u)t(u3

2u ++=r

hay [ ])t(u.a)t(u.a)t(u3

2u ca

2

baabline ++=r

vi0120jea=

V d1.10: Xt bnghch lu trng thi 100:Khi cc in p pha uab=Udc, ubc= 0, uca= -Udc.

Phng php i s: theo phng trnh trn:

[ ] [ ]000 240jdcdc240jca120jbcab1_line eUU3

2e)t(ue)t(u)t(u

3

2u =++=r

[ ] ( )

++=+==

2

3j

2

11U

3

2e1U

3

2eUU

3

2u dc

60j

dc

240j

dcdc1_line

00r

A

sur

B

C

scur

2/3Udc

sa

ur

sbur

scsbsa uuu

rrr

++

U1(100)


13/106

(HTKS-KCM) TB


030j

dcdcdc1_line eU33

2

2

1j

2

3U3

3

2

2

3j

2

3U

3

2u =

+=

+=

r

Phng php hnh hc:c hnh v:

Hnh 1.11: Vector khng gian in p dy stator 1_lineur

ng vi trng thi (100).

trng thi (100), vector khng gian in p dy stator 1_lineur

c ln bng

dcU33

2 v c gc pha l 30o.

V d1.11: Tm ( ln v gc ca) vector khng gian in p stator lineur

ng vi

trng thi (110), 2_lineur

? (Gii theo phng php i sv phng php hnh hc)

Xt tng t cho cc trng thi cn li, rt ra c cng thc tng qut

6)1k2(j

dck_line eU33

2U

= k = 1, 2, 3, 4, 5, 6

AB

BC

CA

bcur

2/3Udc

abur

Uline_1

Ud1

Ud2

Ud3

Ud6

Ud5

Ud4

Ud0

Ud7 Trc uab


14/106

(HTKS-KCM) TB


Hnh 1.12: Cc vector khng gian in p dystator.

V d1.12: Chng minh cc vector in p c gi trnhsau:

a/5

36

2

3

j

pha DCv V e

= b/5

63

23

3

j

day DCv V e

=

iu chbin v gcvector khng gian in p dng bnghch lu ba pha

Hnh 1.13: iu chbin v gc vector khng gian in p.

khng qu iu ch, bin in p phi nm trong

vng trn ni tuyn ca lc gic: 3dc

s

U

u

)U(UT

TU

T

TU

T

Tu 70

PWM

02

PWM

21

PWM

1s ++= hay )U(U.cU.bU.au 7021s ++=

U1(100)

us

T1

T2

U2(110)U3(010)

U6(101)U5(001)

U4(011)

CCW

CW

U0(000)

U7(111)

sur


15/106

(HTKS-KCM) TB


)3

sin(3 =

Udc

ua s ( )sin3

Udc

ub

s= c 1 (a+b)

T1= a.TPWM T2= b.TPWM T0= c.TPWMvi chu kiu rng xung: TPWM(T1+ T2) + T0 hay T0TPWM (T1+ T2)

vi TPWMconstTng qut: us=a.Ux+ b.Ux+60+ c.{U0, U7}

Trong , l gc gia vector Uxv vector in p us.

C thtnh theo:( )

3

2 dc

s

U

u

bacba

+=++ hay ( )

+= 1

u3

U2bac

s

dc

Bng cch iu khin chuyn i trng thi ng ct cc kha ca bnghch lu

thng qua T1, T2v T0, ddng iu khin lnv tc quayca vector khng gianin p. Khi dng in p ng ra bnghch lu c dng PWM sin.

U1(100)

us

T1

T2

U2(110)

U0(000)

U7(111)


16/106

(HTKS-KCM) TB


Hnh 1.14: iu chbin v tn sin p.

Hnh 1.15: Dng in p v dng in PWM sin.

V d1.13: Chng minh

+

= 6j

dc2dc1

j

s eU3

2TU

3

2Teu


17/106

(HTKS-KCM) TB


Bi tp 1.1. Chng minh: 34

j

dc5_phase eU3

2u

=

Bi tp 1.2. Chng minh: 67

j

dc4_line eU33

2u

=

Bi tp 1.3. in p ba pha 380V, 50Hz. Ti thi im t = 6ms. Tnh usa, usb, usc, usvus, |us|? Bit gc pha ban u ca pha A l o= 0.

Bi tp 1.4. Li in 3 pha 4 dy 380V. Tnh Udctrung bnh, Usmax (trong vng trn nitip), Upha, Udyca bin tn:

a. 1 pha.

b.

3 pha.

Bi tp 1.5. in p ba pha cp cho bnghch lu l 380V, 50Hz. Tnh in p pha lnnht m bnghch lu c thcung cp cho ng cni Y.

Bi tp 1.6. in p mt pha cp cho bnghch lu l 220V, 50Hz. Tnh in p dy lnnht m bnghch lu c thcung cp cho ng c.

Bi tp 1.7. Bbin tn dng Vit Nam, 3 pha 380V (ng vo). c cp ngun 3 pha380V, 50Hz.a) Tlc kh nh:

_ Tnh in p trung bnh trn DC Link?_ Tnh bin in p pha ln nht (cha qu iu ch)?_ Tnh in p hiu dng pha ln nht?_ Tnh in p hiu dng dy ln nht?b) Tlc ln: tnh li cu a).

Bi tp 1.8. Tnh li cu trn vi bin tn 1 pha 220V (ng vo).Bi tp 1.9. in p ba pha cp cho bnghch lu l 380V, 50Hz. in p pha bnghch

lu cp cho ng cl 150V v 50Hz. Ti thi im t = 6ms. Tnh T1, T2v

T0? Bit gc pha ban u o= 0 v tn siu rng xung l 20KHz.Bi tp 1.10. Lp bng v v gin vector cc in pdy thnh phn tng ng vi 8 trng thi ca bnghch lu.

Bi tp 1.11. Nu cc chc nng ca kho S7 v ccdiode ngc (mc song song vi cc kho ng ct S1 S6) trong bnghch lu?

Bi tp 1.12. Cho Udc= 309V, trng thi cc kho nhsau: S2, S3, S6: ON; v S1, S4, S5: OFF. Tnh cc in p usa, usb, usc,UAB, UBC?

Bi tp 1.13. Khi tng tn siu rng xung (PWM) ca bnghch lu, nh gi tc

ng ca sng hi bc cao ln dng in ng c. Phng php iukhin no c tn sPWM lun thay i?

V d1.1: Chng minh?a) ( ) ( ) ( )[ ]tsinjtcosutueu)t(u sssss

tj

sss +===

r (1.6)

b) [ ]

+= csbscsbsass u

2

3u

2

3ju5,0u5,0u

3

2u (1.5)

V d1.2: Chng minh ma trn chuyn i htoabc?


18/106

(HTKS-KCM) TB


=

s

s

s

s

cs

bs

as

u

u

2

3

2

1

2

3

2

1

01

u

u

u

(1.11)

V d1.3: Chng minh:

V d1.4: Chng minh cc phng trnh tnh in p pha?

a) ( )CnBnAnNn UUU3

1U ++=

b) CnBnAnAN U3

1U3

1U3

2U =

V d1.5: Tnh in p cc pha trng thi S1, S3, S6 ON v S2, S4, S5 OFF?V d1.6: Tnh cc in p thnh phn us v us tng ng vi 8 trng thi trong

bng 1.1?

V d1.7: Bnghch lu trng thi 110, chng minh060j

dc1_phase eU3

2u =r

V d1.8: Tm (ln v gc ca) vector khng gian in p stator )t(usr

ng vitrng thi (101)? (Gii theo phng php i snh trn hay theo phng

php hnh hc)

V d1.9: Chng minh00j

dc0_phase eU32u =

V d1.10: Bnghch lu trng thi 100, chng minh030j

dc1_line eU33

2u =r

V d1.11: Tm ( ln v gc ca) vector khng gian in p stator lineur

ng vi

trng thi (110), 2_lineur

? (Gii theo phng php i sv phng php hnh hc)

V d1.12: Chng minh cc vector in p c gi trnhsau:

a/5

36

2

3

j

pha DCv V e

= b/5

63

23

3

j

day DCv V e

=

V d1.13: Chng minh

+

= 6

j

dc2dc1

j

s eU3

2TU

3

2Teu


19/106

(HTKS-KCM) TB

Chng 2: Hqui chiu quay II.1

Chng 2: HQUI CHIU QUAY

I. Hqui chiu quayTrong mt phng ca h ta , xt thm mt h ta th 2 c trc

honh d v trc tung q, hta th2 ny c chung im gc v nm lch i mtgc sso vi hta stator (hta ). Trong ,

dt

d aa

= quay trn quanh

gc ta chung, gc a= at + a0. Khi s tn ti hai ta cho mt vectortrong khng gian tng ng vi hai hta ny. Hnh vsau sm tmi lin hca hai ta ny.

Hnh 2.1: Chuyn htocho vector khng gian sur

thta sang hta dq v ngc li.

Thnh 1.5 ddng rt ra cc cng thc vmi lin hca hai ta camt vector ng vi hai hta v dq. Hay thc hin bin i i s:

(1.10a)(1.10b)

Theo pt (1.9a) th: sss fjffrrr

+= (1.11)

v tng tth: sqsddqs fjff

rrr

+= (1.12)

V d2.1: Chng minh ajdqss eff rr

=

Khi thay hpt (1.10) vo pt (1.11) sc:( ) ( )asqasdasqasds cosfsinfjsinfcosff ++=

r

( )( ) ajdqsaasqsd efsinjoscjff r

=++= (1.13)

Hayajdq

ss eff rr

= aj

sdqs eff

=rv

(1.14)

j

fs

0

sfr

fs

d

jq

fsd

fsq

a

dt

d aa

=

s

fs= fsdcosa- fsqsina

fs= fsdsina+ fsqcosa


20/106

(HTKS-KCM) TB


V d2.2: Tnh fsdv fsqtheo fs, fsv a.

Thay pt (1.11) vo pt (1.14), thu c phng trnh:(1.15a)

(1.15b)

Hnh 2.2: Hta quay

sf

Cun dy

pha A

Cun dy

pha B

Cun

dy pha C

0

d

jq

fsd

fsq

a

a

s

fsd= fscosa+ fssina

fsq = - fssina+ fscosa


21/106

(HTKS-KCM) TB


XT KHI 0a =

II. Biu din cc vector khng gian trn hta tthng rotorMc ny trnh by cch biu din cc vector khng gian ca ng ckhng

ng b(CKB) ba pha trn hta tthng rotor. Githit mt CKB bapha ang quay vi tc gc

dt

d= (tc quay ca rotor so vi stator ng

yn), vi l gc hp bi trc rotor vi trc chun stator (qui nh trc cun dypha A, chnh l trc trong hta ).


22/106

(HTKS-KCM) TB


Hnh 2.3: Biu din vector khng gian sir

trn htotthng rotor, cn gi lhtodq.

Trong hnh 1.6 biu din chai vector dng stator sir

v vector tthng rotor

rr

. Vector tthng rotor rr

quay vi tc gc ssr

ra f2

dt

d

=== (tc

quay ca tthng rotor so vi stator ng yn). Trong , fsl tn sca mch instator v rl gc ca trc d so vi trc chun stator (trc ).

chnh lch gia sv (githit si cc ca ng cl p=1) stonn dng in rotor vi tn sfsl, dng in ny cng c thc biu din didng vector ri

r

quay vi tc gc sl= 2fsl, (sl= s- r- ) so vi vectortthng rotor r

r

.Trong mc ny ta xy dng mt h trc ta mi c hng trc honh

(trc d) trng vi trc ca vector tthng rotor rr

v c gc trng vi gc ca hta , hta ny c gi l hta tthng rotor, hay cn gi l htadq. Hta dq quay quanh im gc chung vi tc gc rs, v hp vi hta mt gc r.

sir

is

Cun dypha A

Cun dypha B

Cundy pha C

0is

d

jq

isd

isq

rr

r=a

r

Trc t

thng rotor

Truc rotor

jdt

d rr

=


23/106

(HTKS-KCM) TB


Vy ty theo quan st trn hta no, mt vector trong khng gian scmt ta tng ng. Qui nh chstrn bn phi ca k hiu vector nhn bitvector ang c quan st thta no:

s: ta (stator coordinates). f: ta dq (field coordinates).

Nhtrong hnh 1.6, vector sir

sc vit thnh:

ssir

: vector dng stator quan st trn hta .

f

sir

: vector dng stator quan st trn hta dq.Theo pt (1.8a) v pt (1.11) th:

(1.16a)(1.16b)

Nu bit c gc rth sxc nh c mi lin h:

(1.17a)(1.17b)

Theo hpt (???) v pt (1.17b) th c thtnh c vector dng stator thngqua cc gi trdng iav ibo c (hnh 1.7).

Hnh 2.4: Thu thp gi trthc ca vector dng stator trn hta dq.

C KB

==

3~

Udc

iukhin

M3~

a b c

Nghchlu

2=

3

isa

isbis

isrje

isd

isq

r

pt (2.)pt (2.)

ssir

= is+ jis

fsir

= isd+ jisq

rjfs

ss eii

=rr

rjss

fs eii

=rr


24/106

(HTKS-KCM) TB


Tng t nhi vi vector dng stator, c th biu din cc vector khcca CKB trn hta dq:

(1.18a)(1.18b)

(1.18c)(1.18d)(1.18e)

Tuy nhin, tnh c isdv isqth phi xc nh c gc r, gc rcxc nh thng qua r= + sl. Trong thc tchc l c tho c, trong khi

(tc trt) sl= 2fslvi fsl l tn sca mch in rotor (lng sc) khng oc. V vy phng php iu khin CKB ba pha da trn cc m t trn hta d dq bt but phi xy ng phng php tnh r chnh xc. Ch khi xydng m hnh tnh ton trong h ta dq, do khng th tnh tuyt i chnh xcgc rnn vn gili rq ( rq =0) m bo tnh khch quan trong khi quan st.

III. u im ca vic m tng ckhng ng bba pha trn hta tthng rotor

Trong h ta t thng rotor (h ta dq), cc vector dng stator fsir

v

vector tthng rotorf

r

r

, cng vi hta dq quanh (gn) ng bvi nhau vitc r quanh im gc, do cc phn t ca vector

fsi

r

(isd v isq) l cc ilng mt chiu. Trong chxc lp, cc gi trny gn nhkhng i; trong qutrnh qu , cc gi trny c thbin theo theo mt thut ton iu khin cnh trc.

Hn na, trong h ta dq, rq=0 do vung gc vi vectorfr

r

(trng vi

trc d) nn frr

=rd. (1.19)

i vi CKB 3 pha, trong hta dq, tthng v mmen quay cbiu din theo cc phn tca vector dng stator:

(1.20a)

(1.20b)

(Hai phng trnh trn sc chng minh trong chng sau).vi: Te momen quay (momen in) ca ng c

Lr in cm rotorLm hcm gia stator v rotor

p si cc ca ng cTr hng sthi gian ca rotor

fsi

r

= isd+ jisq

f

sur

= usd+ j

isqf

rir

= ird+ jirq

sqsdfs j+=

r

rqrdfr j+=

r

sdr

mrd isT1

L+

=

dt

d

P

JTip

L

L

2

3T Lsqrd

r

me

==


25/106

(HTKS-KCM) TB


s ton tLaplacePhng trnh (1.20a) cho thy c th iu khin t thng rotor rrd =

r

thng qua iu khin dng stator isd. c bit mi quan hgia hai i lng ny lmi quan htrbc nht vi thi hng Tr.

Nu thnh cng trong vic p t nhanh v chnh xc dng isdiu khin n

nh tthng rd ti mi im lm vic ca ng c. V thnh cng trong vic pt nhanh v chnh xc dng isq, v theo pt (1.20b) th c th coi isq l i lngiu khin ca momen Teca ng c.

Bng vic m tCKB ba pha trn h ta t thng rotor, khng cnquan tm n tng dng in pha ring l na, m l ton b vector khng giandng stator ca ng c. Khi vector si

r

scung cp hai thnh phn: isdiu

khin tthng rotor rr

, isqiu khin momen quay Te, t c thiu khin

tc ca ng c.(1.21a)

(1.21b)

Khi , phng php m tCKB ba pha tng quan ging nhi ving cmt chiu. Cho php xy dng hthng iu chnh truyn ng CKB

ba pha tng tnhtrng hp sdng ng cin mt chiu. iu khin tc CKB ba pha thng qua iu khin hai phn tca dng in si

r

l isdv isq.

Bi tp 2.1. Chng minh:

Bi tp 2.2. Cho in p ba pha:usa= 540cos(100t) (V);

usb= 540cos(100t 2/3) (V);

usc= 540cos(100t 4/3) (V)Ti thi im t = 0,004giy:

a) Tnh usa, usb, usc, us, usv sur

?

b) Bit gc htoquay lc l r= s= 45o.

Tnh usdv usq?

isd rr

isqTe

2 2

r r

r r

r

r r

d d

dt dt

=+


26/106

(HTKS-KCM) TB


C KB

==3~

Udc

iu

khin

M3~

a b c

Nghchlu

2=

3

isa

isbis

isrje

isd

isq

r

pt (2.)pt (2.)


27/106

(HTKS-KCM) TB

Chng 3: M hnh CKB trong hqui chiu quay III.1

Chng 3: M HNH CKB TRONG HQUI CHIU QUAY

I. Mt skhi nim cbn ca ng ckhng ng bba pha

I.1. Mt squi c k hiu dng cho iu khin CKB ba phaxy dng m hnh m tng cKB ba pha, ta thng nht mt squi

c cho cc k hiu cho cc i lng v cc thng sca ng c.

A

B

C

N

stator

Cun dypha A

isa

usa

irA

isc

usc

isbusb

Cun dy

pha C

Cun dypha B

rotor

irC

irB

stator

Trc chun


28/106

(HTKS-KCM) TB


Hnh 2.1: M hnh n gin ca ng cKB ba pha

mL

s

Rr

rLsLsR

sv

si ri

mi

Hnh 2.2: Mch tng ng ca ng cKB ba pha

Trc chun ca mi quan st c qui c l trc ca cun dy pha A nhhnh 2.1. Mi cng thc c xy dng sau ny u tun theo qui c ny. Sau yl mt scc qui c cho cc k hiu:

Hnh thc v vtr cc chs: Chsnhgc phi trn:

s i lng quan st trn hqui chiu stator (hta ).f i lng quan st trn h qui chiu t thng rotor

(hta dq).r i lng quan st trn hta rotor vi trc thc l trc

ca rotor (hnh 1.6).*, ref, gi trt /lnh (reference)e gi trc lng

Chsnhgc phi di:o Chci u tin:

s i lng ca mch stator.r i lng ca mch rotor.

o Chci thhai:d, q phn tthuc hta dq., phn tthuc hta .a, b, c i lng ba pha ca stator.A, B, C i lng ba pha ca rotor, li.

Hnh mi tn () trn u: k hiu vector (2 chiu).

Gch chn (_) di: k hiu vector, ma trn. ln (modul) ca i lng:k hiugia hai du gch ng (| |).

Cc i lng ca CKB ba pha:u in p (V).i dng in (A). tthng (Wb). tthng mc vng (A.vng).Te momen in t(N.m).TL momen ti (momen cn - torque) (hay cn k hiu l MT) (Nm).

tc gc ca rotor so vi stator (rad/s).a tc gc ca mt htobt k(arbitrary)(rad/s).


29/106

(HTKS-KCM) TB


s tc gc ca tthng stator so vi stator (s= + sl) (rad/s).r tc gc ca tthng rotor so vi stator (rs) (rad/s).sl tc gc ca tthng rotor so vi rotor (tc trt) (rad/s). gc ca trc rotor (cun dy pha A) trong hto(rad).s gc ca trc d (htoquay bt k) trong hto(rad).

r gc ca trc d (htoquay bt k) so vi trc rotor (rad).s gc ca tthng stator trong hto(rad).r gc ca tthng rotor trong hto(rad).r

e gc ca tthng rotor c lng (estimated)trong hto(rad).

gc pha gia in p so vi dng in. Cc thng sca CKB ba pha:

Rs in trcun dy pha ca stator ().Rr in trrotor qui i vstator ().Lm hcm gia stator v rotor (H).

Ls in cm tn ca cun dy stator (H).Lr in cm tn ca cun dy rotor qui i vstator (H).P si cc ca ng c.J momen qun tnh c(Kg.m2).

Cc thng snh ngha thm:Ls= Lm+ Ls in cm stator.Lr= Lm+ Lr in cm rotor.

Ts=s

s

R

L hng sthi gian stator.

Tr=r

r

R

L

hng sthi gian rotor.

= 1 rs

2

m

LL

L hsttn tng.

Tsamp chu kly mu. Cc i lng vit bng chthng chhoa:

Chthng: i lng tc thi, bin thin theo thi gian.i lng l cc thnh phn ca cc vector.

Chhoa: i lng vector, module ca vector, ln.

I.2. Cc phng trnh cbn ca CKB ba phaCc phng trnh ton hc ca ng ccn phi thhin r cc c tnh thi

gian ca i tng. Vic xy dng m hnh y khng nhm mc ch m phngchnh xc vmc ton hc i tng ng c. Vic xy dng m hnh y chnhm mc ch phc vcho vic xy dng cc thut ton iu chnh. iu cho

php chp nhn mt siu kin ginh trong qu trnh thit lp m hnh, ttnhin sto ra mt ssai lch nht nh gia i tng v m hnh trong phm vicho php. Cc sai lch ny phi c loi trbng kthut iu chnh.

c tnh ng ca ng ckhng ng bc m tvi mt hphngtrnh vi phn. xy dng phng trnh cho ng c, ginh l tng ha kt cu

dy qun v mch tvi cc githuyt sau: Cc cun dy stator c btr i xng trong khng gian.


30/106

(HTKS-KCM) TB


Bqua cc tn hao st tv sbo ha ca mch t. Dng tha v ttrng phn bhnh sin trong khe hkhng kh. Cc gi trin trv in khng xem nhkhng i.

mL

s

Rr

rLsLsR

sv

si ri

mi

RssI

& jXs

sU&

Mch tng ng ng cKB vi dng tho

mI&

jXm

jXr'rI

&

s

R'r

RssI

& jXs

sU&

'

rI&

'

rR jXr

Mch tng ng ng cKB vi tn hao st t

'

rRs

s1

RFe jXm

mI&

FeI&

RssI& jXs

RmmI

&

sU&

jXm

'

rI&

'

rR jXr

Mch tng ng ca ng cKB

'

rRs

s1


31/106

(HTKS-KCM) TB


r

rj

sR sL rL

r

rR

s

*

sv mLmi

*

r

r

r

rv

si

Rr

Phng trnh in p trn 3 cun dy stator:

usa(t) = Rsisa(t) +dt

)t(d sa (2.1a)

usb(t) = Rsisb(t) +dt

)t(d sb (2.1b)

usc(t) = Rsisc(t) +dt

)t(d sc (2.1c)

Biu din in p theo dng vector:

[ ]00 240jsc120jsbsass e)t(ue)t(u)t(u3

2)t(u ++=

r (2.2)

Thay cc phng trnh in p pha (2.1a),(2.1b),(2.1c) vo (2.2), ta c:V d3.1: Chng minh:

s

sur

(t) = Rs. )t(is

s

r

+dt

)t(d ssr

(2.3)

Trong , tng tnhi vi in p:

[ ]00 240jsc120jsbsass e)t(ie)t(i)t(i3

2)t(i ++=

r

(2.4)

[ ]00 240j

sc120j

sbsass e).t(e).t()t(

32)t( ++=r (2.5)


32/106

(HTKS-KCM) TB


Tng t, ta c phng trnh in p ca mch stotor. Khi quan st trn h quichiu rotor (rotor ngn mch):

( ) ( )

dt

tdtiR0)t(u

r

rr

rr

r

r

+==

rrrr

(2.6)

Cc vector tthng stator v rotor quan hvi cc dng stator v rotor:V d3.2: Chng minh:

CM rmsss iLiLrrr

+= (2.7a)

CM rrsmr iLiLrrr

+= (2.7b)

vi ( )rsmmmm iiLiLrrrr

+== (2.7b)

smrmsss iLiL rrrrr+=+=

rmrrsmr iLiL rrrrr

+=+=

V d3.3: Chng minh: Lm= 3/2LaA?

CKB l mt hin c, c phng trnh momen:

( ) ( )rrsse ixP2

3ixP

2

3T

rrrr == (2.8)

V d3.4: Chng minh:

( ) ( ) ( )srmsmsrr

me ixiPL

2

3ixP

2

3ix

L

LP

2

3T

rrrrrr===

v phng trnh chuyn ng:

Te = TL +dt

d

P

J (2.9)

Vic xy dng cc m hnh cho CKB ba pha trong cc phn sau u phida trn cc phng trnh cbn trn y ca ng c.

II. M hnh lin tc ca CKB trn hta stator (to )

Tng t nh (1.13), t h quy chiu rotor quy v h quy chiu stator theo ccphng trnh:

V d3.5: Chng minh:js

r

r

r eii =rr

(2.10)

V jsr

r

r e=

rr (2.11)

vi

=dt

d, trong l tc quay ca rotor (theo hnh 2.3).

Thay pt (2.10) v pt (2.11) vo pt (2.6), qui pt (2.6) vhquy chiu stator:

V d3.6: Chng minh: srsrs

rr jdt

diR0

rrr+= (2.12)


33/106

(HTKS-KCM) TB


Smch in tng ng ca m hnh ng caCKB trong HTstator

Vy tcc pt (2.3), (2.7), (2.8), (2.9) v(2.12) ta c hphng trnh:

s

sur

= Rs.s

sir

+dt

d ssr

(2.13a)

0 = Rrs

rir

+

dt

ds

rr

- srj r

(2.13b)

s

rm

s

ss

s

s iLiLrrr

+= (2.13c)s

rr

s

sm

s

r iLiLrrr

+= (2.13d)

Te =2

3p( s

rx si

r

)= -2

3p( r

rx ri

r

) (2.13e)

Te = TL +dt

d

p

J (2.13f)

xc nh dng in stator v tthng rotor, tpt (2.13d) v pt (2.13c) c:s

r

ir

=rL

1 ssm

s

r

iLrr

(2.14)

s

s = Ls.s

si +r

m

L

L ssm

s

r iL (2.15)

Thay (2.14) v (2.15) vo (2.13a) v (2.13b), vi cc nh ngha sau:

Ts =s

s

R

L : hng sthi gian stator.

r

rr

R

LT = : hng sthi gian rotor.

rs

2

m

LL

L1= : hsttn tng.

Phng trnh (2.13a) v (2.13b) trthnh:

dt

d

L

L

dt

idLiRu

s

r

r

m

s

ss

s

sS

s

s

++=

rrrr

(2.16)

dt

dj

T

1i

T

L0

s

rs

r

r

s

s

r

m +

+=

rrr

(2.17)

suy ra:

s

rr

s

sr

m

s

r jT

1i

T

L

dt

d

=

rrr

(2.19)

Thay (2.19) vo (2.16):


34/106

(HTKS-KCM) TB


s

s

s

s

r

rm

s

s

rs

s

s uL

1j

T

1

L

1i

T

1

T

1

dt

id rrrr

+

+

+= (2.20)

s

r

r

s

s

r

m

s

r jT

1i

T

L

dt

d

=

rrr

(2.21)

Chuyn sang dng cc thnh phn ca vector trn hai trc to:

+

+

+

+

= ss

r

m

r

mr

s

rs

s uL

1

L

1

LT

1i

T

1

T

1

dt

di (2.22a)

+

+

+

= ss

r

m

r

mr

s

rs

su

L

1

L

1

LT

1i

T

1

T

1

dt

di (2.22b)

=

rr

r

s

r

mr

T

1i

T

L

dt

d (2.22c)

+=

rr

r

s

r

mr

T

1i

T

L

dt

d (2.22d)

Thay pt (2.14) srir

=rL

1 ssm

s

r iLrr

vo pt (2.13e), c:

( ) ( )sssrr

m

r

s

sm

s

r

s

re i.xL

LP

2

3

L

1iLxp

2

3T

rrrrr =

=

Thay cc thnh phn ca vector tthng rotor v dng stator, c:V d3.7: Chng minh:

( ) srsrrm

eii

L

Lp

2

3T =

(2.24)

[ ]Le TTJ

p

dt

d=

M hnh ton ng cDCMch tng ng ca ng cDC:

Phng trnh mch vng in p cho phn ng ca ng c.

U = E + Ruiu+ Lu udi

dt

Trong : E = K

Ikt

U

R

I

E = kE.kt. k.Ikt.Rkt

Ukt kt

L


35/106

(HTKS-KCM) TB


ktkkt.ikt

Phng trnh cn bng moment trn trc ng c:

Te= TL + Jd

dt

+ B

Trong : Te= k..i

u

J - Moment qun tnh ca hthng quy i vtrc ng c.B - Hsma stTL- Moment cn quy i vtrc ng c.

Ap dng bin i laplace, tcc phng trnh trn, c m hnh ng cDC:( ) ( ) ( ) ( )ssILsIRsEsV uuuu ++= ( ) ( )sKsE = ( ) ( ) ( ) ( )sJssBsTsT L ++= ( ) ( )sIksT u=

( ) ( ) ( )

uu

uRsL

sEsVsI

+=

( ) ( ) ( )

BJs

sTLsTs

+

=

Skhi m hnh ng cDC:

III. M hnh ca CKB trn hta tthng rotor (todq)

Theo hpt (1.17), biu din pt (2.3) v pt (2.6) ln htrc ta tthng rotor (h

trc dq):

s

sur

(t) = Rs. )t(is

s

r

+dt

)t(d ssr

(2.3)

( ) ( )

dt

tdtiR0)t(u

r

rr

rr

r

r

+==

rrrr

(2.6)

Vi rrjf

r

tjf

r

s

r eieii

rrr

==

( ) ( )tjfr

tjf

r

r

rss eieii

== rrr

V d3.8: Chng minh:

uu RsL

1

+

.K

.K

BJs

1

+

( )s T (s)

( )sTL

( )V s I (s)

E (s)


36/106

(HTKS-KCM) TB


f

sur

= Rsf

sir

+ js fsr

+dt

d fsr

(2.29a)

0 = Rrf

rir

+ jsl frr

+dt

d frr

(2.29b)

S

mch

in t

ng

ng c

a m hnh

ng c

a

CK

B trong HT

dq

Kt hp vi hai pt trn vi hphng trnh (2.7), c hphng trnh:

f

sur

= Rsf

sir

+ js fsr

+dt

d fsr

(2.30a)

0 = Rrf

rir

+ j(s-) frr

+dt

d frr

(2.30b)

f

rm

f

ss

f

s iLiLrrr

+= (2.30c)

f

rr

f

sm

f

r iLiLrrr

+= (2.30d)

Suy ra

( )fsmfrr

f

r iLL

1i =

f

r

r

mf

s

r

2

ms

f

sL

Li

L

LL +

=

Thc hin tng ti vi vic xy dng m hnh ng ctrn hta , kh

cc bin frir vf

sr , c hsau:

f

s

s

f

r

rm

f

ss

f

s

rs

f

s uL

1j

T

1

L

1iji

T

1

T

1

dt

id rrrr

+

+

+=

f

rsl

r

f

s

r

m

f

r jT

1i

T

L

dt

d

rrr

+=

Chuyn sang dng cc thnh phn ca vector trn hai trc to:

dt

disd =

+

rs T

1

T

1isd + sisq + rd

mrLT

1

+ rq

mL

1

+ sd

s

u

L

1

(2.31a)


37/106

(HTKS-KCM) TB


dt

disq=

+

rs T

1

T

1isqsisd+ rq

mrLT

1

rdmL

1

+ sqs

uL

1

(2.31b)

rqslrdsd

r

mrd

Tr

1i

T

L

dt

d+=

(2.31c)

rdslrq

r

sq

r

mrq

T

1i

T

L

dt

d=

(2.31d)

Trong hta dq, rq=0 do vung gc vi vector frr

nn frr

=rd .

dt

disd =

+

rs T

1

T

1isd + sisq + rd

mrLT

1

+ sds

uL

1

(2.32a)

dt

disq=

+

rs T

1

T

1isqsisd rd

mL

1

+ sq

s

u

L

1

(2.32b)

rd

r

sd

r

mrd

T

1i

T

L

dt

d=

(2.32c)

dt

d rq= 0 (2.32d)

v rdslsqr

m iT

L=

V d3.9: Chng minh: sdr

mrdr i

sT1

L

+==

V d3.10: Chng minh:r

sq

r

msl

i

T

L

=

Phng trnh moment:

Thay frr

mf

s

r

2

ms

f

sL

Li

L

LL +

=

r

(2.33)

Vo: ( ) ( )sdrqsqrdr

mf

s

f

r

r

mf

s

f

se iiL

Lp

2

3i

L

Lp

2

3ip

2

3T =

==

rrrr

(2.34)

c ( )sdrqsqrdr

me ii

L

Lp

2

3T = (2.35)

vi tc trt: sl= r =r

m

T

L

rd

sqi

(2.36)

V d3.11: Chng minh: ( )sdrqsqrdrm

e iiL

L

P2

3

T =

V d3.12:


38/106

(HTKS-KCM) TB


Te = TL +dt

d

p

J hay

dt

d

p

J = Te- TL (2.37)

Trong h ta t thng rotor (h ta dq), cc vector dng stator fsir

v

vector tthng rotor frr

, cng vi hta dq quanh (gn) ng bvi nhau vi

tc s quanh im gc, do cc phn t ca vector fsir

(isdv isq) l cc i

lng mt chiu. Trong chxc lp, cc gi trny gn nhkhng i; trong qutrnh qu , cc gi trny c thbin theo theo mt thut ton iu khin cnh trc.Hn na, trong hta dq, rq=0 do vung gc vi vector fr

rnn fr

r=rd.

i vi CKB 3 pha, trong hta dq, tthng v mmen quay cbiu din theo cc phn tca vector dng stator:

(Hai phng trnh trn c trnh by ta theo phng trnh (2.34c) v phngtrnh (2.34d) trong chng II).

Phng trnh trn cho thy c thiu khin tthng rotor rrd = r

thng

qua iu khin dng stator isd. c bit mi quan hgia hai i lng ny l mi

quan htrbc nht vi thi hng Tr.Nu thnh cng trong vic p t nhanh v chnh xc dng isdiu khin nnh tthng rd ti mi im lm vic ca ng c. V thnh cng trong vic pt nhanh v chnh xc dng isq, v theo pt (1.20b) th c th coi isq l i lngiu khin ca momen Teca ng c.

Bng vic m tCKB ba pha trn h ta t thng rotor, khng cnquan tm n tng dng in pha ring l na, m l ton b vector khng giandng stator ca ng c. Khi vector si

r

scung cp hai thnh phn: isdiu

khin tthng rotor rr

, isqiu khin momen quay Te, t c thiu khin

tc ca ng c.

sdr

m

rdr

isT1

L

+==

dt

d

P

JTip

L

L

2

3T Lsqrd

r

me +==


39/106

(HTKS-KCM) TB


()

()

Khi , phng php m tCKB ba pha tng quan ging nhi ving cmt chiu. Cho php xy dng hthng iu chnh truyn ng CKB ba

pha tng t nh trng hp s dng ng cin mt chiu. iu khin tc CKB ba pha thng qua iu khin hai phn tca dng in si

r

l isdv isq.

u im khi ca m hnh tn ca CKB trong HTdq so vi HT :

1. Cc i lng khng bin thin dng sin theo thi gian.2. Hphng trnh n gin hn (rq=0).

3.

Phn ly iu khin tthng rotor rr

v momen Te(tc ).4. Gn ging vi iu khin ng cmt chiu.

Te =2

3P( s

rx si

r

)= -2

3P( r

rx ri

r

)=2

3P( m

rx si

r

) ??? (2.8)

Bi tp 1.1. Tpt: ( ) ( )

dt

tdtiR0)t(u

r

rr

rr

r

r

+==

rrrr

CM:

Bi tp 1.2. Vskhi ng ctrong HT

isd rr

isq

Te


40/106

(HTKS-KCM) TB


Bi tp 1.3. Vskhi ng ctrong HTdq

Bi tp 1.4. CM: ( ) ( ) ( )sqrr

msr

r

msse i

L

LPi

L

LPiPT

2

3

2

3

2

3===

rrrr

Bi tp 1.5. CM: ( )srme ixiPLTrr

2

3=


41/106

(HTKS-KCM) TB

Chng 4: iu khin nh hng t thng CKB IV.1

Chng 4:

IU KHIN NH HNG TTHNG CKB

I. Hiu chnh PID (PID CONTROL)

Phng trnh vi phn m thi u chnh PID:

u(t) = KPe(t) + K I dt)t(e + K D dt)t(de

KP: hs khu t l .KI: hs khu tch phn.KD:hs khu vi phn.

Bin i Laplace:

++== s.T

s.T

11K

)s(e

)s(u)s(G D

I

p trong :

P

DD

I

PI K

KT,

K

KT ==

Vn thi t kl c n hiu chnh cc gi trK p , K i v K D sao cho h th a t

c cht lng ti u.

Thtc hiu chnh PID

Khu hiu chnh khuch i t l (P) c a vo h th ng nhm lm gimsai sxc l p, vi u vo thay i theo hm nc sgy ra v t lv trong m t strng hp l khng chp nhn c i vi mch ng lc.

Khu tch phn tl (PI) c m t trong hth ng dn n sai lch tnh trit tiu

(hv sai). Mu n tng chnh xc c a h th ng ta phi tng h s khuy ch i,xong vi mi hth ng thc u bh n chv s c m t ca khu PI l bt buc.

Sc m t ca khu vi phn t l (PD) lm gi m v t l, p ng ra bt nhpnh v hth ng s p ng nhanh hn.

Khu hiu chnh vi tch phn t l (PID) k t hp nhng u im ca khu PDv khu PI, c khn ng tng d tr pha t n sc t, khch m pha. Sc m tca khu PID c thd n n sdao ng ca hdo p ng qu b v t lb ihm dirac (t). Cc b hi u chnh PID c ng dng nhiu trong lnh vc cngnghip di dng thit b iu khin hay thut ton phn mm.

e(t) u(t)PID

i tngiu khin

c(t)r(t)


42/106

(HTKS-KCM) TB


Tm tt Vai tr ca mi khu hiu chnh (adjustment) trong b iu khin PID:

Khu khuch i tlKp (Proportional gain):Khi Kp tng

Sai sxc l p gim

Vt l t ngThi gian ln nhanh

Khu tch phn tlKi (Integral gain):Khi Ki tng

Sai lch tnh gim (trit tiu - v sai vi hm nc)p ng chm

Khu vi phn tlKd (Derivative gain):Khi Kd tng

Vt l gi mp ng nhanhBt nhp nh (dao ng)

PI ri rc:

u(k)=u p (k)+uI(k)

u p (k)=K p .e(k)

uI(k)= uI(k-1)+KI.T.e(k) = uI(k-1)+ IK' . e(k)

Trong :T l tn sly muvoid PID_Control(void) /*Khau PI*/{

ss_n = n_dat - RPM; /* tinh sai so toc do hoi tiep ve*/Up = Kp * ss_n; /* hieu chinh khau ti le (P)*/

Ui = Ui + Ki * Tsamp * ss_n; /* hieu chinh khau tich phan (I)*/U_pt = Up + Ui;

if( U_pt < 0) /*Gioi han dien ap >= 0*/U_pt= 0;

if( U_pt> 1.0) /*Gioi han dien ap


43/106

(HTKS-KCM) TB


p ng ca hthng sdng biu khin PID

p ng bc hm bc 1(t)

PID s(Phng php 1):

( ) ( ) ( ) ( )

dt

tdeKdtteKteKtu DIP ++=

Ri rc ha:

( ) ( ) ( ) ( )kukukuku DIP ++=

( ) ( )ke.Kku PP =

( ) ( ) ( ) ( ) ( )ke.K1kuke.T.K1kuku 'IIIII +=+=

( ) ( ) ( )

( ) ( )( )1kekeKT

1keke.Kku 'DDD =

=

Trong :T l tn sly mu.hi

Ri rc ha _ Phng php gn ng (khu I):

( ) ( ) ( ) ( )( )kukukuku DIP ++=

( ) ( )keKku PP .=

( ) ( ) ( )( )1. += kekeKku II

( ) ( ) ( )( )1= kekeKku DD


44/106

(HTKS-KCM) TB


PID s(Phng php 2):

o hm 2 v:( ) ( )

( ) ( )

2

2

dt

tedKteK

dt

tdeK

dt

tduDIP ++=

( ) ( ) ( ) ( ) ( )kukukuT

kukuDIP ++= 1

( ) ( ) ( )

( ) ( )( )1'1

. =

= kekeKT

kekeKku pPP

( ) ( )keKku II .=

( ) ( ) ( ) ( ) ( )

( ) ( ) ( )( )21*2'211

. +=

= kekekeKT

keke

T

kekeKku DDD

Hay: ( ) ( ) ( ) ( ) ( )( )kukukukuku DIP '''1 +++= ( ) ( ) ( )( )1''' ' = kekeKku pP

( ) ( )keKku II .'' =

( ) ( ) ( ) ( )( )21*2''' += kekekeKku DD

Hay:

( ) ( ) ( ) ( ) ( )( )kukukukuku DIP +++= 1

( ) ( ) ( )( )1. = kekeKku PP

( ) ( )keKku II .=

( ) ( ) ( )( )1' = kukuKku PPDD

iu khin tc ng cDC:

Ikt

U

R

I

E = kE.kt. k.I kt.Rkt

Ukt kt

L


45/106

(HTKS-KCM) TB


Nu n > ndatth e < 0. PID s iu khin GIM u n gi m bt. Nu n < ndatth e > 0. PID s iu khin TNG u n t ng thm. Nu n ndat th e 0. PID s GI NGUYN u n N NH.

( ) ( ) ( ) ( ) ( )( ) ( )11 =+++= kukukukukuku DIP

void PID(void){// --------------------------------------------------------------------------------------------------------------------------------------

ek =(int)(((long)(N_ref - N_sensor)*255)/3000);// Tc cao nht ca ng cl 3000, quy i l *255/3000

// ek < 255 (s8 bit cha gi trti a l 255).

Proportional = (ek - ek_1)*Kp;Integral = ek*Ki;Differential = (ek-2*ek_1+ek_2)*Kd;

PID_D = (Proportional + Integral + Differential);uk = uk_1 + PID_D; // uk phi l sinterger (s16 bit)

// --------------------------------------------------------------------------------------------------------------------------------------

uk_1 = uk;ek_2 = ek_1;ek_1 = ek;

// --------------------------------------------------------------------------------------------------------------------------------------

if(uk < 0)uk = 0; // Khng c gi tr(in p) m

if(uk > 255)uk = 255; // Gi trln nht l 255 (s8bit)

// --------------------------------------------------------------------------------------------------------------------------------------

duty_cycle = uk /Umax * T_PWM;}

Mt sp ng ca hthng iu khin sdng biu khin PID

ndatng c

+ u_

n

n

PIDtc e


46/106

(HTKS-KCM) TB


p ng bc Vt l Dao ng


47/106

(HTKS-KCM) TB


II. Phng php iu khin V/f.

Ti moment hng s T i moment thay i theo tc (Thang my, cn cu, bng chuyn) (Bm, qut,)

III. iu khin tip dng - iu khin moment, tthng

f

V, T

Te

Rs*Ili

TL

fm

V/f=constVm

Tm

V/f=const

f

V, T

Te

Rs*Ili

TL

fm

Tm

Vm


48/106

(HTKS-KCM) TB


IV. iu khin tip p - iu khin dng in trong HQC quay

iu khin dng in ng c(hi tip) bng dng in lnh (t) bng cch iuchnh in p u ra (tip p).ed, eq=???

V. Phng php iu khin nh hng trng (FOC)

V.1. M hnh ng cKB 3 pha

M


49/106

(HTKS-KCM) TB


=???

r, Te, r=???

V.2. iu khin trc tipiu khin trc tip tgi tr h i tip o v:iu khin tip dng:


50/106

(HTKS-KCM) TB


r, Te, r=???

iu khin trc tipt gi tr h i tip - tip p:

ed, eq=???

M


51/106

(HTKS-KCM) TB


ed, eq=???

iu khin trc tipt gi tr t - tip p:


52/106

(HTKS-KCM) TB


V.3. iu khin gin tipiu khin gin tipt gi tr t - tip dng:

ids, iqs, sl=???


53/106

(HTKS-KCM) TB


K1, K2=??? , mech=???

r=???


54/106

(HTKS-KCM) TB


V.4. iu khin trc tip - tip p

Cu trc ca h th ng iu khin nh hng trng nh hng trng

(Field Oriented Control -FOC) trong iu khin ng c khng ng b ba phac trnh by trong hnh vsau:

Hnh 4.1: Cu trc ca hth ng iu khin CKB ba pha dng FOC

Bng vic m t CKB ba pha trn h t a t thng rotor, vector sir

s chia thnh hai thnh phn: isd iu khin t thng rotor r

r, isq iu khin

momen quay Te, t c th iu khin tc c a ng c.

(4.1a)(4.1b)

M

TL

b

a

i

i

MTu

BB

c

b

a

u

u

u

ng c

*sdi + Cid

Ciq

MTi

+

*

r

CTi

C

+

*

+

+

*r

*r

s r

*sqi

sdi

sqi

sdi

sqi

dy

qy

sdu

squ

isd rr

isqT e


55/106

(HTKS-KCM) TB


V.4.1. Xy dng thut ton iu khinGii thut ca tng khi trong hth ng iu khin nh hng trng (hnh

4.1) c trnh by nhsau:

Mng tnh dng (MTi)

( )m

*r

r*sd L

sT1i += (4.2a)

*r

m

*rr*

sq L

Ti = (4.2b)

Mng tnh p (MTu)

qs

sdssd ysT1

LyRu

+= (4.3a)

*dr

r

md

s

sqssq L

LysT1

LyRu +++=

(4.3b)

Trong ,s

ms

s

ss R

LL

R

LT ==

Tnh gc r

sr

r

= (4.4)

Chuyn i hta dng in (CTi)

is= i sa (4.5a)is = ( )sbsa i2i

3

1+ (4.5b)

isd = i scosr+ i ssinr (4.6a)isq = - issinr+ i scosr (4.6b)

Bbin i (BB)

o Chuyn i hta dng in (CTi)

us= u sdcosr u sqsinr (4.7a)

us = usdsinr+ u sqcosr (4.7b)o Bbin i in p (biu chvector khng gian)

usa= u s (4.8a)

+= sssb u23

u2

1u (4.8b)

usc= u sa u sb (4.8c)

Khu iu chtc quay (C)L khu hiu chnh PI:

( )

+=

*IP

*

r s

K

K (4.9)


56/106

(HTKS-KCM) TB


Cc khu iu chdng (DCidv DCiq)o Khu iu chdng isd(DCid)

sdId

Pdd is

KKy

+= (4.10)

o

Khu iu chdng isq(DCiq)sq

IqPqq is

KKy

+= (4.11)

Ch : Xt trong hta tthng rotor nn 0rq= , rdr = (4.12)

Cc thng sK Pv K Itrong cc b iu khin PI c hiu chnh sao cho hthng t ti p ng tt nht.

V.4.2. nh gi p ng ca thut ton iu khin FOCHth ng n nh.

Sai sxc l p ca tc nh , sai sxc l p ca tthng rotor l n.

Thi gian p ng ca hth ng tng i nhanh.

Momen ti khng tc ng nhiu n p ng ca tc , v p ng ca t

thng rotor.

Cht lng p ng suy gim khi bnhi u tc ng ln tn hiu hi tip.

Hth ng dm t n nh khi c sai sm hnh hay b tc ng ca nhiu.

Dng in khi ng ln so vi dng in lm vic; dng khi ng tng lnkhi c sai sm hnh.


57/106

(HTKS-KCM) TB


VI. Tnh ton thit k h thng iu khin gin tip CKB theo phngphp nh hng tthng rotor

V d1: M t ng c3 pha, 4 c c, ni Y, 380V, 50Hz, 2,1A, 5,07Nm, J = 0,1kgm2.Rs= 10, Rr = 6,3, Xs= 13,5 , Xr= 12,6 , Xm= 132 .Khi vn hnh nh mc: Tnh (bin ) isd, isq, r, sl v tc c, n?

Tnh K1, K2, KP, KI? Bit chu kx l c a b iu khin l 20s.


58/106

(HTKS-KCM) TB


(Ch , khi vn hnh cng sut nh mc: isdn< isqn)

f2

XL mm

=

f2

XL ss

= sms LLL +=

r

ss R

LT =

f2

XL rr

= rmr LLL += r

rr R

LT =

sT1iL

r

sdmrd += T thng khng i, dsmr iL=

sdm

e

m

r

r

e

m

rsq iL

T

L

L

P3

2T

L

L

P3

2i ==

e2

m

rsdsq TL

L

P3

2ii =

M s2ds

2sqs I2iii =+=

Khi bit momen in Tev dng in Is,T2 ph ng trnh trn tnh c isdv i sqv r.

Ch : khi ng cv n hnh nh mc, th isdth ng nhh n isq.

Thng thng isd b ng 20-60% Is nh mc.

V tnh crr

qsm

slT

iL

= .

T tnh c tc gc tr t c:p

slco_slsl

== v tnh c tc ng c.

m

rrdrdsd L

sTi +=

r

e

r

msq

T

L

L

p2

3i

=

sd

sq

rr

rr

qsr

m

sl i

i

T

1L

T

iL

L

==

RssI& jXs

sU&

'rI&

'rR jXr

Mch tng ng ng cK B vi tn hao st t

'

rRs

s1

RFe jXm

mI&FeI&


59/106

(HTKS-KCM) TB

Chng 4: iu khin nh hng tthng CKB IV.19

Tthng khng i: *dsm*

r iL=

*

e1

*

qs TKi = *ds

2

m

r

*

e

*

qs

1i

1

L

L

p3

2

T

iK ==

*qs2*sl iK= *dsr

*

rr

m*

qs

*

sl2iT

1

T

L

iK =

=

= v *sl*rr*qsm TiL =


60/106

(HTKS-KCM) TB



61/106

(HTKS-KCM) TB



62/106

(HTKS-KCM) TB


Ch :


63/106

(HTKS-KCM) TB


Vi

Biu khin

Bxl

PWM

EP ADC

SCI

I/O

ADC

L3L2L1

NL


64/106

(HTKS-KCM) TB


I

PI T

KK =

Vi Tdom= J/P v = T delay= chu k x l ( t ng thi gian tr).Thm khu smooth:


65/106

(HTKS-KCM) TB


V d1: M t ng c3 pha, 4 c c, ni Y, 380V, 50Hz, 2,1A, 5,07Nm, J = 0,1kgm2.Rs= 10, Rr = 6,3, Xs= 13,5 , Xr= 12,6 , Xm= 132 .Khi vn hnh nh mc: Tnh (bin ) isd, isq, r, sl v tc c, n?

Tnh K1, K2, KP, KI? Bit chu kx l c a b iu khin l 20s.

iu khin trc tip tgi tr h i tip:

iu khin gin tip tgi tr t:


66/106

(HTKS-KCM) TB


Tnh tt c cc gi tr c n thit iu khin gin tip t thng rotor, c minhha trong v d 1 ny. ng cK B 3 pha 4 cc, ni Y, rotor lng sc, cc thngs 50Hz l:Rs= 10, Rr = 6,3, Xs= 13,5 , Xr= 12,6 , Xm= 132 . Dng in nh mc l 2.1A 380V. iu khin gin tip nh hng t thng

rotor (FOC) ng cK B trn.

B iu khin dng cn tnh ton tgi tr dng in o vv gi tr dng in t.Tc ng c c iu khin t0 n tc nh mc. T thng khng i vbng gi trt thng nh mc. Moment nh mc l 5.07Nm v moment qun tnhl 0.1 kgm2. V vy cn tnh dng in nh mc isdn, isqnt moment nh mc Ten,tnh tthng nh mc r nv tnh v n tc gc nh mc sln?

(Ch , khi vn hnh cng sut nh mc: isdn< isqn)


67/106

(HTKS-KCM) TB


V d2: Mt ng ckhng ng bc cc thng s ( quy vstator) nh sau:Rs= 10, Rr = 6,3, Ls= L r= 0,04H, L m= 0,4H.ng c3 pha, 4 c c, cun dy stator ni Y, 50Hz, 380V, 1400rpm. Tnh dngin nh mc isdn, isqn, t thng nh mc r n v dng in nh mc Isn? TnhK1, K2, Kp, Ti? Bit chu k x l c a b iu khin l 20s, moment qun tnhca ng cl 0.5 kgm 2.

RssI& jXs

sE&sU&

Stator vi dng tho

mI&

jXm

s

R 'r

jXr

'rE&

'rI&

Rotor quy vstator

RssI& jXs

sU&

Mch tng ng ng cK B vi dng tho

mI&

jXm

jXr'rI&

'rR

'rRs

s1


68/106

(HTKS-KCM) TB

Chng 4: iu khin nh hng tthng CKB IV.28

*

e1

*

qs TKi = *ds

2

m

r*

e

*

qs1

i1

LL

p32

TiK ==

*

qs2

*

sl iK= *dsr

*

rr

m

*

qs

*

sl2

iT

1

T

L

iK =

=

=

v *sl*

rr

*

qsm TiL =

Vi Tdom= J/P v = Tdelay= chu kxl (tng thi gian tr).


69/106

(HTKS-KCM) TB


Dong stator: isn= 3.549093Momen dien tu: Te = 6.401969

isn= 3.549093 ATe= 6.401969 Nm

isn= 3.549093 A

isdn= 1.984

isqn=2,941

K1= 0.462

K2= 2.888Kp= 6.25*103

Ti = 0.00008


70/106

(HTKS-KCM) TB


Bi tp 4.1. ng c c cc thng s (quy v stator nh hnh v sau) nh sau:Rs=0,105, Rr=0,081, Ls=0,58mH, Lr=0,87mH, Lm= 13,7mH (bqua t n haost t).

a) Tnh dng in nh mc Is?b) Tnh moment in t nh mc Te?

c) Tnh tthng nh mc r?

d) Tnh cc hs K 1, K2?e) Tnh cc h s hi u chnh ca b iu khin PI (nhhnh v ) l Kpv

Ki? Bit chu k x l l 50 s.

Bi tp 4.2. Cho ng ckhng ng bba pha, 4 c c, 380V, 50Hz, ni Y, 2,1A,moment qun tnh 0.2 kgm2. Moment nh mc l 4,**Nm (v i ** l 2 s cu ica m s sinh vin). ng c c cc thng s : Rs = 10 , Rr = 6,3 , Ls =0,043H, Lr= 0,04H, L m= 0,42H. Trong h t a t thng rotor, khi ng cvn hnh nh mc, tnh:

a) Dng in isd v isq ? (Bit nh mc, isd < isq).b) l n tthng rotor | r| ?c) Tc tr t slv t c ng cn ?d) Hs K 1, K2nh hnh v ?e) Tnh cc hs hi u chnh ca b iu khin PI (nhhnh v ) l Kpv K i?

Bit chu kx l l 50 s.

Bi tp 4.3. Mt ng ckhng ng b c cc thng s (t t cquy v stator) nh sau: Rs= 10 , Rr= 6,3 , Ls= 0,043H, L r= 0,04H, L m= 0,42H.

ng c 3 pha, 4 c c, cun dy stator ni Y, 50Hz, 380V, 1400rpm. Tnh dngin nh mc Isn, dng in nh mc isdn, isqn, t thng nh mc r n v momentnh mc Ten? Tnh K1, K2, Kp, Ti? Bit chu k x l c a b iu khin l 20s,moment qun tnh l 0.5 kgm2.

Rs

is L

usLm

Lr

Rr

rRs

s1


71/106

(HTKS-KCM) TB


Bi tp 4.4. Cho ng ckhng ng bba pha, 4 c c, 380V, 50Hz, ni Y, 2,1A.Moment nh mc l 4,** Nm (v i ** l 2 s cu i ca m ssinh vin). Tronght a t thng rotor, khi ng cv n hnh nh mc, tnh:

a) Dng in isd v isq ? (Bit nh mc, isd < isq).b) l n tthng rotor | r| ?c) Tc tr t slv t c ng cn ?d) Hs K 1, K2nh hnh v ?

Cho bit ng cc cc thng s :Rs= 10, Rr= 6,3 ,Ls= 0,043H, L r= 0,04H,Lm= 0,42H.


72/106

(HTKS-KCM) TB



73/106

(HTKS-KCM) TB

Chng 5: Mt sphng php c lng tthng rotor CKB V.1

Chng 5: MT S PHNG PHP C LNGTTHNG ROTOR CKB

Gii thiu cc phng php o dng in tc thi, o tthng khe hkhng kh,

o in p, tc ng c,...

I. c lng tthng rotor tdng hi tip v tthng khe hkhng kh

[ ]( )mbar ,i,i =

mL

s

Rr

rLsLsR

sv

si ri

mi

sR sR rR

rRmL

rj

mi

si

ri

sv

s

r

s

s

s

m iii +=

smrrr iLiL +=

( ) ssmsssmrssmsrrsr iLiiLiLiL +=+=

( ) ( ) ssrsmmm

rs

smr

s

mr

s

r iLiLL

LiLLiL ==

rr

s

sr

s

m

m

rs

r jiLL

L+==

2

r

2

rrr +==


74/106

(HTKS-KCM) TB


r

rrr coscos

==

r

r

rr sinsin

==

hay

r

r

rr tgtg

==

==

r

rrr arctg

dt

d

( ) srsrr

me ii

L

Lp

2

3T =

II. c lng tthng rotor tin p v dng hi tip

[ ] [ ]( )babar i,i,u,u=

Tthng rotor c c lng ttthng stator v dng stator:

rmsss iLiLrr

r

+=

rrsmr iLiLrr

r

+=

( )sssssm

s

r iLL

1i

r

r

r

=

( ) ssm

rs

rs

2

ms

s

m

rs

ss

s

s

m

rs

smr iL

LL

LL

L1

L

LiL

L

LiL

r

r

r

r

rr

=+=

s

s

m

rss

s

m

rs

r iL

LL

L

L =

Trong , tthng stator c c lng tdng stator v p stator nhsau:

dt

diRu

s

ss

ss

s

s

+=

s

ss

s

s

s

s iRudt

d=

Hay

dt

di

L

LL

dt

d

L

L

dt

d ss

m

rs

s

s

m

r

s

r

=

( )dt

di

L

LLiRu

L

L

dt

d ss

m

rss

ss

s

s

m

r

s

r =

=

dt

diLiRu

L

L

dt

d ss

ssss

ss

m

r

s

r


75/106

(HTKS-KCM) TB


vi ( ) ( )2'r2'

r

'

r +=r

r

r

rr coscos

==

r

r

rr sinsin

==

hay

r

r

rr tgtg

==

==

r

rrr arctg

dt

d

( ) srsrr

me ii

L

Lp

2

3T =

III. c lng tthng rotor ttc v dng hi tip

[ ]( )bar i,i,=

s

r

r

s

s

r

msr jT1i

TL

dtd

r

r

r

=

rr

r

s

r

mr

T

1i

T

L

dt

d=

+=

rr

r

s

r

mr

T

1i

T

L

dt

d

vi ( ) ( )2'r2'

r

'

r +=r

r

r

rr coscos

==

r

r

rr sinsin

==

hay

r

r

rr tgtg

==

==

r

rrr arctg

dt

d

( ) srsrr

me ii

L

Lp

2

3T =

S o khoi bo c lng t thong rotor tren he toa o .


76/106

(HTKS-KCM) TB


0 0 .5 1 1.5 2 2.5 3

0

0.2

0.4

0.6

0.8

1

1.2

t ime (s)

Fir

est

(Wb)

tu thong dat

tu thong dap ung

tu thong uoc luong

0 0.5 1 1.5 2 2.5 3

0

0.2

0.4

0.6

0.8

1

1.2

1.4

time (s)

Fir

est

(Wb) tu thong dat

tu thong dap ung

tu thong uoc luong

ap ng cua bo c lng t thong rotor tren toa o .

IV. c lng vtr tthng rotor gin tip ttthng t v Tet

Cc phng trnh c lng v tr vector t thng rotor tcc gi tr lnh ca tthng rotor v moment in tnhsau:


77/106

(HTKS-KCM) TB


V. c lng tthng rotor ttc v dng hi tip trong HT(dq)[ ]( )bar i,i,=

r

r

sd

r

mr

T

1i

T

L

dt

d=

vi tc trt: r = + sl = +r

m

T

L

rd

sqi

c Te =2

3pr

m

L

Lsqrdi

Cc phng trnh sau c dng c lng tthng rotor:

Hay cc phng trnh ny c thc vit li nhsau:

Vtr tc thi ca vector tthng rotor c xc nh nhsau:


78/106

(HTKS-KCM) TB


S o khoi bo c lng t thong rotor tren he toa o dq.

0 0.5 1 1.5 2 2.5 3

0

0.2

0.4

0.6

0.8

1

1.2

time (s)

Fir

est

(Wb)

tu thong dat

tu thong dap ung

tu thong uoc luong

0 0.5 1 1.5 2 2.5 3

0

0.2

0.4

0.6

0.8

1

1.2

1.4

time (s)

Fir

est

(Wb)

tu thong dat

tu thong dap ung

tu thong uoc luong

ap ng cua bo c lng t thong rotor tren toa o dq.

VI. c lng tthng rotor dng khu quan st (observer)

Hnh 5.1: Sbc lng tthng rotor dng khu quan st.

Thut ton c lng tthng rotor cho CKB ba pha dng khu quan st

pt (5.1a)pt (5.1b)pt (5.1c)pt (5.1d)pt (5.1e)

s

sr

ssi

r

r

pt (5.1f)

s

sr

s

sir

s

sur

s

KK ip+

ucomp

ucomp

pt (5.1h)s

rr

s

sir

s

sr

pt (5.1g)s

rr

ssi

r

r

s

sur

s

rr


79/106

(HTKS-KCM) TB


isd= iscosr+ issinr (5.1a)

rd

r

sd

r

mrd

T

1i

T

L

dt

d=

(5.1b)

r= rdcoss (5.1c)r= rdsins (5.1d)

s

r

r

ms

s

r

2

mrss

sL

Li

L

LLL += rr

r

(5.1e)

dt

d ssr

= ssur

Rs.s

sir

+ ucomp (5.1f)

( )ssssipcomp s

KKu

+= rr

(5.1g)

s

s

m

2

mrss

s

r

ms

r iL

LLL

L

L rrr = (5.1h)

( ) ( )2

r

2

rr +=r

(5.1i)

VII. p ng iu khin dng cbng FOCp ng ca bc lng tthng rotor khi cc thng sCKB ba pha c sai s:

0 0.5 1 1.5 2 2.5 3

0

0.2

0.4

0.6

0.8

1

1.2

1.4

time (s)

Fir

est

(Wb) tu thong dat

tu thong dap ung

tu thong uoc luong

Hnh 5.2: p ng cabc lng tthngrotor ttc v dnghi tip trn ta .

0 0.5 1 1.5 2 2.5 3

0

0.2

0.4

0.6

0.8

1

1.2

1.4

time (s)

Fir

est

(Wb)

tu thong dat

tu thong dap ung

tu thong uoc luong

Hnh 5.3: p ng cabc lng tthngrotor ttc v dnghi tip trn ta dq.

0 0.5 1 1.5 2 2.5 3

0

0.2

0.4

0.6

0.8

1

1.2

time (s)

Fir

est

(Wb)

tu thong dat

tu thong dap ung

tu thong uoc luong

Hnh 5.4: p ng ca

bc lng t thngrotor dng khu quan

st.


80/106

Bi ging HThng iu Khin S (CKB) TB


KT QUM PHNG NG CC TIP DNG

Tthng c a n gi trnh mc khi moment vn c gigi trzero.

Sau khi tthng t gi trn nh, ng cc lnh tng tc n mt gitrvn tc dng.

Moment c a n gi trdng mc ti a. Moment c a trvgi trm v sau zero khi vn tc thc bng vn

tc lnh, v moment c gizero vn tc thc bng vn tc lnh.

H truyn ng ban u ang hot ng vi t thng rotor khng i v gi trlnh, moment ti bng zero.

Moment ti sau c tng n gi trnh mc dng theo kiu step-wise.

Sau mt khong thi gian th moment ti c a vzero cng theo kiu

step-wise.


81/106



Tthng c gikhng i gi trnh mc. Vn tc c o ngc t-40% ca vn tc sang 40% vn tc nh mc. Moment ti bng zero trong sut qu trnh m phng trn. Bnghch lu c gisl ngun dng l tng.


82/106



KT QUO C CA NG CC TIP DNG

Dng stator trong trng thi n nh c phn tch. Trsng harmonic bc

nht, cc sng hi bc cao thng tp trung quanh cc di tn s10 kHz, 20

kHz, 30 kHz, 40 kHz

Tthng bng 70% tthng nh mc, ban u ng cang chy khng ti 600rpm, moment ti bng moment nh mc dng c c tng theo kiu step-wise.


83/106



Bin i ca dng stator khi my tng tc t200 rpm n 1500 rpm.

ng c hon ton c tho trc, moment ti bng zero.

Bin i ca dng stator khi my tng tc t200 rpm n 1500 rpm. ng c hon ton c tho trc, moment ti bng zero.


84/106

(HTKS-KCM) TB

Chng 6: Cc phng php iu khin dng VI.1

Chng 6: CC PHNG PHP IU KHIN DNG

I. iu khin dng trong hqui chiu stator

I.1. iu khin vng trdng iniu khin dng, tip dng


85/106

(HTKS-KCM) TB


( ) ( )rrsse ixP2

3ixP

2

3T

r

r

r

r

==

I.1. iu khin so snh dng iniu khin dng, tip dng


86/106

(HTKS-KCM) TB


II. iu khin dng trong hqui chiu tthng rotoriu khin dng (dq), tip p

III. iu khin piu khin in p vng h(V/F, VFF,)

Phn bit:iu khin tip piu khin tip dngiu khin dngiu khin dng trong htotthng rotor (dq) (FOC)iu khin dng trong htostator (abc)Bnghch lu pBnghch lu dng


87/106

(HTKS-KCM) TCB

Chng 7: Mt s ph ng php c lng tc CKB VII.1

Chng 7: MT S PHNG PHP C LNGTC NG CCKB

I. Cc phng php c lng vn tc ng ckhng ng b

I.1. Phng php 1

( )

srsr2rr

m

2

r

2

r

rr

r

r

slr ii1

T

Ldt

d

dt

d

+

==

Trong : ( )[ ] sssssm

rr iLdtiRu

L

L=

( )[ ] sssssm

rr iLdtiRu

L

L=

( )2r2rr +=

Biu khin

Bxl

PWM

EP ADC

SCI

I/O

ADC

L3L2L1

NL


88/106

(HTKS-KCM) TCB


Chng minh (cch 1):

Chng minh:2

r

2

r

rr

r

r

r

dt

d

dt

d

+

=

2

r

r

2

r

rrrr

r

r

srr

1

1dt

d

dt

d

arctgdt

d

dt

d

dt

d

+

=

===

2

r

2

r

rr

r

r

r

dt

d

dt

d

+

=

Chng minh: ( )

srsr2rr

msl ii

1

T

L

=

C slrrsqm TiL = v sqrdr

me i

L

Lp

2

3T =

2

r

e

r

rsl

T

p3

2

T

L

=

m rrsmr iLiL += rr

s

r

mr

L

1i

L

Li +=

nn ( )

+== r

r

s

r

mrrre

L

1i

L

Lxp

2

3xip

2

3T

( )srr

ms

r

mre xi

L

Lp

2

3i

L

Lxp

2

3T =

=

( ) ( ) srsr2

rr

m

2

r

sr

r

m

2

r

e

r

rsl ii

1

T

Lxi

T

LT

p3

2

T

L

=

=

=

Chng minh (cch 2):

C: =

rr

r

s

r

mr

T

1i

T

L

dt

d

+=

rr

r

s

r

mr

T

1iT

L

dt

d

2rrrr

rs

r

mrr

T

1i

T

L

dt

d

=

2

rrr

r

rs

r

mr

rT

1i

T

L

dt

d

+=

( ) ( )2r2rrsrsr

mrr

r

r iiT

L

dt

d

dt

d

++=

( )

rsrsr

mr

r

r

r

2

r iiT

L

dt

d

dt

d

=


89/106

(HTKS-KCM) TCB


( )

2

r

rsrs

r

m

2

r

rr

r

rii

T

L

dt

d

dt

d

=

Chng minh:

dt

diRu

sss

ss

s

s

+=

ssss

s

s

s iRudt

d=

V rmsss iLiLrr

r

+=

rrsmr iLiLrr

r

+=

( )sssssm

s

r iLL

1i

r

r

r

=

( )s

s

m

rs

rs

2

mss

m

rsss

ss

m

rssmr i

LLL

LLL1

LLiL

LLiL

r

r

r

r

rr

=+=

ssm

rss

s

m

rs

r iL

LL

L

L =

dt

di

L

LL

dt

d

L

L

dt

d ss

m

rs

s

s

m

r

s

r

=

=

dt

diLiRu

L

L

dt

d sss

s

ss

s

s

m

r

s

r

=

dtdiLiRu

LL

dtd sssss

m

rr

=

dt

diLiRu

L

L

dt

d sssss

m

rr

Hay ( )[ ]ssssssssm

rs

r iLdtiRuL

L=

I.2. Phng php 2

rrrr

rr

r

r

ii

dt

di

dt

di

+

=

Trong , ( )sssssm

s

r iLL

1i =

( ) ( )

( ) ( )

rsssrsss

rsss

r

sss

iLiL

dt

diL

dt

diL

+

=

Vi ssm

rss

s

m

rs

r iL

LL

L

L =


90/106

(HTKS-KCM) TCB


sm

rss

m

rr i

L

LL

L

L=

sm

rss

m

rr i

L

LL

L

L=

vi ssssss

s iRudt

d =

( )dtiRu ssss = ( )dtiRu ssss =

Chng minh:

s

r

s

rs

rr jdt

diR0 r

r

r

+=

rr

rrdt

diR0 ++=

r

r

rrdt

diR0 +=

rrr

rrrrr idt

diiiRi.0 ++=

rr

r

rrrrr

idt

diiiRi.0 +=


91/106

(HTKS-KCM) TCB


( )

rrrrr

r

r

r iidt

di

dt

di0 +

=

rrrr

rr

r

r

ii

dt

di

dt

di

+

=

II. c lng vn tc vng knDng iu khin thch nghi m hnh (Model Reference Adaptive Control MRAC)

M hnh tham kho:

=

dt

diLiRu

L

L

dt

ds

ss

s

ss

s

s

m

r

s

r

[ ] [ ]( )babar i,i,u,u=

M hnh thch nghi (trong hta stator):

rrr

s

r

mr

T

1i

T

L

dt

d=

(7.1a)

rr

r

s

r

mr

T

1i

T

L

dt

d+=

(7.1b)

Sai s m hnh:= ( ) ( )srsr

rr

= r r rr (7.2)

Hiu chnh sai s:

+=s

KK ip (7.3)

t phthuc vo thng sm hnh v cc i lng hi tip.

Khi

, tc

c

c l

ng theo s

sau:

Hnh 7.1: S nguyn l b c lng tc CKB ba pha.

M hnhthch nghi

h pt (7.1)

srr

s

KK ip+

( ) ( )srsr

rr

pt (7.2)

tthng c lngM hnhtham kho

s

rr

s

sur

s

sir

s

sir


92/106

(HTKS-KCM) TCB


_ Sai lch tc c iu chnh tng._ Kt qut bnh hng bi sai lch thng sng c._ Khi lng tnh ton rt nhiu nn tc c lng chm.

Tch ng 3, c:s

rr

s

sr

m

s

r

jT

1

iT

L

dt

d

=

rrr

s

r

mr

T

1i

T

L

dt

d=

rr

r

s

r

mr

T

1i

T

L

dt

d+=

chs gc trn phi cht thng c tnh trc tip tt c c lng .

Nhn xt:Theo h ph ng trnh (7.1) th t thng rotor (xt trong h t a stator)phthu c vo tc .Mc khc, b c lng t thng cho kt qu t ng i chnh xc vgi trc a vector tthng rotor.Nhv y, nu tc c lng trong ph ng trnh (7.1) khc vi tc thc ca ng c th vector t thng ( r ,

r ) tnh c ph ng trnh

(7.1) s sai l ch vi vector t thng ( r , r ) c lng. Sai lch ny

c nh ngha bng:= ( ) ( )srsr

rr

= r r rr (7.2)

nu sai lch cng nh th t c c lng ca ng cs cng g n bngvi tc th c ca ng c.B c lng tc cho ng cK B ba pha sd ng khu hiu chnh tchphn tl PI gi m thiu sai lch gia hai vector tthng trn:

+=s

KK ip (7.3)


93/106

(HTKS-KCM) TCB


III. iu khin khng dng cm bin(Sensorless Vector Control - SVC)

Scu trc hiu khin nh hng tthng rotor khng dng cm bin vn tc:


94/106

(HTKS-KCM) TCB


p ng m phng:

0 0.5 1 1.5 2 2.5 3

0

20

40

60

80

100

120

time (s)

West

(rad/s)

toc do dat

toc do dap ung

toc do uoc luong

Hnh 7..: p ng ca b c lng

tc v i m hnh l tng.

0 0.5 1 1.5 2 2.5 3

0

20

40

60

80

100

120

time (s)

West

(rad/s)

toc do dat

toc do dap ung

toc do uoc luong

Hnh 7..: p ng ca b c lng

tc v i m hnh c sai lch.


95/106

Chng 7: Mt s ph ng php c lng tc TB


p ng trn hthc:


96/106

(HTKS-KCM) TB

Chng 8: Biu khin ng ckhng ng bba pha VIII.1

Chng 8: IU KHIN SNG C

I. Cu trc mt hthng iu khin ng c

I.1. Skhi mt hthng iu khin ng cKB 3 pha

Biu khin

Bxl

PWM

EP ADC

SCI

I/O

ADC

L3L2L1

NL

TL

ai

MTu

BB

*

sdi + PI

PI

MTi

+

*

r

CT

*

sl

r r

*

sqi

sdi

sqi

sdi

sqi

'

sdu

'

squ

sdu

squ

PI

+*

+

+*

sl

bi

au

bu

cu

r

r

Motor

~3


97/106

(HTKS-KCM) TB



98/106

(HTKS-KCM) TB


I.2. Cc khi chc nngBchnh lu

Bnghch lu 3 pha v mch kch (FPGA, DSP)

Hm

Bxl


99/106

(HTKS-KCM) TB


dsPIC

DSP Texas Instruments

Cm bin o lngMch giao tip, ni mng


100/106

(HTKS-KCM) TB


II. Cm bin o lng

II.1. o in p DCSdng bchuyn i ADC ca biu khin thng qua mch chia p...

II.2. Cm bin o dng in

o in p trn in trShunt

Bin dng

.c KB

==

3~

Vdc

ieukhien

M3~

a b c

Bientan

2= 3

isa

isbis

isrje

isd

isq

r

TL

ai

MTu

BB

*

sdi + PI

PI

MTi

+

*

r

CT

*

sl

r r

*

sqi

sdisqi

sdi

sqi

'

sdu

'

squ

sdu

squ

PI

+*

+

+*

sl

ngc

bi

au

bu

cu

r

r


101/106

(HTKS-KCM) TB


Cm bin Hall

II.3. Cm bin o tc Tachometter


102/106

(HTKS-KCM) TB


Incremental Encoder


103/106

(HTKS-KCM) TB



104/106

(HTKS-KCM) TB


Absolute Encoder (o gc)

III. Mt su im khi sdng biu khin tc ng cBbin tn

7/21/2019 Bai Giang Htdkso All Hk2nh1011 v9

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