Đồ án bộ băm

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n mn hc TCS.

PHN I: TNG QUAN V NG C IN MT CHIUV CC PHNG PHP IU CHNH TC NG C.

ng c in mt chiu l my in mt chiu lm vic ch ng c. Ch my pht khi E > U v ch ng c khi E < U. Vic chuyn t ch my pht sangch ng c xy ra hon ton t ng khng cn thay i g mch ni, c th l khhigim dng in kch thch khin cho E ca my pht h n mc E < U, dng in trong

phn ng s t ng i chiu, nng lng s chuyn theo chiu ngc li v my phtngu nhin tr thnh ng c.

ng c in mt chiu dng rt ph bin trong cng nghip v giao thng vn tiv ni chung nhng thit b cn iu chnh tc quay lin tc trong mt phm vi rng.

Cng nh my pht, ng c in mt chiu c phn loi theo cch kch thch tthnh cc ng c in c kch t c lp, kch t song song, kch t ni tip v kch thn hp.

S ng c in mt chiu:

Hnh 1-1 s ng c in mt chiu.

I.1. c tnh c ca ng c in mt chiu.

Ty theo cch kch t m ng c in mt chiu c nhng tnh nng khc nhauc biu din bng cc ng c tnh lm vic, c tnh c khc nhau. Trong cc ctnh quan trng nht l c tnh c biu th quan h gia tc v m men.

Gi in p t vo ng c l Ud ta c pt:

Ud = E + (R + Rd)I (1)

Trong : E l sc in ng phn ngR in tr mch phn ngI dng in mch phn ng

M ta c E = k

SVTH: Ng Vn Tun Trang3

RKT

CKT

Rf

Ud

IKT

Eu

UKT

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Thay vo (1) ta c:

Iuk

Rf

k

Ud

=

(2)

mt khc ta c: Mdt = kI I = Mdt/k

Mdt: m men in t ca ng c.

(2) ta suy ra tMkRf

k

Ud2)(

=

T biu thc trn ta thy rng vic iu chnh tc ng c c th thc hin bng cchthay i cc i lng U, R, .

-Phng php iu chnh tc bng cch thay i t thng c p dng tngi ph bin, c th thay i tc lin tc v kinh t. Trong qu trnh iu chnh in pth hiu sut teC= v s iu chnh da trn vic tc dng ln mch kch thch c cngsut rt nh so vi cng sut ng c. Bnh thng ng c lm vic ch nh mcvi kch thch ti a nn ch c th iu chnh theo chiu hng gim , tc l iuchnh tc trong vng trn tc nh mc v gii hn iu chnh tc b hn ch bicc iu kin c kh v i chiu ca my.

-Phng php iu chnh tc bngcch thm in tr ph vo mch phn ng tng R ch cho php iu chnh tc

trong vng di tc nh mc v lunkm theo tn hao nng lng trn in tr ph lm gim hiu sut ca ng c.Phng php ny ch p dng cho ng ccng sut nh.

-Phng php iu chnh tc bngcch thay i in p cng ch cho phpiu chnh tc di tc nh mc vkhng th nng cao in p hn in p cang c. Phng php ny khng gy nn

tn hao trong ng c, nhng i hi phic ngun ring c in p iu chnh c.

I.2. nh hng ca in p phn ng.Gi thit t thng dm= = const; in tr R = const, khi thay i n p theo

hng gim so vi Udm ta c tc


0maxmax

min

k

1

ki

Mm MHnh 1-2. c tnh

c nhng in pkhc nhau

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dm

ok

Uu

thay iV cng c tnh c

Ruk

W

M 2)(

=

= = const

Nh vy khi thay i in p t vo phn ng ng c ta c mt h c tnhsong song vi c tnh c t nhin:

I.3. Nguyn l iu chnh in p phn ng.

iu chnh in p phn ng ca ng c in mt chiu cn c thit b ngunmt chiu. M theo s liu ban u ta c ngun l li in xoay chiu 3 pha 220V/380V.V vy ta cn c thit b bin i ngun xoay chiu to thnh in p mt chiu.

Ta c s :

Hnh 1-3:S iu chnh in p.

Ngun mt chiu c sc in ng Eb, tn hiu iu khin Uk c in tr Rb. Ta c pt ctnh c ca h thng ch xc lp:

Eb E = I( Rb + Rd)

Hay Iuk

RudRb

k

Eb

+

=

= o M/

Ta c t thng c gi khng i nn cng c tnh c cng khng i.Tc khng ti l tng th ty thuc vo gi tr in p iu khin Uk ca h thngnn ta thy rng tc ln nht ca ng c c chn bi in p nh mc, t thngnh mc, cn tc nh nht ca ng c b gii hn bi yu cu v sai s v mmenkhi ng.

Khi mmen ti l nh mc ta c cc gi tr ln nht v nh nht ca tc max = omax Mdm/


Lk

MBBU

k

Rb

R

EE

b(u

k)

I

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min = omin - Mdm/ tha mn kh nng qu ti th c tnh thp nht ca ng c phi c mmen

ngn mchMnmmin = Mcmax = KMMdm

Trong :Mcmax : mmen cn ln nhtKM: h s qu ti v mmen

min = KMMdm/ - Mdm/ = (KM -1)Mdm/Phm vi iu chnh tc l:

D =max/min =(omax Mdm/)/(KM -1)Mdm/)Ta c max, KM, Mdm xc nh nn D ch ph thuc tuyn tnh vo cng .

V t thng kch t c gi nguyn trong sut qu trnh iu chnh in p phn

ng nn mmen ti cho php ca h thng s khng i.

Mccp = kdmIdm = MdmI.4. B bm in p mt chiu kiu c o dng iu chnh tc ng c.

I.1.a. S v nguyn l hat ng:B bm o dng l s kt hp gia b bm ni tip v b bm song song. B bm

o dng s cho php truyn nng lng theo hai chiu. Ta c s nguyn l ca b bmo dng c trnh by nh sau:

Hnh 1-4 S nguyn l ca b bm o dng.

I.1.b Cch iu chnh tc :

Khi ti l my in mt chiu, U l ngun in p mt chiu khng i th myin c th lm vic hai trng thi: ng c v my pht.


Ud

Id

E

I2

U

T1

D1

-

+

D2

L

R

T2

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0

1= 1

2= 0

0,750,75 0,5

0,5

0,25

0,25

0

1

n

M, I

n mn hc TCS.

trng thi ng c: T2 kha v T1 m trong khong thi gian l 1T cachu k, lc ny in p ra trn ti s l U d = 1U v sc n ng ca ng c E = Ud RId = 1U- RId vi Id > 0.

trng thi my pht: T1 kha v T2 m trong khong thi gian 2 T ca chuk, lc ny in p ra trn ti s l Ud = (1-2U ) v sc in ng E = Ud - RId = ( 1 2 )

U RId vi Id < 0. Mi quan h gia t s chu k 1 v 2 l: 1 + 2 =1.Nh vy vi b bm o dng, bng cch tc ng vo t s chu k 1 v 2 c th

c c mt h c tnh c in ca ng c in mt chiu ch ng c v hm tisinh.

Hnh 1-5 H c tnh c ca ng c s dng b bm c o dng.

Nh vy, i vi b bm o dng s m bo cho ng c lm vic gc phn tth nht v th t ca mt phng ta U, I.

Gi tr trung bnh ca in p ra trn ti lun lun dng nu thi gian m ca cc

T1 v T2 ln hn thi gian ngt ca chng. Nu ngc li th gi tr trung bnh ca in pra trn ti s c gi tr m.

PHN II.TNG QUAN V B BM IN P MT CHIU.

B bm in p hay cn gi l b bin i DC-DC nhm mc ch to ra n pmt chiu bin i c iu chnh tc ng c. Nh ng hay m sm gc to ra in p trung bnh ln hay nh m ng c thay i tc nhanh hay chm. C mts b bin i sau.

II.1. B bm n p mt chiu kiu ni tip.II.1.a. S mch.


(-)(+)

Id

D0

ID0

LC

DC

Us

EU

+

-

C

+---

Tp

Ta

L

d

Rd

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Hnh 2-1 S b bm ni tip.

Trong : Tp: tiristor chnhTa: tiristor phUs: ngun mt chiuDr: diot hon nng lngLc, Dc, C cc phn t chuyn mch

II.1.b. th v dng sng.

Hnh 2-2 Dng sng ca in p vo v ra ca b bm ni tip.

II.1.c Nguyn l lm vic.

Trng thi ban u Tp v Ta u b kha, t in C c np in, bn cc dng bn trn. Cho xung kch m Tp. Tiristor ny m cho dng in chy qua. Dng in tngun Us chy qua Tp qua ti v tr v ngun. ng thi t C phng in theo mch C-Tp-Lc-Dc-C v c np ngc li. in p trn ti l Ud = Us.


T1

T2

T

Utb

Ud

U

t0

Ud

us

t0

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By gi cho xung kch m Ta, tiristor ny m t in p gia hai bn t in C lnTp khin Tp b kha li, tc l Ud = 0.

Gi T l chu k bm: T = T1 + T2

T1 = T l thi gian Tp dn

T2 = T T1 thi gian Tp kha = T1/T l t s chu k

Tr trung bnh ca in p trn ti l

== T

UsdtT

Ud0

1Us

Nh vy bng cch lm bin i t s chu k ( trong khi gi cho chu k bm T =const) c th iu chnh c tr trung bnh ca in p t trn ti.

Trng hp ti l R+ L.in cm L tch ly mt nng lng in trng W=1/2LI d2 khi dng Id tng.

Ngi ta thng dng diot hon nng lng Dr u song song ngc vi mch ti to ng phng nng lng in trng ni trn khi dng id suy gim.Lc u dng ti id = 0. dng id c xc lp dn dn. Qua mt vi chu k dng ids bin ng gia hai gi tr bin I1 v I2.Khi Tp dn ta c pt:

Ldid/dt + Rid = Us ( id tng t I2 n I1)

Khi Tp kha ta c pt:

Ldid/dt + Rid = 0 ( id gim t I1 n I2) Trng hp ti l R+ L+ E.Khi Tp dn ta c pt:

Ldid/dt + Rid +E = UsKhi Tp kha ta c pt:

Ldid/dt + Rid +E = 0Tr trung bnh ca dng in ti:

REUsEUdId ==

R

II.2. B bm in p mt chiu kiu song song.

II.2.a S mch


U

E

Ud

Id

IT

-

+

D

L

R

T

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Hnh 2-3 S b bm song song.

II.2.b Dng sng.

Hnh 2- 4 th in p v dng in.

II.2.c Nguyn l lm vic.

Ti l mt my in mt chiu ( E+R+L). S b bm song song cho php thchin hm ti sinh. Trong ch hm ti sinh, my in lm vic nh mt my pht intr nng lng v ngun tng nui n khi n lm ch ng c.


U

TT

Ud

t0

Ie

t

0I

T

t0

Id

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L l in cm mch phn ng gi thit ln gi cho id = Id = const trong mtqu trnh chuyn mch.

S hot ng nh sau.Gi thit khi t = 0, ta cho xung iu khin kch m Tp, Ud = 0, is = 0, iH = id, diot D

b kha li. Nng lng t ngun E ( cng sut trn trc ng c) c bin thnh nng

lng t trng tch trong cun cm L, ri khi H h mch, tr nng lng v ngunnui qua iot D.

Khi t = T1 = T, ta cho xung m Ta, lc ny Tp b kha li, phng trnh mch tic dng:

E Rid Ldid/dt = Ud

Tp kha lm cho mch h ra, dng id khng cn bng Id m gim nh i (do L ); L did/dt < 0.

Sc in ng t cm e = -Ldid/dt > 0, cng vi E to ra dng in id = is tr vngun nui. Khi dng id gim n bin tr thp, ngi ta li ng H v chu trnh c lp

li. Tr trung bnh ca in p ti.

Ud = UUdtT

T

=

)1(1

Tr trung bnh ca dng in tr v ngun.

Is = IdddtIT

T

T

)1(1

=

Tr trung bnh ca dng chy qua Tp.

IH = IdddtIT

T

=0

1

Phng trnh mch ti khi my in thc hin hm ti sinh.

E Rid L did/dt = Ud

R

UdEId

=

Nhn xt:Mun hm ti sinh, tr nng lng v ngun, Id >0 th E>Ud c th khng ch

dng in bng cch tc ng vo t s chu k .

II.3 B bm in p mt chiu kiu c o dng.

S v nguyn l lm vic.


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Hnh 3-1 S b bm o dng.

Ghp mt b bm ni tip H1 vi mt b bm song song H2 s c c mt b bmo dng. B bm o dng cho php truyn nng lng theo 2 chiu. Ti l mt my

in mt chiu do c th lm vic ch my pht in mt chiu hay ng c inmt chiu. Ta c th c tnh lm vic ca b bm nh sau.

Hnh 3-2 c tch c ca M s dng b bm o dng.

+ Ch ng c: Id >0, cho b bm ni tip lm vic ( H1 m H2 kha)E = Ud Rid = 1 Us Rid

+ Ch my pht: Id < 0, cho b bm song song lm vic ( H1 kha H2 m) ( cho mt bngh hn).

E = Ud Rid = (1 2)Us RidQuan h gia 1 v 2 l: 1 + 2 = 1.

Nh vy, vi b bm c o dng, bng cch tc ng vo 1 v 2 c th c cmt h c tnh c in ca ng c in mt chiu ch ng c v hm ti sinh.

PHN III. THIT K B CHNH LU.


Ud

Id

E

I2

U

T1

D1

-

+

D2

L

R

T2

0

1= 1

2= 0

0,750,75 0,5

0,5

0,25

0,25

0

1

n

M, I

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B chnh lu l b bin i in p xoay chiu thnh in p mt chiu . C nhius chnh lu, trong c hai loi thng dng l chnh lu hnh cu v chnh lu hnhtia. Cc linh kin in t thng s dng trong mch chnh lu l Trisistor v Diod,Trisistor s dng trong chnh lu c iu khin cho ra dng in p ra thay i c.Do vy theo yu cu ca n ny th ta ch s dng chnh lu dng Diod v yu cu in

p ra mt chiu khng thay i .Chnh lu hnh cu in p ngc t ln van nh hn, cho ra in p t nhp nh

hn nhng m s phn t s dng trong mch ln hn nhiu. Tuy nhin gi thnh ccDiod r nn khng ng k nhng m yu cu sng ra cn sang phng hn do ta chns chnh lu hnh cu.

I/ S v nguyn l hat ng ca chnh lu cu 3 pha.

1/S v dng sng.

Hnh 3-1 S chnh lu cu 3 pha


ib

Id

ic

ia

ud

ud+

uc

+

ub

+

ua

+U

id

Ud

I

d

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Hnh 3-2 Dng sng ra ca chnh lu

2/ Nguyn l lm vic.

Cc van chnh lu 3 pha chia lm 2 nhm :- Nhm Katt chung (l cc van c cc Katt cng ni vo ngun xoay chiu) :

gm cc van D1, D3, D5 hay cn gi l nhm l.- Nhm Ant chung (l cc van c cc Ant cng ni vo ngun xoay chiu) : gm

cc van D2, D4, D6 hay cn gi l nhm chn.Trong nhm Ant van no c pha Katt m hn th dn, ngc li bn nhm Katt

th van no pha Ant dng hn th dn.Nh vy ti mi thi im lun c 2 van thuc hai nhm khc nhau cng dn v

mi van dn trong khong thi gian l = 2/3. Ta xt trong khong 10 van D1 v D6

dn khi in p ti c 1 in p ua qua D1 v ub i qua D6 t vo, ta c

ud = ua ub = uab

Ti thi im 1 in p ub = uc v uc m dn th Diod D6 bt u ngng dn v D2bt u dn y l thi im chuyn mch D6 qua D2 .

Sang thi im 21 th in p trn ti l (D2 dn cn D6 kho hn)

ud = ua uc = uacTng t cho cc qu trnh tip theo.

Nh vy:T /6 3/6 th D1, D6 m cn D2,3,4,5 kha.

T 3/6 5/6 th D1, D2 m cn D1,3,4,5 kha.

T 5/6 7/6 th D2, D3 m cn D1,,4,5,6 kha.

T 7/6 9/6 th D3, D4 m cn D1,2,5,6 kha.


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T 9/6 11/6 th D4, D5 m cn D1,2,3,6 kha.

T 11/6 13/6 th D5, D6 m cn D1,2,3,4, kha.Ta c gi tr in p trung bnh u ra l :

f2f2

6

6

f2d U.34,2U63

dcosU22

6U =

=

=

in p ngc ln nht t ln diod l :

f2

o

f2im U630cosU22U ==

Dng chy qua diod l dng ra ngoi :

IP = id

Gi tr trung bnh ca dng ra l :

Id =R

Ud

R

cosU6

2

6 d6

6

f2 =

Dng qua mi diod trung bnh l :

ID = Id/3

II/ S v nguyn l hot ng tng th ca mch ng lc.

1/ S mch ng lc.

Hnh 3-3 s mch ng lc.


U~ng

Binp

Chnhlu Lc

Bbm

ng c

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Hnh 3-4 s nguyn l mch ng lc.

2/ Nguyn l lm vic.

in p 3 pha t li a vo my bin p bin in p ra hp l, in p nya vo b chnh lu bin in p xoay chiu thnh in p mt chiu. Sau cho qua

b lc nhm n nh dng in 1 chiu ng thi lm suy gim dng in tn s khc.Khi dng in v in p n nh ta a vo b bm in p mt chiu iu khintc ng c.

Phng trnh cn bng in p cho h thng ch xc lp.

Eb E = I( Rb + R)

Trong :Eb l in p ngun thay i c.Rb l in tr b bin iE = k.

phng trnh vn tc gc ca ng c l:

IudmK

RuRb

dmK

Eb

+

=

)(

Hay

MdmK

RuRb

dmK

Eb2)(

)(

+

=

cng c tnh c: =RuRb

dmK

+

2)(


D o n

c o

T a 1

T p 1

C 1

L c 2

C

L

D c 2T a 2 T p 2

C 2

D r 1

D c 1 L c 1

D r 2

D 5D 1 D 3

D 4 D 6 D 2

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Ud

C

L

R

n mn hc TCS.

Trong qu trnh iu chnh in p phn ng ng c th t thng c gikhng i nn cng khng i. Nn tc ng c khng ti l tng ch ph thucvo in p Eb.Ta c tc cc i ca ng c.

Mdmm = 0

c tnh thp nht phi c mmen ngn mch bng mmen ti cc i.

Mmin = Mcmax = KqtMdmvi Kqt l h s qu ti ca my.

Do tc nh nht ng vi mmen ti nh mc:

)1(dmmin = qtKM

Nh vy khi ng c lm vic mmen nh mc th ta c di iu chnh tc ng c bng cch iu chnh gc m ca Tiristor Tp trong b bm in p mt chiu DC-DC.

III/ Thit k b lc.

1/ Gii thiu.

B lc l thit b ni gia ngun chnh lu v ph ti in mt chiu. Chc nngca b lc l cho dng in c tn s no i qua m bin khng b suy gim, ngthi lm suy gim mnhdng in tn s khc. C hai loi b lc l lc bng t in v

b lc LC.2/ Nguyn l hat ng ca b lc LC.

B lc ny cho php thnh phn mt chiu ca in p chnh lu i qua v ngnchn thnh phn xoay chiu.

in p u ra ca b chnh lu c th c trin khai thnh chui Fourier. Nunh ch dng li hai s hng u ta c:

Ud = 22 235

3663UU

+ cos6t.

Biu thc tng qut:

Ud = ud +Uc = Ud + 2 U2cosnt Hnh3-5 s b lc LC.

IV/ Tnh chn cc phn t ca b chnh lu v my bin p.


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in p ra khng ti s l:

Udo = 220.1,005 + 2 = 234 (V)

Tr hiu dng ca in p pha th cp

U2 = 10063

234.14,3

63==do

U(V)

T s bin p.

263,0380

100

1

2 ===U

Um

in p ngc ln nht mi diot phi chu.

Um = 2456 2 =U (V)

Dng chnh lu nh mc: Id = 25,2 (A)Tr trung bnh ca dng in chy trong mi diot l

ID = Id/3 = 25,2/3 = 8,4 (A)

Tr hiu dng ca dng in chy trong mi pha th cp mba

I2 = 57,202,25.3

2

3

2==dI (A)

Tr hiu dng ca dng chy trong mi pha s cp mba

I1 = mI2 = 0,263.20,57 = 5,41 (A)

+ Chn dit:H s d tr in p v dng in l: Ku = 1,6; Ki = 1,2. Ta phi chn Ditt nht chuc in p ngc l: 1,6.245 = 392 (V) v chu c dng trung bnh l: 1,2.8,4 =

10,08 (A). Vy ta chn 6 dit loi B-10; 400 V.Tnh my bin p:

a) Mch t:Cng sut biu kin ca mba

S2 = 3.100.20,57 = 6171 (W); S1 = 3.380.5,41 = 6167,3 (W); S = 6167 (W)


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n mn hc TCS.

Chn mch t ba tr, tit din mi tr c tnh theo cng thc

Q = k fcS. (cm

2)

Trong : k = 4 5 nu l mba du; k = 5 6 nu l mba khS: cng sut biu kin ca mbaC: s trf: tn s ngun in xoay chiu

Q = 6 5,3850.3

6167= (cm2)

Mch t c lm bng tn silic, tn tht 1,3W/kg, trng lng ring 7,5kg/dm 3. Chn t

cm BT = 1,1 tesla.

b) Dy qun.

S vng dy mi pha- s cp

4041,1.10.5,38.50.44,4

380

..44,4 41 === BmQf

Un vng

- th cp

106380

1004042 ==n vng

chn mt dng in: J1 = J2 = 2,75 A/mm2

ng knh dy dn pha s cp

6,175,2.14,3

41,5.41 ==d mm

ng knh dy dn pha th cp

375,2.14,3

57,20.42 ==d mm

Dy qun s cp chia thnh 6 lp, gia hai lp t mt lp giy cch in dy 0,1 mm.Dy qun th cp chia thnh 4 lp, gia hai lp t mt lp giy cch in dy 0,1 mm.


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Gia dy qun s cp v th cp mt khong cch cch in l 8 mm.

V/ Tnh chn cc phn t ca b lc LC.

B lc LC cho php thnh phn mt chiu ca in p chnh lu i qua v ngn

chn thnh phn xoay chiu.in p u ra ca b chnh lu c trin khai theo chui Fourier. Nu ch dng li hais hng u ta c biu thc tng qut ca in p chnh lu:

Ud = ud + uc = ud + 2 U2cosnt.

Trong U2 gi tr hiu dng ca in p pha

N tn s gc ca sng hiUd ga tr trung bnh ca in p chnh lu

i vi chnh lu cu 3 pha hnh cu ta c:

Ud = 22 235

3663UU

+ cos6t

+Tnh chn L-C:Ta c t s nhp nh

KLC =LCn

A

U

Uc22

22 =

Trong = 2 314=f

A = 0,095n = 6

Chn KLC = 0,01 ta c:

LC =6

222210.676,2

314.6.01,0

095,0 ==nK

A

LC

Nu chn L = 2,676 mH th C = 1000 F

VI/ Tnh chn cc phn t ca b bm c o dng.

Tnh chn b bm ni tip.

Chn cc phn t TP v TaTa c cc gi tr nh mc:


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n mn hc TCS.

Udm = 220 VIdm = 25,2 AUmax = Ku.Udm = 1,6.220 = 352 V

V dng in trung bnh chu ng c l:

Itb = KiIdm = 1,2. 25,2 = 30,24 A

Da vo bng s liu ca Tiristor ta chn Tiristor loi C45, C46 vi cc thng s sau:Uim = 25 900 VTtb = 35 AToff = 30 s

Da theo gio trnh TCS ca Nguyn Bnh ta c cng thc tnh t in C nh sau:

C = (1,2 1,5)s

offPd

U

tI .trong toff l thi gian kha ca Tp

C = (1,2 1,5) 666

10.85,410.87,3234

10.30.2,25 =

Xc nh Lc.

Lc = ( ) 32662

10.074,0)10.30.(10.85,4

4,04,0

==offatC

H

Chn Diot chuyn mch Dc c cc thng s sau.

Imax = 20,57 A, Umax = 352 V

Vy ta chn loi Diot B-20 vi cc thng s sau:

Itb = 20 A, Uim = 100 1000 V, U = 0,7 V.Chn Dr ging Dc.Ta chn cc thit b cho b bm song song ging vi b bm ni tip.

Phn IV : THIT K MCH IU KHIN

Mch iu khin to xung iu khin cc tiristor T a v Tp t ta c th thay ic gc m thay c in p, iu chnh tc ng c .

I. S cu to nguyn l hot ng cu mch khuch i thut ton.1. S .


-Vcc

+Vcc

R

C

+

R1

R2

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n mn hc TCS.

Hnh 4-1 mch khuych i thut ton.Dng sng

Vr+E T

T1 T2

t1 t2 t3 t4

-E

Vc

E.R1/(R1+R2)

t

-E.R1/(R1+R2)

Hnh 4-2 dng sng in p.

2. Nguyn l hot ng.

trng thi ban u 0 t t1Ti thi im t = 0 , ng ra ca trng thi bo ho dng Vr = +ETi V+ ta c in p :

21

1.

RR

REV

+=+


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+Vcc

Dr

D2

Q2R2

R

Rc

Rg

n mn hc TCS.

Do V+ = + E nn t C c np theo chiu V RR C Mass. T C cng np in ptrn t cng tng ln, in p trn t C tng n khi Vc =V- V+ th ng ra ca mch itrng thi sang mc bo ho m . Lc ny Vr= -E v:

21

1.

RR

REV

+

=+

Ti t1 t t2Ti t = t1 khi Vc = V- V+ th mch i trng thi sang mc bo ho m:

Vr = -E,21

1.

RR

REV

+

=+

V Vr = -E nn t C s np v x theo chiu ngc li. T C cng c np th inp trn t cng tng theo chiu m V- > V+, th ng ra ca mch i trng thi. Vy mchlun chuyn t trng thi ny sang trng thi khc m khng cn xung kch khi uvo, thi gian x v np l T1 v T2, T = T1 = T2. Do chu k:

2

212

21 ln222R

RRRCTTT

+===

thay i c chu k T1 m T = Const th ta thay i R thnh mt bin tr to ra hai qu trnh np vi hai in tr khc nhau, tc l ta c th thay i c t s chu

kT

T1 thng qua bin tr .

Do dng in chy ra trong mch khuych i thut ton nh nn ta dng my bin pxung .

II. S cu to mba xung

1. S cu to

Hnh 4-3 cu to my bin p xung.

Dng sng

Hnh 4-4 dng sng in p iu khin.


Vc

T1 T2

t0

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in p ng vi mc logic 0 l Vlogic 0 = (-E 0)in p ng vi mc logic 1 l Vlogic 1 = (0 +E)

2. Nguyn l hot ng.

Gi s tn hiu u vo V2 =1 th Trasistor Q bo ho, c ngha la dng t Vcc quaRc, qua cun dy s cp mba xung v chn Emito xung mass, lm cho cun dy L cdng in tng theo hm m:

)1(/t

C

CL eR

Uii == vi

CR

L=

Cun dy th cp s sinh ra mt sut in ng. Nu u dy th cp kn mch thto thnh dng Ig, qua D2, qua Rg vo cng G ca Tiristor iu khin.

Khi V2 = 0 th Trasistor b kho li, lc ny dng in qua cun L c gi tr (t =T1):

1

0

/

11 )1()()(1

REIe

RUTiTi T

C

CL ===

Transistor T b kha li iC = 0

Nu khng c it Dr th nng lng2

02

1W LI= s sinh ra qu in p trn cc cc C, E ,

T c th t n 100V ph hy Transistor T.T t = T1 tr i, iL suy gim theo quy lut:

'

/

0LiteI = vi '

'

R

L=

R : in p ri trn cun cm v in p thun Dr

III. S tng th mch iu khin.

1. S cu to


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n mn hc TCS.

Hnh 4-5 s tng th mch iu khin.

2. Nguyn l hot ng:

Gi s ng ra ca b hnh ch nht OA1, mc bo ha dng Vr = +E th in thti cc nn Q1 > in th ngng ( 0,6V ), lm Q1 dn bo ha. Dng ic chy qua Rc qua

cun dy s cp ca my bin p xung I v cc mit xung mass. cun th cp sinh rasdd dn n c dng iu khin chy xung cc G ca Tp lm cho Tp m.Trong khi in th ti ng ra AO2 mc bo ha m nn in th m ny t

vo nn ca Q2 lm cho Q2 b kha li khng c dng chy qua. Do cun th cp mybin p xung II khng c s nn khng c dng iu khin chy n Ta nn Ta vn bkha.

Khi ng ra ca b to xung OA1 mc bo ha m th ngc li khng c dngchy xung cc G ca Tp lm Tp b kha li. Cn ti ng ra ca b so snh OA2 c in


+Vcc

+Vcc

-Vcc

+Vcc

+Vcc

+Vcc

-Vcc

+Vcc-Vcc

+Vcc

R2

R1

D2

D1

C

R

+

OA1

C

Dr

D

Q2

+

OA2

Dr

D

Q1

C

Dr

D

Q2

+

OA2

Dr

D

Q1

R3

R4

R2

R

Rc

Rg

R2

R

Rc

Rg

R2

R

Rc

Rg

R2

R

Rc

Rg

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n mn hc TCS.

th mc bo ha dng lm cho Q2 dn bo ha do c ngun chy t Vcc qua Rcqua cun dy s cp my bin p xung II xung cc Emit ca my bin p xung II sinhra s dn n c dng chy xung cc G ca Ta lm cho Ta m.

Vy mun thay i gc m ca Tp v Ta ta ch cn thay i bin tr R cu bpht xung hnh ch nht OA1 to ra s thay i chu k in th ng ra ca b to xung

OA1. l lm thay i T1 v T2 nhng T = const.

IV. Tnh chn cc phn t trong mch iu khin.

1. B to xung.

Hnh 4-6 mch pht xung

Vn tc cc i khi ng c mang ti nh mc :

1000;60

2U dm === n

nxma

)/(10560

10002srad

xma==

Ta chn cng c tnh c ca ng c:

=

= 632,0;)( 2 uu

dm RR

k

T phng trnh c tnh c :

dm

dmudmdm

k

IRU

=


R1

R2

+Vcc

-Vcc

D2

D1

C

R

OA1

R3

R4

-Vs

+Vs

T1

T2

V2

t

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94,1105

2.25.632,0220=

=

=

dm

dmudmdm

IRUk

97.5632,0

94,1 2==

Ta c vn tc gc nh nht ng vi mmen ti nh mc :

1

)1(min == qtdm kM

Vi Mdm = k.dm.Idm = 1,94.25,2 = 48,88Kqt : h s qu ti ca ng c ; chn kqt = 2 .

)/(18,897,5

1)12(88,48min srad==

Vy in p tng ng l :

dmdm

MkU += minmin

)(05,24

97.5

88,4894,1.18,8min VU =+=

M Umin = minU

1,0220

05,24minmin ===

U

U

Ta c :f

T1

0 =

Nu chn tn s b bm f = 400 Hz

)(10.5,2400

1 30 sT

==

Ta tnh c 0,1 1; T1 = .T0T

min= 0,1.2,5.10-3 = 2,5.10-4

Tmax = 1.T0 = T0 = 2,5.10-3

Theo cng thc tnh chu k ca b iu khin s dng khuych i thut ton, ta c: ( nuchn R3 = R4 )T0 = 1,1C( R1 + R2 ) = 1,1C .R1 + 1,1C.R2 = T1 + T2Chn t C c C = 0,47F

===

48410.47,0.1,1

10.5,2

1,1 6

4

min1

min1C

TR

Vy ta chn bin tr R c gii hn l : 484 R 48402. Tnh chn my bin p xung.

Tnh ton ta c :Ug = 7VIg = 0,3AChn Rg = 10St p trn Rg : URg = 0,3.10 = 3VChn R1 = 33St p trn R1 : UR1 = Ug - URg = 4V


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n mn hc TCS.

Dng in hiu dng th cp mba khi khng ti

AR

UI 12,0

33

4

1

R1'

2 ==

=

Do st p trn it nh nn ta chn in th cun th cp mba xung l :U2 = UR1 = 4V

V t s mba xung l : m = 0,75

Ta c :2

1

1

2

I

I

U

Um ==

Hiu in th cun s cp mba xung l :

)(33,575,0

421 V

m

UU ===

Dng in s cp mba xung l:

I1 = m.I2 = 0,75.0,42 = 0,315 ( A )

Vi I2 = Ig + I2 = 0,3 + 0,12 = 0,42 ( A )Chn gi tr ca in tr RC = 13,5 St p trn RC l : URc = RC.I1 = 13,5.0.315 = 4,25 ( V )Do st p trn BJT nh nn chn ngun cung cp cho BJT l:

UCC = +12VChn h s khuych i ca BJT l : = 20

IC = .Ib )(01573,020

315,0A

II cb ===

Chn IC khuych i thut ton loi c :Ic = 0,15VVcc = 12V

Hiu in th ra Ur= | E | = 10V

in tr )(6,666015,0

102 ===

b

r

I

UR

Vy chn MBA xung c:U1 = 5,33V I1 = 0,315AU2 = 4V I2 = 0,42A

3.Thit k ngun nui cho mch iu khina) S :


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n mn hc TCS.

Hnh 4-7 s ngun nui

Gm : 1 my bin p in lc3 con it2 con vi mch n p : 7812 v 7912

b)Nguyn l hot ng :

na chu k u ( 0 ) : A dng nht ; B m nht D1 m , D2 kha .

Ud = U21 > 0 ; 222

UUd =

=

id = i21 = iD1 > 0 dD II2

1=

na chu k sau ( < < 2 ) : B dng nht ; A m nht D1 kha , D2 m

Ud = U22 > 0 ; 222

UUd =

+=

id = i22 = iD2 > 0

Dng in mt chiu chy qua ti : dD II2

1=

)(22 2 VUUim =

c_Tnh chn cc gi tr :Ta c :

Ud = 35 VId = 1,5 V

Gi tr trung bnh ca dng chy qua mi it :


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n mn hc TCS.

)(75,02

5,1

2

1AII dD ===

Gi tr hiu dng ca dng th cp ca MBA :

)(18,15,14

14,3

422212 AIIII d =====

Gi tr trung bnh ca in p pha th cp MBA : )(97,38

22

35.14,3

222 V

UU d ===

Suy ra t s MBA :

18,0220

97,38

1

2 ===U

Um

Dng in chy trong MBA l :I1 = mI2 = 0,18.1,18 = 0,21(A)

Tnh cng sut biu kin ca my bin p :S2 = 2.U2.I2 = 2.38,97.1,18 = 91,97 92 W

S1 = U1.I1 = 220.0,21 46 WVy :

W)(692

4692

2

21 =+

=+

=SS

S

Vy ta chn MBA c cng sut biu kin l 69W, vi ngun cung cp l ngun xoaychiu 220V, tn s 50 Hz .

PHN V. BO V MCH IN T CNG SUTPHN V. BO V MCH IN T CNG SUT

Cac phn t ban dn cng sut c s dung ngaycang nhiu do co nhng u im nh gon nhe, lam vic tincy, tac ng nhanh, hiu sut cao, d thc hin t nghoa..

Tuy nhin, cac phn t ban dn cng sut cung oi hoicac iu kin khc khe. Trc ht la phai tn trong nhng trs gii han s dung do nha san xut ra i vi tngphn t nh sau :

in ap ngc ln nht Tr trung bnh cho phep i vi dong in Nhit ln nht cua mt ghep

Tc tng trng ln nht cua in apdt

du

Tc tng trng ln nht cua dong indt

di

Thi gian khoa toff


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n mn hc TCS.

Cac phn t ban dn cng sut cn c bao v chng lai nhng s c xay ra, nhng nhiu loan nguy himnh ngn mach tai, qua in ap hoc qua dong in.

I.Cng sut tn tht va lam mat :Khi tiristor m cho dong chay qua, cng sut tn tht bn

trong se t nong no. Mt ghep la ni b t nong nhiu nht,ngi ta dung k hiu Tj ch nhit mt ghep, Tjm chnhit ln nht cho phep.

i vi ban dn Ge : Tjm = 800 1000Ci vi ban dn Si : Tjm = 1500 2000CCng sut tn tht trong thit b ban dn k hiu la)W(P . Thng c chia thanh tn tht chnh ( 1P ) va tn

tht phu ( 2P ). Tn tht chnh do dong in gy nn. Tn thtphu bao gm tn tht chuyn trang thai(t khoa sang m vangc lai) va tn tht trong mt ghep. Thng tn tht phukhng vt qua 5% cua tn tht P . V vy co th xem nhP 1P .

II. Ngn mach u ra thit b chnh lu va baov ngn mach, qua tai bng dy chay :

Khi mt thit b ban dn b choc thung th no ch nhmt dy dn in. Tiristor hoc diode thng b choc thung khichuyn mach v khi y chung phai chu in ap ngc lnnht. Van nao b choc thung th dong qua no la ln nht. bao v Tiristor va diode tranh c dong in pha hoai, ngita thng dung dy chay tac ng nhanh (khoang vai ms). Loaidy chay nay lam bng bac la, t trong vo bng s co chacat thach anh hoc nc ct.

Hoat ng cua dy chay chia thanh hai giai oan.Giai oan 1: t khi dong in s c tac ng n khi

xut hin h quang thq. Trong giai oan nay dy chay b nong vamm ra.

Giai oan 2 : t khi xut hin h quang n khi ct xongdong in s c tc, trong giai oan nay in ap h quang tngdn va do o dong in s c giam dn n 0.

Cac thng s c trng cho dy chay la : in ap nhmc va dong in nh mc. Khng nn t dy chay vaomach in co in ap cao hn in ap cua dy chay. Dongin nh mc cua dy chay phai bng hoc ln hn dongin bao v no, nhng khng ln hn 10%.

Bao v ring bit tng Tiristor c s dung trongtrng hp khi mt Tiristor b choc thung vn yu cu thit bbin i tip tuc lam vic. Nu co N tirisotr ghep song song


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n mn hc TCS.

ma mt tiristor b choc thung th dong tai chia u cho N - 1tiristor con lai. Khi chon dong in nh mc cho dy chay cnlu y iu nay. Co nhiu cach t dy chay bao v thitb bin i ban dn.

t ni tip vi tng tiristor hoc it t tng pha cua cun dy th cp may bin ap t ni tip vi nhom tiristor ghep song song t u ra cua thit b bin i t pha s cp may bin ap

i vi s cua trng hp thit k ta t: t s cp may bin ap, v tr 1 t th cp may bin ap, v tr 2 t ni tip vi tng it, v tr 3

Hnh 5-1 s b tr thit b bao v.

Tnh chon mach bao v:Dy chay v tr 1: U1 = 380 V(AC); I1 = 5,41 A;Chon h s d tr luc khi ng ki = 2; chon dy

chay.I1dc = 2*5,41 = 10,82 (A);

Dy chay v tr 2 : U2 = 100 V(AC); I2 = 20,57 A;Chon h s d tr in luc khi ng la ki = 2

I2dc = 2*20,55 = 41,14 (A);Dy chay v tr 3 : Um = 245 V (DC); IT = 8,4 AChon h s d tr in luc khi ng la ki = 2

I3dc = 2*8,4 = 16,8 (A);


3

1 2

Z

FFF

FFF

F

F

F

F

F

F

D5D1 D3

D4 D6D2

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n mn hc TCS.

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