đồ án bảo vệ

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TRNG I HC SPKT HNG YN N MN HC

1

LI NI U

Ngy nay vi s pht trin nhanh chng ca k thut bn dn cng sutln, cc thit b bin i in nng dng cc linh kin bn dn cng sut

c s dng nhiu trong cng nghip v i sng nhm p ng cc nhucu ngy cng cao ca x hi. Trong thc t s dng in nng ta cn thayi tn s ca ngun cung cp, cc b bin tn c s dng rng ri trongtruyn ng in, trong cc thit bt nng bng cm ng, trong thit bchiu sng... B nghch lu l b bin tn gin tip bin i mt chiu thnhxoay chiu c ng dng rt ln trong thc tnh trong cc h truyn ngmy bay, tu thu, xe la...

Trong thi gian hc tp v nghin cu, c hc tp v nghin cumn in t cng sut v ng dng ca n trong cc lnh vc ca h thng

sn xut hin i. V vy c th nm vng phn l thuyt v p dng kinthc vo trong thc t, chng em c nhn n mn hc vi ti:Thit k v c to mch nghc lu mt pha. Vi ti c giao,chng em vn dng kin thc ca mnh tm hiu v nghin cu lthuyt, c bit chng em tm hiu su vo tnh ton thit k phc v chovic hon thin sn phm.

Di shng dn ch bo nhit tnh ca thy cng vi s c gng n lc ca cc thnh vin trong nhm chng em hon thnh xong n ca mnh. Tuy nhin do thi gian v kin thc cnhn ch nn khng trnh khi thiu st khi thc hin n ny. V vy

chng em rt mong s nhn c nhiu kin nh gi, gp ca thy cgio, cng bn b ti c hon thin hn.

Chng em xin chn thnh cm n!

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2

NN IO IN

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Hng Yn, ngy 19 thng 07nm 2012.

Gio vin hng dn

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3

K hoch thc hin

TT Ni dung thc hin TunThc

hin

Thngqua

GVHD

Ngithc hin

aim

Thchin

Nhn xtGVHD

1 Lp bng k hoch thchin

Tun1

Tun1

Ch Nam Tinh

2 Phn tch yu cu, mctiu, png n tchin v ng dng.

Tun1

Tun2

V. Vit

Tinh

ng IGii thiu tng quan v

nghc lu1.1 Tng quan v

nghch lu1.1.1 Nghch lu ph

thuc 1 pha 2 nachu k.

1.1.2 Nghch lu phthuc 3 pha na chuk.

1.1.3 Nghch lu phthuc cu 3 pha

1.2 Nghch lu c lp1.2.1 Nghch lu c lp

ngun p

1.2.2 Nghch lu c lpngun dng1.2.3 Nghch lu ngun

dng 3 pha1.2.4 Nghch lu ngun p

3 pha

Tun1

Tun2

ngThanh

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4

3

ng IIThit k mch lc

2.1 Tnh ton My Bin

p2.2 Mch lc2.2.1 Chn van khuch icng sut.22.2.mch cch ly

ng IIIThit k mc iu khin

Tun2

Tun3 Ch

Nam,

VnVit,

nhThanh

Tinh

3.1 thit k mch iukhin3.1.1 Tnh chn gi tr

phn t3.2 Thit k mch bo v

qu dng qu p3.3 Thit ks nguyn

l3.3.1 Phn tch nguyn lv khi chc nng trongmch3.3.2 La chn thit b

4 ng I

Ch to m hnh

Ch to, o kt qu vhiu chnh thng s png vi yu cu ti

Tun3

Tun4

Cham,

V.VitnhThanh

Xngin

5 Hon thnh sn phmHon thnh ni dungquyn thuyt minh. Ktlun

Tun4Tun4 V. Vit

C.NamnhThanh

Tinh

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5

6 Thng qua GVHD tnghp nh gi

Tun4

Tun4

ngin

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7

vung v c khuych i ln bng cc van bn dn l Transistor,IGBT

* Mc tiu ca ti.Nm c mt cch tng quan v cc phn t bn dn cng sut.

Nghin cu v cc mch nghch lu, hiu c nguyn l lm vic camch nghch lu, cc phng php bin i t la chn mt phng nti u nht c p dng trn n ca mnh v ngoi thc tin.

C khnng tnh ton, thit k v ch to mch nghch lu in p mtpha vi cng sut cho trc.

nga ca ti. gip sinh vin c th c th cng c kin thc, tng hp v nng

cao kin thc chuyn nghnh cng nh kin thc ngoi thc t. ti cnthit k ch to thit b, m hnh cc sinh vin trong trng c bit lsinh vin khoa inin t tham kho, hc hi to tin ngun ti liucho cc hc sinh, sinh vin kho sau c thm ngun ti liu nghin cuv hc tp.

Nhng kt quthu c sau khi hon thnh ti ny trc tin l sgip chng em c th hiu su hn v cc b nghch lu, cc phng phpbin i in p. T s tch luc kin thc cho cc nm hc sau vra ngoi thc t.

* Ni dung cn hon thnh:- p k hoch thc hin.

- Gii thiu mt s ng dng v c im ca mch nghch lu mtpha.- Phn tch nguyn l lm vic v cc thng s trong mch nghch lu

mt v ba pha.- Thit k, ch to mch nghch lu mt pha m bo yu cu:

+ in p u vo mt chiu U = 12V ly tc quy.+ in p u ra dng cho cc thit bin xoay chiu U = 220V -

f = 50HZ , P = 300W+ Th nghim, kim tra sn phm, sn phm phi m bo yu

cu k thut, m thut. Quyn thuyt minh.

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8

ng III I M M N L

1.1. Tng quan v nghc luTrong cng ngh, ta thng gp vn bin i in p mt chiu thnh

in xoay chiu v ngc li bng cc thit b nn in. Cc thit b c gil nghch lu.Khi nim: Nghch lu l qu trnh bin i nng lng mt chiu thnhnng lng xoay chiu.Phn loi:Cc s nghch lu c chia lm hai loi.- S nghch lu lm vic ch ph thuc vo li xoay chiu.- S nghch lu lm vic chc lp (vi cc ngun c lp nh cquy, my pht mt chiu ....)

Nghch lu ph thuc c s nguyn l ging nh chnh lu c iukhin. Mch nghch lu ph thuc l mch chnh lu trong c ngun mtchiu c i du so vi chnh lu v gc m ca cc tiristo tho mniu kin (/2 < /2). Lc nyngun mt chiu tr thnh ngun pht in cung cp nng lng cho liin cn b bin i lm vic ch nghch lu ph thuc.c duy tr donng lng tch ly trong cuc cm xd. n thi im 4 ta li a tn hiu

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9

mvan T2, dng in li chuyn t T1 sang T2theo nh qu trnh ni trn.

Thi gian o mnh phi kt thc sm hn thi im 5 mt thi imdng van T1 kp khi phc tnh cht kho ca mnh. Nu thi gian khi

phc khng th sau 5 van T1 vn tip tc dn in, in p U2a > 0 dngtrong mch Id = (E +U2a) / Rtd . V in trtng ng rt b nn dng squ ln. Ta gi y l s clt nghch lu. S clt s xut hin khigc kho qu b hoc gc o mch qu ln.Vy nghch lu lm vic c an ton th gc kho gc =/18 vitiristo thng dng vi khong 100 200 s th trong mi trng hp th vicly = /18 cn c coi trng, do /2 < < - .

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10

u21

u22

~id

uT1

T1

uT2

T2

E

-+L

a,

c,

i1

i2

0

0

iG

0

0

ud

1

u2b

5

u2a

Id

43 2

iG1 iG2

E

-E

2

b,

id

0

ud

u2b

4

2

=+

01

u2a

E

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11

1.2.2. Nghc lu p thuc ba pha na chu k.S nguyn l ca nghch lu c lp ba pha na chu k cho (hnh

1.2).V hnh thc ging nh chnh lu ba pha na chu k.Cho ti khi < /2 bbin i cn lm ch chnh lu, ti = /2 in p m bng in pdng tr trung bnh ca in p bng khng khi > /2 in p trung bnhdng sng l m cho n khi = . Dng sng lc ny tng tnh khi = 0 nhng ngc chiu.

Ti ca chnh lu trn hnh 1.2 l ng c in mt chiu lm vic chng c. Sau khi i chiu S ca ng c, n tr thnh my pht lcny b bin i lm vic ch nghch lu. Chiu dng in khng thay

i c, n do chiu ca tiristo quyt nh. i chiu S ng c ctho cc tnh phn ng hoc o chiu kch tng c. Kt qu ca vici chiu p mt chiu lm cho xut hin dng in chy trong tng pha cabin p khi in p pha m. Ni cch khc ng c pht ra cng sutchuyn vo li in xoay chiu.

cc tiristo chuyn mch c, b bin i cn phi ni vo li xoaychiu, v vy y l b nghch lu ph thuc.Ta khng th chuyn dng in ca tiristo, v d t T1 sang T2 nu U2b tm hn U2a = th U2a = U2bdo = l gii hn ca s lm vic.Nguyn l lm vic nh sau:

Ti thi im 1 (hnh 1.2) lch gc 2/ ta cp xung iu khin voT1, T1 mv in p ca n l dng nht. Khi T1 msc in ng e sphng mt dng in i qua pha a ca bin p. Dng in ny cduy trcho n tn thi im

2 nhsc in ng v sc in ng cm ng ca

mch ti. Ti2 lch gc chng ta cp xung iu khin vo T2, T2 m, T1

kho li. Dng in do sc in ng e sinh ra lc ny li chy qua pha bca bin p...

Tng tnh vy n thi im 3 lch gc ta li cp xung iukhin vo T3. Dng lc ny li chy qua pha c ca bin p. Cc xung iu

khin lch nhau gc 3/2 . Chu k xung l 2 . Trong mt chu kdng inln lt qua cun th cp ca my bin p pha n lch vi pha kia 3/2 vitn s bng tn s li. Do vy trong cun s cp cng cm ng sc inng xoay chiu ba pha c tn s bng tn s ca li. Ni cch khc l s bin i c in p mt chiu ca ngun sc in ng e thnh inp xoay chiu ba pha tr v ngun.

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12

Khi ch ti in cm ca mch ant La(in cm ca bin p) thvic chuyn dng t T1 sang T2, T2 sang T3.... khng tc thi nh m t trn m phi qua giai on trng dn (chuyn mch). ln ca gc chuynmch ph thuc vo tr s La. Trn hnh 2 trnh by cc ng cong dng, p

ca nghch lu khi ch ti La. nghch lu khng b lt cn bo m gc kho ln. Do ch nghch lu s nm trong: /2 < < -

xc nh cn xut hin t thi gian kho ca tofftiristo c dngtrong s nghch lu. Trn hnh 1-2, b c, d biu din sng in p khi ch chnh lu vi cc gc mkhc nhau. hnh 1.2b c c gi tr nh hnh 1-2c chnh lu c in p tc thi c on m. hnh 1.2d gc=900 nn Utb = 0.

Trn hnh 1-2 e, n 1-2n biu din ng cong dng p ch nghch

lu khi c hin tng trng dn v khi khng c hin tng trng dn. C = - c th hin thi gian cn thit khi tiristo ang dn m bo tnh trngkho khi in p m c gi l gc tt, tr s ca n thng khng di50.

Cc h thc tnh ton vrne uuuCosuu

22

63

Trong :

CosuCosu22

17.12

63 : p nghch lu khng ti.

uRr = Id.R2 : gim p trn in trth cp bin p.

d

a

I

x

u

2

3

: gim p do o mch.

d

aI

xu

2

3 : gim p do o mch.

uv = (0.5 1).v : gim p trong van

T c:v

a

dneu

xRICosuu

2

317.1

22

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13

u2b u2cu2auL 900

utb

u2b u2cu2a

0

utb

uL

c

d,

u2b u2c

u2a

0

iG2

iG1

iG3

iT

u2b u2c

u2a

0

utb

uL

e,

u2b u2c

u2a

0

utb

uL

f,

T1u2A

T2u2B

T3u2C

uT

ud

A

B

C

u2b u2cu2a

0

utb

uL

b,

a

uL

iT2 iT3 iT1

h

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14

C B A

i

iG

u2b u2cu2a

0

i3 i5 i1 i3

1

22

acee

2ac ee

2

ba ee

ud

5

4

5

6

1

6

1

2

3

2

e

T1

T3

T5

T4

T6

T2

idLd

ec eb e

La La La

+

1.2.3.Nghc lu p thuc cu ba pha.S nguyn l ca nghch lu ph thuc cu ba pha cho trn hnh 1.3.

V hnh thc ging nh s chnh lu ba pha c iu khin. Khi chuynsang lm vic ch nghch lu th gc m >900v S E ca ng c

c i chiu so vi chng c. Vic i chiu E c th thc hin theonhiu gii php k thut khc nhau. Vy ch nghch lu cc dng cangun E c du vo nhm ant, cn cc m u vo nhm catt.

nghch lu ny cc van ca nhm catot s lm vic khi p th cp mcn nhm anot cc van s lm vic khi p th cp dng. Trong bt k mtthi im no cng c hai van lm vic, v vy a nghch lu vo lmvic (khi khi ng) cn a ng thi hai xung mvan mt vo nhm antchung, mt vo nhm katt chung. Nh vy ta phi a n in cc iukhin van hai xung hp cch nhau 600 in, hoc mt xung rng c thi giant > 600.Vic gii thch nguyn l lm vic ca nghch lu ny cng tng tnhchnh lu ba pha na chu k, v th hin trn trn th dng, p nghch lu(hnh 1.3).

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15

Cc h thc tnh ton:in p trung bnh ca nghch lu cu ba pha:

vrne uuuCosuu

2

63

Trong : cc tn tht in p trn in tr, o mch v trn van u c gitr gp i so vi s ba pha mt chiu na chu k.

ur= 2.Id.R2

d

a Ix

u

2

6

vuv 15.02

va

dne ux

RICosuu

2

6234.2 2

1.3. Nghc lu c lp .Nghch lu c lp vit tt l N l thit b bin i ngun in mtchiu thnh dng xoay chiu vi ti c lp khng ph thuc vo li in.Dng xoay chiu c th bin i c vi tn s mong mun nn cng gil thit b bin tn gin tip. Gin tip v thng c ngun mt chiu phic mt khu chnh lu. mt s ti liu k thut thit bny c gi l nghch lu tonm hay onduleur.

Nh ni 1.1 nghch lu c lp c phn lm ba loi l Ndng, N p v N cng hng. Ngi ta cng c th phn loi theos pha mt pha, ba pha, cu...N dng u vo phi c in cm Ld

ln v to mch c np theo lut khng chu k. Dng u vo Id l lintc, phng khng nhp nh ngha l ngun cung cp thit b ny l ngundng.Trong N cng hng ti c in cm ln, cng vi cc phn t R,L, C ca mch to nn mch vng dao ng RLC v c cng hng p. Tns ring ca mch cng hng phi cao hn hoc bng tn s cng tc ca

N.Trong N p ngun cung cp cho n phi l ngun p (my pht phi

c in trtrong nh)nh vc p dng ch yu ca N dng v p l bini tn s, cung cp in xoay chiu cho cc thit b xoay chiu v phc v cho

cc T.. c iu chnh tn s.Cc N cng hng c dng c li khi tn s ra khong 12 KHz.Cp cho cc thit b nhit in, thit b siu m v cc truyn ng cao tc.

1.3.1 Nghc lu c lp ngun p mt phac my bin p im gia1311 nguyn l:

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S nguyn l b nghch lu p mt pha c my bin p im giac cho trnhnh 1.4vi ngun p mt chiu U, my bin p c im giav hai kho chuyn mch K1 v K2, dng in ti xoay chiu i. im giamy bin p 0 ni vi mt cc ngun p U, cn u kia qua kho K1 ni vi

A v qua kho K1 ni vi B. Gi thit my bin p l tng, in p ccdy qun t l vi s vng dy, ta c:

v1

= v1, u =

2/1

2

n

nv

1

Khi K1ng c: v

1= U , u =

1

22

n

n U

Khi K1ng c: v

1= - U , u = -

1

22

n

nU

Nu bqua dng in tho, dng s cp v th cp lin hqua phng

trnh cn bng sc tng:

2

1n (i 1K - i 1K ) = n 2 i

Hnh 1.4.B nghc lu y bi p im giaHai kho chuyn mch K

1v K2 phi mt kho ng, mt kho m

v nu chai u mngha l mch ti xoay chiu hmch v:i 1K = i 1K = 0 do i = 0

Nu chai kho cng ng th ngn mch pha ngun mt chiu:V

A- Vo = V

B- Vo do V A = V B = Vo

Biu thc vdng sng:

in p ra u c chu k T ta mC tt = 0 n T =2

T v K1

t T =2

T

n T

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- Khi 0 < t 0 K1 v K 1 phi c hai chiu to nn bng mt linh kinng t pht v mc iu khin vi mt diode mc song song ngc. Nu < 0 K

1v K

1phi c hai chiu to nn bng mt linh kin ng c iu

khin v mt pht vi mt diode mc song song ngc. Nu bt k linhkin phi c ng v mtutheo trng hp s dng ta c s sau:

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20

~

U

T1

T1

D1

D1

U

i

n2n1i+-0

A

B

Hnh v7: S b nghc lu p, y bi p im giaHnh 1.6pha trn v khong dn cc linh kin bn dn. Ta nhn thy

mi chuyn mch dng in khng bng khng to nn gia mt linh kinc iu khin v 1 diode. Tnh cht ny chung cho cc b nghch lu p. Domy bin p c im gia, in p trn cc kho chuyn mch hbng 2 lnin p ngun mt chiu U ta gi b nghch lu nhn i in p.33c tnh:

Nu gi thit in p vo khng i, dng in ra hnh sin, mybin p v linh kin bn dn l tng, ta ddng tm c in p ra vdng in vo:

- in p ra: Trong mt na chu kin p ra bng2/1

2

n

n U v -

2/1

2

n

nU

na chu kkia. in p ra u c tr hiu dng bng: U =1

22

n

n U

- Tr hiu dng thnh phn c bn : U1

=

22 .1

22

n

nU

- T siu ho: 'u =1'

1

U1

22 '' UU = 0.483

- Khai trin Fourier c:U =

1

22*4

n

n

U (Sin t +

3

1 Sin3 t +5

1 Sin5 t +7

1 Sin7 t ...) i sng c bn,

ucn cha tt ccc iu ho bc l. Bin iu nghch vi bc ca n.Dng in vo i c chu k bng mt na chu k ca cc i lng ra t < Ta c biu thc:

i = )('2

1

2 tSinIn

nm

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21

Gi tr trung bnh:

I =

cos2

'2

1

2

mI

n

n

Hiu dng:

I hd =2

1'

2

1

2

mI

n

n

Nhp nh:

i = i max - i min = )1('2

1

2 SinIn

nm

T siu ho:221

II

I

hdi

2

2cos

81

cos22i

Tnh cht thun nghchB qua cc tn hao trong b nghch lu, cng sut tc thi u ra v vonh nhau:

P = ui = ui

Cng sut tc dng: P = UI = U1

cos2

'm

I

132 nghc lu c lp ngun in p cu 1pha :3 S nguyn l:

Hnh 1.8

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22

2.3.2.2. Nguyn l hot ng:c im ca nghch lu c lp ngun in p l lun lun nh

dng in p hnh ch nht trn ti vi mi ti , cn dng thdng inti li ph thuc vo tnh cht ti Cd+Ed to ra ngun in p l tng .

a.

xt trng hp ti c tnh cht dung khng:* V ti c tnh cht in dung , nn dng in ti sm pha hn so viin p ti mt gc t .Da vo s nguyn l ta c thdng in , in p trn ti nh hnh v* Xc nh min dn ca cc van :Kt hp thdng in i(t), in p u(t) trn ti vi s nguyn tc tanhn thy :

-T 0 - t c : it > 0 ,Ut > 0 c c iu ny th V1v V4 thng

-T

- t

: it < 0 Ut > 0 D1 v D4 thng-T2- t : Ut < 0

it < 0 V2 v V3 thng-T 2- t2 : it > 0

Ut < 0 D2 v D3 thng* Xc nh thi im chuyn mch gia cc van

Ti - t : c s chuyn mch gia cc van : V1 D1 ; V4 D4

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9: thd i , in p trn ti c tnh cht dung khng- Ti : c s chuyn mch : D1 V3 ; D4 V2- Ti 2- t: c s chuyn mch : V2 D2 ; V3 D3- Ti 2 : c s chuyn mch : D3 V1; D2V4* Xt s chuyn mch dng in gia cc van :- Xt ti : cn c s chuyn mch : t D1 V3 ; D4 V2Cn kho D1, D4 , mV3, V2. Trc mt khonh khc D1, D4ang thngdn dng ti nn UV2= Ed >0UV3 = Ed > 0 (v thng qua D1dng ngun t vo a, thng qua D4 mngun t vo b)V vy n pht xung iu khin vo V2 v V3 th hai van ny mngay.Khi V2 v V3 thng : +Edt vo b ; - Edt vo a.

Do UD1=Ed < 0 ; UD4 = Ed < 0 D1 v D2 khoNh vy qu trnh chuyn mch l s chuyn mch t nhin V1, V2 ,V3 ,

V4 c th chn thyristor thng; xung iu khin cc van ny l xung n.- Xt ti 2- t : cn c s chuyn mch t V2 D2 ; V3 D3

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24

ti 2- tv sau dng in ti iti du t (-) sang (+) dng in ny ingcchiu dng in ca V2,V3 lm cho V2, V3 kho li do dng inngc. Do yu cu cn bng cng sut phn khng nn D2 v D3 phi thng tip tc duy tr hng ca dng in ti ;

Chiu dng in : Zt D3CdD2Zt* V UV1,iV1,UD1,iD1 v idu vo nghch lu :- Khi V1v V4 thng dn dng in ti th UV1=(12 )V >0do st p trnvan,ng thi iV1= it > 0

UD1= U V1= (12 )V < 0 ; iD1=0.id khp mch : + Ed V1Zt V4 - Ed

id= it > 0-Khi D1 v D4 thng :

UD1= U D1= (12 )V > 0

iD1=it > 0UV1= U D1= (1 2 )V < 0 ; iV1= 0id khp mch : Zt D1Cd D4 Zt

id = it= iD1 < 0- Khi V2v V3 thng dn dng in ti th UV1= Ed >0 ; iV1= 0

UD1= Ed < 0 ; iD1=0.id khp mch : + Ed V3Zt V2 - Ed ; id= it > 0- Khi D2 v D3 thng :

UV1=Ed > 0; IV1= 0; UD1=Ed < 0; iD1= 0id khp mch : Zt D1Cd D4 Zt

id = it= iD2 < 0*Tnh cng sut ngun:

Pd = Ed . Id vi Id =

22 .I2 . cost

Pt = Ut.It.cost ; it = 2 . I2. cost

Ut1 =

4 .Ed.cos(t) : thnh phn sng bc 1

Do : Pt =.2

4.Ed.I2.cost =

22 .Ed.I2.cost

Pd = 22

.Ed.I2.costb.Xt trng hp ti c tnh cht cm khng:* V ti c tnh cht in cm, nn dng in ti chm pha hn so vi inp ti mt gc t .* Xc nh min dn ca cc van :

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0: thd i , in p trn ti c tnh cht cm khng

Bng phng php phn tch nh trn,kt hp thdng in i(t) , inp u(t) trn ti vi s nguyn tc ta xc nh c :

-T 0 t : it< 0Ut > 0 nn D1D4 thng

-Tt: it > 0Ut > 0 nn V1V4 thng

-T

+ t : Ut < 0it > 0 nn D2D3 thng-T + t2 : it< 0

Ut < 0 nn V2V3 thng* Xc nh thi im chuyn mch gia cc van

Ti t : c s chuyn mch : D1 V1 ; D4 V4Ti : c s chuyn mch : V1 D3 ; V4 D2Ti +t : c s chuyn mch : D2 V2 ; D3 V3Ti 2: c s chuyn mch : V3 D4 ; V2 D1

* Xt s chuyn mch dng in :-Xt ti : c s chuyn mch : t V1 D3 ; V4 D2Trc mt khonh khc c V1V4ang thng iv1 = iv4 = it >>0

Theo yu cu ca ti th ti phi i du chiu in p ut tc l khoV1V4 v m D2D3. Mun kho V1V4 bng cch duy nht l cng bckho bng xung V1V4l van iu khin hon ton. Nu dng V1V4 l van

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iu khin khng an ton th phi u t hp ca cc van iu khinkhnghon ton tin cm to ra tnh cht nh van iukhin hon ton. Khi V1V4 kho th dng in ti gim vkhng nhngdo mch ti c in cm nn sinh ra sc in ng tip tc duy tr dng

in theo hng dng khi D2D3 thng dng in ti:ZtD3 CdD2Zt- Xt ti + t : D2 V2 ; D3 V3 . Trong thi gian D2D3 thng th Uv2 =UD2 < 0 ; Uv3 = UD3 < 0 ; V2V3 bt in p ngc.n + t : iD2 = iD3 = 0 D2D3 bt u kho, sau thi im ny dng

in ti o chiu t (+) (-) ngc vi chiu dng qua D2D3. Ti + t :

tt ccc van u kho UV2 = UV3 = 2dE

> 0. Nu ti thi im ny c xung

iu khin vo V2V3 V2V3 thng.

*Xc nh cng sut ngun v cng sut ti.Pd = Ed . Id vi Id =

22 .I2 . cost

Pt = Ut.It.cost ; it = 2 . I2. cost

Ut1 =

4 .Ed.cos(t) : thnh phn sng bc 1

Do : Pt =.2

4 .Ed.I2.cost =

22 .Ed.I2.cost

Pd =

22.Ed.I2.cost

1.3.2.3. Nhn xt vs nghc lu c lp ngun in p :- V tnh cht ca nghch lu nn cc im t , + t , 2+ t l thay itheo s thay i ca ti nn ngi ta chn cc im 0, ,2, lm thiim pht xung iu khin cho cc van.- Nghch lu c lp ngun p khng lm vic c ch khng ti .

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133 nghc lu c lp ngun dng in cu 1 pha :33 S nguyn l:

Hnh 2.11.3.3.2. Nguyn l hot ng:- Ed : Ngun in p 1 chiu, in cm Ld to ra ngun dng in1 chiu l tng vi Id- Ngun dng in lun nh dng dng in hnh ch nht trn ti ,dng in p ph thuc vo ti . xt trng hp ti c tnh cht dung khng:V ti c tnh cht in dung nn dng in sm pha hn so vi in p timt gc t .

Da vo s nguyn tc ta c thdng in , in p trn ti nh hnhv 2.2.* Xc nh min dn ca cc van:Kt hp thdng in i(t) ,in p u(t) trn ti vi s nguyn tc ta c:

- t 0: it > 0 V1V4 thng- t2 : it < 0 V2V3 thng- t 23 : it > 0 V1V4 thng

* Xc nh thi im chuyn mch gia cc van :

Ti cn c s chuyn mch t V1 V3 , V4 V2Ti 2 cn c s chuyn mch t V3 V1 , V2 V4* Xt s chuyn mch dng in gia cc van :Ti : c chuyn mch V1 V3 ; V4 V2

Trc mt khong khc , khi V1 v V4ang thng dn dng in ti ,ta thy ut> 0 , c ngha l cc (+) ca utt a , cn (-) ca utt b.

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Nh vy : Utthng qua V1t thun trc tip ln V3Ut thng qua V4t thun trc tip ln V2

do UV3= Ut > 0 ; UV2= Ut> 0, n pht xung iu khin vo V2 V3 thlp tc hai van ny thng , khi V2 v V3 thng dn n cc (+) Utt vo

katot V4 , cc m t vo anot V1 lm cho UV1=Ut < 0 UV4= Ut < 0Do hai van V1 v V4 kho kt thc qu trnh chuyn mch. Qu trnhchuyn mch y l chuyn mch t nhin.V vy cc van t V1 n V4 c th chn thyristor thng xung iu khincc van ny l xung n. thdng in in p cc van nh th

: thd i , in p trn ti c tnh cht dung khng* Tnh cng sut ngun :

Pd = Ud .Id = Pd = Ed . Id vi Ed = Uv=

22 .U2 . cost

Pd =

22 .U2.Id.cost

Ut = 2 U2.sin tPti =Ut.It.cost

Tr hiu dng ca dng in ti sng bc 1 l : I1t = .24

Id = 22

.IdDo Pti =

22 .U2.Id.cost Pt =Pndo cng sut t ngun truyn ti ti

l ln nht .

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b xt trng hp ti c tnh cht cm khng:

3: thd i, in p trn ti c tnh cht cm khng* V ti c tnh cht in cm , nn dng in ti chm pha hn so vi in

p ti mt gc t .* Xc nh min dn ca cc van :0 : it > 0 V1V4 thng2 : it < 0 V2V3 thng23 : it > 0 V1V4 thng

Khi V2V3 thng : UV1=Ut < 0 V1V4 kho lil gc in p ngc ln van chnh l thi gian khi phc li tnh cht iukhin ca van min 2f.tq* Xt s chuyn mch ca cc van

- Xt ti : V1 V3 , V4 V2Trc mt khong khc V1V4ang dn ; Ut< 0 th(+) t b ; (-) t a. Thng qua V1V4ang thng Utt ngc ln V2V3 ; UV2 = UV3 = Ut < 0.

Nh vy n khi pht xung iu khin vo V2V3 th V2V3cha thng c.Theo yu cu ca mch bt buc phi o chiu dng phi cng

bc kho V1V4 . Khi V1V4 kho ; V2V3cha thng th Uv1 =Uv2 = Uv4= 2dE

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> 0. Khi pht xung iu khin vo V2V3 th 2 van ny thng v thc hino chiu dng t (+) (-) Chuyn mch y l chuyn mch cngbc. Khi V1 V4 phi chn van iu khin hon ton.*Cng sut ngun v cng sut ti.

Pd = Ud .Id = Pd = Ed . Id vi Ed = Uv=22 .U2 . cost

Pd =

22 .U2.Id.cost

Ut = 2 U2.sin tPti =Ut.It.cost

Tr hiu dng ca dng in ti sng bc 1 l : I1t =.2

4 Id =

22 .Id

Do Pti =

22 .U2.Id.cost Pt =Pndo cng sut t ngun truyn

ti ti l ln nht1.3.3.3. Nhn xt vs nghc lu c lp ngu d in :Nghch lu c lp ngun dng in khng lm vic c ch ngnmch.1.3.4. Nghc lu dng in ba pha

S nguyn l ca nghch lu ngun dng ba pha cho trn hnh2.10. T hnh v cho thy nu ta thm mt nhnh vo cu mt pha sccu ba pha. Chc nng ca cc phn t, C, D, T trn s ny c cp ti phn trc. Gi thit ti i xng ba pha, u sao, c tng trl Za,

Zb, Zc. Cc tiristor c iu khin m theo th t T1, T2, T3, T4, T5, T6,T1, T2....Cc xung iu khin lch pha nhau mt khong thi gian T/6 (T lchu k ca dng ti).

Gi s ti thi im ban u ta cp xung iu khin cho T1 v T2. Cctiristor ny ang lm vic dn dng nn trong khong 3/0 dng inngun Id chy t + ngun qua T1 Za Zc T2 ngun.

Ta c: ia =Id; ic = - Id; ib = 0Cc t C1, C2, C3, C5 c np in, dng bn tri, m bn phi.

in p trn cc t lc ny l Uc1, Uc2, Uc3, Uc5. Ti thi im 3/ choxung iu khin mT3. Khi T3 mUc1t in p ngc ln T1 lm cho T1

kho li. Dng in Id lc ny chy qua T3 ri r lm hai nhnh, nhnh thnht qua C1, nhnh th hai qua C3, C5. C hai nhnh hp li chy qua D1 vopha A v ra pha C qua D2 v T2.Dng Id phn bnh sau:

iC1 = 2Id / 3iC 3 =iC5 = Id / 3

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Cc dng in ny s np cho cc tlm cho in p ca chng bin i mtcch tuyn tnh.

L

C 1 C 3

C 5

C 2

C 4 C 6

T 1 T 3 T 5

T 4 T 6 T 2

D 1 D 3 D 5

D 4 D 6 D 2

Za

Zb

Zc

Hnh 2.4: B nghc lu d b pT C5c np theo chiu ngc li sau mt khong thi gian ngn

dng pha a s gim v khng khi Uc1 = Uabth it D3 bt u dn in. Saukhong thi gian ngn tip theo Uc1 = -Ucmax. Nh vy khi T1 kho T3 m

dng pha a suy gim v0, dng pha b tng trng n IdNu gi t1 l thi im pht xung mT3t2 l thi im Uc1 v tr s 0t3 l thi im it D3 mt4 l thi im Uc1t gi tr max - Ucmax

Khong thi gian kho T1 l:t = t2 - t1

Khong thi gian dng in trong pha a gim v khng cn dng pha btng n Id l :

t = t4

- t3

Trong khong thi gian lc no ta cng cia + ib = iT = Id

Ti thi im t3 D3 mcn D1cng ang mnn c hai pha a v b b ningn mch hnh thnh mch vng dao ng vi tn s gc ring l

CL3

1

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Khi t = t4 dng ia = 0 cn ib = Id qu trnh chuyn dng t pha a sang pha

b kt thc. Ti thi im3

2 cp xung mT4. Khi T4 mth T2 chu in

p ngc ca C2 l - UC2 th T2 b kho li. Sau thi gian chuyn dng

t pha c vpha a (phn tch tng tphn trn nhng lc ny cc t vit nhm catt). Dng in ngun s khp mch qua cc phn t saungun qua T3 D3 Za D1 D4 T4 v cc m. Dng qua pha atng trng theo chiu m v t gi tr -Id.Ta c: i a= - Id ; ib = Id ; ic = 0

Ti thi im cho xung mT5. Khi T5 mlm cho T3 b kho li vchu in p ngcUc3. Dng iB ca pha b gim v 0 sau mt khong thigian ngn.Dng Id ca ngun lc ny chy theo mch:

Cc dng ngun T5 D5 Zc Za D4 T4 m ngun.Dng iC ca pha C tng tkhng n Id.

Ta c: i a= - Id ; ib = 0; ic = id

Mt cch tng t ti3

4 cho xung m T6 th T4 b kho nn trong

khong3

5

3

4

c : i a= 0; ib = - Id ; ic = id

Ti3

5 cho xung mT1 th T5 b kho nn trong khong 2

3

5 ta

c:

i a= id ; ib = - Id ; ic = 0Cho n thi im kt thc mt chu k = 2 cho xung mT2 th T6kho li.

Qu trnh lm vic ca nghch lu lp li nh thi ku = 0 tri.

Trn hnh 2.5trnh by ng cong dng in ba pha ia, ib, iccng nh inp trn t C1, C3, C5. T hnh v 2.5 ta nhn xt rng cc tiristor T1, T3, T5mhnh thnh na sng dng ca dng ia, ib, ic cn khi T2, T4, T6 mshnh thnh na sng m. Ti mi thi im bao gicng c hai tiristor m(mt nhm catt, mt nhm ant). Gc dn ca tiristor bng nhau v

bng 3/2 . Cc dng ia, ib, ic lch nhau gc 3/2 v c bin bng Id cangun dng. Tn s ca dng in ti c xc nh f = 1/T. iu chnh tns bng cch thay i thi gian mca tiristor nhthay i nhp xung m.

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ia

ib

ic

Uc1

Uc3

ig

Uc5

3/

3

2

3/4 3/5 2

1 2 3

T5 6 1 2

t1 t2 t3t4

4

Hnh 2.5: Cc dng sng ca b nghc lu d b p

T C1, C2, C3, C4, C5, C6 trong thc t c gi trnh nhau v c th tnhgn ng theo biu thc sau:

2

1max1

1

max]202.091.0[666.0

t

mm

m

t

m

nm LIf

UL

fU

fIC

Trong :f1r l tn snh mc ca tif1rmax l tn s cc i trong gii iu chnh mong munI1m l dng t ho ca ti(ti l ng c)Ltl in cm tn mt pha ca ti (l ng c)Uml bin ca p dy.

1.3.5 Nghc lu ngun p 3 phaBng cch thm mt nhnh vo s nghch lu ngun p cu 1 pha ta

c s nghch lu ngun p cu ba pha.S hp thnh in p ba pha hnh 2.6

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-

+

us

T1

T2

T3

T4

T5

T6

D1

D4 D2

D3 D5

D6

Ua Ub Uc

ia ib ia

id

Hnh 2.6

T1D1

T4D4

T1 D1

T6 D6 D6T6

T3D3

T5D5 T5D5

T2D2

TTPa

TTPb

TTPc

D1T1

D4T4

T3D3

D6T6

T2D2

T5D5

2

3us

Ua

Ub

Uc

UAN , UBN, UCN l in p gia cc im A,B,C v im N ca ngun E

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to UAN , UBN, UCN c dng nh hnh vngi ta cho xung kch mcc tiristor theo trnh t T1, T2,T3, T4, T5, T6, T1.....xung n cch xung kiaT/6

UAB = UANUBN ,

UBC = UBNUCN ,UCA = UCNUANi vi im O ca mch ti, ta c

UAB = VAVBUBC = VBVCUCA = VCVA

Ti u hnh Y (hnh sao). Ti im O, tng i scc dng in bng0. Nu ti 3 pha i xng ta c

VA + Vb + Vc = 0

Trong khong

cc tiristor dn dng l T1, T2 v T6 ti ZAni tip vi ZB // ZC

IA = I =

, UA =

.Z =

I B =IC = -

= -

UB = UC = -

. z = -

Dng in pha tiTi gm R,L mang tnh cm khng dng ti chm sau in p ti 1 gc

arc.tg

KT LUNTrong thc t s dng in nng ta cn thay i tn s ca ngun cung

cp, cc b bin tn c s dng rng ri trong truyn ng in, trong ccthit bt nng bng cm ng, trong thit b chiu sng... B nghch lu lb bin tn gin tip bin i mt chiu thnh xoay chiu c ng dng rtln trong thc t.

Trong cc loi nghch lu th nghch lu c lp ngun p c tm quantrng v thc thn trong cuc sng sinh hot hng ngy chnh v vy nhm

chng em quyt nh chn ti: thit k v ch to b nghch lu c lpngun p(Ura =220 v. P = 300 w)

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ng II

THIT K MCH LC

2.1.Tnh chn my bin p 1 pha tn s 50Hz

My bin th c ththay i hiu in th xoay chiu, tng th hoc h

th, u ra cho 1 hiu in thtng ng vi nhu cu s dng. My bin p

c s dng quan trng trong vic truyn ti in nng i xa. Ngoi ra cn

c cc my bin th c cng sut nhhn, my bin p (n p) dng n

nh in p trong nh, hay cc cc bin th, cc xc, ... dng cho cc thit

bin vi hiu in th nh (230 V sang 24 V, 12 V, 3 V, ...). trong mch

ny ta la chn my bin p im gia v so snh v mt kinh t v mt k

thut phng n la chn ny l ti u
http://biendt.biz/index.php?option=com_content&view=article&id=674:may-bien-ap&catid=103:may-bien-ap&Itemid=133http://biendt.biz/index.php?option=com_content&view=article&id=674:may-bien-ap&catid=103:may-bien-ap&Itemid=133http://biendt.biz/index.php?option=com_content&view=article&id=674:may-bien-ap&catid=103:may-bien-ap&Itemid=133http://biendt.biz/index.php?option=com_content&view=article&id=674:may-bien-ap&catid=103:may-bien-ap&Itemid=133http://biendt.biz/index.php?option=com_content&view=article&id=674:may-bien-ap&catid=103:may-bien-ap&Itemid=133http://biendt.biz/index.php?option=com_content&view=article&id=674:may-bien-ap&catid=103:may-bien-ap&Itemid=133http://biendt.biz/index.php?option=com_content&view=article&id=674:may-bien-ap&catid=103:may-bien-ap&Itemid=133http://biendt.biz/index.php?option=com_content&view=article&id=674:may-bien-ap&catid=103:may-bien-ap&Itemid=133http://biendt.biz/index.php?option=com_content&view=article&id=674:may-bien-ap&catid=103:may-bien-ap&Itemid=133http://biendt.biz/index.php?option=com_content&view=article&id=674:may-bien-ap&catid=103:may-bien-ap&Itemid=133

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Dng sng thc to c tn u ra

My bin p c cc thng s: U1 =Uv = 12V, Ur = 220V, f = 50HZ, S=

300VA

Dng in s cp:

I= S/U1 = 300/12 =25 (A)

Thit din day s cp A1 = 2.5/I1 = 2.5/25=0.1 (mm)

(2.5 l mt dng in )

ng knh day s cp : A1 =( d12)/4

>> d1 = = 0.279 cm chng ta chn dy 3mmLi thp S = 1.2 = 20.7 (cm2)

S vng cn qun cho cun s cp

N s = (K x U s) /S +sai s =( 45.12)/ 20.7 = 26 (vng)

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S vng bn th cp:

N th = (K x U th) /S +sai s = (45.220)/20.7 = 478 (vng)

V th ta chn my bin p c

N1=30(vng),

N2=500(vng)

Vi :N1 l s vng dy qun ca cun dy s cp

N2 l s vng dy qun ca con dy th cp

T ta tnh c:N11=N12=N1/2=30/2=15(vng)

Do my bin p im gia nn in p U1 = 2.U11 = 2.12 = 24( V )

Cng sut ca my bin p: S = .U2.I2

Trong : S l cng sut ca my bin p

U2l in p ca cun th cp my bin p

I2l dng in ca cun th cp my bin p

l hiu sut my bin p

Chn = 0,85 ta tnh c dng in th cp ca my bin p

I2 =S/(.U2 )=220.85,0

300 = 1.6 ( A )

p dng t s my bin p

1

2

2

1

I

I

U

U I1 =

1

22 .

U

IU

I1

=24

6.1.220 = 14,67( A )

Cng sut ti a m ti chiu cl:

P1 = U1 . I1cos = 24 .14,67.0.8 = 300 (W)

Mt khc ta c: Pdmax=U2I2

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Chn = 0,85 ta tnh c dng in th cp ca my bin p

(A)

p dng t s my bin p

1

2

2

1

I

I

U

U I1 =

1

22.

U

IU

Do my bin p im gia nn in p s cp c tnh bng U1 = 24(V )

Vy ta chn my bin p c cng sut Sba = 300 VA vi I = 15A

2.2.Mc ng lc

A)S nguyn l

Rt

N11

N22

1N2

i2

GND

GND

IRF 3205 IRF 3205

IRF 3205 IRF 3205D1

D3 D4

D2

+12v

Mch lc gm2 nhm van:Q1 ,Q2 v Q3,Q4 lp theo cch ghp song

song nhm gia tng h s khuch i ca mch

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B)Nguyn l hot ng

Ti thi im ban u khi cha c xung iu khin kch mcc van Q1,

Q2, Q3, Q4 dng trn cc van bng 0, in p trn van mc cao.

Khi c xung iu khin vo vo cc cng G mosfet s kch mQ1, Q2

cho dng in i t cc mng ti cc ngun( mosfet loi N) v m ngun.

Lc ny dng trn Q1, Q2 dn mc cao cn in p trn van bng 0. khi

sc dng in chy trong cun s cp ca my bin p im gia sinh ra t

thng trong li thp my bin p cun th cp sc in p vi na chu k

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dng. Khi c xung kch mcho Q3, Q4, cng tng tnh trn s c dng

chy trong cun scp theo chiu ngc li so vi khi Q1, Q2 dn. in p

trn van Q3, Q4 bng 0 mc thp xut hin dng trn van mc cao. in

p s cp bin p mc cao v ngc chiu so vi na chu ku.Dng it ngun qua s cp bin p, qua Q3, Q4 v m ngun. in p trn cun s

cp ngc chiu so vi ban u nn t thng sinh ra mc vng cuc th cp

lm in p th cp cng ngc chiu so vi na chu ku. Tn hiu u

ra 2 u s cp nhn c c 2 na chu k scho ta ngc chiu.

+)S xung iu khin n cc van

3.2. MOSFET cng sut IRF 540

a) Tnh chn MOSFET

u1=12v, u2=220v, P=300VA

Dng lm vic trn 1 van l l I=14,6A

Chn MOSFET c dng lm vic l6.0

I=25A

Vy ta chn MOSFET c dng lm vic ln hn hoc bng 25A

Nh tnh trn ta s chn MOSFET cng sut l IRF3205 v dng

Thng snh trong bng:

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Hnh dng

Transistor MOSFEST c ba cc :

D - cc mng ( drain ) : cc in tch a s t thanh bn dn chy ra

mng.

S - cc ngun ( source ) : cc in tch a s t cc ngun chy vo thanh

bn dn.

G - cc cng ( gate ) : cc iu khin

b)Thng sv c tnh

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2.3. Mch cch ly:

Ta c in p u ra ti hai chn 10 v 11 ca 4047 l 5,6V .Dng

in v in p lm vic ca IC nh, cn mch ng lc dng lm vic

ln. cch ly gia mch iu khin v mch ng lc ta s dng PC817

c cu trc v thng snh sau

Hnh dng

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in p vo lm vic ca PC ln nht l 6V v th khi ni gia u ra Q

v Q ca 4047 v PC ta ni qua in tr gy st p trn in tr. Suy ra ta

chn R= 1 k .

S lc v hot ng ca mch ny nh sau: Khi in p u ra chn

11ca 4047 l 4,8V qua diode v in trc cp vo chn 1 ca Opto

PC817 khi chn 3Opto c in p l 5,8V tc ng xung n mch lc v

lm cho ti hot ng

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N III

THIT K M IU KHIN

3.1.Thit k mc iu khin

3.1.1. Nhim v v chc nng ca mc iu khin :* Nhim v

- iu chnh c rng xung trong na chu k dng ca inp t ln colector v emitor ca van .

- To ra c xung m c bin cn thit kho van trongna chu k cn li .

- Xung iu khin phi c bin v nng lng mvkho van chc chn .

- To ra c tn s theo yu cu .- D dng lp rp, thay th khi cn thit, vn hnh tin cy, n

nh .- Cch ly vi mch ng lc

* Yu cu chung v mch iu khin l :

- Mch iu khin l khu quan trng trong h thng, n l b phn

quyt nh ch yu n cht lng v tin cy ca b bin i nn cn c

nhng yu cu sau :

+)V ln ca dng in v in p iu khin:Cc gi tr ln nht khng vt qu gi tr cho php. Gi tr nh nht

cng phi m bo c rng cung cp cho cc van mv kho an ton.

Tn tht cng sut trung bnh cc iu khin nhhn gi tr cho php .

+)Yu cu v tnh cht ca xung iu khin :

Gia cc xung mca cc cp van phi c thi gian cht, thi gian cht

ny phi ln hn hoc bng thi gian khi phc tnh cht iu khin ca van.

* Yu cu v tin cy ca mch iu khin :

Phi lm vic tin cy trong mi mi trng nh trng hp nhit thay

i , c t trung...

+)Yu cu v lp rp v vn hnh :

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S dng d dng , d thay th , lp rp . . .

3.1.2 Tnh ton mc iu khin:

to ra khi pht xung ta s dng vi mch CD4047B .Vi mch

CD4047B mc ch chnh trong mch l to dao ng(xung vung),chnh

bi xung to ra t IC ny nn IC CD4047B c chn to dao ng

trong cc mch nghch lu.

in p u ra t c dng sin hay khng l ph thuc phn ln vo

xung tc ng vo cc chn iu khin ca cc van .Cc u ra n nh c

tnh chun v i xng. V cng sut tiu th thp. Chnh nhng u im m ta chn vi mch CD4047B thit k mch iu khin cho mch nghch

lu.

a) Mch to xung

b) chn ca vi mch

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Cu trc bn trong ca vi mc n sau

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+)IC4047 c 14 chn trong :

-Chn 1,2,3 l cc chn u vo(lm thay i c xung u ra)

- Chn 5 astable cho php mch lm b to dao ng a hi qua cng 5

- Chn 6 kch hot (t mc cao)

-Chn 7 ni mass

- Chn 8 kch hot (t mc thp)

-Chn 9 l chn khi ng li(ni mas)

-Chn 10chn 11 l chn u ra

- Chn 12 cho php kch mtrli khi n l xung dng

-Chn 13 hin thsng u ra (bnh thng khng cp g)

-Chn 14 l chn cho ngun (+) vo cung cp cho ICc)Hot ng ca I n sau

Hot ng ca chn astable c php khi t u vo chn 5 mc caohoc mc thp ca chn 4 hoc ca 2 chn.

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rng ca xung vung ca Q v Q l hm ca u vo ph thuc voRC

Chn 5 astable cho php mch lm b to dao ng a hi qua cng 5.

rng xung chn 13 bng 1/2 u ra Q trong ch astable. Tuy nhiniu ny chng 50%

Trong chn nh n khi c sn dng u vo +trigger(8), khi

chn trigger(6) mc thp cc xung u vo c th thuc bt k thi im

no tng ng vi xung u ra

+)Ta tnh ton c c xung ra l 50Hz nh sau:

-Gin thi gian tin hiu vo,ra v cng thc tnh rng xung

Dng sng u ra

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Dng sng thc t

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Thay s: VDD

= 12 V

V TR = 50% V DD

Vi f = 50 (hz) T = 0.02 (s)

4.4RC = 0.02 RC = 4.55*10 3

Chn t c C = 10-4 f ,

Umax Q v Q = 10 V

V thay i tn s khi lm vic ta c th dng bin tr100k vi di

iu chnh rng khi bin tr gi tr min th tn s F gi tr max. Vi

thng s khi s dng my hin sng o c ng vi tn s 50hz tc l vi

thi gian l 0.02s= 20 ms bin trsthay i vi gi tr R= 55k. T nhnggi tr thc t khi dng my hin sng o c tn hiu u ra c tn s

50hz, vi bin l 10v trn 2 chn u ra 10, 11

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3.2. Tnh ton v thit k mch bo v p

a. Tnh ton mch phn p

Ta s dng Lm324 lm mch so snh

Cu phn p

Vss=21

2.

RR

RV

5 =21

2.12

RR

R

=> 12R2 =5(R1 +R2)

7R2 = 5R1 1.5R2 = R1T gi tr thc t ca in trTa chn R1 = 3.3k => R2 = 6.8k

b. Mch bo v hon chnh

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-Nguyn l hot ng:

IC1A bo v qu p

IC1B bo v thp p

IC2 l IC n p.

Khi qu p th led1 s bo

Khi thp p th led2 s bo

R le k ngt in khi xy s c thp p hoc qu p

-Khi cp ngun 12vdc vo th IC 7812 s c nhim vn nh ngun. sau

sa vo mch so snh. Nu ngun vo vt qu 15v th u ra Ic1a

ln mc 1 s cp ngun cho T1 hot ng.do rle k sc in s mtip

im thng ng v ngt ngun cung cp cho mch iu khin

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3.3 nguyn l

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3.3.1 Phn tch nguyn l v khi chc nng tong mch

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Khi in p 12v b bo v cp in p cho mch iu khin. Ic to

xung 4047 hot ng a ra tn hiu ti chn s 10,11.

Ti thi im ban u khi cha c xung iu khin kch mcc van Q1,

Q2, Q3, Q4 dng trn cc van bng 0, in p trn van mc cao.

Khi c xung iu khin vo vo cc cng G mosfet s kch mQ1, Q2 cho

dng in i t cc mng ti cc ngun( mosfet loi N) v m ngun. Lc

ny dng trn Q1, Q2 dn mc cao cn in p trn van bng 0. khi s

c dng in chy trong cun s cp ca my bin p im gia sinh ra t

thng trong li thp my bin p cun th cp sc in p vi na chu k

dng. Khi c xung kch mcho Q3, Q4, cng tng tnh trn s c dng

chy trong cun s cp theo chiu ngc li so vi khi Q1, Q2 dn. in ptrn van Q3, Q4 bng 0 mc thp xut hin dng trn van mc cao. in

p s cp bin p mc cao v ngc chiu so vi na chu ku.Dng i

t ngun qua s cp bin p, qua Q3, Q4 v m ngun. in p trn cun s

cp ngc chiu so vi ban u nn t thng sinh ra mc vng cuc th cp

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lm in p th cp cng ngc chiu so vi na chu ku. Tn hiu u

ra 2 u s cp nhn c c 2 na chu k scho ta ngc chiu.

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KT LUN

Sau qu trnh thc hin bn n chng em thu c mt s kt qu

nh sau:+ Gii thiu mt sng dng v c im ca mch nghch lu mt pha+ Phn tch nguyn l lm vic v cc thng s trong mch nghch lu

mt+ Gii thiu mt sng dng v c im ca mch nghch lu mt pha.+ c bit l chng em hon thin sn phm ca mnh theo yu cu

t ra c th:* in p ra dng cho ti U = 220 V

Tuy nhin bn n ca chng em cn c nhc im l cha n nh

c in p ra khi ti ln.Vi s c gng n lc ca mi thnh vin trong nhm chng em honthnh n ca mnh theo ng thi gian. Mt ln na chng em xin gi licm n ti cc thy c trong khoa in - in T, c bit l thy trc tiphng dn chng em trong vic hon thnh n. Chng em rt mong nhnc nhng kin nhn xt, gp ca cc thy c v cc bn bn nca cng em hon thin hn.

Chng em xin chn thnh cm n!

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ng IGii thiu tng quan v nghc lu1.1 Tng quan v nghch lu ............................................................81.1.1 Nghch lu ph thuc 1 pha 2 na chu k....................................8

1.1.2 Nghch lu ph thuc 3 pha na chu k.......................................111.1.3 Nghch lu ph thuc cu 3 pha....................................................141.2 Nghch lu c lp ........................................................................151.2.1 Nghch lu c lp ngun p.........................................................161.2.2 Nghch lu c lp ngun dng.....................................................271.2.3 Nghch lu ngun dng 3 pha........................................................301.2.4 Nghch lu ngun p 3 pha............................................................33

ng IIThit k mch lc2.1 Tnh ton My Bin p...................................................................362.2 Mch lc..........................................................................................392.2.1 Chn van khuch i cng sut....................................................4122.2.mch cch ly..................................................................................44

ng IIIThit k mc iu khin

3.1 thit k mch iu khin.................................................................46

3.1.1 Tnh chn gi tr phn t..............................................................473.2 Thit k mch bo v qu dng qu p...........................................533.3 Thit ks nguyn l................................................................563.3.1 Phn tch nguyn l v khi chc nng trong mch .....................57

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