Chuong5 - chinh luu khong dieu khien 1 pha

8/3/2019 Chuong5 - chinh luu khong dieu khien 1 pha

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CHNG5: CHNH LU KHONG IEU KHIEN

Phan e cng :5.1 Gii thieu5.2 Chnh lu na song ( chnh lu mot xung)5.2.1 Vi mot tai ien tr5.2.2 Vi mot tai cam ng (RL)5.2.3 Vi mot tai cam ng va mot iot (FWD)5.3 Chnh lu bien ap co iem gia toan song ( chnh lu hai xung)5.3.1 vi mot tai ien tr5.3.2 Vi mot tai cam ng (RL)5.4 Chnh lu cau toan song.5.4.1 Vi mot tai ien tr5.4.2 Vi mot tai cam ng (RL)5.5 Thao luan5.6 Cac cong thc

Muc ch hoc tap Mo ta qua trnh hoat ong dang song cua chnh lu khong ieu khi

vi tai ien tr va tai cam ng Giai thch y ngha cua diot FWD Giai thch s khac nhau gia chnh lu na song va toan song Mo ta hoat ong dang song cua mot chnh lu khong ieu khien t

cac tai la ien tr va cam ng Thao luan u iem cua chnh lu toan song so vi chnh lu na so


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5.1 Gii thieuS chnh lu la qua trnh bien oi ien the hay dong ien xoay chie

chieu. Chnh lu khong ieu khien ch s dung cac diot nh nhng chnh lu. ien ap ra mot chieu DC la on nh co o ln bang biecua ien ap vao xoay chieu AC. Tuy nhien dong ra DC la khong thungha no bao gom cac thanh phan xoay chieu AC c goi la gn sotr s gn song nay, mot s loc c a vao sau chnh lu

Trong chng nay, chung ta se hoc dong mot pha, day chnh lu khkhien t chnh lu na song n gian dung mot diot n en chnhsong phc tap dung mot vai diot. Chung ta se gia s rang cac diotac iem tieu chuan nh nhau tren hnh 2.3

5.2 Chnh lu na song ( chnh lu mot xung)5.2.1 Vi mot tai ien tr :

Chnh lu na song n gian mach ien ch mot tai ien tr trnh b5.1a. ien ap nguon la mot song hnh sin vi gia tr cc ai Vm va chu ky T .Trong na chu ky dng, khi o ien ap tai cc dng la dng so vdiot v tr m, dong ien nay chay qua tai ien tr R. ien ap tao ) theo sauna song sin dng. Trong luc na chu ky am, ien ap cc dng tso vi cc am , diot v tr ong. Khi o khong co dong ien chay qra (vo ) trnh bay tren hnh 5.1b, cung trnh bay dong ien tai.

Chnh lu na song nh vay a thay oi nguon ien AC thanh DC. dao ong mot chieu DC bao gom gn song rong. V vay, dong ien mco gii han gia tr thc te khi s dung cho cong suat cao.

Con so chu y trong dong ien nay la trung bnh hay ien ap tai vtai

ien ap tai trung bnh cho bi :

Vo(avg)= 2Vs

= V m / = 0,318 V u1 5.1

ay :Vs = gia tr hieu dung RMS cua ien ap cung capVm = Gia tr cc ai cua ien ap cung cap


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Hnh 5.1 Chnh lu na song vi tai ien tr (a) cac dang song cua dong ien

Dang song cua dong ien co hnh dang giong nh dang song ien ap, nhong dang tren cong thc 5.1 cho ra dong ien tai trung bnh :

I0 (avg.) = Im / = 0,318 I m 5.2

ay Im = Vm / RDong ien hieu dung cho bi :

IRMS = Im / 2 5.3Ket qua tren phng trnh 5.3 la ac thu cua dong ien na song dang sDang song cua ien ap diot ( hnh 5.1b ) cho thay rang diot phai co khac mot ien ap oi chieu bang nh chop cua ien ap nguon.


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ien ap nay c s dung e chon diot thch hp tren mot dong ien a( hay PRV) oi vi diot c cho bi :

Rai PIV : Vm 5.4Muc ch cua chnh lu la bien oi nguon AC thanh nguon DC. T o tng mot thiet b hoan hao, khong bi mat nang lng khi chnh lu, doxoay chieu AC au vao vi chu ky u phai bang nang lng dong ienau ra. Dong ien mot chieu DC ( trung bnh) ra tren tai c cho bi :

P0 (avg.) = V0 (avg.)* I0 (avg.)= mV

* m

I

= 2**

m mV V R

=2

2 *mV

R 5.5

Dong ien AC au vao cho bi :

PAC = VRMS * IRMS VRMS oi vi mot giai oan t ) ti ( na chu ky ) la Vm / 2, Bi v :

PAC =2mV *

2m I =

*4*m mV V

R=

2

4mV R

5.6

Hieu qua chnh lu c xac nh qua ty so cong suat DC au ra va covao

0( .)avg

AC

P

P = 5.7

Dang thc te o c gan ung hnh dang cua ien ap au ra. Mot caien ap DC au ra cua mot may chnh lu la hang so. Trong thc te, clu cung cap dong ra la nhng dang song hnh sin thieu. Dang thc te qua ty so RMS ien ap ra vi gia tr trung bnh cua ien ap ra :

Dang thc te (FF) =( .)

oRMS

o avg

V V

5.8

Gia tr ly tng cua FF la ong nhat. FF la ong nhat neu ien ap ra lhang so, e ma VoRMS =Vo (avg.)ien ap ra cua mot may chnh lu bao gom ca hai DC va AC ( gn songTan so va o ln cua ien ap gn song la mot nhan to quan trong t

may chnh lu. Dong gn song tan so cao hn va bien o nho hn, dehn trong gii han chap nhan cSo xung la ty so gia tan so gn song c ban cua ien ap DC vi tacung cap

So xung =tantansogonsongcoban

songuonAC 5.9


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Thanh phan gn song la ty so gia gia tr RMS cua thanh phan AC v

He so gn song =oAC oDC

I I

Theo ly tng thanh phan gn song nen la 0. Thanh phan gn song cbi lan au tm ra cong suat a tieu hao trong tai ien tr R

PL = 2 2 2 RMS DC AC I R I R I R= + Hay : 2 2 2 AC RMS DC I I I =

2 2 2

2 2 2

22

2

2

2

1

1

AC RMS DC

DC DC DC

RMS

DC

RMS

DC

I I I I I I

I RF

I I

RF I

=

=

=

5.10

Vi du 5.1 :Mot may chnh lu na song tren hnh 5.1a c noi vi nguon ien ACien tr la 100 , tm

a) ien ap tai cc ai

b) ien ap tai trung bnhc) Dong ien tai cc aid) Dong ien tai trung bnhe) Dong ien tai RMSf) Rai PIV cho diotg) Cong suat ra DC (trung bnh)h) Cong suat vao ACi) Tan so may chnh lu j) Dang thanh phank) So xungl) Thanh phan gn songm) Goc o dan


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Giai ap :a) ien ap tai cc aiVm = 2 . 50 = 70,7 Vb) ien ap tai trung bnhVo(avg) = 0,318 . Vm

= 0,318 * 70,7 = 22,5 Vc) Dong tai cc aiIm = Vm / R = 70,7 / 100 = 707 mAd) Dong tai trung bnh : Io(avg) = 0,318 * 0,707 = 225 mAe) Dong tai RMS : IoRMS = Im / 2 = 0,707 / 2 = 353,5 mAf) Rai PIV cua diot : PIV Vm = 70,7 V

g) Cong suat dong DC ra :2 2

2 2

70,75,1

* *100m

DC

V P W

R = = =

h) Cong suat dong AC vao :2 270,7

12,54 4*100

m AC

V P W

R= = =

i) Hieu suat chnh lu: 24

0, 405 40, 5% DC AC

pP

= = = =

j) He so nh dang:( )

/ 2 / 2 1,57

/ oRMS m

o avg m

V V FF

V V

= = = =

k) So xung : p =0

0

tan 60 3601

tan 60 360sogonsongcoban

songuonAC = = =

l) He so gn song :2 2

22 2

( / 2)1 1 ( / 2) 1 1, 21

( / ) RMS m

DC m

I I RF

I I

= = = =

m) Goc o dan : = 180 0

V du 5.2Trong hnh 5.1a, ien ap nguon la 120V vi tai ien tr la 10 . Tma) Dong tai cc aib) ien ap tai trung bnhc) Dong tai trung bnhd) Dong tai RMSe) Cong suat at tren taif) He so cong suatg) Rai PIV cua diot


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Giai ap :Goi gia tr ien ap nguon la

Vm = 2 V RMS = 1,414 * 120 = 169,7 Va) Dong tai cc ai: Im = Vm / R = 169,7 / 10 = 16,97 Ab) ien ap tai trung bnh : V0(avg) = 0,318 Vm = 0,318 * 169,7 = 54,0 Vc) Dong tai trung bnh : I0(avg) = 0,318 Im = 0,318 * 16,97 = 5,4 Ad) Dong tai RMS : IRMS = Im / 2 = 16,97 / 2 = 8,49 Ae) Cong suat at len tai : P =2 * RMS I R = 8,49 2 * 10 = 720 Wf ) Cong suat bieu kien : S = VRMS * IRMS = 120 * 8,49 = 1018,8 VA

He so cong suat : 720 0,7071018,8

PS

= = o&

g) Rai PIV cua diot : PIV Vm = 169,7 V

5.2.2 vi mot tai cam ng (RL)Dong ien na song vi mot tai cam ng (RL) , thc te trnh b5.2a. Chung ta hay phan tch hoat ong cua mach ien nay

a) Nh trong trng hp cua tai ien tr, diot v tr m, trong khi cua no tr nen dng so vi cc am. ien ap qua tai giong nhdng cua nguon AC.

b) Trong thi gian nay, nang lng truyen tai t nguon AC va la ntch trong vung t trng ngoai vi phan cam.

c) Dong ien i qua bo phan cam khong the thay oi ngay lap tdong ien tang them dan dan cho ti khi no at en gia tr cNen nh rang dong ien khong at en nh chot cua no khicc ai. ieu nay la phu hp vi thc te dong ien o chay qtre vi ien ap.

d) Khi ien ap nguon giam i, dong ien bat au giam dan dan,khi tat ca nang lng d tr bi phan cam c phong thch tVay th dong ien tai tat han oi vi na nho hn toan bo giai

e) cung mot thi gian, t trng gap lien ket vi phan cam vaien ap oi ngc vi s giam cua ien ap cung cap

f) Dong ien cang sm bang 0 , diot la doc ngc. Diot luc o gitr tat oi vi s ngh cua chu ky am. Hnh 5.2b cho thay dang soTrong khoang t 0 en /2 , ien ap nguon vs tang dan t 0 en cc ai

dng cua no, trong khi ien ap cam ng tren phan cam vL ngc vi s thayoi cua dong ien qua tai. Trong khoang thi gian /2 en , ien ap nguongiam dan t gia tr cc ai dng en 0 . Cung thi gian nay, ienco oi chieu phan cc ngc vi s giam cua dong ien, o ladiot trc dong ien.


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Hnh 5.2:chnh lu na song vi tai cam ng (a) bieu o dong ien (b) daien ap va dong ien

Tai , ien ap nguon oi chieu va tang dan en gia tr cTuy nhien ien ap cam ng ngang qua phan cam van con dng vaoi chieu dan cua diot cho en khi ien ap cam ng giam en 0, can. Nen nh rang mac du diot la doc ngc , co mot dong iencho en khi goc = + c at en. o la ket qua cua nang lntr trong t trng la c tr ve nguon.

Dang song cua ien ap ngang qua ien tr cung giong nh dcua io(vR = io R). S khac nhau gia vR va vo la ien ap ngang qua phan cam (

xem hnh 5.2b)g) Neu chung ta tang tai cam ng L, diot se dan dong ien nhiechu ky.

Neu chung ta ap dung L e la vo han, dong ien i qua diot se cobang phang va v vay lien tuc. Trong tnh trang nay, diot se co the la chap ngang qua diot se co the bang 0 va gia tr cua vs va vo se co the bang nhau.


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Dong ien se co the khong hieu chnh dai hn, nh vay dong ixoay chieu. ieu o khong thc hien c, bi v, dong ien au ra phagian t hn na mot chu ky. ien ap ra trung bnh v vay phu thuoc s liecam va ien tr cua tai. Dang song trnh bay tren hnh 5.2b la ve cho ien ap trung bnh ra la khac 0 va dong ien do o la khong lien tuc.

Gia tr trung bnh cua ien ap tai c cho bi :

0( ) (1 cos )2m

avg

V V

= 5.11

= + la goc dan ( phu thuoc vao gia tr cua R va L)0

0( )

00( )

180 , 0

360 , 0

avg

avg

V

V

Tai cam ng giam ien ap tai trung bnh.Nh rang neu = 180 0 th V 0(avg) = V m / ( cong thc 5.1)

0( ) (1 cos )2m

avg

V I R

= 5.12

Ap dung oi vi cong suat cao, dong ien mot xung co gii han thc tra thap va gn song rong trong ien ap DC ra.

5.2.3 Vi mot tai cam ng va diot lp FWDDong ien tren hnh 5.2 co the thay oi ra ket qua dong ien trn

hnh 5.3, ma thc te s dung cong suat thap. No van con la chnh ludong ien tai co the chay trong toan bo chu ky

iot th hai D2 , c mac song song vi tai, c biet nh la diot lp FDiot nay ngan nga ien ap am xuat hien tren tai, ket qua la no latrung bnh V0(avg)tren tai, cung nh vay vi dong ien trung bnh. Trong nam cua ien ap cung cap, FWD dan va chuan b phan xoay chieu oi vTrong khoang thi gian nay FWD dan, diot chnh D1 la doc ngc va ngng dan , vnh the dong ien nguon la bang 0 i vi na giai oan nay. FWD giudong ien tai i ti 0, va nh vay giam bout gn song.

Dang song cua ien ap tai la giong nh trong dong na song vtr. ien ap tai trung bnh la Vm / , cho bi cong thc 5.1

Dang song cua ien ap va dong ien trnh bay trong hnh ve na

cam ng ln. Tai cam ng cang ln, dong ien tai cang tr nen ban


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Hnh 5.3:tai cam ng vi mot diot FWD (a) , s o dong ien (b) kieu dang soncam ng ln

V du 5.3: Trong dong ien trnh bay tren hnh 5.3, Vs = 240 V va R = 10 .Neu tai ien cam la ln, tm

a) ien ap tai trung bnhb) Dong ien tai trung bnhc) Gia tr RMS cua dong ien taid) Cong suat tieu thu tren taie) He so gn song

f) He so cong suatGiai ap:Vm = 2 ( 240 ) = 339,4 V

a) ien ap tai trung bnh : V0(avg) = 0,318 * 339,4 = 108 V


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b) Dong ien tai trung bnh : I0(avg) = 108 / 10 = 10,8 Ac) Theo gia tr ln cua tai cam ng , dong ien tai gan nh la khong cua dong ien tai v vay bang gia tr trung bnh cua no.d) Cong suat tieu thu tren tai :2 2(10,8) *10 1166, 4 L RMSP I R W = = =

e) He so gn song :2 2(339, 4 / 2)

1 1 1, 211108

RMS

DC

V RF

V

= = =

f)He so cong suat :1166,4

0,45240*10,8

PS

= =

5.3 Chnh lu bien ap toan song co iem gia ( chnh lu hai xung)

Mot pha n, chnh lu na song la rat khong thc te , ien ap ra trungthap, hieu suat kem, va he so gn song cao. Nhng nhc iem nay cobang chnh lu toan song. Chnh lu toan song thng c s dung hnsong. Dong ien va ien ap trung bnh cao hn, hieu suat cao hn, va giasong

5.3.1 Vi mot tai ien tr :Hnh 5.4(a) trnh bay gian o cua chnh lu toan song dung mot may biethou cap co iem gia. ien ap nguon va tai ien tr la nh nhau trona song. Trong chu ky na dng ( hnh 5.4b) diot D1 dan va diot D2 la doc ngc.Dong ien chay qua tai la nguyen nhan ri giam ien dng

Hnh 5.4Chnh lu toan song iem gia (a) dong ien (b) ng lng dong ien tky dng (c) ng lng dong ien trong na chu ky am


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Hnh 5.4(d) Dang song cua dong ien va ien ap


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Trong na chu ky am (hnh 5.4c) diot D2 dan va D1 la ong. Dong ien chayqua R, duy tr cung nh phan cc oi vi ien ap ngang qua tai (xem hnh 5v dang song cua ien ap tai khong thay oi tot hn na chu ky song sgia tr trung bnh cao hn va he so gn song cao hn.

Gia tr trung bnh va RMS giong mot chut vi trng hp na songV0(avg) = (2 Vm ) / = 0,636 V m Nh rang trung bnh toan song bang hai lan trung bnh na song, i

hien nhien bi kiem tra hai o th ien ap au. Tng t ien ap tai trcho bi :

0,636( ) )(2 / 0,636

mV o avg m m R I I I = = = 5.14

Dong ien ra RMS cho bi :( ) 2

0,707.m I o avg m I I = = 5.15

o th cua ien ap ngang qua diot tren hnh 5.4d cho thay rang mophai khong co ien ap ngc bang 2 Vm. Rai PIV oi vi diot c s dung trodong ien nay c tnh bi :

Rai PIV cho diot 2 V m 5.16 Dong ien trung bnh qua diot la :

ID1(avg) = ID2(avg) = I0(avg)/ 2 = Im / 5.17Dong ien RMS qua diot la :

IDRMS = Im / 2 5.18 Trung bnh hay cong suat phan phoi tren tai c cho bi :

P0(avg) = V0(AVG)* I0(AVG)

=2

2 2

2 2 4 * 4*

* *

m m m m mV I V V V

R R

= = 5.19

Cong suat vao AC c cho bi :PAC = VRMS* IRMS

=2*

*2* 2*2 2

m m m m mV I V V V R R

= = 5.20


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V du 5.4:Chnh lu toan song trnh bay tren hnh 5.4a, c cung cap t nguon 12Neu tai ien tr la 10 . Tm

a) Dong ien tai cc aib) ien ap tai trung bnhc) Dong ien tai trung bnhd) Dong ien trung bnh diote) Gia tr RMS cua donh ien taif) Cong suat tren taig) Dai PIV cua dioth) Tan so gn song

Giai ap :ien ap nh : Vm = 2 (120) = 169,7 V

a) Dong ien tai cc ai

Im = Vm / R = 169,7 / 10 = 16,97 Ab) ien ap tai trung bnhV0(avg) = 0,636 Vm = 0,636 * 169,7 = 108 V

c) Dong ien tai trung bnhI0(avg) = 0,636 * 16,97 = 10,8 A

d) Dong ien trung bnh diotID1(avg) = ID2(avg) = I0(avg)/ 2 = 10,8 / 2 = 5,4 A

e) Gia tr RMS cua dong ien taiI0RMS = Im / 2 = 16,97 / 1,414 = 12 A

f) Cong suat tren taiI2 R = 122 * 10 = 1440 W

g) Dai PIV cua diot = 2 Vm = 339,4 Vh) T khi hai chu ky dong ra xuat hien oi vi moi mot chu ky cua d

Tan so gn song = 2 * tan so dong vao = 2 * 60 = 120 Hz--------------------------------------------------------------------------------------------------------

V du 5.5:Chnh lu toan song trnh bay tren hnh 5.4a, c cung cap t nguon 12Neu tai ien tr la 10 . Tm

a) ien ap tai trung bnhb) Dong ien tai cc aic) Dong ien tai trung bnhd) Gia tr RMS cua donh ien taie) Rai PIV oi vi diotf) Cong suat trung bnh phan phoi tren taig) Cong suat vao ACh) Hieu suat cua may chnh lu


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i) He so dang j) So xungk) He so gn songl) Goc danm) He so cong suat

Giai ap :ien ap nh : Vm = 2 (50) = 70,7 V

a) ien ap tai trung bnh : V0(avg) = 0,636 Vm = 0,636 * 70,7 = 45 Vb) Dong tai cc ai : Im = Vm / R = 70,7 / 100 = 707 mAc) Dong tai trungbnh : I0(avg) = 0,636 * 0,707 = 450 mAd) Dong tai RMS: I0RMS = Im / 2 = 0,707 / 1,414 = 500 mA e) Rai PIV cho diot : PIV 2 V m = 141,4 V f) Cong suat trung bnh phan phoi tren tai:

P0(avg) =2 2

2 2

4 4(70, 7)20,25

* *100

mV W R

= =

g) Cong suat vao AC :2 270,7

252 2*100

m AC

V P W

R= = =

h) Hieu suat chnh lu :2

2 2 2

2 84 * 0,81 81%

*avg m

AC m

P V RP R V

= = = = =

i) He so hnh dang : FF =0( )0( )

501,11

45 RMS

avg

V

V = =

j) So xung :p = 0 0tan . . . . . 120 / 60 360 /180 2

tan . . .

so gon song co ban

so nguon AC = = =

k) He so gn song :2 2

2 2

0,51 1 0, 484

0,45 RMS

DC

I RF

I = = =

l) Goc dan : = 180 0

m) He so cong suat : 0( )0( )

( / )( / 2)2 / 0,64

( / 2 2)( / 2)m avg

m avg

V I PF

V I

= = =


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5.3.2 Vi mot tai cam ng RLThem mot bo phan cam ng cung vi mot tai ien ch lam thay oien ap va dong ien, nh trnh bay phan 5.2.2. Nen nh rang dong chay mot giai oan cua thi gian sau khi diot doc ngc, va ket qua la lao cua ien ap rat rung bnh.

Hnh 5.5 trnh bay chnh lu toan song co iem gia vi mot tai dang song cua dong ien va ien ap tong hp cua no. Nh co the tha

Hnh 5.5 :Chnh lu co iem gia vi mot tai cam ng (a) s o dong ien (b) dong ien va ien ap


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Dong ien tai at cc ai khi ien ap nguon Vs bang 0. Khi Vs tang dan o lntrong khoang thi gian t ) en / 2 , khi Vs at en cc ai , dong ien tai la nnhat. Trong khoang thi gian gia /2 va , nay ien ap nguon giam dan o lien ap cam ng ngang qua phan cam oi ngc vi s giam dan cubi c them ien ap nguon. V vay dong ien tai tang dan en mokhi vs = 0 . Qua trnh tiep tuc vi moi na chu ky cua chnh lu song sintai khong bao gi giam en 0 t luc nang lc d tr trong t trng dien chay

Cong thc la nh nhau oi vi chnh lu iem gia vi mot tai itrung bnh cua ien ap tai la :

V0(avg) = (2 Vm) / = 0,636 V m 5.21

Hnh 5.6

Dang song cua dong ien oi vi mot tai cam ng cao


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Gia tr trung bnh cua dong ien tai la :

0( )

20,636m mavg

V V I

R R = = 5.22

Neu tai cam ng la u ln, dong ien tai gan nh khong oi, giong nhnh 5.6

Gia tr RMS cua dong ien tai la :I0(RMS) = I0(avg) = V0(avg) / R 5.23

ID(RMS) = 0( )2avg I 5.24

V du 5.6 :Chnh lu toan song trnh bay tren hnh 5.5a, nguon ien cung ng la

tai ien tr la 100 , tm :

a)

ien ap ra DCb) Dong ien tai trung bnhc) Cong suat phan phoi tren taid) Cong suat vao ACe) Hieu suat chnh luf) He so gn songg) He so hnh dang

Giai ap :ien ap nh la : Vm = 2 (115) = 162,6 V

a) ien ap ra DC la :V0(avg) = 0,636 Vm = 0,636 * 162,6 = 103,4 V

b) Dong ien tai trung bnh la :

I0(avg) = 103,4 / 100 = 1,03 A

c) Cong suat phan phoi tren tai :PL = V0(avg) * I0(avg) = 103,4 * 1,03 = 107 Wd) Cong suat au vao AC:

PAC = VRMS* IRMS =2 2162,6

132,22 2*100

mV W R

= =

e) Hieu suat chnh lu :107

0,81 81%132,2

L

AC

PP

= = = =


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f) He so gn song :2 2

1151 1 0, 48

103,4 RMS

dc

V RF

V

= = =

g) He so hnh dang:FF =

0( )

115 1,11103,4

RMS

avg

V V

= =

5.4 Chnh lu cau toan song5.4.1 Vi mot tai ien tr

Chnh lu toan song cung co the at c bi chnh lu cau nh thnh 5.7. Chnh lu toan song cau nay s dung bon diot. Trong na chcua ien ap nguon (hnh 5.8a, diot D2 va D3 la doc ngc do vay co the thay th

bang mot nut ong. Dong ien tai chay qua trong giai oan nay i qua2 va tai Rsau o i qua D3 va tr ve nguon. Nguyen do nay la mot ien ap dngqua R

Hnh 5.7Mach ien chnh lu cau toan song

Hnh 5.8b cho thay dong ien cau toan song trong na chu ky am cnguon. Bay gi diot D1 va D4 la doc ngc va v the co the c thay the banut ong. Dong ien tai bay gi i ngang qua D4 va qua R , sau o qua D1 en nguon.ng i cua dong ien qua R la co hng cung nh trc, bi v co sdng ngang qua R trong ca hai na chu ky . Nh vay, chnh lu cau toan sch la dong ien tai chay trong ca hai na chu ky. Hnh 5.9 trnh bay song thch hp.

Gia tr trung bnh va RMS cua ien ap va dong ien la tng t nhhp chnh lu toan song co iem gia. Tuy nhien dang song cua ien diot trenhnh 5.9 cho thay rang moi mot diot phai chu c mot ien apVm


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Hnh 5.8Chnh lu cau (a) na chu ky dng (b) na chu ky am.

Hnh 5.9Nhng dang song cua chnh lu cau


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Rai PIV oi vi diot Vm Bi v co hai ng dan cho dong ien tai, dong ien diot trung bnhna cua dong ien tai trung bnhID(avg) = Iavg / 2

V du 5.7:Chnh lu cau toan song tren hnh 5.7, c cung cap bi nguon 120V . Ntai la 10,8 , tm :

a) ien ap tai nhb) ien ap DC ngang qua taic) Dong ien tai DCd) Dong ien trung bnh trong moi mot diote) Cong suat ra trung bnh

f) Hieu suat chnh lug) He so gn songh) He so cong suat

Giai ap :a) ien ap nh :Vm = 2 VRMS = 1,414 * 120 = 170 Vb) ien ap DC ngang qua tai :V0(avg) = 0,636 * 170 = 108 Vc) Dong ien tai DCI0(avg) = 108 / 10,8 = 10 Ad) dong ien trung bnh trong moi diot : t luc cac diot mang dong ta

na chu ky xoay chieuID(avg) = I0(avg)/ 2 = 10 / 2 = 5 Ae) Cong suat ra trung bnh :P0(avg) = V0(avg) * I0(avg) = 108 * 10 = 1080 Wf) Hieu suat chnh lu :

28 / = = 0,81 = 81%g) He so dang :

2 2

1201 1 0, 482108

RMS

DC

V RF V = = =

h) He so cong suat

PF = 0( ) 0( )* 108*10 0,9* 120*10

avg avg

RMS RMS

V I PS V I

= = =


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V du 5.8:Chnh lu cau toan song tren hnh 5.7, c cung cap bi nguon ien 1Neu tai ien tr la 10 , tm :

a) ien ap tai trung bnhb) Dong ien tai cc aic) Dong ien tai trung bnhd) Dong ien tai RMSe) Cong suat tren taif) Rai PIV cho cac diotg) Dong ien diot trung bnhh) Tan so gn song

Giai ap :ien ap tai nh la : Vm = 2 VRMS = 1,414 * 120 = 170 Va) ien ap tai trung bnh : V0(avg) = 0,636 * 170 = 108 V

b) Dong ien tai cc ai : Im = 170 1710mV A

R= =

c) Dong ien tai trung bnh : I0(avg) = 0,636 * 17 = 10,8 Ad) Dong ien tai RMS : IRMS = 17 2 = 12 Ae) Cong suat tren tai : P =2 212 *10 1440 RMS I R W = = f) Rai PIV cho cac diot : PIV Vm = 170 Vg) Dong diot trung bnh : ID(vag) = I0(avg)/2 = 10,8 / 2 = 5,4 Ah) T hai chu ky cua au ra xay ra cho moi mot chu ky cua au vao,Tan so gn song = 2 * tan so cua AC vao = 2 * 60 = 120 Hz

V du 5.9:Chnh lu cau toan song tren hnh 5.7, c cung cap bi mot nguon 50tr tai la 100 , tm :

a) ien ap trung bnh ngang qua taib) Dong ien tai trung bnhc) Dong ien RMSd) So xunge) Goc dan

f) Rai PIV cho diotg) He so hnh dangh) He so gn song

Giai ap:ien ap tai nh la : Vm = 2 VRMS = 1,414 * 50 = 70,7 Va) ien ap trung bnh ngang qua tai : V0(avg) = 0,636 * 70,7 = 45 Vb) Dong ien tai trung bnh : I0(avg) = 45 / 100 = 0,45 A


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c) Dong RMS: IRMS = 0,52 2m m I V A

R= =

d) So xung : p = 2e) Goc dan : = 1800 f) Rai PIV : PIV = Vm = 70,7 Vg) He so hnh dang : FF =

0( )

501,11

45 RMS

avg

V V

= =

h) He so gn song : RF =2 2

501 1 0, 483

45 RMS

DC

V V

= =

5.4.2 Vi mot tai ien cam RLThem mot cuon cam cung loai vi mot ien tr lam thay oi d

ien ap va dong ien. Hnh 5.10 cho thay mot chnh lu cau vi mot Hay them mot cuon cam L o chng bang R . Dong ien tai khong na song sin, nhng dong ien trung bnh con cc tnh nh cong thc 5.22

I0(avg) = 2 2m m I V R

=

ng dong ien khong la ham so sin ma gan nh la mot son5.11 trnh bay dang song cua ien ap va dong ien.Hnh 5.10Chnh lu cau dong ien (s o xoay chieu) vi tai cam ng

Neu chung ta tang dan tai cam ng cho en khi no ln hn R , gqua tai tr nen nho. Neu chung ta them vo han tai cam ng , dong khong oi va cach chay cua dong ien nh nhng dang song tren hnh2 va D3 dan dong tai khong oi tren na chu ky dng, trong diot D1 va D4 lam viec nh vay tren na chu ky am. Dong ien nguon c cho bi :

Is = i3 - i1 = i2 - i4 5.27


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Dau cho khong phai la mot song sin, dong nguon AC la mot dang songhnh dang ch nhat. Tai la luon luon noi tieo vi nguon, nhng s nongc tren na chu ky xoay chieu.

ien ap ra V0 la mot dang song chnh lu toan song. Gia tr trung bnco the c xac nh t :

V0(avg) = VL(avg) + VR(avg) 5.28Hnh 5.11:Dang song oi vi hnh 5.10 ( L=R)

Hnh 5.12:Dang song oi vi hnh 5.10 ( L? R )


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ay VR la ien ap ngang qua ien tr va VL la ien ap ngang qua phancam. Trong qua trnh hoat ong VL(avg)ien ap trung bnh ngang qua phan cam , phabang 0, v vay :

VR(avg) = V0(avg) = (2Vm) / = 0,636 Vm 5.29ien ap tai trung bnh la giong nh oi vi trng hp ien tr. Dongbnh co the c xac nh t :

I0(avg) = VR(avg) / R = 0,636 ( Vm / R ) 5.30T nay dong ien tai bay gi ve ban chat la khong oi, gia tr RMS, cbnh cua no la nh nhau :

IRMS = I0(max) = I0(avg) 5.31T nay cac diot trong cau dan tren na chu ky xoay chieu, dong ien trunmoi mot diot la :

ID(avg) = I0(avg) / 2 5.32Va dong ien RMS trong moi mot diot la :

ID(RMS) = 0( )2avg I 5.33

V du 5.10:Mot chnh lu cau toan song vi mot tai RL noi tiep vi mot nguonien tr la 10 va L? R, tm :

a) ien ap tai trung bnhb) Dong ien tai trung bnhc) Dong ien tai cc aid) Gia tr RMS cua dong taie) Dong trung bnh tren moi diotf) Dong RMS trong moi diotg) Cong suat at tren tai

Giai ap :ien ap tai nh la Vm = 2 VRMS = 1,414 * 120 = 170 Va) Dong ien tai trung bnh : V0(avg) = 0,636 * 170 = 108 Vb) Dong ien tai trung bnh V0(avg) / R = 108 / 10 = 10,8 Ac) Dong tai cc ai = dong tai trung bnh = 10,8 Ad) Gia tr RMS dong tai = dong tai trung bnh = 10,8 A

e)

Dong trung bnh trong moi diot : ID(avg) = I0(avg) / 2 = 10,8 / 2 = 5,4 Af) Dong RMS trong moi diot id(RMS) = 0( ) 10,8 7,6

2 2avg I A= =

g) Cong suat tren tai : 2 210, 8 *10 1167 RMS I R W = =


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5.5 thao luan :5.1 nh ngha chnh lu5.2 Phac hoc s o mot chnh lu na song va giai thch hoat ong 5.3 Mot may chnh lu na song co ien ap vao trung bnh 120rai PIV yeu cau cho diot5.4 Mot chnh lu na song co ien ap nguon 120 V. Neu ien tr tm :a) ien ap tai trung bnhb) Dong diot cc aic) Dong diot trung bnhd) Rai PIV cho diote) cong suat tai trung bnhf) he so gn songg) he so dang

5.5 Chnh lu na song vi mot tai cam ng va FWD co ien amot tai ien tr 5 va mot tai cam ng 20 mH. Tma) ien ap tai trung bnhb) Dong tai trung bnhc) Dai ien ap va dong diot5.6 u iem chnh cua chnh lu toan song so vi chnh lu na son5.7 Trong mot chnh lu toan song co iem gia, Vs = 208 V , va ien tr taiR = 100 , tm :a) ien ap tai cc aib) ien ap tai trung bnh

c) Dong diot cc aid) Dong diot trung bnhe) Rai PIV cho diotf) Gia tr RMS cua ien ap rag) He so gn song5.8 Bien ap cung cap cho chnh lu toan song co iem gia t moV . Neu ien tr tai R = 500 va L = 1 H. Tm :a) ien ap tai trung bnhb) ien ap tai trung bnhc) Dong diot cc aid) Cong suat phan phoi tren taie) Rai PIV cho diot5.9 Cho mot chnh lu toan song mot pha vi mot tai RL , ien acuon th cap la 400 V. Neu tai ien tr la 100 . Tm :a) ien ap tai trung bnhb) ien ap tai RMSc) Dong diot cc ai


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d) Dong diot trung bnhe) Rai PIV cho diot5.10 Phac hoa dang song au vao cua hnh 5.7 tat ca cac diot ngc5.11 Trong mot chnh lu cau toan song, Vs = 240 V , va tai ien tr R = 10 . Tm :a) ien ap tai trung bnhb) Dong tai trung bnhc) Dong tai RMSd) Dong diot trung bnhe) Cong suat ra trung bnhf) Rai PIV cho diot5.12 Phac hoa dang song au ra cua hnh 5.13. Tm ien ap ra trungrai PIV cho diot

Hnh 5.13:cho cau 5.12

5.6 Cac cong thc :V0(avg) = 0,318 Vm 5.1I0(avg) = Im / 5.2I0RMS = Im / 2 5.3PIV Vm 5.4

P0(avg) =2

2

*

mV

R 5.5

PAC =2

4mV R

5.6

avg

AC

P

P = 5.7


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FF = 00( )

RMS

avg

V V

5.8

So xung =tan . . . . .tan . . .so gon song co ban

so nguon AC 5.9

RF =2

2 1 RMS

DC

I I

5.10

V0(avg) = (1 cos )2

mV

5.11

I0(avg) = (1 cos )2

mV R

5.12

V0(avg) = (2Vm ) / = 0,636 Vm 5.13I0(avg) = ( 2 Im ) / = 0,636 Im = 0,636 Vm / R 5.14I0(RMS) =

2m I = 0,707 Im 5.15

PIV 2 Vm 5.16ID1avg = ID2avg = I0avg/ 2 = Im / 5.17ID(RMS) = Im / 2 5.18

P0(avg) =2

2

4*

mV R

5.19

PAC =2

2mV R

5.20

V0(avg) = (2Vm ) / = 0,636 Vm 5.21

I0(avg) =2

0,636m mV V

R R =

5.22IRMS = I0(avg) = V0(avg)/ R 5.23

ID(RMS) = 0( )2

avg I 5.24

PIV Vm 5.25ID(avg) = I0(avg)/ 2 5.26Is = i3 - i1 = i2 - i4 5.27V0(avg) = VL(avg) + VR(avg) 5.28VR(avg) = V0(avg) = (2Vm ) / 5.29

I0(avg) = VR(avg) / R 5.30IRMS = I0(nax) = I0(avg) 5.31ID(avg) = I0(avg)/ 2 5.32

ID(RMS) = 0( )2

avg I 5.33

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