giao_trinh_ky_thuat_do_luong_5435

8/7/2019 giao_trinh_ky_thuat_do_luong_5435

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TS. NGUYN HU CNG

K THUT O LNG

NH XUT BN I HC QUC GIA H NI


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LI GII THIU

Quyn sch ny nhm mc ch cung cp kin thc cbn vthit bv phng php o lng cc i lngin. Ni dung gio trnh phcv cho sinh vin cc ngnh in -in t- My tnh ca cc trngihc. ng thi cng gip ch cho sinh vin cc chuyn ngnh khc v

cc cn b kthut c quan tm n lnh vc o in.Khi vit gio trnh ny chng ti c tham kho kinh nghim ca cc

nh gio ging dy nhiu nm cc trngi hc, ng thi cpnht nhng ni dung mi, va p ng yu cu nng cao cht lngoto phc v snghip cng nghip ho - hin i ho, va m bo tnhst thc ca cc thit bo cng nhphng php o m cc cn b kthutang vn hnh trong thc t.

Tuy cc tc gi c nhiu cgng khi bin son, nhng gio trnh

skhng trnh khi nhng khim khuyt. Chng ti mong nhn c sng gp kin ca qu ng nghip v cc bn sinh vin gio trnhny c hon thin.

Sau ht chng ti xin chn thnh cm n sng gp ng kcaThc sNguyn Vn Ch, cm n Khoa in t , Trngi hc Kthut Cng nghip,i hc Thi Nguyn gip v to iu kin chochng ti hon thnh quyn sch ny.

Tc gi


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Chng 1KHI NIM CHUNG VO LNG

1.1. nh ngha v phn loi thit b

1.1.1. nh ngha

o lng l mt qu trnh nh gi nh lngi tng cn o c kt qu bng sso vi n v.

Vi nh ngha trn th o lng l qu trnh thc hin ba thao tcchnh: Bin i tn hiu v tin tc.

- So snh vi n v o hoc so snh vi mu trong qu trnh olng.

- Chuyn n v, m ho c kt qu bng s so vi n v.

Cn c vo vic thc hin cc thao tc ny ta c cc phng php v

h thng o khc nhau.Thit bo v thit bmu

Thit bo l mt h thng m i lng o gi l lng vo, lngra l i lng ch trn thit b (l thit bo tc ng lin tc) hoc lcon s km theo n vo (thit bo hin s). i khi lng ra khnghin th trn thit b m a ti trung tm tnh ton thc hin ccAlgorithm k thut nht nh.

- Thit b mu dng kim tra v hiu chnh thit bo v n v

o.Theo quy nh hin hnh thit b mu phi c chnh xc ln hn t

nht hai cp so vi thit b kim tra.

V d: Mun kim nh cng tcp chnh xc 2 th bn kim nhcng tphi c cp chnh xc t nht l 0,5.

1.1.2. Phn loi

1.1.2.1. Thit bo lng

C nhiu cch phn loi song c th chia thit bo lng thnh hai

loi chnh l thit bo chuyn i thng v thit bo kiu so snh.


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Thit bo chuyn i thng

i lng cn o a vo thit b di bt k dng no cng cbin thnh gc quay ca kim ch th. Ngi o c kt qu nh thangchia v nhng quy c trn mt thit b, loi thit b ny gi l thit bo cin. Ngoi ra lng ra cn c th bin i thnh s, ngi o ckt qu ri nhn vi h s ghi trn mt my hoc my tng lm vic

, ta c thit bo hin s.Thit bo kiu so snh

Thit b so snh cng c th l ch th cin hoc l ch th s. Tutheo cch so snh v cch lp i lng b (b m ho s tng t) ta ccc thit b so snh khc nhan nh: thit b so snh kiu tung (ilng o x v i lng b x lun bin i theo nhau); thit b so snhkiu qut (i lng b x bin thin theo mt quy lut thi gian nhtnh v s cn bng ch xy ra ti mt thi im trong chu k).

Ngoi ra cng cn c vo vic lp i lng b ngi ta chia thnhdng c m ho s xung, tn s xung, thi gian xung. Cn c vo iukin cn bng ngi ta chia thnh dng c b khng lch (zero) v dngc b c lch (vi sai).

Cn c vo quan h gia lng ra v lng vo, ngi ta chia thnh:thit bo trc tip (i lng ra biu th trc tip i lng vo), thit bo gin tip (i lng ra lin quan ti nhiu i lng vo thng quanhng biu thc ton hc xc nh), thit bo kiu hp b (nhiu ilng ra lin quan ti nhiu i lng vo thng qua cc phng trnh

tuyn tnh).

1.1.2.2. Chuyn io lng

C hai khi nim:

- Chuyn i chun ho: C nhim v bin i mt tn hiu in phi

tiu chun thnh tn hiu in tiu chun (thng thng U = 0 10V;

I = 4 20mA).

Vi loi chuyn i ny ch yu l cc b phn p, phn dng, binin p, bin dng in, cc mch khuch i... c nghin cu kcc gio trnh khc nn ta khng xt.


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- Chuyn i s c p (S: Sensor): C nhim v bin mt tn hiukhng in sang tn hiu in, ghi nhn thng tin gi tr cn o. C rtnhiu loi chuyn i scp khc nhau nh: chuyn i in tr, incm, in dung, nhit in, quang in...

1.1.2.3. Thp thit bo

Vi mt thit b c th (mt knh):

Hnh 1.1. Cu trc h thng o mt knh

+ Chuyn i o lng: bin tn hin cn o thnh tn hiu in.

+ Mch o: thu nhn, x l, khuch i thng tin.... bao gm: ngun,cc mch khuch i, cc b bin thin A/D, D/A, cc mch ph...

+ Ch th: thng bo kt qu cho ngi quan st, thng gm ch ths v ch th cin, ch th t ghi, v.v...

1.1.2.4. Vi hthngo lng nhiu knh

Trng hp cn o nhiu i lng, mi i lng o mt knh,nh vy tn hiu o c ly t cc sensor qua b chuyn i chun hoti mch iu ch tn hiu mi knh, sau sa qua phn knh(multiplexer) c sp xp tun t truyn i trn cng mt h thngdn truyn. c s phn bit, cc i lng o trc khi a vo mch

phn knh cn phi m ho hoc iu ch (Modulation - MOD) theo tns khc nhau (th d nh f10, f20...) cho mi tn hiu ca i lng o.

Ti ni nhn tn hiu li phi gii m hoc gii iu ch(Demodulation - DEMOD) ly li tng tn hiu o. y chnh l hnhthc o lng t xa (TE1emety) cho nhiu i lng o.


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Hnh 1.2. H thng o lng nhiu knh

1.2. S cu trc thit bo lng

1.2.1. H thng o hin i thng

Trong h thng o bin i thng, i lng vo x qua nhiu khubin i trung gian c bin thnh i lng ra z. Quan h gia z v xc th vit:

z = f(x)

trong f() l mt ton t th hin cu trc ca thit bo.


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Trong trng hp quan h lng vo v lng ra l tuyn tnh ta cth vit:

z = S.x (1.1)

y S gi l nhy tnh ca thit b.

- Nu mt thit b gm nhiu khu ni tip th quan h gia lngvc v lng ra c th vit:

trong Si l nhy ca khu th i trong thit b.

1.2.2. H thng o kiu so snh

Trong thit bo kiu so snh i lng vo x thng c bin ithnh i lng trung gian yX qua mt php bin i T:

yX = T.x.

Hnh 1.3. H thng o kiu so snh

Sau yXc so snh vi i lng b yk

Ta c: y = yX - yk

C th cn c vo thao tc so snh phn loi cc phng php okhc nhau.

1.2.2.1. Phn loi phng php o cn cvo iu kin cn bng

a) Phng php so snh kiu cn bng (Hnh 1.4)

Trong phng php ny, i lng vo so snh: yX = const; ilng b yk= const.

Ti im cn bng:

b) Phng php so snh khng cn bng (Hnh 1.5)


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Cng ging nh trng hp trn song y 0

1.2.2.2. Phn loi phng php o cn cvo cch to in p b

a) Phng php m ho thi gian

Trong phng php ny i lng vo yX = const cn i lng b ykcho tng t l vi thi gian t:

yk= y0.t (y0 = const)

Hnh 1.6. Phng php m ha thi gian

Ti thi im cn bng yX = yk= y0.tX

i lng cn o yXc bin thnh khong thi gian tXy phpso snh phi thc hin mt b ngng


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b) Phng php m ho tn sxung

Trong phng php ny i lng vo yX cho tng t l vi i lngcn o x v khong thi gian t: yX = t.x, cn i lng b ykc gikhng i.

Hnh 1.7. Phng php m ho tn s xung

Ti im cn bng c:

i lng cn o x c bin thnh tn sfX.y php so snhcng phi thc hin mt b ngng.

c) Phng php m ho sxung

Trong phng php ny i lng vo yX = const, cn i lng byk cho tng t l vi thi gian t theo quy lut bc thang vi nhng bcnhy khng i vo gi l bc lng t.

T = const cn gi l xung nhp.

Ta c:


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Ti im cn bng i lng vo yx c bin thnh con s NX

yX NX.y0 (1-6)

xc nh c im cn bng, php so snh cng phi thc hinmt b ngng:

Ngoi ra cn phng php m ho s xung ngc, phng php mxung, phng php trng phng.

1.3. Cc c tnh ca thit bo

1.3.1. nhy, chnh xc v cc sai s ca thit bo

1.3.1.1. nhy v ngng nhy

Ta bit phng trnh cbn ca thit bo l z = f(x). c mt s

nh gi v quan h gia lng vo v lng ra ca thit bo, ta dngkhi nim v nhy ca thit b:

trong : z l bin thin ca lng ra v x l bin thin ca lngvo.

Ni chung S l mt hm ph thuc x nhng trong phm vi x nhth S l mt hng s. Vi thit b c quan h gia lng vo v lng ra

l tuyn tnh, ta c th vit: z = S.x, lc S gi l nhy tnh ca thit


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bo.

Trong trng hp thit bo gm nhiu khu bin i ni tip th

nhy c tnh =

=n

1iiSS , vi Si l nhy ca khu th i trong thit b.

Theo l thuyt khi xt ti quan h gia z v x th x c th nh bao

nhiu cng c, song trn thc t khi x < no th z khng th

thy c.V d 1.1: Khi ph ti tiu th qua mt cng tmt pha 10A nh hn

10W (chng hn) th cng tkhng quay na.

Nguyn nhn ca hin tng ny rt phc tp, c th do ma st, do

hin tng tr... c gi l ngng nhy ca thit bo.

C th quan nim ngng nhy ca thit bo l gi tr nh nhtm thit bo c th phn bit c.

Tuy nhin ngng nhy ca cc thit bo khc nhau rt khc

nhau n cha c trng cho tnh nhy ca thit b. V vy so snhchng vi nhau ngi ta phi xt ti quan h gia ngng nhy vthang o ca thit b.

Thang o (D) l khong t gi tr nh nht ti gi tr ln nht tuntheo phng php o lng ca thit b

T a ra khi nim v kh nng phn ly ca thit bo:

v so snh cc R vi nhau.

1.3.1.2. chnh xc v cc sai sca thit bo

- chnh xc l tiu chun quan trng nht ca thit bo.. Bt kmt php o no u c sai lch so vi i lng ng

trong xi l kt qu ca ln o th

x l gi trng ca i lng o


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i l sai lch ca ln o th i

- Sai s tuyt i ca mt thit bo c nh ngha l gi tr lnnht ca cc sai lch gy nn bi thit b trong khi o:

- Sai s tuyt i chn nh gi c tnh chnh xc v yu cu cngngh ca thit bo. Thng thng chnh xc ca mt php o hoc

mt thit bo c nh gi bng sai s tng i:+ Vi mt php o, sai s tng i c tnh

+ Vi mt thit bo, sai s tng i c tnh

Gi tr, % gi l sai s tng i quy i dng sp xp cc thit

bo thnh cc cp chnh xc.

Theo quy nh hin hnh ca nh nc, cc dng co cin ccp chnh xc: 0,05; 0,1; 0,2; 0,5; 1; 1,5; 2; 2,5; v 4.

Thit bo s c c p chnh xc: 0,005; 0,01; 0,02; 0,05; 0,1; 0,2;0,5; 1.

Khi bit cp chnh xc ca mt thit bo ta c th xc nh c sais tng i quy i v suy ra sai s tng i ca thit b trong cc

php o c th.

Ta c:

trong l sai s tng i ca thit bo, ph thuc cp chnh xc vkhng i nn sai s tng i ca php o cng nh nu D/x dn n 1.

V vy khi o mt i lng no c gng chn D sao cho: D x.

1.3.2. in trvo v tiu th cng sut ca thit bo

Thit bo phi thu nng lng ti tng o di bt k hnh thc


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no bin thnh i lng u ra ca thit b. Tiu th nng lng nyth hin phn tc dng ca thit bo ln i tng o gy ra nhngsai s m ta thng bit c nguyn nhn gi l sai s ph v phng

php. Trong khi o ta c gng phn u sao cho sai s ny khng lnhn sai s cbn ca thit b.

- Vi cc thit b o c hc sai s ch yu l phn tc dng ca

chuyn i. Vi cc thit bo dng p, sai s ny ch yu l do nhhng ca tng trvo v tiu th cng sut ca thit b.

Tn hao nng lng vi mch o dng p l:

Vy ta tm tnh sai s ph do nh hng ca tng trvo l:

vi RA l in trca ampemet hoc phn t phn ng vi dng;

RV l in trca volmet hoc phn t phn ng vi p;

Rt l in trti.

V d 1.2: Phn tch sai s ph khi o p trn Hnh 1.9.

+ Gi s cn kim tra in p UA0.

Theo l lch [ UA0] = 50 2 (V).

+ Xt khi cha o (k m), ta c ngay:

UA0 = 50 V.

+ Xt khi o (kng).


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Gi s RV = 100 k. Vy in p o c: Uv = UA0 = 33,3 V.

Sai s t 33 V tr ln 50 V chnh l sai s ph v phng php donh hng in trca V sinh ra.

1.3.3. Cc c tnh ng ca thit bo

Khi o cc i lng bin thin ta phi xt n c tnh ng cadng co c tnh ng ca dng co th hin cc c trng sau:

- Hm truyn t ca thit bo hay nhy ng ca thit boK(p) tc l quan h gia i lng ra v i lng vo trng thi ng

c tnh ny th hin di cc dng sau:

+ c tnh qu ng vi tn hiu vo c dng bc nhy:

+ c tnh xung hay tn hiu vo l xung hp:

+ c tnh tn lc tn hiu vo c dng hnh sin:

+ c tnh tn th hin hai dng: c tnh bin tn A() v c tnh

pha tn ().

c tnh cn th hin di dng sai s tn s, sai s ny th hin

sai s bin tn A v sai s pha tn :

trong : A() l bin u ra ph thuc tn s;

A0 l bin ca khu l tng khng ph thuc tn s;

() l gc pha u ra ph thuc tn s;

0 l gc pha l tng khng ph thuc tn s.

Trong dng co cc sai s ny phi nh hn mt gi tr cho php


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quy nh bi nh nc. Gii tn ca dng co l khong tn s ca ilng vo cho sai s khng vt qu gi tr cho php.

Thi gian n nh hay thi gian o ca thit b l thi gian k t khit tn hiu vo ca thit b cho ti khi thit bn nh c th bit ckt qu.

Chnh da vo thi gian o ca thit b ny cho php ta tng ri

rc ho i lng cn o o gi tr tc thi, sau dng cc php giacng ton hc hoc dng phng tin phc hi li hon ton hintng xy ra.

1.4. Gia cng kt quo lng

Gia cng kt quo lng l da vo kt qu ca nhng php o cth ta xc nh gi trng ca php o v sai s ca php o y.

Dng co no cng c sai s v nguyn nhn sai s rt khc nhau,

v vy cch xc nh sai s phi ty theo tng trng hp m xc nh.Hin nay dng nhiu phng php khc nhau php o m bo yucu k thut ra.

1.4.1. Tnh ton sai s ngu nhin

- xc nh sai s ngu nhin ta da vo phng php thng knhiu kt quo lng. Sai s ngu nhin ca ln o th i c tnh

trong : xi l kt qu ln o th i;

M[x] l k vng ton hc ca v s ln o i lng x.

Hnh 1.10. Lut phn b chun


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- Theo ton hc thng k th s phn b ca sai s ngu nhin xungquanh gi tr k vng ton hc theo mt quy lut nht nh gi l lut

phn b xc sut.

Trong cc thit bo lng v iu khin thng theo quy lut phnb chun:

trong l lch qun phng hay phng sai ca sai s ngu nhin.

Ta c cng thc:

vi D l tn x.

Trong k thut ta thng dng khi nim phng sai D = v n

c cng th nguyn vi i lng cn o.

Hnh 1.11. K vng v tn x ca lut phn b chun

Qu trnh gia cng kt qu nh sau:

a) Khi sln o l rt ln (n > 30)

Sai s ngu nhin c tnh:

x = k. (l-14)

trong k l h s, c tra trong s tay k thut (bng hoc ngcong).

b) Khi sln o c hn (n 30)

Qu trnh gia cng c tin hnh nh sau:


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+ K vng ton hc c ly l trung bnh cng ca n ln o

+ Phng sai ca sai s ngu nhin c tnh theo cng thc BessE1

Nu ta ly kt qu l gi tr trung bnh ca n ln o th phng sai s

gim i n ln

+ Sai s ngu nhin c tnh:

trong kst l h s Student, n ph thuc vo s ln thu thp n v xcxut yu cu p. H s kstc tra trong cc s tay k thut: kst = f(n,p).

+ Kt quo c tnh:

( )1nn

xn1

x

kn

xxxx

n

1i

2n

1iii

sti

== = = (1-18)

Ch : Trong thc t c nhng ln thu thp s liu cho kt qu khngng tin cy (v ta thng gi l nhiu ca tp s liu), ta phi loi bln o ny nhthut ton sau:

Sau khi tnh ta so snh cc |i| vi 3 vi i = 1 n n, nu ln o

no c |i| 3 th phi loi b ln o v tnh li tu vi (n - 1)php o cn li. C th chng minh rng vic loi b m bo tin cy 99,7%.

V d 1.3: Tnh kt quo v sai s ngu nhin vi mt xc sut


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ng tin p = 0,98 ca mt php o in trbng cu kp vi kt qu nh

sau (n v tnh = m):

140,25; 140,5; 141,75; 139,25; 139,5; 140,25; 140; 126,75; 141,15;142,25; 140,75; 144,15; 140,15; 142,75. Bit sai s ngu nhin c phn

b chun.

Bi lm:

So snh cc i = Ri - R vi 3. Ta thy ln o th 8 phm phi sai

lm ln (8 = R8 - R 3) nn ta b qua ln o ny v tnh li tu vi13 ln o cn li. Ta lp bng sau:

Bng 1.1. V d v tnh ton sai s ngu nhin

STT Ri i i21 140,25 -0,73 0,5329

2 140,5 -0,48 0,2304

3 141,75 0,77 0,5929

4 139,25 -1,73 2,9929

5 139,5 1,48 2,1904

6 140,25 -0,73 0,5329

7 1 40 -0,98 0,96248 141,15 0,17 0,0289

9 142,25 1,27 1,6129

10 140,75 -0,23 0,0529

11 144,15 3,71 13,7641

12 140,15 -0,8t3 0,6889

13 142,75 1,77 3,1329

Tng: 0 Tng: 23,64


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1.4.2. Tnh ton sai s gin tip

Trong thc t c nhiu phng php o m kt quc tnh t phpo trc tip khc ngi ta gi php o l php o gin tip.

Gi s c mt php o gin tip i lng y thng qua cc php otrc tip x1, x2, xn: y =f(x1, x2, xn)

Ta c:

Sai s tuyt i ca php o gin tip c nh gi

x1,x2,xn: sai s tuyt i ca php o cc i lng trc tipx1, x2, xn

Sai s tng i ca php o gin tip c tnh l:


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trong : xl, x2,.xn l sai s tng i ca cc php o trc tip x1,x2, xn.

Bng 1.2. Bng tnh sai s tuyt i v sai s tng ica mt s hm y thng gp

V d 1.4:Ngi ta s dng ampemet v volmet o in trbngphng php gin tip. Ampemet c thang o l lA, cp chnh xc l 1.Volmet c thang o l 150V, cp chnh xc 1,5. Khi o ta c s chca hai ng h l: I = lA, U =100V.

Hy tnh sai s tuyt i v tng i ca php o in trtrn.

Hnh 1.12. V d v tnh ton sai s gin tip

Bi lm:


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+ Sai s tuyt i ca ampemet l:

+ Sai s tuyt i ca volmet l:

+ Gi trin trtheo php o l:

+ Sai s tuyt i ca php o in trl:

+ Sai s tng i ca php o in tr


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Chng 2CC CCU CH TH

2.1. Ccu ch th cin

2.1.1. Cschung

2.1.1.1. Khi nimCcu ch th l dng co m s ch ca n l i lng t l vi

i lng o lin tc. Ch th cin l ccu ch th c tn hiu vo ldng in v tn hiu ra l gc quay ca kim ch th. i lng cn o strc tip bin i thnh gc quay ca kim ch th, tc l thc hin vicchuyn i nng lng in t thnh nng lng chc lm quay kin

ch thi mt gc : = f(x), x l i lng vo.

C cu ch th cin bao gm hai phn: phn tnh v phn quay.

Ty theo phng php bin i nng lng in t ngi ta chia thnhccu ch th kiu tin, in t, in ng, cm ng v tnh in.

2.1.1.2. Cc chi tit ckh chung ca chthcin

a) Trc v tr

Trc v tr l b phn quan trng trong cc chi tit ckh ca cc ccu ch th cin, m bo cho phn ng quay trn trc c gn cc chitit ca phn ng nh kim ch th, l so phn, khung dy.

b) B phn phn khng

B phn phn khng bao gm l so phn khng hoc dy cng hocdy treo. Mc ch to ra mmen phn khng.

c) Kim chth gc quay

Kim ch th gc quay c gn vi trc quay. di chuyn ca

kim trn thang chia t l vi gc quay . Ngoi ra c th ch th gcquay bng nh sng.


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d) Thang chia

Thang chia l mt khc thang o, xc nh gi tro

e) B phn cn du

B phn cn du c tc dng rt ngn qu trnh dao ng ca phnng, xc lp v tr nhanh chng trong ccu ch th. Thng thng c

hai loi cn du c s dng, l cn du kiu khng kh v cn dukiu cm ng.

2.1.2. Phng trnh c tnh ca ccu cin


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2.1.2.1. Cc mmen tc ng ln phn ccu

a) Mmen quay

Khi c dng in chy trong ccu ch th, th trong n s tch lumt nng lng in t, nng lng ny c bin thnh cnng lmquay phn ng i mt gc no , c ngha l thc hin mt cng chc:

dA = Mqdtrong : dA l lng vi phn ca cng chc;

Mq l mmen quay;

d l lng vi phn ca gc quay.

Theo nh lut bo ton nng lng:

dWe = dA

dWe l lng vi phn ca nng lng in t

Vy ddWMe

q =

b) Mmen phn

Di tc dng ca mmen quay, nu khng c g cn li th phnng ca c cu s quay i mt gc ln nht c th c c. V vy

ngi ta to ra cc mmen phn t l vi gc quay nhcc b phnphn khng l l so xon, dy cng hoc dy treo.

Ta c: Mp = D;

vi D l h s ph thuc vo kch thc vt liu ch to l so, dy cnghoc dy treo.

Khi mmen quay cn bng vi mmen phn th phn ng ng yn

Mq = Mp = D.

c) Mmen ma st

i vi cc dng c dng trc quay ta phi xt n nh hng calc ma st gia trc v , mmen ma st c tnh theo cng thc kinhnghim

Mms = K.Gn


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trong K l h s t l, G l trng lng phn ng, n = (1,3 l,5)

d) Mmen cn du

Khi trc quay dn n kim ch th quay theo cho ti v tr cn bngri mi dng li, do phn ng c qun tnh v l so b ko nn kim sdao ng ri mi ng yn cho nn phi c b phn n nh dao ngkim hay b phn cn du.

Mmen cn du c ch to sao cho c tr s t l vi tc quayca phn ng

vi p l h s t l ph thuc vo c imcu to ca b phn cn du. T biu thctrn ta thy khi phn ng v tr cn bng

th 0

dt

d= , nh vy mmen cn du khng

lm nh hng n kt quo.

2.1.2.2. Phng trnh cn bng phn ng ca ccu o

Theo nh lut chc i vi mt chuyn ng quay, o hm bcnht ca mmen ng lng theo thi gian bng tng cc mmen tcng ln vt quay y.

trong : J l mmen qun tnh phn ng;

Mi l tng cc mmen tc ng ln phn ng ca ccu,bao gm:

Thay cc i lng trn vo phng trnh, ta c:


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Phng trnh ny chnh l phng trnh mmen chuyn ng ca c

cu Gii phng trnh ny ta tm c (t). Tu theo quan h gia J, P,

D m ccu dao ng hay khng dao ng v quyt nh tnh n nh vthi gian o ca ccu.

2.1.3. Ccu o tin

2.1.3.1. Loi c mt khung dy ng

1. Cu to

Phn tnh gm: nam chm vnh cu, cc t, li st non, trong khehkhng kh gia cc t v li st l u nhau.

Phn ng gm: khung dy, l so phn, kin ch th.

2. Nguyn l lm vic

- Khi ta cho dng in mt chiu I chy vo khung dy, di tc

dng ca t trng nam chm vnh cu trong khe hkhng kh, cc cnh


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ca khung dy s chu tc dng mt lc:

F = BlWI

trong : B l tr s cm ng t trong khe hkhng kh; ll chiu di tcdng ca khung dy; W l s vng dy; I l tr s dng in.

Ta thy hai cnh ca khung dy cng chu tc dng ca lc F nhngngc chiu nhau nn s to ra mmen quay:

trong : d l kch thc ngang ca khung dy;

S = dll thit din b mt khung dy.

Mmen phn ca l so: MP = D..

Vy phn ng s cn bng khi:

y SI =D

B.S.W= const l nhy ca ccu theo dng in.

Ta thy t l bc nht vi I.

3.c im v ng dng

c im:

- u im:

+ Dng c c nhy cao v khng i trong ton thang o;

+ chnh xc cao, t chu nh hng ca t trng ngoi, tiu thnng lng t;

+ V t l bc nht vi I nn thang chia ca ccu u.

- Nhc im:

+ Ch to kh khn, gi thnh t;

+ Do khung dy phn ng nn phi qun bng dy c kch thcnh nn kh nng qu ti km;

+ Cho c dng mt chiu. Tht vy, khi ta cho dng xoay chiu

i = Imsint vo khung dy, ta c mmen quay tc thi theo thi gian:


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mq(t) = B.S.W.i

Vy mun o cc i lng xoay chiu ta phi kt hp vi b chnhlu.

ng dng:

Dng ch to ampemet, volmet, ommet, in k c nhy cao,dng lm ccu ch khng trong cc ng h vn nng, trong cc cuo...

2.1.3.2. Loi c hai khung dy ng (Logomet tin)

1. Cu to

Phn tnh ging nh ccu mt khung dy nhng khe hkhng khgia cc t v li st non l khng u nhau.

- Phn ng ta t hai cun dy cho nhau 60

o

, gn cng trn trcquay v ln lt cho dng in I1 v I2 chy qua sao cho chng sinh rahai mmen quay ngc chiu nhau. Phn ng khng c l so phn.

2. Nguyn l lm vic


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Khi ta cho cc dng mt chiu I1, I2 chy vo cc cun dy ng,di tc dng ca t trng nam chm vnh cu s to ra cc mmenquay M1, M2 vi:

M1 = B1.S1.W1.I1

M2 = B2.S2.W2.I2

V khe hkhng kh l khng u nn cm ng t B ph thuc v tr

ca khung dy ng.

V khng c l so phn nn phn ng s cn bng khi M1 = M2. Tac:

Vy

Gii phng trnh ny ta tm c quan h:

=

2

1

I

If

3.c im v ng dng

c im:

Tng t nh ccu mt khung dy trn khng c chnh xccao hn, cng sut tn tht thp, nhy rt cao, t bnh hng ca t

trng ngoi. Gc lch t l vi t s hai dng in i qua cc khungy, iu ny thun li khi o cc i lng vt l thng phi chothm ngun ngoi. Nu ngun cung cp thay i nhng t s hai dngin vn c gi nguyn do vy m trnh c sai s.

ng dng:

c dng ch to cc ommet, megommet.

2.1.4. Ccu o in t


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2.1.4.1. Cu to

Ccu gm hai loi chnh: kiu cun y dt (ccu ch thin tloi ht) v kiu cun dy trn (ccu ch thin t loi y). Ccucun dy dt c phn tnh l cun dy dt cho dng in cn o i qua,cn phn ng l mt l thp t lch tm c th quay trong khe hcundy tnh. Kiu cun dy trn c phn tnh l cun dy trn bn trong gn

mt l thp. Phn ng cng l mt l thp gn trn trc. Ngoi ra cn cb phn cn du, l so phn, kim ch th


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2.1.4.2. Nguyn l lm vic

Khi c dng in chy vo cun dy tnh, trong lng cun dy s cmt t trng. i vi cun dy dt t trng ny ht l thp vo tronglng cun dy tnh, cn i vi cun dy trn th t trng s t ho hail thp, khi hai l thp c cng cc tnh nn y nhau. C hai trng

hp trn s lm cho phn ng quay i mt gc .


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- Khi cho dng in mt chiu chy vo cun dy:

Ta c mmen quay:

vi We l nng lng in t trng tch lucun dy

trong L ph thuc .

Vy mmen quay:

- Khi cho dng in xoay chiu vo cun dy:

Gi s i = ImaXsint. Lc mmen quay Mq theo t s l:

Mmen quay trung bnh:

vi I l tr hiu dng ca dng hnh sin.

Ti v tr cn bng Mq = MP;

Vy ccu ch thin t c tho c c dng mt chiu v dngxoay chiu.

2.1.4.3. c im v ng dng

c im:

- u im:


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+ C cun dy phn tnh nn c th qun bng dy kch thc lnnn kh nng qu ti tt.

+ D ch to, gi thnh h.

+ C tho c ci lng mt chiu v xoay chiu.

- Nhc im:

+ Gc quay t l vi bnh phng ca dng in v thang o chia

khng u (hnh dng l thp c ch to sao choddL gim theo gc

quay thang chia c th tng i u).

+ chnh xc thp do c tn hao trong li thp.

ng dng:

Ch yu o dng, p xoay chiu tn s cng nghip.

2.1.5. Ccu o in ng

2.1.5.1. Loi c mt khung dy ng

a) Cu to

Ccu gm hai cun y. Cun dy tnh c tit din ln, t vng dyv thng chia lm hai phn on. Phn ng l mt khung dy c nhiuvng dy v tit din nh. Ngoi ra cn c kim ch th, b phn cn du,l so phn.

b) Nguyn l lm vic


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- Xt khi cho cc dng in mt chiu I1 v I2 vo cc cun dy phntnh v ng, trong lng cun dy tnh s tn ti mt t trng. Ttrng ny s tc ng ln dng in chy trong cun dy ng v to rammen quay:

Nng lng t trng tch lu trong lng cun dy l:

trong L1, L2 l din cm ca cc cun dy v chng khng ph thuc

vo gc quay ;M12 l h cm ca hai cun dy, thay i khi phn ngquay. Mmen quay

- Xt khi hai dng in a vo cc cun dy l dng in xoaychiu th:

Do phn ng c qun tnh m khng k p thay i theo gi tr tcthi cho nn thc t ly theo gi tr trung bnh trong mt chu k:

Vi l gc lch pha gia hai dng in; I1, I2 l cc gi tr hiudng ca dng in ln lt chy trong cc cun dy tnh v ng.

Tm li, trong mi trng hp ta u c:


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c)c im v ng dng

c im:

- u im:

+ chnh xc cao v khng c tn hao trong li thp.

+ C tho c ci lng mt chiu v xoay chiu.

- Nhc im:

+ D chu nh hng ca t trng ngoi.

+ Kh nng qu ti km v khung dy phn ng kch thc nh.

+ Cu to phc tp, t tin.

+ Thang chia khng u (tr khi ch to wattmet).

ng dng:

+ Ch to cc ng ho dng, p xoay chiu c tn s cao hocyu cu chnh xc cao.

+ Ch yu ch to ng ho cng sut tc dng v phn khng.

2.1.5.2. Loi c hai khung dy ng (logometin ng)a) Cu to

Phn tnh gm mt cun dy c chia lm hai na. Trong lng cundy tnh c hai cun dy ng gn trn trc quay cng vi kim ch th,khng c l so phn.

b) Nguyn l lm vic

Khi cho hai dng in xoay chiu i, i1, i2 ln lt chy vo cun dytnh v cc cun dy ng, trong lng cun dy tnh s c mt t trng.

T trng ny s tc ng ln dng in chy trong cc cun dy ng


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sinh ra cc mmen Mq1, Mq2

vi M1, M2 l h s h cm gia cun dy tnh v ln lt cc cun dyng.

Ngi ta b tr sao cho cc mmen ny ngc chiu nhau, vy khicn bng phn ng, ta c Mq1 = Mq2

vi I1, I2 l cc gi tr hiu dng ca cc dng in i1, i2; 12 l gc lch

pha gia dng in i, i1 v i, i2


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c)c im v ng dng

Ging nh c cu mt khung dy ng nhng ch yu ch to

ng ho cos 1 pha, 3 pha cho li in xoay chiu.

2.1.5.3. Ccu stin ng v logomet stin ng

a) Ccu stin ng

Gm cun dy tnh, mch t nhm to ra t trng trong khe h

khng kh. Khung dy ng c gn vi trc quay cng kim ch th, lso phn v b phn cn du.

Gc quay c tnh:

b) Logomet stin ng

Gm mch t c cu to sao cho to nn khe hkhng kh khngu, phn ng gm hai khung dy t cho nhau 60o v gn trn trcquay cng vi kim ch th. Gc quay c tnh:

c)c im ng dng

- C tho dng mt chiu hoc xoay chiu. T trng qua khungdy ln nn t chu nh hng ca t trng ngoi.

- Tn hao st t ln, chnh xc khng cao.

- Thng dng ch to cc dng co dng, o p, cng sut v

gc lch pha.


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2.1.6. Ccu o cm ng

2.1.6.1. Cu to

Ccu cm ng c cu to nh hnh 2.1 1.

Hnh 2.11. Ccu ch th cm ng

1.Cun dy 1 ; 2. Cun dy 2; 3. Ccu cn du ; 4. a nhm v trc quay


Khi cho dng in i1 vo cun dy 1 th cun dy 1 to ra t thng 1

xuyn qua a nhm, dng in i2 vo trong cun dy 2 to ra t thng 2cng xuyn qua a nhm.

T thng 1 cm ng trn a nhm

sc in ng e1 chm pha hn 1 mt

gc /2.

T thng 2 cm ng trn a nhm

sc in ng e2 chm pha hn 2 mtgc /2.

V a nhm c coi nh rt nhiuvng dy t st nhau, cho nn E1, E2 sto ra trn a nhm cc dng in xoyiX1 v iX2 chm pha hn so vi e1 v e2

cc gc 1 v 2 v ngoi in trthuncn c thnh phn cm ng, tuy nhin

do cc thnh phn cm ng rt nh nn ta gi thit cc gc 1 v


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2 0.

Do c s tng h gia t thng 1, 2 vi cc dng in iX1 v iX2m sinh ra cc lc F1 v F2 v cc mmen tng ng lm quay a nhm.Ta xt cc mmen thnh phn nh sau:

M11 l mmen sinh ra do 1 tc ng ln iX1




Gi tr tc thi ca mmen quay M1t do s tc ng tng h gia 1v dng tc thi iX1 l:

M1t = C1iX1

vi C l h s t l.

vi l gc lch pha gia 1 v iX1, ta c:

V phn ng c qun tnh cho nn ta c mmen l i lng trungbnh trong mt chu k T:

Nh vy mmen quay s l tng cc mmen thnh phn:

Mq = M12 + M21


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M12 v M21 c du ngc nhau do vy mmen tng s ko a nhmv mt pha duy nht:

Nu dng in to ra 1 v 2 l hnh sin v a nhm l ng nht(ch c in trthun) th cc dng in xoy IX1 v IX2 s t l vi tn sv t thng sinh ra n, tc l:

vi C = C12C4+ C21C3 l hng s ca ccu ch th cm ng.

2.1.6.3. c im v ng dng

iu kin c mmen quay l phi c hai t trng, mmen quaycc i khi sin = 1, c ngha l gc lch pha gia hai t thng 1 v 2

l /2.

Ccu ph thuc tn s, chnh xc thp v khi lm vic dng inxoy trong a nhm gy tn hao cng sut.

Ccu c ng dng ch yu ch to cng to nng lng tcdng v phn khng trong li in xoay chiu.

2.2. Ccu ch th s

2.2.1. Khi nim v nguyn l cbn ca ccu ch th sTrong nhng nm gn y xut hin v s dng rng ri cc ch th

s, u vit ca ccu ch th s l thun li cho vic c ra kt qu, phhp vi cc qu trnh o lng xa, qu trnh tng ho sn. xut, thunli cho nhng i thoi gia my v ngi

S khi ca ccu ch th s c th tm tt nh sau:

Hnh 2.13. S khi ca ccu ch th s


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i lng o xin qua b bin i thnh xung (BX), s xung N t lvi ln x(t) c a vo b m ho (MH), b gii m (GM) v bhin s. Cc khu m ho, gii m, b hin s to thnh b ch th s.

2.2.2. Ch th s

C nhiu loi ch th s khc nhau nhng ph bin hin nay vn dngch th sn phng in nhiu cc v ch th s ghp 7 thanh bng mt

pht quang hoc tinh th lng.2.2.2.1. Chthn phngin nhiu cc

Ch th l mt n n ng c mt qut v 10 katot. Anot thng t in p 220V - 250V. Katot c ch to bng dy Cr-Ni un thnh hnhcc ch s t 0 - 9. Mi katot l mt con s.

Khi c in p gia qut v mt katot no n s phng in,katot s sng ln v con s xut hin.

- u im ca ch th ny l hnh dng cc con sp.

- Nhc im: Kch thc cng knh, ngun in p cung cp cao,ch ph hp trong cng nghip.

2.2.2.2. Chthsghp 7 thanh

Ch th ny c ghp bng 7 thanh dng mt pht quang (LED:Light Emitting Diode) hoc tinh th lng (LCD: Liqui Crystal Display).


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it pht quang l nhng cht bn dn m pht ra nh sng di tcdng ca dng in mt chiu. Tinh th lng l nhng mng mng lm

bng cht tinh th lng. l nhng cht di tc dng ca in p mtchiu chuyn pha t dng lng sang dng tinh th v ngc li. Khi dng tinh th thanh ny trnn trong sut, ta c th nhn thy mu sc nn ng sau. Mt u im cbn tinh th lng tiu th dng in rtnh: 0,1A/thanh, trong khi mt pht quang c: 10mA/thanh.

Trong thc t cn ch th s 16 thanh, ma trn im...

2.2.3. M v cc mch bin i m

2.2.3.1. M

M s l nhng k hiu v mt tp hp s, t t hp ca cc k hiuta c th m tc cc con s khc nhau. C cc loi m s sau:

- M cs 10, l hm thp phn c 10 k t t 0, 1, 2,..., 9.

- M cs 2 l loi m c hai trng thi c k hiu t 0 v 1 (cn

gi l m nh phn).- M 2 - 10 (cn gi l m BCD) l s lin h gia m cs 2 v m

cs 10 d quan st v dc.

i vi ccu ch th s th hin nay ch yu ngi ta s dng m cs 2.

2.2.3.2. Cc mch bin i m

Hnh 2.15. Mch gii m tm nh phn sang ch th 7 thanh


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Mch bin i m l thit b dng bin i t m cs 2 hoc m2 - 10 thnh m cs 10, ngha l th hin di dng s thp phn. Ngynay cc b gii m c ch to di dng vi mch. V d nh vi mchSN74247 c cc u ra hcc gp dng iu khin LED c chungant 5V. Cc in trR1, R2,, R7 hn ch dng.

Phn sau y s trnh by nguyn l mt s mch bin i t m.

Da vo nguyn l ca cc mch bin i m ny m ngi ta ch tothnh cc vi mch chuyn dng.

a) Mch bin i tm thp phn sang nh phn

Tng qut c m u vo tng ng vi m s thp phn t 0, 1, 2...m-1 v n u ra tng ng vi n bt ca m s nh phn. Ngi ta thngtng hp b bin i m vi su vo m = 10 tc l gm x0, x1,... x9ng vi cc s thp phn t 0, 1, 2,... 9. Nh vy b bin i m s c

bn u ra tng ng y8, y4, y2, y1ng vi bn bt ca m nh phn ctrng s 8, 4, 2, 1. Ta c bng trng thi nh sau:

Bng 2.1. Bng trng thi bin i ts thp phn sang nh phn

M nh phnS thp phn

Y8 Y4 Y2 Y1

X0 (0) 0 0 0 0

X1 (1) 0 0 0 1

X2 (2) 0 0 1 0

X3 (3) 0 0 1 1

X4 (4) 0 1 0 0

X5 (5) 0 1 0 1

X6 (6) 0 1 1 0

X7 (7) 0 1 1 1

X8 (8) 1 0 0 0

X9 (9) 1 0 0 1

T bng trng thi ta c:


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Vy ta c th thnh lp mch bin i m t thp phn sang nh phn

nh sau:

Hnh 2.16. Mch tun i m tthp phn sang nhi phn

b) Mch bin i m tnh phn sang thp phnNhim v ca mch ny ngc vi mch trn. Vi bng trng thi

2.1 ta c X0 X9 l cc bin ph thuc cn Y1 Y8 l cc bin c lp.V vy ta c cc phng trnh logic v s mch logic tng ng:


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c) Mch bin i tsthp phn sang chth 7 thanh

u vo l cc s t nhin t 0 9, u ra l trng thi cc thanhsng ca ch th 7 thanh bng mt pht quang hoc tinh th lng. Xut

pht t thc t ta c bng trng thi nh sau:

Bng 2.2. Bng trng thi bin i ts nhi phn sang thp phn

Trng thi cc phn tS thp phn

Y1 Y2 Y3 Y4 Y5 Y6 Y7X0 (0) 1 1 1 1 1 1 0

X1 (1) 0 1 1 0 0 0 0

X2 (2) 1 1 0 1 1 0 1

X3 (3) 1 1 1 1 0 0 1

X4 (4) 0 1 1 0 0 1 1

X5 (5) 1 0 1 1 0 1 1

X6 (6) 1 0 1 1 1 1 1


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X7 (7) 1 1 1 0 0 0 0

X8 (8) 1 1 1 1 1 1 1

X9 (9) 1 1 1 1 0 1 1

T bng trng thi ta c th vit c phng trnh nh sau (vi sth t cc thanh nh phn trc)

Ty ta c th thit lp mch logic sau:


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d) Mch bin i m tm nh phn sang chth 7 thanh

u vo l m s nh phn (8 4 2 1) ta gn cc tn bin l X8, X4, X2,

X1. u ra l trng thi cc thanh sng ca ch th 7 thanh. Ta c bng

trng thi sau:


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Bng 2.3. Bng trng thi bin i ts nhi phn sang ch th 7 thanh

S th p phn S nh phn Trng thi cc thanh sng

X8 X4 X2 X1 Y1 Y2 Y3 Y4 Y5 Y6 Y7

0 0 0 0 0 1 1 1 1 1 1 0

1 0 0 0 1 0 1 1 0 0 0 0

2 0 0 1 0 1 1 0 1 1 0 0

3 0 0 1 1 1 1 1 1 0 0 1

5 0 1 0 0 0 1 1 0 0 1 1

5 0 1 0 1 1 0 1 1 0 1 1

6 0 1 1 0 1 0 1 1 1 1 1

7 0 1 1 1 1 1 1 0 0 0 0

8 1 0 0 0 1 1 1 1 1 1 1

9 1 0 0 1 1 1 1 1 0 1 1

T bng trng thi ta vit c cc phng trnh logic quan h giau ra Y1,,Y7 vi cc u vo X8,X4, X2, X1. Tuy nhin cc phngtrnh ny phc t p v i hi phi ti gin bng ba cc n (ti ginhm).

V d:


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Chng 3O DNG IN V IN P

3.1. Nhng yu cu cbn ca vic o dng in v in p

3.1.1. Yu cu vin tr

3.1.1.1. Khio dnginAmpemet l mt phn tc trng cho

nhm cc phn t phn ng vi dng innh: cun dng ca cng t, wattmet; ccrle dng in... nn khi xt ti yu cu ivi ampemet l xt chung cho c nhm.

Khi o dng, ampemet c mc ni tipvi ti (nh Hnh 3.1) nn in trampemetsnh hng n kt quo nh sau:

Gi s ph ti c in trl Rt, trc khi mc A vo mch th dng

din c tnh:tR

UI = .

Khi mc A ni ti p vo mch, do nh hng ca in trA, dng

in c tnh:At

A RR

UI

+= .

Vi IA l dng in ch bi ampemet, RA l in trca ampemet, Rt

l in trti. Sai s ph trong qu trnh o lng sc tnh:

Ta thy sai s do A gy ra i vi mch ti cng nh nu in trcaampemet cng nh so vi in trti. V th yu cu i vi ampemet odng in l in trca ampemet cng nh cng tt.

Vi mt ph ti c in trl Rt cp chnh xc ca ampemet s dngl y (hoc chnh xc yu cu ca mch ly tn hiu dng l lth in

trca ampemet phi m bo iu kin sao cho:


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Nu khng m bo iu kin trn, sai s ph gy ra s ln hn sai

s yu cu lc ta phi s dng cng thc hiu chnh:

Trong trng hp in trtrong ca ngun cung cp ng k so viin trti, th Rtc tnh l in trti cng thm vi in trngun.

V d 3.1: Tnh in trca (A) khi th nghim o in trmt chiucun dy th cp ca MBA 560KVA, 10/0,4 KV nh Hnh 3.2, bit chnh xc yu cu % = 0,5%.

Theo l lch, in trmt chiu ca cun dy th cp l Rt=50 (m).

Nh vy iu kin cn ca vic ly tn hiudng qua ti m bo sai s nh hn 0,5% l

RA 0,25 (m).

Trong thc t khng c (A) no tho mn nn s th nghim nykhng c ngha.

3.1.1.2. Khio in p

Volmet l mt phn tc trng cho nhm cc phn t phn ng viin p nh: cun p ca cng t, wattmet; cc rle in p, cc mchkhuch i in p... nn khi xt ti yu cu i vi volmet l xt chungcho c nhm.

Khi o in p, volmet c mc song song vi ti nh Hnh 3.3.Nh vy ta thy in trca ti c mc song song thm vi in trca volmet v lm thay i in p trn ti v gy ra sai s ph trong qutrnh o lng. Xt khi cha mc volmet vo mch, in p trn ti c

tnh:


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trong : E l sc in ng cangun, Rt l in tr ti, Rn l ni trca ngun.

Xt khi mc volmet vo mch, in

p Uv do volmet o c s l:

Sai s phP do volmet gy ra c tnh:

Ta thy sai s ph do volmet gy ra cng nh nu in tr ca ncng ln so vi in tr ti. V th yu cu i vi volmet l in trcng ln cng tt. Thc t trn cc thit bo hin i hoc trn ng hvn nng ngi ta ghi tng trvo ca n.

Vi mt ph ti c in trRtt trong mch c in trngun Rnnu dng volmet cp chnh xc (hoc chnh xc yu cu ca mchly tn hiu p l ) th in trca volmet phi m bo iu kin saocho P < hay ta c:

Nu khng m bo iu kin trn, sai s ph do vomet gy ra lnhn sai s ca bn thn c cu ch th v ta phi dng cng thc hiuchnh.

V d 3.2: Tnh tng trvo yu cu ca mch khuch i ca mt

my in tim nh Hnh 3.4. Bit u1 = 7mv, Rd = 100k. (in trtrungbnh da ngi), chnh xc yu cu % = 1%.


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Bi lm:

Ta c % = 1% nn = 0,01.

3.1.2. Yu cu vc tnh tn

Ngoi yu cu vin trcc ampemet v volmet xoay chiu phi cc tnh tn thch hp vi di tn s cn o. Lm vic ngoi di tn s s gy sai s ph do tn s. Sai s ny phi tnh n nh hng cacc mch o lng i theo ch th nh Shunt, bin dng, bin p, chnh

lu, khuch i v.v. Cng v vy trong nhiu ampemet v volmet, lc cnm bo sai s do tn s nh hn gi tr quy nh (thng l b hn cpchnh xc quy nh cho dng c) ta phi s dng trong mch o cnhng nhng khu b tn s. C trng h p ngi ta phi s dngnhng linh kin c bit m bo tn s lm vic ca dng c. Trncc dng co dng v p xoay chiu c ghi tn s hay gii tn s lmvic.

3.2. o dng in trung bnh v ln bng cc loi ampemet

3.2.1. Phng php sdngNgi ta s dng mt s ccu ch th cin ch to ampemet

o trong mch mt chiu v xoay chiu.

Ampemet t in: Ch to da trn c cu ch th tin, c cim l rt nhy, tiu th t nng lng nn thng dng ch to

ampemet c c p chnh xc t (0,5 2). i vi ampemet tin, khinhit thay s lm cho in trca cun dy thay i dn ti sai s. gim sai s ngi ta thng dng phng php b nhit, tc l dng mt

nhit in tr c h s nhit in tr m mc ni ti p trong mch ca


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ampemet, v vy s lm cho in trca ampemet gn nh khng thayi theo nhit . Ampemet tin ch c tho dng in mt chiu.

Ampemetin t. c ch to da trn ccu ch thin t. Loiny c chnh xc thp hn nhng n bn chc, d s dng v r tinnn c s dng rng ri hn trong cng nghip. Ampemet in t ctho c c dng mt chiu v dng xoay chiu nhng ch yu l o

dng xoay chiu. C nhiu loi ampemet in t, chng ging nhau vnguyn l lm vic song ch khc nhau v hnh thc, s vng dy v kchthc cun dy t phn tnh.

Ampemetin ng: C cu to phc t p v t tin nn ch dngtrong nhng trng hp cn chnh xc cao, hoc tn hiu o c tn scao hn. Sai s tn s trong di t mt chiu ti 3000Hz c xem nhkhng ng k.

Vi cc ampemet in ng khi dng nh mc I 0,5A th cun dyng v cun dy tnh ni tip nhau, cn khi dng nh mc ln hn thcun dy ng v cun dy tnh mc song song vi nhau nh hnh v:

Ampemet chnh lu: Khi odng c tn s cao hng kHz hocmch o dng trong cc ng hvn nng ngi ta thng dng ccampemet t in chnh lu. Ccampemet chnh lu c th s dngchnh lu mt na hay hai na chuk. Tuy nhin s ch ca ampemetchnh lu l gi tr trung bnh ca dng xoay chiu, nhng thng thng

cc dng co in t hoc in ng li ch gi tr hiu dng ca ng


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xoay chiu. V th thng nht s dng ngi ta quy c khc vch ccdng c chnh lu theo cc gi tr hiu dng, vi iu kin dng in lhnh sin. Vy nu em dng c chnh lu o dng khng sin s phmthm sai s v hnh dng, ta phi xc nh hiu chnh.

Nu chnh lu mt na chu k th gi tr dng in trung bnh qua ccu l:

vi I l gi tr hiu dng ca dng in xoay chiu. Nu chnh lu haina chu k th

3.2.2. Cc phng php mrng thang o

3.2.2.1. i vi ampemet mt chiu

Ta bit ccu ch th tin dng ch to cc ampemet cho mchmt chiu. Khung dy c qun bng dy ng c kch thc nh t

0,02 0,04 mm. V vy dng in chy qua khung dy thng thng nhhn hoc bng 20mA. V vy khi cn o dng in ln hn ta phi dngRs (in trShunt) l in trc ch to bng hp kim ca maganc n nh cao so vi nhit . in trShunt c mc song songvi ccu o nh Hnh 3.7 (Shunt = r nhnh).

Ta gi I l dng in cn o, I0 l dng in chy qua ccu, Is ldng chy qua in trShunt Rs, R0in trca ccu o.

Ta c:


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Khi bit R0 dng in nh mc lch ton thang o I0 dng cn o I,ta c th tnh c:

Mt ampemet mt chiu c th c nhiu gii hn o, thay i gii

hn o bng cch thay i gi tr RsCn ch rng trn Shunt c cp chnh xc,

c ghi gi tr dng nh mc, gi trin trvthng phn thnh cc cc dng v cc pring nh Hnh 3.8.

V d 3.3: Tnh in trShunt cho mt bin phn c dng cn o l I = 10kA. Bitdng nh mc qua ccu l I0 = 20mA, in

trccu l R0 = 1.Bi lm:


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3.2.2.2. i vi ampemet xoay chiu

a) Phng php chia nh cun dy

Vi anlpemet xoay chiu mrng thang o ngi ta khng dngRs, v nh th s cng knh, t tin, gy tn tht nng lng, mt anton. Thng thng cun dy tnh c cu to thnh nhiu phn onc s vng nh nhau, thay i gii hn o bng cch i ni cc phn

on y theo kiu song song hoc ni tip, tuy nhin phi m bo iukin sc tng tng trong thit b bng hng s.

b) Phng php dng bin dngin

Bin dng in (BI) l mt my bin p

c bit c cun scp rt t vng cho dngph ti trc ti p chy qua. Cun th cpqun rt nhiu vng, dy nh v c nikn mch vi mt ampemet (hoc cundng ca cng t, wattmet...). V in trca ampemet rt nh cho nn c th coimy bin dng lun lm vic ch ngnmch.

Ta c:


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KI gi l h s my bin dng.

Thng thng, d dng cho vic ch to v s dng, W1 ch cmt vng, ng vi dng in I1chnh mc theo mt dy su

tin no ; W2 nhiu vng hn ng vi dng I2chnh mc l:I2m = 1A hoc I2m = 5A.

V d: my bin dng: 100/5 ; 200/5; 300/5...

Trong trng hp ampemet ni hp b vi bin dng in th s chca ampemet c khc theo gi tr dng in I1 pha scp.

Cn ch rng bin dng in l phn t c cc tnh, c cp chnhxc, v phi c kim nh trc khi lp t.

3.3. o dng in nh

o dng in nhc t ra khi dng in cn o nh hn dngnh mc ca ccu. Cho ti nay vic nng cao nhy ca dng c vh th p ngng nhy ca dng c v cc mch khuch i l rt khkhn, sau y l mt s phng php.

Ngi ta s dng phng phpckh tng nhy ca cc ink, ng ch nht l in k tin. in k tin s dng ccu

ch th tin c nhy cao. Bin php nng cao nhy l tng tcm trong khe hkhng kh v gimh s phn khng ca dy treo.

Tng t cm trong khe hkhngkh bng cch dng nam chm vnhcu c kch thc ln, tuy nhin ti nay t cm trong khe hkhngkh ca ccu ch th tin vn cha vt qu 0,1T.

Gim hng s phn khng ca dy treo, tuy nhin nu gim qu dn


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n ko di thi gian dao ng ca c cu gy kh khn cho vic ongi ta tm cch dung ho gia hai yu t trn.

Bin php quang hc: L s dng khong cch t thang chia ngng quay ca in k tng nhy, gng gn trn trc ca phnng, c mt ngun sng chiu vo gng, lia phn x ca gng chiuln thang o nh hnh v

a l khong di chuyn ca vt sng trn khong chia ,

ll khong cch t gng ti thang chia .

S dng gng quay s tng quay ca tia phn chiu khi gng

quay i mt gc so vi tia ti, lc tia phn x quay i mt gc 2.

Nh vy nhy c tng ln gp hai ln.

3.4. o in p trung bnh v ln bng cc loi volmet

3.4.1. Phng php sdng

Ngi ta s dng cc ch th cin ch to cc loi volmet oin p nh volmet tin, volmet in t, volmet in ng.

Volmet tin: Volmet tin c cu to t ccu ch th tin,loi ny thng dng o cc in p mt chiu, c nhy cao, cho

php dng nhi qua, cng c th s dng km vi b chnh lu oin p trong mch xoay chiu (trong trng hp cn nng cao chnhxc hoc nng cao di tn s ca tn hiu o). Tuy nhin ging nhampemet ta phi ch ti h s hnh dng ca dng hnh sin.

Volmetin t: Volmet in t c cun dy b tr phn tnh nn cth qun nhiu vng dy to nn in tr ln kh d dng, tuy nhinnu qun nhiu vng dy qu m khi o mch xoay chiu th xut hindng in cm ng sinh ra bi tn s ca dng in, do snh hngn tr s trn thang o ca volmet. Khc phc iu ny bng cch mcsong song vi cun dy mt tin b.

Volmet in ng: Khi o in p tn s cao hn tn s cngnghip hoc khi cn nng cao chnh xc ca php o ta dng volmet

in ng, trong volmet in ng bao gi cun dy tnh v cun dy


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ng cng c mc ni tip nhau.

Khi o in p c tn s qu cao, s c nhng sai s ph do tn s, vvy phi b tr thm t b cho cc cun dy tnh v ng.

3.4.2. Phng php mrng gii hn o

3.4.2.1. Phng php dngin trph

Yu cu cbn ca volmet l in trca n cng ln cng tt v th m rng thang o trong cc volmet cch n gin nht l ni thmin trph vo ccu o nh Hnh 3.13.

Hnh 3.13. Mrng thang o cho volmet

vi: R0in trca ccu o

RP l in trphU0in p t ln ccu

UXin p cn o.

Ta c:


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Cc in trph thng c ch to bng hp kim ca ma ngan c

chnh xc cao v t thay i theo nhit . ch to volmet nhiuthang o th ngi ta dng nhiu in trph mc ni tip vi ccu cno.

V d: Sin ca mt volmet c ba gii hn o

3.4.2.2. Phng php dng bin in p

Khi cn o in p cln hng KV trln, nu dng in trph scng knh v t tin, tn hao cng sut v mt an ton, do ta phidng bin in p o lng BU. Bin in p o lng l mt my binp c bit m cun scp qun rt nhiu vng c ni vi in p cno, cun th cp qun t vng hn c ni vi volmet in t hoc inng (hoc cun p ca cng t, wattmet...).

V volmet c in trln nn c th coi bin in p lun lm vic ch khng ti.

Ta c:


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tin trong qu trnh s dng vch to ngi ta quy c in p nhmc ca bin in p pha th c p baogicng l 100V. Cn pha scp c

ch to tng ng vi cc cp ca inp li Khi lp hp b gia bin in pv volmet ngi ta khc volmet theogi trin p pha scp.

Ging nh bin dng in, bin inp l phn t c cc tnh, c c p chnhxc, v phi c kim nh trc khilp t.

3.5. o dng in v in p bng phng php so snh3.5.1. Khi nim

Cc bin php o dng v p k trn s dng ch th cui cng lnhng ccu cin lm quay kim ch trn thang chia , nh vy sais khng th nh hn sai s ca cc ch th dng vo dng c v cha kn sai s gy ra do cc mch o s dng. Cp chnh xc cao nht cacc dng co cin hin nay cha vt qu 0,01 nn php o trctip trn cng khng vt qua cp chnh xc y.

nng cao chnh xc vphp o in p, tng tng trvo, ngi ta dng phng phpso snh hay cn gi l phng

php b tc l so snh in pcn o vi in p mu. y lnguyn l ca tt c cc in thk, cc volmet s c chnh xccao nht hin nay.

Nguyn tc cbn ca phng php so snh c tm tt nh sau:


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in p cn o UXc so snh vi in p b Uk l in p ri trnin trRk. Rk l in trmu c chnh xc rt cao v rt t thay i

theo nhit . Trong qu trnh so snh nu U = 0 ta c so snh cn

bng, nu U 0 ta c so snh khng cn bng hay l so snh kiu visai. U c xc nh bng dng c c nhy cao hay dng c tng

pht hin s chnh lch hay cn gi l cquan zero.

Cc loi phng php so snh khc nhau ch khc nhau cch toi lng b Uk. chnh xc ca in p b v cc yu cu khc cngvi nhy, ngng nhy ca dng c cn bng hay cquan zerou do sai s yu cu ca php o quyt nh.

Sau y ta xe tm hiu mt s dng co dng phng php so snh.

3.5.2. in th k mt chiu

3.5.2.1. in thkmt chiu in trln

S ca in th kmt chiu in trln nh Hnh 3.17.

Rk, Rc l cc bin tr, EN l ngun pin mu, Ucc l in p cung cpcho mch, UX l in p cn o, G in k ch khng.

in th k mt chiu in tr ln gm hai mch chnh l mch todng cng tc v mch o. Khi o ta tin hnh hai thao tc:

+ iu chnh dng cng tc

Khi iu chnh dng cng tc ta ng kho K sang v tr 1,1 ni


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in k vo mch to dng cng tc, ta iu chnh Rc in k G chkhng, khi xy ra quan h:

EN = URN = IP.Rc

Gi tr dng cng tc:

+ Tin hnh o in p cn o UX

Ta ng kho K sang v tr 2,2 ni in p cn o UX vo mcho, sau ta iu chnh con trt trn in trRk cho n kh in k Gch khng.

Lc ta c quan h sau:

Vy in p UXc xc nh theo quan h trn.Trn s ngun pin mu ENc ch to vi chnh xc cc

0,001% 0,01% v c h s nht nh (EN = 101863V). Tuy nhin gi tca pin mu bnh hng bi nhit ca mi trng xung quanh.

Quan h gia gi tr ca phi mu vi nhit ca mi trng nhsau:

trong EN20oC l gi tr ca pin mu nhit chun 20oC, thng

EN20oC = 1,0186V, t l nhit ti ni s dng in th k.

Ch : Thng thng ngi ta iu chnh sao cho RN = 10186dng cng tc IP = 0,1A, thun li cho qu trnh tnh in p cn o UX.

Sin th k mt chiu loi ny gi trin trRk tng i ln,in p cn o UX cV cho nn nh hng ca in trtip xc v scin ng tip xc khng ng k, ngc li nu o in p crt nh ta

phi dng in th k mt chiu in trnh.

3.5.2.2. in thkmt chiu in trnh


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in th k mt chiu in trnhc ch to da trn nguyn tcgi nguyn gi trin trmu Rk thay i dng cng tc IP qua Rkthay i in p Uk= IPRkb li vi in p UX.

S nguyn l ca in th k mt chiu in trnh nh hnh v:

Hnh 3.18. Sin thkmt chiu in trnh

Ngi ta to ngun dng mu IP qua in trmu Rk bng khuchi thut ton.

t u vo khuch i thut ton mt gn mu EN b vi inp ri trn cc in trmc song song u vo khuch i. Nu EN vin p ri trn cc in trmc song song Ug b hon ton nhau:

Mt khc tu ra ca khuch i thut ton ta c:

vi gg ==

n

1iig , gi l cc in dn mc song song u vo khuch i.

Vy:

Ta iu chnh cng tc K thay i cc gi tr dng cng tc IP chon khi kim in k ch khng, ta c:


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Trong mch to in p b khng c u tip xc cho nn loi trc sai s do sc in ng tip xc v in trtip xc. Sai s ch yul do ngng vo v h s khuch i ca b khuch i quyt nh.

3.5.3. in th k xoay chiu

V nguyn l th in th k xoay chiu cng so snh in p cn ovi in p ri trn in trmu khi c dng in cng tc chy qua.Song i vi tn hiu xoay chiu th vic to mu v iu chnh cn bngkh khn v phc tp.

hiu chnh dng cng tc trong mch xoay chiu ngi ta khngdng gn mu (v khng c pin xoay chiu) m phi chnh nh nhampemet c chnh xc cao. Do cp chnh xc ca in th k xoaychiu khng th vt qu cp chnh xc ca ampemet, mt khc muncho UX v Uk Cn bng th phi iu chnh cn bng c v modun v v

gc pha, tc l tho mn ba iu kin l in p UX v in p Uk phicng tn s, cng bng nhau v tr s v UX v Ukphi ngc pha nhau.

thc hin iu kin th nht ngi ta mc in p UX v Uk vongun cng tn s. Dng b ch th khng thc hin iu kin th haiv phi tch Uk thnh hai thnh phn lch nhau 90

o to UX ngc Uk

C hai loi in th k xoay chiu l:

- in th k xoay chiu to cc;

- in th k xoay chiu to vung gc ( cc).

3.5.3.1. in thkxoay chiu to cc


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Trong in th k xoay chiu loi ny, in p cn o UXc cn

bng vi in p ri trn in trR (xc nh bi cc con trt D1 v D2)Mun UX = IPR. Dng cng tc IPc xc nh nhampemet chnhxc cao v in trRiu chnh (Rc). Biu chnh pha dng cn

bng v pha, ng thi cng lm ngun cung c p cho mch to dngcng tc Ip, biu chnh pha ny chnh l nhc im ca in th kxoay chiu v kh xc nh chnh xc v tr n nh ca phn quay ngvi gc quay khi iu chnh pha v dng IP thay i lm cho vic iuchnh cn bng kh khn.

3.5.3.2. in thkxoay chiu to vung gc

Trong in k s dng hai cun dy t gn nhau v dng h cm Mca chng to Uk thnh hai thnh phn lch nhau 90

o v UX s cn bng

vi tng hai vc tthnh phn.


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S gm hai mch cng tc v mt mch o. Mch cng tc thnht gm bin trdy qun c chun ho AB c im gia l O,cun scp w1 ca bin p khng li thp, ampemet A v in tr(Rc).Dng in I1 t ngun cung c p xoay chiu (c xc nh nhampemet) to trn bin trAB mt in p UAB. in p Uk1c xcnh bi dng I1 v v tr con trt D1 trn bin trAB. V dng I1 khngthay i trong qu trnh o nn thang chia c khc theo gi trinp trn bin trAB.

Mch cng tc th hai gm bin tr dy qun c chun hoA'B' c im gia O' ni vi im O gia ca bin trAB, cun thcp w2 ca bin p khng li v hp in trRf b tn s. Dng inI2 trong mch cng tc lch pha vi I1 gc 90

o (v in cm L2 khng lnlm nn c th coi nh I2 trung pha vi E2 m E2 lch pha vi E1 mt gc90o). Trong mch th nht I1 c gi tr xc nh nn I2 cng c gi tr xcnh:


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M l h cm ca w1 v w2

Ta xc nh Uk2 = I2R2 (R2 l mt phn in tr ca AB c xcnh nhv tr ca con trt D2 trn A'B'). V Ukl = I1R1 v Uk2 = I2R2 mI1 v I2 lch nhau mt gc 90

o nn Ukl v Uk2 cng lch pha nhau 90o.

Ch rng khi tn s f thay i = 2f, nh vy khi thay i dnti I2 thay i v gi tr khc trn AB cng thay i. khc phc

ngi ta dng hp in trph Rf b cho tn s khng i (tc l Rfthay i ph thuc vo s thay i ca tn s ngun cung cp).

Mch o l mch ch yu ca in th k bao gm ngun tn hiucn o UX, in th k ch khng G, cc phn ca bin tr dy qunchun D1O, D2O'.

th biu din cc gi tr Uknh Hnh 3.19.

iu chnh cc con trt Uk1 v Uk2 trn cc bin trdy qun AB vAB thng qua tnh ton ta sc tr hiu dng v gc pha ca in p

UX cn o

Sai s ch yu ca in th k xoay chiu l sai s ca ampemet (nhnht l 0,1)

3.5.3.3. in thktng tghi

Loi ny thng dng o nhit l ti, ram, nhit luyn, dng nhn


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dng cc i tng l l gia nhit.

- S tm tt nguyn l nh Hnh 3.23.

S gm cc khi nh sau:

+ Cp nhit in c nhim v bin i t nhit tx sang sut inng mt chiu Ex. Vi h thng thc thng c thm mch b nhit u t do.

+ Cu so snh gm EP, RP v cc in p mu khc nh: R0, R1, R2,R3, R4. Nhim v l to ra cc in p mu mt chiu vi chnh xccao (Trong thit b thc t EP c ly t ngun in p xoay chiu220V qua b chnh lu, qua n p mt chiu vi cht lng cao).

+ B bin i mt chiu, xoay chiu c nhim v bin i in p

mt chiu U thnh in p xoay chiu tn s 50Hz. Mch ny c th lcon rung chc hoc rung in t.

+ Mch khuch i c nhim v khuch i tn hiu xoay chiu vicng sut ln cung c p cho cun dy iu khin ng cKB.Tng cui ca mch khuch i s l khuch i cng sut nhy pha

+ Hai ng cgm mt ng ckhng ng b c nhim v kocon trt trn cc bin trRP, R0 v mt ng cng b c nhim v


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ko bng giy chuyn ng trong ch t ghi.

Qu trnh o c chia lm hai bc:

Kim tra chnh xc ca cc in p mu

Lc ny kho Kv tr H trc hp gim tc ca ng cKB ca vo n khp vi u bin trRP. Khi ngun sut in ng chunEcc so snh vi in p U4 l in p ri trn in trR4

Ta c:U= Ec - U4

trong Ec l mt ngun chun vi chnh xc rt cao c sn trongthit b.

U c a vo mch bin i mt chiu, xoay chiu sau ca ti mch khuch i v tn hiu c khuch i ln vi cng sut ln cung cp cho cun dy iu khin ca ng cKB. V tngcui ca mch khuch i l khuch i cng sut nhy pha nn pha ca

in p trn cun dy iu khin s ph thuc vo du ca U.

Tm li, khi U 0, ng cKB s quay ko con trt trn u

bin trRP thay i in p Uk theo chiu hng sao cho U 0.Lc mt tn hiu iu khin v dng li. Cc in p mu trn ccnhnh ca cu coi nht yu cu v chnh xc.

Qu trnh o nhit

Lc ny kho Kv tr X, trc hp gim tc ca ng cKB ca vn khp vi u bin trR0

Nhcp nhit in, nhit cn o bin thnh sut in ng mtchiu Ex. Khi o Ex ta so snh vi U12 l in p ri trn cc in trmu

R1, R2 v mt phn R0. Ta c: U= Ex - U12

Khi U 0 th theo nguyn l phn trn, ng cKB s quay,ko con trt trn u bin trR0 thay i U12 c xu hng sao cho

U 0 th mt tn hiu iu khin v dng li. Lc ta xc nh cEx = U12. Vy cn c vo v tr ca con trt trn bin trR0 ta xc nhc U12 ri ta suy ra Ex. Thc t trn R0 ngi ta c sn cc vch chiatheo n v nhit nn ta c c kt qu.


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Qu trnh tghi

Lc ny trn u bin trR0 ta gn sn mt ngi ghi, ngi ghi t lnbng giy (mt cch lin tc hoc gin on hoc bng nhit).

Trong ch t ghi bng giy c ng cng b ko chuynng vi tc khng i. Nh vy s to ra trc thi gian t.

Ta thy khi to thay i, ngi ghi s chuyn ng t tri sang phi nh

ng cKB, cn bng giy th chuyn ng vi tc khng i tdi ln trn nhng cng b nn ngi ghi s v trn bng giy biu nhit theo thi gian.

3.6. o in p bng cc volmet ch th s

Ngy nay volmet sc s dng rng ri trong o lng v khnng chnh xc kh cao, gn nh, thun tin cho ngi s dng. Tu theocch bin i in p thnh cc i lng ch th s m ngi ta chiara thnh ba loi volmet s nh sau:

- Volmet s chuyn i thi gian;

- Volmet s chuyn i tn s;

- Volmet s chuyn i trc tip (chuyn i b).

3.6.1. Volmet s chuyn i thi gian

Nguyn l chung l bin i in p cn o thnh khong thi gian,sau lp y khong thi gian bng cc xung c tn s chun (f0) sau dng bm m s lng xung (N) t l vi Ux suy ra Ux

S cu trc chung ca volmet s nh sau:


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Hnh 3.25. S cu trc volmet s chuyn i thi gian mt nhp

Biu thi gian:

Nguyn l lm vic:

Khi mmy ti thi im t1, my pht xung chun qua b chia tnkhi ng my pht xung rng ca, u ra my pht xung rng ca cUrc (Uk) i tin b so snh so snh vi in p Ux cn o u vo.ng thi cng tu ra ca b pht in p rng ca c xung th nhtn trigv t trigtrng thi kch hot mkho K cho php ccxung mang tn s chun (f0) t pht xung qua kho Kn bm v chth s. Ti thi im t2 khi Ux = Urc thit b so snh pht xung th 2 tcng vo trigv kho kho K, thi gian t t1n t2 tng ng vi tx. Ty ta c mi quan h:


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Vi mt my pht in p rng cn cnh th t v tr l hng s, vvy Ux t l vi tx s lng xung n bm trong khong thi gian tx sl:

S lng xung i qua kho K trong mt chu k ca in p rng cat l vi in p cn o.

Sai s ch yu l do my pht in p rng ca gy ra, tc l t v trkhc hng s, tip theo l sai s lng t.

Ch : Khi Ux bin thin vi mt tc no th khng tho

c v ng cong in p rng ca khng ct Ux. Do vy s bin thinca in p Ux Phi m bo iu kin sau:

3.6.2. Volmet s chuyn i tn s

Nguyn l lm vic ca volmet s chuyn i tn s da trn nguyntc bin in p thnh tn s ri dng cc my o tn s ch th theoin p.

S cu trc ca volmet s chuyn i tn s nh sau


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in p cn o Uxc a n u vo, qua khu tch phn sc

in p U1. U1c so snh vi in p U2 (in p U2 c n nh caoKhi U1 = U2 thit b so snh pht xung qua khuch i 2 (ti thi im tthng kho K1 v Kn bm, ch th s. Khi K1 thng, in p U0(ngc du vi U1) qua K1n b vi in p U1 (y l mch phngin ca t C) trong khong thi gian T

k(t t

1n t

2) ti t

2in p U

0b

hon ton U1


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Vy tn s fx t l vi in p cn o Ux

ch th s ta dng phn to gc thi gian v cc kho, bm vch th s ging nh mt my o tn s nhng hin th theo in p. Cth b to gc thi gian l my pht xung chun T0 to thi gianTc = kT0iu khin kho cho cc xung mang tn s fx qua n. S lngxung mang tn s fx qua kho K trong khong thi gian Tcn ch th sc xc nh nh sau:

m bo chnh xc th ngun to in p U0 phi n nh, sai sca volmet loi ny c hai loi l sai s do chuyn tin p sang tns khong 0,2% v sai s lng t khong 0,01%.

3.6.3. Volmet s chuyn i trc tip


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Ta so snh in p cn o Ux vi in p chun Uk ph thuc vovic gia cng i lng b Uk v quy trnh so snh vi Ux m ngi ta

phn ra thnh volmet s chuyn i trc tip kiu b qut v volmet sbin i trc tip kiu tung.

3.6.3.1. Volmet schuyn i trc tip kiu b qut

in p b Uk thay i lp li theo chu k, trong mi chu k bin

thin ca Uk ta ly s do mt ln tc l ti thi im Ux, Uk ta c ktqu ca php o. in p b Uk c th thay i tuyn tnh hoc thay itheo bc thang. Nu thay i theo bc thang th c bc thang bng nhauv bc thang khng bng nhau.

S cu trc gm hai phn: phn chuyn i in p Ux thnhkhong thi gian Tx v phn o khong thi gian. Thc cht gm hai

phn l phn bin i in p cn o thnh s lng xung N1 v phntip theo lm nhim v bin i s lng xung N1 thnh m thp phn

N10iu khin cc phn t hin s.


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Nguyn l lm vic:

Khi mmy, b pht xng chun bt u lm vic, cc xung f0nbm 1 v kho K; sau mt tp xung f0 tng ng vi thi im t0 thbm 1 pht xung n thng kho K, trong thi gian K thng cc xungmang f0 qua Kn bm 2 v ch th s. ng thi c mi xung f0n

D/A s tng in p ra ca n Uk ln mt mc U (cc mc U bng

nhau). Qu trnh tip tc cho n khi Uk Ux (ti thi im tr b so snhtc ng vo b khuch i to tn hiu kho kho K. Qu trnh o kt

thc v b phn ch th hin kt qu. Nu tt c cc mc in p U tonn Uk liu bng nhau th s lng xung N1 s t l vi in p cn o

Ux tc l Ux Uk = N1U. y l gi tr tc thi ca in p cn o tithi im t1. Nu mun o Ux ti thi im khc th qu trnh o s lpli tu.

i vi volmet ch th b qut vi i lng Uk thay i theo cc bc

thang khng bng nhau. Trong cc volmet ny cc mc bc thang Ukhng nh nhau, c th to cc U theo tng hng m ca con shm nht nh. Do c th da vo hm nh phn, nh thp phn vthp phn gia cng in p b Uk.

V d 3.4: Qu trnh gia cng in p b Uk theo hm thp phn.


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Hnh 3.31. Qu trnh gia cng in p b

Qu trnh so snh t hng ln nht, vi Ux = 43V. Con s thp phnc hai hng m l hng chc v hng n v. Nguyn l ca qu trnh sosnh nh sau:

+Nu Uk> Ux th m s ghi l 0.

+Nu Uk Ux th m s ghi l mt s tng ng vi hng m ca

Uk v khi hiu |Uk - Ux| < U (mc ca hng m) th qu trnh so snhs chuyn sang hng m nh hn). C thy ta bt u so snh Uxvi Uk= 90 ta sc m ra l 0, Uk = 80 ta sc m ra l 0,... chon khi Uk= 40 tc l:

Lc ny m ra s l 4 (hng chc nn ghi l 40) tip tc qu trnh

so snh s din ra hng n v vi gi tr ln nht ca hng l 9 v mimc U = 1

Qu trnh gia cng Ukkt thc ta sc tng gi tr

y Uk10 l m hng chc, Uk1 l m hng n v.

3.6.3.2. Volmet schuyn i trc tip kiu tung

Trong cc volmet ny i lng b Uk thay i lun bm theo sbin thin ca i lng cn o Ux. V vy trong cu trc ca n c bchuyn i A/D, D/A tc ng theo hai chiu thun nghch. c im c

bn ca dng co l kh nng cho kt qu lin tc ti thi im bt k.

Volmet s chuyn i trc tip kiu tung c hai loi bao gm loigia cng i lng b Uk thay i theo bc thang bng nhau v loi gia


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cng i lng b Uk thay i theo bc thang khng bng nhau.

a) Volmet schuyn i trc tip kiu tung c Uk thay i theo bcthang bng nhau.

Nguyn l cbn

in p Uxc so snh vi in p b Uk bt u t thi im t1

in p Uk tng lin tc, mi mc tng l Uk (l nhng bc thang bngnhau) cho n thi im t2 khi Ux Uk. Xut hin bt phng trnh

Ux - Uk< Uks kt thc qu trnh o v cho ra kt quch th s.

Thi gian gia cng c xc nh bi s mc lng t ln nht (Ndm)

v thi gian t ca mt mc lng t.

t0 = Ndmt

Da vo sai s lng t yu cu xc nh Ndm.

+ Volmet s chuyn i trc ti p kiu tung vi bm thunnghch c cu trc nh sau:


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Khi bt u lm vic b pht xung chun pht lin tc n ch

kho (K). Ti thi im Ux = 0 hoc Ux = Uk th kho (K) kho, cc xungmang tn s f0 khng thn bm thun nghch. Khi Ux > Uk tc l

Ux - Uk = U > 0, tn hiu U qua khuch i c lch n thng kho Kv iu khin bm lm vic ch cng. M ra ca bm iu

khin b chuyn i D/A tng dn Uk cho n khi Ux Uk th kho K skho, kt thc qu trnh o, b phn ch th s cho kt qu o. Khi

Ux < Uk tc l Ux - Uk = U < 0 th khuch i c lch to xung thngkho K, iu khin bm lm vic ch tr. M ra ca bm iu

khin chuyn i A/D gim Uk cho n khi Ux Uk th kho K s kho,b phn ch th s cho kt quo.

+ Volmet s chuyn i trc ti p kiu tung vi ng c thunnghch.

S khi nh sau:

Ta m ho gc quay ca ng c (tc l U c bin thnh


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gc ca ng c). Dng c thng c hai u ra, mt u l m s,mt u khc l tn hiu tng t (sau ng c) c th ghi hoc ch th

bng kim trn thang chia . Khu A/D ca dng c l chuyn i khng

gian dng mt n hoc thc m ho bin i gc quay thnh mGray ri t m Gray thnh m nh phn, gii m, ch th s.

b) Volmet schuyn i trc tip kiu tung c Uk thay i theo bc

thang khng bng nhau.Volmet gm hai loi vi hai php gia cng Uknh sau:

+ Gia cng Uk t hng m ln nht

Trng thi ban u, tt c cc hng m ( cc) u bng 0 tc lUk= 0. Trong mi hng bt u t s nh nht ca hng m tng dn Uk

cho n khi hiu Ux - Uk < Uk ca hng th chuyn sang hng mnh hn v qu trnh lp li nh trn. Qu trnh o (gia cng) kt thckhi:

hng nh nht, thit b so snh s thng bo iu ny.

Nu Ux = const th Uk s tng lin tc hoc gim lin tc, s mclng t khng ln lm. Nu Ux bin thin, Uk s thay i cho ph hpvi s bin thin ca Ux siu khin s phc tp hn, s lng nhpthc hin gia cng Ukc xc nh:

trong : n l s lng nhp; a1, a2 a3, l s mc ca cc cc tothnh gi tr s ca i lng cn o.

Thi gian cc i gia cng theo phng php ny:

t0 = k.9.t

k l s cc, 9 l ch s trong mt cc.

+ Gia cng Uk t hng nh nht

Trng thi ban u Uk= 0 v bt u t gi tr nh nht ca hng nh

nht. V d hng n v: Uk = 0, 1, 2, 3,..., 9. Nu gia cng ht hng nh


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m hiu Ux - Uk > Uk1 (Uk1 l mc gi tr ca hng nh nht) th tiptc gia cng n hng ln hn khi xut hin Uk > Ux tc l hiu Ux - Uk

i du th quay trv hng m nh nht v gim dn tng mc Uk1

gim Uk cho n khi Uk Ux. Qu trnh o kt thc v kt qu hin rach th s. u im ca phng php ny l siu khin tng in gin, nhc im l thi gian gia cng di, nht l trng hp dng

bn cc m s 9090 phi thc hin 90 nhp. Thi gian gia cng sc bn ch s:

t0 = 90.t.


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Chng 4O CNG SUT V NNG LNG

4.1. o cng sut v nng lng trong mch mt pha

4.1.1. o cng sut tc dng bng wattmet in ng

4.1.1.1. o cng sut trong mch mt chiuo cng sut ngi ta thng dng wattmet in ng, wattmet in

ng c ch to da trn ccu ch thin ng, gc quay ca ccuch thin ng c tnh nh sau:

vi l gc lch pha gia cc dng I1 v I2

S mc wattmet in ng nh Hnh 4.1.

Wattmet in ng c hai cun dy, cun dy tnh cn gi l cundng c cun bng dy c kch thc ln, t vng, cho dng ph titrc tip chy qua hoc ni vi th cp ca bin dng in, n ng vaitr nh mt ampemet. Cun dy ng hay cn gi l cun p thngc ni tip vi RP, c ot trc tip ln in p ca ph ti hoc nivi th cp ca bin in p o lng, n ng vai tr nh mt volmet.

Xt vi mch mt chiu ta c:

cos = 1, I1 I


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vi Ru l in trmt chiu ca cun dy ng.

Thay gi tr I2 vo (4-1) ta c:

vi P l cng sut tc dng m ph ti tiu th qua W vup

1 RR

KK

+=

Kt lun: Gc quay t l bc nht vi cng sut tiu th trn ti,vy c th dng wattmet in ng o cng sut trong mch mtchiu.

4.1.1.2. o cng sut trong mch xoay chiu

Gi s mch xoay chiu c in p u = Umsint v dng ph ti

i = Imsin(t - ) = i1y l gc ti.

V ccu khng c mch t nn dng i2ch chm pha hn so vi in p u mt gckh nh no . Ta c th vc t nhHnh 4.2.

Vn t cng thc (4-1) ta c:

vi u l gc lch pha gia in p v dng in trong cun dy ng.

Cui cng ta tnh c:


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Ta xt hai trng hp:

- Coi gc u rt nh: u 0 (Xu


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trong Unk, Ink l in p v dng in nh mc ng vi thang o

th k no . an l s vch trn chia trn ton thang o. Wattmet inng c th c nhiu gii hn o (Ti sao?) mi gii hn c mt h sCw tng ng. Cng sut o c tnh bng tch ca h s Cw trn thang

o tng ng vi s vch chia m kim ch th th hin.

4.1.1.3. o cng sut phn khng

Ta s dng wattmet inng cng vi in tr, cuncm. S mc nh sau: Cundy dng in c mc ni tipvi ph ti. Cun dy in pc mc song song vi mtin trR1, sau c mc nitip vi mt cun cm L v intrR, ta iu chnh tr s R1, L,R sao cho U v I vung gc vi

nhau. Khi o gc quay cawattmet l:

4.1.2. Wattmet sdng nhng phn tphi tuyn

4.1.2.1. Wattmet nhitin

a) Csl lun chung

Wanmet in ng cho cng sut trong mch in tn s thp v mt di tn nht nh. Khi cn o cng sut tn s cao hoc c trongmt di tn rng no ngi ta dng wattmet nhit in. Phn t c

bn c s dng trong wattmtt nhit din l hai c p nhit in ging


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nhau A, B c mc nh Hnh 4.4.

Gi cng sut sinh ra trn cc in trnhit r l: pa, pb. R rng sutin ng trn cc cp nhit ngu s t l vi pa, pb

Ea = K.pa; Eb= K.pb

vi K l h s t l.

Gi thit ngi ta b tr sao cho dng in qua in trr1 bng tng

ca hai ng i1, i2 cn dng in qua ra bng hiu i1, i2ia = i1+i2, ib =i1 - i2

Khi c th tnh c cng sut nhn c tin cc in tr r nhsau:

Vi cch ni cc cp nhit ngu nh hnh v s ch ca m s bng:


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R rng s ch ca mV t l vi T

0

21 dtiiT

1.Vn y ta cn phi

o cng sut tiu th trn ti. M ta bit cng sut tiu th trn ti s

bng T

0

u.idtT1 .

V vy ta phi xy dng s sao cho cc dng in ia, ib l tng vhiu ca cc dng i1, i2 Mt khc cc dng i1, i2 li phi t l vi dngin v in p trn ti tc l

S ch ca m s l:

vi C = 4abKr.

Tc s ch ca m t l vi cc cng sut tc dng trn ph ti b

b) Wattmet nhitin

Trn csl thuyt nu trn, ta xy dng s nguyn l nh hnh

v


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Trn s A, B l cc cp nhit in, r1 l in trc gi tr rt nh,

l dng in ph ti, ta c: iu i.

Vi cch b tr mch nh trn ta c:

Tng t

Trn s thng chn ra = rb = r. Kt hp vi csl lun ban us ch ca m trong s ny s t l vi cng sut tc dng P ln phti.

4.1.2.2. Wattmet sdng phn tbnh phng

a) Csl lun chung

Ta bit trong thit bin c nhng phn t m u ra (dng, p) tl vi bnh phng u vo v nh vy gi t trung bnh u ra cng tl vi bnh phng gi tr trung bnh u vo. Nhng phn t nh vy cth s dng o cng sut tc dng P trong mch. Loi thng dng ldit bn dn.

Gi thit i lng u ra A t l bnh phng vi in p vo u

A = n.u2.


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S cu trc tng hp s dng hai phn t phi tuyn B1, B2 nhhnh v:

Ngi ta tng hp sao cho

vi U, I l in p v dng in cn s dng o cng sut P. Lc tac:

vi l gc lch pha gia in p v dng in trn ph ti.

T c th vit:

vi C l h s t l, P l cng sut cn o.

Vy c tho A ri suy ra cng sut cn o.

b) Warttmet sdng phn tbnh phng - it bn dn

S c tho cng sut s dng B1, B2 nh sau:


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Trn s r1 l in trShunt. Ta bit vi mt bn dn dng in tl vi bnh phng in p tc l

T ta c:

Lc gi tr trung bnh ca in p ri trn in trra l:

Tng t, nu trn phn t B2 xut hin in p Ub vi gi tr hiudng Ub lc ta cng c:

T hnh v ta c:


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Trn thc t thng chn ra = rb = r, lc in p trn mV s bng

Ch : Wattmet ny s dng trong di tn rt rng, ti hng nghn

Hz c sai s t 1 3 % v tiu th mt cng sut rt nh.

4.3.1. o nng tng tc dng bng cng tcm ng mt pha

C rt nhiu cch o nng lng, song cng t cm ng mt phac ng dng rng ri nht trong k thut v mmen quay ln, lmvic tin cy, sai s nm trong phm vi cho php.

4.1.3.1. Cu to

Cu to ca cng tmt pha nh Hnh 4.8 gm hai nam chm in Av B.

- Nam chm in A gi l cun dng, thng c cun bng dy ckch thc ln, t vng v cho dng ph ti trc tip chy qua hoc nivi th cp ca my bin dng in.

- Nam chm in B c gi l cun p, thng c cun bng yc kch thc nh, rt nhiu vng, t trc tip ln in p li hoc ni


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vi th cp ca bin in p o lng.

- a nhm c kp cng trn trc quay, ngoi ra cn nam chmvnh cu M, thanh dn t G v h thng ccu m.


Xt khi cun dng c dng in xoay chiu i chy qua s xut hin t

thng i xuyn qua a nhm hai ln, khi t in p xoay chiu u ln

cun p s to ra dng in iu chm pha hn so vi in p mt gc 90o.Dng iu sinh ra t thng u. T thng u gm hai thnh phn:

+ up ch khp mch qua mch t cun p gi l t thng ph;

+ uc xuyn qua a nhm gi l t thng lm vic.

i v uc s cm ng trn a nhm nhng dng in xoy. Theonguyn l ca c cu ch th cm ng, a nhm s chu tc dng cammen quay c xc nh:

vi l gc lch pha gia hai t thng i v uc

Ta coi mch t cha bo ho, nn t thng i t l vi I:

i =c1.I

vi c1 = const.

Ta coi tn s l khng i nn uc t l vi U:

uc =c2.U

vi c2 = const.vy mmen quay c tnh:

Mq = Kfc1c2UIsin = K1UIsin vi K1 = Kfc=1c2

Ta xt hai trng hp:

* Trng hp l tng

Coi cc t thng trng pha vi dng in kch thch tng ng, ta c th vc tnh Hnh 4.9.


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T th vc tta thy:

vi l gc lch pha gia dng in v in p trn ti. Vy:

* Trng hp thc t

Cc t thng ny u chm pha hn so vi dng in kch thchtng ng mt gc no (tuy kh nh). Ta c th vc tnh Hnh4.10

Ta xt gc:


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vi 1 l gc lch pha gia dng in v 1 v I.Vy

Do vy ta phi iu chnh gc ai sao cho tho mn iu kin trn.

Khi c mmen quay a nhm s gia tc ti tc rt ln nu khngc g cn li, v vy ngi ta t nam chm vnh cu M to ra mmenhm.

Khi a nhm quay ct ngang t trng ca nam chm vnh cu, trn

a nhm xut hin nhng dng in xoy, nhng dng in ny li tcdng vi chnh t trng ca nam chm vnh cu to ra mmen hm:

a nhm quay tc n nh khi cn bng hai mmen, do tac:

Tch phn hai v ta c:

V tri ca phng trnh t l vi nng lng m ph ti tiu th quacng t trong khong thi gian t t1n t2 cn v phi t l vi lnggc quay ca a nhm cng trong khong thi gian . Ta c:


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(N: s vng quay ca a nhm)

Vy

W' = Cm N

vi Cm l h snh mc ca cng t.

Kt lun: Nh vy ta chng minh c rng s vng quay ca a

nhm t l bc nht vi nng lng in m ph ti tiu th qua cng t.

4.1.3.3. Ccu m v cc thng scbn ca cng t

- C cu m: Gm h thng bnh vt, trc vt, cc con ln v ccbnh rng ch th s.

- Thng s cbn ca cng t:

+ H s truyn ti ca cng t

l lng in nng truyn ti qua cng t khi a nhm quay ht mtvng.

+ H snh mc ca cng t

l s vng quay ca a nhm khi truyn ti qua cng t1 kWh in.

4.1.3.4. Sai sv cch khc phc

Do tn ti ca ma st, do nh hng ca t thng ph, do sai lchhng s ca cng t (mmen cn ln hoc nh) do cng tsai s tnhiu.

Trc khi s dng bt buc phi hiu chnh li tc l tin cch khcphc sai s.

a) B ma st

- Khi ph ti nh, mmen ma st sng k so vi mmen quay.V vy ngi ta phi ch to b phn b ma st trn c s nguyn l

chung l phn chia t thng cun p thnh cc t thng ph bng cc vt


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chia t thng hoc vng ngn mch khng i xng (cha th hin trnhnh v).

- Khi iu chnh v tr vng ngn mch khng i xng hoc vt chiat thng ta s b c ma st (tuy nhin nu iu chnh qu sang trihoc sang phi th cng ts t quay thun hoc quay ngc khi khngc ti).

b) Chng hin tng tquay ca cng tKhc phc hin tng t quay khi mmen b ln hn mmen ma st

ngi ta ch to b phn chng t quay bng cch trn mch t cacun p v trn trc quay ngi ta gn hai l thp non T1 v T2. Khi anhm quay ti thi im hai l thp i din nhau th chng s tc ngtng h v to ra mmen hm (tuy nhin ch vi mmen kh nh).

c)iu chnh gc lch pha 1gia 1 v I

Ta c:

Mong mun rng

coi nh khng i i vi mi loi cng tsau khi ch to. V vyta phi iu chnh gc ai bng cch trn mch t ca cun dng ngi ta

cun vi vng dy ni qua mt in trR c thiu chnh c. Khiiu chnh gi tr R s lm thay i tn hao t trong mch t cun dng,tc l ai thay i.

d) Kim tra hng sca cng t

Ta iu chnh sao cho cos = 1, cho dng in I = In, U = Un lc ta c P = UnIn; o thi gian quay ca cng tbng ng h bm giy,m s vng quay N ca cng tquay trong khong thi gian t.

Ta tnh c hng s ca cng tnh sau:


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Ta so snh Cp vi gi trnh mc ghi trn cng t, nu khc nhau ta

phi iu chnh v tr ca nam chm vnh cu tng hay gim mmencn cho n khi Cp bng gi tnh mc ca cng t. Thc t hin nay,vic hiu chnh cng tthng da vo cng tmu.

4.2. o cng sut v nng lng trong mch ba pha4.2.1. o cng sut tc dng trong mch ba pha i xng

i vi mch ba pha i xng ta c cng sut tng ca c mch l:

- Theo i lng pha:

PA, PB, PC l cng sut tng pha A, B, C.

- Theo i lng dy:

Ud, Id l in p v dng in dy.

4.2.1.1. Mch ba pha bn dy - Phng php mt wattmet

Theo (4-28) ta ch cn o cng sut mt pha bng mt wattmet rily ch s ca wattmet nhn 3 ta sc cng sut ca c ba pha: Gis wattmet mc vo pha A nh sau:

S ch ca wattmet l:


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Do vy cng sut ca ba pha l:

Tng t c th mc wattmet vo pha B hoc pha C.

4.2.1.2. Mch ba pha ba dy - Phng php dng kho chuyn i

S mc wattmet nh sau:

Cun dng c dng in khi kho Kv tr 1 cun p c in p UAC;khi kho Kv tr 2 cun p c in p U

AB.

Vy khi ng kho K v pha 1, s ch ca wattmet l:

Khi ng kho K v pha 2, s ch ca wattmet l:

Hnh 4.12. th vc tca phng php o cng sut

dng kho chuyn i


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Theo th vc tta c:

Tng t ta cng c th mc wattmet pha B hoc C o cngsut theo cch trn.

4.2.2. o cng sut tc dng trong mch ba pha khng i xng

4.2.2.1. Mch ba pha bn dy - phng php ba wattmet

Vi mch ba pha khng i xng, ta c

Do vy ta dng ba wattmet mt pha hoc mt wattmet ba pha baphn to cng sut cc pha A, B, C. Sau cng i s cc s chca ba wattmet (hoc ba phn t) ta c cng sut ca mch ba pha.

Ta c:


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Trong thc t ngi ta ch to wattmet ba pha ba phn t. N gm bac p cun dy tnh tng ng c ba phn ng gn trn cng mt trcquay. Mmen lm quay phn ng l tng mmen ca ba phn t

4.2.2.2. Mch ba pha ba dy Phng php dng hai wattmet

Xt cng sut tc thi trong mch ba pha l:

i vi mch ba pha ba dy, v khng c dy trung tnh nn dngin trung tnh bng khng ngha l:

Vy cng sut tc dng ca ba pha l:

Nh vy ta c th dng hai wattmet mt pha c s nh Hnh 4.14o cng sut trong mch ba pha. Thc t cng da trn nguyn tcny ngi ta ch to wattmet ba pha hai phn t. Cch mc nh sau:

4.2.3. o nng lng tc dng trong mch ba pha


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- i vi mch ba pha bn dy c th dng cng tba pha ba phn thoc ba cng tmt pha. S mc ging nh mc wattmet o cngsut tc dng.

- i vi mch ba pha ba dy c th dng cng tba pha hai phn thoc hai cng tmt pha. S mc ging nh mc wattmet o cngsut tc dng.

- Vi mch h p cng sut ln ta kt h p gia bin dng in vcng tba pha o nng lng tc dng.

V d 4.1: S kt hp gia BI v cng to nng lng tc dngpha h th.

- Vi mch cao p, ta kt hp gia BU, BI v cng tba pha onng lng tc dng

4.2.4. o nng lng phn khng trong mch ba pha

4.2.4.1. Dng cng tphn khng ba pha ba phn t

S mc cng tnh sau:


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Hnh 4.16. Su dy v th vc tca cng tphn khng ba pha ba phn t

im o m thng l u ngun nn ta coi mch ba pha c nguni xng, ph ti mang tnh cht cm.

Ta c mmen quay tng ca cng tl:

Ta thy mmen quay t l vi cng sut phn khng trong mch bapha cho nn s ch ca cng ts t l vi nng lng phn khng tiuth trong mch ba pha.

4.2.4.2. Dng cng tphn khng ba pha hai phn tc cun dy nitip ph

S mc nh Hnh 4.17.

im o m l u ngun nn ta coi mch ba pha c ngun ixng, ph ti mang tnh cht cm. Xt tng phn t, ta tnh c momen quay nh sau:


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Mmen quay t l vi cng sut phn khng trong mch ba pha vys ch ca cng tt l vi nng lng phn khng trong mch ba pha.

4.2.4.3. Dng cng tphn khng ba pha hai phn tc R0 to gc lchpha 60o


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Trong s cng tny, cc cun p c mc ni tip vi in trmu R0. in trny c tnh ton sao cho dng in trong cun p chchm pha so vi in p tng ng mt gc 60o. Ta c th vc tnh

hnh vTa c m men quay ca cc phn t l:

Hn na ta c:

Thay vo ta c:Tng t

Vy m men quay tng l:


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Vy: M men quay tng t l vi cng sut phn khng trong mch

ba pha nn s ny thng c dng o nng lng phn khngtrong mch ba pha. Nu vi mch ba pha khng i xng th c sai s

nht nh.4.2.5. V d so m cao th

Thc t c rt nhiu so m cao th: Tc l s kt hp BU,BI v cng tba pha o nng lng tc dng v phn khng cho mch ba

pha cao th.

+ Cng ttc dng ba pha hai phn t c cun dng cc pha A, B.+ Cng tphn khng ba pha ba phn t.+ Cc cun dng ca cng ttc dng v phn khng u ni pha

th cp ca my bin dng, vy dng nh mc qua cc cun dng l 5A.+ Cc cun p ca cng t tc dng v phn khng u ni pha

th cp ca bin in p, vy in p nh mc trn cc cun p l 100V


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Chng 5O GC PHA V TN S

5.1. o gc pha v h s cng sut cos5.1.1. Phng php o cos gin tip5.1.1.1. Phng php V - A - W

H s cng sut cos quan h vi dng in v in p trong mchqua cng thc:

Do :

Vy dng cc ng h V, A, W o U, I, P trn ti ta tnh c cos.Sai s:

5.1.1.2. Phng php xc nh cosTBTa c:

Vi Wpk l in nng phn khng ch bi cng t phn khng trongkhong thi gian xt;Wtd l in nng tc dng ch bi cng t tc dng trong khong

thi gian xt.Dng cng t o nng lng tc dng v phn khng trong mt

khong thi gian no (thng l mt thng) ta xc nh c cosTBca ph ti theo cng thc (5.2).5.1.2. Phng php o cos trc tip

Thng dng cos met in ng v st in ng.


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5.1.2.1. Cos metin ng mt phaNgi ta s dng ccu ch th logomet in ng ch to dng

co cos trong mch mt pha.Cun tnh ca cosin ng c mc ni tip vi mch cn o

cos (hoc ni vi th cp ca my bin dng), hai cun dy ng cmc ni tip vi R, L v c t ln in p trn ti (hoc ni vi th

cp ca bin in p o lng).V ccu khng c mch t nn vic ni cc cun dy ng nh vy

s to nn cc dng i1 v i2 l vung pha vi nhau. Ta c su dyv th vc tnh Hnh 5.1.

Theo cng thc ca ccu logomet in ng ta c:

vi gc:

Vy

Ch : Trn thc t khi tn s thay i dn ti L thay i vy I2


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thay i do t s1

2

II

khc hng s nn s c sai s. khc phc, nh

sn xut ci tin s nh sau (xemHnh 5.2):

Cun dy ng c chia lm hainhnh 2 v 2. Hai nhnh ny mc

ngc cc tnh nhau, mt nhnh ni viL, mt nhnh ni vi C. Ta thy ccdng i2' v i2" ngc pha nhau, mt khchai cun dy li mc ngc cc tnh nns to ra m men ca cun dy ng thhai l tng ca hai m men cng du:M2 = M2" + M2". V vy khi tn s thayi lm XL, XC thay i ngc nhau hayI2' v I2" thay i ngc nhau. Vy M2 = const. Tc l nhnh ny khng

ph thuc tn s5.1.2.2. Cos metin ng ba pha

S mc nh Hnh 5.3.Cun tnh c mc ni tip vo pha A, hai cun dy ng c mc

vi hai in trR v c t vo cc in p UAB v UAC.

Gc quay ca ccu l:


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Ch :+ Trong s ny, cc cun p u ni tip vi in tr R nn

khng ph thuc tn s, hay1

2

I

I= const.

+ Cun dng c th mc vo cc pha B, C ty .5.1.3. Phazomet in t5.1.3.1. Csl thuyt

Xt hai in p cng bin , tn s nhng lch pha nhau mt gc :

Vy

Ta xt tr hiu dng ca u:

V vy bit U, o U ta xc nh c gc .5.1.3.2. Phazometin t

Ta a ra sn gin ca phazomet in t nh sau:


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Hai tn hiu in p cn so snh gc pha c a vo hai u cahai mch khuch i qua hai bin trRl v R2. Khi o, ta iu chnh ccv tr con trt trn cc bin trR1 v R2 sao cho in p u ra ca haimch khuch i l bng nhau, v c kim tra bng cc volmet V1 V2.

Sau khi kim tra UV1 = UV2 = U, ta o U bng volmet V ri suy ragc theo (5.5).

trnh phi so snh hai in p u1 v u2 ngi ta thng binchng thnh nhng xung vung sau a vo b cng i sin p

hay dng in nh Hnh 5.5. Gin thi gian nh Hnh 5.6.

Tu theo gc lch pha gia hai tn hiu, in p hay dng in ra tmch cng thay i. in p ny c o bng dng co chnh lu


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Da trn nguyn tc ny nhiu hng trn th gii ch to dng c

o gc lch pha trong khong t (0 180o) vi sai lch nh hn 1%.5.1.4. Phazomet ch th s

Da trn nguyn tc bin i gc lch pha thnh m, c ngha l gclch pha cn o gia hai tn hiu c bin thnh khong thi gian, sau lp y khong thi gian bng cc xung vi tn s bit trc.

Cu trc bao gm: b bin i gc pha thnh khong thi gian, bto xung TX1, TX2, TX3, bm, ch th s, my pht xung chun, khoK1, K2

S cu trc nh sau:


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Cc tn hiu u1, u2 c dng hnh sin cng bin , tn sc a

vo b to xung TX1, TX2. Cc xung xut hin khi tn hiu i qua mc

"0", cc xung ny sc a n cc u vo ca trigto ra u ramt xung m di ca n t l vi gc lch pha cn o x.Kho Kc mtrong khong thi gian tx. T my pht xung chun f0 c tn sn nh (hay T0 = 1/f0) c a qua K1 khi K1 m trong khong thigian tx. Mt khc b to xung TX3 pht ra xung c di cnh l Tnv kho K2c m trong khong thi gian . Vy cc xung t cckhong thi gian Tn si qua K2 vo bm v ch th s.

S xung m c l:


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vi Tn =KT0.Vy s xung N t l vi gc lch pha xNhc im :- Nu tn s nh, v T

xcha trong khong T

nnh, do vy ta phi m

rng Tn.- Nu tn s ln, dn n sai s lng t ho trong khong Tx tng

ln, dn n sai s tng.Thng thng lm vic trong khong mt vi Hz n vi MHz, c sai

s= 0,1 0,2%.5.2. o tn s5.2.1. Phng php gin tip

Dng volmet, ampemet, wattmet kt hp vi in cm mu, ta c thxc nh c tn s:

Bit L0, cn c vo s ch ca cc ng ho, ta xc nh c tns.

5.2.2. Tn s met cng hng


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Nguyn l hot ng:Tn s met cng hng gm mt nam chm in, to ra bi cun dy

in qun trn li st t hnh ch U, mt ming thp nm trong t trngca nam chm in, gn cht vo thanh l cc l thp rung c tn s daong ring khc nhau. Tn s dao ng ring ca hai l thp k nhau hnkm nhau l 0,25 hoc 0,5Hz. in p ca tn hiu cn o tn s sca vo cun dy ca nam chm in s to ra s dao ng ca tt c ccl thp. Tuy nhin l thp no c tn s dao ng ring bng tn s f ths dao ng cc i do cng hng ring, cn cc thanh khc khngcng hng th khng dao ng cc

giao_trinh_ky_thuat_do_luong_5435

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