10. Ky Thuat Do Luong Dien _20 Tiet

7/25/2019 10. Ky Thuat Do Luong Dien _20 Tiet

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TRNG CAO NG NGH DU KHKHOA INTNGHO

---------------------------------------------

GIO TRNH

K THUT O LNG IN(TI LIU LU HNH NI B)

Vng Tu, 2011


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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 lng i tng cn o c kt qubng 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 qubng sso vi n v.

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

hthng o khc nhau.Thit bo v thit bmu

Thit bo l mt hthng m i lng o gi l lng vo, lngra l i lng chtrn thit b(l thit bo tc ng lin tc) hoc lcon skm 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 kthut nht nh.

- Thit bmu dng kim tra v hiu chnh thit bo v n v

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

nht hai cp so vi thit bkim tra.

V d: Mun kim nh cng t cp 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 thchia 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 bdi bt kdng 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 bny gi l thit bo cin. Ngoi ra lng ra cn c thbin i thnh s, ngi o ckt quri nhn vi hsghi trn mt my hoc my tng lm vic

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

Thit bso snh cng c thl chthcin hoc l chths. Tutheo cch so snh v cch lp i lng b (bm ho stng t) ta ccc thit bso snh khc nhan nh: thit b so snh kiu tung (ilng o x v i lng b x lun bin i theo nhau); thit bso snhkiu qut (i lng b x bin thin theo mt quy lut thi gian nhtnh v scn bng chxy ra ti mt thi im trong chu k).

Ngoi ra cng cn cvo vic lp i lng b ngi ta chia thnhdng cm ho sxung, tn sxung, thi gian xung. Cn cvo iukin cn bng ngi ta chia thnh dng cb khng lch (zero) v dngcb c lch (vi sai).

Cn cvo quan hgia lng ra v lng vo, ngi ta chia thnh:thit bo trc tip (i lng ra biu thtrc 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 i o lng

C hai khi nim:

- Chuyn i chun ho: C nhim vbin 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 chyu l cc bphn 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 cp (S: Sensor): C nhim v bin mt tn hiukhng in sang tn hiu in, ghi nhn thng tin gi trcn o. C rtnhiu loi chuyn i s cp khc nhau nh: chuyn i in tr, incm, in dung, nhit in, quang in...

1.1.2.3. Thp thit bo

Vi mt thit bcth(mt knh):

Hnh 1.1. Cu trc hthng o mt knh

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

+ Mch o: thu nhn, xl, khuch i thng tin.... bao gm: ngun,cc mch khuch i, cc bbin thin A/D, D/A, cc mch ph...

+ Chth: thng bo kt qucho ngi quan st, thng gm chthsv chthcin, chthtghi, v.v...

1.1.2.4. Vi hthng o lng nhiu knh

Trng hp cn o nhiu i lng, mi i lng o mt knh,nhvy tn hiu o c ly tcc sensor qua bchuyn 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 hthngdn truyn. c sphn bit, cc i lng o trc khi a vo mch

phn knh cn phi m ho hoc iu ch(Modulation - MOD) theo tnskhc nhau (th dnhf10, 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 txa (TE1emety) cho nhiu i lng o.


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Hnh 1.2. Hthng o lng nhiu knh

1.2. Scu trc thit bo lng

1.2.1. Hthng 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 hgia z v xc thvit:

z = f(x)

trong f() l mt ton tthhin cu trc ca thit bo.


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Trong trng hp quan hlng vo v lng ra l tuyn tnh ta cthvit:

z = S.x (1.1)

y S gi l nhy tnh ca thit b.

- Nu mt thit bgm nhiu khu ni tip th quan h gia lngvc v lng ra c thvit:

trong Sil nhy ca khu thi trong thit b.

1.2.2. Hthng o kiu so snh

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

yX = T.x.

Hnh 1.3. Hthng o kiu so snh

Sau yXc so snh vi i lng b yk

Ta c: y = yX- yk

C thcn cvo 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 nhtrng 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 tlvi 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 bngng


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b) Phng php m ho tn sxung

Trong phng php ny i lng vo yXcho tng tlvi 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 sxung

Ti im cn bng c:

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

c) Phng php m ho sxung

Trong phng php ny i lng vo yX= const, cn i lng bykcho tng tlvi 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 sNX

yXNX.y0 (1-6)

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

Ngoi ra cn phng php m ho sxung 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 sca 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 vquan hgia lng vo v lng ra ca thit bo, ta dngkhi nim vnhy ca thit b:

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

Ni chung S l mt hm phthuc x nhng trong phm vi x nhth S l mt hng s. Vi thit bc quan hgia lng vo v lng ra

l tuyn tnh, ta c thvit: 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 Sil nhy ca khu thi trong thit b.

Theo l thuyt khi xt ti quan hgia z v x th x c thnhbao

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

thy c.V d1.1: Khi phti tiu thqua mt cng tmt pha 10A nhhn

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

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

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

C thquan nim ngng nhy ca thit bo l gi trnhnhtm thit bo c thphn 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 tgi trnhnht ti gi tr ln nht tuntheo phng php o lng ca thit b

T a ra khi nim vkhnng 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 xil kt quca ln o th

xl gi trng ca i lng o


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il sai lch ca ln o thi

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

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

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

+ Vi mt thit bo, sai stng i c tnh

Gi tr, %gi l sai stng 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 cp 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 thxc nh c sais tng i quy i v suy ra sai s tng i ca thit b trong cc

php o cth.

Ta c:

trong l sai stng i ca thit bo, phthuc cp chnh xc vkhng i nn sai stng i ca php o cng nhnu D/x dn n 1.

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

1.3.2. in trvo v tiu thcng 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 thnng lng nythhin phn tc dng ca thit bo ln i tng o gy ra nhngsai sm ta thng bit c nguyn nhn gi l sai sphvphng

php. Trong khi o ta c gng phn u sao cho sai sny khng lnhn sai scbn 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 thcng sut ca thit b.

Tn hao nng lng vi mch o dng p l:

Vy ta tm tnh sai sphdo nh hng ca tng trvo l:

vi RAl in trca ampemet hoc phn tphn ng vi dng;

RVl in trca volmet hoc phn tphn ng vi p;

Rtl in trti.

V d1.2: Phn tch sai sphkhi o p trn Hnh 1.9.

+ Giscn 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 (k ng).


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

Sai s t33 V trln 50 V chnh l sai sphvphng 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 thhin cc c trng sau:

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

c tnh ny thhin 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 thhin hai dng: c tnh bin tn A() v c tnh

pha tn ().

c tnh cn thhin di dng sai s tn s, sai sny thhin

sai sbin tn Av sai spha tn :

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

A0l bin ca khu l tng khng phthuc tn s;

() l gc pha u ra phthuc tn s;

0l gc pha l tng khng phthuc tn s.

Trong dng co cc sai sny phi nhhn mt gi trcho php


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quy nh bi nh nc. Gii tn ca dng co l khong tn sca ilng vo cho sai skhng vt qu gi trcho php.

Thi gian n nh hay thi gian o ca thit bl thi gian ktkhit tn hiu vo ca thit bcho ti khi thit bn nh c thbit ckt qu.

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

rc ho i lng cn o o gi trtc 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 quca nhng php o cthta xc nh gi trng ca php o v sai sca php o y.

Dng co no cng c sai sv nguyn nhn sai srt khc nhau,

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

1.4.1. Tnh ton sai sngu nhin

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

trong : xil kt quln o thi;

M[x] l kvng ton hc ca v sln o i lng x.

Hnh 1.10. Lut phn bchun


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- Theo ton hc thng k th sphn bca sai sngu nhin xungquanh gi tr kvng ton hc theo mt quy lut nht nh gi l lut

phn bxc sut.

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

trong l lch qun phng hay phng sai ca sai sngu 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 thnguyn vi i lng cn o.

Hnh 1.11. Kvng v tn xca lut phn bchun

Qu trnh gia cng kt qunhsau:

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

Sai sngu 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 nhsau:


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

+ Phng sai ca sai sngu nhin c tnh theo cng thc BessE1

Nu ta ly kt qul gi trtrung bnh ca n ln o th phng sai s

gim i n ln

+ Sai sngu nhin c tnh:

trong kstl hsStudent, n phthuc vo sln thu thp n v xcxut yu cu p. Hskstc tra trong cc stay kthut: kst= f(n,p).

+ Kt quo c tnh:

( )1nn

xn

1x

kn

xxxx

n

1i

2n

1iii

sti

== = = (1-18)

Ch : Trong thc tc nhng ln thu thp sliu cho kt qukhngng 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 3vi i = 1 n n, nu ln o

no c |i| 3th phi loi b ln o v tnh li tu vi (n - 1)php o cn li. C thchng 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 qunh

sau (n vtnh = 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 sngu nhin c phnbchun.

Bi lm:

So snh cc i= Ri- Rvi 3. Ta thy ln o th8 phm phi sai

lm ln (8= R8- R3) nn ta bqua ln o ny v tnh li tu vi13 ln o cn li. Ta lp bng sau:

Bng 1.1. V dvtnh ton sai sngu nhin

STT Ri i i2

1 140,25 -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 sgin tip

Trong thc tc nhiu phng php o m kt quc tnh tphpo trc tip khc ngi ta gi php o l php o gin tip.

Gisc 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 stuyt 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 stng i ca php o gin tip c tnh l:


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trong : xl, x2,.xnl sai stng i ca cc php o trc tip x1,x2, xn.

Bng 1.2. Bng tnh sai stuyt i v sai stng ica mt shm y thng gp

V d1.4:Ngi ta sdng 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 schca hai ng hl: I = lA, U =100V.

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

Hnh 1.12. V dvtnh ton sai sgin tip

Bi lm:


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+ Sai stuyt i ca ampemet l:

+ Sai stuyt i ca volmet l:

+ Gi trin trtheo php o l:

+ Sai stuyt i ca php o in trl:

+ Sai stng i ca php o in tr


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Chng 2CC CCU CHTH

2.1. Ccu chthcin

2.1.1. Cschung

2.1.1.1. Khi nimCcu chthl dng co m schca n l i lng tlvi

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

chthi 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 tngi ta chia thnhccu chthkiu 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 trl bphn quan trng trong cc chi tit ckh ca cc ccu chthcin, m bo cho phn ng quay trn trc c gn cc chitit ca phn ng nhkim chth, l so phn, khung dy.

b) Bphn phn khng

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

c) Kim chthgc quay

Kim ch thgc quay c gn vi trc quay. di chuyn ca

kim trn thang chia tlvi gc quay . Ngoi ra c thchthgcquay bng nh sng.


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d) Thang chia

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

e) Bphn cn du

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

hai loi cn du c sdng, 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 stch 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;

dl lng vi phn ca gc quay.

Theo nh lut bo ton nng lng:

dWe= dA

dWel 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 lvi gc quay nhcc bphnphn khng l l so xon, dy cng hoc dy treo.

Ta c: Mp= D;

vi D l hsphthuc vo kch thc vt liu chto 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 cdng 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 hstl, G l trng lng phn ng, n = (1,3 l,5)

d) Mmen cn du

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

Mmen cn du c chto sao cho c trstlvi tc quayca phn ng

vi p l hstlphthuc vo c imcu to ca bphn cn du. Tbiu thctrn ta thy khi phn ng vtr cn bng

th 0

dt

d= , nhvy 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;

Mil 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). Tutheo quan hgia 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 tv li st l u nhau.

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

2. Nguyn l lm vic

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

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


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ca khung dy schu tc dng mt lc:

F = BlWI

trong : B l trscm ng ttrong khe hkhng kh; l l chiu di tcdng ca khung dy; W l svng dy; I l trsdng in.

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

trong : d l kch thc ngang ca khung dy;

S = dll thit din bmt khung dy.

Mmen phn ca l so: MP= D..

Vy phn ng scn bng khi:

y SI =D

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

Ta thy tlbc nht vi I.

3.c im v ng dng

c im:

- u im:

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

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

+ V tlbc nht vi I nn thang chia ca ccu u.

- Nhc im:

+ Chto kh khn, gi thnh t;

+ Do khung dy phn ng nn phi qun bng dy c kch thcnhnn khnng 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 bchnhlu.

ng dng:

Dng ch to ampemet, volmet, ommet, in k c nhy cao,dng lm c cu chkhng trong cc ng hvn nng, trong cc cuo...

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

1. Cu to

Phn tnh ging nhccu mt khung dy nhng khe hkhng khgia cc tv 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 I1v I2chy 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, M2vi:

M1= B1.S1.W1.I1

M2= B2.S2.W2.I2

V khe hkhng kh l khng u nn cm ng tB phthuc vtr

ca khung dy ng.

V khng c l so phn nn phn ng scn 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 tnh 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 tlvi tshai 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 tshai dngin vn c ginguyn do vy m trnh c sai s.

ng dng:

c dng chto 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 chthin tloi ht) v kiu cun dy trn (ccu chthin tloi 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 thquay 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 cbphn cn du, l so phn, kim chth


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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 scmt ttrng. i vi cun dy dt ttrng ny ht l thp vo tronglng cun dy tnh, cn i vi cun dy trn th ttrng stho hail thp, khi hai l thp c cng cc tnh nn y nhau. Chai trng

hp trn slm 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 Wel nng lng in ttrng tch lucun dy

trong L phthuc .

Vy mmen quay:

- Khi cho dng in xoay chiu vo cun dy:

Gisi = ImaXsint. Lc mmen quay Mqtheo t sl:

Mmen quay trung bnh:

vi I l trhiu dng ca dng hnh sin.

Ti vtr cn bng Mq= MP;

Vy ccu chthin tc tho c cdng 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 thqun bng dy kch thc lnnn khnng qu ti tt.

+ Dchto, gi thnh h.

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

- Nhc im:

+ Gc quay t lvi 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 thtng i u).

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

ng dng:

Chyu o dng, p xoay chiu tn scng 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 chth, bphn cn du,l so phn.

b) Nguyn l lm vic


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- Xt khi cho cc dng in mt chiu I1v I2vo cc cun dy phntnh v ng, trong lng cun dy tnh s tn ti mt t trng. Ttrng ny stc ng ln dng in chy trong cun dy ng v to rammen quay:

Nng lng ttrng tch lutrong lng cun dy l:

trong L1, L2l din cm ca cc cun dy v chng khng phthuc

vo gc quay ;M12l hcm 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 kp thay i theo gi tr tcthi cho nn thc tly theo gi trtrung 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:

+ Dchu nh hng ca ttrng ngoi.

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

+ Cu to phc tp, t tin.

+ Thang chia khng u (trkhi chto wattmet).

ng dng:

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

+ Chyu chto ng ho cng sut tc dng v phn khng.

2.1.5.2. Loi c hai khung dy ng (logomet in 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, i2ln lt chy vo cun dytnh v cc cun dy ng, trong lng cun dy tnh sc mt ttrng.

Ttrng ny stc ng ln dng in chy trong cc cun dy ng


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sinh ra cc mmen Mq1, Mq2

vi M1, M2l hshcm 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, I2l cc gi trhiu dng ca cc dng in i1, i2; 12l gc lch

pha gia dng in i, i1v 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 cos1 pha, 3 pha cho li in xoay chiu.

2.1.5.3. Ccu st in ng v logomet st in ng

a) Ccu st in 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 chth, lso phn v bphn cn du.

Gc quay c tnh:

b) Logomet st in ng

Gm mch t c cu to sao cho to nn khe h khng kh khngu, phn ng gm hai khung dy t cho nhau 60o v gn trn trcquay cng vi kim chth. 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 ttrng ngoi.

- Tn hao st tln, chnh xc khng cao.

- Thng dng chto 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 nhhnh 2.1 1.

Hnh 2.11. Ccu chthcm ng

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


Khi cho dng in i1vo cun dy 1 th cun dy 1 to ra tthng 1

xuyn qua a nhm, dng in i2vo trong cun dy 2 to ra tthng 2cng xuyn qua a nhm.

T thng 1cm ng trn a nhm

sc in ng e1 chm pha hn 1 mt

gc /2.

T thng 2cm ng trn a nhm

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

V a nhm c coi nhrt nhiuvng dy t st nhau, cho nn E1, E2sto ra trn a nhm cc dng in xoyiX1v iX2 chm pha hn so vi e1v e2

cc gc 1v 2v 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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20.

Do c stng hgia tthng 1, 2vi cc dng in iX1v iX2m sinh ra cc lc F1v F2v cc mmen tng ng lm quay a nhm.Ta xt cc mmen thnh phn nhsau:

M11l mmen sinh ra do 1tc ng ln iX1




Gi trtc thi ca mmen quay M1tdo stc ng tng hgia 1v dng tc thi iX1l:

M1t= C1iX1

vi C l hstl.

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

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

Nhvy mmen quay sl tng cc mmen thnh phn:

Mq= M12 + M21


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M12v M21c du ngc nhau do vy mmen tng sko a nhmvmt pha duy nht:

Nu dng in to ra 1v 2l hnh sin v a nhm l ng nht(chc in trthun) th cc dng in xoy IX1v IX2stlvi tn sv tthng sinh ra n, tc l:

vi C = C12C4+ C21C3l hng sca ccu chthcm ng.

2.1.6.3.c im v ng dng

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

l /2.

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

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

2.2. Ccu chths

2.2.1. Khi nim v nguyn l cbn ca ccu chthsTrong nhng nm gn y xut hin v sdng rng ri cc chth

s, u vit ca ccu chthsl 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

Skhi ca ccu chthsc thtm tt nhsau:

Hnh 2.13. Skhi ca ccu chths


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i lng o xin qua bbin i thnh xung (BX), sxung N tlvi ln x(t) c a vo bm ho (MH), bgii m (GM) v bhin s. Cc khu m ho, gii m, bhin sto thnh bchths.

2.2.2. Chths

C nhiu loi chthskhc nhau nhng phbin hin nay vn dngchthsn phng in nhiu cc v chthsghp 7 thanh bng mt

pht quang hoc tinh thlng.2.2.2.1. Chthn phng in nhiu cc

Chthl mt n n ng c mt qut v 10 katot. Anot thng t in p 220V - 250V. Katot c chto bng dy Cr-Ni un thnh hnhcc chst0 - 9. Mi katot l mt con s.

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

- u im ca chthny l hnh dng cc con sp.

- Nhc im: Kch thc cng knh, ngun in p cung cp cao,chph 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 thlng (LCD: LiquiCrystal 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 thlng l nhng mng mng lm

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

Trong thc tcn chths16 thanh, ma trn im...

2.2.3. M v cc mch bin i m

2.2.3.1. M

M sl nhng k hiu vmt tp hp s, tthp ca cc k hiuta c thm tc cc con skhc nhau. C cc loi m ssau:

- M cs10, l hm thp phn c 10 k tt0, 1, 2,..., 9.

- M cs2 l loi m c hai trng thi c k hiu t0 v 1 (cn

gi l m nhphn).- M 2 - 10 (cn gi l m BCD) l slin hgia m cs2 v m

cs10 dquan st v dc.

i vi ccu chthsth hin nay chyu ngi ta sdng m cs2.

2.2.3.2. Cc mch bin i m

Hnh 2.15. Mch gi

i m t

m nh

phn sang ch

th

7 thanh


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Mch bin i m l thit bdng bin i tm cs2 hoc m2 - 10 thnh m cs10, ngha l thhin di dng sthp phn. Ngynay cc bgii m c chto di dng vi mch. V dnhvi mchSN74247 c cc u ra h cc gp dng iu khin LED c chungant 5V. Cc in trR1, R2,, R7hn chdng.

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 nhphn

Tng qut c m u vo tng ng vi m s thp phn t0, 1, 2...m-1 v n u ra tng ng vi n bt ca m snhphn. Ngi ta thngtng hp bbin i m vi su vo m = 10 tc l gm x0, x1,... x9ng vi cc sthp phn t0, 1, 2,... 9. Nhvy bbin i m sc

bn u ra tng ng y8, y4, y2, y1ng vi bn bt ca m nhphn ctrng s8, 4, 2, 1. Ta c bng trng thi nhsau:

Bng 2.1. Bng trng thi bin i tsthp phn sang nhphn

M nhphnSthp 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

Tbng trng thi ta c:


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Vy ta c ththnh lp mch bin i m tthp phn sang nhphnnhsau:

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

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

2.1 ta c X0X9l cc bin phthuc cn Y1Y8l cc bin c lp.V vy ta c cc phng trnh logic v smch logic tng ng:


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c) Mch bin i tsthp phn sang chth7 thanh

u vo l cc s tnhin t 0 9, u ra l trng thi cc thanhsng ca ch th7 thanh bng mt pht quang hoc tinh th lng. Xut

pht tthc tta c bng trng thi nhsau:

Bng 2.2. Bng trng thi bin i tsnhi phn sang thp phn

Trng thi cc phn tSthp 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

Tbng trng thi ta c thvit c phng trnh nhsau (vi sthtcc thanh nhphn trc)

Ty ta c ththit lp mch logic sau:


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d) Mch bin i m tm nhphn sang chth7 thanh

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

X1. u ra l trng thi cc thanh sng ca ch th7 thanh. Ta c bng

trng thi sau:


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Bng 2.3. Bng trng thi bin i tsnhi phn sang chth7 thanh

Sthp phn Snhphn 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

Tbng trng thi ta vit c cc phng trnh logic quan hgiau ra Y1,,Y7vi cc u vo X8,X4, X2, X1. Tuy nhin cc phngtrnh ny phc tp 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. Khi o dng inAmpemet 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 cnhm.

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

Gisphti c in trl Rt, trc khi mc A vo mch th dng

din c tnh:tR

UI= .

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

in c tnh:At

A RR

UI

+= .

Vi IAl dng in chbi ampemet, RAl in trca ampemet, Rt

l in trti. Sai sphtrong qu trnh o lng sc tnh:

Ta thy sai sdo A gy ra i vi mch ti cng nhnu in trcaampemet cng nhso vi in trti. V thyu cu i vi ampemet odng in l in trca ampemet cng nhcng tt.

Vi mt phti c in trl Rtcp chnh xc ca ampemet sdngl 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 sphgy ra sln hn saisyu cu lc ta phi sdng cng thc hiu chnh:

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

V d3.1: Tnh in trca (A) khi th nghim o in trmt chiucun dy thcp ca MBA 560KVA, 10/0,4 KV nhHnh 3.2, bit chnh xc yu cu % = 0,5%.

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

Nhvy iu kin cn ca vic ly tn hiudng qua ti m bo sai snhhn 0,5% l

RA0,25 (m).

Trong thc tkhng c (A) no thomn nn sth nghim nykhng c ngha.

3.1.1.2. Khi o in p

Volmet l mt phn tc trng cho nhm cc phn tphn 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 cnhm.

Khi o in p, volmet c mc song song vi ti nh Hnh 3.3.Nhvy 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 sphtrong 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 Uvdo volmet o c sl:

Sai sphPdo volmet gy ra c tnh:

Ta thy sai s phdo 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 ttrn cc thit bo hin i hoc trn ng hvn nng ngi ta ghi tng trvo ca n.

Vi mt phti 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 sphdo vomet gy ra lnhn sai s ca bn thn c cu ch thv ta phi dng cng thc hiuchnh.

V d3.2: Tnh tng trvo yu cu ca mch khuch i ca mt

my in tim nhHnh 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 scn o. Lm vic ngoi di tn s sgy sai sphdo tn s. Sai sny phi tnh n nh hng cacc mch o lng i theo ch thnhShunt, bin dng, bin p, chnh

lu, khuch i v.v. Cng v vy trong nhiu ampemet v volmet, lc cnm bo sai sdo tn snhhn gi trquy 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 hp ngi ta phi s dngnhng linh kin c bit m bo tn slm vic ca dng c. Trncc dng co dng v p xoay chiu c ghi tn shay gii tn slmvic.

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

3.2.1. Phng php sdngNgi ta sdng mt sccu chthcin chto 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 cp chnh xc t (0,5 2). i vi ampemet tin, khinhit thay slm cho in trca cun dy thay i dn ti sai s. gim sai sngi ta thng dng phng php b nhit, tc l dng mt

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


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ampemet, v vy slm cho in trca ampemet gn nhkhng thayi theo nhit . Ampemet tin chc tho dng in mt chiu.

Ampemet in t. c chto da trn ccu chthin t. Loiny c chnh xc thp hn nhng n bn chc, dsdng v rtinnn c sdng rng ri hn trong cng nghip. Ampemet in tctho c cdng mt chiu v dng xoay chiu nhng chyu l o

dng xoay chiu. C nhiu loi ampemet in t, chng ging nhau vnguyn l lm vic song chkhc nhau vhnh thc, svng dy v kchthc cun dy t phn tnh.

Ampemet in ng: C cu to phc tp v t tin nn ch dngtrong nhng trng hp cn chnh xc cao, hoc tn hiu o c tn scao hn. Sai stn strong di tmt 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 nhhnh 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 sdngchnh lu mt na hay hai na chuk. Tuy nhin s ch ca ampemetchnh lu l gi trtrung bnh ca dng xoay chiu, nhng thng thng

cc dng co in thoc in ng li chgi trhiu dng ca ng


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Khi bit R0dng in nh mc lch ton thang o I0dng cn o I,ta c thtnh c:

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

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

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

V d3.3: Tnh in trShunt cho mt bin phn c dng cn o l I = 10kA. Bitdng nh mc qua c cu 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 nhcun 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 svng nhnhau, 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 bbng hng s.

b) Phng php dng bin dng in

Bin dng in (BI) l mt my bin p

c bit c cun scp rt t vng cho dngph ti trc tip 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 chngnmch.

Ta c:


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KIgi l hsmy bin dng.

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

tin no ; W2nhiu 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 bvi bin dng in th schca ampemet c khc theo gi trdng in I1pha scp.

Cn ch rng bin dng in l phn tc 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 nhhn dngnh mc ca ccu. Cho ti nay vic nng cao nhy ca dng cvh thp ngng nhy ca dng c v cc mch khuch i l rt khkhn, sau y l mt sphng php.

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

ch th tin c nhy cao. Binphp nng cao nhy l tng tcm trong khe hkhng kh v gimhsphn khng ca dy treo.

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

Gim hng sphn 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 ttrn.

Bin php quang hc: L sdng khong cch tthang chia ngng quay ca in ktng nhy, gng gn trn trc ca phnng, c mt ngun sng chiu vo gng, lia phn xca gng chiuln thang o nhhnh v

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

l l khong cch tgng ti thang chia .

Sdng gng quay s tng quay ca tia phn chiu khi gng

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

Nhvy 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 sdng cc ch thcin ch to cc loi volmet oin p nhvolmet tin, volmet in t, volmet in ng.

Volmet tin: Volmet tin c cu to tccu chthtin,loi ny thng dng o cc in p mt chiu, c nhy cao, cho

php dng nhi qua, cng c thsdng km vi bchnh 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 hshnh dng ca dng hnh sin.

Volmet in t:Volmet in tc cun dy btr phn tnh nn cthqun nhiu vng dy to nn in trln kh ddng, tuy nhinnu qun nhiu vng dy qu m khi o mch xoay chiu th xut hindng in cm ng sinh ra bi tn sca dng in, do snh hngn trstrn 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 squ cao, sc nhng sai sphdo tn s, vvy phi btr thm tb cho cc cun dy tnh v ng.

3.4.2. Phng php mrng gii hn o

3.4.2.1. Phng php dng in trph

Yu cu cbn ca volmet l in trca n cng ln cng tt v thm rng thang o trong cc volmet cch n gin nht l ni thmin trphvo ccu o nhHnh 3.13.

Hnh 3.13. Mrng thang o cho volmet

vi: R0in trca ccu o

RPl in trphU0in p t ln ccu

UXin p cn o.

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 cp baogicng l 100V. Cn pha scp c

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

Ging nhbin dng in, bin inp l phn t c cc tnh, c cp 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 chtrn thang chia , nhvy saiskhng thnhhn sai sca cc chthdng vo dng cv cha kn sai sgy ra do cc mch o sdng. 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 sc chnh xccao nht hin nay.

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


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in p cn o UXc so snh vi in p b Ukl in p ri trnin trRk. Rkl 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 cc nhy cao hay dng ctng

pht hin schnh lch hay cn gi l cquan zero.

Cc loi phng php so snh khc nhau chkhc 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 syu cu ca php o quyt nh.

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

3.5.2. in thkmt chiu

3.5.2.1.in thkmt chiu in trln

Sca in thkmt chiu in trln nhHnh 3.17.

Rk, Rcl cc bin tr, ENl ngun pin mu, Uccl in p cung cpcho mch, UXl in p cn o, G in kchkhng.

in thkmt chiu in trln 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 vtr 1,1 ni


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in kvo mch to dng cng tc, ta iu chnh Rc in kG chkhng, khi xy ra quan h:

EN= URN= IP.Rc

Gi trdng cng tc:

+ Tin hnh o in p cn o UX

Ta ng kho K sang vtr 2,2 ni in p cn o UXvo mcho, sau ta iu chnh con trt trn in trRkcho n kh in kGchkhng.

Lc ta c quan hsau:

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

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

Quan hgia 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 sdng in thk.

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 thkmt chiu loi ny gi trin trRktng i ln,in p cn o UXcV cho nn nh hng ca in trtip xc v scin ng tip xc khng ng k, ngc li nu o in p crt nhta

phi dng in thkmt chiu in trnh.

3.5.2.2.in thkmt chiu in trnh


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in thkmt chiu in trnhc chto da trn nguyn tcginguyn gi trin trmu Rkthay i dng cng tc IPqua Rkthay i in p Uk= IPRkb li vi in p UX.

Snguyn l ca in thkmt chiu in trnhnhhnh v:

Hnh 3.18. Sin thkmt chiu in trnh

Ngi ta to ngun dng mu IPqua in trmu Rkbng khuchi thut ton.

t u vo khuch i thut ton mt gn mu ENb vi inp ri trn cc in trmc song song u vo khuch i. Nu ENvin p ri trn cc in trmc song song Ugb hon ton nhau:

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

vi gg==

n

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

Vy:

Ta iu chnh cng tc K thay i cc gi trdng cng tc IPchon khi kim in kchkhng, ta c:


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Trong mch to in p b khng c u tip xc cho nn loi trc sai sdo sc in ng tip xc v in trtip xc. Sai schyul do ngng vo v hskhuch i ca bkhuch i quyt nh.

3.5.3. in thkxoay chiu

Vnguyn l th in thkxoay chiu cng so snh in p cn ovi in p ri trn in tr mu 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 thkxoaychiu khng thvt qu cp chnh xc ca ampemet, mt khc muncho UXv UkCn bng th phi iu chnh cn bng cvmodun v v

gc pha, tc l thomn ba iu kin l in p UXv in p Ukphicng tn s, cng bng nhau vtrsv UXv Ukphi ngc pha nhau.

thc hin iu kin thnht ngi ta mc in p UXv Ukvongun cng tn s. Dng bchthkhng thc hin iu kin thhaiv phi tch Ukthnh hai thnh phn lch nhau 90

oto UXngc Uk

C hai loi in thkxoay chiu l:

- in thkxoay chiu tocc;

- in thkxoay chiu tovung gc (cc).

3.5.3.1.in thkxoay chiu tocc


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Trong in thkxoay chiu loi ny, in p cn o UXc cnbng vi in p ri trn in trR (xc nh bi cc con trt D1v D2)Mun UX= IPR. Dng cng tc IPc xc nh nhampemet chnhxc cao v in trR iu chnh (Rc). Biu chnh pha dng cn

bng v pha, ng thi cng lm ngun cung cp cho mch to dngcng tc Ip, biu chnh pha ny chnh l nhc im ca in thkxoay chiu v kh xc nh chnh xc vtr n nh ca phn quay ngvi gc quay khi iu chnh pha v dng IPthay i lm cho vic iuchnh cn bng kh khn.

3.5.3.2.in thkxoay chiu tovung gc

Trong in ksdng hai cun dy t gn nhau v dng hcm Mca chng to Ukthnh hai thnh phn lch nhau 90

ov UXscn bng

vi tng hai vc tthnh phn.


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Sgm 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 w1ca bin p khng li thp, ampemet A v in tr(Rc).Dng in I1 t ngun cung cp xoay chiu (c xc nh nhampemet) to trn bin trAB mt in p UAB. in p Uk1c xcnh bi dng I1v vtr con trt D1trn bin trAB. V dng I1khngthay 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 w2ca bin p khng li v hp in trRfb tn s. Dng inI2trong mch cng tc lch pha vi I1gc 90

o(v in cm L2khng lnlm nn c thcoi nhI2trung pha vi E2m E2lch pha vi E1mt gc90o). Trong mch thnht I1c gi trxc nh nn I2cng c gi trxcnh:


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M l hcm ca w1v w2

Ta xc nh Uk2= I2R2 (R2 l mt phn in tr ca AB c xcnh nhvtr ca con trt D2trn A'B'). V Ukl= I1R1v Uk2= I2R2mI1v I2lch nhau mt gc 90

onn Uklv Uk2cng lch pha nhau 90o.

Ch rng khi tn sf thay i = 2f, nhvy khi thay i dnti I2 thay i v gi trkhc trn AB cng thay i. khc phc

ngi ta dng hp in trphRfb cho tn skhng i (tc l Rfthay i phthuc vo sthay i ca tn sngun cung cp).

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

thbiu din cc gi trUknhHnh 3.19.

iu chnh cc con trt Uk1v Uk2trn cc bin trdy qun AB vAB thng qua tnh ton ta sc trhiu dng v gc pha ca in p

UXcn o

Sai schyu ca in thkxoay chiu l sai sca 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.

- Stm tt nguyn l nhHnh 3.23.

Sgm cc khi nhsau:

+ Cp nhit in c nhim vbin i tnhit txsang sut inng mt chiu Ex. Vi hthng thc thng c thm mch b nhit u tdo.

+ Cu so snh gm EP, RPv cc in p mu khc nh: R0, R1, R2,R3, R4. Nhim vl 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 bchnh lu, qua n p mt chiu vi cht lng cao).

+ Bbin i mt chiu, xoay chiu c nhim vbin i in p

mt chiu U thnh in p xoay chiu tn s50Hz. Mch ny c thlcon rung chc hoc rung in t.

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

+ Hai ng c gm mt ng c khng ng b c nhim vkocon trt trn cc bin trRP, R0v mt ng cng bc nhim v


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ko bng giy chuyn ng trong chtghi.

Qu trnh o c chia lm hai bc:

Kim tra chnh xc ca cc in p mu

Lc ny kho K vtr 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 U4l 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 sutln 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 sphthuc vo du ca U.

Tm li, khi U 0, ng cKB squay 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 vchnh xc.

Qu trnh o nhit

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

Nh cp nhit in, nhit cn o bin thnh sut in ng mtchiu Ex. Khi o Exta so snh vi U12l in p ri trn cc in trmu

R1, R2v mt phn R0. Ta c: U= Ex- U12

Khi U 0 th theo nguyn l phn trn, ng cKB squay,ko con trt trn u bin trR0thay i U12c xu hng sao cho

U 0 th mt tn hiu iu khin v dng li. Lc ta xc nh cEx= U12. Vy cn cvo vtr ca con trt trn bin trR0ta xc nhc U12ri ta suy ra Ex. Thc ttrn R0ngi ta c sn cc vch chiatheo n vnhit nn ta c c kt qu.


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Qu trnh tghi

Lc ny trn u bin trR0ta gn sn mt ngi ghi, ngi ghi tlnbng 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. Nhvy sto ra trc thi gian t.

Ta thy khi to thay i, ngi ghi schuyn ng ttri sang phi nh

ng cKB, cn bng giy th chuyn ng vi tc khng i tdi ln trn nhng cng bnn ngi ghi svtrn bng giy biunhit theo thi gian.

3.6. o in p bng cc volmet chths

Ngy nay volmet sc s dng rng ri trong o lng v khnng chnh xc kh cao, gn nh, thun tin cho ngi sdng. Tutheocch bin i in p thnh cc i lng chthsm ngi ta chiara thnh ba loi volmet snhsau:

- Volmet schuyn i thi gian;

- Volmet schuyn i tn s;

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

3.6.1. Volmet schuyn 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 schun (f0) sau dng bm m slng xung (N) tlvi Uxsuy ra Ux

Scu trc chung ca volmet snhsau:


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Hnh 3.25. Scu trc volmet schuyn i thi gian mt nhp

Biu thi gian:

Nguyn l lm vic:

Khi mmy ti thi im t1, my pht xung chun qua bchia tnkhi ng my pht xung rng ca, u ra my pht xung rng ca cUrc(Uk) i tin bso snh so snh vi in p Uxcn o u vo.ng thi cng tu ra ca bpht in p rng ca c xung thnhtn trigv t trigtrng thi kch hot mkho K cho php ccxung mang tn schun (f0) tpht xung qua kho K n bm v chths. Ti thi im t2khi Ux= Urcthit bso snh pht xung th2 tcng vo trigv kho kho K, thi gian tt1n t2tng ng vi tx. Ty ta c mi quan h:


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Vi mt my pht in p rng cn cnh th tv trl hng s, vvy Uxtlvi txslng xung n bm trong khong thi gian txsl:

Slng xung i qua kho K trong mt chu kca in p rng catlvi in p cn o.

Sai schyu l do my pht in p rng ca gy ra, tc l tv trkhc hng s, tip theo l sai slng 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 sbin thinca in p UxPhi m bo iu kin sau:

3.6.2. Volmet schuyn i tn s

Nguyn l lm vic ca volmet schuyn i tn sda trn nguyntc bin in p thnh tn sri dng cc my o tn schththeoin p.

Scu trc ca volmet schuyn i tn snhsau


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in p cn o Uxc a n u vo, qua khu tch phn scin p U1. U1c so snh vi in p U2(in p U2c n nh caoKhi U1= U2thit bso snh pht xung qua khuch i 2 (ti thi im tthng kho K1v K n bm, chths. Khi K1thng, in p U0(ngc du vi U1) qua K1n b vi in p U1 (y l mch phngin ca tC) trong khong thi gian T

k(tt

1n t

2) ti t

2in p U

0b

hon ton U1


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Ta so snh in p cn o Uxvi in p chun Ukph thuc vovic gia cng i lng b Ukv quy trnh so snh vi Uxm ngi ta

phn ra thnh volmet schuyn 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 sdo mt ln tc l ti thi im Ux, Uk ta c ktquca php o. in p b Ukc ththay 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 N1v phntip theo lm nhim vbin i s lng xung N1 thnh m thp phn

N10iu khin cc phn thin s.


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Nguyn l lm vic:

Khi mmy, bpht xng chun bt u lm vic, cc xung f0nbm 1 v kho K; sau mt tp xung f0tng ng vi thi im t0thbm 1 pht xung n thng kho K, trong thi gian K thng cc xungmang f0qua K n bm 2 v chths. ng thi cmi 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 UkUx(ti thi im trbso snhtc ng vo bkhuch i to tn hiu kho kho K. Qu trnh o kt

thc v bphn chthhin kt qu. Nu tt ccc mc in p U tonn Ukliu bng nhau th slng xung N1stlvi in p cn o

Uxtc l UxUk= N1U. y l gi trtc thi ca in p cn o tithi im t1. Nu mun o Uxti thi im khc th qu trnh o slpli tu.

i vi volmet chthb qut vi i lng Ukthay i theo cc bc

thang khng bng nhau. Trong cc volmet ny cc mc bc thang Ukhng nhnhau, c thto cc U theo tng hng m ca con shm nht nh. Do c thda vo hm nhphn, nhthp phn vthp phn gia cng in p b Uk.

V d3.4: Qu trnh gia cng in p b Uktheo hm thp phn.


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Hnh 3.31. Qu trnh gia cng in p b

Qu trnh so snh thng ln nht, vi Ux= 43V. Con sthp phnc hai hng m l hng chc v hng n v. Nguyn l ca qu trnh sosnh nhsau:

+Nu Uk> Uxth m sghi l 0.

+Nu UkUxth m sghi l mt stng ng vi hng m ca

Ukv khi hiu |Uk- Ux| < U (mc ca hng m) th qu trnh so snhschuyn sang hng m nhhn). Cthy 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 sl 4 (hng chc nn ghi l 40) tip tc qu trnh

so snh sdin ra hng n vvi gi trln nht ca hng l 9 v mimc U = 1

Qu trnh gia cng Ukkt thc ta sc tng gi tr

y Uk10l m hng chc, Uk1l 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 khnng cho kt qulin tc ti thi im bt k.

Volmet schuyn i trc tip kiu tung c hai loi bao gm loigia cng i lng b Ukthay i theo bc thang bng nhau v loi gia


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cng i lng b Ukthay i theo bc thang khng bng nhau.

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

Nguyn l cbn

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

in p Uktng 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< Ukskt thc qu trnh o v cho ra kt quchths.

Thi gian gia cng c xc nh bi smc lng tln nht (Ndm)

v thi gian t ca mt mc lng t.

t0= Ndmt

Da vo sai slng tyu cu xc nh Ndm.

+ Volmet s chuyn i trc tip kiu tung vi bm thunnghch c cu trc nhsau:


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Khi bt u lm vic b pht xung chun pht lin tc n chkho (K). Ti thi im Ux= 0 hoc Ux= Ukth kho (K) kho, cc xungmang tn sf0khng thn bm thun nghch. Khi Ux> Uktc 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 bchuyn i D/A tng dn Ukcho n khi UxUkth kho K skho, kt thc qu trnh o, b phn ch th s cho kt qu o. Khi

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

khin chuyn i A/D gim Ukcho n khi UxUkth kho K skho,bphn chthscho kt quo.

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

Skhi nhsau:

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 thghi hoc chth

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

gian dng mt nhoc thc m ho bin i gc quay thnh mGray ri tm Gray thnh m nhphn, gii m, chths.

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

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

+ Gia cng Ukthng m ln nht

Trng thi ban u, tt ccc hng m (cc) u bng 0 tc lUk= 0. Trong mi hng bt u tsnhnht ca hng m tng dn Uk

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

hng nhnht, thit bso snh sthng bo iu ny.

Nu Ux= const th Uk s tng lin tc hoc gim lin tc, s mclng tkhng ln lm. Nu Uxbin thin, Uksthay i cho ph hpvi sbin thin ca Uxsiu khin sphc tp hn, slng nhpthc hin gia cng Ukc xc nh:

trong : n l s lng nhp; a1, a2a3, l smc ca cc cc tothnh gi trsca i lng cn o.

Thi gian cc i gia cng theo phng php ny:

t0= k.9.t

k l scc, 9 l chstrong mt cc.

+ Gia cng Ukthng nhnht

Trng thi ban u Uk= 0 v bt u tgi trnhnht ca hng nh

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


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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 chto da trn ccu chthin ng, gc quay ca ccuchthin ng c tnh nhsau:

vi l gc lch pha gia cc dng I1v I2

Smc wattmet in ng nhHnh 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 thcp 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 phti hoc nivi thcp ca bin in p o lng, n ng vai tr nhmt volmet.

Xt vi mch mt chiu ta c:

cos= 1, I1I


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vi Rul in trmt chiu ca cun dy ng.

Thay gi trI2vo (4-1) ta c:

vi P l cng sut tc dng m phti tiu thqua 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 tnn dng i2chchm pha hn so vi in p u mt gckh nh no . Ta c th vc t nhHnh 4.2.

Vn tcng thc (4-1) ta c:

vi ul 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 urt nh: u 0 (Xu


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nhau A, B c mc nhHnh 4.4.

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

Ea= K.pa; Eb= K.pb

vi K l hstl.

Githit ngi ta btr sao cho dng in qua in trr1bng tng

ca hai ng i1, i2cn 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 trr nhsau:

Vi cch ni cc cp nhit ngu nhhnh vschca msbng:


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Trn sA, B l cc cp nhit in, r1l in trc gi trrt nh,

l dng in phti, ta c: iui.

Vi cch btr mch nhtrn ta c:

Tng t

Trn sthng chn ra= rb= r. Kt hp vi csl lun ban uschca mtrong sny stlvi 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 tm u ra (dng, p) tlvi bnh phng u vo v nhvy gi ttrung bnh u ra cng tlvi bnh phng gi trtrung bnh u vo. Nhng phn tnhvy cthsdng o cng sut tc dng P trong mch. Loi thng dng ldit bn dn.

Githit i lng u ra A tlbnh phng vi in p vo u

A = n.u2.


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Trn sr1l in trShunt. Ta bit vi mt bn dn dng in tlvi bnh phng in p tc l

T ta c:

Lc gi trtrung bnh ca in p ri trn in trra l:

Tng t, nu trn phn tB2xut hin in p Ubvi gi trhiudng Ublc ta cng c:

Thnh vta c:


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vi thcp ca bin in p o lng.

- a nhm c kp cng trn trc quay, ngoi ra cn nam chmvnh cu M, thanh dn tG v hthng ccu m.


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

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

cun p sto ra dng in iu chm pha hn so vi in p mt gc 90o.Dng iusinh ra tthng u. Tthng ugm hai thnh phn:

+ upchkhp mch qua mch tcun p gi l tthng ph;

+ ucxuyn qua a nhm gi l tthng 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 tthng iv uc

Ta coi mch tcha bo ho, nn tthng itlvi I:

i =c1.I

vi c1= const.

Ta coi tn sl khng i nn uctlvi U:

uc =c2.U

vi c2= const.vy mmen quay c tnh:

Mq = Kfc1c2UIsin= K1UIsinvi K1= Kfc=1c2

Ta xt hai trng hp:

* Trng hp l tng

Coi cc tthng trng pha vi dng in kch thch tng ng, ta cthvc tnhHnh 4.9.


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chia tthng hoc vng ngn mch khng i xng (cha thhin trnhnh v).

- Khi iu chnh vtr 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 tstquay thun hoc quay ngc khi khngc ti).

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

ngi ta ch to bphn chng tquay bng cch trn mch tcacun p v trn trc quay ngi ta gn hai l thp non T1v T2. Khi anhm quay ti thi im hai l thp i din nhau th chng stc ngtng hv to ra mmen hm (tuy nhin chvi mmen kh nh).

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

Ta c:

Mong mun rng

coi nhkhng i i vi mi loi cng tsau khi chto. V vyta phi iu chnh gc ai bng cch trn mch tca cun dng ngi ta

cun vi vng dy ni qua mt in trR c thiu chnh c. Khiiu chnh gi trR slm thay i tn hao ttrong mch tcun 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 = Unlc ta c P = UnIn; o thi gian quay ca cng tbng ng hbm giy,m svng quay N ca cng tquay trong khong thi gian t.

Ta tnh c hng sca cng tnhsau:


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Ta so snh Cpvi gi trnh mc ghi trn cng t, nu khc nhau taphi iu chnh vtr ca nam chm vnh cu tng hay gim mmencn cho n khi Cpbng gi tnh mc ca cng t. Thc thin 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 cmch l:

- Theo i lng pha:

PA, PB, PCl cng sut tng pha A, B, C.

- Theo i lng dy:

Ud, Idl in p v dng in dy.

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

Theo (4-28) ta chcn o cng sut mt pha bng mt wattmet rily chsca wattmet nhn 3 ta sc cng sut ca cba pha: Giswattmet mc vo pha A nhsau:

Schca wattmet l:


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Do vy cng sut ca ba pha l:

Tng tc thmc wattmet vo pha B hoc pha C.

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

Smc wattmet nhsau:

Cun dng c dng in khi kho K vtr 1 cun p c in p UAC;khi kho K vtr 2 cun p c in p U

AB.

Vy khi ng kho K vpha 1, schca wattmet l:

Khi ng kho K vpha 2, schca wattmet l:

Hnh 4.12. thvc tca phng php o cng sut

dng kho chuyn i


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Theo thvc tta c:

Tng t ta cng c thmc 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 scc schca ba wattmet (hoc ba phn t) ta c cng sut ca mch ba pha.

Ta c:


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Trong thc tngi ta chto wattmet ba pha ba phn t. N gm bacp 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:

Nhvy ta c thdng hai wattmet mt pha c snhHnh 4.14o cng sut trong mch ba pha. Thc t cng da trn nguyn tcny ngi ta chto wattmet ba pha hai phn t. Cch mc nhsau:

4.2.3. o nng lng tc dng trong mch ba pha


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Hnh 4.16. Su dy v thvc 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, phti mang tnh cht cm.

Ta c mmen quay tng ca cng tl:

Ta thy mmen quay tlvi cng sut phn khng trong mch bapha cho nn schca cng tstlvi nng lng phn khng tiuthtrong mch ba pha.

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

Smc nhHnh 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 nhsau:


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Trong scng 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 thvc tnh

hnh v

Ta c m men quay ca cc phn tl:

Hn na ta c:

Thay vo ta c:

Tng t

Vy m men quay tng l:


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5.1.2.1. Cosmet in ng mt pha

Ngi ta sdng ccu chthlogomet in ng chto dngco costrong mch mt pha.

Cun tnh ca cosin ng c mc ni tip vi mch cn ocos(hoc ni vi thcp 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 tnn vic ni cc cun dy ng nhvysto nn cc dng i1v i2l vung pha vi nhau. Ta c su dyv thvc tnhHnh 5.1.

Theo cng thc ca ccu logomet in ng ta c:

vi gc:

Vy

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


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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 trRlv R2. Khi o, ta iu chnh ccvtr con trt trn cc bin trR1v R2sao cho in p u ra ca haimch khuch i l bng nhau, v c kim tra bng cc volmet V1V2.

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 bcng i sin p

hay dng in nhHnh 5.5. Gin thi gian nhHnh 5.6.

Tutheo 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 thgii chto dng co gc lch pha trong khong t(0 180o) vi sai lch nhhn 1%.

5.1.4. Phazomet chths

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 sbit trc.

Cu trc bao gm: bbin i gc pha thnh khong thi gian, bto xung TX1, TX2, TX3, bm, chths, my pht xung chun, khoK1, K2

Scu trc nhsau:


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vi Tn=KT0.

Vy sxung N tlvi gc lch pha x

Nhc im :

- Nu tn snh, v Txcha trong khong T

nnh, do vy ta phi m

rng Tn.

- Nu tn s ln, dn n sai s lng tho trong khong Tx tngln, dn n sai stng.

Thng thng lm vic trong khong mt vi Hz n vi MHz, c sais= 0,1 0,2%.

5.2. o tn s

5.2.1. Phng php gin tip

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

Bit L0, cn cvo schca cc ng ho, ta xc nh c tns.

5.2.2. Tn smet cng hng


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Nguyn l hot ng:

Tn smet cng hng gm mt nam chm in, to ra bi cun dyin qun trn li st thnh chU, mt ming thp nm trong ttrngca nam chm in, gn cht vo thanh l cc l thp rung c tn sdaong ring khc nhau. Tn sdao ng ring ca hai l thp knhau hnkm nhau l 0,25 hoc 0,5Hz. in p ca tn hiu cn o tn ssca vo cun dy ca nam chm in sto ra sdao ng ca tt cccl thp. Tuy nhin l thp no c tn sdao ng ring bng tn sf ths dao ng cc i do cng hng ring, cn cc thanh khc khngcng hng th khng dao ng cc i. Nh vy chng ta sc kt

quti trstng ng vi thanh rung cc i.


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5.2.3. Tn smet in t

5.2.3.1. Nguyn tc chung

Tn s met loi ny da trnnguyn tc chung l sdng phng

php m xung n gin bng cchphng np mt t in C t mt

ngun in p khng i U0no .

Tn hiu cn o c tn sfxca vo khng chmt kho in tK, kho ny c thit ksao cho trong mt chu kca in p uk, khoK ng t1 sang 2 mt ln.

Xt khi kho K vtr 1, in tch np vo ttnh nhsau:

q = C.U0

in tch np vo ttrong thi gian mt giy l:

Q = q.fx= C.U0.fx.

in tch ny chy qua chthkhi kho K vtr 2 to ra dng intrung bnh

(K1= const)

ITBc chbng c cu tin G. Thang chia c khc trctip theo n vtn sv ta c thc ngay tn strn chthG. Munmrng gii hn o, ta thay i gi trca tC.

5.2.3.2. Tn smet in t

Tn smet in tc thit knhHnh 5.12. Kho i ni K thchin bng mt n bn dn T. in p uxcn o tn sc a vo ccgc ca T.

na chu km ca in p Ux(so vi cc gc ca T), n T kho,tC c np tngun U0qua D1, qua chthg cho ti khi Uc= U0.

na chu kdng ca in p Uxn T m, tC phng qua n,

qua D2cho ti khi UC= UB.


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in tch m tin np trong mt ln ng mca T l:

Lng in tch phng np trong thi gian mt giy chnh l dngin i qua chth

Vy dng in trung bnh chy qua chthtlbc nht vi fx. Ta cthkhc vch thang chia theo n vtn s.

5.2.4. Tn skchths

Nguyn l: m sxung N tng ng vi schu kca tn scno fxtrong khong thi gian gi l thi gian o T0.

Trong khong T0 ta m c N xung t lvi tn so fx. S

khi ca mt tn skchthsnhsau:

Mch to xung c nhim vbin tn hiu hnh sin hoc tn hiu xungc chu k thnh mt dy xung c bin khng i (khng ph thucvo bin ca tn hiu vo) nhng tn sbng tn sca tn hiu vo.

My pht xung chun f0= 1MHz.

B chia tn svi cc nc c h s chia l 10n. Tn s chun f0=1MHz c chia n 0,01 Hz. Ngha l u ra ca mch iu khintheo 10n(n = l,2,,8) ta c thnhn c khong thi gian T0= 10

-6,

10-5, 10-4, 10-3, 10-2, 10-1, 1, 10, 100s.


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Khong thi gian ny siu khin mkho K (kho c hai u

vo). Tn hiu fxtheo u vo thhai si vo bm ra ccu chth.

Sxung m my m m c sl:

Nu thi gian o c gi trl 1s th sxung N (tc l scc chu k)schnh l tn scn o fxngha l: fx= N.

Mch iu khin ph trch vic iu khin qu trnh o: Bo m

thi gian biu thkt quo ct0,3 5s trn chths, xo kt quo


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Phn tip theo ca ti liu chng ta sni ti my hin sng tngt, loi dng phbin trong kthut o lng in t.

5.3.2. Skhi ca mt my hin sng thng dng

Tn hiu vo c a qua b chuyn mch AC/DC (kho K ngkhi cn xc nh thnh phn DC ca tn hiu cn khi chquan tm nthnh phn AC th mK). Tn hiu ny squa bphn p (hay cn gi l

b suy gim u vo) c iu khin bi chuyn mch nm xoayVOLTS/DIV, ngha l xoay nm ny cho php ta iu chnh t l casng theo chiu ng. Chuyn mch Y-POS xc nh vtr theo chiung ca sng, ngha l c thdi chuyn sng theo chiu ln hoc xung

tu bng cch xoay nm vn ny. Sau khi qua phn p, tn hiu vo sc bkhuch i Y khuch i lm lch a ti iu khin cp lmlch ng. Tn hiu ca b K Y cng c a ti trigo (khi ng

b), trng hp ny gi l ng b trong, kch thch mch to sngrng ca (cn gi l mch pht qut) v a ti iu khin cp lm lchngang (tng hiu quiu khin, mt smch cn sdng thm cc

bkhuch i X sau khi to in p rng ca). i khi ngi ta cngcho mch lm vic chng bngoi bng cch ct ng tn hiutKY, thay vo l cho tn hiu ngoi kch thch khi to sng rng

ca.


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+ a nt gia ca VOLTS/DIV, TIME/DIV, HOLD OFF vvtr CAL (cn chnh)

2. Vn VOLTS/DIV v TIME/DIV vvtr 1V/DIV v 2s/DIV.

3. Bt ngun.

4. Xoay Y-POS iu chnh im sngtheo chiu ng (im sng s chy ngangqua mn hnh vi tc chm). Nu vnTIME/DIV ngc chiu kim ng h(theochiu gim) th im sng s di chuynnhanh hn v khi vtr cs trn mn hnh s l mt vch sng thaycho im sng.

5. iu chnh INTENS thay i chiv FOCUS thay i nt ca vch sng trnmn hnh.

6. a tn hiu chun kim tra chnhxc ca my.

a u o ti vtr ly chun (hoc l tmy pht chun hoc ngaytrn my hin sng vtr CAL 1Vpp, lkHz). Vi gi trchun nhtrnnu VOLTS/DIV vtr 1V/DIV v TIME/DIV vtr 1ms/DIV th trnmn hnh sxut hin mt sng vung c bin nh mt trn mnhnh v rng xung cng l mt trn mn hnh (xoay Y-POS v X-POS m mt cch chnh xc).

Sau khi ly li cc gi trchun trn, tuthuc chlm vic mta sdng cc nt iu khin tng ng nhsni phn tip theo.

5.3.3.2. Cc phn iu khin chnh

a)iu khin mn hnh


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Phn ny bao gm:

+ iu chnh sng - INTENSITY - ca dng sng. Thng thngkhi tng tn squt cn tng thm sng tin quan st hn. Thccht y l iu chnh in p li.

+ iu chnh nt - FOCUS - ca dng sng. Thc cht l iuchnh in p cc anot A1, A2 v A3.

+ iu chnh lch ca trc ngang - TRACE - (khi vtr ca my nhng im khc nhau th tc dng ca ttrng tri t cng khc nhaunn i khi phi iu chnh c vtr cn bng).

b)iu khin theo trc ng

Phn ny siu khin vtr v tlca dng sng theo chiu ng.Khi tn hiu a vo cng ln th VOLTS/DIV cng phi vtr ln vngc li.

Ngoi ra cn mt sphn nh:

INVERT: o dng sng;

DC/AC/GD: hin thphn mt chiu/xoay chiu/t ca dng sng;

CH I/II: chn knh 1 hoc knh 2;

DUAL: chn chai knh;

ADD: cng tn hiu ca chai knh.


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Khi bm nt INVERT dng sng ca tn hiu s bo ngc li(o pha 180o).

Khi gt cng tc vvtr GD trn mn hnh sxut hin mt ngngang, dch chuyn v tr ca ng ny xc nh v tr t ca tnhiu.

Gt cng tc vv tr DC ngha l trong tn hiu bao gm c thnhphn mt chiu v xoay chiu, gt vvtr AC l hin dng sng tchthnh phn mt chiu. Xem hnh di y: (bn tri l chDC v

bn phi chAC).

Khi n nt DUAL chn chai knh th trn mn hnh sxut hin

hai thca hai dng sng ng vi 2 u o. ADD cng cc sngvi nhau. Ni chung vtr ca ba nt CH I/II, DUAI v ADD scho ccchhin thkhc nhau tuthuc vo tng loi my.

c)iu khin theo trc ngang

Phn ny iu khin v tr v t lca dng sng theo chiu ngang.Khi tn hiu a vo c tn scng caoth TIME/DIV phi cng nhv ngcli. Ngoi ra cn mt sphn sau:

X-Y: ch ny knh th 2 s


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lm trc X thay cho thi gian nhchthng.

Ch : khi my hot ng chnhiu knh th cng chc mtphn iu khin theo trc ngang nn tn squt khi sl tn squtchung cho chai dng sng.

5.3.4. ng dng ca my hin sng trong kthut o lng

My hin sng hin nay c gi l my hin sng vn nng v

khng n thun l hin thdng sng m n cn thc hin c nhiukthut khc nh thc hin hm ton hc, thu thp v x l sliu vthm ch cn phn tch cphtn hiu...

Trong phn ny chng ta chni ti nhng ng dng cbn nht camt my hin sng.

5.3.4.1. Quan st tn hiu

quan st c tn hiu ch cn thit lp my chng btrong v iu chnh tn squt v trigo dng sng ng yn trn mn

hnh. Khi ny c thxc nh c sbin thin ca tn hiu theo thigian nhthno. Cc my hin sng hin i c thcho php cng mtlc hai, bn hoc tm tn hiu dng bt kcng mt lc v tn squanst c thln ti 400MHZ.

5.3.4.2.o in p

Vic tnh gi trin p ca tn hiu c thc hin bng cch ms trn mn hnh v nhn vi gi trVOLTS/DIV.

V d: VOLTS/DIV ch1V th tn hiu cho hnh trn c:

Vp = 2,7 x 1V = 2,8V

Vpp = 5,4 x 1V = 5,4V

Vrms = 0,707Vp = l,98V.


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vi tn schun f0. Tn hiu cn o a vo cc Y, tn hiu tn schuna vo cc X. Chlm vic ny ca my hin sng gi l chX-Y mode v cc sng u c dng hnh sin. Khi trn mn hnh shinra mt ng cong phc tp gi l ng cong Lissajou.

iu chnh tn schun ti khi tn scn o l bi hoc c nguynca tn schun th trn mn hnh sc mt ng Lissajou ng yn.

Hnh dng ca ng Lissajou rt khc nhau tu thuc vo ts tn sgia hai tn hiu v lch pha gia chng (xem hnh di).

Ta c:

vi n l smi theo chiu ngang v m l smi theo chiu dc (hoc c

th ly sim ct ln nht theo mi trc hoc sim tip tuyn vihnh Lissajou ca mi trc).

Phng php hnh Lissajou cho php o tn strong khong t10Hzti tn sgii hn ca my.

Nu mun o lch pha ta cho hai tn s ca hai tn hiu bngnhau, khi ng Lissajou c dng elip. iu chnh Y-POS v X-POSsao cho tm ca elip trng vi tm mn hnh (gc to). Khi gclch pha c tnh bng:


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vi A, B l ng knh trc di v ng knh trc ngn ca elip.

Nhc im ca phng php ny l khng xc nh c du cagc pha v sai sca php o kh ln (5 - 10%).


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Nguyn tc

Dng ampemet v volmet o dng v p trn in trri suy ra Rx' =

A

V

I

Uthng qua hai s:

Phn tch sai sph

- Xt Hnh 6.1a

Vy sai sphtrong qu trnh o:

Nhn xt: Nu RA cng nh th p cng nh cho nn phng phpny dng o in trln.

- Xt Hnh 6.1b

Vy sai sphtrong qu trnh o l:

V d6.1: Tnh sai sphkhi tin hnh o in trmt chiu ca


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Th

Nhvy nu trong qu trnh o lun gic tsR1/R2= R3/R4thta stnh c Rxthng qua tstrn.

Ch :

- Cc in trRl, R2, R3, R4, R0 l cc in trmu lm bng hpkim ca mangan c chnh xc cao; R0 l in trcc c ththayi t0 9999,9bc 0,1; Rxl in trcn o.

- Cc in trRl, R2, R3, R4c gi tr10. TsRl/R2c ththayi v thng bng: 10-4, 10-3, 10-2, 10-1, 1, 10. Cn t tsR1/R2saocho ph hp nht vi Rxcn o.

- Cu kp c mt u im ni bt l c thloi trc in trdyni, nhng c nhc im l kh cn bng nu Rxl cc cun dy my

in.

6.1.5. o in trcch in ca li v thit bin

6.1.5.1. Nhn xt vin trcch in

in trcch in l cc gi trin trln (vo khong vi Mtrln v dnhin trcch in ca vt liu cch in), do vy phng

php o in trcch in l cc phng php o c th in trln.in trcch in cng ln tng ng vi cp in p lm vic ca thit

bin cng ln.

Khi o in trc trsln thng thng sc hai thnh phn intr:

+ in trkhi Rv(Volume Resistance), y l thnh phn in trcn o.

+ in trr bmt Rs(Surface Leakage Resistance).

Hai thnh phn in trny xem nhsong song vi nhau, nhvyhai in trny c thso snh c th snh hng ng kn intrkhi cn o.

6.1.5.2. Phng php o in tr cch in dng volmet v


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trnh nh hng ca Rs bng cch loi b dng in Is quamicroampemet, ngi ta dng dy dn in (khng c vbc cch in)qun quanh lp vcch in v ni trc microampemet. Nhvy dngin Isi qua Rslc trc si qua dy dn ny o nh hng ca Rsvo Rvbloi b. Vng dy ny gi l dy bo v.

Ch : Tuy nhin khng phi trng hp no khi o in tr cchin chng ta cng sdng vng dy bo v, khi o in trcch inno chng ta phi xc nh xem in trcch in c bnh hng

bi in trbmt hay khng, nu c mi sdng vng dy bo v.

6.1.5.3. Phng php o in trcch in dng megommet chuyndng

Megommet l thit b chuyn dng o in tr cch in, ccu to bi ccu chthlogomet tin.

Smegommet nhsau:


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Thay i thang o bng cch thay i trsR2.

Trong megommet c u G dng ni dy bo v loi bintrr bmt.

6.1.5.4.o in trcch in ca li in v thit bin

a) Nhn xt

Mi li in c thxem nhhng lot on dy c chiu di mt

n vmc ni tip vi nhau. Cc thng sca chng l cc thng sdimc song song vi nhau nhHnh 6.12.

Trong tnh ton ngi ta thng coi chng nhnhng thng s tptrung. in trcch in ca ng dy thng bthay i hoc c thng dy bsc. V vy khi vn hnh, hoc khi th nghim, nghimthu cc ng dy v cc thit bcch in th bt buc phi o in tr

cch in. Gi trin trny khng c nhhn mt gi trno theo quy trnh, quy phm hin hnh (v din trcch in c o vimegommet kc E = 1000V hoc 2000V v in trcch in ti thiuc quy nh l 1M).

in trcch in ca li trn on c xt thng c o giahai u dy dn in vi nhau hoc tng dy dn in vi dy trung tnh.

b)o in trcch in ca li v thit bin khi tt ngun in

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