CHNG 3 O CC I LNG IN V KHNG INPHN 1: O CC I LNG IN$3.1 ODNG IN V
IN P3.1.1. O DNG IN1.Khi nim chungC th o dng in bng cc phng php o
khc nhau nh: phng php trc tip, gin tip v so snh.Trong phng php o
trc tip, ngi ta dng cc dng c o dng nh: Ampemt, miliAmpemt, micro
ampemt v.v... o dng v trc tip c kt qu trn thang chia . Khi o dng
in, ampemt s c mc ni tip vi ti v s gy nh hng n php o. Nu in tr
trong ca ampemt rt nh so vi in tr ti th sai s do nh hng ca ampemt l
khng ng k.Trong phng php o gin tip, ngi ta c th dng vnmt o in p ri
trn mt in tr mu, thng qua tnh ton (nh lut m) ta s xc nh c dng cn
o.Trong phng php so snh, ngi ta o dng in bng cch so snh dng in cn o
vi dng in mu chnh xc. trng thi cn bng ca dng cn o ta s c kt qu trn
mu.2 o dng in mt chiu odnginmtchiu s dng cc Ampe mt mt chiu, c ch
to t c cu o t in, tuy nhin c cu o t in ch o c dng in mt chiu rt nh
(vi chc mA). o c dng in mt chiu ln hn dng in cc i m c cu c th o c,
ngi ta u song song vi c cu vi mt in tr c bit c tr s nh (in tr
sun).Nu cn o dng in mt chiu I ln gp n ln dng in nh mc ca c cu, ta
mc sun theo s hnh 3.1.Khi :S CCS CCcc ccR RR RI R I+SS CCCCRR RIIn+
1 nRRCCSC th to ra ampemt t in nhiu thang o bng cch mc nhiu sun qua
mt kho chuyn mch c bit. Hnh 3.2 m t ampemt t in 4 thang o (I1, I2,
I3, I4).Cc in tr sun tng ng c mc vo l RS1, RS2, RS3 v Rs4. Tnh cc
in tr sun ny bng cch thnh lp h phng trnh. 65-Hnh 3.1 Mc sun m rng
gii hn o dng in+RCC ICC
Ti
IICCRSHnh 3.2. Ampemt t in 4 thang oCCCCSIInnRR1111;1CCCCS
SIInnRR R2122 1;1 +CCCCS S SIInnRR R R3133 2 1;1 + +CCCCS S S
SIInnRR R R R4144 3 2 1;1 + + +Gii h phng trnh s tm c cc gi tr RS1,
RS2, RS3 v Rs4.Ch : Vi ampemt nhiu thang o, trn thang o ch nh du cc
vch. Mun tnh c tr s dng in cn o phi s dng cng thc: I = CI. :Trong :
l s vch m kim ch CI l hng s thang o: maxdmCCIIC Vi max l s vch ln
nht ca thang o ImCC l dng in ng vi vch ln nht ca thang o.Sai s ca
ampemt c tnh nh sau :S = RS + Rcc + ccTrong : RS sai s hay bin ng
ca gi tr sun. Thng thng ch to Sun bng vt liu c in tr n nh v h s
nhit rt nh. V ngi ta thng chn Sun c cp chnh xc cao hn cp chnh xc ca
c cu hai ln.cc - sai s ca c cu.Rcc- Sai s hay bin ng v gi tr ca
RCC,ch yu l nh hng ca gi tr nht .Nh vy sai s ch yu y l sai s ca c
cu t in.3. o dng in xoay chiu Ty theo phm vi v mc ch s dng m c th c
cc loi ampemt xoay chiu c bn sau :- o dng in xoay chiu min tn s cng
nghip thng dng cc : ampemt in t, ampemt in ng v ampemt st in ng.- o
dng in xoay chiu min tn s m tn thng dng ampemt t in c chnh lu.- o
dng in xoay chiu tn s cao v siu cao thng dng ampemt nhit in.a)
Ampemt t in c chnhlu. Mun o c dng xoay chiu bng ampemt t in ngi ta
mc thm b chnh lu nn in cho dng in qua c cu o.66Hnh 3.3 l s ampemt t
in chnh lu c chu k, vi h s chnh lu cao.Ampemt chnh lu c chnh xc
khng cao do chnh lu to ra sai s cho php o. Kimca ampemt chnhlukhng
quay theo tr hiu dng m theo tr trung bnh ca dng in. Tuy vy ng vi mi
tr trung bnh c mt tr hiu dng tng ng v vy trn thang o ngi ta ghi cc
gi tr hiu dng tng ng vi vch.b) Ampemt in tC cu o in t dng lm ampemt
in t c cun dy vi s vng nh, tit din ln.Thng thng kim ca c cu quay ht
thang o, sc t ng ca cun dy c dng in to ra phi ln F=IW. Thng thng F
=200Ampe.vng.Nhvylc t ho s quyt nh s vng dy trn c s dng in nh mc ca
dng c o.M rng gii hn o cho ampemt in t: Ngi ta khng dng sun m rng
gii hn o cho ampemt in t m bng cch chia cun dy ca n thnh nhiu phn
on ging nhau. Dng kho chuyn mch i ni cch u cc phn on ny (ni tip hoc
song song), ta s c mt ampemt nhiu thang o.Phng php phn on cun dy
tnh ca c cu o in t cng ch p dng ch to ampemt in t nhiu nht l ba
thang o, v tng s lng thang o b tr mch v chuyn thang o phc tp khng
th thc hin c.c) Ampemt in ngC hai loi s ampe mt in ng :-Khi o dng
nh hn hoc bng 0,5A, ngi ta u ni tip cun tnh v cun ng (hnh 3.8a).
Khi :67Cun dyL thpHnh 3.4. Ampemt in tN SHnh 3.3. Ampemt t in c
chnh luHnh 3.5: S u cc phn oncun dy ampemt in t2 1121I IddMD 2
121IddMD vi I1 =I2= I; = 0; cos= 1- Khi o dng ln hn 0,5A, ngi ta u
cun tnh v cun ngsong song vi nhau , v mc thm cc phn t vo mi nhnh b
sai s do tn s gy ra v dng trong hai cun dy ng pha vi nhau (hnh
3.8b). Cch m rng thang o cho ampe mt in ng ging nh ampemt in t.d)
Ampe mt nhit inCng l ampemt chnh lu v nh cp nhit ngu bin i dng xoay
chiu v dng mt chiu c cu to nh hnh 3.94. o dng in lna) o dng in mt
chiu ln: C th s dng phng php o t trng sinh ra xung quanh dn dn.Mun
o dng in I phi to ra mch t tr hnh xuyn v lng dy dn qua mch t ny
(hnh 3.8)T trng sinh ra xung quanh dy dn c t cm: H B0 Ta c: rIH Nh
vy bng cch o t cm B ta c th suy ra dng in cn o.b) o dng in xoay
chiu ln68Hnh 3.6: Ampemt in nga) o I 0, b) o I 0,5A//a)b)Hnh 3.7.
Ampe mt nhit inHnh 3.8: o t trng sinh ra xung quanh dn dn. o dng in
xoay chiu ln th phng php thng dng nht l s dng cc ampe mt xoay chiu
kt hp vi cc bin dng xoay chiu.Bin dng xoay chiu c li hnh xuyn bng
thp k thut in trn c qun hai cun dy s cp W1 v cun th cp W2 (hnh
3.9)Bin dng lm vic ch ngn mch, nnta cIkWWIIW I W I 21212 2 1 1Vi kI
gi l h s bin dng.Thng bin dng thng c ch to sn vi dng nh mc I2 v h s
bin dng kI thay i ph thuc vo dng s cp I1.II. O IN P1.Khi nim chung
Vn mt l mt dng c o in p, khi vnmt c mc vo phn t cn o in p, vnmt c
ni song song vi ti trong mch o (hnh 3.11). Khi s dng vnmt o in p cn
lu n sai s sinh ra trong qu trnh oKhi chamcvnmt vomch, dng in qua
ti l :T ngR REI+in p ri trn ti l:TT ngT TRR REIR U+ C rt nhiu cc
phng php v thit b o in p khc nhau,cc vn mt c phn loi
theonguynllmvic. Trc tin phi k n cc vn mt c in nht, l cc vn mt c in
bao gm : vn mt t in, in t v vn mt in ng. Cc vn mt ny lm vic da trn
s tng tc Ngy nay vi s pht trin ca k thut bn dn, dn n s ra i ca cc
vn mt in t v ngy cng c ng dng rng ri. Cc vn mt in t c phn loi thnh
vn mt tng t v s.69Hnh 3.11: Mc vn mtPh ti inBIHnh 3.10 Ampemt xoay
chiu kt hp vi bin dngAmpemtxoay chiuHnh 3.9: Bin dng in1. Cc vn mt
c inCc vn mt c in o in p da trn nguyn l bin i in p thnh m men c hc.
iu nay c th c thc hin theo nhiu cch khc nhau nh : s tng tc gia cc
dng in (vn mt in ng), s tng tc gia dng in v t trng (vn mt t in). Cc
vn mt c in c s dng ph bin nht l : vn mt t in, vn mt in t v vn mt in
ng.a) Vnmt t in o in p mt chiuC cu o t in (hnh 3.12a)l dng c o in p
mt chiu, nhng n ch o c cc in p rt nh (vi chc mV). Mun o c cc in p
ny ch vic u song song trc tip c cu vo mch cn o. Tuy nhin, o c in p
mt chiu ln hn in p cc i m c cu c th o c, ngi ta u ni tip vi c cu vi
mt in tr ph c bit c tr s ln (hnh 3.12b)Nu cn o in p mt chiu U lngp
m ln in p nhmcca c cu, tamcmt intr ph Rp ni tip vi cun dy ca c cu,
khi :ImCC=CCdmCCp CCRUR RU+Suy ra CCp CCdmCCRR RUUm+ CC pR m R ) 1
( Mun to ra vnmt t in nhiu thang o ta mc ni tip nhiu in tr ph vi c
cu o. Hnh 3.13 m t vnmt t in 3 thang o.Ch : Cng ging nh ampemt nhiu
thang o, vnmt nhiu thang o trn thang o ch nh du cc vch. Mun tnh c
tr s in p cn o phi s dng cng thc: U = CU. ;Trong : l s vch m kim
chCU l hng s thang o: maxdmCCIUC70Hnh 3.13: Vn mt t in nhiu thang
oHnh 3.12: Vn mt t inC cu o t inM rng thang o cho c cu o t inVi max
l s vch ln nht ca thang oUmCC l dng in ng vi vch ln nht ca thang
o.Ta cng c th ch to ng h a nng o dng v p t c cu o t in, thng qua mt
kha chuyn i I/U ch o dng hay p s c la chn (hnh 3.14).Ta c h phng
trnh' + + + + ) ( ) ( ) () ( ) ( ) (1 1 1 11n d n nn d n nR R R r I
R I IR R R r I R I IMun o in p xoay chiu bng vnmt t in, ngi ta u
thm mch chnh lu cho c cu. Hnh 3.15 l s vnmt chnh lu. Cng ging nh
ampemt t in c chnh lu, vnmt t in c chnh lu c chnh xc khng cao do
chnh lu to ra sai s ph. Kim ca vnmt chnh lu khng quay theo tr hiu
dng m theo tr trung bnh ca dng in. Tuy vy ng vi mi tr trung bnh c
mt tr hiu dung tng ng v vy trn thang o ngi ta ghi tr cc gi tr hiu
dng.b)Vnmt in t lm vnmt in t ngi ta s dng ccuintccundytit dinnh,
nhng s vng rt ln (thng t 1000 n 6000 vng).m rng gii hn o v to ra
vnmt in tnhiuthang o ngi ta cng lm tng t nh vnmt t in. Tc l cng mc
ni tip cun dy vi cc in tr ph.c) Vnmt in ngVnmt in ng c cu to ging
nh ampemt in ng o dng in nh hn hoc bng 0,5A, tc l c cun dy tnh v ng
lun mc ni tip vi nhau (hnh 3.17).71RCC RP1 RP2 RP3
UCC U1U2U3Hnh 3.16 : Vnmt in t 3 thang oHnh3.15: Vnmt t in c
chnh luHnh 3.14: ng h o a nngVnmtyu cu c tng tr ln nn cun tnh ca
vnmt in ng c s vng ln, c tit din nhhn rt nhiu so vi cun tnh
caampemt.Phngtrnhgcquaycavnmt in ng :2 1121I IddMD Ta c I1 =I2 =I
VZU;ZV l tng tr vo ca vnmt.Suy raddMDZUV1222C th ch to Vmt in ng
nhiu thang obng cch thay i cch mc song song hoc ni tip hai phn on
ca cun tnh v ni tip vi cc in tr phHnh 3.18 th hin s nguyn l ca vnmt
in ng hai thang o : Khi kho chuyn mch ang v tr 1, hai phn on A1 v
A2 ca cun tnh mc song song vi nhau v mc ni tip vi cun ng B, tng ng
vi gii hn o 150V. Khi kho chuyn mch ang v tr 2, hai phn on A1, A2 v
cun ng B c mc ni tip vi nhau, tng ng vi gii hn o 300V. Cc t in C to
mch b tn s cho vnmt.Ch : Cn o in p li in cao p ta phi mc vnmt kt hp
vi my bin p o lng BU (hnh 3.19)2. Cc vn mt ch th s 72A1 Rf Rf C 1
300V
2
12
12
150VRf Rf
BA2C Hnh 3.18 : S Vmt in ng nhiu thang oBUHnh 3.19 : Mc vmtkt hp
vi bin p
1
2Hnh 3.17: Vn mt in ngVn mt s (digital voltmeter -DVM) l thit b
o in p ch th kt qu o trc tip di dng s (hin th bng tinh th lng hay
bng LED. Ngy nay th cc vn mt s c s dng rng ri do c cc u im nh chnh
xc cao hn, phn gii tt hn, khng c sai s do ch th c, c kh nng t ng
chnh 0 v t ng chuyn thang o, C kh nng x l s o bng my tnh, thit b o
gn v kt cu chc chn hn. Vic phn loi cc vn mt da vo b chuyn i tn hiu
tng t sang s m n s dnga) Vn mt s chuyn i thi gian mt nhp y l b
chuyn i gin tip in p - thi gian, trong phng php ny ta cn c mt in p
hay dng in hay mt thi gian chun tng ng vi mc lng t ho b nht biu din
i lng cn o th cc mc chun ny c cng li vi nhau cho n khi tng gn ng vi
i lng cn chuyn i.Qu trnh ny khng nht thit theo tng bc bc thang m c
th theo ng thng ca xung rng caTrong phng php m n gin in p chun dng
so snh l in p rng ca tuyn tnh, sn tng ca xung rng ca tuyn tnh s c
so snh vi in p tng t bng b so snh. B so snh s thc hin chc nng so
snh in p rng ca UR.Cvi in p tng t cn o UX S khi ca vn mt s chuyn i
thi gian mt nhp c m t nh hnh 3.20aKhi m my xung khi ng s tc ng ln
my pht xung (MFX) rng ca. in p ca xung rng ca URC c a vo 2 b so
snh. Ti b so snh 1, in p rng ca URC c so snh vi tn hiu in p cn o
UX. KhiUX = UR.C (thi im t=t1), b so snh 1 pht xung iu khin m cng
G. ng thi khi cng G m, my pht xung chun a cc xung chun qua cng G n
b m. Sau xung rng ca tip tc gim, in p UR.Cc so snh vi in th t (0V)
b so snh 2. Ti thi im t = t2, khi URC= 0V b so snh 2 s pht xung iu
khin ng cng G, b m ngng m. Gi t l thi gian cng m, th in p cn o Ux s
c ln t l vi t73Hnh 3. 20: S vn mt theo phng php tch phn mt nhpS
khiBiu thi gianBin i in p thi giano thi gianB so snh 1B so snh
2URCUxa bUXS xung m xut hin trong khong thi gian t tng ng vi 0V
UR.C UX, v c ln l N t l vi in p cn o. ViT0 l chu k ca xung chun th
s xung m c l:00tfTtN T cng thc trn ta thy N t l vi UX nh mong mun,
nhng N cn ph thuc U, t, f0. Nu nhng tham s ny khng n nh th kt qu m
c sai s. Do vy trong phng php ny yu cu f0 phi ln m bo chnh xc. th
thi gian c minh ho trn hnh 3.20b.b) Vn mt s chuyn i thi gian hai
nhp Nhc im ca vn mt chuyn i thi gian l sai s ln, khc phc nhc im ny
ngi ta s dng vn mt s tch phn 2 nhp. S khi v nguyn l lm vic ca vnmt
s theo phng php tch phn hai sn dc c biu din trn hnh 3.21.Vnmts
lmvic theo nguynlchuyn i gin tipinp -thigian,trong phng php ny ta
cn c mt in p hay dng in hay mt thi gian chun tng ng vi mc lng t ho
b nht biu din i lng cn o th cc mc chun ny c cng li vi nhau cho n
khi tng gn ng vi i lng cn chuyn i.Nguyn l lm vic ca vn mtKhi m my
xung iu khin khi ng b m (reset b m), i ni S1 ng v S2 m, in p cn o
Ux c a vo b tch phn vt C bt u np. Khi tn hiu ra Vout ca mch tch phn
tng dn, in p ra ca b so snh thay i trng thi ca cng G cc xung chun t
my pht xung cung cp cho b m. B m tng n s ln nht (trong khong thi
gian t1).Ht thi gian t1, S1 mvK2 ng , khi in p Uk c ni vi b tch
phn, do -Uk ngc chiu vi Ux nn t C phng. Tn hiu ra Vout gim tuyn tnh
n 0V, b so snh chuyn trng thi v cng G kha.Thi gian t C phng t2 t l
vi in p vo Uk. Trong khong thi gian t1 in p tch phn c tnh tb t t UC
Rdt UC RU0.1 111 1 in p h trn t sau thi gian t2 l:74iu khinlogicMFX
chunHT sB SSUx-UkS1S2Gt1t2tHnh 3.21 S vn mt theo phng php tch phn
hai nhpa)S khib)Biu thi gian1.1.tRCUUb Xtta c U1 t l vi UXt.b Ht
thi gian t1kho K1ng, kho K2c m ra ng thi cc xung t b pht xung qua
kho K3 thng a n b m (m thun). Khi K2 thng in p U2 np cho t C theo
chiu ngc li UX: tUU .C R202 SSau khong thi giant2: 2202.C RtUU Sau
khong thi gian t2 th|U2|=|U1|, hay ni cch khc in p trn t Uc=0. Khi
thit b so snh pht xung kha K2 , K3 kt thc qu trnh o.Ta c : Hoc
2221..tRUtC RUXt b XtGi T l chu k ca cc xung chun, ta c th xc nh c
s xung a n mch m trong khong thi gian t1 l:n1 = t1 . fo Trong fx:
tn s xung nhp do ta c :t1 =ofn1 Thay vo ta xc nh c t2:t2 = 010
12.fnUURRXDo vy ta c xung nhp m c nh mch m trong khong thi gian t2
l:n2 = t2 .f0 = 10.12nUURRb XtSau thi gian t2mch m u ra b ngt v UC
=0 v mch logic ng cng mch AND qu trnh li c lp li trong chu k chuyn
i tip theo. Nh vy s xung m c u ra t l vi in p tng t UX cn chuyn i.
Kt qu m khng ph thuc vo cc thng s R, C ca mch v khng ph thuc vo tn
s nhp do vy kt qa kh chnh xc. th in p ra trn mch tch phn c biu din
trn hnh 3.21b.Ngun sai s ch yu ca vnmt ny l s khng n nh ca Uo, s
khng n nh ca thit b so snh, p d ca K1; K2, khuch i thut ton.$3.2 O
CC THNG S CA MCH INCc thng s ca mch in bao gm: in tr R, in dung C,
in cm L, gc tn hao (tg) ca t in v h s phm cht (Q) ca cun dy. Cc
thng s ny c th c o theo cc phng php o khc nhau3.2.1 Cc phng php o
in tr1. Cc phng php gin tipa) o in tr dng vn mt v ampemt: C hai cch
mc o in tr (hnh 3.22)Davosch caampemt vvnmtxcnhcgitr caintrRx: IURx
'75Gi tr thc Rx ca in tr cn o c xc nh theo cch mc ampemt v vnmt
trong mch nh sau:Hnh 3.22a:Theo s I= IV + IX vi IV l dng qua vnmt.
Nu dng in qua vnmt IV rt nh so vi dng in qua RX(tng tr vo ca vmt rt
ln so vi RX) th sai s do nh hng ca vnmt khng ng k).Hnh 3.22b: Khi
in tr cn o Rx c xc nh bi:IURx Theo s U= UA + UXVi UA l in p ri trn
ampemt, UX: in p ri trn RX .Nu in tr ca ampe mt RA rt nh so vi RX,
th UX >> UA. Sai s do nh hng ca ampemt khng ng
k.b)ointrbngvnmthocampemt v in tr muo in tr bng vnmt v in tr muin
tr cn o Rx c mc ni tip vi in tr mu R0 (c chnh xc cao) v c ni vo
ngun in U (hnh 3.23). Dng vn mt o in p ri trn in tr Rxl Uxv in p ri
trn in tr mu R0l U0.Ta c:0000RUURRURUI Ixxo xxx o in tr bng ampemt
v in tr muin tr cn o Rx c mc song song vi in tr mu R0 v c ni vo
ngun in U (hnh3.24).Dngampemt odnginqua in tr Rx l Ix v dng in qua
in tr mu R0 l I0.Ta c:000 0 0RIIR R I R I U Uxx x x x 76Hnh 3-22: S
o in tr bng Vnmt v ampemtHnh 3.23: o in tr bng vnmt v in tr muHnh
3.24: o in tr bng ampemt v in tr mu2. Cc phng php o trc tip dng m
mta) m mt mc ni tipC th s dng c cu o t in ch to mmt o in tr. Mch o
c mc theo hnh 3.25. y l mmt mc ni tip bao gm: mt ngun in p mt chiu
(pin) mc ni tip vi in tr ph v c cu o t in c dng nh mc nh. Mch o ny
c hai u ra A, B ni vo in tr cn o.Dng qua c cu ch th: x CC P MR R R
REI+ + +Kim ca c cu o ch mt gc:x CC P MR R R REk kI+ + + Khi Rx = 0
(ni ngn mnh hai u A, B) thCC P MR R REk+ + . Lcnykimlch mt gc ln
nht (dng qua c cu cc i). Khi cc u ra A, B h mch ( Rx=) th = 0 (khng
c dng qua c cu).Thang o ca mmt kiu ny l thang o ngc.Trong qu trnh s
dng sc in ng cangunpingimdndnnphpo km chnh xc. Mun php o ng, trc
khi o phi iu chnh v tr 0 ca kim o bng cch ni ngn mch hai u A, B iu
chnh RM kim ch 0 ri mi tin hnh ommt s ni tip nhiu thang o: Ch to
mmt nhiu thang o bng cch dng nhiu ngun cung cp hoc dng mt ngun cung
cp vi cc in tr phn nhnh cho cc thang o khc nhau.77RMRPHnh 3.25: mmt
mc ni tipHnh 3.26: mmt ni tip nhiu thang oHnh 3.26 l s m mt 6 thang
o dng hai ngun cung cp ng vi kho chuyn mch S1 hai v tr a v b. Vi mi
mt ngun cung cp ta li mc ni tip mmt vi cc in tr ph khc nhau.Khi
thay i thang o dng qua c cu ch th ICC khng thay i nhng tr s in tr c
c trn thang o c nhn vi h s nhn ng vi tng thang o tng ng.b) mmt mc
song songm mt mc song song c in tr cn o mc song song vi c cu ch th
ca mmt. Hai u A, B ni vo in tr cn oS mch o nh hnh 3. 27. m mt cng
bao gm mt ngun in p mt chiu mc ni tipviccintrph. intrRMdng iu chnh
thang o ca mmt khi ngun cung cpthayi.V intrcnomcsongsong vi c cu ch
th nn khi Rx = th dng qua c cu l ln nht, cn khi Rx = 0 th hu nh
khng c dng qua c cu. Do thang o ca mmt kiu ny l thang o thun. Ngi
ta cng to ra mmt song song nhiu thang o vi mt ngun cung cp duy nht
mc ni tip vi cc in tr ph khc nhau.3. o in tr dng cu o mt chiuCu o
(bridge circuit) l mt thit b ocin dngointrvi chnh xc cao. Cu o c
phnbit thnh 2 loi chnh da vo ngun cung cp l ngun p hay ngun dngT s
trn ta nhn thy s ph thuc in p ra ca cu vo cc thng sca cuc mt nh sau
: Hnh a:02 14 34 32 14 3 2 13 2 4 1) ( ) ( ) )( (UZ ZZZ ZZ ZZZ ZZ Z
Z ZZ Z Z ZUout outout+ + + + + +Hnh b:04 2 3 14 3 2 13 2 4 1) )( ()
(IZZ Z Z ZZ Z Z ZZ Z Z ZUoutout+ ++ + + +78RMHnh 3.27: mmt mc song
songHnh 3.28: S nguyn l chung ca cu oNgun pNgun dngKhi in tr Zout =
(khng ti), ta s c tng ng nh sau04 3 2 13 2 4 1) )( (UZ Z Z ZZ Z Z
ZUout+ +;04 3 2 13 2 4 1) (IZ Z Z ZZ Z Z ZUout+ + +iu kin cu cn bng
l: Z1Z4 = Z2Z3Cu thng lm vic hai trng thi chnh l cn bng (cu ch
khng) v khng cn bng. Cu mt chiu o in tr thng c hai loi: cu n v cu
kp.a) Cu n Wheaston o in tr trung bnhCu n mt chiu gm 4 nhnh,cn gi l
4 vai.Trn cc vai t cc in tr. Trong mt vai t in tr Rx cn o, cc vai
cn li t cc in tr mu (hnh 3.29). Mt ng cho ni vi ngun in mt chiu U0,
mt ng cho ni vi ch th cn bng. Ch th cn bng l in k G l c cu o t in c
nhy v chnh xc cao. Nguyn l lm vic ca cu l phng php so snh cn bng,
theo .. ta c iu kin cu cn bng l: 432RRR Rx Trong : 43RRc gi l t s
vai, c gi tr l 0.001;0,01;0,1;1;10.Trongqu trnh o, ngi ta s iu chnh
in tr2R cu cn bng, v gi tr ca in tr cn o s c tnh trc tip theo cng
thc.Kt qu o in tr Rx khng ph thuc vo ngun cung cp, m ph thuc vo
chnh xc ca cc in tr vai.Sai s ca php o in tr:242322) ( ) ( ) ( R R
R Rx + + chnh xc ca php o ph thuc vo nhy ca in k G. nhy ca in k cng
cao th s cn bng cu cng ng. b) Cu o in tr nh Khi o in tr nh m dng cu
n th sai s s ln do in tr tip xc v in tr dy ni nh hng n php o. Dng
cu kp s khc phc c sai s .S nguyn lC th loi b nh hng ca in tr r ca
dy ni bng cch mc 3 dy nh hnh 3.30a. Nu c 3 dy ni ccngintr (cng chiu
di), ta c th vit:) ( ) (3 2 4r R R R r Rx+ +V ) (4 2 3 2 4R R r R R
R Rx + 79Hnh 3.29: Cu WheastonHnh 3.30: o in tr nhLoi 3 dyLoi 4 dy:
Cu kp KenvinNu t R2 = R4, s loi b c hon ton nh hng ca in tr dy ni n
php o.Trong trng hp o in tr rt nh ta s phi s dng cu kp Kenvin (hnh
3.30b)Ta c iu kin cu cn bng l: ) ' ' (' '4 3 44 3 4 3432pxR R R RR
R R RrRRR R+ ++ Trong thc t ngi ta b tr cc intr mu vo cc vai sao
cho 4443' ' RRRRKhi 432RRR Rx Cu Kenvin thng dng o cc in tr c tr s
t 0,0001-10.3.2.2 o in tr c tr s lnC th o in tr ln c 105-1010 (v d
nh in tr cch in) bng phng php vn- ampe1. Mgmt o in tr lnNgi ta s
dng t l k t in ch to mgmmt, l mt dng c o in tr ln nh in tr cch
in.Cu to: Mgmmt bao gm my pht in mt chiu quay tay E (magnheto) c
kch t bng nam chm vnh cu. in p pht ra t 100- 5000V tu tng loi
mgmmt. Ch th kt qu o l t l k t in. Trong cun ng a cmc ni tip vi in
tr mu R1 v in tr cn o Rx. Cun ng b c mc ni tip vi in tr R2 . Hai
cun ng c t cho nhau mt gc v gn trn cng mt trc quay. Trn trc khng c
l xo cn. Cun b chuyn ng ngc chiu kim ng h, cn cun a
chuynngtheochiukim ngh. Haicc1, 2ni vi in tr o. (hnh 3.31)Ngoi
ratrongmchcn c vng bo v lai tr dnginrquacuna nh hngnkt qucaphp
o.Dng I1qua cun a: a xR R REI+ +11Dng I2 qua cun b: bR REI+22Trong
Ra, Rb l in tr ca cun a v b.Suy ra gc quay ca c cu
,_
,_
+ ++
,_
X X abRfR R RR RfII 11221 Vy gc quay t l nghch vi in tr Rxcn o
do thang o c ghi theo chiu ngc.80Quay tayThang oMagnhetoVng bo vHnh
3.31 : Mgmt o in tr ln A)S mch oB) Dng c o12- Khi hai cc 1,2 h mch
(Rx= ), s khng c dng in qua cun a I10. Tuy nhin vn c mt dng in I2
chy trong cun b, lm cho kim ch th lch v pha tri ca thang o (v tr
).- Nu ni ngn mch hai cc 1 v 2 (Rx =0), dng qua cun a s rt ln I1
I1max, lm cho kim ch th lch v pha phi ca thang o (v tr 0).2. ng dng
o in tr cch in v ch dy b chm tin tr cch in ca ng dy bao gm in tr
cch in gia cc pha vi nhau v gia cc pha vi t. N c phn b theo chiu di
ca ng dy. Ta c th coi mi n v chiu di ca ng dy l mt in tr cch in. o
in tr cch in ca cc ng dy ti in hoc phn phi in trong cng nghip. in
tr cch in c o gia hai u dy dn in vi nhau hoc tng dy dn in vi dy
trung tnh.a) o in tr cch in ca ng dy khi cha c in p cng tcNu ng dy
khng c in p cng tc th c th dng mgmt o in tr cch in ca ng dy.Gi s ta
c ng dy hai pha. o in tr cch in ca pha A, mt u ca mgmt ni vi
phaA,mtu ni vi t. Khi o in tr cch in ca pha B ta ch cni ni mt
ucamgmt t pha A sang pha B (hnh 3.32). Nh vy vi cch o ny th ta khng
o c in tr cch in ca pha A vi t RAmintroclRA//(RAB+ RB), do tr s in
tr o c lun nh hn RA. V vy nu kt qu o t yu cu th cch in thc t ca pha
A cng t. Tng t in tr o c khi o in tr cch in gia pha B v t l RB //
(RAB + RA). Cn in tr cch in gia hai dy A, B o c l RAB// (RA + RB).o
in tr cch in ca ng dy 3 pha cng tng t nh vy. Nu kt qu t yu cu th
cch in thc cng t yu cub) o in tr cch in ca ng dy khi c in p cng
tc.C th dng hai hoc ba vnmt kim tra cch in ca ng dy hai pha hoc ba
pha nh hnh 3. 33.Khi cch in tt, mi vnmt s cho kt qu in p ngun cung
cp dy dn. Bt k s gim in tr cch in no ca 1 trong 2 dy dn cng s lm
gim s ch ca vnmt ni gia pha vi t, c th ch ti 0. Cn s ch ca vnmt cn
li s tng ln.Vi mch in 3 pha, in tr cch in ca 3 pha c ch bi 3 vnmt.
Nu nh intrcchinca pha nogim 81Hnh 3.32: o in tr cch in dng mgmt khi
cha c in p cng tcRBRARABHnh 3.33: o in tr cch in bng vnmtkhi c in p
cng tcVVRARBxung, th s ch ca vnmt mc vi pha gim cn vnmt mc vi hai
pha cn li s c s gia tng tr s.3.2.3. o in dung v in cm1 oin dung v
in cm dng vnmt v ampemta) o in dung Mch o c mc nh hnh 3.34.Tng tr
ca in dung c xc nh bi cng thcXCxC IUZ1 Suy ra UICx Ngun tn hiu cung
cp cho mch ol ngun tn hiu xoay chiu hnh sinb) o in cmMch o in cm L
c mc nh hnh 3.35Tng tr ca in dung c xc nh bi cng thc2 2X X LxL RIUZ
+ Suy ra 2 21X xR Z L 2 o in dung v in cm dng cu o1. Cu o xoay
chiuCu o xoay chiu ging nh cu o mt chiu nhng c nhng c im khc nhau
nh sau (hnh 3.36)- Cu lm vic vi in p xoay chiu hnh sin- Cu c t nht
hai vai l tng tr mang tnh in cm hoc in dung.- C cu ch th dng xc nh
cu cn bng c th l my hin sng hoc tai nghe.- Tng tr trn cc vai ca
cu:Z1 = r1 + jx1 = z11 jeZ2 = r2 + jx2 = z22 jeZ3 = r3 + jx3 = z33
jeZ4 = r4 + jx4 = z44 jePhng php o dng cho cu xoay chiu l phng php
cn bng.iukincnbngcu: ngkhoK, iu chnh tng tr cc vai cu cn bng:Z1Z4 =
Z2Z382CXAVUHnh 3-34: S o in dung bng Vnmt v ampemtAVUHnh 3-35: S o
in cm bng Vnmt v ampemtLXHnh 3.36:S nguyn lca cu xoay chiuNh vy cu
cn bng cn c hai iu kin:* z1z4 = z2z3.; tch cc mul ca hai vai i nhau
phi bng nhau*1 + 4 = 2 + 3; tng cc acgument ca hai vai i nhau phi
bng nhauNu c 4 vai u l nhng tng tr th vic ch to v cn bng cu s rt
phc tp. V vy ngi ta ch to hai vai cu l thun tr, hai vai cn li chn
theo nguyn tc sau:- Nu hai vai k nhau l thun tr, th hai vai cn li
cng tnh cht (cng mang tnh in cm hoc in dung).- Nu hai vai i nhau l
thun tr, th hai vai cn li phi khc tnh cht (mt vai mang tnh in cm cn
vai kia mang tnh in dung).Thit b ch th s cn bng ca cu o xoay chiu:
Tai nghe(headphone): githnhr, tng inhy, cdngphbinckhnngphn bit
cscnbngcumt cchtngi chnh xc. Tuy nhin cn ph thuc vo thnh tai ca ngi
lm th nghim.2. Cu o in dung v gc tn hao ca t ina) T in thc v s thay
th ca chngT in thc khi c u vo mch in bao gi cng gy nn mt tn hao nng
lng. Tn hao l nng lng tiu tn trn in tr cch in ca t. Do mch tng ng
ca mt t in thc l mt t in l tng c in dung bng in dung ca t thc mc mc
ni tip hoc song song vi mt in tr R xp x bng in tr cch in ca t (hnh
3.37)Hnh 3.37a m t t in thc C nh mt mch gm t in thun dung mc ni tip
vi mt in tr R (s ny dng khi tn hao trong tin nh tc l khi in tr cch
in ca t in ln). Cng sut tn hao trong t l:P = UIcos=UIsinDo nh nn c
th coi sin= tgtgc trng cho tn hao ca t in, C RXRIItgC CXRHnh 3.37b
m t t in thc C nh mt mch gm t in thun dung mc song song vi mt in tr
R (s ny dng khi tn hao trong t in ln tc l khi in tr cch in ca t in
nh).Ta c C R UUtg1X /R /C 83Thit b ch th cn bng ca cu o xoay chiu
tai ngheb) Cu o in dung v gc tn hao ca t in* CuRC mc ni tip:o t in
c tn hao nh.S cu to ca cu nh hnh 3. 38. Cu bao gm 4 nhnh, trong
nhnh R2 v R3l thun tr, Cx v RX l gi tr in dung v in tr ca t cn o.
C1 v R1 l gi tr in dung v in tr ca t mu.Cu cn bng ta c: Z1Z3 =
Z2Zx2 311)1( )1( RC jR RC jRxx + +Cn bng phn thc v phn o hai vca
phng trnh: R1R3 = R2Rx; xC jRC jR 213Suy ra 231RRR Rx ; 321RRC Cx ;
tg= RxCx = R1C1Tng t, khi t c tn hao ln ta coi t l mt mch gm t l
tng cng in dung v in tr cch in mc song song (hnh 3.39 )th kt qu o s
l:42RRR RM x ; 24RRC CM x ; M M x xC R C Rtg 1 1 84CRI CUICRUCUIa)
Tn hao tb) Tn hao nhiuHnh 3.37: S thay th ca t in thcv th vc tHnh
3.39:S nguyn l ca RC mc song songHnh 3.38: CuRC mc ni tipNgoi ra cn
c th s dng cc cu o in dung sau:3. Cu o in cma) Cu toS nguyn l ca cu
c m t nh hnh 3.24.Cu gm 4 nhnh: Nhnh 1 l cun dy cn o c in cm Lx v
in tr RX. Nhnh 3 mc cun dymucincmmuLMvintrRM. Nhnh 2v nhnh4 l thun
tr gm cc in tr R2, R4. in tr R trong mch c th mc ni tip vi cun dy
cn o hoc cun dy mu, tu theo v tr ca kho chuyn i CD ng sang v tr 1
hay 2. Mc ch mc R vo cu o nhm mc ch d dng cn bng cu. Khi Rx nh, kho
chuyn i c ng sang 2, in tr R mc ni tip vi cun dy cn o. Khi Rx ln,
kho chuyn i c ng sang 1, in tr R mc ni tip vi cun dy mu. b) Nguyn l
lm vicXt trng hp kho chuyn i CDng sang 1, R mc ni tip vi cun
dymu.Khi cu cn bng ta c:2 4) ( ) ( R L j R R R L j RM M x x + + +2
2 2 4 4R L j R R RR R L j R RM M x x + + +Suy ra : 422) (RRR R RM
x+ 42RRL LM x 85Hnh 3.42S nguyn l cacu o in cmCDR2R4KLx, RxLM,
RM12Hnh 3.40: Cu WienHnh 3.41: Cu scheringXt trng hp kho chuyn i CD
ng sang 2, R mc ni tip vi cun dy cn o.Khi cu cn bng ta c:2 4) ( ) (
R L j R R L j R RM M x x + + +2 2 4 4 4R L j R R L j R R RRM M x x
+ + +Suy ra : 422) (RRR R RM x+ 42RRL LM x c) Cch s dng cuKhi o ta
ng kho K, t kho chuyni CD v tr ph hp. iu chnh cu theophng thc: lun
phin iu chnh RM v t s R2/R4.Ch : - trnh hin tng h cm gia cc cun dy
o v mu, ta phi t chng xa nhau v trc cc cun dy ny phi vung gc
vinhau.- Vic ch to cun cm mu kh khn hn t inmu. Vycthdngtinmuthay
cho cun cmmu trong cu o incm, nhng t mu phi t vai i din (hnh
3.43).$3.3 O CNG SUT V IN NNG3.3.1. o cng sut trongmch mt pha1. o
cng sut trong mch mt chiu o cng sut trong mch in mt chiu, ta c th s
dng vn mt v ampemt, vi s mch o c m t nh hnh 3.44Cng sut tiu th ca
ti c xc nh l tch ca in p UL ri trn ti vi dng in IL qua ti:
P=ULIL86Hnh 3.43 : Cu o in cm dng t muHnh 3.44: o cng sut trong mch
in mt chiu dng vn mt v ampemt Trong :I: dng in m ampemt o cU: in p
m vn mt o cIV: dng qua vn mtUA: in p ri trn ampemtRL: in tr tiRA,
RV: ln lt l in tr trong ca ampe mt v vn mt.Ty theo cch mc sai s ca
php o s khc nhau:Hnh 3.44a: VL VL LRR RUI I U P Hnh 3.44b: LA LL
LRR RUI I U P 2. o cng sut trong mch xoay chiu mt pha Cng sut tc
dng trong mch xoay chiu mt phac xc nh nh l gi tr trung bnh ca cng
sut trong mt chu k T: T TuidtTpdtTP0 01 1Trong : p, u, i l cc gi tr
tc thi ca cng sut, p v dng. Trong trng hp khi dng v p c dng hnh sin
th cng sut tc dng c tnh l : P = U.I.cos h s cos c gi l h s cng sut.
Cng sut phn khng c tnh theo : Q = U.I.sin Cn i lng S = U.I gi l cng
sut ton phn c coi l cng sut tc dng Di y s gii thiu mt phng php o
cng sut trong mch xoay chiu mt pha tn s thp v trung bnha) o bng 3
vn mtS mch o c m t nh hnh 3.Khi cng sut tiu th ca ti s c xc nh theo
biu thc:RU U UPBC AB ACL22 2 2 b) Wattmet in ng: s c mc nh hnh
3.46a.Cu to: Wattmet tc dng mt phn t c lm t c cu o in ng. Cun tnh
ca c cu c mc ni tip vi ph ti nn c gi l cun dng ca wattmet.Cun ng ca
c cu c mc ni tip vi mt in tr ph v c u song songvi ph ti nn cn c gi
l cun in p.87UI2IHnh 3.46: Wattmet tc dng 1 phn tCun dngCun dngCun
pa) S mch o b) th vctHnh 3.45: o cng sut bng 3 vn mtTiNguyn l lm
vic: - Khi o trong mch mt chiu: Khi c dng in I1 chy trong cun tnh v
dng I2 chy trong cun ng, th gc quay ca c cu o c xc nh theo biu thc:
P KR RUKI I IddMDPf B+ 2 1121Trong :1121 ddMDK v f BPR RKK+Vi RB l
in tr ca cun ng.-Khi o trong mch xoay chiu: Gi s ti mang tnh in cm,
dng in chm pha sau in p mt gc . Dng in trong cun p chm pha sau in p
mt gc . Ta c th vc t nhhnh 3.46b.Gc lch pha gia hai dng cun tnh v
ng l: = - Gc quay ca c cu cos12 112I IddMDNu chn Rf ln hn rt nhiu
so vi cm khng ca cun ng th = 0, khi = . Ta c :P K UI K IIddMDP P
cos cos1212Ch :- Trn wattmet bao gi cng c cc k hiu * nh du cc u cun
dy tnh v ng gi l u pht. Khi mc mch phi ni cc u c k hiu * vi nhau,
nu u sai wattmet s quay ngc- xc nh c cng sut m wattmet ch,phi tnh c
hng s wattmetmdm dmPI UCTrong : Um, Im l in p v dng in ln nht ng vi
thang o ny88m l s vch ln nht trn thang o.Cng sut trn wattmet l: P =
CP.Trongtrng hptic in pcaov dng in ln, phi phi hp bin pv bindng vi
wattmet o cng sut cho ti (hnh 3.47). Cun in p cawattmetmchaiuthcp
ca bin p, mt u ca cun th cpv v ca bin p c ni vi t. Nh vy cng sut o
c bi wattmet lP = UI cosNhnvi tscabinpvbin dng ta c cng sut ca
ti:PT = KUKIUI cosc) Wattmt nhit inWattmet nhin in c cu to nh hnh
3.48.Hiu in th ng sinh ra cc u t do (u lnh) ca cc chuyn i c o bi mt
milivnmet t in. in p ny t l vi cng sut trung bnh tiu th trn mt ph
ti. d) o cng sut dng chuyn i HallChuyn i Hall l mt mng bn ca c ch
to di dng mt tm mng bng bn dn, c cu to nh hnh 3.49, gm hai cc dng v
hai cc p. Hai cc dng ca chuyn i c mc vo ngun in mt chiu hoc xoay
chiu. Khit vung gc vi b mt chuyn i mt t trng th xut hin hai ucc p
mt th in ng gi l th in ng HallVH c tnh nh sau: VH = kH. B.i(t)Vi
kH:l h s m gi tr ca n ph thuc vo vt liu, kch thc v hnh dng ca chuyn
i, ngoi ra cn ph thuc vo nhit ca mi trng xung quanh v gi tr ca t
trng. B: l t cm ca t trng. i(t) : dng in qua chuyn i89 BUBIHnh
3.47: Mc Wattmet kt hp vi BU v BIHnh 3.48: Wattmet nhin inTrong s
3.49a , cng sut tc dng P s c xc nh bng cch o th in ng VH.Thc vy, vi
vx=ai(t) v ix=bi(t), ta c cng sut tc dng P s c xc nh theo :H HT Tx
xV abk dt t B t iTab dt i uTP 0 0) ( ) (1 1Trong : T l chu k o,
V(H) l gi tr trung bnh ca vH(t)Nh vy th in ng Hall s t l vi cng sut
ti.Thc hin mt watmet bng chuyn i Hall bng cch t chuyn i vo khe h ca
mt nam chm in. Dng in i qua cun ht L ca n chnh l dng in i qua ph ti
L. Cn hai cc dng c dng in t l vi in p t ln ph ti. in tr ph RV hn ch
dng. Th in ng Hall lc s c tnh: VH = k.ui=k.PVi k l h s t l.c)
Wattmet nhnWatt met s dng b nhn cs khi nh hnh 3.50.Tn hiu in p u(t)
v dngin i(t) c a vo b nhn, c p(t)=ku(t)i(t).Tn hiu cng sut tc thic
ly trung bnh chnh l gitr cng sut tc dng cn o. Kt qu s c hin th trn
b ch th s. d) o cng sut bng phng php iu ch tn hiu : Phng php iu ch
tn hiu da trn vic nhn cc tn hiu uu (t l vi in p trn ti cn o) v ui
(t l vi dng in trn ti cn o) trn c s iu ch hai ln tn hiu xung. Cc tn
hiu tng t uu v ui c bin i thnh tn s, chu k, bin , rng ca tn hiu
xung sau ly tchphn.90B biniHnh 3.50:Wattmet nhn ch th sHnh 3.49:
Wattmet bng chuyn i HallXtwattmet da trn phng php iu ch rng xung vi
iu ch bin xung (RXBX): c s cu trc nh hnh 3.51 v th thi gian nh hnh
3.52Dng in ix qua b bin i I/U thnh in p vy t l vi i. Tn hiu in p ny
c a vo b so snh sosnh vi in p vg c pht ra t my pht tn s chun. u ra
ca iu ch RX c cc xung vi rng t2 = k.vy, tn hiu ny s c t
vobiuchbinxungBXvc iuchbinbngtnhiuvx(t). Tnhiu xung vung vm u ra
caiu ch BX s c gi tr Vm= +Vx trong khong thi gian t2 v Vm= -Vx
trong khong thi gian t1.V tn hiu vmc a qua b tch phn to thnh in p
ra c gi tr :tm outdt vRCV01Nh vy in p ra ca b tch phn (TP) s c gi
tr t l vi cng sut trung bnh P.Sai s ca cc atmet s dng cc cp iu ch l
ch di ca chu k iu ch b hn ch, dn n di tn b hn ch. b) o cng sut phn
khng mt phaNgi ta s dng Wattmet phn khng mt phn t o cng sut phn
khng mt pha.Cu to:Wattmet phn khng mt phn t cng c lm t c cu o in
ng. Cun tnh ca c cu c mc ni tip vi ph ti nh hnh 3.53 . 91MFXI/Uiu
ch RXiu ch BXHnh 3.51: S cu trc ca watt mt iu ch xungHnh 3.52: th
in pCun ng ca c cu c mc song song vi mt in tr ph v ni tip vi mt cun
cm L.Nguyn l lm vic: iu chnh tr s ca in tr Rp cho dng in cun ng I2
v in p U lch pha nhau 900.Gc quay ca c cu o in ng= KI1I2 cos(900 -
)= KII2 sin= KQ UI sin=KQ Q92UI2900-Ia) th vctHnh 3.53: Wattmet phn
khng mt phn tA2
A1
B
XL
I2
I1= I a) S mch o3.3.2. o cng sut tc dng trong mch 3 pha1. o bng
1 wattmet mt phn tNu mch 3 pha 3 dy ph ti i xng c th dng 1 wattmet
mt phn t o cng sut tc dng mt pha, sau nhn 3 kt qu o c ta c cng sut
tc dng ton mch.P3fa = 3P1faKhi mch 3 pha c ph ti hnh sao i xng ta c
im trung tnh ca ph ti mc wattmet (hnh 3.54a).Khi mch 3 pha c ph ti
hnh tam gic i xng cn phi to im trung tnh gi bng cch ni vi hai pha
khc hai in tr ph RB, RCc gi tr bng in tr rUca cun in p ca wattmet
(hnh 3.54b).2. o bng 2 wattmet mt phn tTrong trng hp mch 3 pha 3 dy
ph ti by k (i xng hay khng i xng, tam gic hay hnh sao) ta c th dng
hai wattmet mt phn t o cng sut tc dng ca mch. S u nh hnh 3.55 , ly
pha C lm pha chung, ta cng c th mc theo cc cch khc nh ly pha B hoc
pha C lm pha chungGi p1, p2 ln lt l cng sut tc thi ca wattmet th
nht v th hai. Ta c:p1=iAuAC =iA(uA-uC); p2=iBuBC =iB(uB-uC)Suy ra
p1+ p2 =iAuA+ iBuB -uC(iA+iB)93a) Ph ti ni saob) Ph ti tam gicHnh
3.54: o cng sut tc dng mch 3 pha ti i xngTiABCHnh 3.55: o cng sut
tc dng 3 pha dng 2 wattmetTheo nh lut Kichp 1 ta c:iA+ iB + iC = 0;
hay iC = -(iA+iB);Suy ra p1+ p2 =iAuA+ iBuB +iCuC = pA+ pB + pC
=p3faVy cng sut tc dng ton mch l:P3fa =+ +TC B Adt p p pT0) (1 =
+Tdt p pT02 1) (1 = P1 + P2Trong trng hp ti i xng, ta c: P1 = UdId
cos(300 - )P2 = UdId cos(300 + )Khi ta cng d dng xc nh c cng sut
phn khng 3 pha v gc pha ca ti2 12 13P PP Ptg+ ; Q= 3 (P1-P2)Ch:
Trncsphngphp o dng hai wattmet 1 phn t, ngi ta ch to ra wattmet 3
pha 2 phn t dng o cng sut ti 3 pha 3 dy. Cutocawattmet gmc: 2cun in
p c cng trc quay v hai cun dng in.Hnh3.56: Trnhbysmc Wattmet hai
phn t qua BU v BI3. o bng 3 wattmet mt phn tTrong trng hp mch 3 pha
4 dy khng i xng ngi ta dng 3 wattmet mt phn t o cng sut tc dng, kt
qu php o s l tng s ch ca 3 dng c o (hnh 3.57).P3fa =P1+ P2+P3Trong
thc t ngi ta s dng loi wattmet 3 phn t o cng sut tc dng mch 3 pha 4
dy. Thc cht ca loi wattmet ny l t hp ca 3 wattmet mt pha trong mt
v.94BUBIHnh 3.56:Wattmet 3 pha 2 phn t Hnh 3.57 : o cng sut mch 3
pha 4 dy bng 3 wattmet 1 phn tMch 3 pha 4 dyMch 3 pha 3 dyHnh 3.58
trnh by cch mc wattmet 3 pha 3 phn t o cng sut tc dng trong mch 3
pha 4 dy c s dng kt hp bin p v bin dng.Wattmet 3 pha thng c dng
trongcngnghip, ccutogm c: hai cunpc cngmt trc quayvbacundngin. Khi
cng sut ca ti c xc nh:P = PW.KI.KUVi KIl t s bin ca BI; KUl t s bin
ca BU.3.3.3. o cng sut phn khng mch 3 pha1. Mch 3 pha i xng: o bng
1 wattmet mt phn tTrong trng hp mch 3 pha ph ti i xng c th ch s dng
1 wattmet mt phn t cng sut phn khng ca mch (hnh 3.59).Cng sut trn
wattmet l: P = UBCIA cos(900 - )P=UdId sin = 33 faQ2. Mch 3 pha
khng i xnga) o bng 3 wattmet 1 phn tKhi mch 3 pha khng i xng ngi ta
dng 3 wattmt o cng sut phn khng ca mch. S c mc nh hnh 3. 60Gi P1,
P2, P3 ln lt l cng sut trn wattmet th nht, th hai, th ba.Ta c: P1 =
UBCIA cos(900 - A) =3UAIA sin A=3QA95BUBIHnh 3.58: Wattmet 3 pha 2
phn t ABCTI**W900 - UAUBCUCIAUBHnh 3.59: o cng sut phn khng mch 3
pha i xnga) S mch o b) th vctTng t ta c P2 = UCAIB cos(900 - B)
=3QBP3 = UABIC cos(900 - C) =3QCVyP1 + P2 +P3 =3 (QA + QB + QC) =
Q3fab) o bng wattmet in ng 2 phn t o cng sut phn khng mch 3 pha
khng i xng ngi ta cn dng wattmetin ng 2 phn t. Hnh 3.61 l s mc
wattmet 2 phn t o cng sut phn khng trong mch 3 pha. Cun dy dng in
ca c hai phn t c hai phnoncsvngbngnhauvc tch ri ra. Mt phn on ca cc
phn t ny c u vo pha A v C. Hai phn on cnli cuni tipvi nhauvu chung
vo pha B nhng ngc cc.Cun in p ca phn t th nht c u vo gia hai pha B
v C. Cun in p ca phn t th hai c u vo gia hai pha A v B.Mmen quay tc
ng ln phn ng do phn t th nht gy ra:]) + ]) + (30 0B B AB A
sin21cos23sin [I KUcos ) - cos(90 [I KU MA d0A BC q1Mmen quay tc ng
ln phn ng do phn t th hai gy ra:]) + + ]) + 0 ( 0B B CB C
sin21cos23sin [I KU15 cos ) - cos(90 [I KU MC d0C AB q2Mmen tng tc
ng ln phn ng s lMq = Mq1 + Mq2 = KUd [IA sinA + IB sinB +IC sinC] =
K( 3QA + QB +QC) = K Q3pNh vy mmen quay ca phn ng t l thun vi cng
sut phn khng 3 pha.Hnh 3.62 trnh by s o cng sut phn khng 3 pha dng
wattmet phn khng 3 pha96ABCTIHnh 3.61: oo cng sut phn khng mch 3
pha bng wattmet 2 phn tHnh 3.60 : o cng sut phn khng mch 3 pha bng
3 wattmetS mch o th vc t(Varmeter) kt hp vi bin p v bin dng.3.3.4.
o in nng1. o in nng trong mchxoay chiu mt pha o in nng ca ti mt
chiu, phng php n gin nht l o p v dng qua ti. in nng tiu th ca ti
trong thi gian t s c xc nh theo biu thc: E= V.I.ta) Cng t cm ng mt
phaCng t mt pha c c cu hot ng trn nguyntc c cu o cm ng, c cu to nh
hnh, gm c nam chm in 1 mc song song vi ti(cun in p) c s vng dylntit
dindynhchuc in p cao. Nam chm in 2 ca c cu c cun dy c s vng nh, tit
din dy to chuc dnginln vc mc ni tip vi ti nn c gi l cun dngin.
anhm3cmpanm trongkhehcamchtcunpv cun dng. cho a nhm quay u c
namchmvnhcu5, vbchcc bnh rng n khp vi trc quay 4 ca a nhm. Gn trn
trc a nhm c hp s c kh v b m nng lng 6 (thc cht lbmsvngquaycaanhm).
Cngtcbckntrongmt lpv bng thu tinh hoc cht do chng li nh hng ca bn
ngoi tc ng (hnh 3.63)Khi cng t hot ng, dng in I i qua ti to nn t
thng ic xc nh bng: i = KIIin p U cung cp cho ti to nn dng IU trong
cun dy in p v t thng L, u.Trong thnh phn t thng Lkhng qua a nhm m
ni tt theo sun t v 97BUBIHnh 3.62:Wattmt phn khng kt hp vi BU v
BIHnh 3.63: S nguyn l ca cng t cm ng 1 phathnh phn t thng u xuyn
qua a nhm.T thng ca nam chm in p i xuyn qua a nhm bng:u = KuIU = U
KZUKUUU' T thng u v i xuyn qua a nhm to ra dng in xoy trna nhm. Do
c tc dng tng tc gia dng in xoy v t thng to nn mmen ngu lc quay a
nhm: Mq = Cf u i sin= KUIsinNu nh li ca cun dng khng b bo ho,th =
/2 - Khi Mq =KUIcos= KP Theo gin vc t(hnh 3.64), ta c: = - - Nu- =
2 th= 2- Suy ra: Mq =KUIcos= KP Nh vy mmen quay t l vi cng sut.C th
iu chnh - = 2bng cch thay i hoc . iu chnh , tc l iuchnh t thng Ubng
cch thay i v tr mn chn t ca nam chm in p. Hoc thay i , tc l iu chnh
I bng cch thay i tr s ca vng in tr ca cun dy ngn mch qun trn nam
chm dng in.Mmenquaylmchoanhmquay. a nhm quay trong t trng ca nam
chmvnh cu6 bt trng ny hm li vi mmen hm: dtdD MCTrong : D l hng s t
l;dtdl vn tc gc ca a nhmKhi mmencnbngmmenquay th a nhm quay u:dtdD
KPSuy ra d= KW PdtGissaumt thi giant =t2t1 cng t quay c =n2 n1 vng
th21ttkPdt = n2 n1 =KWWTrong WKw[vng/kWh] c gi l hng s cng t.Hnh
3-65 l s mc cng t o nng lng tc dng trong mch mt pha. Hai cc ngun ca
cun in p v dng in c nh ch to ni sn vi nhau v ni vi98UIuLiHnh 3.64:
Gin vctTiCun pCundngHnh 3.65: Cch u cng t mt pha dy pha.Sai s v cch
khc phcDo tn ti ca ma st, do nh hng ca t thng ph, do sai lch hng s
ca cng t (mmen cn ln hoc nh) do cng t sai s t nhiu.Trc khi s dng bt
buc phi hiu chnh li tc l tin cch khc phc sai s.a) B ma st- Khi ph
ti nh, mmen ma st s ng 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 chia t thng hoc vng ngn mch khng i xng (cha th
hin trn hnh v).- Khi iu chnh v tr vng ngn mch khng i xng hoc vt
chia t thng ta s b c ma st (tuy nhin nu iu chnh qu sang tri hoc
sang phi th cng t s t quay thun hoc quay ngc khi khng c ti).b) Chng
hin tng t quay 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 ca cun p v
trn trc quay ngi ta gn hai l thp non T1 v T2. Khi a nhm quay ti thi
im hai l thp i din nhau th chng s tc ng tng h v to ra mmen hm (tuy
nhin ch vi mmen kh nh).c) iu chnh gc lch pha gia v ITa c: - = +
Mong mun rng- = + = 2Vy phi iu chnh- = 2coi nh khng i i vi mi loi
cng t sau khi ch to. V vy ta phi iu chnh gc ai bng cch trn mch t ca
cun dng ngi ta cun vi vng dy ni qua mt in tr R c th iu chnh c. Khi
iu chnh gi tr R s lm thay i tn hao t trong mch t cun dng, tc l i
thay i.d) Kim tra hng s ca cng tTa iu chnh sao cho cos= 1, cho dng
in I = In, U = Un lc ta c P = UnIn; o thi gian quay ca cng t bng ng
h bm giy, m s vng quay N ca cng t quay trong khong thi gian t. Ta
tnh c hng s ca cng t nh sau:t PNt I UNCn n n Ta so snh Cp vi gi tr
nh mc ghi trn cng t, nu khc nhau ta phi iu chnh v tr ca nam chm vnh
cu tng hay gim mmen cn cho n khi Cp bng gi t nh mc ca cng t. Thc t
hin nay, vic hiu chnh cng t thng da vo cng t mu.b) Cng t in t ch to
cng t in t, ngi ta bin i dng in I thnh in p U1 t l vi n: U1 = k1I
99Hnh 3.68: Cch mc cng t 3 pha 3 phn tBUBICngtmt in p khc t l vi in
p t vo U: U2 = k2U qua b phn in t (nhn analog) s nhn c in p U3t l
vi cng sut P: U3 = k3.P Tip theo in p ny s ln lt qua cc khu: qua b
bin i inp-tns(hocbbini A/D), vobm, rach th s. S ch ca c cu ch th s
s t l vi nnglngN=CWtrong khong thi gian cn o nng lng . 2) o in nng
ca ti ba phaTng t nh o cng sut, o nng lng trong mch 3 pha cng s dng
phng php 1 cng t, 2 cng t hay 3 cng t mt pha.a)obng1cngtkhiphtii
xng: nng lng tng bng 3 ln nng lng mt pha.b) o bng 2 cng t khi mch 3
pha 3 dy phti bt k: nnglngtngbngtng nng lng ca hai cng t.c) o bng 3
cng t khi mch 3 pha 4 dy phti bt k: nnglngtngbngtng nng lng ca ba
cng t.Tuy nhin trong thc t ngi ta s dng cng t 3 pha. Cng t 3 pha c
hai loi: loi 2 phn t hoc loi 3 phn t. l hai hoc ba cng t mt phn t c
t hp li trong mt v. - Cng t 3 pha 2 phn t: Cch mc mch ging nh
wattmet ba pha 2 phn t. Hnh 3.67: l s mc cng t vo mch 3 pha 3 dy.
Phn tnh c hai nam chm in p v hai nam chm dng in u vo ba pha. Phn ng
gm 2 a nhm gn trn cng mt trcquay. Mmenquayctorabngtngcahaimmen quay
do hai phnttora, vnnglngo cchnhlnnglngtngca mch 3 pha.- Cng t 3 pha
3 phn t: Hnh 3.68 l cch mc cng to nng lngtrong mch3pha 4 dy.
Cbaphntlmquay3anhm cng mt trc quay. Do trc quay s 100Hnh 3.66: Cng
t in tU UUHnh 3.67: Cch mc cng t 3 pha 2 phn tTiCng tquay theo mmen
tng hp ca ba phn t, vng quay can s biu th nng lng tc dng ton mch.V.
o cng sut, nng lng trong mch cao p. o cng sut v nng lng trong mch
cao p ta phi s dng bin p v bin dng o lng. Khi mc dng c o trong h
thng in c in p cao v dng ln qua bin p Tu v bin dng TI o lng.
101TuTIPh tiHnh 3.69: Mc watt mt trong mch cao p$3.4 O TN S GC PHA
V KHONG THI GIAN3.4.1. o tn sTn s: l mt trong cc thng s quan trng
nht ca qu trnh dao ng c chu k. Tn s c xc nh bi s cc chu k lp li ca
s thay i tn hiu trong mt n v thi gian.Chu k: L khong thi gian nh
nht m gi tr ca tn hiu lp li ln ca n.o tn s ca tn hiu in ta c hai
phng php l phng php bin i thng v phng php so snho tn s bng phng php
so snh c thc hin nh osiloscope, tn s k cng hng v.v..o tn s bng phng
php bin i thng bao gm: tn s k c in tng t (tn s k in t, in ng, st in
ng); tn s k in dung tng t, tn s k ch th s.1.Tn s k in nga) Cu to: y
l thit b o dng c cu ch th t l k t in. S o c mc nh hnh . Cun tnh A c
mc ni tip vi cun ng B1 v mc ni tip vi cc phn t R1, L1, C1. Cun ng
B2 c mc ni tip vi t C2. Hai nhnh ny c mc song song vi nhau v u vo
ngun in p cn xc nh tn s (hnh 3.70).b) Nguyn l lm vic Cc thng s R1,
L1, C1 c iu chnh cng hng tn s gia ca thang o, tc l:1 01 01CLxx -
Khi tn s cn o fx < fxoc hai nhnh mang tnh in dung, hai dng i1v
i2u vt pha trc in p U.Gi s i1 vt pha trc u mt gc 1, i2 vt pha trc i
mt gc 2, 1=0, ta c:
,_
,_
121021 12 2sin ) 90 cos()coscos(IIFIIFIIF 102I1 = II2C2R1L1 C1
UfxHnh 3.70: Tn s k in ng121 =0UfxI1 = II2b) th vcta) S nguyn l
Nhng 11sinzx ; z2=x2 v 2112zzII nn
,_
,_
,_
21112112sinzxFzxzzFIIFThay vo biu thc trn cc gi tr 11 1212C fL f
xxx v z2 = 2221C fxx.Ta c:) (1 4212122211 12 2211xxxxxf F CCC L fFC
fC fL fF
,_
,_
Gc quay ca t l k in ng l mt hm ca tn s nn n c th s dng lm tn s
k.2. Tn s k ch th sHnh 3. 71 l s khi ca tn s k ch th s hay l hecmt
ch th s.Nguyn l chung ca tn s ch th s l m s xung N tng ng vi s chu
k ca tn s cn o fx trong mt khong thi gianTn hiu cn o trc khi vo my
m phi i qua b vo. B vo bao gm mt b khuch i hay suy gim tn hiu ho hp
tn s k vi ngun c tn hiu cn o. Trig Smit (mch to xung) c nhim v bin
tn hiu hnh sin hoc tn hiu xung c chu k thnh mt dy xung c bin khng i
nhng tn s bng tn s ca tn hiu vo.My pht tn s chun c n nh bng thch
anh c tn s. Tn hiu c tn s chun f0 c a qua b chia tn s theo cc nc vi
h s chia l 10n ( n = 1, 2, 3, ..., 8), ta nhn c cc khong thi gian
To=10-6; 10-5; 10-4; 10-3; 10-2; 10-1; 1; 10; 100s. Flip Flop (FF)
ng chc nng iu khin ng m kho K trong khong thi gian To. Kt qu trn li
ra s thu c mt chm xung c tn s bng tn s ca chui xung do b to sng
(trig) to ra, nhng s lng xung ch xut hin trong khong thi gian To.
Mt b m c dng m s xung trn.S xung m b m m c s l:103Hnh 3.71: Tn s k
hin s00ffKTKTTTNxx xdo Nu thi gian o c gi tr l 1s th s xung N chnh
l tn s cn o fx: fx = N Khi o tn s cao sai s ch yu ca php o l do
khng n nh ca my pht tn s chun. Cn khi o tn s thp th sai s c bn l
sai s lng t theo thi gian, sai s ny s tng khi tn s cn o gim. gim
sai s khi o tn s thpth phi tng thi gian To, nhng iu ny khng phi lc
no cng thc hin c. V vy trong tn s ch th s ngi ta phi s dng b nhn tn
s nhn tn s o ln 10n ln hoc l khng o tn s na m chuyn sang o sang o
thi gian mt chu k ca tn hiu o. Dng c o s dng k thut m nghch o c mch
to xung nhp l sng xung vung c tn s bng tn s ca tn hiu cn o.Hnh 3.73
l s khi ca tn s k khi o tn s thp.Tn hiu vo c tn s cn o fxqua b vo v
trig s to ra tn hiu Tx chnh l chu kca tn hiu c tn s cn o. Qua Flip
Flop, tn hiu Tx vo m kho K, nh vt thi gian Tx chnh bng thi gian To.
ng thi khi kho K m, tn hiu f0 t my pht tn s chun i vo b m v qua c
cu ch th.104TxT0UB mHnh 3.72: th thi gianHnh 3.73: Tn s k o tn s
thpS khiGin thi gianS xung m c l: xxffTTN00 Suy raNffx0Cc tn s k
ngoi vic o tn s n c th o chu k o t s gia hai tn s,khong thi gian v
di cc xung v.v..V d: Tn s k ng dng o khong thi gian3.4.2 o gc pha1.
Phazmt in ngTa c th s dng lg mt in ng ch to ra mt dng c o h s cng
sut rong mch mt pha gi l cos -mt mt pha.Cu to: Cun ng B1 mc ni tip
vi cun cm c cm khng XL, cun ng B2 mc ni tip vi in tr R. Hai cun ng
c u song song vi nhau v u vo ph ti. Cun tnh A mc ni tip vi ti (hnh
3.75a). Chn R, XL c tr s ln(b qua tng tr cun dy ng). Dng i2 s trng
pha vi in p u, dng i1 s chm pha 900 so vi u. Gi s ti mang tnh in
cm, i s chm pha sau u mt gc , ta s c th vct nh hnh 3.75b105B ma)
b)Hnh 3.74:Tn s k o khong thi giana) S khi b) Gin thi gianI2XL
Hnh 3.75 : cosmt mt pha I1 B2IA1 A2B1RZt= 21UI2 I1Ia) S nguyn
lb) th vc tNguyn l lm vic:Gc quay phn ng ca t l k in ng l:)coscos(1
12 2IIF Nu ta chn R v XL dng in trong hai cun dy ng c tr hiu dng
bng nhau (I1 =I2 ) th:)coscos(12 F Theo th vc t ta c: = 2; 1 = 900
- ) (cot) 90 cos(cos)coscos(012 g F F F
,_
Nu =F(cotg ) th cng l mt hm ca cos . Bng khc c khc theo n v ca
gc hay h s cos .2. Phazmt in tNguyn l hot ng: da trn vic phn bin i
gc lch pha trc tip thnh dng hay p. o gc lch pha gia hai in p hnh
sin ta thc hin theo s hnh 3.76:Tn hiu hnh sin x1 v x2 qua cc b to
xung TX1 v TX2. Khi tn hiu i qua mc "0" to ra cc xung U'1 v U'2
(H.11.3b), cc xung ny c a n u vo ca Trig (S-R).Nh vy cc tn hiu hnh
sin u vo nh cc b to xung bin lch pha thnh khong thi gian gia cc
xung. Khi c s tc ng ca cc xung ny ln u vo ca trig xut hin tn hiu
ITr u ra, qua c cu ch th t in ta s c mt dng trung bnh:0360m m
tbITtI I vi Im : bin dng in u ra trigT : chu k ca tn hiuT c:
0360mtbII 106Hnh 3.76: Phazmt in ta) S nguyn lb) Gin thi gianNu Im
= const, th i lng gc pha cn o t l vi Itb, o dng trung bnh Itb th
suy ra gc pha.c im: sai s ca php o ny c (12%) ch yu l do s bin ng
ca Im v sai s ca php o dng trung bnh Itb.Loi Phazmt ny thng c s dng
o gc pha 0 1800 di tn s 20Hz 200kHz.107
