sổ tay cdt chuong 11- kthuat dien

8/6/2019 s tay cdt Chuong 11- kthuat dien

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11K thut in

Giorgio Rizzonii hc bang Ohio

11.1Gii thiu ...........................................................................1

11.2C s ca mch in ........................................................ ...1

11.3Phn tch mng in tr ............................................ ........11

11.4Phn tch mng AC ..................................................... ......16

11.1 Gii thiu

Vai tr ca k thut in v in t trong cc h thng c kh tng ln t ngt trong hai thp k qua, nh nhng tin

b trong cc lnh vc in t v vt liu cho php tch hp cc thit b o, tnh ton v cc c cu chp hnh vo cc h thngcng nghip v cc sn phm dn dng. Cc v d ca cuc cch mng tch hp ny, lin quan ti mt lnh vc mi gi l Cin t, c th tm thy trong in t dn dng (camera t ng chnh tiu c, my in, cc thit b dn dng iu khin bngvi x l), trong cc h thng t ng ho cng nghip, v giao thng vn ti, ng k nht l trong cc xe khch. Mc chca chng ny l xem xt li v tng kt nhng kin thc c s ca k thut in lm ti liu tham kho nhanh v hu chti cc lnh vc khc nhau ca k thut in cho cc k s thc hnh c kh. C s nhn mnh c bit i vi cc ch ckh nng lin quan n vic thit k sn phm.

11.2 C s ca mch in

Phn ny trnh by cc nh lut c bn phn tch mch in v l c s phc v vic nghin cu v mch in. Cckhi nim c bn c gii thiu trong phn ny s c s dng cho cc phn tip theo.

i lng in c bn l in tch, v lng in tch nh nht l in tch m mt electron mang, c gi tr l:

191.602 10eq coulomb= (11.1)

Nh chng ta thy, lng in tch ca mt electron kh nh v n v o in tch l culng (C - coulomb), t theotn Charles Coulomb. nh ngha ca Coulomb rt ph hp khi nh ngha dng in, v dng in l mt dng ca mt slng rt ln cc ht electron. Cc ht mang in tch khc trong nguyn t l proton. in tch ca proton c du dng vcng ln vi in tch ca electron:

191.602 10Pq coulomb= + (11.2)

Cc ht electron v proton thng c gi l cc in tch s cp.

Dng in c nh ngha l tc thay i ca in tch chuyn qua mt din tch c xc nh trc. Khi chng taxem tc ng ca mt s lng ln cc in tch s cp nh to thnh dng chy thc, chng ta c th vit mi quan h ny dng khc:

( / s)dqi Cdt

= (11.3)

n v ca dng in c gi l ampe (A), trong 1 A = 1 C/s. Trong k thut in, ngi ta quy c rng chiudng ca dng in l chiu ca cc in tch dng. Tuy nhin, trong cc kim loi dn in, dng in li to thnh t ccht in tch m l cc in t t do trong lp dn, trong cu trc nguyn t ca cc phn t kim loi, cc lp ny d b kchthch do cc in t d b y ra khi v tr bnh thng ca chng trong in trng.

tn ti dng in cn phi c mt mch in kn. Hnh 11.1 v mt mch in n gin, gm c mt ngun (v d, pinkh hoc pin kim 1.5-V) v mt bng n si t.

Ch rng trong mch in trn hnh 11.1, dng in i chy t ngun ti in tr bng dng in chy t bng n tingun in. Ni mt cch khc, dng in (v v vy in tch) khng b "mt" trong mch in kn. Nguyn l ny c nhkhoa hc c G. R. Kirchhoff a ra v c tn l nh lut dng Kirchhoff(KCL). KCL pht biu rng: tng cc dng inti mt nt phi bng khng(trong mt mch in, mt nt l mt im giao gia hai hoc nhiu dy dn):

1

0N

n

n

i=

= (11.4)

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S tay C in t

Hnh 11.2 minh ho cho nh lut KCL, y mch in n gin trn hnh 11.2 c tng thm 2 bng n (ch tihai nt trn mch in c t mu). Khi p dng KCL, thng nh ngha dng in i vo mt nt l du m v ra khint l mang du dng. V vy phng trnh biu din mch in hnh 11.2 l:

1 2 30i i i i+ + + =

Nh nu trn, in tch di chuyn trong mt mch in to ra mt dng in. Hin nhin cn phi c mt cng, haynng lng lm cho in tch di chuyn gia hai im trong mch in, v d t im a ti im b. Cng tng cng trnmt n v in tch lin quan ti chuyn ng ca in tch gia hai im c gi l in p. Do , n v in p l nnglng trn mt n v in tch:

11

joulevolt

coulomb= (11.5)

HNH 11.1 Mt mch in n gin HNH 11.2 Minh ho cho nh lut HNH 11.3 in p trn mt mch in

dng Kirchhoff

in p, hay l s chnh in th gia hai im trong mt mch in th hin nng lng cn di chuyn in tch tmt im ti im khc. Nh s ch ra ngay sau y, hng, hoc cc ca in p lun gn vi nng lng b tiu th hocc pht sinh trong qu trnh. Khi nim c v tru tng ca cng trn c th c s dng trc tip phn tch ccmch in; li xt mch in n gin bao gm mt ngun v mt n si t. V li mch nh hnh 11.3, tin cho vic

phn tch, nh ngha cc nt bng cc ch a v b. Quan st k in th trong mt mch in Kirchhoff xy dng cngthc cho nh lut th 2 ca Kirchhoff, nh lut in th Kirchhoff, hay l KVL. Nguyn tc c bn ca KVL l khng cnng lng no b mt i hoc sinh ra trong mt mch in; Trong mi mch in, tng tt c cc in th ca cc ngunphi bng tng cc in th ca ti, do in th trong mt mch kn l bng 0. Nu iu kin ny khng c tho mn,chng ta cn tm mt s gii thch c tnh vt l cho s tha hoc thiu nng lng ny. KVL c th c pht biu theo cchtng ng dng cho KCL nh sau:

1

0N

n

n

v=

= (11.6) y vn l cc in th ring trong mch kn. Hy xem hnh 11.3, chng ta c th thy rng theo KVL, cng sinh ra bi

ngun cn bng vi nng lng b tiu th trn bng n si t duy tr dng in v chuyn nng lng in thnh nhitv nh sng:

ab bav v= hoc 1 2v v=

HNH 11.4 Ngun v ti trong mt mch in

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HNH 11.5 Quy c v du

K thut in

C th coi cng di chuyn mt in tch t im a ti im b v cng in tch ny di chuyn ngc li t b v achnh l in p ca tng phn t mch. Gi Q l in tch tng di chuyn trong mch trong mt n v thi gian, sinh radng in i. Khi cng di chuyn Q t b n a (i qua ngun) l:

1.5baW Q V= (11.7)Tng t cng di chuyn Q t a n b, i qua bng n. Ch rng khi nim in th cng ngha vi khi nim in

p, trong in p biu din th nng gia hai im trong mch in: khi tho bng n khi mch in (khng ni vingun) vn tn ti mt in p gia hai u b v a.

Cn phi xc nh du ca i lng ny. Xt li trng hp ngun pin kh hoc pin kim, phn ng in ho sinh ra schnh in th l 1.5-V. in th m ngun to ra c th di chuyn in tch trong mch in. Lc ny, dng di chuyn ccin tch trong mch in kn c thit lp (chnh l dng in sinh ra khi ni mch in vi ngun) ph thuc vo phn tmch in m chng ta chn ni vo ngun. Do , in p ngun th hin in th cp nng lngcho mch in, cnin p trn bng n l cng tiu th nng lng. trng hp u tin, nng lng c sinh ra, trng hp th 2, lnng lng b tiu th (ch rng nng lng cng c th c tch tr bng cc phn t mch thch hp s c gii thiusau). S khc bit c bn ny i hi phi ch khi xc nh du (hay cc) ca in th.

Mt cch tng qut, chng ta s coi cc phn t cung cp nng lng l ngun, v cc phn t tiu th nng lng l ti.Cc k hiu chun cho mt mch in vi ngun v ti c a ra trn hnh 11.4. Cc nh ngha thng dng s tip tcc nu phn sau.

Quy c v du v cng sut in

nh ngha in p l cng trn mt n v in tch ph hp nh ngha cng sut. Nh rng cng sut c nh nghal cng thc hin trong mt n v thi gian. Do , cng sut P, hoc l c sinh ra hoc l b tiu th bi cc phn t trnmt mch in c biu din bng mi quan h sau y:

cng cng in tchCng sut in p dng in

thi gian n vin tch thi gian= = = (11.8)

Do , nng lng in c to ra bi mt phn t tch cc hoc b tiu th hoc c lu gi bi mt phn t tiu th,s bng vi i lng l tch ca in p trn phn t v dng in chy qua n nh sau:

P VI = (11.9)D dng thy rng n v ca in p (jun/culng) nhn vi n v ca dng in (culng/giy) chnh l n v ca nng

lng (jun/giy, hay cn gi l ot (W)).

Cn ch rng, cng ging nh in p, cng sut l mt i lng cdu, v cn phi phn bit gia cng sut dngv cng sut m. hiu c s phn bit ny hy tham kho hnh 11.5, trn th hinmt ngun v mt ti tch thnh hai hnh t cnh nhau. Cc tnh ca

in p ngun v chiu dng in ch ra rng ngun in p c tc dngdi chuyn in tch t im c in th thp hn n im c in th cao

hn. Mt khc, ti tiu th nng lng, bi v chiu ca dng in ch rarng in tch ang c chuyn di t im c in th cao hn ti

im c in th thp hn. trnh nhm ln chiu ca cng sut, trongk thut in, ngi ta a ra quy c chung v du, quy c ny quy

nh cng sut tiu th trn mt ti l mt i lng dng(hoc, ngcli, nng lng c sinh ra bi mt ngun l mt i lng dng). Cth pht biu theo cch khc nh sau: khi dng in chy t mt ni c

in th cao n mt ni c in th thp (+ ti -), cng sut tiu th s lmt i lng dng.

Cc phn t ca mch v cc c tnh i-v ca chng

Mi quan h gia dng in v in p trn cc cc ca mt phn t mch in xc nh ng x caphn t ny trong mch in. Phn ny s gii thiu s biu din c tnh cc ca cc phn t mchin. Hnh 11.6 l mt phn t mch thng thng c s dng trong sut chng ny: bin i biu dindng in i qua phn t, bin v l chnh lch in th, hay in p trn phn t.

Gi s t mt in p bit ln mt phn t mch. in p ny v in p ca bnthn n s to ra mt dng in. Nu in p cp bin i th dng in cng thay itheo. Quan h gia in p v dng in c gi l c tnh i-v (hoc c tnh vn-ampere). Quan h ny s xc nh phn t mch. S dng c tnh i-v ca mt phn t

chng ta c th xc nh c dng in (hoc in th) tng ng vi mt in th(hoc dng in) bit trc i qua phn t. T c tuyn i-v ta cng xc nh cnng lng b tiu th hoc c pht ra bi phn t.

HNH 11.6 Biu din tng ng ca cc phn t mch in

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S tay C in t

c tnh i-v ca ngun p v ngun dng l tng cng c th c dng m t ng x ca chng. Mt ngunp l tng s to ra mt in p quy nh c lp vi dng in ti; do , c tnh i-v ca n l ng thng ng ct trcin p ti v tr tng ng vi gi tr in p ca ngun. Tng t, c tnh i-v ca mt ngun dng l tng l mt ngnm ngang ct trc dng in ti v tr tng ng vi gi tr ngun dng. Hnh 11.7 m t cc c tnh trn.

in tr v nh lut Ohm

Khi dng in chy qua mt dy dn kim loi hoc cc phn t mch in khc, n s gp mt tr khng no , ln

ca tr khng ny ph thuc vo cc thuc tnh in ca vt liu. Tr khng i vi dng in c th l khng mong munnh trong trng hp ca dy dn v cp ni, hoc c khai thc mt cch c ch. Tuy nhin, trong thc t tt c cc phnt mch in u c mt tr khng no , v vy, khi dng in i qua mt phn t, nng lng s b tiu tn di dngnhit. Theo nh lut Ohm, mt in tr l tng l mt thit b c cc thuc tnh tr khng tuyn tnh, pht biu nh sau:

V IR= (11.10)

HNH 11.7 c tnh i-v ca cc ngun l tng

HNH 11.8 Phn t in tr

iu c ngha l in p trn mt phn t t l thun vi dng in i qua n.R l gi tr tr khng vi n v l Ohm ( ), y:

1 1 /V A = (11.11)Tr khng ca vt liu ph thuc vo mt thuc tnh c gi l sut in tr, k hiu l ; thuc tnh tri vi sut in

tr l sut dn in, k hiu l . i vi mt phn t in tr dng tr (biu din trn hnh 11.8), tr khng t l vi chiudi lca n, v t l nghch vi din tch mt ct ngangA v sut dn in ca n:

lv i

A= (11.12)

thun tin, thng nh ngha sut dn in ca mt phn t l nghch o tr khng ca n. G c dng k hiusut dn in ca mt phn t:

1siemens( ), trong 1 S =1 A/VG SR

= (11.13)

Bng 11.1 Mt s gi tr in tr thng s dng (Trong di 1/8-,1/4-,1/2-, 1-,2-,W)

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K thut in

Do , c th pht biu nh lut Ohm theo sut dn in nh sau:

I GV = (11.14)nh lut Ohm l mt quan h da trn kinh nghim, c ng dng rng ri trong

k thut in bi tnh n gin ca n. Tuy nhin, nh lut ny ch gn ng v mt vt

l i vi cc vt liu dn in. Thng th, mi quan h tuyn tnh gia in p vdng in trong cc cht dn in khng p dng c khi in p v dng in ln.Hn na, thc t khng phi tt c cc cht dn in u c quan h p dng tuyntnh ngay c trng hp in p v dng in nh. Tuy nhin, hu ht cc phn t mchin c c tnh i-v l tuyn tnh trong mt s di in p v dng.

Hnh 11.8 th hin cu trc in hnh v k hiu ca in tr. in tr c lm tcc ming cc bon hnh tr (vi sut in tr = 3.5 x 10-5 m) rt ph bin v c sntrn th trng vi mt di gi tr rng ng vi mt s di cng sut (s c gii thchsau).

HNH 11.9 M mu ca in tr

Mt k thut thng dng na l dng mng kim loi. Di cng sut thng dng cho cc in tr thng c s dngtrong cc mch in t (v d trong hu ht cc ng dng in t gia dng nh l radio, v tuyn) l W. Bng 11.1 lit k

cc gi tr tiu chun thng dng ca cc in tr v m mu lin quan ti cc gi tr ny (c ngha l cc t hp cc con sb1b2b3 nh c nh ngha trn hnh 11.9). Th d, khi ba di mu u tin trn mt in tr l cc mu (b 1=2), mu tm(b2=7), v mu vng (b3=4), th gi tr tr khng l nh sau:

427 10 270,000 270R k= = = bng 11.1, ct th nht biu din m mu y ; cc ct tip theo ch biu din mu th ba, do ch c mu th ba

thay i. V d, in tr 10 c m nu-en-en, trong khi in tr 100 l nu-en-nu.Ngoi tr khng, cng sut tiu th ti a cho php (hay cng sut nh mc) cng c ghi r trn cc in tr thng

mi. Nu vt qu cng sut nh mc s dn ti hin tng qu nhit v c th l nguyn nhn gy chy in tr. Vi mtin tr R, cng sut tiu th s l:

22 VP VI I R

R= = = (11.15)

y, cng sut tiu th trn mt in tr t l vi bnh phng dng in i qua n, v cng t l vi bnh phng inp i qua n. V d sau s minh ho mt ng dng k thut thng thng ca cc phn t in tr: o sc cng bng in tr.

V d 11.1 o sc cng bng in tr

M Nhnbi

k Nhnbi

k Nhnbi

k Nhnbi

10 Nu-en-en 100 Nu 1.0 Vng 10 Da cam 100 Vng

12 Nu--en 120 Nu 1.2 Vng 12 Da cam 120 Vng

15 Nu-xanh lcy-en

150 Nu 1.5 Vng 15 Da cam 150 Vng

18 Nu-xm-en 180 Nu 1.8 Vng 18 Da cam 180 Vng

22 --en 220 Nu 2.2 Vng 22 Da cam 220 Vng

27 -tm-en 270 Nu 2.7 Vng 27 Da cam 270 Vng

33 Da cm-dacam-en


39 Da cam-trng-en


47 Vng-tm-en 470 Nu 4.7 Vng 47 Da cam 470 Vng

56 Xanh l cy-xanh lc-en 560 Nu 5.6 Vng 56 Da cam 560 Vng

68 Xanh lc-xm-en


82 Xm--en 820 Nu 8.2 Vng 82 Da cam 820 Vng

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S tay C in t

Mt ng dng ph bin ca in tr trong k thut o lng l o sc cng. Cc u o sc cng l thit b c gncht vo b mt ca mt i tng, v tr khng ca n s thay i nh l mt hm ca sc cng b mt ca i tng. Ccu o sc cng c th c s dng o sc cng, ng sut, lc, mmen v p sut. Nh rng tr khng ca mt vt dnin hnh tr c din tch mt ct ngangA, chiu diL v dn in c dng sau:

LR

A=

Khi vt dn b nn hoc b ko gin bi tc dng ca mt ngoi lc, cc kch thc ca n s thay i v do in trca chng cng thay i theo. c bit, khi vt dn b ko gin, din tch mt ct ngang ca n s gim v do in tr stng. Khi vt dn b nn, in tr ca n s gim, do chiu di,L, gim. Mi quan h gia s thay i in tr v s thay ichiu di c th hin bi h s o G nh sau:

/

/

R RG

L L

=

V do sc cng c nh ngha l t s ca gia s di trn di nh sau:

L

L

=

S thay i ca in tr theo sc cng c biu din nh sau:

0R R G =

yR0l tr khng ca u o khi khng c sc cng v c gi l tr khng khi sc cng bng khng. Gi tr G cacc u o sc cng kiu l kim loi khong bng 2.

Hnh 11.10 v mt u o o sc cng in hnh. Sc cng ln nht c th o bng mt u o loi ny khong 0,4 -0,5% tc l /L L = 0.004 n 0.005. i vi u o 120- , gi tr ny tng ng vi s thay i in tr t 0,96 - 1,2 .Mc d s thay i tr khng ny rt b, nhng n vn c th c pht hin bng mch thch hp. Cc u o in tr theonguyn l sc cng thng c ni trong mt mch gi l mch cu Wheatstone, s c phn tch k hn phn tip theo.

Ngn mch v h mch

Hai trng hp l tng ho ca phn t tr khng l nhng trng hp gii hn ca nh lut Ohm khi tr khng tindn n khng hoc v cng. Mt phn t mch in c tr khng tin ti 0 th gi l phn t ngn mch. D thy, phn tngn mch khng c tc dng cn dng in chy qua. Trong thc t, cc cht dn in kim loi (v d, cc dy dn ngn cng knh ln) c ng x gn ging vi ca phn t ngn mch. Phn t ngn mch c nh ngha l mt phn t mch

in m in p trn n bng khng, y khng ch n dng in qua n. Hnh 11.11 th hin k hiu mch cho phn tngn mch l tng.

Bng 11.2 in tr ca dy dn ng

C AWG S si dy ng knh dy in tr trn mi feet dy ( )

24 c 0,0201 28,4

24 7 0,0080 28,4

22 c 0,0254 18,0

22 7 0,0100 19,0

20 c 0,0320 11,3

20 7 0,0126 11,9

18 c 0,0403 7,2

18 7 0,0159 7,5

16 c 0,0508 4,5

16 19 0,0113 4,7

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7/26

K thut in

HNH 11.10 u o sc cng HNH 11.11 Phn t ngn mch HNH 11.12 Phn t h mch

Thng thng, bt k dy dn kim loi no u c tr khng, d l nh. Tuy nhin, di cc iu kin nht nh, nhiuphn t s gn ging phn t ngn mch cho cc mc ch thc t. V d, mt ng dn bng ng ng knh ln thc t sl phn t ngn mch vi mt ngun cp in, trong khi trong mt mch vi in t cng sut thp (v d, i FM) mton dy dn ngn c 24 (tham kho bng 11.2 v tr khng ca dy dn c 24) s l mt phn t ngn mch thch hp hn.

Mt phn t mch in c tr khng tin ti v cng c gi l mt phn t h mch. C th thy ngay rng, s khngc dng in i qua mt phn t h mch, v n s a ra tr khng l v cng vi mi dng in. Trong mt phn t hmch, dng in lun bng khng vi mi in th bn ngoi cp ln mch. Hnh 11.12 minh ho khi nim ny.

HNH 11.13 Nguyn tc ca b phn p

Trong thc t, to mt phn t h mch l khng kh ch cn ngt ng dn. Mt phn t h mch l tng, nh trnhnh 11.12 khng c dng in, tuy nhin in p rt cao. Vt liu cch in gia hai cc s ngt mch in p ln. Khicht cch in l khng kh, cc phn t trong vng ln cn gia hai phn t dn in b ion ho, c th dn ti hin tngphng in; ni cch khc, mt xung dng c th c to ra tc thi khe gia hai phn t dn in (nh nguyn l ny,chng ta c th t hn hp kh-nhin liu trong mt ng c t trong kiu nh la bng cng tc nh la). H mch vngn mch l tng l nhng khi nim hu ch v chng c ng dng rt nhiu trong vic phn tch mch in.

Cc in tr ni tip v lut phn p

Mc d cc mch in c th rt phc tp, nhng vn c th c n gin ha thnh cc t hp song songv ni tipca cc phn t mch. Do , vic quan trng u tin l lm quen cc mch song song v ni tip n gin trc khi tipcn vic phn tch mng in. Cc mch song song v ni tip c mi quan h mt thit vi nh lut ca Kirchoff. Mc chca phn ny v phn tip theo l minh ho hai mch hay gp l cc b phn dng v phn p da trn cc t hp in trsong song v ni tip. Cc dng mch ny l c s phn tch mng in;

Mt v d v mch in ni tip c ch ra hnh 11.13, y c mt ngun in ni vi cc in trR1,R2 vR3. nhngha sau s c p dng.

nh ngha

Hai hoc nhiu phn t c gi l ni tip khi cng mt dng in chy qua mi phn t. V vy 3 in tr mc ni tip

c th thay bng mt in tr c gi tr REQ m khng lm thay i dng in qua chng. T kt qu ny chng ta c th suyrng cho trng hp tng qut hn vi N in tr mc ni tip:

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S tay C in t

1

N

EQ n

n

R R=

= (11.16)iu ny cng c minh ho trn hnh 11.13. Mt khi nim gn lin vi trng hp in tr mc ni tip l b phn

p.

HNH 11.14 Cc mch song song

Dng tng qut ca lut phn p cho mch in c N in tr mc ni tip v vi mt ngun p l:

1 2 ... ...

nn s

n N

Rv v

R R R R

=

+ + + + +(11.17)

Cc in tr mc song song v lut phn dng

Bng cch p dng nh lut v dng in ca Kirchhoff, c th pht trin mt khi nim tng t cho mch in ch ccc in tr mc song song.

nh ngha

Hai hoc nhiu phn t mch in c gi l song song vi nhau khi c cng mt in p trn mi phn t. (Xem hnh11.14)

N in tr mc song song c th c thay th bng mt in tr tng ng, REQ, c cng thc nh sau:

1 2

1 1 1 1...

EQ N R R R R= + + +

(11.18)hoc

1 2

1

1/ 1/ .... 1/EQ NR

R R R=

+ + +(11.19)

Trong quyn sch ny cn rt ch rng chng ta k hiu v mt b (hai hoc nhiu) cc in tr mc song song nh sau:

1 2...R R

y k hiu c ngha l "song song vi".

Cng thc tng qut cho b phn dng trn mt mch in c N in tr mc song song l:

1 2

1/

1/ 1/ .... 1/ ... 1/

n

n sn N

R

i iR R R R= + + + + + (11.20)

V d 11.2 Cu Wheatstone

Cu Wheatstone l mt mch thng c s dng trong rt nhiu mch o. Dng tng qut ca cu c th hin trnhnh 11.15(a), yR1,R2 vR3 bit,Rx l in tr cha bit v cn phi xc nh. Cng c th biu din mch cu theonh trn hnh 11.15(b). Mch in ny s c s dng minh ho cho vic s dng b phn p trong mt mch hn hpsong song-ni tip.

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K thut in

HNH 11.15 Cc mch cu Wheatstone

Mc ch chnh l xc nh in tr cha bitRx.

1. Tm gi tr inp vad = vad - vbd theo 4 in tr bit trc v in p ca ngun, vs. Ch rng do im quy chiu dl nh nhau cho hai c hai in p, chng ta c th vit vab = va - vb.

2. NuR1 =R2 =R3 = 1 k , vs = 12 V, v vab = 12 mV, thRx = ?

Li gii

1. u tin, chng ta quan st thy rng mch in bao gm 3 mch ph nh mc song song vi nhau: ngun in p,mch 2 in trR1 vR2 mc ni tip, v mch 2 in trR3 vRx mc ni tip. Do 3 mch ph ny mc song songvi nhau, qua chng s xut hin cng mt in p, chnh l in p ca ngun vs.

V vy, in p ca ngun trn mi cp in trR1-R2 v R3-Rx, c xc nh theo lut phn p l: va l mtphn ca ngun in p t trnR2, v vb l phn cn li trnRx:

2

1 2 3

xa s b s

x

RRv v v v

R R R R= =

+ +Cui cng, c th tnh c hiu in th gia hai im a v b nh sau:

2

1 2 3

xab a b S

x

RRv v v v

R R R R

= = + +

y l mt kt qu tng qut v rt hu dng, kt qu ny c ng dng rt nhiu trong cc mch in thc t.

2. c th tm c in tr cha bit, chng ta thay cc gi tr s vo phng trnh nu trc:

10000.012 12

2000 1000x

x

R

R

= +

T c th tm cRx

966xR =

Ngun p v ngun dng thc t

Cc m hnh l tng ho ca ngun p v ngun dng s khng hp l khi xem xt bn cht ca cc ngun dng v ptrong thc t vi mc nng lng gii hn. Mc ch ca phn ny l m rng m hnh l tng c th m t c ccgii hn vt l ca cc ngun dng v p s dng trong thc t. Th d, xt mt m hnh ca ngun p l tng. Khi in trti (R) gim, dng in ca ngun cung cp tng khi duy tr mc in p vs(t) qua cc cc ca n khng i:

( )( ) S

v ti t

R= (11.21)

HNH 11.16 Ngun p thc t

9


10/26

S tay C in t

HNH 11.17 Ngun dng thc t

Vi mch in ta c th thy rng trong trng hp in tr ti tin ti 0, ngun in p s cung cp mt dng in lnv cng ( ), qu ti.

Hnh 11. 16 v m hnh ngun p thc t; bao gm mt ngun p l tng, vs, ghp ni tip vi in trrs. C th tnh

c gi tr gii hn ln nht ca dng in nh sau:

max

SS

S

vi

r= (11.22)

D thy rng, tnh cht mong mun ca mt ngun p l tng l c in tr trong rt b tho mn yu cu v dngin cp cho mt ti bt k.

Tng t, m t mt ngun dng thc t ngi ta cng dng m hnh ngun dng l tng Mch in hnh 11.17 mt mt ngun dng thc t n gin, bao gm mt ngun l tng ghp song song vi mt in tr. Ch rng, khi tr khngti tin ti v cng (c ngha l mch h), in p ra ca ngun dng s tin ti gii hn ca n,

maxS S Sv i r= (11.23)Mt ngun dng tt phi c ng x gn ging ng x ca mt ngun dng l tng. V vy c tnh tr khng trong

mong mun ca mt ngun dng l cng ln cng tt.

Cc thit b o

Ampe k

Ampe k l mt thit b m khi c ni ni tip vi mt phn t mch in, n c th o c dng in chy qua phnt . Hnh 11.18 minh ho mt ampe k. T hnh 11.18, ta thy cn phi m bo c 2 yu cu c th o dng inchnh xc:

1. Phi ni ampe k ni tip vi phn t mun o dng in qua n (v d in trR2).

2. Ampe k khng cn dng in trn mch. Mt ampe k l tng phi c in tr trong bng khng.

Vn k

Vn k l mt thit b o in p trn phn t mch in. Do in p l hiu in th gia hai im khc nhau trn mtmch in, vn k cn phi c ni vi ng phn t mch cn o in p. Mt vn k cn phi tho mn y hai yucu sau:

1. Vn k cn phi c ni song song vi phn t cn o in p.

2. Vn k phi khng nh hng ti gi tr dng in qua phn t cn o in p, nu khng in p qua phn t oc s khng chnh xc. Do , mt vn k l tng phi c in tr trong l v cng.

Hnh 11.19 minh ho hai yu cu ny.

Nhc li mt ln na rng, cc nh ngha trn y ch pht biu cho ampe k v vn k l tng, lm c s nh giv cc gii hn thc t ca cc thit b ny. Mt ampe k trong thc t s c trong n mt vi in tr mc ni tip vi mchin m n s o dng in; mt vn k thc t khng hon ton nh l tng l mt mch h nhng lun theo nguyn tc tnh hng nht ti gi tr dng in trong mch cn o. Hnh 11.20 v cc m hnh mch in ca vn k v ampe k trongthc t.

10


11/26

K thut in

HNH 11.18 o dng in

HNH 11.19 o in p

HNH 11.20 Cc m hnh vn k v ampe ktrong thc t

HNH 11.21 o cng sut

Nhng g gii thiu v cc ampe k v vn k trong thc t c th c ng dng lm watt k - mt dng c dng o cng sut tiu th trn mt phn t mch in, do watt k l s kt hp vn k v ampe k.

Hnh 11.21 v s ghp ni in hnh ca mt watt k trong cng mt mch in ni tip trnh by phn trc.Theo , watt k o ng thi dng in v in p qua ti, sau nhn 2 gi tr s thu c gi tr cng sut tiu th trnti.

11.3 Phn tch mng in tr

Phn ny s gii thiu cc phng php phn tch mch in c bn. Cc phng php c gii thiu l da trn nhlut Ohm v Kirchhoff. Cc phng php phn tch mch in ny s c s dng trong sut quyn sch ny.

Phng php in p nt

Phn tch in p nt l phng php phn tch mch in tng qut nht. Trong phn ny, s minh ho ng dng caphng php in p nt phn tch cc mch in tr tuyn tnh. Phng php in p nt da trn nh ngha in pti mi nt l mt bin c lp. Mt trong s cc nt s c chn lm nt quy chiu (thng l t nhng khng bt buc),v in p ca cc nt khc s c tham chiu so vi nt ny. Khi xc nh c in p trn mi nt, nh lut Ohm cth c p dng gia 2 nt k tip xc nh c dng in chy trong mi nhnh. Vi phng php in p nt, midng in nhnh c biu din theo mt hoc nhiu in p nt; do , cc dng in khng th hin tng minh trong ccphng trnh. Hnh 11.22 minh ho cch xc nh cc dng in nhnh theo phng php ny.

11


12/26

S tay C in t

HNH 11.22 Phng trnh dng in nhnh khi phn tch nt HNH 11.23 S dng KCL khi phn tch nt

Khi mi dng in nhnh c xc nh theo cc in p nt, c th p dng nh lut dng Kirchhoff ti mi nt. Theonh lut dng Kirchhoff KCL, ti mi nt, tng cc dng in i vo bng tng cc dng in i ra khi nt:

in out i i= (11.24)Hnh 11.23 Minh ho cho pht biu ny.

Vic p dng c h thng phng php ny cho mt mch in n nt c th vit c n phng trnh tuyn tnh. Tuy

nhin, mt trong cc in p nt l in p quy chiu thng c gi thit bng khng. Do , chng ta c th vit n-1phng trnh c lp tuyn tnh theo n-1 bin c lp (cc in p nt). Phng php phn tch nt s a ra ti thiu phngtrnh cn thit gii mch in, do c th xc nh c in p hoc dng in ca nhnh bt k t cc in p nt bit.

Phng php phn tch in p nt

Cng c th c nh ngha gm cc bc sau:

1. Chn mt nt quy chiu (thng l t). Tt c cc in p nt khc s c tham chiu ti nt ny.

2. t n-1 in p nt cn li nh l cc bin c lp.

3. p dng KCL ti mi nt, biu din cc dng in nhnh theo cc in p nt lin k nhau.

4. Gii h n-1 phng trnh tuyn tnh vi n-1 bin.

Vi mt mch in c n nt, chng ta c th vit c nhiu nht n-1 phng trnh c lp.

Phng php dng in mch vngTrong phng php dng in mch vng, chng ta thy rng mt dng in chy qua mt in tr theo mt hng xc

nh s to ra s phn cc in p trn in tr, nh c minh ho trn hnh 11.24, tng in p trn mt mch in vngkn phi bng khng, theo KVL. Khi quy c chiu dng in chy trong mt mch vng, KVL s cho php xy dngc phng trnh mong mun. Hnh 11.25 minh ho cch xc nh ny.

S phng trnh xy dng c theo phng php ny bng vi s mch vng trong mch in. T c th c c ttc in p v dng in ca cc mch vng. Do c th d dng ch ra cc mch vng trong mt mch in, nn phng phpny rt hiu qu cho vic phn tch cc mch in. Phn tip theo s trnh by cc bc phn tch mch tuyn tnh dngphng php mch vng.

Phng php phn tch dng in mch vng1. Xc nh tng dng in mch vng mt cch thch hp. thun tin, s lun quy c chiu dng in mch vng

l chiu kim ng h.

12

HNH 11.24 Nguyn l c bn caphng php phn tch mch vng

HNH 11.25 S dng KVLtrongphn tch mch vng

HNH 11.26 Mng mt cng


13/26

K thut in

2. p dng KVL trong tng mch vng, biu din tng in p theo mt hoc nhiu dng in mch vng.

3. Gii h phng trnh tuyn tnh thu c vi cc dng in mch vng l cc bin c lp.

Khi phn tch mt mch vng, vic chn chiu dng in mt cch thch hp l rt quan trng. trnh nhm ln trongkhi vit cc phng trnh mch in, nn nh ngha dng in mch nhnh theo chiu kim ng h khi nh s dng phngphp ny.

Cc mng mt cng v mch tng ng

Hnh 11.26 biu din mt mch in tng qut. Cu trc ny c gi l mng mt cng v c bit thng c sdng biu din cc mch in tng ng. Ch rng mng trn hnh 11.26 c din t hon ton bng c tnh i-vca n.

Cc mch Norton v Thvenin tng ng

Phn ny gii thiu mt trong s cc vn quan trng nht trong phn tch mch in: mch tng ng. C th chra rng mt mch in phc tp lun c th biu din thnh nhiu mch ngun v ti tng ng n gin hn, v cc phpbin i s to ra cc mch tng ng d qun l. Khi phn tch phng php in p nt v phng php dng nhnh,chng ta thy rng c mt s tng ng (c gi l i ngu) no , mt mt l gia cc ngun dng vi cc ngun p, vmt khc l cc mch in ni tip, song song. Tnh i ngu ny c th hin rt r rng khi phn tch cc mch in tngng. Ni tm li l cc mch in tng ng s l mt trong hai loi: ngun p hoc ngun dng (tng ng) v ccin tr ni ni tip hoc song song, th hin cng nguyn l i ngu ny. Vic trnh by cc mch tng ng s bt uvi hai nh ngha rt quan trng, minh ho trn hnh 11.27 v 11.28.

HNH 11.27 Minh ho nh l Thvenin

HNH 11.28 Minh ho nh l Norton

HNH 11.29 Tnh in tr Thvenin

nh l Thvenin

Vi mt ti ni xa, bt k mng no gm cc ngun p, ngun dng l tng v cc in tr tuyn tnh, lun c thc biu din bi mt mch in tng ng bao gm mt ngun p l tng vT mc ni tip vi mt in tr tngngRT.

nh l Norton

Vi mt ti ni xa, mi mng li in c cc ngun p l tng, ngun dng l tng, v cc in tr tuyn tnh,

lun c th c biu din bi mt mch in tng ng bao gm mt ngun dng l tng iNmc song song vi mt intr tng ngRN.

Xc nh tr khng Norton hoc Thvenin tng ng

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14/26

S tay C in t

tnh mt mch Thvenin hoc Norton tng ng, u tin phi tnh tr khng tng ng ti cc cc ca n. Vicny c thc hin bng cch t tt c cc ngun trong mch in bng 0 v tnh ton cc tr khng gia cc cc. Ccngun p v dng hin c trong mch c t ti 0 nh sau: thay th cc ngun p bng cc ngn mch, cc ngun dngbng cc h mch. Chng ta c th a ra mt tp hp cc lut n gin nh l mt s tr gip tnh tr khng Thvenin(hoc Norton) tng ng cho mt mch tr khng tuyn tnh.

Tnh tr khng tng ng ca mt mng mt cng nh sau:

1. B ti.

2. t cc ngun p v ngun dng v gi tr 0.3. Tnh tr khng tng gia cc cc ca ti, khi b ti. Tr khng ny s tng ng vi tr khng c to ra bi

mt ngun dng ni vi mch v tr ca ti.

V d, tr khng tng ng ca mch in trn hnh 11.29 l:

((2 2) 1 2 1eqR = + =

HNH 11.30 Tng ng ca in p h mch v in p Thvenin

HNH 11.31 Minh ho mch tng ng Norton

Tnh in p Thvenin

in p tng ng Thvenin c nh ngha nh sau: in p tng ng (Thvenin) bng vi in p h mchgia cc cc ca ti khi b ti.

Trong trng hp ny tnh ton vT, cn b ti v tnh in p h mch ti cc cc ca mt cng. Hnh 11.30 minh hoin p h mch vOC v in p Thvenin vT, chng phi l mt khi tun theo nh l Thvenin. iu ny l ng bi v trongmch in c vT vRT, in th vOCphi ngang bng vi vT, v khng c dng in chy qua RT v do in th trn RT bng0. nh lut in p ca Kirchhoff xc nhn iu ny:

(0)T T OC OC v R v v= + = (11.25)

Tnh dng in Norton

Tnh dng Norton tng ng cng tng t nh vic tnh in p Thvenin.

nh ngha

Dng in Norton tng ng bng dng in h mch khi thay th ti bng mt phn t h mch.

Hnh 11.31 minh ho cho nh ngha dng Norton bng cch xt mt mng mt cng bt k, y a ra c mng mtcng cng vi mch tng ng Norton ca n.

Cn nh rng dng in iSC chy qua phn t ngn mch chnh l dng Norton iN, do tt c cc dng in ngun ca mchin trn hnh 11.31 phi chy qua phn t ngn mch ny.

Xc nh bng thc nghim cc i lng Norton v Thvenin tng ng

Hnh 11.32 minh ho php o in p h mch v dng in ngn mch ca mt mng bt k c ni vi ti bt k, v

cng ch ra rng phng php ny yu cu mt s ch ring, bi v bn cht khng l tng ca cc dng c o thc t.Hnh cng minh ha s tn ti ca tr khng hu hn rmca thit b o nn cn phi a i lng ny vo qu trnh tnhton dng in ngn mch v in p h mch; vOCv iSC c t trong du ngoc kp ch ra rng, trong thc t "in p

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15/26

K thut in

h mch" v "dng in ngn mch" o c b nh hng bi tr khng trong ca thit b o v y khng phi l cc ilng ng.

HNH 11.32 o in p mch h v dng in ngn mch

HNH 11.33 c tnh i-v ca in tr tun theo hm m

Di y l cc cng thc chnh xc cho dng in ngn mch v in p h mch.

1

1

mSC

T

TT OC

m

ri

R

Rv vr

+

= +

(11.26)

y iN l dng in Norton l tng, vT l in p Thvenin l tng, vRT l in tr Thvenin thc.

Cc phn t mch phi tuyn

Biu din cc phn t phi tuyn

Trong mt s trng hp, tn ti mi quan h hm n gin gia in p v dng in ca mt phn t mch phi tuyn.V d, hnh 11.33 miu t mt phn t vi ng cong c tnh i-v c dng hm m, c m t bi cc phng trnh sau:

0

0

, 0

, 0

= >=

vi I e v

i I v(11.27)

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16/26

S tay C in t

HNH 11.34 Biu din mt phn t phi tuyn trong mch in tuyn tnh

Trong thc t, tn ti mt phn t mch in (it bn dn) gn nh tho mn hon ton mi quan h n gin ny. Cikh trong quan h i-v ca phng trnh (11.27) l khng th, ni chung, nhn c li gii gii tch dng khp kn, thm chngay c vi mch in rt n gin.

Mt cch tip cn phn tch mt mch in cha phn t phi tuyn l c th coi phn t phi tuyn nh l mt ti, vsau tnh ton lng tng ng Thvenin ca mch cn li, nh th hin trn hnh 11.34. p dng KVL, nhn cphng trnh sau:

T T x xv R i v= + (11.28) tm c phng trnh th hai phc v cho vic gii bi ton vi 2 n s l in th vx, v dng in ix, chng ta cn

xt n c tnh i-v ca phn t mch in phi tuyn, phng trnh (11.27). Nu ti mt thi im ch xt cc in p dng,c th biu din ton b mch in di dng h phng trnh sau:

0, 0= >

= +

xv

x

T T x x

i I e v

v R i v(11.29)

H phng trnh (11-29) gm 2 phng trnh vi 2 n s. C th la chn bt k mt phng php gii s no gii hphng trnh .

11.4 Phn tch mng AC

Phn ny s gii thiu cc phn t tch ly nng lng, cc mch ng, v phn tch cc mch b kch ng bi cc inp v dng in hnh sin. Cc tn hiu hnh sin (hoc AC) l lp tn hiu quan trng nht khi phn tch cc mch in. ngin v thc cht tt c cc ngun nng lng in s dng trong sinh hot v trong cng nghip u dng in p v dng

in hnh sin.

Cc phn t tch ly nng lng (ng)

in tr l tng c gii thiu qua nh lut Ohm phn 11.2 l mt s l tng ho nhiu thit b in thc t.Tuy nhin, ngoi c tnh cn dng in chy trong mch cn gi l hin tng tiu th (mt nng lng), cc thit b incng c th tch ly nng lng, hin tng ny ging nh mt l xo hoc mt bnh c th tch ly nng lng c hc. Chai c ch tch ly nng lng khc bit trong mch in l: in dung v in cm, c hai u lu gi nng lng trong mttrng in t.

T in l tng

Mt t in vt l l mt thit b m c th tch nng lng in bng cch phn cch in tch khi c phn cc bi mtin trng thch hp (mt in p). Cu trc ca mt t in n gin nht bao gm hai tm dn song song c din tch mtct ngang l A, c cch ly bi khng kh (hoc mt cht in mi1khc, v nh mica hoc Teflon). Hnh 11.35 ch ra mtcu trc in hnh v k hiu ca mt t in.

HNH 11.35 Cu trc ca t in dng bn song song

1 Vt liu in mi cha mt s lng ln cc lng cc in s b phn cc trong mt in trng16


17/26

K thut in

Cht cch in t gia hai tm dn in s khng cho php to ra dng in DC, do vy, tc ng ca mt t in gingnh mt phn t h mch trong s hin din ca cc dng in DC. Tuy nhin, khi in p trn cc cc ca t in thay inh mt hm ca thi gian, in tch c tch lu ti hai mt ca t in s thay i, v phn cc l hm ca intrng cp, m n li thay i theo thi gian. Trong mt t in, s phn cch in tch do s phn cc ca cht in mi tl vi in th ngoi, tc in trng cp:

Q CV= (11.30) y tham s C c gi l in dung ca phn t v l s o kh nng np hoc tch ly in tch ca thit b. n v

ca in dung l culng/vn v c gi l farad (F). Farad l mt n v rt ln hu nh khng tn ti, do vy thng sdng microfarad (1 610F F = ) hoc picofarad ( 121 10pF F = ). T phng trnh (11.30) d dng thy rng khi in pngoi cp ln cc bn t in thay i theo thi gian, s lm thay i in tch bn trong ca t in:

( ) ( )q t Cv t = (11.31)V vy, mc d khng c dng in no chy qua t in khi in p trn t khng i, nhng khi in p thay i theo

thi gian th in tch cng thay i theo thi gian. S tch in ca t thay i theo thi gian ging nh mt dng in. Miquan h gia dng in v in p trn mt t in c biu din nh sau:

( )( )

dv ti t C

dt= (11.32)

Khi tch phn cc phng trnh vi phn trn, chng ta c th tm c mi quan h in p trn mt t in nh sau:01

( )t

C Cv t i dt C = (11.33)

Phng trnh (11.33) ch ra rng in pca t in ph thuc vo dng in i qua t in t trc n thi imang xt t. Tt nhin, thng khng c cc thng tin chnh xc v dng in qua t in trong ton b khong thi gian trc v vy cn phi nh ngha in p ban u (hoc in kin u) ca t in nh sau:

0 0

1( )

t

C CV v t t i dt C = = = (11.34)

trong t0 l mt thi im ban u bt k. V in th ca t in l

00 0

1( )

t

C Ct

v t i dt V t t C

= + (11.35) ngha ca in p ban u V0 n gin l ti mt thi im t0 lng in tch c lu gi trong t in s to ra mt

in th vC (t0), theo quan h Q = CV. iu kin u bit gii thch ton b qu trnh bin i qua trc cadng in trn t. (Xem hnh 11.36).

T quan im phn tch mch, ch ra cc t in c mc ni ni tip v song song c th c gp thnh mt in dungtng ng. Nguyn tc xc nh in dung tng ng c minh ho trn hnh 11.37 nh sau:

Cc t in vt l t khi c cu trc dng hai bn song song c cch ly bi khng kh v c gi tr in dung rt thp, trtrng hp din tch ca hai bn t ny rt ln. tng in dung (hay kh nng tch lu nng lng), ngi ta thng lmcc t in vt l bng cch cun cht cc cc tm kim loi mng, cng vi mt cht in mi (giy hoc Mylar) gia.

Bng 11.3 a ra cc gi tr in hnh, vt liu, in p nh mc ti a, v cc di tn s c ch cho cc loi t in. in pnh mc c bit quan trng, v cc cht cch in s b ph hu khi c mt in p cao t vo n. Nng lng tchtrong mt t in c tnh nh sau:

17

HNH 11.36 Phng trnh ca mt t in l tng vh lc-khi lng tng ng

HNH 11.37 Ghp cc t in trong mt mchin


18/26

S tay C in t

21( ) ( ) ( )2C C

W t Cv t J =

BNG 11.3 Cc t in

Vt liu Di in dung in th ln nht (V) Di tn s (Hz)

Mica 1 pF n 0.1 F 100-600 103-1010

Gm 1 pF n 0.1 F 50-1000 103-1010

Mylar 0.001 n 10 F 50-500 102-108

Giy 1000 pF n 50 F 100-105 102-108

in phn 0.1 F n 0.2 F 3-600 10-104

V d 11.3 u o dch chuyn kiu in dung v Microphone

Nh trn hnh 11.26, in dung ca mt t in dng bn song song c biu din nh sau:

AC

d

=

Trong l hng s in mi ca cht in mi, A l din tch mi bn t, v dl khong cch. Xt mt t in vihng s in mi ca khng kh l 120 8,854 10 /x F m

= , din tch ca hai bn t song song l 1 m2, khong cch gia hai

bn t l 1 mm, in dung ca t l 38,854 10x F , y l mt gi tr rt b so vi din tch rt ln ca bn t. S khng hiuqu ny lm cho cc t in dng bn song song khng mang tnh thc t khi s dng trong cc mch in. Tuy nhin, cc tin kiu ny li c ng dng trong cc u o chuyn ng, l cc thit b o chuyn ng hoc dch chuyn ca mti tng. Trong mt u o chuyn ng kiu in dung, khe khng kh gia cc bn t c thit k c th thay i,thng bng cch c nh mt bn v ni bn cn li vi mt i tng chuyn ng. Gi tr in dung ca mt t in dngbn song song, c th biu din bi phng trnh:

38.854 10C

x

=

trong Cl in dung c picofarad,A l din tch cc bn t tnh theo millimt vung, v x l khong cch (c th thay

i) tnh theo millimt. Mt iu quan trng chng ta cn thy l s thay i in dung do s dch chuyn ca mt trong haibn t l phi tuyn, v in dung bin thin t l nghch vi dch chuyn. Tuy nhin, i vi cc dch chuyn nh, in dungbin i gn nh tuyn tnh.

nhy S, ca u o chuyn ng ny c nh ngha l dc ca s thay i in dung theo s thay i dchchuynx theo quan h:

3

2

8.854 10( / )

2

dC AS pF mm

dx x

= =

Do vy, nhy s tng i vi cc dch chuyn nh. ng x ny c th c xc nh bng cch v in dung nh lhm ca x v ch rng khi x tin ti 0, dc ca ng cong phi tuyn C(x) tr nn dc hn ( nhy ln hn). Hnh11.38 th hin ng x ny ca mt u o chuyn ng c din tch l 10 mm 2.

HNH 11.38 p ng ca mt u o chuyn ng kiu in dung

18


19/26

K thut in

HNH 11.39 u o p sut kiu in dung v mch cu tng ng

Trong thc t u o dch chuyn kiu in dung n gin ny c ng dng ph bin trong micr in dung (hocmicr t in), y p sut sng m thanh s to ra dch chuyn ca mt bn t. S thay i in dung sau s cchuyn thnh s thay i in p hoc dng in bi mt mch thch hp. Mt ng dng na l o chnh p sut c th

hin nh trn hnh 11.39. Trong hnh v mt t in 3 cc c lm t hai bn mt c nh (in hnh l mch kiu cu nit vi cc a knh v c bao ph bi mt vt liu dn in) v mt tm lch tm (thng c lm t thp) gia cc aknh. Cc ca vo s c cung cp p sut, cho tm lch tm c th di chuyn tip xc vi dng cht lng m chng tacn o p sut. Khi p sut trn c hai mt ca tm lch tm bng nhau, in dung gia hai cc b v dl Cbd, s bng vi indung Cbc gia hai cc b v c. Khi tn ti bt k chnh p sut no, hai in dung ny s thay i, vi vic tng mt phani m tm lch tm tin gn ti b mt c nh v s gim tng ng pha cn li.

ng x ny ph hp mt cch l tng cho ng dng mch cu, tng t nh mch cu Wheatstone c minh ho trongv d 11.2, v cng c th hin trn hnh 11.39. Trong mch cu, in p u ra vout c cn bng mt cch chnh xc khi chnh p sut trn u o bng 0, nhng n s lch khi 0 khi hai in dung khng bng nhau do c chnh p sut trnu o. Chng ta s phn tch mch cu trong v d 11.4.

Cun cm l tng

Mt cun cm l tng l mt phn t c kh nng tch lu nng lng trong mt t trng. Cc cun cm in hnh lcun dy qun quanh li, li c th lm bng vt liu cch in hoc st t, c minh ho trn hnh 11.40. Khi c dng inchy qua cun dy, s to ra mt t trng, bn c th xem li cc th nghim vt l v in t. Trong cun cm l tng,in tr ca dy dn bng khng, do khi dng in khng i chy qua cun cm s khng gy ra tn hao in p. Nimt cch khc, tc ng ca cun cm l tng ging nh mt phn t ngn mch trong i vi ca cc dng in DC.Trng hp nu in p thay i theo thi gian, trn cun cm s xut hin mt dng in tng ng theo mi quan h sau:

( ) LLdi

v t Ldt

= (11.36)

HNH 11.40 Cun cm li st

trong L l in cm ca cun dy v c n v l henry (H), y:

1 1 sec/H V A= (11.37)

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S tay C in t

Henry, millihenry (mH) v microhenry ( H) l cc n v ca cun cm thng c s dng trong thc t.

Dng in qua cun cm c tm thy bng cch tch phn in p trn cun cm:

1( )

t

L Li t v dt L = (11.38)

Nu dng in chy qua cun cm ti thi im t=t0 lI0 vi:

0

0 0

1

( )

t

L LI i t t v dt L = = = (11.39)th c th tm c dng in qua cun cm theo phng trnh sau:

0

00 0

1( )

t

L Lt

i t v dt I t t L

= + (11.40)Cun cm tng ng ca t hp cc cun cm mc ni tip bng tng cc cun cm. Cun cm tng ng ca t

hp cc cun cm mc song song c xc nh theo cng nguyn tc xc nh in tr tng ng ca t hp cc in trni song song. Xem hnh 11.41-11.43.

Bng 11.4 v cc hnh 11.36, 11.41 v 11.43 m t s tng t gia cc phn t c kh vi cc phn t in l tng.

Bng 11.4 S tng t gia cc bin c v bin in

H thng c H thng in

Lc,f(N) Dng in, i (A)

Vn tc, (m/sec) in th, v (V)

gim chn,B (N sec/m) in dn, 1/R (S)

mm do, 1/k(m/N) in cm,L (H)

Khi lng, M(kg) in dung, C(F)

HNH 11.41 Phng trnh m t cun cm l tng tng t vi h thng lc-l xo

HNH 11.42 Cun cm trong mt mch in HNH 11.43 S tng t gia cc phn t c kh v cc phn t in

Cc ngun tn hiu ph thuc thi gian

Hnh 11.44 a ra k hiu ca cc ngun tn hiu ph thuc thi gian.

Mt trong cc lp tn hiu ph thuc thi gian quan trng nht l cc tn hiu tun hon. Cc tn hiu ny thng xuthin trong cc ng dng thc t v l cc tn hiu c ch gn ging vi cc tn hiu ca nhiu hin tng vt l. Mt tn hiutun honx(t) l tn hiu tho mn phng trnh sau:

( ) ( ) 1, 2,3,...x t x t nT n= + = (11.41) y T l chu k cax(t). Hnh 11.45 minh ho mt s dng sng tun hon thng c s dng khi nghin cu mch

in. Cc dng sng nh sin, tam gic, vung, xung v rng ca c cung cp di dng in p (hoc i khi l dng in)

bng cch s dng cc my pht tn hiu (hoc sng) c sn. Cc thit b nh vy cho php chn bin nh sng v chuk ca n.

Nh nu trong phn gii thiu, cc sng hnh sin l lp tn hiu ph thuc thi gian quan trng nht. Hnh 11.46 thhin cc tham s ca mt sng hnh sin. Mt sng hnh sin c nh ngha nh sau:

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K thut in

( ) cos( )x t A t = + (11.42)

HNH 11.44 Cc ngun tn hiu ph thuc thi gian

HNH 11.45 Cc dng sng ca tn hiu tun hon yA l bin , l tn s o theo radian, v l gc pha. Hnh 11.46 tng hp li cc nh ngha v A, v chocc dng sng.

1 2( ) cos( ) ( ) cos( )x t A t x t A t = = +

y:

1tn s ring (chu k/s hoc Hz)f

T= =

tn s gc 2 (rad/s)f = =

2 (rad) 360 ()T T

T T

= = (11.43)

lch pha cho php biu din mt tn hiu hnh sin bt k. Do , c th biu din mt hm cosin bng cc tn hiuhnh sin bt k. V d, c th biu din mt sng sin di dng sng cosin n gin bng cch a vo mt gc lch pha

/ 2 radian:

sin( ) cos2

A t A t

=

(11.44)

iu quan trng phi ch rng, mc d chng ta lun s dng bin (c tnh theo rad/s) k hiu tn s sin,nhng n li thng lin quan ti tn s ring f c tnh theo s chu k trong mt giy, hoc hertz (Hz). Mi quan h giahai i lng trn l:

2 f = (11.45)

Gi tr trung bnh v RMS

21

HNH 11.46 Cc dng sng hnh sin


22/26

S tay C in t

trn nh ngha mt s dng sng khc nhau, cn nh ngha cc i lng o tng ng xc nh ln ca mttn hiu bin i theo thi gian. Cc kiu i lng o thng gp nht l gi tr trung bnh (hoc DC) ca sng tn hiu,tng ng vi vic o in p hoc dng in trung bnh trong mt chu k thi gian, v gi tr cn bc hai - trung bnh -bnh phng (rms), l gi tr c s dng nh gi s dao ng ca tn hiu quanh gi tr trung bnh ca n. V nguyntc, vic tnh ton gi tr trung bnh ca mt tn hiu chnh l tch phn sng tn hiu trong mt vi chu k thi gian (c lachn cho ph hp). Chng ta nh ngha gi tr trung bnh theo thi gian ca mt tn hiu x(t) nh sau:

0

1

( ) ( )

T

x t x t dt T= (11.46) y Tl chu k tch phn. Hnh 11.47 minh ho cho qu trnh tnh ton, trong thc t, chnh l vic tnh bin trung bnhcax(t) trong mt chu k Tgiy.

cos( ) 0A t + =

HNH 11.47 Tnh trung bnh ca mt sng tn hiu HNH 11.48 Mch in cha phn t tch nng lng

Mt s o in p quan trng ca mt sng AC l gi tr rms ca tn hiu x(t) c nh ngha nh sau:

2

0

1( )

T

rmsx x t dt T (11.47)Cn ch rng, khix(t) l mt in p th xrms tng ng cng phi c n v l vn. Nu phn tch phng trnh (11.47),

c th thy rng, thc cht, gi tr rms l cn bc hai ca gi tr trung bnh bnh phng ca tn hiu. Do , thut ng rms chra chnh xc cc php ton c thc hin vix(t) tnh c gi tr rms ca n.

Gii cc mch in cha cc phn t ng

im khc nhau chnh gia vic phn tch cc mch thun tr vi cc mch c cha t in v cun cm l cc phngtrnh c dn ra t vic p dng nh lut Kirchhoff khng phi l cc phng trnh i s nhn c khi gii cc mchthun in tr. Xt v d mch in trn hnh 11.48 l mch bao gm mt ngun p mc ni tip vi mt in tr, v mt tin. p dng KVL cho mch vng ny, chng ta c th suy ra c phng trnh sau:

( ) ( ) ( )S R Cv t v t v t = + (11.48)Thy rng iR = iC, kt hp phng trnh 11.48 vi phng trnh biu din t in (phng trnh 4.6.6), ta c:

1( ) ( )

t

S C Cv t Ri t i dt C = + (11.49)

Phng trnh (11.49) l mt phng trnh tch phn, c th chuyn n v mt phng trnh vi phn quen thuc hn bng

cch ly vi phn c hai v ca phng trnh v s dng:

( ) ( )t

C C

di dt i t

dt = (11.50)

ta c c phng trnh vi phn sau:

1 1C SC

di dvi

dt RC R dt + = (11.51)

y i s (t) bin mt.

T phng trnh (11.51) c th thy rng, bin c lp l dng in chy trong mch, v cng khng phi l phngtrnh duy nht m t mch inRCni tip. Nu trong v d ny, p dng KCL ti nt ni in tr vi t in, chng ta s cmi quan h sau:

S C CR C

v v dvi i CR dt = = = (11.52)

hoc:

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23/26

K thut in

1 1CC S

dvv v

dt RC RC + = (11.53)

Ch ti s ging nhau gia cc phng trnh (11.51) v (11.53). V tri ca c hai phng trnh l ng nht, ngoi trcc bin ph thuc, trong khi v phi th khc nhau. Tuy nhin, gii phng trnh trn l xc nh c tt c ccin th v dng in trong mch.

Chng ta c th tng qut ha cc kt qu trn bng cch nhn xt rng bt k mt mch in no cha mt phn t tchly nng lng u c th m t bi phng trnh vi phn c dng:

1 0

( )( ) ( )

dy ta a t F t

dt+ = (11.54)

yy(t) biu din in p ca t in trong mch trn hnh 11.48 v cc hng s a0 v a1 l t hp cc tham s ca cc phnt mch in. Phng trnh (11.54) l mt phng trnh vi phn thng bc nht vi cc h s hng.

Tip theo xt mt mch in cha hai phn t tch ly nng lng trn hnh 11.49. p dng KVL, ta c phng trnh sau:

( ) 1( ) ( ) ( )

t

S

di tRi t L i t dt v t

dt C + + = (11.55)

Phng trnh (11.55) c gi l phng trnh vi phn-o hm (integro-differential) v n c c php ton o hm v viphn trong . Phng trnh ny c th c bin i thnh mt phng trnh vi phn bng cch ly vi phn c hai v nhn c:

2

2( )( ) ( ) 1 ( ) Sdv tdi t d i t R L i t

dt C dt dt+ + = (11.56)

HNH 11.49 Mch in bc 2

hoc tng t, bng vic nhn xt rng dng in chy trong mch in ni tip lin quan ti in p ca t in nh sau: i(t)= CdvC/dtv phng trnh (11.55) c th c vit li nh sau:

2

2

( )( ) ( )C C C S

dv d v t RC LC v t v t

dt dt + + = (11.57)

Ch rng, mc d c cc bin khc nhau xut hin trong cc phng trnh vi phn trc, nhng c phng trnh (11.55)v (11.57) u c th c t chc li c cng mt dng tng qut nh sau:

2

2 1 02

( ) ( )( ) ( )

d y t dy t a a a y t F t

dtdt+ + = (11.58)

y bin tng quty(t) biu din hoc l dng in ca mch ni tip trn hnh 11.49 hoc l in p t in. Tng t nhvy vi phng trnh (11.54), chng ta gi phng trnh (11.58) l phng trnh vi phn thng bc 2 vi cc h s hng.

Do , nu tng s cc phn t tch ly nng lng trong mch in, c th c cc phng trnh vi phn bc cao hn.Pha v tr khng

Phn ny s gii thiu k hiu dng biu din cc tn hiu hnh sin di dng cc s phc, v rt ra cc yu t cn thitgii cc phng trnh vi phn.

Pha

Xin nhc li rng c th biu din mt tn hiu hnh sin tng qut di dng phn thc ca mt vc t phc vi i shoc gc ca n c a ra bi ( t + ) v chiu di hoc ln ca n bng bin nh hnh sin. Pha phc ca mt tnhiu hnh sinAcos( t + ) c nh ngha l s phc jAe :

jAe = k hiu pha phc ca cos( )A t + (11.59)

1. Mt tn hiu hnh sin bt k c th c biu din ton hc bng mt trong hai cch sau: dng min thi gian( ) cos( )v t A t = +

v dng min tn s (pha)

23


24/26

S tay C in t

( ) jV j Ae =2. Pha l mt s phc, c biu din dng cc, c ln bng vi bin nh ca tn hiu hnh sin v gc pha

bng vi lch pha ca tn hiu sin theo mt tn hiu cosin.

3. Khi s dng k hiu pha, cn ch ch thch cho tn s ring ca tn hiu sin, v n s khng xut hin mtcch tng minh trong biu din pha.

Tr khng

By gi chng ta s phn tch mi quan h i-v ca ba phn t mch in l tng bng cch s dng k hiu pha. Kt qus l mt cng thc mi vi cc in tr, t in v cun cm s c m t theo cng mt k hiu. iu ny s m rng ccnh l mch in phn 11.3 cho cc mch in AC. Trong khi nguyn cu v cc mch in AC, bt c phn t no trong3 phn t mch in l tng c nh ngha cc phn trc cng s c miu t bi mt tham s gi l tr khng,v c th thy nh mt tr khng phc.

HNH 11.50 Phn t tr khng

Khi nim tr khng tng ng vi vic coi t in v cun cm l cc in tr ph thuc tn s, tc l in tr cachng l mt hm ca tn s kch tn hiu sin. Hnh 11.50 v cng mt mch in c biu din dng thng thng (trn)v dng tr khng-pha (di); cch biu din sau th hin tng minh in p pha v dng in theo pha v coi cc phn tmch in l mt tr khng suy rng. iu ny th hin rng mi phn t trong 3 phn t mch in l tng u c th cbiu din bng mt phn t tr khng trnh by.

Khng mt tnh tng qut, nh ngha ngun p trong mch in hnh 11.50 nh sau:0( ) cos ( ) jS Sv t A t V j Ae = = (11.60)

Khi dng in i(t) c nh ngha bng mi quan h i-v cho mi phn t mch in. Chng ta s cng nghin cu ccthuc tnh ph thuc tn s ca in tr, cun cm v t in.

Tr khngca in tr c nh ngha l t s ca in p pha trn in tr vi dng in pha chy qua n, v k hiuZR c dng k hiu n:

( )( )

( )S

R

V jZ j R

I j

= = (11.61)

Tr khng ca cun cm c nh ngha nh sau:

90

( )( ) ( )

jSL

V jZ j Le j LI j

= = =o

(11.62)

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25/26

K thut in

HNH 11.51 Cc tr khng ca R, L, v C trong mt phng phc

Ch rng, lc ny cun cm s ging nh mt in tr phc ph thuc tn s, v ln L ca in tr phc ny t l

vi tn s gc ca tn hiu. Do , cun cm s cn dng in t l vi tn s ca tn hiu ngun. iu ny c ngha rngti cc tn s tn hiu thp, mt cun cm c tc dng nh phn t ngn mch, trong khi ti cc tn s cao n c xu th gingphn t h mch hn. Mt im quan trng na l tr khng ca mt cun cm lun lun dng, bi v c L v u l ccs dng. n v ca i lng ny l Ohm.

Tr khng ca t in l tng, ZC( j ) c nh ngha nh sau:

90( ) 1 1( )( )

jSC

V j jZ j e

I j C C j C

= = = =o

(11.63)

y chng ta dng0

901/ jj e j= = . V th, tr khng ca mt t in cng l mt i lng phc ph thuc tn s,v tr khng ca t in l hm nghch o ca tn s, do vy t in tc ng nh phn t ngn mch tn s cao, khi tns thp chng ging mt phn t h mch nhiu hn. Mt im quan trng khc l tr khng ca mt t in lun lun m,do c C v u l cc s dng. Cng nn nh rng n v tr khng ca mt t in l Ohm. Hnh 11.51 v Z C( j )

trong mt phng phc, cng vi ZR( j ) v ZL( j ).Cc tham s tr khng c nh ngha trong phn ny rt hu ch khi phn tch mch in AC, bi v nh chng m hu

ht cc nh l c pht trin cho mch in DC c dng trong mch AC bng cch thay th in tr bi cc tr khngphc. Dng tng qut nht, tr khng ca mt phn t mch in c nh ngha l tng ca mt phn thc v mt phn o:

( ) ( ) ( )Z j R j jX j = + (11.64) yR c gi l tr khng AC vXc gi l in khng. S ph thuc vo tn s caR vX c ch ra rt rrng. Cc v d sau s minh ha cch tnh tr khng gm c phn thc v phn o trong mt mch in.

V d 11.4 u o dch chuyn kiu in dung

Trong v d 11.3, ta xt mt u o dch chuyn kiu in dung c cu to bi mt t in c 2 bn song song trong c mt bn c nh v mt bn chuyn ng. in dung ca t in ny bin i v l mt hm phi tuyn theo v tr x ca

bn chuyn ng (xem hnh 11.39). Trong v d ny, chng ti mun ch ra rng trong cc iu kin no , tr khng ca tin s thay i nh mt hm tuyn tnh ca dch chuyn, tc l t in c bn di ng c th dng nh u o dch chuyntuyn tnh.

Nhc li cng thc trong v d 11.3:

38.854 10 AC

x

=

y Cl in dung c n v picofarad, A l din tch cc bn t n v mt vung, v x l khong dch chuyn (thay i)n v milimt. Khi t in ni vo mt mch in AC, tr khng ca n s c xc nh bng cng thc sau:

1CZ j C

=

v vy

8.854Cx

Zj A

=

25


26/26

S tay C in t

Do vy, ti mt tn s khng i , tr khng ca t in s thay i tuyn tnh theo dch chuyn. C th thy cthuc tnh ny qua cc mch cu v d 11.3, trong v d ny trnh by mt u o chnh p sut c lm t hai t inc bn t c th chuyn ng, thy rng nu tr khng ca mt t in tng do c s chnh p sut i qua u o, tr khngca t cn li phi gim i mt lng tng ng (nht l i vi trng hp cc dch chuyn nh). Mch in trn hnh 11.52gm c hai in tr v hai t in c th thay i (c biu din l C(x)) ni thnh mch cu. Mch cu ny c kch bimt tn hiu hnh sin.

S dng k hiu pha, chng ta c th biu din in p u ra nh sau:

( ) 2

( ) ( ) 1 2

( ) ( ) bc

db bc

C x

out S

C x C x

Z RV j V j

Z Z R R

= + +

in dung danh ngha ca mi t in vi mn ngn v tr chnh gia l:

AC

d

=

y dl khong cch danh ngha (khng dch chuyn) gia mn ngn v mt c nh ca t in (mm), in dung cacc t in s thay i theo lut sau:

db bc

A AC C

d x d x

= =

+

Hnh 11.52 Mch cu cho u o dch chuyn theo kiu in dung

Khi xut hin s chnh p sut trn u o, tr khng ca cc t in thay i theo dch chuyn:

8.854 8.854db bcC Cd x d xZ Z

j A j A += =

V chng ta c c cng thc tnh in p u ra dng pha nh sau, nu chn R1 =R2.

2

1 2

2

1 2

8.854( ) ( )

8.854 8.854

1( )

2 2

( ) 2

out S

S

S

d xRj A

V j V jd x d x R R

j A j A

RxV j

d R R

x

V j d

+ =

+ + +

= + +

=V vy, in p u ra s bin i tng ng vi in p u vo v t l vi dch chuyn.

Ti liu tham kho

[1] Irwin, J.D., 1989. Basic Engineering Circuit Analysis, 3rd ed., Macmillan, New York.

[2] Nilsson, J.W., 1989. Electric Circuits, 3rd ed., Addison-Wesley, Reading, MA.

[3] Rizzoni, G., 2000. Principles and Applications of Electrical Engineering, 3rd ed., McGraw-Hill, BurrRidge, IL.

[4] Smith, R.J. and Dorf, R.C., 1992. Circuits, Devices and Systems, 5th ed., John Wiley & Sons, New York.

[5] 1993. The Electrical Engineering Handbook, CRC Press, Boca Raton, FL.

[6] Budak, A., Passive and Active Network Analysis and Synthesis, Houghton Mifflin, Boston.[7] Van Valkenburg, M.E., 1982,Analog Filter Design, Holt, Rinehart & Winston, New York.

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