CHƯƠNG I NHỮNG KHÁI NIỆM CƠ BẢN - VNU-UETuet.vnu.edu.vn/~tantd/circuit/baigiangLTMach.pdf · (H 1.2) là một số trường hợp khác nhau của hàm nấc đơn v

_______________________________________________Chng 1 Nhng khi nim c bn - 1

CHNG I

NHNG KHI NIM C BN

DNG SNG CA TN HIU Hm m

Hm nc n v Hm dc

Hm xung lc Hm sin

Hm tun hon PHN T MCH IN

Phn t th ng Phn t tc ng MCH IN Mch tuyn tnh

Mch bt bin theo thi gian Mch thun nghch

Mch tp trung MCH TNG NG

Cun dy T in

Ngun c lp ________________________________________________________________

L thuyt mch l mt trong nhng mn hc c s ca chuyn ngnh in t-Vin

thng-T ng ha.

Khng ging nh L thuyt trng - l mn hc nghin cu cc phn t mch in nh t in, cun dy. . . gii thch s vn chuyn bn trong ca chng - L thuyt mch ch quan tm n hiu qu khi cc phn t ny ni li vi nhau to thnh mch in (h thng).

Chng ny nhc li mt s khi nim c bn ca mn hc.

1.1 DNG SNG CA TN HIU

Tn hiu l s bin i ca mt hay nhiu thng s ca mt qu trnh vt l no theo qui lut ca tin tc.

Trong phm vi hp ca mch in, tn hiu l hiu th hoc dng in. Tn hiu c th c tr khng i, v d hiu th ca mt pin, accu; c th c tr s thay i theo thi gian, v d dng in c trng cho m thanh, hnh nh. . . .

Tn hiu cho vo mt mch c gi l tn hiu vo hay kch thch v tn hiu nhn c ng ra ca mch l tn hiu ra hay p ng.

Ngi ta dng cc hm theo thi gian m t tn hiu v ng biu din ca chng trn h trc bin - thi gian c gi l dng sng.

Di y l mt s hm v dng sng ca mt s tn hiu ph bin.

___________________________________________________________________________ Nguyn Trung Lp L THUYT MCH


1.1.1 Hm m (Exponential function) t)( = Ketv K , l cc hng s thc.

(H 1.1) l dng sng ca hm m vi cc tr khc nhau

(H 1.1)

1.1.2 Hm nc n v (Unit Step function)

(H 1.4b).

Vi cc tr khc nhau ca ta c cc tr khc nhau ca f0(t) nhng phn din tch gii hn gia f0(t) v trc honh lun lun =1 (H 1.4c).

Khi 0, f1(t) u(t) v f0(t) (t). Vy xung lc n v c xem nh tn hiu c b cao cc ln v b rng cc nh v

din tch bng n v (H 1.4d). Tng qut, xung lc n v ti t=a, (t-a) xc nh bi:

0 v A l s thc dng (H 1.5a) Tch hai hm sin c tn s khc nhau

v(t)=Asin1t.sin2t (H 1.5b)

(a) (H 1.5) (b)

1.1.6 Hm tun hon khng sin Ngoi cc tn hiu k trn, chng ta cng thng gp mt s tn hiu nh: rng ca,

hnh vung, chui xung. . . . c gi l tn hiu khng sin, c th l tun hon hay khng. Cc tn hiu ny c th c din t bi mt t hp tuyn tnh ca cc hm sin, hm m v cc hm bt thng.

(H 1.6) m t mt s hm tun hon quen thuc

(H 1.6)

1.2 PHN T MCH IN

S lin h gia tn hiu ra v tn hiu vo ca mt mch in ty thuc vo bn cht v ln ca cc phn t cu thnh mch in v cch ni vi nhau ca chng.

Ngi ta phn cc phn t ra lm hai loi: Phn t th ng: l phn t nhn nng lng ca mch. N c th tiu tn nng

lng (di dng nhit) hay tch tr nng lng (di dng in hoc t trng). Gi v(t) l hiu th hai u phn t v i(t) l dng in chy qua phn t. Nng lng

ca on mch cha phn t xc nh bi:

=t

(t)dt(t).W(t) iv

- Phn t l th ng khi W(t) 0, ngha l dng in i vo phn t theo chiu gim ca in th.


_______________________________________________Chng 1 Nhng khi nim c bn - 5 in tr, cun dy v t in l cc phn t th ng.

Phn t tc ng: l phn t cp nng lng cho mch ngoi. Nng lng ca on mch cha phn t W(t)


1.2.1.3 T in

(a) (H 1.9) (b)

- K hiu (H 1.9a)

- H thc: dt

(t)dC(t) vi =

- Hay =t

(t)dtC1(t) iv

n v ca t in l F (Farad) Do t in l phn t tch tr nng lng nn thi im t0 no c th n tr

mt nng lng in trng ng vi hiu th v(t0) Biu thc vit li:

)(t(t)dtC1(t) 0

t

t 0viv +=

V mch tng ng ca t in c v nh (H 1.9b)

Nng lng tch tr trong t in

=t

(t)dt(t).W(t) iv

Thay dt

(t)dC(t) vi =

0(t)C21](t)C

21(t)dCW(t) 2t2

t=== vvvv (v v(-)=0)

Ch : Trong cc h thc v-i ca cc phn t R, L, C nu trn, nu i chiu mt trong hai lng v hoc i th h thc i du (H 1.10): v(t) = - R.i(t)

(H 1.10)

1.2.2 Phn t tc ng y ch cp n mt s phn t tc ng n gin, l cc loi ngun.

Ngun l mt phn t lng cc nhng khng c mi quan h trc tip gia hiu th v hai u v dng in i i qua ngun m s lin h ny hon ton ty thuc vo mch ngoi, do khi bit mt trong hai bin s ta khng th xc nh c bin s kia nu khng r mch ngoi.



1.2.2.1 Ngun c lp

L nhng phn t m gi tr ca n c lp i vi mch ngoi - Ngun hiu th c lp: c gi tr v l hng s hay v(t) thay i theo thi gian. Ngun hiu

th c gi tr bng khng tng ng mt mch ni tt - Ngun dng in c lp: c gi tr i l hng s hay i(t) thay i theo thi gian. Ngun

dng in c gi tr bng khng tng ng mt mch h

(H 1.11)

1.2.2.2 Ngun ph thuc

Ngun ph thuc c gi tr ph thuc vo hiu th hay dng in mt nhnh khc trong mch. Nhng ngun ny c bit quan trng trong vic xy dng mch tng ng cho cc linh kin in t.

C 4 loi ngun ph thuc:

- Ngun hiu th ph thuc hiu th (Voltage-Controlled Voltage Source, VCVS) - Ngun hiu th ph thuc dng in (Current-Controlled Voltage Source, CCVS) - Ngun dng in ph thuc hiu th(Voltage-Controlled Current Source, VCVS) - Ngun dng in ph thuc dng in (Current-Controlled Current Source, CCCS)

(a)VCVS (b) CCVS

(c)VCCS (d) CCCS (H 1.12)

1.3 MCH IN

C hai bi ton v mch in: - Phn gii mch in: cho mch v tn hiu vo, tm tn hiu ra. - Tng hp mch in: Thit k mch khi c tn hiu vo v ra.

Gio trnh ny ch quan tm ti loi bi ton th nht. Quan h gia tn hiu vo x(t) v tn hiu ra y(t) l mi quan h nhn qu ngha l tn

hiu ra hin ti ch ty thuc tn hiu vo qu kh v hin ti ch khng ty thuc tn hiu


_______________________________________________Chng 1 Nhng khi nim c bn - 8 vo tng lai, ni cch khc, y(t) thi im t0 no khng b nh hng ca x(t) thi im t>t0 .

Tn hiu vo thng l cc hm thc theo thi gian nn p ng cng l cc hm thc theo thi gian v ty thuc c tn hiu vo v c tnh ca mch.

Di y l mt s tnh cht ca mch da vo quan h ca y(t) theo x(t).

1.3.1 Mch tuyn tnh Mt mch gi l tuyn tnh khi tun theo nh lut: Nu y1(t) v y2(t) ln lt l p ng ca hai ngun kch thch c lp vi nhau x1(t)

v x2(t), mch l tuyn tnh nu v ch nu p ng i vi x(t)= k1x1(t) + k2x2(t)

l y(t)= k1y1(t) + k2y2(t) vi mi x(t) v mi k1 v k2.

Trn thc t, cc mch thng khng hon ton tuyn tnh nhng trong nhiu trng hp s bt tuyn tnh khng quan trng v c th b qua. Th d cc mch khuch i dng transistor l cc mch tuyn tnh i vi tn hiu vo c bin nh. S bt tuyn tnh ch th hin ra khi tn hiu vo ln.

Mch ch gm cc phn t tuyn tnh l mch tuyn tnh.

Th d 1.1 Chng minh rng mch vi phn, c trng bi quan h gia tn hiu vo v ra theo h

thc:

dtdx(t)y(t) = l mch tuyn tnh

Gii

Gi y1(t) l p ng i vi x1(t): dt

(t)dx(t)y 11 =

Gi y2(t) l p ng i vi x2(t): dt

(t)dx(t)y 22 =

Vi x(t)= k1x1(t) + k2 x2(t) p ng y(t) l:

dt(t)dxk

dt(t)dxk

dtdx(t)y(t) 2211 +==

y(t)=k1y1(t)+k2y2(t) Vy mch vi phn l mch tuyn tnh

1.3.2 Mch bt bin theo thi gian (time invariant)

Lin h gia tn hiu ra v tn hiu vo khng ty thuc thi gian. Nu tn hiu vo tr t0 giy th tn hiu ra cng tr t0 giy nhng ln v dng khng i.

Mt hm theo t tr t0 giy tng ng vi ng biu din tnh tin t0 n v theo chiu dng ca trc t hay t c thay th bi (t-t0). Vy, i vi mch bt bin theo thi gian, p ng i vi x(t-t0) l y(t-t0) Th d 1.2

Mch vi phn th d 1.1 l mch bt bin theo thi gian Ta phi chng minh p ng i vi x(t-t0) l y(t-t0). Tht vy:

)x1ty(td(t)

)td(tx)td(t)tdx(t

dt)tdx(t

00

0

00 =

=



minh ha, cho x(t) c dng nh (H 1.13a) ta c y(t) (H 1.13b). Cho tn hiu vo tr (1/2)s, x(t-1/2) (H 1.13c), ta c tn hiu ra cng tr (1/2)s, y(t-1/2) c v (H 1.13d).

(a) (b)

(c) (d) (H 1.13)

1.3.3 Mch thun nghch Xt mch (H 1.14)

___________________________________________________________________________

(H 1.14)

+

v1

Nu tn hiu vo cp cc 1 l v1 cho p ng cp cc 2 l dng in ni tt i2 . By

gi, cho tn hiu v1 vo cp cc 2 p ng cp cc 1 l i2. Mch c tnh thun nghch khi i2=i2.

1.3.4 Mch tp trung Cc phn t c tnh tp trung khi c th coi tn hiu truyn qua n tc thi. Gi i1 l dng in vo phn t v i2 l dng in ra khi phn t, khi i2= i1 vi mi t ta ni phn t c tnh tp trung. (H 1.15)

Mt mch ch gm cc phn t tp trung l mch tp trung.. Vi mt mch tp trung ta c mt s im hu hn m trn c th o nhng tn

hiu khc nhau. Mch khng tp trung l mt mch phn tn. Dy truyn sng l mt th d ca mch

phn tn, n tng ng vi cc phn t R, L v C phn b u trn dy. Dng in truyn trn dy truyn sng phi tr mt mt thi gian n ng ra.

Mch

i2

+

v1

Mch

i2

Phnt

i1i2

Nguyn Trung Lp L THUYT MCH


1.4 MCH TNG NG

Cc phn t khi cu thnh mch in phi c biu din bi cc mch tng ng. Trong mch tng ng c th cha cc thnh phn khc nhau

Di y l mt s mch tng ng trong thc t ca mt s phn t.

1.4.1 Cun dy

(H 1.16)

Cun dy l tng c c trng bi gi tr in cm ca n. Trn thc t, cc vng dy c in tr nn mch tng ng phi mc ni tip thm mt in tr R v chnh xc nht cn k thm in dung ca cc vng dy nm song song vi nhau

1.4.2 T in

(a) (b) (c)

(H 1.17) (H 1.17a ) l mt t in l tng, nu k in tr R1 ca lp in mi, ta c mch

tng (H 1.17b ) v nu k c in cm to bi cc lp dn in (hai m ca t in) cun thnh vng v in tr ca dy ni ta c mch tng (H 1.17c )

1.4.3 Ngun c lp c gi tr khng i

1.4.3.1 Ngun hiu th

Ngun hiu th cp n trn l ngun l tng. Gi v l hiu th ca ngun, v0 l hiu th gia 2 u ca ngun, ni ni vi mch

ngoi, dng in qua mch l i0 (H 1.18a). Nu l ngun l tng ta lun lun c v0 = v khng i. Trn thc t, gi tr v0 gim khi i0 tng (H 1.18c); iu ny c ngha l bn trong ngun c mt in tr m ta gi l ni tr ca ngun, in tr ny to mt st p khi c dng in chy qua v st p cng ln khi i0 cng ln. Vy mch tng ng ca ngun hiu th c dng (H 1.18b)



(a) (b) (c)

(H 1.18)

1.4.3.2 Ngun dng in

Tng t, ngun dng in thc t phi k n ni tr ca ngun, mc song song vi ngun trong mch tng ng v in tr ny chnh l nguyn nhn lm gim dng in mch ngoi i0 khi hiu th v0 ca mch ngoi gia tng.

(H 1.19)

BI TP

-- --

1. V dng sng ca cc tn hiu m t bi cc phng trnh sau y:

a. vi T=1s =

10

nnT)(t

1

b. u(t)sinT

t2 v u(t-T/2)sinT

t2

c. r(t).u(t-1), r(t)-r(t-1)-u(t-1) 2. Cho tn hiu c dng (H P1.1).

Hy din t tn hiu trn theo cc hm:

___________________________________________________________________________

a. u(t-a) v u(t-b) b. u(b-t) v u(a-t) c. u(b-t) v u(t-a)

(H P1.1) 3.Vit phng trnh dng sng ca cc tn hiu khng tun

hon (H P1.2) theo tp hp tuyn tnh ca cc hm bt thng (nc, dc), sin v cc hm khc (nu cn)



(a) (b) (H P1.2) 4. Cho tn hiu c dng (H P1.3)

(H P1.3) (H P1.4) a. Vit phng trnh dng sng ca cc tn hiu theo tp hp tuyn tnh ca cc hm sin v

cc hm nc n v. b. Xem chui xung c dng (H P1.4) Chui xung ny c dng ca cc cng, khi xung c gi tr 1 ta ni cng m v khi tr ny =0 ta ni cng ng. Ta c th din t mt hm cng m thi im t0 v ko di mt khong thi gian T bng mt hm cng c k hiu:

T)tu(t)tu(t(t) 00T,t 0 = Th din t tn hiu (H P1.3) bng tch ca mt hm sin v cc hm cng. 5. Cho kin v tnh tuyn tnh v bt bin theo t ca cc tn hiu sau:

a. y =x2

b. y =tdtdx

c. y =xdtdx

6. Cho mch (H P1.6a) v tn hiu vo (H P1.6b) Tnh p ng v v dng sng ca p ng trong 2 trng hp sau (cho vC(0) = 0): a. Tn hiu vo x(t) l ngun hiu th vC v p ng l dng in iC. b. Tn hiu vo x(t) l iC ngun hiu th v p ng l dng in vC. Bng di y cho ta d kin ca bi ton ng vi cc (H 5a, b, c...) km theo. Tnh p ng v v dng sng ca p ng

(a) (b) (c)

(H P1.6)



(a) (b) (c)

(d) (e) (f)

(H P1.5)

Cu Mch hnh Kch thch x(t) Dng sng p ng a b c d e f g h

a a a a b b b b

vc vc ic ic vL vL iL iL

d f c d c d e f

ic ic vc vc iL iL vL vL


_________________________________________Chng2nhlutvnhlmchi n

1

___________________________________________________________________________

CHNG 2

NH LUT V NH L MCH IN

NH LUT KIRCHHOF

IN TR TNG NG

NH L MILLMAN

NH L CHNG CHT

NH L THEVENIN V NORTON

BIN I Y (NH L KENNELY)

_______________________________________________________________________________________________

Chng ny cp n hai nh lut quan trng lm c s cho vic phn gii mch, l cc nh lut Kirchhoff.

Chng ta cng bn n mt s nh l v mch in. Vic p dng cc nh l ny gip ta gii quyt nhanh mt s bi ton n gin hoc bin i mt mch in phc tp thnh mt mch n gin hn, to thun li cho vic p dng cc nh lut Kirchhoff gii mch.

Trc ht, n gin, chng ta ch xt n mch gm ton in tr v cc loi ngun, gi chung l mch DC. Cc phng trnh din t cho loi mch nh vy ch l cc phng trnh i s (i vi mch c cha L & C, ta cn n cc phng trnh vi tch phn)

Tuy nhin, khi kho st v ng dng cc nh l, chng ta ch ch n cu trc ca mch m khng quan tm n bn cht ca cc thnh phn, do cc kt qu trong chng ny cng p dng c cho cc trng hp tng qut hn.

Trong cc mch DC, p ng trong mch lun lun c dng ging nh kch thch, nn n gin, ta dng kch thch l cc ngun c lp c gi tr khng i thay v l cc hm theo thi gian.

2.1 nh lut kirchhoff

Mt mch in gm hai hay nhiu phn t ni vi nhau, cc phn t trong mch to thnh nhng nhnh. Giao im ca hai hay nhiu nhnh c gi l nt. Thng ngi ta coi nt l giao im ca 3 nhnh tr nn. Xem mch (H 2.1).

(H 2.1)

- Nu xem mi phn t trong mch l mt nhnh mch ny gm 5 nhnh v 4 nt. - Nu xem ngun hiu th ni tip vi R1 l mt nhnh v 2 phn t L v R2 l mt

nhnh (trn cc phn t ny c cng dng in chy qua) th mch gm 3 nhnh v 2 nt. Cch sau thng c chn v gip vic phn gii mch n gin hn.

Nguyn Minh Lun K THUT IN T


2

___________________________________________________________________________

Hai nh lut c bn lm nn tng cho vic phn gii mch in l:

2.1.1. nh lut Kirchhoff v dng in : ( Kirchhoff's Current Law, KCL )

Tng i s cc dng in ti mt nt bng khng .

(2.1) 0j

j = iij l dng in trn cc nhnh gp nt j.

Vi qui c: Dng in ri khi nt c gi tr m v dng in hng vo nt c gi tr dng (hay ngc li).

(H 2.2)

Theo pht biu trn, ta c phng trnh nt A (H 2.2):

i1 + i 2 - i 3 + i 4=0 (2.2) Nu ta qui c du ngc li ta cng c cng kt qu:

- i 1 - i 2 + i 3 - i 4 =0 (2.3) Hoc ta c th vit li:

i 3 = i 1 + i 2 + i 4 (2.4)

V t phng trnh (2.4) ta c pht biu khc ca nh lut KCL: Tng cc dng in chy vo mt nt bng tng cc dng in chy ra khi nt

.

nh lut Kirchhoff v dng in l h qu ca nguyn l bo ton in tch:

Ti mt nt in tch khng c sinh ra cng khng b mt i.

Dng in qua mt im trong mch chnh l lng in tch i qua im trong mt n v thi gian v nguyn l bo ton in tch cho rng lng in tch i vo mt nt lun lun bng lng in tch i ra khi nt .

2.1.2. nh lut Kirchhoff v in th: ( Kirchhoff's Voltage Law, KVL ). Tng i s hiu th ca cc nhnh theo mt vng kn bng khng

(2.5) 0(t)K

K =v p dng nh lut Kirchhoff v hiu th, ta chn mt chiu cho vng v dng qui

c: Hiu th c du (+) khi i theo vng theo chiu gim ca in th (tc gp cc dng trc) v ngc li.

nh lut Kirchhoff v hiu th vit cho vng abcd ca (H 2.3).



3

___________________________________________________________________________

- v1 + v 2 - v 3 = 0

(H 2.3)

Ta cng c th vit KVL cho mch trn bng cch chn hiu th gia 2 im v xc nh hiu th theo mt ng khc ca vng:

v1 = vba = vbc+ vca = v2 - v3nh lut Kirchhoff v hiu th l h qu ca nguyn l bo ton nng lng: Cng

trong mt ng cong kn bng khng. V tri ca h thc (2.5) chnh l cng trong dch chuyn in tch n v (+1) dc

theo mt mch kn.

Th d 2.1 . Tm ix v vx trong (H2.4)

(H 2.4)

Gii: p dng KCL ln lt cho cc cho nt a, b, c, d

- i1 - 1 + 4 = 0 i1 = 3A

- 2A + i1 + i2 = 0 i2 = -1A

- i3 + 3A - i2 = 0 i3 = 4A

ix + i3 + 1A = 0 ix = - 5A

p dng nh lut KVL cho vng abcd:

- vx - 10 + v2 - v3 = 0 Vi v2 = 5 i2 = 5.( - 1) = - 5V v3 = 2 i3 = 2.( 4) = 8V vx =- 10 - 5 - 8 = -23V

Trong th d trn , ta c th tnh dng ix t cc dng in bn ngoi vng abcd n cc nt abcd. Xem vng abcd c bao bi mt mt kn ( v nt gin on).

nh lut Kirchhoff tng qut v dng in c th pht biu cho mt kn nh sau: Tng i s cc dng in n v ri khi mt kn bng khng.

Vi qui c du nh nh lut KCL cho mt nt. Nh vy phng trnh tnh ix l:



4

___________________________________________________________________________

- ix - 4 + 2 - 3 = 0 Hay ix = - 5 A nh lut c th c chng minh d dng t cc phng trnh vit cho cc nt abcd

cha trong mt kn c dng in t cc nhnh bn ngoi n. Th d 2.2:

L v R trong mch (H 2.5a) din t cun lch ngang trong TiVi nu L = 5H, R = 1 v dng in c dng sng nh (H 2.5b). Tm dng sng ca ngun hiu th v(t).

(a) (b)

(H 2.5) Gii:

nh lut KVL cho :

- v(t) + v R(t) + v L(t) = 0 (1)

hay v (t) = v R + v L(t) = Ri(t) + ( )

dttdL i

Thay tr s ca R v L vo:

v L(t) = ( )

dttd5 i (2)

v R(t) = 1. i(t) (3)

V v (t) = i(t) + ( )dt

td5 i (4)

Da vo dng sng ca dng in i(t), suy ra o hm ca i(t) v ta v c dng sng

ca vL(t) (H 2.6a) v v(t) (H 2.6b) t cc phng trnh (2), (3) v (4).

(a) (H 2.6) (b)



5

___________________________________________________________________________

2.2 in tr tng ng

Hai mch gi l tng ng vi nhau khi ngi ta khng th phn bit hai mch ny bng cch o dng in v hiu th nhng u ra ca chng.

Hai mch lng cc A v B (H 2.7) tng ng nu v ch nu:

ia = ib vi mi ngun v

(H 2.7)

Di y l pht biu v khi nim in tr tng ng: Bt c mt lng cc no ch gm in tr v ngun ph thuc u tng ng

vi mt in tr. in tr tng ng nhn t hai u a & b ca mt lng cc c nh ngha:

Rt = iv (2.6)

Trong v l ngun bt k ni vo hai u lng cc.

(H 2.8)

Th d 2.3: Mch (H 2.9a) v (H 2.9b) l cu chia in th v cu chia dng in. Xc nh cc

in th v dng in trong mch.

(a) (H 2.9) (b) Gii:

a/ (H 2.9a) cho v = v1+ v2 = R1 i + R2 i= (R1 + R2) i

Rt = iv = R1 + R2

T cc kt qu trn suy ra : i 21 RR +

=v



6

___________________________________________________________________________

v1 = R1 i v21

1

RRR+

= v v2 = R2 i v21

2

RRR+

=

b/ (H 2.9b) cho

i = i1+ i2 hay 21t RRR

vvv+=

21t R

1R1

R1

+= hay Gt = G1+ G2

T cc kt qu trn suy ra: v i21 GG

1+

=

i1 = G1v = ii21

2

21

1

RRR

GGG

+=

+ v i2 = G2v = ii

21

1

21

2

RRR

GGG

+=

+

Th d 2.4: Tnh Rt ca phn mch (H 2.10a)

(a) (b)

(H 2.10) Gii:

Mc ngun hiu th v vo hai u a v b nh (H2.10b) v ch i = i1.

nh lut KCL cho i1 = i3 + 131 i i3 = 13

2 i

Hiu th gia a &b chnh l hiu th 2 u in tr 3

v = 3i3 = 2i1 = 2i Rt = iv = 2

2.3. nh l Millman

nh l Millman gip ta tnh c hiu th hai u ca mt mch gm nhiu nhnh mc song song.

Xt mch (H 2.11), trong mt trong cc hiu th Vas = Va - Vs ( s = 1,2,3 ) c th trit tiu.

(H 2.11)

nh l Millman p dng cho mch (H 2.11) c pht biu:



7

___________________________________________________________________________

vab =

s

s

s

sas

G

G.v (2.7)

Vi Gs = sR

1 l in dn nhnh s.

Chng minh: Gi vsb l hiu th hai u ca Rs: vsb = vab - vasDng in qua Rs:

is = sasabs

asab

s

sb GRR

)( vvvvv ==

Ti nt b : = 0 s

Si

G( ) s

asab vv s = 0

Hay =s

sass

sab GG vv

vab =

ss

ssas

G

Gv

Th d 2.5 Dng nh l Millman, xc nh dng in i2 trong mch (H 2.12).

(H 2.12)

ta c vab =

51612,88

2511

0,56,4

18

+=

++

+

vab = 6,5 V

Vy i2 = 5

6,5 = 1,3 A

2.4. nh l chng cht ( superposition theorem)

nh l chng cht l kt qu ca tnh cht tuyn tnh ca mch: p ng i vi nhiu ngun c lp l tng s cc p ng i vi mi ngun ring l. Khi tnh p ng i vi mt ngun c lp, ta phi trit tiu cc ngun kia (Ni tt ngun hiu th v h ngun dng in, tc ct b nhnh c ngun dng in), ring ngun ph thuc vn gi nguyn. Th d 2.6

Tm hiu th v2 trong mch (H 2.13a).



8

___________________________________________________________________________

(a) (b) (c)

(H 2.13) - Cho ngun i3 = 0A ( h nhnh cha ngun 3A), ta c mch (H 2.13b):

v'2 = 1,8V64

61 =+

v (dng cu phn th)

- Cho ngun v1 = 0V (ni tt nhnh cha ngun 3V), mch (H 2.13c).

Dng in qua in tr 6: 246

4+

= 0,8A (dng cu phn dng)

v''2 = - 0,8 x 6 = - 4,8 V Vy v2 = v'2 + v''2 = 1,8 - 4,8 = - 3V

v2 = - 3V Th d 2.7 Tnh v2 trong mch (H 2.14a).

(a) (b)

(c)

(H 2.14)

Gii: - Ct ngun dng in 3A, ta c mch(H 2.14b).

i1 = A21

42=

i3 = 2i1 = 1A v'2 = 2 - 3i3 = -1 V - Ni tt ngun hiu th 2 V, ta c mch (H 2.14c). in tr 4 b ni tt nn i1 = 0 A Vy i3 = 3A v''2 = - 3 x 3 = - 9 V Vy v2 = v'2 + v''2 = -1 - 9 = -10 V



9

___________________________________________________________________________

2.5. nh l Thevenin v Norton

nh l ny cho php thay mt phn mch phc tp bng mt mch n gin ch gm mt ngun v mt in tr.

Mt mch in gi s c chia lm hai phn (H 2.15)

(H 2.15)

nh l Thevenin v Norton p dng cho nhng mch tha cc iu kin sau: * Mch A l mch tuyn tnh, cha in tr v ngun. * Mch B c th cha thnh phn phi tuyn. * Ngun ph thuc, nu c, trong phn mch no th ch ph thuc cc i lng nm

trong phn mch . nh l Thevenin v Norton cho php chng ta s thay mch A bng mt ngun v

mt in tr m khng lm thay i h thc v - i hai cc a & b ca mch . Trc tin, xc nh mch tng ng ca mch A ta lm nh sau: Thay mch B

bi ngun hiu th v sao cho khng c g thay i lng cc ab (H2.16).

(H 2.16)

p dng nh l chng cht dng in i c th xc nhbi:

i = i1 + isc (2.8) Trong i1 l dng in to bi ngun v mch A trit tiu cc ngun c lp

(H2.17a) v isc l dng in to bi mch A vi ngun v b ni tt (short circuit, sc) (H2.17b).

(a) (H 2.17) (b)

- Mch th ng A, tng ng vi in tr Rth, gi l in tr Thevenin, xc nh bi:

i1 =- thR

v (2.9)

Thay (2.9) vo (2.8)

i = - thR

v + isc (2.10)



10

___________________________________________________________________________

H thc (2.10) din t mch A trong trng hp tng qut nn n ng trong mi trng hp.

Trng hp a, b h (Open circuit), dng i = 0 A, phng trnh (2.10) thnh:

0 = th

oc

Rv

+ isc

Hay voc = Rth . isc (2.11) Thay (2.11) vo (2.10):

v = - Rth . i + voc (2.12)

H thc (2.12) v (2.10) cho php ta v cc mch tng ng ca mch A (H 2.18)

v (H 2.19)

(H 2.18) (H 2.19)

* (H 2.18) c v t h thc (2.12) c gi l mch tng ng Thevenin ca

mch A (H 2.15). V ni dung ca nh l c pht biu nh sau: Mt mch lng cc A c th c thay bi mt ngun hiu th voc ni tip vi

mt in tr Rth. Trong voc l hiu th ca lng cc A h v Rth l in tr nhn t lng cc khi trit tiu cc ngun c lp trong mch A (Gi nguyn cc ngun ph thuc).

Rth cn c gi l in tr tng ng ca mch A th ng.

* (H 2.19) c v t h thc (2.10) c gi l mch tng ng Norton ca mch A (H 2.15). V nh l Norton c pht biu nh sau:

Mt mch lng cc A c th c thay th bi mt ngun dng in isc song song vi in tr Rth. Trong isc l dng in lng cc khi ni tt v Rth l in tr tng ng mch A th ng. Th d 2.8

V mch tng ng Thevenin v Norton ca phn nm trong khung ca mch (H2.20).

(H 2.20)

Gii: c mch tng ng Thevenin, ta phi xc nh c Rth v voc.

Xc nh Rth Rth l in tr nhn t ab ca mch khi trit tiu ngun c lp. (H 2.21a).



11

___________________________________________________________________________

T (H 2.21a) :

Rth = 2 + 363x6

+ = 4

(a) (b)

(H 2.21) Xc nh voc

voc l hiu th gia a v b khi mch h (H 2.21b). V a, b h, khng c dng qua in tr 2 nn voc chnh l hiu th vcb. Xem nt b lm chun ta c

vd = - 6 + vc = - 6 + voc /L KCL nt b cho :

2A6

63

ococ =

+vv

Suy ra voc = 6 V Vy mch tng ng Thevenin (H2.22)

(H 2.22) (H 2.23)

c mch tng ng Norton, Rth c, ta phi xc nh isc. Dng isc chnh l dng

qua ab khi nhnh ny ni tt. Ta c th xc nh t mch (H 2.20) trong ni tt ab. Nhng ta cng c th dng h thc (2.11) xc nh isc theo voc:

isc = 1,5A46

Rthoc ==

v

Vy mch tng ng Norton (H 2.23) Th d 2.9

V mch tng ng Norton ca mch (H 2.24a).

(a)



12

___________________________________________________________________________

(b) (c)

(H 2.24) Ta tm isc t mch (H 2.24c) KCL nt b cho:

i1 = 10 - i2 - iscVit KVL cho 2 vng bn phi:

-4(10 - i2 - isc) - 2i1 + 6i2 = 0

- 6i2 + 3isc = 0 Gii h thng cho isc = 5A

tnh Rth (H 2.24b), do mch c cha ngun ph thuc, ta c th tnh bng cch p vo a,b mt ngun v ri xc nh dng in i, c Rth = v/i ( in tr tng ng ).

Tuy nhin, y ta s tm voc ab khi a,b h (H 2.25).

(H 2.25)

Ta c voc = 6i2Vit nh lut KVL cho vng cha ngun ph thuc :

-4(10 - i2) - 2 i1+ 6i2 = 0 Hay i2 = 5 A v voc = 6 x 5 = 30 V

Vy Rth = == 6530

sc

oc

iv

Mch tng ng Norton:

(H 2.26)

Th d 2.10: Tnh vo trong mch (H 2.27a) bng cch dng nh l Thevenin



13

___________________________________________________________________________

(a) (b)

(c) (H 2.27) (d)

c mch th ng, ni tt ngun v1 nhng vn gi ngun ph thuc 1/3 i1, ta c

mch (H 2.27c). Mch ny ging mch (H 2.10) trong th d 2.4; Rth chnh l Rt trong th d 2.4.

Rth = 2 tnh voc, ta c mch (H2.27b)

voc = v5 + v1 v5 = 3i5 i4 = 0 A ( mch h ) nn:

i5= A32

24x

31

2x

31

31 1

1 ===vi voc = 3

32 + 4 = 6 V

voc = 6 V Mch tng ng Thevenin v (H 2.27d).

v vo = V51012610

102oc ==+v

vo = 5 V

2.6. Bin i - Y ( nh l Kennely ).

Coi mt mch gm 3 in tr Ra, Rb, Rc ni nhau theo hnh (Y), ni vi mch ngoi ti 3 im a, b, c im chung O (H 2.28a). V mch gm 3 in tr Rab, Rbc, Rca ni nhau theo hnh tam gic (), ni vi mch ngoi ti 3 im a, b, c (H 2.28b).



14

___________________________________________________________________________

(H 2.28)

Hai mch v Y tng ng khi mch ny c th thay th mch kia m khng nh hng n mch ngoi, ngha l cc dng in ia, ib, ic i vo cc nt a, b, c v cc hiu th vab,vbc, vca gia cc nt khng thay i.

- Bin i Y l thay th cc mch bng cc mch Y v ngc li. Ngi ta chng minh c : Bin i Y :

Rab =R R R R R R

Ra b b c c a

c

+ +

Rbc = R R R R R R

Ra b b c c a

a

+ + (2.13)

Rca = R R R R R R

Ra b b c c a

b

+ +

Bin i Y:

Ra =R R

R R Rab ca

ab bc ca

.+ +

Rb = R R

R R Rab bc

ab bc ca

.+ +

(2.14)

Rc = R R

R R Rbc ca

ab bc ca

.+ +

Nn thn trng khi p dng bin i Y. Vic p dng ng phi cho mch tng ng n gin hn. Th d 2.11:

Tm dng in i trong mch (H 2.29a).



15

___________________________________________________________________________

(a) (b) (c) (d)

(H 2.29) - Bin i tam gic abc thnh hnh sao, ta c (H 2.29b) vi cc gi tr in tr:

Raf = ==++

0,854

1222x2

Rbf = == 0,452

52x1

Rcf = == 0,452

52x1

- in tr tng ng gia f v d:

2,41,41,4x2,4

+= 0,884

- in tr gia a v e: Rac = 0,8 + 0,884 +1 = 2,684

v dng in i trong mch :

i = 2,684Rac

vv= A

2.7 Mch khuch i thut ton ( Operation amplifier, OPAMP )

Mt trong nhng linh kin in t quan trng v thng dng hin nay l mch khuch i thut ton ( OPAMP ).

Cu to bn trong mch s c gii thiu trong mt gio trnh khc. y chng ta ch gii thiu mch OPAMP c dng trong mt vi trng hp ph bin vi mc ch xy dng nhng mch tng ng dng ngun ph thuc cho n t cc nh lut Kirchhoff .

OPAMP l mt mch a cc, nhng n gin ta ch n cc ng vo v ng ra (b qua cc cc ni ngun v Mass...). Mch c hai ng vo (a) l ng vo khng o, nh du (+) v (b) l ng vo o nh du (-), (c) l ng ra.



16

___________________________________________________________________________

(H 2.30)

Mch c nhiu c tnh quan trng , y ta xt mch trong iu kin l tng: i1 v i2 dng in cc ng vo bng khng (tc tng tr vo ca mch rt ln) v hiu th gia hai ng vo cng bng khng .

Lu l ta khng th dng nh lut KCL tng qut cho mch (H 2.30) c v ta b qua mt s cc do mc d i1 = i2 = 0 nhng i3 0.

Mch OPAMP l tng c li dng in nn trong thc t khi s dng ngi ta lun dng mch hi tip.

Trc tin ta xt mch c dng (H 2.31a), trong R2 l mch hi tip mc t ng ra (c) tr v ng vo o (b), v mch (H 2.31b) l mch tng ng .

(a) (b) (c)

(H 2.31)

v mch tng ng ta tm lin h gia v2 v v1. p dng cho KVL cho vng obco qua v2

vbc + v2 - vbo = 0 Hay vbc = vbo - v2 = v1 - v2 (vbo = v1)

p dng KCL nt b:

0RRRR 2

21

1

1

2

bc

1

bo =

+=+vvvvv

Gii phng trnh cho: v2 = Av v1 vi Av = 1 + 1

2

RR

Ta c mch tng ng (H 2.31b), trong Av l li in th. Xt trng hp c bit R2 = 0 v R1 = , Av = 1 v v2 = v1 (H 2.31c) mch khng

c tnh khuch i v c gi l mch m ( Buffer ), c tc dng bin i tng tr. Mt dng khc ca mch OP-AMP v (H 2.32a) Ap dng KCL ng vo o.

0RR 2

2

1

1 =vv hay v2 = 1

1

2

RR v

Ta thy v2 c pha o li so vi v1 nn mch c gi l mch o. Mch tng ng v (H 2.31b), dng ngun hiu th ph thuc hiu th .



17

___________________________________________________________________________

Nu thay 1

1

Rv = i1 , ta c mch tng ng (H 2.32c), trong ngun hiu th ph

thuc hiu th c thay bng ngun hiu th ph thuc dng in .

(a) (b) (c)

(H 2.32)

BI TP --o0o--

2.1. Cho mch (H P2.1)

(H P2.1)

Chng minh: v3 =

+

2

2

1

10 RR

R vv

Lu l v3 khng ph thuc vo thnh phn mc a, b. y l mt trong cc mch lm ton v c tn l mch cng. 2.2. Cho mch (H P2.2a)

(H P2.2a) (H P2.2b)



18

___________________________________________________________________________

Chng minh rng ta lun c: v1 = v2 v i1 = 21

2

RR i

Vi bt k thnh phn ni vo b,d. p dng kt qu trn vo mch (H P2.2b) xc nh dng in i. 2.3. Tm dng in i trong mch (H P2.3).

(H P2.3)

2.4. Cho mch (H P2.4) a/ Tnh vo. b/ p dng bng s v1 = 3 V, v2 = 2 V, R1 = 4K, R2 = 3K, Rf = 6K v R = 1K. 2.5. (H P2.5) l mch tng ng ca mt mch khuch i transistor. Dng nh l Thevenin hoc Norton xc nh io/ii ( li dng in).

(H P2.4) (H P2.5)

2.6. Cho mch (H P2.6a). Tm cc gi tr C v R2 nu vi(t) v i(t) c dng nh (H P2.6b) v (H P2.6c).

(a) (b) (c)

(H P2.6)

2.7 Tnh ( )( )tt

1

1

iv trong mch (H P2.7) v th t tn cho phn mch nm trong khung k nt

gin on.

2.8. Tnh Rtd ca (H P2.8).



19

___________________________________________________________________________

(H P2.7) (H P2.8) 2.9. Cho mch (H P2.9), tm iu kin vo = 0. 2.10. Thay th mch in trong khung ca (H P2.10) bng mch tng ng Thevenin sau tnh io.

(H P2.9) (H P2.10)

2.11. Dng nh l chng cht xc nh dng i trong mch (H P2.11). 2.12 Tm mch tng ng ca mch (H P2.12).

(H P2.11) (H P2.12) 2.13. Dng nh l Thevenin xc nh dng i trong mch (H P2.14).

(H P2.13) (H P2.14) 2.14. Dng nh l Norton xc nh dng i ca mch (H P2.1). 2.15. Dng nh l Norton ( hay Thevenin ) xc nh dng i trong mch (H P2.16).

(H P2.15)



20

___________________________________________________________________________


_______________________________________________Chng 3 Phng trnh mch in - 1

Chng3

PHNG TRNH MCH IN

KHI NIM V TOPO Mt s nh ngha

nh l v topo mch PHNG TRNH NT Mch cha ngun dng in

Mch cha ngun hiu th PHNG TRNH VNG

Mch cha ngun hiu th Mch cha ngun dng in

BIN I V CHUYN V NGUN Bin i ngun

Chuyn v ngun __________________________________________________________________________________________

Trong chng ny, chng ta gii thiu mt phng php tng qut gii cc mch in tng i phc tp. l cc h phng trnh nt v phng trnh vng. Chng ta cng cp mt cch s lc cc khi nim c bn v Topo mch, phn ny gip cho vic thit lp cc h phng trnh mt cch c hiu qu.

3.1 Khi nim v Topo MCH

Trong mt mch, n s chnh l dng in v hiu th ca cc nhnh. Nu mch c B nhnh ta c 2B n s v do cn 2B phng trnh c lp gii. Lm th no vit v gii 2B phng trnh ny mt cch c h thng v t c kt qu chnh xc v nhanh nht, l mc ch ca phn Topo mch.

Topo mch ch n cch ni nhau ca cc phn t trong mch m khng n bn cht ca chng.

3.1.1. Mt s nh ngha Gin thng

v gin thng tng ng ca mt mch ta thay cc nhnh ca mch bi cc on thng (hoc cong) v cc nt bi cc du chm.

(a) (b)

(H 3.1)



Trong gin cc nhnh v nt c t tn hoc nh s th t. Nu cc nhnh c nh hng (thng ta ly chiu dng in trong nhnh nh hng cho gin ), ta c gin hu hng.

(H 3.1b) l gin nh hng tng ng ca mch (H 3.1a). Gin con

Tp hp con ca tp hp cc nhnh v nt ca gin . Vng

Gin con khp kn. Mi nt trong mt vng phi ni vi hai nhnh trong vng . Ta gi tn cc vng bng tp hp cc nhnh to thnh vng hoc tp hp cc nt thuc vng .

Th d: (H 3.2a): Vng (4,5,6) hoc (a,b,o,a). (H 3.2b): Vng (1,6,4,3) hoc ( a,b,o,c,a).

(a) (b)

(H 3.2) Cy

Gin con cha tt c cc nt ca gin nhng khng cha vng. Mt gin c th c nhiu cy. Th d: (H 3.3a): Cy 3,5,6 ; (H 3.3b): Cy 3,4,5 . . ..

(a) (b)

(H 3.3)

* Cch v mt cy: Nhnh th nht c chn ni vi 2 nt, nhnh th hai ni 1

trong hai nt ny vi nt th 3 v nhnh theo sau li ni mt nt na vo cc nt trc. Nh vy khi ni N nt, cy cha N-1 nhnh.

Th d v cy ca (H 3.3b) ta ln lt lm tng bc theo (H 3.4).

___________________________________________________________________________ (H 3.4)



phn bit nhnh ca cy vi cc nhnh khc trong gin , ngi ta gi nhnh ca cy l cnh v cc nhnh cn li gi l nhnh ni. Cnh v nhnh ni ch c ngha sau khi chn cy.

Gi L l s nhnh ni ta c:

B = (N - 1) + L

Hay L = B - N +1 (3.1) Trong B l s nhnh ca gin , N l s nt.

Trong gin trn hnh 3.1 : B = 6, N = 4 vy L = 6 - 4 + 1 = 3 Nhn thy, mt cy nu thm mt nhnh ni vo s to thnh mt vng c lp ( l

vng cha t nht mt nhnh khng thuc vng khc ). Vy s vng c lp ca mt gin chnh l s nhnh ni L.

3.1.2. nh l v Topo mch Nhc li, mt mch gm B nhnh cn 2B phng trnh c lp gii, trong B

phng trnh l h thc v - i ca cc nhnh, vy cn li B phng trnh phi c thit lp t nh lut Kirchhoff . nh l 1:

Gin c N nt, c (N -1) phng trnh c lp do nh lut KCL vit cho (N-1) nt ca gin .

Tht vy, phng trnh vit cho nt th N c th suy t (N-1) phng trnh kia. nh l 2

Hiu th ca cc nhnh (tc gia 2 nt) ca gin c th vit theo (N-1) hiu th c lp nh nh lut KVL.

Tht vy, mt cy ni tt c cc nt ca gin , gia hai nt bt k lun c mt ng ni ch gm cc cnh ca cy, do hiu th gia hai nt c th vit theo hiu th ca cc cnh ca cy. Mt cy c (N - 1) cnh, vy hiu th ca mt nhnh no ca gin cng c th vit theo (N-1) hiu th c lp ca cc cnh.

Trong th d ca (H 3.1), cy gm 3 nhnh 3, 4, 5 c bit quan trng v cc cnh ca n ni vi mt nt chung O, O gi l nt chun. Hiu th ca cc cnh l hiu th gia cc nt a, b, c (so vi nt chun). Tp hp (N - 1) hiu th ny c gi l hiu th nt.

Nu mch khng c c tnh nh trn th ta c th chn mt nt bt k lm nt chun. nh l 3

Ta c L = B - N +1 vng hay mt li c lp vi nhau, trong ta c th vit phng trnh t nh lut KVL. nh l 4

Mi dng in trong cc nhnh c th c vit theo L = B - N +1 dng in c lp nh nh lut KCL.

Cc vng c lp c c bng cch chn mt cy ca gin , xong c thm 1 nhnh ni vo ta c 1 vng. Vng ny cha nhnh ni mi thm vo m nhnh ny khng thuc mt vng no khc. Vy ta c L = B - N + 1 vng c lp. Cc dng in chy trong cc nhnh ni hp thnh mt tp hp cc dng in c lp trong mch tng ng .

Th d: Trong gin (H 3.1b), nu ta chn cy gm cc nhnh 3,4,5 th ta c cc vng c lp sau y:



(H 3.5) Mt phng php khc xc nh vng c lp l ta chn cc mt li trong mt

gin phng (gin m cc nhnh ch ct nhau ti cc nt). Mt li l mt vng khng cha vng no khc. Trong gin (H 3.1b) mt li l cc vng gm cc nhnh: (4,5,6), (2,3,4) & (1,2,6).

Mt mt li lun lun cha mt nhnh khng thuc mt li khc nn n l mt vng c lp v s mt li cng l L.

Cc nh l trn cho ta B phng trnh gii mch : Gm (N-1) phng trnh nt v (L = B - N + 1) phng trnh vng.

V tng s phng trnh l:

(N-1) + L = N - 1 + B - N + 1 = B

3.2 Phng trnh Nt

3.2.1 Mch ch cha in tr v ngun dng in Trong trng hp ngoi in tr ra, mch ch cha ngun dng in th vit phng

trnh nt cho mch l bin php d dng nht gii mch. Chng ta lun c th vit phng trnh mt cch trc quan, tuy nhin nu trong mch c ngun dng in ph thuc th ta cn c thm cc h thc din t quan h gia cc ngun ny vi cc n s ca phng trnh mi iu kin gii mch. Ngun dng in c lp:

Nu mi ngun trong mch u l ngun dng in c lp, tt c dng in cha bit c th tnh theo (N - 1) in th nt. Ap dng nh lut KCL ti (N - 1) nt, tr nt chun, ta c (N - 1) phng trnh c lp. Gii h phng trnh ny tm hiu th nt. T suy ra cc hiu th khc. Th d 3.1:

Tm hiu th ngang qua mi ngun dng in trong mch (H 3.6)

(H 3.6)

Mch c 3 nt 1, 2, O; N = 3 vy N - 1 = 2, ta c 2 phng trnh c lp. Chn nt O lm chun, 2 nt cn li l 1 v 2 . v1 v v2 chnh l hiu th cn tm.

Vit KCL cho nt 1 v 2.



Nt 1: 024

5 211 =++ vvv (1)

Nt 2: 02632

2212 =+++ vvvv (2)

Thu gn:

521

21

41

21 =

+ vv (3)

261

31

21

21

21 =

+++ vv (4)

Gii h thng (3) v (4), ta c : v1 = 8 (V) v v2 = 2 (V) Thit lp phng trnh nt cho trng hp tng qut

Xt mch ch gm in tr R v ngun dng in c lp, c N nt. Nu khng k ngun dng in ni gia hai nt j v k, tng s dng in ri nt j n nt k lun c dng:

Gjk (vj - vk ) (3.2) Gjk l tng in dn ni trc tip gia hai nt j , k ( j k ) gi l in dn chung gia hai nt j , k ; ta c:

Gjk = Gkj (3.3)

Gi ij l tng i s cc ngun dng in ni vi nt j. nh lut KCL p dng cho nt j:

( ) =k

jkjjkG ivv (ij > 0 khi i vo nt j )

Hay jk k

kjkjkj GG ivv = ( j k ) ( 3.4)

Gjkk : L tng in dn ca cc nhnh c mt u ti nt j. Ta gi chng l in

dn ring ca nt j v k hiu:

(3.5) =k

jkjj GG

Phng trnh (3.4) vit li:

(3.6) ( kjGG jk

kjkjjj = ivv )Vit phng trnh (3.6) cho (N - 1) nt ( j = 1, ..., N - 1 ), ta c h thng phng trnh Nt 1: G11v1 - G12v2 - G13v3 . . . - G1(.N-1)vN-1 = i1 Nt 2: - G21 v1 + G22 v 2 - G23 v 3 . . . - G2.(N-1) v N-1 = i2 : : : Nt N -1: - G(N-1).1 v 1 - G(N-1).2 v 2 . . . +G(N-1)(.N-1) v N-1 = iN-1 Di dng ma trn:



=

1N

2

1

1N

2

1

11.NN1.2N1.1N

12.N2221

11.N1211

:::

:::

G ...............GG-:::::::::

G...............GG-G...............GG

i

ii

v

vv

Hay

[G][V] = [I] (3.7) [G]: Gi l ma trn in dn cc nhnh, ma trn ny c cc phn t i xng qua ng cho chnh v cc phn t c th vit mt cch trc quan t mch in . [V]: Ma trn hiu th nt, phn t l cc hiu th nt. [I]: Ma trn ngun dng in c lp, phn t l cc ngun dng in ni vi cc nt, c gi tr dng khi i vo nt. Tr li th d 3.1:

G11 = 21

41+ ; G22 =

61

31

21

++ ; G12 = 21

i1 = 5A v i2 = - 2A H phng trnh thnh:

=

++

+

2

5

61

31

21

21

21

21

41

2

1

v

v

Ta c kt qu nh trn. Ngun dng in ph thuc :

Phng php vn nh trn nhng khi vit h phng trnh nt tr s ca ngun dng in ny phi c vit theo hiu th nt gii hn s n s vn l N-1. Trong trng hp ny ma trn in dn ca cc nhnh mt tnh i xng.

Th d: 3.2 Tn hiu th ngang qua cc ngun trong mch (H 3.7).

(H 3.7)

Ta c th vit phng trnh nt mt cch trc quan:



=

+++

=

+

321

21

361

31

21

21

521

21

41

ivv

vv (1)

H thng c 3 n s, nh vy phi vit i3 theo v1 v v2.

2

213

vvi = (2)

Thay (2) vo (1) v sp xp li

521

43

21 = vv & 021

21 = vv

v1 = - 20 (V) v v2 = - 40 (V) Th d 3.3 Tnh v2 trong mch (H 3.8).

(H 3.8)

Chn nt chun O, v1 & v2 nh trong (H 3.8) H phng trnh nt l:

=

++

+=

+

321

321

911

4121

ivv

ivv (1)

Vi i3 = 5v1 (2) Ta c :

=+

=

09104

427

21

21

vv

vv (3)

Suy ra : v2 = - 114 (V)

3.2.2 Mch ch cha in tr v ngun hiu th Ngun hiu th c lp



Nu mt nhnh ca mch l 1 ngun hiu th c lp, dng in trong nhnh khng th tnh d dng theo hiu th nt nh trc. V hiu th ca ngun khng cn l n s nn ch cn (N-2) thay v (N-1) hiu th cha bit, do ta ch cn (N-2) phng trnh nt, vit nh nh lut KCL gii bi ton. c (N-2) phng trnh ny ta trnh 2 nt ni vi ngun hiu th th dng in chy qua ngun ny khng xut hin.

Cui cng, tm dng in chy trong ngun hiu th, ta p dng nh lut KCL ti nt lin h vi dng in cn li ny, sau khi tnh c cc dng in trong cc nhnh ti nt ny. Th d 3.4 Tnh v4 v in tr tng ng nhn t 2 u ca ngun hiu th v1 trong (H 3.9).

(H 3.9)

Mch c N = 4 nt v mt ngun hiu th c lp. Chn nt chun O v nt v1 ni vi ngun v1 = 6 V nn ta ch cn vit hai phng trnh cho nt v2 v v3. Vit KCL ti nt 2 v 3.

=+

+

=

++

024

61

0121

6

3323

3222

vvvv

vvvv

(1)

Thu gn:

=+

=

23

47

625

32

32

vv

vv (2)

Gii h thng (2):

v2 = 932 V v v3 =

926 V

v4 = v2 - v3 = 32 V

Dng i1 l tng cc dng qua in tr 1 v 4.

929

97

922

46

16 32

1 =+=

+

=vvi A

in tr tng ng:

Rt = 2954

9296

=

Rt = 2954



Chng ta cha tm c mt phng php tng qut vit thng cc phng trnh nt trong nhng mch c cha ngun hiu th.

Trong thc t ngun hiu th thng c mc ni tip vi mt in tr (chnh l ni tr ca ngun) nn ta c th bin i thnh ngun dng in mc song song vi in tr (bin i Thevenin, Norton).

Nu ngun hiu th khng mc ni tip vi in ta phi dng phng php chuyn v ngun trc khi bin i ( cp trong mt phn sau ).

Sau cc bin i, mch n gin hn v ch cha ngun dng in v ta c th vit h phng trnh mt cch trc quan nh trong phn 3.2.1.

Trong th d 3.3 trn, mch (H 3.9) c th v li nh (H 3.10a) m khng c g thay i v bin cc ngun hiu th ni tip vi in tr thnh cc ngun dng song song vi in tr ta c (H 3.10b).

(H 3.10)

V phng trnh nt:

61211 32 =

++ vv

- v2 + 1,5121

41

3 =

++ v

Gii h thng ta tm li c kt qu trn. Ngun hiu th ph thuc :

Ta cn mt phng trnh ph bng cch vit hiu th ca ngun ph thuc theo hiu th nt. Th d 3.5

Tm hiu th v1 trong mch (H 3.11)

(H 3.11)

Mch c 4 nt v cha 2 ngun hiu th nn ta ch cn vit 1 phng trnh nt cho nt b. Chn nt O lm chun, phng trnh cho nt b l:

0432

124 1bb =+ vvv (1)



Vi phng trnh ph l quan h gia ngun ph thuc v cc hiu th nt: 1b -24 vv = (2)

Thay (2) vo (1), sau khi n gin: v1=2 (V)

3.3 Phng trnh Vng

Mch c B nhnh, N nt c th vit L = B - N + 1 phng trnh vng c lp . Mi dng in c th tnh theo L dng in c lp ny.

3.3.1 Mch ch cha in tr v ngun hiu th Ngun hiu th c lp :

Nu mch ch cha ngun hiu th c lp, cc hiu th cha bit u c th tnh theo L dng in c lp.

p dng KVL cho L vng c lp (hay L mt li) ta c L phng trnh gi l h phng trnh vng. Gii h phng trnh ta c cc dng in vng ri suy ra cc hiu th nhnh t h thc v - i. Th d 3.6: Tm cc dng in trong mch (H 3.12a).

(a) (b) (c)

(H 3.12) Mch c N = 5 v B = 6 Vy L = B - N + 1 = 2

Chn cy gm cc ng lin nt (H 3.12b). Cc vng c c bng cch thm cc nhnh ni 1 v 2 vo cy.

Dng in i1 v i2 trong cc nhnh ni to thnh tp hp cc dng in c lp. Cc dng in khc trong mch c th tnh theo i1 v i2.

Mt khc, thay v ch r dng in trong mi nhnh, ta c th dng khi nim dng in vng. l dng in trong nhnh ni ta tng tng nh chy trong c vng c lp to bi cc cnh ca cy v nhnh ni (H 3.12c). Vit KVL cho mi vng:

(1)

0 = 24+ 4+ ) - 6( + 20 = 60 - 3 + ) - 6(

2122

121

iiiiiii

Thu gn:

(2) ( )

=+++ 246426-60 = 6 - 3) + 6 (

21

21

iiii

Gii h thng ta c :

i1 = 8A v i2 = 2A Dng qua in tr 6: i1 - i2 = 6 (A) ___________________________________________________________________________ Nguyn Trung Lp L THUYT MCH

_______________________________________________Chng 3 Phng trnh mch in - 11 Thit lp phng trnh vng cho trng hp tng qut

Coi mch ch cha in tr v ngun hiu th c lp , c L vng. Gi ij, ik ...l dng in vng ca vng j, vng k ...Tng hiu th ngang qua cc in tr chung ca vng j v k lun c dng:

Rjk ( ij ik) (3.8) Du (+) khi ij v ik cng chiu v ngc li. Rjk l tng in tr chung ca vng j v vng k. Ta lun lun c:

Rjk = Rkjvj l tng i s cc ngun trong vng j, cc ngun ny c gi tr (+) khi to ra dng in cng chiu ij ( chiu ca vng ). p dng KVL cho vng j:

( ) =k

jkjjkR vii (3.10)

Hay (3.11) jvii = kk

jkk

jkj RR

Rjkk chnh l tng in tr chung ca vng j vi tt c cc vng khc tc l tng in tr c trong vng j. t = RRjk

k jj v vi qui c Rjk c tr dng khi ij v ik cng chiu v m khi ngc li,

ta vit li (3.11) nh sau:

Rjjij + (3.12) jkk

jkR vi =i vi mch c L vng c lp : Vng 1 : R11i1 + R12i2 + . . . . R1LiL = v1Vng 2 : R21i1 + R22i2 + . . . . R2LiL = v2 : : : : : : : : : : Vng L: RL1i1 + RL2i2 + . . . . RLLiL = vL Di dng ma trn

=

L

2

1

2

1

.2L.1

2.2221

1.1211

:::

:::

.....R..........RR:::::::::

.....R..........RR

.....R..........RR

v

vv

i

ii

LLLL

L

L

H phng trnh vng vit di dng vn tt:

[R] .[I] = [V] (3.13) [R]: Gi l ma trn in tr vng c lp. Cc phn t trn ng cho chnh lun lun dng, cc phn t khc c tr dng khi 2 dng in vng chy trn n cng chiu, c tr m khi 2 dng in vng ngc chiu. Cc phn t ny i xng qua ng cho chnh. [I] : Ma trn dng in vng [V]: Ma trn hiu th vng ___________________________________________________________________________ Nguyn Trung Lp L THUYT MCH

_______________________________________________Chng 3 Phng trnh mch in - 12 Tr li th d 3.6 ta c th vit h phng trnh vng mt cch trc quan vi cc s liu sau:

R11 = 3 + 6 = 9 ,

R22 = 2 + 4 + 6 = 12 ,

R21 = R12 = - 6 ,

v1 = 60 V

v

v2 = - 24 (V) Ngun hiu th ph thuc

Nu mch c cha ngun hiu th ph thuc, tr s ca ngun ny phi c tnh theo cc dng in vng. Trong trng hp ny ma trn in th mt tnh i xng.

Th d 3.7 Tnh i trong mch (H 3.13)

(H 3.13)

Vit phng trnh vng cho cc vng trong mch 6i1- 2 i+ 4i2=15 (1) 4i1+ 2 i+ 6i2= 2 i (2) -2i1+ 8 i+ 2i2=0 (3)

(2) cho 21 23 ii = (4)

(3) cho 4

21 iii = (5)

Thay (5) vo (1) 11i1+ 9i2=30 (6)

Thay (4) vo (6) ta c i2=- 4 A i1= 6 A

V i = 2,5 (A)

3.3.2. Mch cha ngun dng in Ngun dng in c lp

Nu mt nhnh ca mch l mt ngun dng in c lp, hiu th ca nhnh ny kh c th tnh theo dng in vng nh trc. Tuy nhin nu mt dng in vng duy nht c v qua ngun dng in th n c tr s ca ngun ny v ch cn (L-1) n s thay v L (bng cch khng chn nhnh c cha ngun dng lm cnh ca cy).

Th d 3.8: Tnh dng in qua in tr 2 trong mch (H3.14a)



(a) (H 3.14) (b) Mch c B = 8, N = 5, cy c 4 nhnh v 4 vng c lp . Chn cy nh (H 3.14b) (nt lin), cnh ca cy khng l nhnh c cha ngun dng c lp. Ta c:

i3 = 10 A v i4 = 12 A Vit phng trnh vng cho hai vng cn li. Vng 1: ( 4 + 6 + 2 )i1 - 6i2 - 4i4 = 0 (1) Vng 2: - 6i1 + 18i2 + 3i3 - 8i4 = 0 (2) Thay i3 = 10 A v i4 = 12 A vo (1) v (2)

12i1 - 6i2 = 48

- 6i1 + 18i2 = 66 Suy ra i1 = 7 (A)

Th d trn cho thy ta vn c th vit c h phng trnh vng cho mch cha ngun dng in c lp. Tuy nhin ta cng c th bin i v chuyn v ngun (nu cn) c mch cha ngun hiu th v nh vy vic vit phng trnh mt cch trc quan d dng hn.

Mch (H 3.14a) c th chuyn di v bin i ngun c mch (H 3.15) di y.

(a) (H 3.15) (b) Vi mch (H 3.15b), ta vit h phng trnh vng. Vng 1: 12i1 - 6i2 = 48 Vng 2: - 6i1 + 18i2 = 96 - 30 Ta c li kt qu trc. Ngun dng in ph thuc

Tm v1 trong mch (H 3.16)



(a) (b) (c)

(H 3.16) Mch c B = 5, N = 3 cy c hai cnh v 3 vng c lp . Chn cy l ng lin nt ca (H 3.16b). Cc ngun dng in nhnh ni

Vit phng trnh cho vng 3 26i3 + 20i2 + 24i1 = 0 (1)

Vi i1 = 7A v i2= 31 41

8iv = (2)

Thay (2) vo (1) 26i3 - 5i3 + 168 = 0 i3 = - 8 (A) v v1= 16 (V)

3.4 Bin i v chuyn v ngun

Cc phng php bin i v chuyn v ngun nhm mc ch sa son mch cho vic phn gii c d dng. Mch sau khi bin i hoc phi n gin hn hoc thun tin hn trong vic p dng cc phng trnh mch in .

3.4.1. Bin i ngun Ngun hiu th ni tip v ngun dng in song song (H 3.17).

(H 3.17)

Ngun hiu th song song v ngun dng in ni tip. Ta phi c: v1 = v 2 v i1 = i2.

(H 3.18)

Ngun hiu th song song vi in tr v ngun dng in ni tip in tr : C th b in tr m khng nh hng n mch ngoi.



(H 3.19) Ngun hiu th mc ni tip vi in tr hay ngun dng mc song song vi in tr. Ta c

th dng bin i Thevenin Norton bin i ngun hiu th thnh ngun dng in hay ngc li cho ph hp vi h phng trnh sp phi vit.

(H 3.20)

3.4.2. Chuyn v ngun : Khi gp 1 ngun hiu th khng c in tr ni tip km theo hoc 1 ngun dng in

khng c in tr song song km theo, ta c th chuyn v ngun trc khi bin i chng. Trong khi chuyn v, cc nh lut KCL v KVL khng c vi phm. Chuyn v ngun hiu th :

(H 3.21) cho ta thy mt cch chuyn v ngun hiu th . Ta c th chuyn mt ngun hiu th " xuyn qua mt nt " ti cc nhnh khc ni vi nt v ni tt nhnh c cha ngun ban u m khng lm thay i phn b dng in ca mch, mc d c s thay i v phn b in th nhng nh lut KVL vit cho cc vng ca mch khng thay i. Hai mch hnh 3.21a v 3.21b tng ng vi nhau.

(a) (b)

(H 3.21) Th d 3.9: Ba mch in ca hnh 3.22 tng ng nhau:

(H 3.22)

Chuyn v ngun dng in:

Ngun dng in i mc song song vi R1 v R2 ni tip trong mch hnh 3.23a c chuyn v thnh hai ngun song song vi R1 v R2 hnh 3.23b.

(H 3.23)



nh lut KCL cc nt a, b, c ca cc mch (H 3.23) cho kt qu ging nhau. Hoc mt hnh thc chuyn v khc thc hin nh (H 3.24a) v (H 3.24b). nh lut

KCL cc nt ca hai mch cng ging nhau, mc d s phn b dng in c thay i nhng hai mch vn tng ng .

(a) (H 3.24) (b)

Th d 3.10: Tm hiu th gia a b ca cc mch hnh 3.25a

(a) (b) (c)

(H 3.25)

Suy ra vab =855

311

815

= V

vab = 855 V



Tm li, khi gii mch bng cc phng trnh vng hoc nt chng ta nn sa son cc mch nh sau:

- Nu gii bng phng trnh nt, bin i ch c cc ngun dng in trong mch. - Nu gii bng phng trnh vng, bin i ch c cc ngun hiu th trong mch.

BI TP --o0o--

1. Dng phng trnh nt, tm v1 v v2 ca mch (H P3.1) 2. Dng phng trnh nt , tm i trong mch (H P3.2).

(H P3.1) (H P3.2) 3. Dng phng trnh nt tm v v i trong mch (H P3.3). 4. Dng phng trnh nt, tm v trong mch (H P3.4)

(H P3.3) (H P3.4) 5. Dng phng trnh nt, tm v v v1 trong mch (H P3.5) 6. Cho vg = 8cos3t (V), tm vo trong mch (H P3.6)

(H P3.5) (H P3.6)

7. Tm v trong mch (H P3.7), dng phng trnh vng hay nt sao cho c t phng trnh nht.



(H P3.7)

8. Tm Rin theo cc R, R2, R3 mch (H P3.8). Cho R1 = R3 = 2K. Tm R2 sao cho Rin = 6K v Rin = 1K

(H P3.8)

9. Cho mch khuch i vi sai (H P3.9)

- Tm vo theo v1, v2, R1, R2, R3, R4.

- Tm lin h gia cc in tr sao cho: vo = ( )121

2

RR vv

10. Tm hiu th v ngang qua ngun dng in trong mch (H P3.10) bng cch dng phng trnh vng ri phng trnh nt.

(H P3.9) (H P3.10)

11. Tnh li dng in i

0

ii ca mch (H P.11) trong 2 trng hp.

a. R2 = 0 b. R2 = 1 12. Tm ix trong mch (H P.12)



(H P.11) (H P.12)


___________________________________________Chng 4 Mch in n gin- RL & RC -

1

CHNG 4

MCH IN N GIN: RL V RC

MCH KHNG CHA NGUN NGOI - PHNG TRNH VI PHN THUN NHT Mch RC khng cha ngun ngoi Mch RL khng cha ngun ngoi

Thi hng MCH CHA NGUN NGOI - PHNG TRNH VI PHN C V 2.

TRNG HP TNG QUT

Phng trnh mch in n gin trong trng hp tng qut Mt phng php ngn gn

VI TRNG HP C BIT p ng i vi hm nc Dng nh l chng cht

Chng ny xt n mt lp mch ch cha mt phn t tch tr nng lng (L hoc C) vi mt hay nhiu in tr.

p dng cc nh lut Kirchhoff cho cc loi mch ny ta c cc phng trnh vi phn bc 1, do ta thng gi cc mch ny l mch in bc 1.

Do trong mch c cc phn t tch tr nng lng nn p ng ca mch, ni chung, c nh hng bi iu kin ban u ca mch. V vy, khi gii mch chng ta phi quan tm ti cc thi im m mch thay i trng thi (th d do tc ng ca mt kha K), gi l thi im qui chiu t0 (trong nhiu trng hp, n gin ta chn t0=0). phn bit thi im ngay trc v sau thi im qui chiu ta dng k hiu t0-(trc) v t0+ (sau).

4.1 MCH KHNG CHA NGUN NGOI - PHNG TRNH VI PHN THUN NHT

4.1.1 Mch RC khng cha ngun ngoi Xt mch (H 4.1a). - Kha K v tr 1 ngun V0 np cho t. Lc t np y (hiu th 2 u t l

V0) dng np trit tiu i(0-)=0 (Giai on ny ng vi thi gian t=- n t=0-). - Bt K sang v tr 2, ta xem thi im ny l t=0. Khi t>0, trong mch pht sinh dng

i(t) do t C phng in qua R (H 4.1b). Xc nh dng i(t) ny (tng ng vi thi gian t0).

(a) (b) (H 4.1)


___________________________________________Chng 4 Mch in n gin- RL & RC - 2

Gi v(t) l hiu th 2 u t lc t>0 p dng KCL cho mch (H 4.1b)

0Rdt

dC =+ vv

Hay

0RC1

dtd

=+ vv

y l phng trnh vi phn bc nht khng c v 2. Li gii ca phng trnh l:

RCt

Ae(t)

=v A l hng s tch phn, xc nh bi iu kin u ca mch. Khi t=0, v(0) = V0 = Ae0 A=V0

Tm li: RCt

eV(t)

= 0v khi t 0 Dng i(t) xc nh bi

RC-t

0 eRV

R==

v(t)i )t( khi t 0

RV

0 0=+)(i

T cc kt qu trn, ta c th rt ra kt lun: - Dng qua t C thay i t ngt t tr 0 t=0- n V0/R t=0+. Trong lc - Hiu th hai u t khng i trong khong thi gian chuyn tip t t=0- n t=0+:

vC(0+)=vC(0-)=V0. y l mt tnh cht c bit ca t in v c pht biu nh sau: Hiu th 2 u mt t in khng thay i tc thi

Dng sng ca v(t) (tng t cho i(t)) c v (H 4.2)

(a) (b)

(H 4.2) - (H 4.2a) tng ng vi V0 v R khng i, t in c tr C v 2C ( dc gp i) - (H 4.2b) tng ng vi V0 v C khng i, in tr c tr R v 2R Ch : Nu thi im u (lc chuyn kha K) l t0 thay v 0, kt qu v(t) vit li:

RCt-t 0

eV(t))(

0

=v khi t t0

4.1.2 Mch RL khng cha ngun ngoi Xt mch (H 4.3a).

___________________________________________________________________________



3

(a) (H 4.3) (b)

- Kha K v tr 1, dng qua mch tch tr trong cun dy mt nng lng t trng. Khi mch t trng thi n nh, hiu th 2 u cun dy v(0-)=0 v dng in qua

cun dy l i(0-) = I0 = RV0

- Bt K sang v tr 2, chnh nng lng t trng tch c trong cun dy duy tr dng chy qua mch. Ta xem thi im ny l t=0. Khi t>0, dng i(t) tip tc chy trong mch (H 4.3b). Xc nh dng i(t) ny.

p dng KVL cho mch (H 4.3b)

0RdtdL =+ ii

Hay 0LR

dtd

=+ ii

Li gii ca phng trnh l: t

LR

Ae(t)

=i A l hng s tch phn, xc nh bi iu kin u ca mch

Khi t=0, i(0) = I0 =RV0 = Ae0 A = I0

Tm li: t

LR

eI(t)

= 0i khi t 0

t

LR

0L eRI(t)R(t)

== iv khi t 0

T cc kt qu trn, ta c th rt ra kt lun: - Hiu th hai u cun dy thay t ngt i t vL(0-)=0 n vL(0+)=-RI0. - Dng qua cun dy khng i trong khong thi gian chuyn tip t t=0- n t=0+:

iL(0+) = iL(0-) = I0 = V0/R. y l mt tnh cht c bit ca cun dy v c pht biu nh sau: Dng in qua mt cun dy khng thay i tc thi

Dng sng ca v(t) (tng t cho i(t)) c v (H 4.4)

(a) (H 4.4) (b) - (H 4.4a) tng ng vi V0 v R khng i, cun dy c tr L v 2L - (H 4.2b) tng ng vi V0 v L khng i, in tr c tr R v 2R



4.1.3 Thi hng Trong cc mch c cha cc phn t tch tr nng lng v cc in tr, khi mch

hot ng nng lng ca phn t c th gim dn theo thi gian do s tiu hao qua in tr, di dng nhit. o mc gim nhanh hay chm ca cc i lng ny, ngi ta dng khi nim thi hng.

Trong hai th d trn, p ng c chung mt dng:

=t

eY(t) 0y (4.1)

i lng trong biu thc chnh l thi hng. Vi mch RL: =L/R (4.2) Vi mch RC: =RC (4.3) tnh bng giy (s).

Khi t = 0100 0,37YeYeY(t) ===

y

Ngha l, sau thi gian , do phng in, p ng gim cn 37% so vi tr ban u Bng tr s v gin (H 4.5) di y cho thy s thay i ca i(t)/I0 theo t s t/

t/ 0 1 2 3 4 5 y(t)/Y0 1 0,37 0,135 0,05 0,018 0,0067

(H 4.5)

Ta thy p ng gim cn 2% tr ban u khi t = 4 v tr nn khng ng k khi t = 5. Do ngi ta xem sau 4 hoc 5 th p ng trit tiu.

Lu l tip tuyn ca ng biu din ti t=0 ct trc honh ti im 1, tc t = , iu ny c ngha l nu dng in gim theo t l nh ban u th trit tiu sau thi gian ch khng phi 4 hoc 5.

Thi hng ca mt mch cng nh th p ng gim cng nhanh (th d t in phng in qua in tr nh nhanh hn phng in qua in tr ln). Ngi ta dng tnh cht ny so snh p ng ca cc mch khc nhau.

4.2 MCH CHA NGUN NGOI-PHNG TRNH VI PHN C V 2

4.2.1 Mch cha ngun DC Chng ta xt n mch RL hoc RC c kch thch bi mt ngun DC t bn ngoi.

Cc ngun ny c gi chung l hm p (forcing function). Xt mch (H 4.6). Kha K ng ti thi im t=0 v t tch in ban u vi tr V0.

Xc nh cc gi tr v, iC v iR sau khi ng kha K, tc t>0.



5

(H 4.6)

Khi t>0, vit KCL cho mch:

0IRdtdC =+ vv

Hay

CI

RC1

dtd 0=+ vv

Gii phng trnh, ta c:

0RC

t

RIAe(t) +=

v Xc nh A nh iu kin u. t=0+: v(0+) = v(0-) = V0 V0=A+RI0 Hay A=V0-RI0

)( RCt

0RC

t

00RC

t

00 e1RIeVRI)eRI-(V(t)

+=+=v Hng s A by gi ty thuc vo iu kin u (V0) v c ngun kch thch (I0) p ng gm 2 phn:

Phn cha hm m c dng ging nh p ng ca mch RC khng cha ngun ngoi, phn ny hon ton c xc nh nh thi hng ca mch v c gi l p ng t nhin:

vn= RCt

00 )eRI-(V

l vn 0 khi t

Phn th hai l mt hng s, ty thuc ngun kch thch, c gi l p ng p vf=RI0 . Trong trng hp ngun kch thch DC, vf l mt hng s. (H 4.7) l gin ca cc p ng v, vnv vf

(H 4.7)

Dng iC v iR xc nh bi: RCt

00 eRRI-V

dtdC(t)

==

viC

Re

RRI-VI-I(t) RC

t00

00vii CR =+==

Lu l khi chuyn i kha K, hiu th 2 u in tr thay i t ngt t RI0 t=0- n V0 t=0+ cn hiu th 2 u t th khng i.

V phng din vt l, hai thnh phn ca nghim ca phng trnh c gi l p ng giao thi (transient response) v p ng thng trc (steady state response).



p ng giao thi 0 khi t v p ng thng trc chnh l phn cn li sau khi p ng giao thi trit tiu.

Trong trng hp ngun kch thch DC, p ng thng trc l hng s v chnh l tr ca p ng khi mch t trng thi n nh (trng thi thng trc)

4.2.2 iu kin u v iu kin cui (Initial and final condition)

4.2.2.1 iu kin u

Trong khi tm li gii cho mt mch in, ta thy cn phi tm mt hng s tch phn bng cch da vo trng thi ban u ca mch m trng thi ny ph thuc vo cc i lng ban u ca cc phn t tch tr nng lng.

Da vo tnh cht: Hiu th ngang qua t in v dng in chy qua cun dy khng thay i tc thi:

vC(0+)=vC(0-) v iL(0+)=iL(0-) - Nu mch khng tch tr nng lng ban u th:

vC(0+)=vC(0-) = 0, t in tng ng mch ni tt. iL(0+)=iL(0-) = 0, cun dy tng ng mch h.

- Nu mch tch tr nng lng ban u: * Hiu th ngang qua t ti t=0- l V0=q0/C th t=0+ tr cng l V0 , ta thay bng

mt ngun hiu th. * Dng in chy qua cun dy ti t=0- l I0 th t=0+ tr cng l I0 , ta thay bng

mt ngun dng in. Cc kt qu trn c tm tt trong bng 4.1

Phn t vi iu kin u Mch tng ng Gi tr u

Mch h IL(0+)=IL(0-)=0

Mch ni tt VC(0+)=VC(0-)=0

IL(0+)=IL(0-)=I0

VC(0+)=VC(0-)=V0

Bng 4.1

4.2.2.2 iu kin cui

p ng ca mch i vi ngun DC gm p ng t nhin 0 khi t v p ng p l cc dng in hoc hiu th tr khng i.

Mt khc v o hm ca mt hng s th bng 0 nn:

vC =Cte 0dt

dCC == Cvi (mch h) v iL =Cte 0

dtdL LL ==

iv (mch ni tt)

Do , trng thi thng trc DC, t in c thay bng mt mch h v cun dy c thay bng mt mch ni tt.

Ghi ch: i vi cc mch c s thay i trng thi do tc ng ca mt kha K, trng thi cui ca mch ny c th l trng thi u ca mch kia.

Th d 4.1



7

Xc nh hiu th v(t) trong mch (H 4.8a). Bit rng mch t trng thi thng trc trc khi m kha K.

(a)

(b) (c)

(H 4.8) (H 4.8b) l mch tng ca (H 4.8a) t=0-, tc mch (H 4.8a) t trng thi thng

trc, t in tng ng vi mch h v in tr tng ng ca phn mch nhn t t v bn tri:

=+++

+= 104)(23

4)3(28Rt

v hiu th v(0-) xc nh nh cu phn th 10 v 15

v(0-)= 40V1510

10100 =+

Khi t>0, kha K m, ta c mch tng ng (H 4.8c), y chnh l mch RC khng cha ngun ngoi.

Ap dng kt qu trong phn 4.1, c:

t

0 eV(t)

=v

vi =RC=10x1=10 s v V0= v(0+)= v(0-)=40 (V)

10t

40e(t)

=v (V)

4.3 TRNG HP TNG QUT

4.3.1 Phng trnh mch in n gin trong trng hp tng qut

Ta c th thy ngay phng trnh mch in n gin trong trng hp tng qut c dng:

QPydtdy

=+ (4.4)

Trong y chnh l bin s, hiu th v hoc dng in i trong mch, P l hng s ty thuc cc phn t R, L, C v Q ty thuc ngun kch thch, c th l hng s hay mt hm theo t.



Ta c th tm li gii tng qut cho phng trnh (4.4) bng phng php tha s tch phn: nhn 2 v phng trnh vi mt tha s sao cho v th nht l o hm ca mt hm v sau ly tch phn 2 v Nhn 2 v ca (4.4) vi ept

ptpt QePy)edtdy

=+( (4.5)

V 1 ca phng trnh chnh l )( ptyedtd v (4.5) tr thnh:

ptpt Qeyedtd

=)( (4.6)

Ly tch phn 2 v:

+= AdtQeye ptpt Hay (4.7) += -ptpt-pt AedtQeeyBiu thc (4.5) ng cho trng hp Q l hng s hay mt hm theo t. Trng hp Q l hng s ta c kt qu:

PQAey pt += (4.8)

p ng cng th hin r 2 thnh phn : - p ng t nhin yn=Ae-pt v - p ng p yf = Q/P. So snh vi cc kt qu phn 4.1 ta thy thi hng l 1/P

Th d 4.2

Tm i2 ca mch (H 4.9) khi t>0, cho i2(0)=1 A

(H 4.9)

Vit phng trnh vng cho mch

Vng 1: 8i1-4i2=10 (1)

Vng 2: -4i1+12i2+dtd 2i =0 (2)

Loi i1 trong cc phng trnh ta c:

dtd 2i +10i2=5 (3)

Dng kt qu (4.6)

i2(t)=Ae-10t + 21 (4)

Xc nh A:

Cho t=0 trong (4) v dng iu kin u i2(0)=1 A

i2(0)=A + 21 =1 A=

21



9

i2(t)=21 e-10t +

21

4.3.2 Mt phng php ngn gn Di y gii thiu mt phng php ngn gn gii nhanh cc mch bc 1 khng

cha ngun ph thuc. Ly li th d 4.2. Li gii i2 c th vit: i2 = i2n + i2f- xc nh i2n, ta xem mch nh khng cha ngun (H 4.10a) in tr tng ng nhn t cun dy gm 2 in tr 4 mc song song (=2), ni

tip vi 8, nn Rt = 2+8 = 10

(a) (b) (H 4.10)

V 101

RL

t

== (s) i2n =Ae-10t

- p ng p l hng s, n khng ty thuc thi gian, vy ta xt mch trng thi thng trc, cun dy tng ng mch ni tt (H 4.10b).

in tr tng ng ca mch: Rt=4+ 844.8+

=320

i1f = 23

20/310

= (A)

i2f = 21 (A)

Vy i2(t)=Ae-10t + 21 (A) v A c xc nh t iu kin u nh trc y.

Th d 4.3

Tm i(t) ca mch (H 4.11) khi t>0, cho v(0)=24 V

(H 4.11)

Ta c

i = in + if

xc nh in ta lu n c cng dng ca hiu th v 2 u t in.



Tht vy, tt c cc p ng t nhin khc nhau trong mt mch th lin h vi nhau qua cc php ton cng, tr, vi tch phn; cc php ton ny khng lm thay i gi tr trn m m n ch lm thay i cc h s ca hm m.

Thi hng ca mch l: =RC=10x0,02=0,2 s

in =Ae-5t

trng thi thng trc, t in tng ng mach h: if = i = 1A

Vy i(t) =Ae-5t + 1 (A)

xc nh A, ta phi xc nh i(0+) Vit phng trnh cho vng bn phi

-4 i(0+) +6[1- i(0+)] +24 = 0 i(0+) = 3 A 3=A+1 A=2

Vy i(t) =2e-5t + 1 (A)

Th d 4.4

Xc nh i(t) v v(t) trong mch (H 4.12a) khi t>0. Bit rng mch t trng thi thng trc t=0- vi kha K h.

(H 4.12a)

(H 4.12b)

trng thi thng trc (t=0-), t in tng mch h v cun dy l mch ni tt.

Hiu th 2 u t l hiu th 2 u in tr 20 v dng in qua cun dy chnh l dng qua in tr 15

Dng cu chia dng in xc nh d dng cc gi tr ny: i(0-)=2A v v(0-) = 60 V Khi ng kha K, ta ni tt 2 nt a v b (H 4.12b). Mch chia thnh 2 phn c lp vi nhau, mi phn c th c gii ring.

* Phn bn tri ab cha cun dy l mch khng cha ngun: i(t) = Ae-15t (A)

Vi i(0-) = i(0-)=2 A=2 i(t) = 2e-15t (A)

* Phn bn phi ab l mch c cha ngun 6A v t .15F Hiu th v(t) c th xc nh d dng bng phng php ngn gn:



11

v(t) = 20e-t+40 (V)

4.4 VI TRNG HP C BIT

4.4.1 p ng i vi hm nc Xt mt mch khng cha nng lng ban u, kch thch bi mt ngun l hm nc

n v. y l mt trng hp c bit quan trng trong thc t. Mch (H 4.13), trong vg=u(t)

(H 4.13)

Ap dng KCL cho mch

0Ru(t)

dtdC =

+

vv

Hay

u(t)RCRCdt

d 1=+

vv

* Khi t < 0, u(t)=0, phng trnh tr thnh:

0RCdt

d=+

vv v c nghim l: v(t)=Ae-t/RC

iu kin u v(0-) = 0 A = 0 v v(t)=0 * Khi t 0 , u(t) = 1, pt thnh:

RCRCdtd 1

=+vv

v(t) = vn+vfvf c xc nh t mch trng thi thng trc: vf = vg=u(t) = 1 V

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