Bài giảng môn Tín hiệu và Mạch - Ths Ngô Hán Chiêu [Tuần 4]

8/8/2019 Bi ging mn Tn hiu v Mch - Ths Ng Hn Chiu [Tun 4]

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GV: ThS Ng Hn ChiuEmail: [email protected]

HP: 0908.978.988

TN HIU V MCH

I HC CNG NGH THNG TIN


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Quy tc phn n & phn dng

a) Quy tc phn p

Quy tc phn p dng tnh cho cc mchch cha cc in tr mc ni tip.

Nu bit trcEv ix = 0, ta c:

Do u1 = i1R1, u2 = i2R2 nn:

E

R1

R2

u1

+

u2

i1

i2

iX

1 21 2

1 2 1 2

; R R

u E u E R R R R

Tng qut: Khi c nhiu in tr mc ni tip vbit in p E trn tonb cc in tr th in p ri trn mt in tr bt k s bng in p E nhn vi gi tr in tr v chia cho tng tt c cc in tr.

E

R1 R2 Rn

+

1

.

k

kR n

j

j

E Ru

R

1 2

1 2

Ei i

R R


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Quy tc phn n & phn dng (2)

Lu : Quy tc phn p ch c th p dng khi khng c phn t no mc viin tr m ngun cung cp nng lng (Hay dng ixphi bng 0)

E

R1

R2

u1

+

u2

i1

i2

iX

Hnh B1.11

Bi tp 1.16Tm in p u1 v u2 trn hnh B1.11 khi:E = 12V, R

1= 22k

; R2

= 33k .

(p s u1 = 4,8V, u2 = 7,2V)

Bi tp 1.17Tm in p u1 v u2 trn hnh B1.11 khi:E = -6V, R1 = 18k

; R2 = 27k .(p s u1 = -2,4V, u2 = -3,6V)


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b) Quy tc phn dng

Tng t nh quy tc phn p, quy tcphn dng dng cho cc mch cha haiin tr mc song song.

Nu bit trcI0 ti ntNta c:

in p trnR1 vR2phi bng nhau:

Ta c:

2 0 1i I i

2

1 1 2 2 1 21

Ri R i R i i

R

N

R1 R2

i1 i2

I0

22 0 2

1

Ri I i

R

Hay:

0 12

1 2

0 21

1 2

I Ri

R R

I Ri

R R


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Bi tp 1.18Tm dng in i4 qua R4 trn hnh B1.12 nu: I0 = 12mA, R1 = 2k ; R2= 1k , R3 = 1k

v R4 = 4k . (p s 2,4mA).

Bi tp 1.19Tm in p trn R4 trn hnh B1.12 nu: I0 = 10mA, R1 = 2k ; R2 =

5k ,R3 = 1k v R4 = 2k . (p s 8,6V).

Hnh B1.12

R2

R4

i4

I0

R1

R3

I0

Bi tp 1.20Tm dng in i1 qua R1 trn hnhB1.12 nu: I0 = 6mA, R1 = 2k ; R2= 250 ,R3 = 750k v R4 = 2k .(p s 2,4mA).


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BIN I V CHUYN V NGUN

BIN I NGUN


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BIN I NGUN


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CHUYN V NGUN P

V d:

Ta c th chuyn mt ngun hiuth "xuyn qua mt nt" ticc nhnh khc ni vi nt vni tt nhnh c cha ngun banu m khng lm thay i phn

bdng in ca mch.


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CHUYN V NGUN DNG

Ngun dng in imc songsong vi R1 v R2 ni tiptrong mch c chuyn v thnh hai ngun song songviR1 v R2.


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V D CHUYN V NGUN DNG

Tm hiu in th ab ca mch sau:

Gii: Tin hnh chuyn v ngun


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V D CHUYN V NGUN DNG (tt)


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MCH CHA KHUCH I THUT TON(OP-AMP)

OP-AMP l mt mch a cc, nhng n gin ta ch n ccng vo v ng ra (b qua cc cc ni ngun v Mass...). Mch c haing vo: gm (a) l ng vo khng o, nh du (+) v (b) l ngvo o nh du (-), (c) l ng ra.

Mch c nhiu c tnh quan trng , y ta xt mch trongiu 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 ngvo cng bng khng.


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Mch tng ng

Mch khng c tnhkhuych i (Buffer)


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BI TP


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MNG 4 CC

Mng bn cc (cn gi l mch hai ca) l m hnh ca ccphn t v ccphn mch in thng gp trong thc t (nh m hnhbin p, transistor...).

Cc nh lut tng qut dng cho mch tuyn tnh u c th p dng chobn cc tuyn tnh, nhng l thuyt mng bn cc ch yu i su vo phntch mch in theo h thng, lc y c th khng cn quan tm ti mch cth na m coi chng nh mt hp en v vn ngi ta cn n l miquan h dng v p hai ca ca mch.

L thuyt mng bn cc cho php nghin cu cc mch in phc tp nh ls ghp ni ca ccbn cc n gin theo nhiu cch khc nhau, n l mttrong nhng phng php hu hiu dng phn tch v tng hp mch.

U1, I1: in p v dng in ti ca 1U2, I2: in p v dng in ti ca 2

I1 I2

U1U2

Mng bncc


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MNG 4 CC (2)

Cc h phng trnh c tnh ca bn cc.

H phng trnh c tnh tr khng[Z]

Ma trn tr khng:

i vi trng hp bn cc tng h ta c:

z12 =z21

I1 I2

U1U2

Mng bncc

1 11 1 12 2

2 21 1 22 2

U z I z I

U z I z I

1 1

2 2

=[Z].U I

U I

2

111

1 0I

Uz

I1

222

20I

Uz

I1

112

20I

Uz

I2

221

1 0I

Uz

I

11 12

21 22

[Z]z z

z z


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MNG 4 CC (2)

V D 1:

Gii:

2

111

1 0I

Uz

I1

222

20I

Uz

I1

112

20I

Uz

I2

221

1 0I

Uz

I


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MNG 4 CC (2)

V D 2:

Gii:

2

111

1 0I

Uz

I

1

222

2 0I

Uz

I

1

112

2 0I

Uz

I

2

221

1 0I

Uz

I


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MNG 4 CC (3)


H phng trnh c tnh dn np [Y]

Ma trn dn np:


y12 =y21

I1 I2

U1U2

Mng bncc

1 11 1 12 2

2 21 1 22 2

I y U y U

I y U y U

1 1

2 2

=[Y].I U

I U

11 12

21 22

[Y]y y

y y

2

1

111 0U

I

y U1

2

222 0U

I

y U1

1

122 0U

I

y U2

2

211 0U

I

y U


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MNG 4 CC (4)


H phng trnh c tnh truyn t[A]

Ma trn truyn t:


A = -1

I1 I2

U1U2

Mng bncc1 11 2 12 2

1 21 2 22 2

U a U a I

I a U a I

1 2

1 2

[A]U U

I I

11 12

21 22

[A]a a

a a

2

111

2 0I

U

a U2

122

2 0U

Ia I

2

112

2 0U

Ua I

2

121

2 0I

Ia U


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MNG 4 CC (5)


H phng trnh c tnh truyn t ngc [B]

Ma trn truyn t ngc:


B = -1

I1 I2

U1U2

Mng bncc2 11 1 12 1

2 21 1 22 1

U b U b I

I b U b I

2 1

2 1

[B]U U

I I

11 12

21 22

[B]b b

b b

1

211

1 0I

U

b U1

222

1 0U

I

b I1

212

1 0U

U

b I1

221

1 0I

I

b U


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MNG 4 CC (6)


H phng trnh c tnh hn hp [H]

Ma trn hn hp:


h12 = - h21

I1 I2

U1U2

Mng bncc

1 11 1 12 2

2 21 1 22 2

U h I h U

I h I h U

1 1

2 2

[H]U I

I U

11 12

21 22

[H]h h

h h

2

111

1 0U

Uh I

1

222

2 0I

Ih U

1

112

2 0I

Uh U

2

221

1 0U

Ih I


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MNG 4 CC (7)

H phng trnh c tnh hn hp ngc [G]

Ma trn hn hp ngc:


g12 = - g21

I1 I2

U1U2

Mng bncc

1 11 1 12 2

2 21 1 22 2

I g U g I

U g U g I

1 1

2 2

[G]I U

U I

11 12

21 22

[G]g g

g g

2

111

1 0I

Ig U

1

222

2 0U

Ug I

1

112

2 0U

Ig I

2

221

1 0I

Ug U


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MNG 4 CC (8)

Cc phng php ghp ni mng bn cc .a) Ghp ni tip ni tip

Ghp ni tip c hai ca I v II

Mng 4 cc mi, c ma trn

tr khng [Z] nh sau:

Tng qut khi c kmng 4 cc mc ni tip ni tip:

I1

I2

I1 I2

U1 U2

I

U1 U2

II

I2

U2

I1

U1

[Z] [Z ] + [Z ]

1

[Z] [Z ]n

k

k


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MNG 4 CC (8)


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MNG 4 CC (9)

Cc phng php ghp ni mng bn cc .

b) Ghp song songsong song

Ghp song song c hai ca I v II


dn np [Y] nh sau:

Tng qut khi c kmng 4 cc mc song songsong song:

[Y] [Y ] + [Y ]

1

[Y] [Y ]n

k

k


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MNG 4 CC (9)


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MNG 4 CC (10)


c) Ghp ni tip song song

Ghp ni tip ca I v song song ca II


hn hp [H] nh sau:

Tng qut khi c kmng 4 cc mc ni tip song song:

[H] [H ] + [H ]

I2

I1

U2U1

I2I1

U1 U2

I2

U2

I1

U1

I

II

1

[H] [H ]n

k

k


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MNG 4 CC (11)


d) Ghp song song ni tip

Ghp song song ca I v ni tip ca II


hn hp ngc [G] nh sau:

Tng qut khi c kmng 4 cc mc song song ni tip:

[G] [G ] + [G ]

I2

I1

U2

U1

I2I1

U1 U2

I1U1

I2

U2

I

II

1

[G] [G ]n

k

k


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MNG 4 CC (12)


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MNG 4 CC (13)


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MNG 4 CC - BNG BIN I CC THNG S


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MNG 4 CC - BNG BIN I CC THNG S


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MNG 4 CC V D

Cho mng bn cc hnh v, hy xc nh cc thng s dn np ngnmch yij v cc thng s truyn t aij ca mng. Cho bit R1 = 10,R2 = 2, R3 = 3, R4 = 5, R5 = 5, R6 = 10.


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MNG 4 CC V D

Nhn vo s ta nhn thy mch

in c th phn tch thnh hai mngbn cc thnh phn hnh T v mcsong song-song song

Y = YT + Y

Nh vy ta s phi tnh cc thng syij ca tng bn cc thnh phn


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MNG 4 CC V D


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MNG 4 CC V D


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MNG 4 CC V D

Bài giảng môn Tín hiệu và Mạch - Ths Ngô Hán Chiêu [Tuần 4]

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