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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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Quy tc phn n & phn dng (3)
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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Quy tc phn n & phn dng (4)
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 V CHUYN V NGUN
BIN I NGUN
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BIN I V CHUYN V NGUN
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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BIN I V CHUYN V NGUN
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 CHA KHUCH I THUT TON(OP-AMP)
Mch tng ng
Mch khng c tnhkhuych i (Buffer)
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MCH CHA KHUCH I THUT TON(OP-AMP)
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MCH CHA KHUCH I THUT TON(OP-AMP)
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)
Cc h phng trnh c tnh ca bn cc.
H phng trnh c tnh dn np [Y]
Ma trn dn np:
i vi trng hp bn cc tng h ta c:
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)
Cc h phng trnh c tnh ca bn cc.
H phng trnh c tnh truyn t[A]
Ma trn truyn t:
i vi trng hp bn cc tng h ta c:
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)
Cc h phng trnh c tnh ca bn cc.
H phng trnh c tnh truyn t ngc [B]
Ma trn truyn t ngc:
i vi trng hp bn cc tng h ta c:
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)
Cc h phng trnh c tnh ca bn cc.
H phng trnh c tnh hn hp [H]
Ma trn hn hp:
i vi trng hp bn cc tng h ta c:
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:
i vi trng hp bn cc tng h ta c:
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
Mng 4 cc mi, c ma trn
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)
Cc phng php ghp ni mng bn cc .
c) Ghp ni tip song song
Ghp ni tip ca I v song song ca II
Mng 4 cc mi, c ma trn
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)
Cc phng php ghp ni mng bn cc .
d) Ghp song song ni tip
Ghp song song ca I v ni tip ca II
Mng 4 cc mi, c ma trn
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