ky-thuat-so__kts1_c3_he-to-hop - [cuuduongthancong.com].pdf

bai giang ky thuat so chuong 3

1. Chöông 3: HEÄ TOÅ HÔÏP
I. Giôùi thieäu –Caùch thieát keá heä toå hôïp:
Maïchlogicñöôïcchialaøm2loaïi:
-Heä toå hôïp (Combinational Circuit)
-Heä tuaàn töï (Sequential Circuit).
Heä toå hôïp laø maïch maø caùc ngoõ ra chæ phuï thuoäc vaøo giaù
trò cuûa caùc ngoõ vaøo. Moïi söï thay ñoåi cuûa ngoõ vaøo seõ laøm ngoõ ra
thay ñoåi theo.
Ngoõ vaøo
(INPUT)
Ngoõ ra
(OUTPUT)
COÅNG
LOGIC
NguyenTrongLuat 1

2. * Caùc böôùc thieát keá:
-Phaùt bieåu baøi toaùn.
-Xaùc ñònh soá bieán ngoõ vaøo vaø soá bieán ngoõ ra.
-Thaønh laäp baûng giaù trò chæ roõ moái quan heä giöõa ngoõ vaøo
vaø ngoõ ra.
-Tìm bieåu thöùc ruùt goïn cuûa töøng ngoõ ra phuï thuoäc vaøo
caùc bieán ngoõ vaøo.
-Thöïc hieän sô ñoà logic.
Ngoõ vaøo
X
n-1
… X
1
X
0
Ngoõ ra
Y
m-1
… Y
1
Y
0
0 … 0 0
1 … 1 1
NguyenTrongLuat 2

3. Vd: Thieát keá heä toå hôïp coù 3 ngoõ vaøo X, Y, Z; vaø 2 ngoõ ra F, G.
-Ngoõ ra F laø 1 neáu nhö 3 ngoõ vaøo coù soá bit 1 nhieàu hôn soá bit
0; ngöôïc laïi F = 0.
-Ngoõ ra G laø 1 neáu nhö giaù trò nhò phaân cuûa 3 ngoõ vaøo lôùn
hôn 1 vaø nhoû hôn 6; ngöôïc laïi G = 0.
X Y Z F G
0 0 0
0 0 1
0 1 0
0 1 1
1 0 0
1 0 1
1 1 0
1 1 1
0
0
0
1
0
1
1
1
0
0
1
1
1
1
0
0
XY
Z
F
0
1
00011110
11
Y Z
1
1
X Z
X Y
F = X Y + Y Z + X Z
XY
Z
G
0
1
00011110
11
11
X Y X Y
G = X Y + X Y = X

Y
NguyenTrongLuat 3

4. F
F = X Y + Y Z + X Z G = X Y + X Y = X

Y
X
Y
Z
G
NguyenTrongLuat 4

5. Tr
ườ
ngh
ợ
pheätoåhôïpkhoângsöûduïngtaátcaû2
n
toåhôïpcuûangoõ
vaøo,thìtaïicaùctoåhôïpkhoângsöûduïngñoùngoõracoùgiaùtròtuøyñònh.
Vd:Thieát keá heä toå
hôïp coù ngoõ vaøo bieåu
dieãn cho 1 soá maõ BCD.
Neáu giaù trò ngoõ vaøo
nhoû hôn 3 thì ngoõ ra coù
giaù trò baèng bình
phöông giaù trò ngoõ
vaøo; ngöôïc laïi giaù trò
ngoõ ra baèng giaù trò ngoõ
vaøo tröø ñi 3.
A B C D
0 0 0 0
0 0 0 1
0 0 1 0
0 0 1 1
0 1 0 0
0 1 0 1
0 1 1 0
0 1 1 1
1 0 0 0
1 0 0 1
1 0 1 0
1 0 1 1
1 1 0 0
1 1 0 1
1 1 1 0
1 1 1 1
F
2
F
1
F
0
X X X
X X X
X X X
X X X
X X X
X X X
0 0 0
0 0 1
1 0 0
0 0 0
0 0 1
0 1 0
0 1 1
1 0 0
1 0 1
1 1 0
F2 = A + B C D + B C D
F1 = A D + B C D + B C D
F0 = A D + B D + A B C D
NguyenTrongLuat 5

6. II.Boäcoäng-tröønhòphaân:
1. Boä coäng (Adder):
a. Boä coäng baùn phaàn (Half Adder –H.A):
Boä coäng baùn phaàn laø heä toå hôïp coù nhieäm vuï thöïc hieän
pheùp coäng soá hoïc x + y (x, y laø 2 bit nhò phaân ngoõ vaøo); heä
coù 2 ngoõ ra: bit toång S (Sum) vaø bit nhôù C (Carry).
x y C S
0 0
0 1
1 0
1 1
0 0
0 1
0 1
1 0
S = x y + x y = x

y
C = x y
x
y
S
C
x
y
S
C
H.A
NguyenTrongLuat 6

7. b. Boä coäng toaøn phaàn (Full Adder –F.A):
Boäcoängtoaønphaànth
ự
chi
ệ
npheùpcoängsoáhoïc3bitx+y+z
(zbieåudieãnchobitnhôùtöøv
ị
trícoùtroïngsoánhoûhôngôûitôùi)
x
y
S
C
F.A
z
x y zC S
0 0 0
0 0 1
0 1 0
0 1 1
1 0 0
1 0 1
1 1 0
1 1 1
0 0
0 1
0 1
1 0
0 1
1 0
1 0
1 1
xy
z
S
0
1
00011110
1
1
1
1
S = x y z + x y z + x y z + x y z
xy
z
C
0
1
00011110
11 1
1
C = x y + x z + y z
NguyenTrongLuat 7

8. S = x y z + x y z + x y z + x y z
= z (x y + x y) + z (x y + x y)
= z (x

y) + z (x

y)
S = z

(x

y)
C = x y + x z + y z
= x y + x y z + x y z + x y z
= x y (1 + z) + z (x y + x y)
C = x y + z (x

y)
x
y
z
S
C
NguyenTrongLuat 8

9. 2. Boä tröø (Subtractor):
a. Boä tröø baùn phaàn (Half Subtractor –H.S):
Boä tröø baùn phaàn coù nhieäm vuï thöïc hieän pheùp tröø soá
hoïc x -y (x, y laø 2 bit nhò phaân ngoõ vaøo); heä coù 2 ngoõ ra: bit
hieäu D (Difference) vaø bit möôïn B (Borrow).
x
y
D
B
H.S
x y B D
0 0
0 1
1 0
1 1
0 0
1 1
0 1
0 0
D = x y + x y = x

y
B = x y
x
y
D
B
NguyenTrongLuat 9

10. b. Boä tröø toaøn phaàn (Full Subtractor –F.S):
Boätröøtoaønphaànth
ự
chi
ệ
npheùptröøsoáhoïc3bitx-y-z
(zbieåudieãnchobitmöôïntöøvítròcoùtroïngsoánhoûhôn)
x
y
D
B
F.S
z
x y zB D
0 0 0
0 0 1
0 1 0
0 1 1
1 0 0
1 0 1
1 1 0
1 1 1
0 0
1 1
1 1
1 0
0 1
0 0
0 0
1 1
xy
z
D
0
1
00011110
1
1
1
1
xy
z
B
0
1
00011110
11 1
1
S = x y z + x y z + x y z + x y z
C = x y + x z + y z
S = z

(x

y)
C = x y + z (x

y)
NguyenTrongLuat 10

11. 74283
3. Boä coäng/tröø nhò phaân song song:
a. Boä coäng nhò phaân:
M: M3 M2 M1 M0
N: N3 N2 N1 N0
S0S1S2S3
C1C2
+
C3
C4
xy
zC
S
F.A
xy
zC
S
F.A
xy
zC
S
F.A
xy
zC
S
F.A
M0 N0M1 N1M2 N2M3 N3
S0
C0
= 0
C1 C2 C3
S1 S2 S3 C4
NguyenTrongLuat 11

12. b. Boä tröø nhò phaân:
-SöûduïngcaùcboätröøtoaønphaànF.S
-Thöïchieänbaèngpheùpcoängvôùibuø2cuûasoátröø
M –N = M + Buø_2(N) = M + Buø_1(N) + 1
M0 N0M1 N1M2 N2M3 N3
C0
= 1
xy
zC
S F.A
xy
zC
S F.A
xy
zC
SF.A
xy
zC
SF.A
C1 C2 C3
S0 S1 S2 S3 C4
Keát quaû:-C4 = 1 keát quaû laø soá döông
-C4 = 0 keát quaû laø soá aâm
NguyenTrongLuat 12

13. c. Boä coäng/tröø nhò phaân:
M0 N0M1 N1M2 N2M3 N3
C0
xy
zC
S
F.A
xy
zC
S
F.A
xy
zC
S
F.A
xy
zC
S
F.A
C1 C2 C3
S0 S1 S2 S3 C4
Pheùp toaùnC
0
y
i
0 N
i
COÄNG
TRÖØ 1 N
i
T = 0:Coäng
T = 1:Tröø
Ngoõ vaøo ñieàu khieån
C
0
= T
y
i
= T

N
i
T
NguyenTrongLuat 13

14. III.Heächuyeånmaõ(CodeConversion):
-Heä chuyeån maõ laø heä toå hôïp coù nhieäm vuï laøm cho 2 heä thoáng
töông thích vôùi nhau, maëc duø moãi heä thoáng duøng maõ nhò
phaân khaùc nhau.
-Heä chuyeån maõ coù ngoõ vaøo cung caáp caùc toå hôïp maõ nhò phaân A
vaø caùc ngoõ ra taïo ra caùc toå hôïp maõ nhò phaân B. Nhö vaäy, ngoõ
vaøo vaø ngoõ ra phaûi coù soá löôïng töø maõ baèng nhau.
Maõ
nhò phaân B
Heä
chuyeån
maõ
Maõ
nhò phaân A
NguyenTrongLuat 14

15. Vd:Thieát keá heä chuyeån maõ töø maõ BCD thaønh maõ BCD quaù 3.
A B C D
0 0 0 0
0 0 0 1
0 0 1 0
0 0 1 1
0 1 0 0
0 1 0 1
0 1 1 0
0 1 1 1
1 0 0 0
1 0 0 1
1 0 1 0
1 0 1 1
1 1 0 0
1 1 0 1
1 1 1 0
1 1 1 1
W X Y Z
X X X X
X X X X
X X X X
X X X X
X X XX
X X X X
0 0 1 1
0 1 0 0
0 1 0 1
0 1 1 0
0 1 1 1
1 0 0 0
1 0 0 1
1 0 1 0
1 0 1 1
1 1 0 0
W = A + B (C + D)
X = B

(C + D)
Y = C

D
Z = D
A
B
C
D
W
X
Z
Y
NguyenTrongLuat 15

16. IV.Boägiaûimaõ(DECODER):
1. Giôùi thieäu:
-Boä giaûi maõ laø heä chuyeån maõ coù nhieäm vuï chuyeån töø maõ nhò
phaân cô baûn n bitôû ngoõ vaøo thaønh maõ nhò phaân 1 trong môû
ngoõ ra.
Maõ
1 trong m
X
0
X
1
X
n-1
Maõ
nhò phaân
Y
0
Y
1
Y
m-1
m = 2
n
-Coù2daïng:ngoõratíchcöïccao(möùc1)vaøngoõratíchcöïc
thaáp(möùc0).
-V
ớ
igiaùtr
ị
ic
ủ
at
ổ
h
ợ
pnh
ị
phaân
ở
ngoõvaøo,thìngoõraY
i
s
ẽ
tíchc
ự
cvaøcaùcngoõracoønl
ạ
is
ẽ
khoângtíchc
ự
c.
NguyenTrongLuat 16

17. a. Boä giaûi maõ ngoõ ra tích cöïc cao:
X
0
(LSB)
X
1
Y
0
Y
1
Y
2
Y
3
X
1
X
0
Y
3
Y
2
Y
1
Y
0
0 0
0 1
1 0
1 1
0 0 0 1
0 0 1 0
0 1 0 0
1 0 0 0
Y
0
= X
1
X
0
= m
0
Y
1
= X
1
X
0
= m
1
Y
2
= X
1
X
0
= m
2
Y
3
= X
1
X
0
= m
3
X
0
X
1
Y
0
Y
1
Y
2
Y
3
Ngoõ ra:Y
i
= m
i
(i = 0, 1, .., 2
n
-1)
NguyenTrongLuat 17

18. b. Boä giaûi maõ ngoõ ra tích cöïc thaáp:
X
1
X
0
Y
3
Y
2
Y
1
Y
0
0 0
0 1
1 0
1 1
1 1 1 0
1 1 0 1
1 0 1 1
0 1 1 1
X
0
X
1
Ngoõ ra:Y
i
= M
i
(i = 0, 1, .., 2
n
-1)
X
0
(LSB)
X
1
Y
0
Y
1
Y
2
Y
3
Y
0
= X
1
+X
0
= M
0
= m
0
Y
1
= X
1
+X
0
= M
1
= m
1
Y
2
= X
1
+X
0
= M
2
= m
2
Y
3
= X
1
+X
0
= M
3
= m
3
Y
0
Y
1
Y
2
Y
3
NguyenTrongLuat 18

19. c. Boä giaûi maõ coù ngoõ vaøo cho pheùp:
-Ngoaøi caùc ngoõ vaøo döõ lieäu, boä giaûi maõ coù theå coù 1 hay
nhieàu ngoõ vaøo cho pheùp.
-Khi caùc ngoõ vaøo cho pheùp ôû traïng thaùi tích cöïc thì maïch
giaûi maõ môùi ñöôïc hoaït ñoäng. Ngöôïc laïi, maïch giaûi maõ seõ khoâng
hoaït ñoäng; khi ñoù caùc ngoõ ra ñeàu ôû traïng thaùi khoâng tích cöïc.
Y
0
Y
1
Y
2
Y
3
X
0
(LSB)
X
1
EN
EN X
1
X
0
Y
3
Y
2
Y
1
Y
0
0 X X
1 0 0
1 0 1
1 1 0
1 1 1
0 0 0 0
0 0 0 1
0 0 1 0
0 1 0 0
1 0 0 0
X
0
X
1
Y
0
Y
1
Y
2
Y
3
EN
NguyenTrongLuat 19

20. 2. IC giaûi maõ:
a. IC 74139:goàm 2 boä giaûi maõ 2 sang 4 ngoõ ra tích cöïc thaáp
1Y
0
1Y
1
1Y
2
1Y
3
1A
(LSB)
1B
1G
2Y
0
2Y
1
2Y
2
2Y
3
2A
(LSB)
2B
2G
1
2
3
15
14
13
4
5
6
7
12
11
10
9
G B AY
3
Y
2
Y
1
Y
0
1 X X
0 0 0
0 0 1
0 1 0
0 1 1
1 1 1 1
1 1 1 0
1 1 0 1
1 0 1 1
0 1 1 1
NguyenTrongLuat 20

21. b. IC 74138:boä giaûi maõ 3 sang 8 ngoõ ra tích cöïc thaáp
Y
0
Y
1
Y
2
Y
3
A
(LSB)
B
C
Y
4
Y
5
Y
6
Y
7
G1
G2A
G2B
1
2
3
4
6
5
9
12
11
10
7
15
14
13
G
1
G
2A
G
2B
CB A Y
7
Y
6
Y
5
Y
4
Y
3
Y
2
Y
1
Y
0
0 X X X X X
X 1 X X X X
X X 1 X X X
1 0 0 0 0 0
1 0 0 0 0 1
1 0 0 0 1 0
1 0 0 0 1 1
1 0 0 1 0 0
1 0 0 1 0 1
1 0 0 1 1 0
1 0 0 1 1 1
1 1 1 1 1 1 1 1
1 1 1 1 1 1 1 1
1 1 1 1 1 1 1 1
1 1 1 1 1 1 1 0
1 1 1 1 1 1 0 1
1 1 1 1 1 0 1 1
1 1 1 1 0 1 1 1
1 1 1 0 1 1 1 1
1 1 0 1 1 1 1 1
1 0 1 1 1 1 1 1
0 1 1 1 1 1 1 1
NguyenTrongLuat 21

22. 3. Söû duïng boä giaûi maõ thöïc hieän haøm Boole:
Ngoõ ra cuûa boä giaûi maõ laø minterm (ngoõ ra tích cöïc cao)
hoaëc maxterm (ngoõ ra tích cöïc thaáp) cuûa n bieán ngoõ vaøo. Do
ñoù, ta coù theå söû duïng boä giaûi maõ thöïc hieän haøm Boole theo
daïng chính taéc.
z
y
x
0
1
0
F1 (x, y, z) =

(2, 5, 7)
= m
2
+ m
5
+ m
7
= M
2
+ M
5
+ M
7
= M
2
M
5
M
7
F2 (x, y, z) =

(0, 1, 4)
= M
0
M
1
M
4
F
1
F
2
Y
0
Y
1
Y
2
Y
3
A
(LSB)
B
C
Y
4
Y
5
Y
6
Y
7
G1
G2A
G2B
74138
NguyenTrongLuat 22

23. V.Boämaõhoùa(ENCODER):
1. Giôùi thieäu:
-Encoderlaøheächuyeånmaõthöïchieänhoaïtñoängngöôïclaïivôùi
decoder.Nghóalaøencodercoùmngoõvaøotheomaõnhòphaân1
trongmvaønngoõratheomaõnhòphaâncôbaûn(vôùim
≤
2
n
).
-VôùingoõvaøoI
i
ñöôïctíchcöïcthìngoõrachínhlaøtoåhôïpgiaùtrò
nhòphaânitöôngöùng.
I
0
I
1
I
2
I
3
(LSB)
Z
0
Z
1
I
3
I
2
I
1
I
0
Z
1
Z
0
0 0
0 1
1 0
1 1
0 0 0 1
0 0 1 0
0 1 0 0
1 0 0 0
Z
1
= I
3
+ I
2
Z
0
= I
3
+ I
1
Z
1
Z
0
I
3
I
2
I
1
NguyenTrongLuat 23

24. * Boä maõ hoùa coù öu tieân (Priority Encoder):
Boämaõhoùacoùöutieânlaømaïchmaõhoùasaochoneáucoùnhieàu
hôn1ngoõvaøocuøngtíchcöïcthìngoõraseõlaøgiaùtrònhòphaân
cuûangoõvaøocoùöutieâncaonhaát.
I
0
I
1
I
2
I
3
(LSB)
Z
0
Z
1
V
I
3
I
2
I
1
I
0
Z
1
Z
0
V
X X 0
0 0 1
0 1 1
1 0 1
1 1 1
0 0 0 0
0 0 0 1
0 0 1 X
0 1 X X
1 X X X
Z
1
= I
3
+ I
2
Z
0
= I
3
+ I
2
I
1
V = I
3
+ I
2
+ I
1
+ I
0
I
3
I
2
I
1
I
0
Z
1
Z
0
V
Thöù töï öu tieân:I
3

I
2

I
1

I
0
NguyenTrongLuat 24

25. 2.ICmaõhoùaöutieân8

3(74148):
EI I
7
I
6
I
5
I
4
I
3
I
2
I
1
I
0
A
2
A
1
A
0
GS EO
1 X X X X X X X X
0 0 X X X X X X X
0 1 0 X X X X X X
0 1 1 0 X X X X X
0 1 1 1 0 X X X X
0 1 1 1 1 0 X X X
0 1 1 1 1 1 0 X X
0 1 1 1 1 1 1 0 X
0 1 1 1 1 1 1 1 0
0 1 1 1 1 1 1 1 1
1 1 1 1 1
0 0 0 0 1
0 0 1 0 1
0 1 0 0 1
0 1 1 0 1
1 0 0 0 1
1 0 1 0 1
1 1 0 0 1
1 1 1 0 1
1 1 1 1 0
EI
I
7
I
6
I
5
I
4
I
3
I
2
I
1
I
0
1
5
3
12
14
9
A
2
A
1
(LSB)
A
0
GS
EO
7
6
15
4
2
13
11
10
NguyenTrongLuat 25

26. VI.Boädoànkeânh(Multiplexer-MUX):
1. Giôùi thieäu:
-MUX 2
n

1 laø heä toå hôïp coù nhi
ề
u ngoõ vaøonhöng chæ coù
1 ngoõ ra. Ngoõ vaøo goàm 2 nhoùm: mngoõ vaøo döõ lieäu (data input)
vaø nngoõ vaøo löïa choïn (select input).
-Vôùi 1 giaù trò icuûa toå hôïp nhò phaân caùc ngoõ vaøo löïa choïn,
ngoõ vaøo döõ lieäu D
i
seõ ñöôïc choïn ñöa ñeán ngoõ ra. (m = 2
n
)
D
0
D
1
:
D
m-1
S
0
(LSB)
S
1
:
S
n-1
Y
Ngoõ vaøo döõ lieäu
(Data Input)
Ngoõ vaøo löïa choïn
(Select Input)
NguyenTrongLuat 26

27. * Boä MUX 4

1:
D
0
D
1
D
2
D
3
S
0
(LSB)
S
1
Y
S
1
S
0
Y
0 0
0 1
1 0
1 1
D
0
D
1
D
2
D
3
= m
0
D
0
+ m
1
D
1
+ m
2
D
2
+ m
3
D
3
=

m
i
D
i
(i = 0, 1, 2, 3)
Y = S
1
S
0
D
0
+ S
1
S
0
D
1
+ S
1
S
0
D
2
+ S
1
S
0
D
3
S
1
S
0
D
0
D
1
D
2
D
3
Y
Toångquaùt:Y=

m
i
D
i
(vôùii=0,1,..,2
n
-1)
NguyenTrongLuat 27

28. 2. IC doàn keânh:
a. 74LS153:goàm 2 boä MUX 4

1
1G
1C
0
1C
1
1C
2
1C
3
A
(LSB)
B
1Y
2G
2C
0
2C
1
2C
2
2C
3
2Y
14
15
10
11
12
13
2
1
6
5
4
3
7
9
G BA Y
1 X X
0 0 0
0 0 1
0 1 0
0 1 1
0
C
0
C
1
C
2
C
3
NguyenTrongLuat 28

29. b. 74151:boä MUX 8

1
EN
A
(LSB)
B
C
YD
0
D
1
D
2
D
3
D
4
D
5
D
6
D
7
Y
15
14
13
12
9
4
3
2
1
5
6
11
10
7
EN C BAY
1 X X X
0 0 0 0
0 0 0 1
0 0 1 0
0 0 1 1
0 1 0 0
0 1 0 1
0 1 1 0
0 1 1 1
0
D
0
D
1
D
2
D
3
D
4
D
5
D
6
D
7
NguyenTrongLuat 29

30. 3. Söû duïng boä MUX thöïc hieän haøm Boole:
a.BoäMUX2
n
thöïchieänhaømBoolenbieán:
EN
A
(LSB)
B
C
YD
0
D
1
D
2
D
3
D
4
D
5
D
6
D
7
Y
F(x, y, z) =

(0, 1, 4, 7)
= m
0
+ m
1
+ m
4
+ m
7
= m
0
1 + m
1
1 + m
2
0 + m
3
0
+ m
4
1 + m
5
0 + m
6
0 + m
7
1
Y =

m
i
D
i
= m
0
D
0
+ m
1
D
1
+ m
2
D
2
+ m
3
D
3
+ m
4
D
4
+ m
5
D
5
+ m
6
D
6
+ m
7
D
7
D
0
= D
1
= D
4
= D
7
= 1
D
2
= D
3
= D
5
= D
6
= 0
z
y
x
0
1
0
F
NguyenTrongLuat 30

31. b. Boä MUX 2
n
thöïc hieän haøm Boole n+1 bieán:
F(x, y, z) =

(0, 1, 4, 7)
= x y z+ x y z + x y z + x y z
= x y .1+ x y .0 + x y .z + x y .z
Y = m
0
D
0
+ m
1
D
1
+ m
2
D
2
+ m
3
D
3
D
0
= 1; D
1
= 0; D
2
= z; D
3
= z
1G
1C
0
1C
1
1C
2
1C
3
A
(LSB)
B
1Y
2G
2C
0
2C
1
2C
2
2C
3
2Y
y
x
0
1
0
z
F
= m
0
.1+ m
1
.0 + m
2
.z + m
3
.z
x y zF
0 0 0
0 0 1
0 1 0
0 1 1
1 0 0
1 0 1
1 1 0
1 1 1
1
1
0
0
1
0
0
1
D
0
= 1
D
1
= 0
D
3
= z
D
2
= z
NguyenTrongLuat 31

32. VII.Boäphaânkeânh(DEMUX):
1. Giôùi thieäu:
-BoäDEMUX1

2
n
coùchöùcnaêngthöïchieänhoaïtñoängngöôïc
laïivôùiboäMUX.Maïchcoù1ngoõvaøodöõlieäu,nngoõvaøolöïa
choïnvaø2
n
ngoõra.
-Vôùi 1 giaù trò icuûa toå hôïp nhò phaân caùc ngoõ vaøo löïa choïn,
ngoõ vaøo döõ lieäu Dseõ ñöôïc ñöa ñeán ngoõ ra Y
i
.
Y
0
Y
1
:
Y
m-1
S
0
(LSB)
S
1
:
S
n-1
D
Ngoõ vaøo döõ lieäu
(Data Input)
Ngoõ vaøo löïa choïn
(Select Input)
Ngoõ vaøo döõ lieäu
(Data Input) Ngoõ ra
NguyenTrongLuat 32

33. * Boä DEMUX 1

4:
Y
0
Y
1
Y
2
Y
3
D
S
0
(LSB)
S
1
S
1
S
0
Y
3
Y
2
Y
1
Y
0
0 0
0 1
1 0
1 1
0 0 0 D
0 0 D 0
0 D 0 0
D 0 0 0
Y
0
= S
1
S
0
D = m
0
D
Y
1
= S
1
S
0
D = m
1
D
Y
2
= S
1
S
0
D = m
2
D
Y
3
= S
1
S
0
D = m
3
D
Y
0
Y
1
Y
2
Y
3
S
1
S
0
D
NguyenTrongLuat 33

34. B A1G 1C1Y
0
1Y
1
1Y
2
1Y
3
2. IC phaân keânh 74LS155:goàm 2 boä phaân keânh 1

4
1Y
0
1Y
1
1Y
2
1Y
3
A
(LSB)
B
2Y
0
2Y
1
2Y
2
2Y
3
2G
2C
1
2
15
13
3
7
6
5
4
12
10
11
9
14
1G
1C
X X
X X
0 0
0 1
1 0
1 1
1 X
X 0
0 1
0 1
0 1
0 1
1 1 1 1
1 1 1 1
0 1 1 1
1 0 1 1
1 1 0 1
1 1 1 0
B A2G 2C2Y
0
2Y
1
2Y
2
2Y
3
X X
X X
0 0
0 1
1 0
1 1
1 X
X 1
0 0
0 0
0 0
0 0
1 1 1 1
1 1 1 1
0 1 1 1
1 0 1 1
1 1 0 1
1 1 1 0
NguyenTrongLuat 34

35. VIII.Boäsosaùnh
đ
oälôùn(Comparator):
1. Giôùi thieäu:
-Boä so saùnh laø heä toå hôïp coù nhieäm vuï so saùnh 2 soá nh
ị
phaân
khoâng daáu A vaø B (moãi soá n bit).
-Boä so saùnh coù 3 ngoõ ra (A>B), (A=B) vaø (A<B); chæ coù 1 ngoõ ra
tích cöïc theo keát quaû so saùnh.
* Boä so saùnh 3 bit:
A: A
2
A
1
A
0
B: B
2
B
1
B
0
Söû duïng bieán trung gian:
x
i
= A
i

B
i
(i = 0, 1, 2)
(A = B) = x
2
x
1
x
o
(A > B) = A
2
B
2
+ x
2
A
1
B
1
+x
2
x
1
A
0
B
0
(A < B) = A
2
B
2
+ x
2
A
1
B
1
+x
2
x
1
A
0
B
0
(A>B)
(A=B)
(A<B)
A
B
= (A=B) + (A>B)
NguyenTrongLuat 35

36. x
0x
1
x
2
(A=B)
B
0
A
0
B
1
A
1B
2
A
2
(A>B)
(A<B)
NguyenTrongLuat 36

37. 2. IC so saùnh 74LS85:
3
4
9
ALTBIN
AEQBIN
AGTBIN
B
0
B
1
B
2
B
3
10
12
13
15
11
14
1
7
5
A
0
A
1
A
2
A
3
ALTBOUT
AEQBOUT
AGTBOUT 6
2
AGTBOUT = (A>B) + (A=B)AGTBIN
AEQBOUT = (A=B) AEQBIN
ALTBOUT = (A<B) + (A=B)ALTBIN
NguyenTrongLuat 37