bo don kenh mux

8/3/2019 bo don kenh mux

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KhoaKhoa CNTTCNTTBoBo monmon KyKy thuathuatt MaMayy ttnhnh

Pham Tng Hai

oan Minh Vng

Phan nh The Duy


2/18

Logic Design 1 - Chapter 4 2

TTii liliuu thamtham khkhoo

Digital Logic Design Principles, N. Balabanian &B. Carlson John Wiley & Sons Inc., 2004

Digital Design, 3

rd

Edition, J.F. Wakerly,Prentice Hall, 2001

Digital Systems

, 5

th

Edition, R.J. Tocci, PrenticeHall, 1991


3/18


ChngChng 4.4.


4/18Logic Design 1 - Chapter 4 4

DDnn nhnhpp

Mch s c cc ng ra chph thuc vo gi tr/trng thi ca ccng vo thi im hin hnh c gi l mch lun l t hp(combinational logic circuits) hay gi tt l mch t hp

C th c nhiu mch t hp c thit k p ng cng 1 chcnng ra. Cc mch s ny c nh gi (nhm la chn mchno thch hp hn) da trn nhiu yu t khc nhau. Tc hot ng

phc tp Gi thnh phn cng

Nng lng tiu tn

S p ng v mt linh kin

Thit k ch trng tng yu t ny c th dn n s gim st

yu t khc



MMchch ccngng nhnh phnphn

Mch thc hin tc v cng i vi 2 gi tr nh phn

Hiu sut ca mch nh gi theo tc thc hin php ton C th da trn cc cng lun l ch to theo cng ngh thin v tc

Tc c th tng ng k ty theo cch thit k mch m khng quph thuc vo cng ngh ch to cng lun l

Cn nhc la chn gia thit k u tin cho tc v thit k thinu tin cho chi ph phn cng

S khi ca mch cng nh phn

Y

S

n

n n+1

Binary

Adder



MMchch ccngng (MC)(MC) totonn phphnn

C th xy dng mch cng 2 s nh phnn-bitt cc mch cng nh phn 1-bit

S khi ca mch cng

ton phn (full adder) Bng s tht Ba Karnaugh

xi

yi

Full

Adder

Si

Ci+1

Ci

Ci yi xi Si Ci+1

0

0

0

01

1

1

1

0

0

1

10

0

1

1

0

1

0

10

1

0

1

0

1

1

01

0

0

1

0

0

0

10

1

1

1

00 01 11 100 1 1

1 1 1Ci

yixiSi

00 01 11 10

0 1

1 1 1 1Ci

yixiCi+1

Dng hm ca cc ng ra

Si = xi yi Ci + xiyi Ci

+ xi yi Ci + xiyi Ci

= xi yi CiCi+1 = xiyi + xi Ci + yi Ci

= xiyi + Ci (xi + yi)

= xiyi + Ci (xiyi + xi yi)= xiyi + Ci (xi yi)


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MCMC bbnn phphnn vv MCMC rippleripple--carrycarry

Mch cng ton phnSi = xi yi CiCi+1 = xiyi + Ci (xi yi)

Mch cng bn phn(half adder)

xiyi

Si

Ci+1

Ci

xi

yi

Half

Adder

Si

Ci+1

xi

yi

Si

Ci+1

Mch cng ripple-carry

Gii hn do thi gian tr ca cc tnhiu carry !

Fu

llAd

der

A3

B3

C3

S3

Fu

llAd

der

A2

B2

C2

S2

Fu

llAd

der

A1

B1

C1

S1

Ha

lfAd

der

A0

B0

C4

S0


8/18


MMchch ccngng CarryCarry--LookaheadLookahead

Tnh carry t cc bit ca tonhngA,B v Co

nh nghaGenerated Carry Gi = AiBi

Propagated CarryPi = Ai Bi Ta tnh c

Ci = Ai-1Bi-1 + Ci-1 (Ai-1 Bi-1)= Gi-1 + Pi-1 Ci-1

= Gi-1 + Pi-1 (Gi-2 + Pi-2 Ci-2)

= Gi-1 + Pi-1 Gi-2 + Pi-1Pi-2 Ci-2

Tnh ln ltC1 = G0 + P0 C0

C2 = G1 + P1 G0 + P1P0 C0

C3 = G2 + P2 G1 + P2P1 G0 + P2P1P0 C0

C4 = G3 + P3 G2 + P3P2 G1+ P3P2P1 G0

+ P3P3P1P0 C0

P1

P2

P3

G1

G2

C0

P0

G0

P1

P2

P3

P2

P3

G3

P3

C4


9/18


MMchch ccngng CarryCarry--LookaheadLookahead

Tng qut

Ci+1 = Gi + Pi Gi-1 + PiPi-1 Gi-2 + PiPi-1Pi-2 Gi-3 +

+ PiPi-1Pi-2 P1 G0 + PiPi-1Pi-2 P1P0 C0Mch cng Carry-Lookahead

1

2

3

S0

1

2

3

S1C1

C2

C3

C4

P1

P2

P3

S2G

1

G2

G3

S3

0

0

C0

P0

G0

P1

P2

P3

Carry

-Look

ah

ea

d


10/18


MMchch trtr nhnh phnphn

Biu din s nguyn m nh phn di dng b 2

Mch cng 2 s di dng b 2 c khc g so vi mch

cng nh phn xem xt ?Mch tr c thay th bng mch chuyn i b 2 v

mch cng

A1A2A3

S0

B1B2B3

S1

C1C2C3C4

S2

M

S3

A0

B0

Overflow


11/18


BB ddnn knhknh

D liu sinh ra v tr A nhng c s dng v tr B truyn d liu t A n B qua knh truyn thng

Lm sao c th truyn d liu t nhiu ngun khc nhau trn

cng mt knh truyn duy nht ?

C ch cho php chn d liu no truyn trn knh truyn gi l kthut dn knh (multiplexing)

Thit b thc hin dn knh gi l b dn knh (multiplexer)

Pha thu, u bn kia ca knh truyn thng, cn b phn knh(demultiplexer) phn phi d liu trn knh truyn n cc ng ra

communication

channeldatain

dataout

Mu

ltiple

xer

Demu

ltip

lexer


12/18


BB ddnn knhknh

B dn knh s l mch c 2n ng d liu vo

1 ng d liu ra

n ng vo selecthay selector

B dn knh vi n = 3

Mu

ltiple

xer

D0

D1

D2

D3

D4

D5

D6

D7

r

s0

s1

s2

D0

r

s0

D1

D2

D3

D4

D5

D6

D7

s1

s2


13/18


XyXy ddngng mmchch tt hhpp tt bb ddnn knhknh

Tn ti cc mch dn knhc thng mi ha didng MSI

Dng b dn knh hinthc 1 mch t hp bt k ?

B dn knh c dng 2 lpAND-OR

Cng AND c n+1 ng nhp

Dng s-o-p chnh tc ca 1hm chuyn mch n+1 bin

B dn knh m-1 selectorcth c s dng hinthc mch t hp ca hm mbin

Th df(x, y, z) = (1, 2, 4, 7)

= z y x + z y x + z y x + z y x

gn s0 = y v s1 = z

f = s1 s0 x + s1 s0x + s1 s0 x + s1 s0x

= s1 s0 D0 + s1 s0D1 + s1 s0 D2 + s1 s0D3

suy ra D0 = D3 = x v D1 = D2 = x

V mch ?

Th df(w, x, y, z) = (0, 4, 9, 13, 14)

Thit k ?

V mch ?


14/18


BB gigiii mm BB mm hhaa

Mch t hp nhn n ng nhp (n 1) v nh tuyn dliu t cc ng nhp n mt trong s ti a 2n ng ra

gi l b gii m (decoder)B m ha (encoder), mch ngc li vi b gii m, l

mch nhn d liu t mt s rt ln cc ng nhp

ri bin i thnh d liu xut ra trn mt s nh hncc ng xut (khng nht thi t ch1 ng xut)

C s gn ging gia

B m ha vi b dn knh B gii m vi b phn knh

Hy chra s khc bit gia cc mch trn ?


15/18


BB phnphn knhknh

B phn knh vi 8ng xut Mch lun l

Bng s tht

D0

D1

D2

D3

D4

D5

D6

D7

C0

C1

C2

Datainput

x

Control inputs Data outputs

C2 C1 C0 D0 D1 D2 D3 D4 D5 D6 D7

0 0 0 x 0 0 0 0 0 0 0

0 0 1 0 x 0 0 0 0 0 0

0 1 0 0 0 x 0 0 0 0 0

0 1 1 0 0 0 x 0 0 0 0

1 0 0 0 0 0 0 x 0 0 0

1 0 1 0 0 0 0 0 x 0 0

1 1 0 0 0 0 0 0 0 x 0

1 1 1 0 0 0 0 0 0 0 x


16/18


BB gigiii mm ngng nn rara 22nn

B gii m ng n ra 2n (n-to-2n line decoder) cxy dng t b phn knh 2n ng xut bng cch: B bt ng nhp d liux

Mi cng AND chcn li n ng nhp

B gii m ng 3 ra 8Control inputs Data outputs

C2 C1 C0 D0 D1 D2 D3 D4 D5 D6 D7

0 0 0 1 0 0 0 0 0 0 0

0 0 1 0 1 0 0 0 0 0 0

0 1 0 0 0 1 0 0 0 0 0

0 1 1 0 0 0 1 0 0 0 0

1 0 0 0 0 0 0 1 0 0 0

1 0 1 0 0 0 0 0 1 0 0

1 1 0 0 0 0 0 0 0 1 0

1 1 1 0 0 0 0 0 0 0 1

Deco

der

3x

8

D0

D1

D2

D3

D4

D5

D6

D7

s0

s1

s2


17/18


BB gigiii mm ngng nn rara 22nn

MSI gii m ng thng dng 2 4 , 3 8 , 4 16

Gii m ma trn cng AND

Gii m cy Xy dng mch t hp t cc

b gii m ng

74LS138

A2A1A0

E3

E2E1

Q7Q6Q5Q4Q3Q2

Q1Q0

74LS154

E1E0

A3A2A1A0

1514

1312111098765432

10

74LS139

A1aA0aEa

A1bA0bEb

Q3aQ2aQ1aQ0aQ3bQ2bQ1bQ0b


18/18


BBii ttpp

Problem 4.4

Problem 4.7

Problem 4.10

Problem 4.11

Problem 4.12

Thy Phan nh Th [email protected]

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