chuong04 thiet ke mach to hop

Thit k mch t hpBi Vn Hiu [email protected]

Dn nhpMch s c cc ng ra ch ph thuc vo gi tr/trng thi ca cc ng 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 chc nng. Cc mch s ny c nh gi 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 (trade-off)Khoa Khoa hc v K thut my tnh - B mn k thut my tnh 2

Mch cng nh phnMch 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 X C th tng ng k tc da trn cch thit k ph hp m khng qu ph thuc vo cng Y ngh ch to cng lun l Cn nhc la chn gia thit k u tin cho tc v thit k thin u tin cho chi ph phn cngKhoa Khoa hc v K thut my tnh - B mn k thut my tnh

n

n

Binary adder

n+1

S

S khi3

Thit k mch cngThit k theo phng php lp bng thc tr B cng 2 bitX1 X0 Y1 Y0 S2 S1 S0 D C B A0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 0 0 0 0 1 1 1 1 0 0 0 0 1 1 1 1 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 0 0 0 0 0 0 1 0 0 1 1 0 1 1 1 0 0 1 1 0 1 1 0 1 1 0 0 1 0 0 1 0 1 0 1 1 0 1 0 0 1 0 1 1 0 1 0

Khoa Khoa hc v K thut my tnh - B mn k thut my tnh

4

Thit k mch cngThit k theo phng php lp X1 X0 Y1 Y0 S2 S1 S0 D C B A bng thc tr 0 0 0 0 0 0 0 B cng 2 bit 0 0 0 1 0 0 1 0 0 1 0 0 1 0 S2 = DB + DCA + CBA 0 0 1 1 0 1 1 0 1 0 0 0 0 1 S1 = DCB + DBA + DCB + 0 1 0 1 0 1 0 DBA + DCBA + DBCA 0 1 1 0 0 1 1 0 1 1 1 1 0 0 S0 = CA + CA 1 0 0 0 0 1 01 1 0 0 0 1 1 1 1 0 1 1 0 0 1 1 1 0 1 0 1 0 1 0 1 1 0 1 1 1 1 0 0 1 0 0 1 1 0 1 1 0 1 0

B cng n bit?

1 1 1 1 1


5

Php cng hai s nh phn+ 1010 11 + 1010 1110 000 11 + 1010 1110 00 01 + 1010 1110 0 0 + 1010 1110 1110 11000


6

Php cng hai s nh phn+ 1010 11 + 1010 1110 000 x3 c3 s4 s3 y3 x2 c2 s2 11 + 1010 1110 00 y2 x1 c1 s1 01 + 1010 1110 0 y1 x0 c0 s0 0 + 1010 1110 1110 11000

y0


6

Mch cng bn phn (half adder)A C SA S B A B C

B

A B C S0 0 1 1 0 1 0 1 0 0 0 1 0 1 1 0

S = BA + BA = B C = AB

A

A B A B

S S C7


Mch cng ton phn (full adder)BA Co B

A Ci

Ci B A Co S0 0 0 0 1 0 1 0 1 0 1 0 0 0 1 0 1 1 1 0 0 0 1 1 1 1

SCi0 1

BA00 01 1 1 00 0 1 1 01 1 11 10 1

S

A0 B11 0 0 1 1

S

1 1 0 1 0 0 1

BA

SA B Ci B A C

CoCi

11 1 1

10

1

S Co


8


A Ci

Ci B A Co S0 0 0 0 1 0 1 0 1 0 1 0 0 0 1 0 1 1 1 0 0 0 1 1 1 1

SCi0 1

BA00 01 1 1 00 0 1 1 01 1 11 10 1

S

A0 B11 0 0 1 1

S

1 1 0 1 0 0 1

BA

SA B Ci B A C

CoCi

11 1 1

10

1

S = CiBA + Ci BA + Ci BA + CiBA

S Co


8


A Ci

Ci B A Co S0 0 0 0 1 0 1 0 1 0 1 0 0 0 1 0 1 1 1 0 0 0 1 1 1 1

SCi0 1

BA00 01 1 1 00 0 1 1 01 1 11 10 1

S

A0 B11 0 0 1 1

S

1 1 0 1 0 0 1

BA

SA B Ci B A C

CoCi

11 1 1

10

1


S Co

= Ci

B

A


8


A Ci

Ci B A Co S0 0 0 0 1 0 1 0 1 0 1 0 0 0 1 0 1 1 1 0 0 0 1 1 1 1

SCi0 1

BA00 01 1 1 00 0 1 1 01 1 11 10 1

S

A0 B11 0 0 1 1

S

1 1 0 1 0 0 1

BA

SA B Ci B A C

CoCi

11 1 1

10

1


S Co

= Ci

B

A

C0 = BA + CiB + CiA


8


A Ci

Ci B A Co S0 0 0 0 1 0 1 0 1 0 1 0 0 0 1 0 1 1 1 0 0 0 1 1 1 1

SCi0 1

BA00 01 1 1 00 0 1 1 01 1 11 10 1

S

A0 B11 0 0 1 1

S

1 1 0 1 0 0 1

BA

SA B Ci B A C

CoCi

11 1 1

10

1


S Co

= Ci

B

A

C0 = BA + CiB + CiA = BA + Ci (B + A)


8


A Ci

Ci B A Co S0 0 0 0 1 0 1 0 1 0 1 0 0 0 1 0 1 1 1 0 0 0 1 1 1 1

SCi0 1

BA00 01 1 1 00 0 1 1 01 1 11 10 1

S

A0 B11 0 0 1 1

S

1 1 0 1 0 0 1

BA

SA B Ci B A C

CoCi

11 1 1

10

1


S Co

= Ci

B

A

C0 = BA + CiB + CiA = BA + Ci (B + A) = BA + Ci (BA + BA)


8


A Ci

Ci B A Co S0 0 0 0 1 0 1 0 1 0 1 0 0 0 1 0 1 1 1 0 0 0 1 1 1 1

SCi0 1

BA00 01 1 1 00 0 1 1 01 1 11 10 1

S

A0 B11 0 0 1 1

S

1 1 0 1 0 0 1

BA

SA B Ci B A C

CoCi

11 1 1

10

1


S Co

= Ci

B

A

C0 = BA + CiB + CiA = BA + Ci (B + A) = BA + Ci (BA + BA) = BA + Ci(B A)8


Mch cng ripple-carryS3 S2 S1 S0 B3 C3 B2 C2 B1 C1

Half adderS0

Full adder

Full adder

Full adder

A3

A2

A1

A0 B0

C4

S3

S2

S1

Full adder

Full adder

Full adder

B3 C3

B2 C2

B1 C1

Full adder

A3

A2

A1

A0 B0

C4

GND


9

Mch cng ripple carryThi gian tr ca mch cng Gi s cc cng u c thi gian tr l tp Thi gian tr ca mt tng full-adder: 2.tp Thi gian tr ca mch cng n bit trng hp xu nht l: 2.n.tp C phng php no gim thi gian tr ca b cng?Khoa Khoa hc v K thut my tnh - B mn k thut my tnh 10

Mch cng carry lookaheadThm phn mch cng tnh c nh S input cho mch ln hn 2 v nh hn 2n phc tpMch hai lp Carry lookahead Ripple carry

Thi gian trKhoa Khoa hc v K thut my tnh - B mn k thut my tnh 11

Mch cng carry-lookaheadnh ngha Generated Carry Gi = Ai Bi Propagated Carry Pi = Ai Bi Ci = Ai-1Bi-1 + Ci-1(Ai-1 Bi-1)

= Gi-1 + Pi-1Ci-1 = Gi-1 + Pi-1(Gi-2 + Pi-2Ci-2) = Gi-1 + Pi-1Gi-2 + Pi-1Pi-2Ci-2 Ci+1 = Gi + PiGi-1 + PiPi-1Gi-2 + PiPi-1Pi-2Gi-3 + + PiPi-1Pi-2P1G0 + PiPi-1Pi-2P1P0C0Khoa Khoa hc v K thut my tnh - B mn k thut my tnh 12

Carry-lookahead mch cng 4 bitC1 = G0 + P0C0G0 P0 C0

G0 P0 C0

G0 P0 C0

C1

C1

C1

C2 = G1 + P1G + P P0C0P0 C0

G1 0 P1 1 G0

G1 P1 G0 P0 C0G2 P2 G1 P1

G1 P1 G0 P0 C2 C0

C2

C2

C3 = G2 + P2G1 + G2 + P2P1P0C0 P2G1 P1 G0 P0

G2 P2 P2P1G0 G1 P1 G0 P0 C0

C3

G0 P0 C3 C0

C3

C4 = G3 + P3G2 + P3P2G1 + P3P2P1G0 + P3P2P1P0C0C0Khoa Khoa hc v K thut my tnh - B mn k thut my tnh 13

4-bit Adder with Carry Lookahead Mch cng Carry-Lookahead (tt)! Complete adder:" Same number of stages for each bit

! Drawback?" Increasing complexity of lookahead logic for more bits

10/10/2007

Engin 112 - Intro to ECE

15


14

4-bit Adder with Carry Lookahead Mch cng Carry-Lookahead (tt)! Complete adder:

Mch "cng 4 bit of stages for Same numbereach bit

MchDrawback? bit, 64 bit? ! cng 16http://fourier.eng.hmc.edu/e85/ of " Increasing complexity lectures/arithmetic_html/node7.html

lookahead logic for more bits

10/10/2007

Engin 112 - Intro to ECE

15


14

Mch cng b 2 (s c du)Mch cng 2 s di dng b 2 c khc g so vi mch cng nh phn xem xt ?


15

Mch cng b 2 (s c du)Mch cng 2 s di dng b 2 c khc g so vi mch cng nh phn xem xt ?

Overow

S3

S2

S1

S0

Full adder

Full adder

Full adder

XORC4

B3 C3

B2 C2

B1 C1

Full adder

A3

A2

A1

A0 B0

GND


15

Mch tr b 2Mch tr c thay th bng mch chuyn i b 2 v mch cng Mch chuyn i b 2 ?


16

Mch tr b 2Mch tr c thay th bng mch chuyn i b 2 v mch cng Mch chuyn i b 2 ?


16

Mch cng tr b 2Dng mt mch thc hin chc nng cng v tr?


17



17


Mch pht hin trn ? (sv t c)Khoa Khoa hc v K thut my tnh - B mn k thut my tnh 17

B dn knh


18

B dn knhD liu sinh ra v tr A nhng c s dng v tr B


18

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


18

B dn knhD liu sinh ra v tr A nhng c s dng v tr B Cn 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 ?


18



18



18



18


C ch cho php chn d liu no truyn trn knh truyn gi l k thut dn knh / hp 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 raKhoa Khoa hc v K thut my tnh - B mn k thut my tnh 18

B dn knhB dn knh s l mch c 2n ng d liu vo (input) 1 ng d liu ra (output) n ng vo select hay selector Cc ng vo Select s chn ng d liu vo no xut ra ng d liu ra B dn knh vi n = 3D0 D2 D3 D4 D5 D6 D7

Multiplexers0 s1 s2

D1

r


19

B dn knh n = 1Cn gi l b chn (selector) Bng thc tr? Biu thc? Mch?D0 D1

MUXs0

r

D0 r D1 s0


20

B dn knh n = 1 (tt)s00 0 0 0 1 1 1 1

D1 D00 0 1 1 0 0 1 1 0 1 0 1 0 1 0 1

r0 1 0 1 0 0 1 1

D 1D 0 s01 1 1 1

r = s0D0 + s0D1D0 r D1 D2

D0 D1 s0Khoa Khoa hc v K thut my tnh - B mn k thut my tnh

D321

B dn knh n = 2Bng thc tr (64 t hp) r = s1s0D3 + s1s0D2 + s1s0D1 +s1s0D0D0 r D1 D2 D3 s0 s1Khoa Khoa hc v K thut my tnh - B mn k thut my tnh 22

D0 D1 r D2 D3

r

s0

s1

B dn knh n=3s1 D0 D1 D2 D3 D4 D5 D6 D7 s0 s1 s2Khoa Khoa hc v K thut my tnh - B mn k thut my tnh 23

r

B dn knh n=3s1 D0 D1 D2 D3 D4 D5 D6 D7 s0 r

D0 D1 s0 r

D0 D1 D2 D3 s0 s1 r

D0 D1 D2

D0 D1 s0r

D0 r

D1

s1

D2

D3s2Khoa Khoa hc v K thut my tnh - B mn k thut my tnh

D323

D4

B dn knh n=3s1 D0 D1 D2 D3 D4 D5 D6 D7 s0

D0 D1 s0 r

D0 D1 D2 D3 r

s0 Mch dn knh (hin thc r bng mch hai lp)s1tng qut? D0 D1 D2

D0 D1 s0r

D0 r

D1

s1

D2

D3s2Khoa Khoa hc v K thut my tnh - B mn k thut my tnh

D323

D4

Mch dn knh c Enabler D1 D2 r

D0

Thm tn hiu Enable D3 Khi tn hiu Enable khng tch cc th ng ra gi nguyn mc 0 (hoc mc 1) D0s0 s1 D1 D2 D3 s0 s1 EN rKhoa Khoa hc v K thut my tnh - B mn k thut my tnh 24

r

ng dng khc ca mch dn knhTn ti cc mch dn knh c thng mi ha di dng MSI Dng b dn knh hin thc 1 mch t hp bt k B dn knh m-1 Selector c th c s dng hin thc mch t hp ca hm m bin f(x,y,z) = (1, 2, 4, 7) vi z l MSB = zyx + zyx + zyx + zyx Gn s1 = z, s0 = y: s1s0D0 + s1s0D2 + s1s0D1 + s1s0D3 Suy ra: D0 = D3 = x, D1 = D2 = x Mch ?Khoa Khoa hc v K thut my tnh - B mn k thut my tnh 25

Bi tpThit k mch f (w, x, y, z) = (0, 4, 9, 13, 14) f (w, x, y, z) = (1, 3, 5, 6, 12,14, 15) z l MSB w l LSB


26

B phn knhB phn knh s l mch c 1 ng d liu vo 2n ng d liu ra x n ng vo select hay selector Cc ng vo Select s chn ng d liu vo s xut ra ng d liu ra no Cc ng ra cn li c gi tr 0 (hoc 1) B phn knh vi n = 3

Demultiplexers0 s1 s2

D0 D1 D2 D3 D4 D5 D6 D7


27

B phn knh n=1D0 x

DE MUXs0

Bng thc trs0 x D0 D10 0 1 1 0 1 0 1 0 1 0 0 0 0 0 1 1 a

D1

s0 x D0 D10 a

a0

0

a

D0 x D1 s0

Biu thc D0 = s0x D1 = s0x Mch

x

D0 D1

x

s028


B phn knh n=2Bng thc tr

s1 s0 D 0 D 1 D 2 D 30 0 1 1 0 1 0 1

x0 0 0

0

0 0

0 0 0

x0 0

x0

xD0 D1 D2 D3

Biu thc x D0 = s1s0x D1 = s1s0x s0 D2 = s1s0x D3 = s1s0x Mch

D0 D1

x

s0 s1


29

B phn knh n=3D0 x D0 D1 D2 D1

X

D0 D1 D2

Bng thc tr

D3 s1 s0 s2 Ds0D1 D2 D3 D4 D5 D6 0

D70 0 0 0 0 0 0 xS0 S1 S2

D3 D4 D5 D6 D7

0 0 0 0 1 1 1 1

0 0 1 1 0 0 1 1

0 1 0 1 0 1 0 1

x 0 s1 0 x 0 0 0 0 0 0 0 0 0 0 0 0

0 0 x 0 0 0 0 0

0 0 0 x 0 0 0 0

0 0 0 0 x 0 0 0

0 0 0 0 0 x 0 0

0 0 0 0 0 0 x 0

MchKhoa Khoa hc v K thut my tnh - B mn k thut my tnh

30

B phn knh n=3D0 x D0 D1 D2 D1

X

D0 D1 D2

Bng thc tr

D3 s1 s0 s2 Ds0D1 D2 D3 D4 D5 D6 0

D70 0 0 0

D3 D4 D5 D6 D7

0 0 0 0 1 1 1 1

0 0 1 1 0 0 1 1

0 1 0 1 0

x 0 s1 0 x 0 0 0 0

0 0 x 0

0 0 0 x

0 0 0 0

0 0 0 0

0 0 0 0

Mch0phn knh (hin thc 0 0 0 x 0 0 0 1 0 0 0 0 0 x 0 0 bng0mch 0hai lp) tng qut? 0 0 0 0 0 x 01 0 0 0 0 0 0 0 xS0 S1 S2

MchKhoa Khoa hc v K thut my tnh - B mn k thut my tnh

30

B m ha - B gii mB m ha (encoder) l mch t hp nhn d liu t cc ng nhp mt dng ri bin i thnh d liu thnh mt dng khc xut ra cc ng xut B gii m (decoder) l mch t hp thc hin chc nng ngc vi b m ha, bin i d liu tr li dng ban u Cho hai mch t hp: binary BCD; BCD binary Mch no l mch m ha, mch no l mch gii m


31

B m ha - B gii mB m ha (encoder) l mch t hp nhn d liu t cc ng nhp mt dng ri bin i thnh d liu thnh mt dng khc xut ra cc ng xut B gii m (decoder) l mch t hp thc hin chc nng ngc vi b m ha, bin i d liu mli dng Khi no mch c gi l tr ban u

ha, gii m?

Cho hai mch t hp: binary BCD; BCD binary Mch no l mch m ha, mch no l mch gii m


31

M one-hotMt m kh ph bin c im C n ng d liu trng thi hot ng, ch c 1 trong n ng l tch cc, cn cc ng khc khng tch cc VD: m one-hot 4 ng d liu, tch cc l 1. Cc gi tr ca m l: 0000, 1000, 0100, 0010, 0001


32

Mch bin i binary one-hotCn gi l n-to-2n line decoder Xy dng da trn b phn knh B bt ng nhp d liu x Mi cng AND ch cn li n ng nhp B gii m 3 ra 8


33

Mch bin i binary one-hotCn gi l n-to-2n line decoder Xy dng da trn b phn knh B bt ng nhp d liu x Mi cng AND ch cn li n ng nhp B gii m 3 ra 8


33

Mch bin i binary one-hotCn gi l n-to-2n line decoder Xy dng da trn b phn knh B bt ng nhp d liu x Mi cng AND ch cn li n ng nhp B gii m 3 ra 8 Control inputsS2 0 0 0 0 1 1 1 1Khoa Khoa hc v K thut my tnh - B mn k thut my tnh

Data outputs 0 1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 133

S1 S0 0 0 1 1 0 0 1 1 0 1 0 1 0 1 0 1

D0 D1 D2 D3 D4 D5 D6 D7 1 0 0 0 0 0 0 0

B gii m n ra

n 2

(tt)

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

Cc b gii m ln hn (6 x 64, 8 x 256 ...) ?


34

B gii m n ra

n 2

(tt)


Cc b gii m ln hn (6 x 64, 8 x 256 ...) ?


34

B gii m n ra

n 2

(tt)


Cc b gii m ln hn (6 x 64, 8 x 256 ...) ?


34

B gii m n ra

n 2

(tt)


Cc b gii m ln hn (6 x 64, 8 x 256 ...) ?


34

Gii m ma trn cng AND


35

Gii m ma trn cng AND


35

Gii m cy (tree decoder)


36

Gii m cy (tree decoder)


36

B gii m lm mch a nng


37

B gii m lm mch a nng


37

ROM (read-only memory)C th dng b gii m n-to-2n lm ROM B gii m n-to-2n (n input v 2n output) Cc cng OR 2n ng nhp (ng ra ca mch) Mng kt ni ng ra ca b gii m vi ng vo ca cc cng OR Hai loi ROM n gin Mask programmable ROM Field programmable ROM (PROM)


38

PLA (programmable logic array)Gim s lng cng AND Da vo tnh cht s lng term ti a ca mt biu thc n bin l 2n-1 p cng AND 2n ng nhp Cc cng OR p ng nhp Mng kt ni n ng nhp vi cc ng nhp ca cng AND Mng kt ni ng ra ca cc cng AND vi ng vo ca cc cng OR


39

PAL (programmed array logic)Gim s kim kt ni so vi PLA C nh ng kt ni p cng AND vi cc ng vo ca cc cng OR


40

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