Kythuatso

H ni 7/ 2005

o Thanh Ton Phm Thanh Huyn

----- -----

Bi ging K thut s

Chuyn ngnh: KTVT, KTTT, KH-THGT

74LS73J1K1CP1RD1J2K2CP2RD2

Q1Q1__

Q2Q2__

U2

U1A74LS73J1K1CP1RD1J2K2CP2RD2

Q1Q1__

Q2Q2__

U2

U1A

1100110000111111000000

PTH-DTT

2

Li ni u:

K thut s l mn hc nghin cu v cc mc logic s phng php biu din ti thiu ho bi ton v tn hiu s, nghin cu cc mch s c bn: mch t hp, mch dy.

Bi ging K thut s c bin son da trn cc gio trnh v ti liu tham kho mi nht hin nay, c dng lm ti liu tham kho cho sinh vin cc ngnh: K thut Vin thng, K thut Thng tin, T ng ho, Trang thit b in, Tn hiu Giao thng.

Trong qu trnh bin son, cc tc gi c cc ng nghip ng gp nhiu kin, mc d c gng sa cha, b sung cho cun sch c hon chnh hn, song chc chn khng trnh khi nhng thiu st, hn ch. Chng ti mong nhn c cc kin ng gp ca bn c

Xin lin h: [email protected]

BomonKTDT-HGTVT

3

Phn 1

i s boolean v vi mch s

PTH-DTT

4

Chng 1:

H thng m v m

I. Biu din s trong cc h thng m 1. Khi nim c bn

+ H thng m l t hp cc quy tc gi v biu din cc con s c gi tr xc nh

+ Ch s l nhng k hiu dng biu din mt con s

+ Phn loi h thng m gm 2 loi l h thng m theo v tr v h thng m khng theo v tr

. H thng m theo v tr l h thng m trong gi tr v mt s lng ca mi ch s ph thuc vo v tr ca ch s nm trong con s

V d: trong h m thp phn: Con s 1278 c s 8 ch 8 n v

Con s 1827 c s 8 ch 8.103 n v

Nh vy tu vo v tr khc nhau trong con s m ch s biu din gi tr khc nhau. . H thng m khng theo v tr l h thng m gi tr v mt s lng ca mi ch s khng ph thuc vo v tr ca ch s nm trong con s.

V d: trong h m La m trong cc con s IX, XX hay XXXIX u c X biu din gi tr 10 trong h thp phn m khng ph thuc vo v tr ca n trong con s.

Nhn xt: h thng m khng theo v tr cng knh khi biu din gi tr ln do t s dng. Do vy, khi ni ti h thng m ngi ta hiu l h thng m theo v tr v gi tt l h m. 2. Cc h m thng dng

Nu mt h m c c s l N th mt con s bt k trong h m s c gi tr trong h thp phn thng thng nh sau:

00

11

22

11 ....... NaNaNaNaA

nn

nn ++++=

Trong ak l cc ch s lp thnh con s (k = 0, 1 n-1) v 0 < ak < N-1

Sau y l mt s h m thng dng:

+ H m mi (thp phn): c c s l 10, cc ch s trong h m ny l: 0, 1, 2, 3, 4, 5, 6, 7, 8 v 9.

v d: con s 1278 = 1.103 + 2.102 + 7.101 + 8.100 biu din mt nghn hai trm by mi tm n v theo ngha thng thng + H m hai (nh phn): c c s l 2, cc ch s trong h m ny l 0 v 1

v d: 1011 trong h nh phn s biu din gi tr

A = 1.23 + 0.22 + 1.21 + 1.20 = 11 trong h m 10 thng thng + H m mi su (thp lc phn hexa): c c s l 16 vi cc ch s: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C, D, E v F

BomonKTDT-HGTVT

5

v d: 8E trong h m hexa s biu din gi tr

A = 8.161 + 14.160 = 142 trong h m 10 thng thng + H m tm (bt phn octa): c c s l 8 vi cc ch s 0, 1, 2, 3, 4, 5, 6 v 7

vd: con s 12 trong h octa biu din gi tr

A = 1.81 + 2.80 = 10 trong h m thng thng Bng i chiu 16 con s u tin trong cc h m trn

H 10 H 2 H 16 H 8

0 0000 0 0

1 0001 1 1

2 0010 2 2

3 0011 3 3

4 0100 4 4

5 0101 5 5

6 0110 6 6

7 0111 7 7

8 1000 8 10

9 1001 9 11

10 1010 A 12

11 1011 B 13

12 1100 C 14

13 1101 D 15

14 1110 E 16

15 1111 F 17

3. Biu din s trong cc h m

Mt s trong h 10 c biu din vi cc thnh phn: du ( + hoc - ), phn nguyn, du phy ( , ) v phn l

Khi cc con s c x l bi cc mch s th cc con s ny phi c biu din di dng h 2 hoc dng m no to thnh t cc s h 2 nh m BCD, m Gray ). Do vy, cc con s c th biu din theo s sau:

PTH-DTT

6

Du phy tnh:

Dng nguyn: du phy lun sau ch s cui bn phi. v d: 1001,

Dng l: du phy lun trc ch s u bn tri. v d: ,1001 Du phy ng:

Chuyn s thnh dng chun ho dng lu tha

v d: 12,78 chuyn thnh (,1278).102

Du : quy c ly gi tr 1 ch du m v gi tr 0 ch du dng v d: 1 0101 trong h 2 ch s -5 trong h m 10

0 1001 trong h 2 ch s +9 trong h m 10

Tuy nhin, ngi ta cng cn thng s dng s b biu din s m nh sau:

S b 1: dng s 1 biu din du m v phn gi tr thc hin php ly phn b cho mi ch s (chuyn 1 thnh 0 v 0 thnh 1 cho mi ch s)

v d: s b 1 ca 0101 l 1 1010

S b 2: dng 1 biu din du m cn phn gi tr i ra s b 1 sau cng thm 1 vo hng n v

v d: s b 2 ca -0101 l 1 1011

S b 9: dng 1 biu din du m cn phn gi tr tr thnh mt s sao cho tng ca s mi v s c mi hng bng 9

v d: s b 9 ca 0011 0100 0010 (bng 342 theo h mi) l 1 0110 0101 0111 (bng 657 theo h mi)

S b 10: ly s b 9 cng thm 1 n v

v d: s b 9 ca 0011 0100 0010

l 1 0110 0101 1000 (bng -658 theo h mi)

II. h m hai (nh phn)

1. Cc php tnh s hc trong h m 2 (module 2)

Con s

Du phy tnh

Du phy ng

Dng l

Dng nguyn

H 2

H BCD

C s

28 10 16

Dng l

Dng nguyn

H 2

H BCD

BomonKTDT-HGTVT

7

+ Php cng: Da trn cc nguyn tc sau

0 + 0 0

1 + 0 1

0 + 1 1

1 + 1 10 (0 nh 1)

+ Php tr: Da trn cc nguyn tc sau

0 - 0 0

1 - 0 1

1 + 1 0

10 - 1 1

+ Php nhn: Da trn cc nguyn tc sau

0 . 0 0

1 . 0 0

0 . 1 0

1 . 1 1

+ Php chia: thc hin nh vi h thp phn 2. Chuyn i gia h 2 v h 10

Trong khi con ngi s dng h m 10 th cc mch gia cng v x l s liu li s dng h m 2 nn vic chuyn i gia hai h m ny l rt quan trng.

a. Chuyn i t h 2 sang h 10

Mt con s trong h 2 c gi tr trong h 10 l:

00

11

22

11 2.2....2.2. aaaaA

nn

nn ++++=

trong ak = 0 hoc 1 (vi k = 0, 1, 2, n-1)

v d: chuyn i con s 1001 trong h 2 sang h 10 nh sau:

A = 1.23 + 0.22 + 0.21 + 1.20 = 9

b. Chuyn i s t h 10 sang h 2

Chuyn i tng phn nguyn v phn l sau gp li

Chuyn i phn nguyn theo nguyn tc chia v ly phn d

v d: chuyn i s 17 h mi sang h hai nh sau

PTH-DTT

8

Phn nguyn chia cho 2

0 1 2 4 8 17 s h 10

S d 1 0 0 0 1 S h 2

Chuyn i phn l theo nguyn tc nhn 2 tr 1nh sau:

t s 10 (phn l) tn cng bn tri. Nhn s h mi ny vi 2, nu tch s ln hn 1 th ly tch s tr i 1, ng thi ghi 1 xung hng di (hng t h s cn tm), nu tch s nh hn 1 t 0 xung hng di, ghi sang ct 2 v tip tc ti khi hiu s bng 0 hoc t s l theo yu cu

v d: chuyn i s 0,525 h mi sang h hai. p dng quy tc trn ta c:

H 10

0,525

0,525 x 2 = 1,05

1,05 1 = 0,05

0,05 x 2 = 0,1 0,1 x 2 = 0,2 0,2 x 2 = 0,4

H 2 1 0 0 0

Vy s h 2 thu c l 0,1000 T 2 kt qu trn ta tm c s h 2 tng ng vi s h 10 bng cch gp phn nguyn v phn l vi nhau

v du:

S h 10 S h 2

17 10001

0,525 0,1000

17,525 10001,1000

III. M ho h s 10 1. Khi nim v m ho h s

thc hin vic chuyn i cc con s gia 2 h thng m 2 v 10 ngi ta s dng phng php biu din 2 10. Phng php ny gi l m ho cc con s trong h m 10 bng cc nhm m h 2 (BCD Binary Coded Decimal).

Cc ch s trong h 10 gm cc s t 0 ti 9 do s c biu din bng cc h s hai c 4 ch s. Ngha l thc hin chuyn i mt s h 2 sang h 10 ta phi thc hin chuyn i vi n = 4

0123

00

11

22

11

12482.2....2.2.

aaaaAaaaaA nn

nn

+++=++++=

Trong , 8-4-2-1 gi l trng s v m c quy lut trn gi l m BCD c trng s t nhin hay m BCD 8421

v d:

BomonKTDT-HGTVT

9

H 10 M BCD 8421

12 0001 0010

1278 0001 0010 0111 1000

Tuy nhhin, trn thc t ngi ta cn s dng cc m BCD vi trng s khc nhau nh: 7421, 5421, 2421

Ch : Cc con s biu din bng m BCD 8421 v 7421 l duy nht trong khi cc

m BCD 5421 hay 2421 l khng duy nht.

2. Cc m thng dng

Khi s dng 4 ch s h 2 ta s c 16 t hp khc nhau nhng m BCD ch s dng 10, do d 6 t hp. Bng cch chn 10 trong s 16 t hp khc nhau ngi ta s c nhiu loi m khc nhau. Thng dng nht l: M BCD, M tha 3,M

Gray Ngoi ra c th s dng 5 ch s h 2 m ho, v d: M Johnson, M 2

trn 5

+ M BCD: c trnh by trn

+ M tha 3: c to thnh bng cch cng thm 3 n v vo m BCD 8421. Loi m ny c s dng rng ri trong thit b tnh ton s hc ca h thng x l hoc gia cng cc tn hiu s.

+ M Gray: c c im l khi chuyn t mt m s ny sang m s khc tip

theo th t m ch thay i ti cng 1 v tr ca k hiu m

+ M 2 trn 5: s dng 5 ch s h 2 biu din cc ch s h 10. Mi t

hp lun c 2 ch s 1 v 3 ch s 0.

+ M Johnson: s dng 5 ch s h 2 vi c im l khi chuyn sang m s

k tip s thay 0 bng 1 bt u t phi sang tri ti khi t 11111 ( ng vi 5 trong

h 10) s bt u thay 1 bng 0 v cng theo chiu t phi sang tri.

PTH-DTT

10

Bng biu din cc ch s h 10 theo cc loi m khc nhau

S h 2

(BCD 8421)

M tha 3 M Gray M 2 trn 5 MS h 10

B3 B2 B1 B0 A3 A2 A1 A0 G3 G2 G1 G0 D4 D3 D2 D1 D0 J4 J

0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 1 1 0 0

1 0 0 0 1 0 1 0 0 0 0 0 1 0 0 1 0 1 0 0

2 0 0 1 0 0 1 0 1 0 0 1 1 0 0 1 1 0 0 0

3 0 0 1 1 0 1 1 0 0 0 1 0 0 1 0 0 1 0 0

4 0 1 0 0 0 1 1 1 0 1 1 0 0 1 0 1 0 0 1

5 0 1 0 1 1 0 0 0 0 1 1 1 0 1 1 0 0 1 1

6 0 1 1 0 1 0 0 1 0 1 0 1 1 0 0 0 1 1 1

7 0 1 1 1 1 0 1 0 0 1 0 0 1 0 0 1 0 1 1

8 1 0 0 0 1 0 1 1 1 1 0 0 1 0 1 0 0 1 1

9 1 0 0 1 1 1 0 0 1 1 0 1 1 1 0 0 0 1 0

BomonKTDT-HGTVT

11

Chng 2:

i S Boolean I. Khi nim chung 1. M u

K thut in t ngy nay c chia lm 2 nhnh ln k thut in t tng t v k thut in t s. K thut in t s ngy cng th hin nhiu tnh nng u vit v tc x l, kch thc nh gn, kh nng chng nhiu cao, tiu th in nng t . Do , in t s c ng dng rng ri trong nhiu lnh vc v ngy cng tr thnh mt phn thit yu hn trong cc h thng v thit b hu ht cc lnh vc c ng dng khoa hc k thut v cng ngh mi (c kh, ho hc, y hc...).

Hn na, vi s pht trin ca mch tch hp to nn s thc y cng mnh m trong vic to ra nhng mch s c phc tp cng tng. Nn cng ngh ban u ch to c cc mch tch hp c nh (S.S.I) nhng, ngy nay, vic s dng cc mch tch hp c va (M.S.I), c ln (L.S.I) v cc ln (VLSI) ngy cng tr nn ph bin.

Trong mch s, tn hiu u vo 1 trong 2 trng thi logic 0 hoc 1 v u ra cng 1 trong 2 trng thi 0 hoc 1tu theo tn hiu u vo v cc phn t trong mch gi l cc cng logic. m t mch s ngi ta s dng cng c ton hc l i s Boolean (i s logic). y l c s ton hc cho mi lnh vc c lin quan n k thut s. 2. Mt s khi nim c bn

+ i s logic: l mt tp hp S ca cc i tng A, B, C trong xc nh 2 php ton cng logic v nhn logic vi cc tnh cht sau:

Tnh cht Tn gi

S cha (A + B) v (A.B) tnh ng kn

A + B = B + A

A.B = B.A

Lut giao hon

(A + B).C = A.C + B.C

A + B.C = (A + B).(A + C)

Lut phn phi

(A + B) + C = A + (B + C)

(A.B).C = A.(B.C)

Lut kt hp

A + A = A

A.A = A

A + B = B A.B = A tnh nht qun

PTH-DTT

12

A + 0 = A

A . 0 = 0

A + 1 = 1

A . 1 = A

0.

1

==+

AA

AA

A. (A + B) A + A.B A Lut hp th

BABA

BABA

+==+

.

.

Lut De Morgan

CBCACBCABA

BAABA

BABAA

.....

.

+=+++=++=+

10

01

== AA

+ Gin Venn: y l cch biu din trc quan cc php ton trong i s logic. Trn gin Venn tp hp S c biu din bng 1 vung cn cc phn t A, B, C c biu din bng cc min nm trong vung . Min khng c trn gin c coi bng 0 v min ln nht (ton b vung) c coi bng n v 1. v d: tp hp S l mt nhm cc sinh vin v c biu din bi ton b min trong

hnh vung; trong nhm sinh vin c 2 nhm ph A v B, vi sinh vin thuc nhm A c tc nu trong khi cc sinh vin ca nhm B c mt xanh.

Khi , phn giao ca A v B bao gm cc sinh vin c c mt xanh v tc nu (A.B). H l thnh vin ca c nhm A v nhm B.

Nhm cc sinh vin m c tc nu hoc mt xanh c th c biu din: A+B (c xem nh hp ca cc nhm)

A.B hay BA A+B hay BA

BomonKTDT-HGTVT

13

II. Bin v hm logic 1. Khi nim v bin v hm logic

+ Bin logic l mt khi nim dng thay cho thut ng mnh tu , mnh ny c th ng hoc sai v khng c kh nng mt mnh va ng va sai, ngha l bin logic ch nhn mt trong hai gi tr l ng hoc sai

V d, cu: Hm nay l th Nm v tri ang ma c th c biu din nh sau:

C = A.B.

vi A : hm nay l th Nm.

B: tri ang ma. C: ton b cu.

Khi no th ton b cu l ng?

C th thit lp mt bng lit k cc trng hp ng(True) hay sai(False) cho A v B:

A B C

sai

sai

ng

ng

sai

ng

sai

ng

sai

sai

sai

ng

Nu 1 c s dng thay th cho pht biu ng v 0 cho pht biu sai th bng trn c th c biu din li nh sau:

A B C

0

0

1

1

0

1

0

1

0

0

0

1

Nh vy, ton b cu l ng khi A v B u ng cn cc trng hp khc C sai.

+ Mt mnh phc tp c to thnh t cc mnh n gin ban u, n nhn mt trong 2 gi tr l ng hoc sai. Khi , k hiu l F(A, B, C ) hay F(x1, x2, x3 ), ngi ta gi l hm logic ca cc bin A, B, C hay ca x1, x2, x3 + Trong k thut s cc gi tr ng v sai ca bin logic hay hm logic c k hiu l 1 v 0 (y n thun l k hiu m khng phi l ch s ca h hai). Thm na vic thc hin cc gi tr logic cn ph thuc vo vic chn cc tr s vt l biu din.

V d: vi vi mch thuc h TTL ngi ta a ra 2 cch k hiu cho mc logic

PTH-DTT

14

. Mc logic dng: Xi = 1 ng vi mc in p cao 5V

Xi = 0 ng vi mc in p thp 0V

. Mc logic m:

Xi = 1 ng vi mc in p thp 0V

Xi = 0 ng vi mc in p cao 5V 2. Cc hm logic s cp

a. Hm logic s cp mt bin

A F(A) Fi

0 1 Biu thc Tn gi

F1 0 0 0 Hng s 0

F2 0 1 A Lp li A YES

F3 1 0 A o bin A NOT

F4 1 1 1 Hng s 1

b. Hm logic hai bin

A 0 0 1 1

B 0 1 0 1 K hiu v biu thc i s

ca hm Tn gi ca hm

F0 0 0 0 0 F0 = 0 Hng s 0

F1 0 0 0 1 F1 = A.B Nhn logic AND

F2 0 0 1 0 F2 = BA. Cm B

F3 0 0 1 1 F3 = A Lp li A

YES / BUFFER

F4 0 1 0 0 F4 = AB. Cm A INHIBITION

F5 0 1 0 1 F5 = B Lp li B

YES / BUFFER

F6 0 1 1 0 F6 = BA. + AB. = BA Khc du / cng module 2 XOR

F7 0 1 1 1 F7 = A + B Cng logic OR

F8 1 0 0 0 F8 = BABA += Hm Pierce NOR

BomonKTDT-HGTVT

15

F9 1 0 0 1 F9 = A ~ B = BABA .. + ng du F10 1 0 1 0 F10 = B B ca B

NOT B

F11 1 0 1 1 F11 = BAAB += Ko theo A IMPLICATION

F12 1 1 0 0 F12 = A B ca A NOT B

F13 1 1 0 1 F13 = BABA += Ko theo B IMPLICATION

F14 1 1 1 0 F14 = A/B = BA. Hm Sheffer NAND

F15 1 1 1 1 F15 = 1 Hng s 1

Cc hm logic s cp

+ Hm F(A,B) = A.B

Hm ny thc hin php nhn logic ca hai bin A v B. Phn t thc hin chc nng ca hm trn l phn t AND (cn gi l cng AND). Mt cng AND c hai hay nhiu u vo v ch c mt u ra. u ra c mc logic 1 ch khi tt c cc u vo mc 1; v c mc 0 khi mt trong cc u vo mc 0. Hnh di y ch ra k hiu v bng chn l ca cng AND vi 2 u vo.

Tng qut: Hm AND ch mang ga tr 1 khi cc u vo ng thi bng 1 + Hm F(A,B) = A + B

Hm ny thc hin php cng logic. Phn t thc hin l phn t OR (cn gi l cng OR). Cng OR c mc logic cao khi c t nht mt u vo mc 1; v ch khi c 2 u vo mc logic 0 u ra cng OR mi c mc logic 0. Hm OR c k hiu v bng chn l nh hnh di y:

Tng qut: Hm OR ch mang gi tr 0 khi tt c cc u vo ng thi bng 0

PTH-DTT

16

+ Hm F(A) = A Hm ny thc hin php ly phn t b ca A. Phn t thc hin hm l phn

t NOT, thng c gi l cng o, c mt u vo v mt u ra. Trng thi ca u ra lun ngc vi u vo. K hiu ca mch v bng chn l nh sau: + Hm F(A,B) = BA.

Hm ny cn gi l hm Sheffer. Phn t mch in thc hin hm l phn t NAND (cng NAND). V c bn, y l mt cng AND theo sau l cng NOT. u ra c mc logic 0 ch khi tt c u vo c mc logic 1. Di y l k hiu v bng trng thi (bng chn l) ca cng NAND 2 u vo.

Tng qut: Hm NAND ch mang gi tr 0 khi tt c cc u vo u c mc logic 1

+ Hm F(A,B) = BA + Hm ny cn gi l hm Pierce. Phn t mch in thc hin hm l phn t

NOR (cng NOR). y l cng OR theo sau bi cng NOT. u ra c mc logic thp khi mt hay nhiu u vo mc logic cao; v u ra c mc logic cao ch khi tt c u vo mc thp. Di y l k hiu v bng chn l ca hm. Tng qut: hm NOR ch mang gi tr 1 khi tt c cc u vo u c mc logic 0

+ Hm F(A,B) = BA BABA .+= Phn t thc hin hm ny l phn t Exclusive OR (hay cng XOR). Cng

ny c 2 u vo. Cng ny l thnh phn c bn ca php so snh. Khi 2 u vo ging nhau, u ra mc logic 0; cn khi 2 u vo khc nhau, u ra c mc logic 1. Di y l k hiu v bng trng thi. Tng qut: hm XOR cho gi tr 1 khi s cc ch s 1 trong t hp l mt s l.

A Y 1 0 0 1

BomonKTDT-HGTVT

17

y chnh l tnh cht ca hm cng module n bin

+ Hm F(A,B) = BA = BABA =~ = BABA .. + Hm ny gi l hm tng ng. Cng logic thc hin hm ny l cng

XNOR. y l s kt hp ca hm XOR v theo sau bi hm NOT. Khi 2 u vo ging nhau u ra mc logic 1; cn khi 2 u vo khc nhau, u ra c mc logic 0. Di y l bng chn l v k hiu hm

Tng qut: hm XNOR s mang gi tr 1 khi s cc ch s 1 trong t hp l mt s chn (k c 0)

Ch : Vi cng mt phn cng nh nhau nhng nu s dng vi cc mc logic khc nhau th chc nng ca cc cng s thay i. Cc cng logic trn c thc hin vi kiu logic dng. Nu dng logic m th ta c tng ng nh sau:

3. H hm y d

Mt hm logic bt k lun c biu din di dng t hp ca cc hm s cp trn. Tuy nhin, trn thc t khng nht thit phi s dng ht cc hm s cp m ch cn mt b phn ca cc hm s cp.

PTH-DTT

18

Mt h hm s cp c gi l y nu c th biu din mt hm logic bt k bng cch thc hin cc php ton ca i s logic ln cc phn t ca h hm ny.

Cc h hm sau c chng minh l cc h hm y :

+ H hm 1: gm cc hm AND, OR, NOT

+ H hm 2: gm cc cng AND, NOT

+ H hm 3: NOR

+ H hm 4: NAND

+ H hm 5: AND, NOT

Gii thch chi tit hm NOR v hm NAND to thnh cc hm khc nh th no v trnh by phng php thit k mch dng cng NOR v cng NAND III. Phng php biu din hm logic 1. Phng php dng bng gi tr ca hm

Phng php ny s dng bng ghi mi t hp c th ca bin v gi tr hm tng ng. Bng ny cn gi l bng hm hay bng chn l (bng s tht)

v d: Cho mt hm 3 bin c gi tr nh trong bng ng vi cc t hp ca bin nh sau:

X3 X2 X1 F 0 0 0 0 0 0 1 1 0 1 0 X 0 1 1 1 1 0 0 1 1 0 1 0 1 1 0 0 1 1 1 X

X l k hiu m ti gi tr ca hm khng xc nh (c th l 0 v c th l 1)

Nhn xt: Phng php trn c u im l trc quan v r rng nhng n t ra cng knh v qu rm r khi s bin tng ln. Do phng php ny ch dng biu din cho cc hm s cp hay cc hm c s bin nh. 2. Phng php hnh hc

Trong phng php ny ngi ta biu din n bin ng vi khng gian n chiu. Mi t hp ca bin c biu din bi mt im trong khng gian

Nh vy, n bin s biu din bi 2n im vi quy c 2 im trn cng mt cnh ch khc nhau 1 bin duy nht.

BomonKTDT-HGTVT

19

v d: trng hp 1, 2 v 3 bin biu din nh trong hnh di y 3. Phng php biu thc i s

nh l: Mt hm logic n bin bt k lun c th biu din di dng chun tc tuyn y hoc chun tc hi y

Dng chun tc tuyn y l tuyn ca nhiu thnh phn, mi thnh phn l hi gm y n bin

Dng chun tc hi y l hi ca nhiu thnh phn, mi thnh phn l tuyn gm y n bin

a. Cch vit hm s d-i dng chun tc tuyn ( CTT ) y : + S ln hm bng 1 s l s tch ca n bin

+ Trong mi tch cc bin c gi tr 1 c gi nguyn, cc bin c gi tr 0 c ly ph nh

+ Hm F bng tng cc tch trn

b. Cch vit hm s d-i dng chun tc hi ( CTH ) y : + S ln hm bng 0 s l s tng ca biu thc n bin

+ Trong mi tng cc bin c gi tr 0 c gi nguyn, cc bin c gi tr 1 c ly ph nh

+ Hm F bng tch cc tng trn

v d: Xy dng hm logic ca cc bin A, B ,C c cc gi tr nh sau: F (0,0,0) = F( 1, 0,0) = F(1,1,0) = 1

Cc trng hp khc bng 0 Thc hin cc bc nh trn ta c hm F vit di dng CTT v CTH nh sau:

F(A, B, C) = =++ 6,4,0...... CBACBACBA F(A, B, C) =

=++++++++++ 7,5,3,2,1))()()()(( CBACBACBACBACBA 4. Phng php dng bng Karnaugh

Quy tc xy dng bng:

+ Bng c 2n biu din hm n bin, mi cho mt t hp bin

010 011

001

101100

110

000

111

10 11

0100

10

PTH-DTT

20

+ Cc cnh nhau hay i xng nhau ch khc nhau 1 bin (ghi theo th t ca m Gray). Cc hng v ct ca bng c ghi cc t hp gi tr bin sao cho hng v ct cnh nhau hay i xng nhau ch khc nhau 1 bin

+ Ghi gi tr ca hm ng vi t hp ti

Ch : i vi CTT gi tr hm bng 0 c trng i vi CTH gi tr hm bng 1 c trng Hm khng xc nh ti t hp no th nh du X vo

v d: biu din hm sau bng bng Karnaugh

F(A, B, C) = 5,2,0 vi N = 1, 4 (cch vit theo CTT) F(A, B, C) = 7,6,3 vi N = 1, 4 (cch vit theo CTH) Vi N l tp hp ca t hp bin m ti gi tr ca hm khng xc nh.

Thc hin nh cc bc trn ta c bng Karnaugh biu din cho hm F theo CTT nh sau:

A \ BC 00 01 11 10

0 1 X 1

1 X 1

Hoc c th biu din hm F theo CTH nh sau:

A \ BC 00 01 11 10

0 X 0

1 X 0 0

BomonKTDT-HGTVT

21

Chng 3

Ti thiu ho hm Boolean

I. Phng php ti thiu ho 1. Khi nim ti thiu ho

Ti thiu ho l tm dng biu din i s n gin nht ca hm. Khi s gim c ti a s cng thc hin hm. y l yu cu rt cn quan tm v n gip cho vic thc hin mch c n gin v hiu qu. V d: Cho hm c dng CTT v CTH y nh sau:

))()()((

......

123123123123

123123123

XXXXXXXXXXXXF

XXXXXXXXXF

++++++++=++=

Khi s cng thc hin hm s c dng:

U4A

U3B

U3A

U2C

U2B

U1C

U1B

U1A

U2A

Tuy nhin nu s dng bng chn l ca hm ta c:

X3 X2 X1 F

0 0 0 0

0 0 1 0

0 1 0 1

0 1 1 X

1 0 0 0

1 0 1 0

1 1 0 1

1 1 1 1

T bng chn l d dng thy F = X2. R rng biu thc ny n gin hn rt nhiu so vi biu thc trn, v th mch lc ny cng ch l mt b m cho X2 m thi

PTH-DTT

22

FX2

Cng c mt s yu t khc ngoi gi thnh nh hng n phc tp ca mch cn c quan tm. Mt trong cc yu t l thi gian tr truyn t, l khong thi gian tnh t lc c s thay i ti u vo ti khi c s thay i kt qu ti u ra. Cng nhiu cng c mc ni tip vi nhau th thi gian tr ny cng ln.

V d vi hm : f = A*B*C + A*B*C+A*D 1)

l mt dng ti thiu v u ra c mc tr ca cng AND thm vi mc tr ca cng OR.

Tuy nhin, cng vi hm ny theo lut phn phi, ta c: f = A*(B*C+ B*C +D). 2).

Hm ny c thi gian tr ln hn hm trc v n gm mc tr ca 3 cng. Bi th, d r hn, n c thi gian tr ln hn.

Mt yu t ng quan tm khc l ti ca u vo. Xt 1). tn hiu A phi iu khin 3 ti (3 cng), trong khi vi 2). ti ch c mt cng.

Ti nay vn cha c phng php ti u no c th thc hin vic ti thiu ho mt cch ti u. Vic ti thiu ho hm logic c th thc hin bng mt trong hai cch c bn l:

+ Bin i i s

+ Thut ton 2. Phng php ti thiu ho hm logic bng bin i i s

Trong trng hp s bin t v hm c biu din bng phng php gii tch ngi ta c th thc hin bin i trc tip hm theo cc tnh cht ca i s

V d: dng phng php bin i i s ta thc hin rt gn hm f nh sau:

AXfXXAAAXf

XAXAXAXAf

XAXAXAf

+=+++=+++=

++=

)()(

....

...

r rng l hm f c n gin i rt nhiu thay v mt hm phc tp

fAX

U8AfU3C

U7C

U7B

U6BX

AU7A

U6A

BomonKTDT-HGTVT

23

3. Nhm cc phng php ti thiu ho theo thut ton

Mt s khi nim:

nh: nh l mt tch gm y cc bin ca hm ban u (nu hm c n bin th nh l tch n bin)

nh 1 l nh m ti hm s bng 1

nh 0 l nh m ti hm s bng 0

nh khng xc nh l nh ti hm khng xc nh (k hiu l X)

Thng thng khi cho mt hm s dng CTT ngi ta cho tp cc nh 1 v cc nh khng xc nh (N) ca hm ban u.

Tch cc tiu l mt tch m ti hm bng 1 hoc khng xc nh vi thnh phn cc bin khng b bt c n. Tch cc tiu l biu din ca 1 nhm 2k nh. Tch cc tiu ny ph cc nh hay cc nh cha trong tch cc tiu, ngha l dng tch cc tiu biu din ti a s nh vi s bin t nht. C s ton hc ca vic tm tch cc tiu l p dng php dn: AXAXA =+ ..

Tch quan trng l mt tch cc tiu ph t nht 1 nh 1. N nht thit phi xut hin trong biu thc cui cng ca bi ton. Tp hp cc tch quan trng chnh l ph ti thiu, kt qu cui cng ca bi ton.

Ch : Khi tin hnh vi hm vit di dng CTH y th thay cc nh 1 bng nh 0. Cc khi nim tng v tch cng i ch cho nhau. Ngha l:

nh l tng y n bin

Biu din hm bng tch cc tng

Tng cc tiu

Tng quan trng

Ph ti thiu l s tng quan trng t nht m ph ht c s nh 0 Gi tr ca bin s gi nguyn nu c gi tr 0 v o nu c gi tr 1

Qu trnh ti thiu ho gm cc bc nh sau:

+ Biu din hm s di dng CTT y vi tp cc nh 1 v nh khng xc nh hoc CTH y vi tp cc nh 0 v nh khng xc nh

+ Tm cc tch cc tiu

+ Tm cc ph ti thiu

+ a ra cch biu din mi ca hm

a. Phng php dng bng Karnaugh.

Bng Karnaugh l mt bng c 2n , mi tng ng vi mt t hp trong bng trng thi v cha cc gi tr u ra tng ng. Mt c trng ca biu ny l lun sp xp sao cho ch c s thay i ca mt bin khi chuyn t ny sang k cn.

Trong bng ta ch n 2 du hoa th, ta s vit c:

PTH-DTT

24

A = 1L . 1R .L2.R2 + 1L .R1.L2.R2

S dng cc nh l ca i s Boolean, c th vit li:

A = 1L .L2.R2.( 1R +R1)

= 1L .L2.R2. 1

= 1L .L2.R2.

Nh vy, hm c ti thiu ho gm mt cng AND 3 u vo. Nguyn l thit lp biu Karnaugh

chnh l ti cc k nhau, gi tr 1 c nhm li vi nhau. Kch thc ca nhm l lu tha ca 2 (v d: 2 , 4 , 8 , 16 , 32 ...). V d 4 ca ct th t trong bng hnh bn c th c nhm. Nh vy, ton b nhm s c ti gin thnh A.B , chnh l cc phn t chung ca c nhm. Cc phn t c gi tr khc nhau (C v D) s khng xut hin. Kt qu ny cng nhn c nu ta p dng cc nh l ca i s Boolean cho 4 ny nh sau:

f = A. DCB .. + A.B .C.D + A.B .C.D

= A. CB . .(D +D) + A.B .C.(D+D )

= A. CB . +A.B .C = A.B .(C+C )

= A.B

L1

BomonKTDT-HGTVT

25

Ch : Bng Karnaugh, ging nh bn th gii, pha bn phi s tip lin pha bn tri, nn c th nhm cc nm i din nhau. Nguyn l ny cng c p dng cho bn trn v bn di. (tc l chng ta nhm theo kiu i xng hoc lin k)

V d, c th nhm 4 4 gc ca biu nh hnh di y

T cc nhn xt trn ta rt ra c cc bc tin hnh ti thiu ho bng bng Karnaugh cho dng CTT l:

1, Biu din hm cho trn bng Karnaugh

2, Xc nh cc tch cc tiu ca hm (tch cc tiu tm c bng cch dn 2k c gi tr 1 hoc X vi k ti a, cc ny gn k hoc i xng nhau)

3, Tm ph ti thiu l chn mt s t nht cc nhm tch cc tiu sao cho ph ht c cc nh 1 ca hm Ch : . Qu trnh hon ton tng t khi hm biu din dng CTH . Khi lp bng Karnaugh vi CTT nhng bng 0 nn trng cn dng CTH th b trng nhng c gi tr 1.

b. Ti thiu ho bng phng php Quine - Mc.Cluskey

Phng php ny c thc hin cho hm biu din di dng CTT Cc bc tin hnh:

Bc 1: Tm tch cc tiu . Xc nh nh 1 v X

. Sp xp cc t hp bin theo s lng ch s 1 c trong chng

. So snh mi t hp thuc nhm i vi t hp thuc nhm (i + 1). Nu 2 t hp ch khc nhau 1 ct s th kt hp 2 t hp thnh mt t hp mi, trong s dng du thay cho ct s khc nhau. nh du vo 2 t hp va kt hp

. Loi b cc t hp ging nhau v lp li bc trn cho n khi ht cc t hp c kh nng kt hp

. Tp hp cc t hp trong bng cui v cc t hp khng b nh du chnh l tp cc tch cc tiu

Bc 2: Tm ph ti thiu

PTH-DTT

26

. Lp bng c ct l cc gi tr c nh l 1 (cc gi tr ny thng ghi theo h m 10 cho tin theo di), hng l cc tch cc tiu

. nh du X vo m tch cc tiu hng ph nh ct. Ct c 1 du X chnh l tch quan trng

. Loi b cc ct c ph trong tch quan trng

. Loi cc tch quan trng khi hng

. Lp bng mi v tip tc qu trnh n khi tt c cc nh u c ph

V d: Ti thiu ho hm sau bng phng php Quine Mc. Cluskey nh sau:

Hm f = )11,10,9,8,5,2,0( Sp xp li Thc hin php dn

H 10 H 2 H 10 H 2

0 0000 0 0000 (0,2) 00-0 (0,2,8,10) -0-0

2 0010 2 0010 (0,8) -000 (0,8,2,10) -0-0

5 0101 8 1000 (2,10) -010 (8,9,10,11) 10- -

8 1000 5 0101 (8,9) 100- (8,10,9,11) 10- -

9 1001 9 1001 (8,10) 10-0

10 1010 10 1010 (9,11) 10-1

11 1011 11 1011 (10,11) 101-

Vy kt qu cui cng l: -0-0 v 10- -

Hay f = BADB +.

BomonKTDT-HGTVT

27

chng 4:

Gii thiu vi mch s

I. nh ngha v phn loi 1. nh ngha

Vi mch l nhng linh kin in t c mt chc nng xc nh v c ch to bng mt cng ngh ring. Vi mch hin i thng a nng v c th s dng linh hot trong nhiu thit b in t khc nhau

Ngi ta phn loi theo mt s tiu ch sau: + Phn loi theo bn cht ca tn hiu in vo / ra ca vi mch

+ Phn loi theo mt tch hp

+ Phn loi theo cng ngh ch to 2. Phn loi vi mch theo bn cht ca tn hiu vo / ra

Nh bit, tn hiu in c phn thnh 2 loi l tn hiu tng t v tn hiu s.

+ Tn hiu tng t (analog) l tn hiu c bin bin thin lin tc theo thi gian

+ Tn hiu s (digital) l tn hiu c bin mt trong hai ga tr hu hn mang ngha logic 0 hoc 1, ng vi 2 mc thp v cao. Tn hiu s gin on theo thi gian.

Nu k hiu X, Y l tn hiu vo v ra ca vi mch, theo bn cht ca tn hiu

vo / ra ny ta s c cc loi vi mch sau:

Tn hiu vo Tn hiu ra Loi vi mch

Tng t Tng t Tng t

S S S

Tng t S ADC / analog-digital converter

S Tng t DCA / digital-analog converter

Trong phm vi ca mn k thut s chng ta ch xt ti vi mch s, ngha l c u vo ln u ra u l tn hiu s.

PTH-DTT

28

Cc vi mch s ny bao gm t cc cng logic n gin nh AND, OR, NOR, NAND flip-flop, MUX, DEMUX, Memory n cc loi mch cc k phc tp nh cc b vi x l

3. Phn loi theo mt tch hp

Mt tch hp c nh ngha l tng cc phn t tch cc (transistor) hoc cng logic cha trn mt n v din tch ca mng tinh th bn dn trong vi mch

v d: B vi x l Pentium III ca Intel c mt tch hp l 9triu transistor trn 1 inch vung

Mc tch hp c nh ngha l tng s nhng phn t tch cc hoc cng logic trn mng tinh th bn dn ca vi mch

Nhng thng s trn phn no cho thy phc tp ca mch. Phn loi theo mc tch hp ta c

Loi mch S transistor S cng logic V d

SSI Vi mch c nh

Hng chc 1 - 10 Gate, flip-flop

MSI Vi mch c trung bnh Hng trm 10 - 100 Gate, counter, shift-register, encoder, small memory

LSI Vi mch c ln Hng nghn 100 - 1000 Larger Memory, microprocessor 4 / 8bit

VLSI - Vi mch c rt ln Hng vn > 1.000 MP 16/32bit, console i/o 8086, Z8000

ULSI Vi mch c cc ln

Hng triu > 10.000 MP 64bit

4. Phn loi theo cng ngh ch to

IC c th chia ra lm 4 loi: IC mng mng/ mng dy; IC khi rn; v IC lai

Di y l cc hng pht trin vi mch theo cng ngh ch to

BomonKTDT-HGTVT

29

a. Vi mch mng mng / mng dy

Cc IC loi ny c ch to bng cch lng ng nhng vt liu nht nh trn mt cch in (v d nh gm, s..). Sau hng lot cc qu trnh to mask trn to thnh in tr, in dung hay in cm. Cc linh kin tch cc nh diode, transistor s c ch to theo cch thng thng vi kch thc nh (thng l FET). Mch ny cho tch hp kh cao nhng khng bng loi n khi, tuy nhin li c kh nng chu ng in p v nhit tt hn. IC mng mng v mng dy c s dng cho cc mch i hi chnh xc cao

b. Vi mch bn dn khi rn

IC monolithic c to ra hon ton trn mt n v tinh th cht bn dn nn l Si, cc cht bn dn khc s c khuch tn vo trong cht nn to ra nhiu loi mt ghp khc nhau. Nhng mt ghp ny c th to thnh in tr, in dung, diode hay transistor.

Nhng vt liu bn dn c khuch tn vo trong cht nn di dng hi v ng li trn cht nn sau hng lot cc qu trnh to mask nhit cao.

Qu trnh to mask l qu trnh trong ngi ta tin hnh oxy ho b mt cht bn dn, tc l lp kn b mt ca n bng SiO2. Sau ph mt lp cm quang ln trn b mt SiO2. Dng mch thu nh, chp ln phim to thnh khun sng. t khun sng ln b mt cht cm quang, chiu nh sng vo ta s thu c dng mch theo yu cu. Dng ho cht n mn cc rnh, loi b cht cm quang

Vi mch

Thick / thin film Monolithic Hybrid

Digital AnalogDigital Analog (BJT)

UJT MOS

BJT

P / N chanel

CMOS

RTL

DTL

TTL

ECL

PTH-DTT

30

thc hin khuch tn cht vo. Mask c to thnh bng phng php nh trn gi l phng php quang khc.

Vi mch monolithic c 2 loi l mch lng cc v mch MOS, ngy nay vi mch MOS tr nn ph bin do d ch to, din tch nh nn kh nng tch hp cao.

c. Vi mch lai

y l s kt hp ca 2 loi vi mch trn. IC lai c th bao gm nhiu tinh th monolithic c ghp vi nhau thnh khi, cng c th l s kt hp gia mch monolithic vi mch mng mng th ng.

IC lai mang y u im ca 2 loi vi mch monolithic v mng mng / mng dy nh kch thc nh gn m cng sut li ln, chnh xc cao II. Cc thng s chnh ca vi mch s 1. Mc logic

Mc logic l gi tr in p vo / ra c quy nh cho cc s nh phn 0 v 1. Mc logic l thng s quan trng nht ca vi mch s, nh thng s ny m ta c th d dng nhn bit c nhng trng thi logic ra v vo bng cch o nh vn k hoc oscilloscope.

Gia cc thng s khc nhau (in p, dng, thi gian...) c trng cho mt h logic th cc tham s in tnh c bit quan trng bi v chng xc nh gii hn dng v p ti u ra v u vo.

Mi trng thi logic ca linh kin (High hay Low) c xc nh bi di in p cho php.

Tng cng s c 4 di in p, mi di c xc nh bi 2 gii hn in p; nh vy s c 8 gi tr in p c trng cho mi h logic. Cc mc v di in p cho php.

Ta c quan h in p u vo v ra sau:

Volmax =VilMin .

VohMax = VihMin . 2. c tnh truyn t

ng c tuyn truyn t (transfer characteristic) l ng cong ch ra mi quan h gia in p vo v ra.

BomonKTDT-HGTVT

31

Di y l ng c tuyn truyn t ca mt cng o (trn) v mt cng khng o (di).

Vi cng o, di in p v cc mc gii hn s c xc nh nh hnh di y:

Trong thc t, in p vo v ra ln nht c cho bi gi tr in p cung cp

Vcc v cc gi tr nh nht l bng zero tc bng in p t.

Nh vy, s ch cn 4 gi tr in p gii hn v c quan h: Volmax = VihMin .

Bng so snh gi tr in p vo v ra ca cc h logic TTL, CMOS, HCT v h ECL.

Ch : in p cung cp khc nhau.

PTH-DTT

32

3. Cc thng s v dng in.

Cc di gii hn v dng in cng c nh ngha tng t nh di gii hn in p. Cc gi tr dng ra l cao hn cc gi tr vo.

Chiu ca dng in c quy c nh sau: chiu dng l chiu dng i vo cc ca linh kin cn chiu m l chiu dng i ra khi linh kin.

V d: cc thng s c tnh l tng ca cng NAND c ch ra hnh a. Tuy nhin, trong thc t cc ng c tnh ny l cc ng cong nh trong hnh b.

a)

b)

Ni chung, vi cc h logic ta u c:

VolMin = 0.

VohMax = Vcc.

bi th, ch cn 4 gi tr gii hn v dng in:

IilMax .

IihMax .

IolMin.

IohMin .

ta c quan h:

IolMin >= IilMax .

IohMin >= IihMax .

BomonKTDT-HGTVT

33

Bng sau s ch ra cc gi tr dng ra v vo tng ng vi cc h logic TTL, CMOS v ECL.

Dng tiu th trong trng thi tnh.

Ti trng thi tnh, dng cung cp l tng dng tiu th ca tng linh kin khi cc cng ca n l n nh, khng xy ra s chuyn trng thi.

Cc nh sn xut s cung cp cc thng tin v dng tnh quiescent cho tng linh kin v di cc iu kin th c bit.

Bng sau ch ra cc gi tr dng max cho mt vi linh kin ca cc h logic.

4. Cng sut tiu th.

Cng sut tiu th bi cc linh kin logic chia thnh 2 loi: tnh v ng.

Thnh phn cng sut tnh to nn do dng tnh.

Thnh phn ng to nn do dng in yu cu tch v phng cho in dung ti u ra; do dng in yu cu bi cc in dung ni; v do dng in cn thit to trng thi dn cho cc Transistor u ra.

Vi cc linh kin ECL, cng sut tiu th ch yu do hot ng trong min tch cc.

Cng sut tiu th c tnh theo cng thc:

P = Icc*Vcc + Cpd*Vcc2*fi + (CL*Vcc2*fo). vi: Icc: dng tnh.

Vcc : in p cung cp.

fi : tn s tn hiu vo.

fo : tn s tn hiu ra.

Cpd : in dung tng ng u vo. CL : in dung ti.

PTH-DTT

34

Thnh phn cng sut tnh tiu th ca cc linh kin LS-TTL cao hn rt nhiu so vi linh kin CMOS nhng li nh hn so vi linh kin h ECL.

Tng cng sut ng ca 1 linh kin h CMOS ph thuc ch yu vo tn s, khng ging nh linh kin h TTL.

Bng cng sut tiu th ca cc linh kin trong mt s h logic. 5. H s ti FAN-IN; FAN-OUT

H s ti u vo FAN-IN.

FAN-IN l t s gia dng vo ca 1 linh kin c th v dng vo ca 1 mch chun.

Thng thng, mch c ly lm chun s l 1 cng logic c bn ca cng h logic. H s ny c dng nhiu trong qu kh khi cc h logic mi c gii thiu. Ngy nay, h s FAN-IN khng c nhc n trong cc gii thiu sn phm data-sheet ca cc nh sn xut.

H s ti FAN-OUT quan trng hn v c dng nhiu hn. H s ti u ra FAN-OUT.

FAN-OUT l t s gia dng ra nh nht ca 1 linh kin logic v dng ra ca 1 linh kin c th c ly lm chun.

FAN-OUT cng c th c nh ngha l s ln nht cc cng c th c iu khin t 1 u ra, m khng lm vt qu cc gii hn ra ca linh kin.

H s FAN-OUT s c tnh vi c mc in p cao cng nh mc thp v h s nh hn s c chn.

Trong trng hp cng LS-TTL, ta c:

2020

400)( ===AA

IihIohHOUTFAN

204.0

8)( ===mAmA

IilIolLOUTFAN

Bng sau ch ra h s FAN-OUT ca cc h logic:

TTL-LS CMOS HCT ECL

FAN-OUT 20 100 100 34

BomonKTDT-HGTVT

35

6. Khong l chng nhiu (Noise Margin).

Nu u ra ca 1 cng logic c ni vi u vo ca 1 cng logic cng h, bt k nhiu chng ln no cng khng th gy ra li nu bin ca n nh hn khong l chng nhiu.

Khong l chng nhiu (biu din bi NM) c n v l Volts.

Tham s ny c nh ngha cho mc logic thp (NML) cng nh mc logic cao (NMH).

Ta c phng trnh biu din mi quan h ca NM vi cc mc in p NML = VilMax -VolMax .

NMH = VohMin VolMin .

Thng thng, c mt vi ngun nhiu ac, v nh hng ca n ph thuc vo cc nhn t sau:

tr khng vo v ra, in dung nh hng trn ng vo cng nh bn thn nhiu ca ng dy.

nhiu t ngun cung cp.

nhiu t.

Nhng yu t ny to nn nhiu nh hnh sau; nhiu c biu din nh cc ngun in p.

Cc nhiu xung thng thng kh loi b v chng c to nn bi cc s c m rt kh pht hin v chng c truyn i bi cc thnh phn k sinh.

PTH-DTT

36

Bng di y so snh cc loi nhiu vi cc h logic khc nhau S liu trong bng ch ra rng khong l chng nhiu ca h logic CMOS cao

hn nhiu so vi cc h logic khc. Nh vy, ta nn dng h CMOS trong mi trng nhiu chng hn trong mi trng cng nghip.

TTL-LS (+5V) CMOS (+15V) HCT [+5V] ECL [-

5,2V] NMH 0.7V 5V 2.4V 0.3V NML 0.3V 5V 0.7V 0.3V

7. Thi gian truyn t v thi gian qu

C hai khong thi gian c trng cho tng h logic, trong thi gian truyn t l tham s quan trng hn. N l khong thi gian gia thi im thay i mc logic vo v thi im xut hin thay i mc logic ra tng ng. N s xc nh tc ln nht ca ton mch. Thi gian qu xc nh tc chuyn mc ca tn hiu ra.

Thng thng, mi linh kin s s phi ch r cc thi gian truyn t sau: tPHL : thi gian tr vi u ra chuyn t mc cao xung thp.

tPLH : thi gian tr vi u ra chuyn t mc thp ln cao.

Cc thi gian tr ny, phi c o gia cc mc ngng c th, trong hu ht cc trng hp, trng vi 50% khong thay i tn hiu.

Tham s ny ch yu dng cho vic thit k cc h thng logic v khi kt qu thay i, thi gian s phi xc nh theo mt cch c bit cho mi thay i chng li cc xung khng mong mun.

Thi gian qu (transition time)

Thng thng, vi mi linh kin s cng phi nu r cc thi gian qu sau: tTHL : thi gian qu vi u vo chuyn t cao xung thp

tTLH : thi gian qu vi u vo chuyn t thp ln cao.

Thi gian c o trong khong 10-90% thay i ca tn hiu. Thi gian ny cn khi thit k cc mch logic tun t, vi cc u vo kch,

bi v tn hiu kch khng nhanh linh kin s khng lt trng thi.

Hnh sau biu din thi gian tr truyn t cng nh thi gian qu ca mt cng o.

BomonKTDT-HGTVT

37

Bng so snh cc gi tr thi gian ca cc h logic.

Bng ny ch ra rng h ECL c tc cao nht, h CMOS c tc thp nht.

8. Dng v IC

C 3 phng php ng b cho tinh th silic l: phng php T05, ng v dng hp v ng v hai hng chn song song.

+ ng v dng T05, hnh dng ny ging nh ca transistor, ngha l dng m c nhiu chn. Kiu ng rn ny hin nay t c s dng nhng do c kh nng tiu tn nhit tt nn ch yu c dng cho IC tuyn tnh.

+ ng v 2 hng chn song song / DIP, y l cch ph bin nht ng v IC. N ln hn kiu ng rn nhng c u im l d lp rp v s dng. Cc loi IC ng v kiu c s chn t 8 ti hng trm chn.

C nhiu kiu vt liu c s dng ng rn, thng dng v r nht l ng gi cht do. IC c t vo khung kim loi sau ton b mch c bao ph bng k thut c cht do. Ngoi ra tng kh nng chu nhit ngi ta cn dng k thut ng rn bng gm.

+ ng v dng hp / flat pack, y l kiu ng v cho cc IC c mt tch hp cao, thng gi l IC dn.

IC flat pack thng c s dng cho cc h thng yu cu tin cy cao. 9. Gii hn nhit

Hu ht c IC u c th hot ng trong mt di nhit kh rng t -55 ti +1250C. Cc mch c bit c th lm vic ngoi di trn tu theo cu to ca chng.

Vi loi IC ng rn bng cht do th gii hn nhit nh hn (t 0 ti +700C) so vi loi ng rn bng gm v thng c gn thm cc cnh tn nhit hay thm ch c c qut gi. III. Cng ngh IC s 1. Cng ngh n cc (cng ngh MOS Metal Oxide Semiconductor)

Cng ngh MOS c u im l d ch to v cng on thc hin t quy trnh hn, mt tch hp cao do transistor n cc c kch thc nh v c bit l tiu th in nng rt t.

Di y ta s xem xt mt s h logic MOS thng dng nht

a. H logic PMOS

PTH-DTT

38

Cc transistor MOSFET y c dng knh P nn gi l PMOS. Do cc ht mang in l l trng nn PMOS c tn s lm vic kh nh (khong 1MHz) v l trng di chuyn kh hn in t. PMOS c mt tch hp cao, cng sut tiu th nh v d ch to. Tuy nhin h ny khng tng hp vi TTL (h logic rt ph bin m ta s ni c th phn sau) do mch i hi nhiu in p ngun nui khc nhau.

Cng ngh PMOS thng ch to cc b vi x l tc chm nh NEC com 43/44/45 hay TMS 1000

b. H logic NMOS

MOSFET c s dng l MOSFET knh N c ht dn in l in t nn t c tc cao hn PMOS hng chc ln.

NMOS cho mt tch hp rt ln, cng sut tiu th cng ch tng ng PMOS, khong 0,2mW/cng

NMOS c kh nng tng thch vi TTL nn ch cn mt ngun nui duy nht.

H NMOS c mt s ci tin thnh cc h HMOS, XMOS hay VMOS c mt tch hp cao hn, cng sut tiu th nh hn nhng tn s lm vic li cao hn.

Mt s b vi x l c ch to theo cng ngh NMOS nh 8080 / 8085 / 8086, Z80 / Z80000, MC 6800 / 68000

c. H logic CMOS.

H CMOS s dng cc cp MOSFET knh N v knh P ch ti tch cc do cng sut tiu th nh, 10 W/cng. Ngng i trng thi bng khong 1/2 in p ngun nui.

v d: hnh di y l s ca cng NOT s dng cng ngh CMOS. Mch ny gm 2 Transistor trng

khc loi, NMOS (T1) v PMOS (T2).

u vo c ni ti cc ca G v u ra ni ti cc mng D.

in p cung cp trong cc mch logic CMOS thng c k hiu Vdd.

Hot ng

Khi u vo mc logic thp, NMOS s ngt (v VGS 0V) v PMOS dn (v VGS -Vdd ). Bi th, in p u ra c mc cao

thc t bng Vdd (khi khng ti).

Tng t, khi u vo c mc logic cao, dn n u ra c mc logic thp bng 0V (khng ti).

u im ca vic s dng mch 2 T khc loi (b).

Vic s dng 2 T b, khin cng ngh CMOS c nhng u im so vi cc h logic khc:

BomonKTDT-HGTVT

39

Gim cng sut tiu th trong iu kin tnh xung khong vi W (khng c dng ti mch ra v khi 1 T dn, T kia s ngt).

Khi chuyn trng thi, sn xung s dc hn v c thi gian i xng hn, tc: tTHL = tTLH .

Mc logic 0 v 1 ti u ra s xp x 0V v Vdd .

Gim dng u vo trong iu kin tnh, thm ch v 0A do cc G c cch ly i vi MOS.

Tuy nhin, u im ca vic gim cng sut tiu th do cc ca G c cch ly i vi cng ngh MOS s dn n nhc im l: cc u vo c th lu tr cc in tch tnh in to nn mt lp mng cht cch in ng li trn knh. Do , cn c mch chng tnh in ti u vo, nm bn trong mch tch hp. Mch ny, v c bn l mt nhm cc Diode c ni vi nhau nh hnh di y bi th in p VGS khng th ln hn Vdd hay gim xung 0V.

Khng ging cc h logic khc, cng sut tiu th ca CMOS tng nhanh khi tn s hot ng tng v 2 l do chnh:

+ S ln np v phng trn mt giy ca cc in dung k sinh (to bi cc ca G) tng ln.

+ Trong khong thi gian chuyn mc logic, c hai MOS u dn.

V cc l do ny, cng sut tiu th,

m c b qua di iu kin tnh, s tng khi tn s tng, cho n tn s khong vi MHz th cng sut tiu th ca h CMOS

s xp x nh cc h lng cc.

Seri CMOS loi HC v HCT.

Seri HC (CMOS tc cao High Speed) c gii thiu vo nhng nm bt u thp k 80. Loi ny c tc v dng cao hn CMOS chun khong 10 ln, s

chn tng thch vi h TTL; khong l chng nhiu cao hn TTL v Vdd t 2 n 6V. Khi lm vic vi in p 5V nh TTL tc ca cc h trn gim i rt nhiu

Seri mi ny c cng sut tiu th thp hn h TTL; kh nng chng nhiu

cao hn; kh nng iu khin u ra cao hn v in p hot ng t 2 6V.

V in p ra ca HC khng tng thch vi TTL nn seri HCT c pht trin, vi cng tnh nng nh HC nhng c kh nng tng thch TTL vi in p cung cp Vdd = 5V.

Mt s ch tiu k thut ca CMOS:

PTH-DTT

40

Thi gian tr 30 100ns

Cng sut tiu tn 0,01mW (1mW tn s 1MHz)

Kh nng ti 50

n nh nhiu ~ 45%Vdd

Mc logic Mc 0 bng 0V; mc 1 bng Vdd

Ngun cung cp 3 15V

Cc cng logic c bn NOR; NAND

2. Cng ngh lng cc

Thnh phn c bn ca cc vi mch cng ngh lng cc l s dng cc transistor lng cc. Cng ngh ny c mt s h c bn sau:

a. H logic TTL (Transistor Transistor Logic)

y l h vi mch c s dng rng ri trong mi lnh vc v tr thnh tiu chun tng hp TTL cho cc h logic khc. c tnh in ca cng logic TTL.

Xt cng logic c bn ca h TTL l cng NAND c cho nh hnh di Transistor T1 l loi nhiu emiter.

Transistor T2 lm nhim v cung cp 2 tn hiu ngc pha; tn hiu ny iu khin tng ra gm T3, D1 v T4.

Transistor T3 c gi l transistor ni ngun (pull-up) v hot ng nh mt mch lp E khi u ra mc cao n s khin cho tr khng ra rt thp.

Nu c hai u vo mc cao, mch s tiu th dng ca mi u vo khong 40A.

Collector ca T1 c ni vi base ca T2 v c mc in p 2VBE, tc l khong 1,4V. Diode tng ng ca tip

gip base-collector ca T1 ly ngun qua in tr R1, do vy c phn cc thun;

BomonKTDT-HGTVT

41

nh th Transistor T2 ri vo trng thi bo ho. Dng Emitter ca T2 mt phn chy qua R3, mt phn chy vo base ca T4 do , a T4 vo trng thi bo ho. in th base ca T3, Vb3 c gi tr bng vi tng Vbe ca T4 cng vi VceSat ca T2.

in p qua Emitter ca T3 l:

Ve3 = VceSat + Vd1 .

Do , Vb3 = Ve3 v transistor T3 trng thi ngt (OFF). Lc ny, Transistor T4 s thng (ON), c dng in khong 16mA chy qua v u ra c mc logic 0 tc in p t khong 400mV.

Gi tr logic 0 in hnh ti u ra l 220mV, vi dng in t 16mA

Gi tr dng ny iu khin 10 u vo logic TTL trng thi 0.

Tr khng ra Rout do T4 t khong 12Ohm.

Trng thi OFF (mt u vo ti mc thp, u ra mc cao).

Xt trng hp ti thiu mt u vo mc thp (u vo khng vt qu 400mA).

Gi tr dng ln nht khi u vo c mc logic 0 l khong 1,6mA, bi th mt cng vi u ra mc thp c th iu khin khong 10 cng khc.

Lc ny, Transistor T1 s dn, T2 v T4 ri vo trng thi ngt. in p trn collector ca T2 l cao do vy, T3 t bo ho.

Di nhng iu kin ny, dng a qua u ra t 400A, iu khin 10 cng khc.

in p u ra Vo, khi ngun cung cp, khng nh hn 2,4V.

Thc t, vi gi tr p vo thp hn 800mV, in p ra in hnh t 3,3V.

Dng in ra trong iu kin ngn mch c gi tr nh nht l 18mA v gi tr max l 58 mA, c gii hn ch yu bi R4.

Tr khng ra mc cao l khong vi trm Ohm.

Qu trnh chuyn trng thi ca cng TTL.

chuyn trng thi t 1 (OFF) v 0 (ON) mt u vo s c mc th t cn u vo kia ni vi Vcc. Khi in p u vo mc thp tng, dng in u vo s gim v khi tng t ti 0,8V T2 bt u dn v in p trn collector ca n gim. Kt qu, in p u ra gim cho n khi in p u vo t khong 1,4-1,5V, lc ny in p u ra c gi tr khong 2V. in p trn base ca T2 l khong 1,4V v do vy, c T2 v T4 u dn.

Bt u t thi im ny, in p u ra nhanh chng gim xung gi tr VceSat ca T4 , tc l T2 t bo ho cn T3 chuyn sang trng thi ngt (OFF).

C mt khong thi gian rt ngn khi m c T3 v T4 u dn; trong khong thi gian ny c dng chy qua R4, T3, D1 v T4.

Dng ny c hn ch ch yu bi R4. Khi chuyn t trng thi thp (low) ln cao (High), ban u cc u vo

trng thi cao.

PTH-DTT

42

Khi in p ca mt (hay nhiu) u vo gim xung 1,4V, T1 bt u dn khin cho T2 v T4 cng ri vo trng thi dn. Dng qua T2 gim in p trn collector ca T2 tng khin T3 ri vo trng thi dn v th u ra s mc thp.

D cho cng TTL mc ON hay OFF tr khng ra lun thp, cho php cng TTL c th iu khin ti dung khng cao.

Cc nhnh ph ca h TTL.

H logic chun TTL (STD) c thay i qua nhiu nm c cc tnh nng tt hn, to nn cc nhnh ph (sub-families) ca h TTL.

Thc t, cc nhnh ph ca h TTL chun hot ng nhanh hn hay tiu th cng sut t hn so vi h TTL chun.

Chng gm:

S TTL (Schottky TTL) : tc tng gp 3 ln nhng cng sut tiu th tng ln ti 20mW/cng.

AS TTL (Advanced Schottky): tc gn bng ECL (1 n 2ns)

LS TTL (Low Power Schottky TTL) : cng tc nhng cng sut tiu th gim 5 ln. 10ns, 2mW/cng

F TTL (Fast TTL) : tc gp 4 ln, cng sut tiu th gim mt na.

ALS TTL: 3ns, 1.25mW/cng

Mt s k hiu ca TTL cho bit di nhit cng tc

74: 00C - +700C

84: -250C- +850C

54: -550C - +1250C

Mt s ch tiu k thut ca TTL chun:

Thi gian tr 10ns

Cng sut tiu tn 10mW

Kh nng ti 10

n nh nhiu Cao

Mc logic mc 0 bng +0,4V;

mc 1 bng +3,6V

Ngun cung cp 5V 10%

Cc cng logic c bn NOR; NAND

b. H logic ECL.

H logic ECL (Emitter Coupled logic) c to ra s dng cng ngh lng cc (ging nh h TTL).

BomonKTDT-HGTVT

43

y l h logic c tc hot ng nhanh nht trn th trng. N t c tc v 2 l do:

+ Trnh vic a cc linh kin tch cc vo trng thi bo ho. + Cho php tiu th cng sut cao hn trn mi cng so vi cc h logic khc.

Mt thnh phn chnh trong h ECL l b khuch i vi sai, trong 2 Transistor c ghp Emitter chung nh trong hnh di y

Cc c im ca b khuch i vi sai:

Dng emitter khng i.

Dng s chy t Transistor ny sang T kia, khi in p Vin a ti u vo ca T th nht nm trong khong:

VBB 0,1V

PTH-DTT

44

y l cu trc c bn ca cng ECL vi 3 u vo. in p chun VBB c to ra t mch b nhit (khng c ch ra hnh v). Mch ny to ra cc mc in p :

V(0) = -1,7V.

V (1) = -0,9V.

Khong l chng nhiu ca h logic ny rt hp v iu ny gii thch ti sao cc cng c cp

ngun gia t v -VEE (-5.2V) lm gim tr khng trong.

3. Giao tip TTL-CMOS v CMOS-TTL.

Nh gii thiu phn trn ta thy h TTL v h CMOS l 2 h logic ln nht v c s dng nhiu nht. TTL c u im v tc cn CMOS li c u im v cng ngh ch to n gin v tiu th in nng t. V vy, vic ghp ni gia 2 h logic l rt quan trng d rng cc nh sn xut khuyn ngh nn dng cng mt h logic trong mt mch in t.

a. Giao din TTL-CMOS.

Trng hp n gin nht ca giao tip gia linh kin TTL v CMOS l khi ta ch c 1 ngun cung cp duy nht l 5V.

Nh ta thy trong hnh bn, mt in tr ni ngun (pull-up-c gi tr khong

vi KOhm) c s dng ko u ra c mc logic cao ca cng TTL (m c gi tr nh nht l 2,4V, cha c xem nh l mc cao i vi CMOS) ln xp x 5V.

Khi cc linh kin c cc ngun cung cp khc nhau giao tip vi nhau ta phi s dng mt linh kin tng thch TTL vi u ra h collector hay h cc mng D nh hnh sau:

BomonKTDT-HGTVT

45

B m in hnh l 7407 hay 7417 c cp ngun +5V, vi in tr ni

ngun khong vi KOhm gia u ra v VDD (c th iu khin nhm thu c in p t +3V n +18V).

b. Giao tip CMOS-TTL.

Vic ghp ni trc tip linh kin CMOS-TTL s dng cng ngun cung cp +5V yu cu vic xem xt dng r ca linh kin CMOS gi mc in p u ra thp ca CMOS nm trong phm vi cho php ca linh kin TTL.

Trong Seri CMOS CD4000 B tt c cc linh kin u c th iu khin ti

thiu mt linh kin TTL nhng ch thuc h LS. iu khin mt hay nhiu linh kin ca h TTL STD, TTL S hay TTL F...,

yu cu c mt b m (chng hn nh CD4049 hay CD 4050). Khi cc linh kin c ngun cung cp ring khc nhau ghp ni vi nhau, mt

b m CMOS vi u ra h cc mng D c s dng nh hnh di.

Tnh hung ny cn gp phi khi cn truyn d liu t phn thu thp d liu s dng linh kin CMOS (chng hn trong khu vc cng nghip yu cu khong l chng nhiu cao-ch c vi cc linh kin h CMOS) ti h thng x l dng cc linh kin h TTL. Linh kin c s dng nn l b m CMOS vi u ra h cc mng D- MM 74C906. in tr R nn nm trong khong vi KOhm.

PTH-DTT

46

Phn II

Mch t hp

BomonKTDT-HGTVT

47

Chng 5:

phn tch v Thit k mch t hp

Mch s c chia lm 2 loi l : + Mch t hp / Combinational Circuit

+ Mch dy / Sequential Circuit

Mch t hp l mch m tn hiu ra ch ph thuc vo tn hiu vo. Phng trnh xc nh tn hiu ra ca mch l:

Yi = fi(X1, X2, , Xn) vi mi = 1 Yi l tn hiu ra u ra th i, c m u ra

Xj l tn hiu vo u vo th j, c n u vo

Ngi ta cn gi mch t hp l mch khng c nh Mch dy l mch c tn hiu ra ph thuc vo trng thi trong ca mch v c th ph thuc hoc khng ph thuc vo tn hiu vo. Phng trnh c trng ca mch dy l: Yi = fi(X1, X2,Xn, S1, S2 ,. Sk) vi mi = 1

Yi l tn hiu ra u ra th i, c m u ra

Xj l tn hiu vo u vo th j, c n u vo

St l trng thi trong ca mch

Mch dy c kh nng lu tr d liu nn cn c gi l mch c nh. C th coi mch t hp l mt trng hp ring ca mch dy vi s trng

thi trong ca mch l 1. I. M hnh ton hc ca mch t hp

X = { X1, X2, Xn}: tp hp cc tn hiuu vo

Y = { Y1, Y2, Ym}: tp hp cc tn hiu u ra

Khi mch t hp c th c m t bi h m phng trnh i s Boolean nh sau:

Yi = fi( X1, X2, Xn) vi mi = 1 V mt ton hc c th ni m hnh ton hc ca mch t hp chnh l otomat

khng c nh, m t bng phng trnh: O = (X, Y, f)

X1

X2 X3

Xn

Y1

Y2Y3

Ym

Mch

t hp

Mch

t hp

X Y

PTH-DTT

48

Vi X, Y l b ch vo, ra v f l nh x t X vo Y II. Phn tch mch t hp

Bi ton phn tch la bi ton t s logic cho trc vit hm logic ca cc u ra theo cc vo v nu cn th cn phi ch ra dng sng ca tn hiu ra tng ng vi tn hiu vo, xc nh gi tr tnh hiu tng im trong s .

Cc bc phn tch mch t hp nh sau:

+ t cc bin ph vo mi mch u ra ca mi mch logic

+ Vit phng trnh ca cc bin ph (vit ln lt t u vo cho n u ra)

+ Trong biu thc cui cng, thay th cc bin ph bng cc gi tr tng ng rt ra c hm logic cho cc u ra cho s . v d: phn tch mch t hp cho hnh di y:

III. thit k mch t hp 1. Bi ton thit k v cc bc thc hin

y l bi ton ngc vi bi ton phn tch, l t yu cu cho trc nh chc nng, dng sng ta phi xy dng s mch thc hin nhng yu cu .

Trong phm vi ca chng ny ta ch xt n vic s dng cc vi mch c nh (SSI), thc hin theo cc bc sau:

+ M t bi ton di dng chc nng + Ti thiu ho

+ Ch ra s logic dng cho cc cng cho 2. Thit k mch t hp 2 tng v nhiu tng

a. Mch 2 tng

u im:

+ C th thc hin c mi hm logic + C tc cao

+ Vic phn tch v thit k mch n gin

Nhc im: + Trong mt s trng hp thit k khng nhn c s n gin nht + Thng yu cu cc phn t c s u vo ln

U2C

U1C

U1B

U2B

U2A

U1A

BomonKTDT-HGTVT

49

Cc cch thit k mch hai tng vi cc phn t cho trc Tng1 / tng 2 AND OR NAND NOR

AND X CTT X 1. CTH

2. f , D

OR CTH X 1. CTT

2. f , D X

NAND 1. CTH

2. tp , D X

1. CTT

2. f , D X

NOR X 1. CTT

2. tp , D X

1. CTH

2. f , D

Ghi ch:

f : ph nh hai ln hm f

tp : ph nh hai ln tng thnh phn

D: p dng lut Demoorgan

Cc gi tr tn hiu vo Xi v Xi c sn Trn cng mt tng ch s dng mt loi phn t (AND, OR, NAND, v

NOR)

Nhng phn t ny c s u vo khng hn ch

v d: Cho hm logic f = 7,6,5,1,0 Trc khi xy dng s ta cn thc hin ti thiu ho hm trn theo dng CTT v CTH

Biu din hm f trn bng Karnaugh

C / AB 00 01 11 10

0 1 1

1 1 1 1

T bng Karnaugh d dng vit c:

))((

...

CACBAf

CACBCAf

+++=++=

Da vo bng kt hp u vo v u ra ta c th xc nh c s mch cho f nh sau: 1. Tng 1 dng mch AND, tng 2 dng mch OR

CACBCAf ... ++=

PTH-DTT

50

2. Tng 1 dng mch OR , tng 2 dng mch AND

))(( CACBAf +++= 3. Tng 1 dng mch OR, tng dng mch NAND

+ Vit f di dng CTT CACBCAf ... ++= + Ph nh hai ln hm f, sau p dng 2 ln lut D

))().((

...

CACBCAf

CACBCAf

+++=++=

4. Tng 1 dng mch NAND, tng 2 dng mch AND

+ Vit f di dng CTH ))(( CACBAf +++= + Ph nh 2 ln cc thnh phn v p dng De Morgan

CACBAf

CACBAf

....

)()(

=+++=

5. Tng 1 dng mch NAND, tng 2 dng mch NAND

+ Vit hm di dng CTT CACBCAf ... ++= + Ph nh hai ln hm f v p dng De Morgan

CACBCAf

CACBCAf

.....

...

=++=

6. Tng 1 dng mch NOR, tng 2 dng mch OR

+ Vit hm di dng CTT f = CACBCA ... ++ + Ph nh 2 ln cc thnh phn sau p dng D

CACBCAf

CACBCAf

+++++=++= ...

7. Tng 1 dng NOR, tng 2 dng mch NOR

+ Vit hm di dng CTH ))(( CACBAf +++= + Ph nh 2 ln f v p dng D

)()(

))((

CBACAf

CACBAf

++++=+++=

8. Tng 1 dng mch AND v tng 2 dng mch NOR

+ Vit hm f di dng CTH f = ))(( CACBA +++

BomonKTDT-HGTVT

51

+ Ph nh 2 ln hm s f v p dng D

)..().(

)()(

))((

CBACAf

CBACAf

CACBAf

+=++++=

+++=

b. Mch nhiu tng

Khi s u vo ln hn s u vo cho php ca phn t cho trc lc phi tng s rng ca mch. S dng cc s thay th nh sau:

U7A

U6B

U6A

U5A

U11

U1D

U9B

U9A

U2D

U10B

U10AU4

U13A

U12B

U12A

U8A

PTH-DTT

52

3. Thit k mt h hm t hp

C hai cch thit k mt hm t hp l thit k ring tng hm hoc thit k c phn chung hn ch s u vo.

BomonKTDT-HGTVT

53

Chng 6:

Mt s mch t hp thng gp

I. B cng nh phn mt ct s 1. Phn tch bi ton

M hnh ton hc ca b cng y 1 bit (FA Full adder)

trong Ai v Bi l cc s nh phn th i ca A, B a vo cng Ci-1 l s nh ca ct c trng s nh hn bn cnh ( ca php tnh trc )

Si l l ch s ca tng ct th i 1= iCBiAiSi Ci l s nh a n ct c trng s ln hn bn cnh

)(. 1 BiAiCBiAiCi i ++= Ch : Php cng 2 s nh phn lun bt u t ct s c trng s nh nht

Bng chn l ca php cng y mt bit

Ai Bi Ci-1 Si Ci

0 0 0 0 0

0 0 1 1 0

0 1 0 1 0

0 1 1 0 1

1 0 0 1 0

1 0 1 0 1

1 1 0 0 1

1 1 1 1 1

Ai

Bi

Ci-1

Si

Ci

B cng 1 bit FA

PTH-DTT

54

2. Xy dng s

C th xy dng b cng theo 1 trong 2 cch nh sau: + Xy dng trc tip t h phng trnh ca Si v Ci + Xy dng t cc b bn tng (HA Half Adder). y l phng php c

s dng nhiu trong thc t v di y ta s xem xt ti phng php ny. B bn tng l b c bng chn l sau:

A B S C

0 0 0 0

0 1 1 0

1 0 1 0

1 1 0 1

T bng chn l rt ra c: BACBAS

.==

Nh vy s ca b bn tng nh sau: T phng trnh ca b tng y

1= iCBiAiSi )(. 1 BiAiCBiAiCi i ++=

ta xy dng c s ca b tng FA bng 2 b HA v 1 cng OR nh sau:

chng minh:

BACSCSSi ii === 11 12

)()()().().()(

)(1.21

111

111

11

BACABCABCBACiCAABCBBABABACABCi

BACABSCABCCCi

iii

iii

ii

++=+++=+++=++=

+=+=+=

HA 1

A

B

C1

S1

HA2

Ci-1 Si

Ci

C2

S2

HA

C

SBA

U2A

U1A

BomonKTDT-HGTVT

55

Nguyn l hot ng ca b cng nh phn.

Rt nhiu mch logic cn cc thit b c kh nng cng 2 s nh phn. Mt b cng c th tnh ton mt php cng nh phn. V u ra ph thuc u vo ti mt thi im xc nh theo yu cu, nn s s dng mch logic t hp.

Hnh trn l s ca 1 b bn tng 1-bit v mch ton tng. S d c gi l b bn tng v n khng cng bit nh ti u vo, mt vic thng yu cu khi cng nhng s c nhiu s hng.

cng cc s vi nhiu s hng, mch phi c kh nng x l thm 1 u vo na. u vo ny l kt qu ca php cng t tng trc. Mch nh vy, c gi l mch ton tng (Full Adder).

Ghp ni tip cc b cng.

B ton tng s l phn t c s cho vic xy dng b cng n-bit. Hnh bn ch ra lm th no m cc u vo v ra ca mt phn t cng n l c th c ni vi nhau to thnh b cng 3-bit.

Hon ton tng t vi cc b cng nhiu bit khc.

PTH-DTT

56

II. B tr nh phn mt ct s Cch lm hon ton tng t nh khi xy dng b cng nh phn 1 ct s.

Ngha l xy dng bng chn l, ti thiu ho , sau xy dng trc tit hot dng b bn tr to ra b tr y

Bng chn l ca b tr y (FS Full Substructor) nh sau: Ai

Ch s th i ca s b tr

Bi Ch s th i ca s tr

Ci S nh t ct c trng s

nh hn

Hi Ch s th i ca php tr

Ci S nh a ti ct c trng s

ln hn 0 0 0 0 0 0 0 1 1 1 0 1 0 1 1 0 1 1 0 1 1 0 0 1 0 1 0 1 0 0 1 1 0 0 0 1 1 1 1 1

T bng chn l ta ti thiu ho hm Hi v Ci bng bng Karnaugh ta nhn c:

)(. 1

1

BACBACi

CBiAiHi

i

i

++==

Bng chn l ca b bn tr (h s Half Substructor) A B H C 0 0 0 0 0 1 1 1 1 0 1 0 1 1 0 0

BAC

BAH

.==

Khi ta c s b bn tr v b tr y nh sau:

Chng minh:

HS 1

A

B

C1

H1

HS2

Ci-1 Hi

Ci

C2

S2

BomonKTDT-HGTVT

57

)(

)()(

)().(

).(.

.121

1

1

11

11

11

1

11

BACBACi

CABCBACi

ACABCBBACi

CABBABACBABACi

CHBACCCi

BACHCHi

i

ii

ii

ii

i

ii

++=+++=+++=

++=+=+=+=

==

Ch : trn thc t t khi ngi ta s dng b tr m thng bin thnh php cng hay dng m b s dng cc b cng III. B so snh comparator

Mt b so snh s thc hin php so snh 2 s nh phn v kt qu s c th hin ti u ra.

B so snh nh phn thng thng c ba u ra: A=B, A>B, A B A < B

0 0 1 0 0 0 1 0 0 1

1 0 0 1 0 1 1 1 0 0

T bng chn l trn ta thy:

Hm (A = B) = ABBABA += Hm NXOR Hm (A > B) = BA. Hm cm B (inhibition)

Hm (A < B) = BA Hm cm A Di y l s ca b so snh 1 bit theo nguyn l v s c th dng

cc cng NAND.

U5C

U5B

U5A

f3

f1

f2A

B

U4D

U4C

U4B

U4A

vi f1 l hm (A = B)

f2 l hm (A > B)

C

HBA

U3A

U2A

U1A

HS

PTH-DTT

58

f3 l hm (A < B) 2. B so snh n bit

Gi s c 2 s nh phn n bit A v B c biu din nh sau:

11

11

.....

....BBBAAA

nn

nn

trong An, Bn l ct s c trng s ln nht v A1, B1 l ct s c trng s nh nht

xy dng s b so snh ny c 2 cch nh sau: + Xy dng trc tip cc hm f1, f2 v f3 (thc cht l xy dng 1 h 3 hm

logic, mi hm 2n bin)

+ Xy dng s gin tip t cc b so snh 1 bit c sn

Xt v d vi n bng 3

A = A3A2A1

B = B3B2B1

Khi ta thy:

(A = B) (A3 = B3)(A2 = B2)(A1 = B1) (A > B) (A3 > B3) + (A3 = B3)(A2 > B2) + (A3 = B3)(A2 = B2)(A1 > B1) (A < B) (A3 < B3) + (A3 = B3)(A2 < B2) + (A3 = B3)(A2 = B2)(A1 < B1) S dng cc b so snh 1 bit (A1, B1), (A2, B2) v (A3, B3), ta s c s mch

thc hin so snh 3 bit nh sau:

f3

f2

f1

U8B

U1D

U8AU6A

U7C

U1C

U7B

U5D

U1BU7A

BomonKTDT-HGTVT

59

Vi mch so snh nh phn 4-bit.

Vi mch SN74LS85 c mt b so snh cc s nh phn 4-bit.

Di y l s chn v s c th bn trong ca vi mch thc hin

PTH-DTT

60

IV. B to v kim tra chn l Parity Generator and Checker

Phng php kim tra chn l l mt phng php n gin nht xc nh li trong vic truyn d liu. Phng php ny c thc hin bng cch thm 1 bit d liu c truyn i sao cho s ch s 1 trong d liu lun l mt s chn hoc s l. Bit thm vo gi l bit chn l.

Bit chn: nu bit thm vo c gi tr sao cho s ch s 1 trong d liu l mt s chn (Even)

Bit l: nu bit thm vo c gi tr sao cho s ch s 1 trong d liu l mt s l (Odd)

thc hin c vic truyn d liu theo kiu a thm bit chn , l vo d liu cn xy dng:

+ S to c bit chn, l thm vo n bit d liu + S kim tra c h l h chn hay l vi (n+1) bit u vo (n bit d

liu v 1 bit chn / l) 1. Mch to bit chn l

Xt v d trong trng hp d liu 3 bit, trng hp s bit ln hn c th thc hin hon ton tng t.

Gi 3 bit d liu l d1, d2, d3 v Xe, Xo l 2 bit chn, l thm vo d liu. Xe l gi tr bit phi thm vo h l h chn , Xo l gi tr bit phi thm vo h l h l. Bng chn l ca mch to bit chn l cho trng hp d liu 3 bit nh sau:

Vo Ra

d1 d2 d3 Xe Xo

0 0 0 0 1

0 0 1 1 0

0 1 0 1 0

0 1 1 0 1

1 0 0 1 0

1 0 1 0 1

1 1 0 0 1

1 1 1 1 0

T bng chn l ta c:

321

321

dddXeXo

dddXe

===

2. Mch kim tra chn l

Bng chn l ca mch kim tra tnh chn l ca h cho

BomonKTDT-HGTVT

61

Vo Ra

d1 d2 d3 X Fe Fo

0 0 0 0 1 0

0 0 0 1 0 1

0 0 1 0 0 1

0 0 1 1 1 0

0 1 0 0 0 1

0 1 0 1 1 0

0 1 1 0 1 0

0 1 1 1 0 1

1 0 0 0 0 1

1 0 0 1 1 0

1 0 1 0 1 0

1 0 1 1 0 1

1 1 0 0 1 0

1 1 0 1 0 1

1 1 1 0 0 1

1 1 1 1 1 0

T bng chn l ta thy

FeFo

XdddFe

== 321

Fe ch ra tnh chn ca h, ngha l nu h chn Fe = 1

Fo ch ra tnh l ca h, ngha l nu h l Fo = 1

Ch : phng php kim tra chn l ch pht hin c li n hoc s li l l m khng pht hin c li chn v khng c kh nng sa li

PTH-DTT

62

V. Mch phn loi ngt Mch phn loi ngt l mch c nhim v:

+ Ti thi im t nu c t nht 1 trong cc thit b ngoi vi cng gi yu cu ngt ti b VXL th mch phi to ra tn hiu yu cu ngt IR (interrupt request) ti b VXL

+ Ti thi im t nu c nhiu thit b ngoi vi cng g yu cu ngt ti VXL th mhc phi ch ra cc thit b ngoi vi no cn u tin gii quyt ngt ti thi im .

S khi ca b phn loi ngt c cho trong hnh di y Trong (A1, A2 ..An) l cc t hp tng ng vi cc thit b ngoi vi

(TBNV)

IR = khi c t nht 1 trong 2n TBNV c yu cu ngt ti b VXL

Nh vy b phn loi ngt c 2n u vo v n+1 u ra

V d: xy dng s b phn loi ngt cho 4 thit b ngoi vi, vi gi thit cc thit b ngoi vi c u tin ngt theo th t P3, P2, P1, P0

Nh vy b phn loi ngt s c 4 u vo v 3 u ra ( IR v AB l a ch ca cc TBNV). Khi ta c bng chn l ca b phn loi ngt nh sau:

Vo Ra

P3 P2 P1 P0 IR A B

1 x x x 1 1 1

0 1 x x 1 1 0

0 0 1 x 1 0 1

0 0 0 1 1 0 0

0 0 0 0 0 0 0

T bng chn l xc nh c cc hm u ra nh sau: IR = P1 + P2 + P3

A = 3.23 PPP + B = 3213 PPPP +

B phn loi ngt

B

VXL

TBNV 1

TBNV 2

TBNV 2n

I1

I2

I2n

I2IR

A1

An

BomonKTDT-HGTVT

63

VI. B chn knh v phn knh (Multiplexer and Demultiplexer) 1. B chn knh

B dn knh l mch c 2n u vo d liu X, n u vo a ch A, 1 u cho php En v 1 u ra Y (c th c mch c thm u ra Y )

Nhim v ca b chn knh l chuyn thng tin t mt u vo d liu c a ch c xc nh nh cc u vo a ch n u ra khng o khi u vo cho php trng thi tch cc.

Tu theo gi tr ca n u vo a ch m u ra s bng mt trong nhng gi tr u vo Xj. C th l nu gi tr thp phn ca t hp (An-1An-2A0) bng j th Y =Xj nu khi En = 1.

Hnh bn l s khi ca mt b MUX chn 1 u vo t 2n u vo

Hin nay b MUX c dng nh mt phn t vn nng xy dng nhng mch t hp khc, c th l:

+ To hm logic

+ To cc dy xung

+ Truyn d liu

+ MUX c dng nh b chuyn thng tin dng song song u vo thnh ni tip u ra

+ Gii m a ch

+. 2. B phn knh

B chn knh l mch c 1 u vo d liu X, n u vo a ch A, 1 u cho php En v 2n u ra Y

Hnh bn l s khi ca b DEMUX

Nhim v ca b phn knh thc cht l c chc nng gii m t mt a ch ca knh cho to tn hiu iu khin ca knh l chuyn thng tin t mt u vo d liu c a ch c xc nh nh cc u vo a ch n u ra khng o khi u vo cho php trng thi tch cc.

Tu theo gi tr ca n u vo a ch m u ra th i (Yi) s bng gi tr u vo X. C th l nu gi tr thp phn ca t hp (An-1An-2A0) bng i th Yi = X nu khi En = 1.

MUX 2n 1

X0

X1

X2n - 1

An-1 An-2 A0

En

Y

MUX 1 2n

Y0

Y1

Y2n - 1

An-1 An-2 A0

En

X

PTH-DTT

64

VII. B chuyn m Trong cc h thng in t dng mch s, d liu c x l v truyn i

di dng t nh phn n bit, mt t n bit c th biu din cho 2n phn t tin khc nhau. T nh phn n bit gi l m (code) ca phn t tin tc. C rt nhiu loi m khc nhau c s dng cho tng mc ch khc nhau, di y ta s xt n cc m tiu biu v cc mch t hp thc hin chuyn m, gm 2 loi l m ho v gii m (ENCODER v DECODER) 1. Cc loi m tiu biu

a. M k t

+ M ASCII (American Standard Code for Information Interchange): dng 8 bit m ho cho bng ch ci v mt s k t c bit (m ny c s dng rng ri nht m ho k t cho cc h thng x l vn bn)

+ M EBCDI (Extended Binary Coded Decimal Interchange): dng 8 bit m ho cho k t

+ M BAUDOT: dng 5 bit biu din cho 1 k t, thng dng cho teletype v bu in

b. M s

Cc loi m thng s dng l nh phn, d 3, Gray, BCD Phn ny c m t c th trong chng 1. Ngoi ra cn mt s loi m c bit nh m sa sai l m ngoi cc bit mang

thng tin cn c mt s bit thm vo pht hin v sa li, vi d: m chn l, m CRC . M 7 vch c di bng 7 dng biu din ch s thp phn bng n 7 thanh. 2. Mch m ho - lp m (ENCODER)

Gi s t m c n bit, khi s c 2n b gi tr khc nhau biu din cho cc k hiu hoc lnh. Nh vy mi loi m ch c s k hiu hoc lnh

BomonKTDT-HGTVT

65

S h 10 A B C D 0 0 0 0 0 1 0 0 0 1 2 0 0 1 0 3 0 0 1 1 4 0 1 0 0 5 0 1 0 1 6 0 1 1 0 7 0 1 1 1 8 1 0 0 0 9 1 0 0 1

T bng chn l ta c:

A = 8 + 9

B = 4 + 5 + 6 + 7

C = 2 + 3 + 6 + 7

D = 1 + 3 + 5 + 7 + 9

Nh vy mch thc hin m ho 10 BCD 8421 c s nguyn l nh sau:

Khi mt trong cc u vo D0 D9 c mc in p cao th cc ng ra ABCD s c tn hiu tng ng. V d D5 c mc in p cao cn cc ng khc c mc in p thp , ngha l ta mun m ho s 5, khi cc ng ra B v D c mc in p cao cn A v C c mc in p thp, tc ta c ABCD = 0101 nh mong mun.

Ch : v cu to, n gin mch OR thng ch to theo kiu DL (diode logic).

Khi mch tr thnh nh sau:

D8 D9

A

B

C

D

U3A

U2A

U1B

U1A

D4 D5 D7D6D2 D3D1D0

PTH-DTT

66

Ch : Mch in ca b m ho khng c mc u tin (tc khng c nhiu ng

vo cng mc cao) c ch ra nh hnh D09.3. B m ho khng c u vo 0 v, thng thng, n khng cn n trong cc mch logic. Cng c th thc hin b m ho trn theo s di y:

3. Mch gii m (DECODER)

Mch ny c chc nng ngc vi b m ho,ngha l t b bit n bit h 2 cn tm li c 1 trong N k hiu hoc lnh tng ng. B gii m BCD sang thp phn.

B gii m BCD sang h thp phn l mt mch t hp c 4 u vo nh phn v 10 u ra thp phn. u vo l m BCD v s kch hot u ra tng ng vi u vo.

D1 D3D7

A

D9

B

C

D8D6D5D4

D

D2D0

BomonKTDT-HGTVT

67

Di y l bng chn l ca b gii m BCD-thp phn, vi u ra tch cc mc logic m, ng vi vi mch SN 74LS42. C th thy rng cc s ln hn 9 s khng kch hot bt k u ra no.

SN74LS42 l mt vi mch gii m BCD thp phn c s dng rng ri,

di y l s bn trong ca vi mch ny minh ho cho vic chuyn i m.

B gii m BCD sang 7 vch.

n 7 vch c s dng hin th d liu c x l bi thit b in t s. Chng c th hin th cc s t 0 n 9 v cc ch ci t A n F v mt vi k t khc.

Thit b hin th ny c th c iu khin bi b gii m m s chiu sng cc vch (on-segment) ca n ph thuc vo s BCD ti u vo. Cc b gii m

PTH-DTT

68

ny cng cha cc b m cng sut cp dng cho n, do vy, n cn c gi l b iu khin-gii m (Decoder-Driver).

B m ho ny c 4 u vo tng ng vi 4 bit m BCD v 7 u ra, mi u s iu khin mt vch ca n 7 vch. Hnh di ch ra m hnh ca cc vch trong thit b hin th (n) 7 vch v cc s c th hin th.

Hin th 7 vch

n hin th 7 vch bao gm cc vch (on sng segment) nh. Chng c th biu din ti 16 k t trong c 10 s v 6 ch ci nh hnh di y:

Cc m u vo t 0 -9 hin th cc ch s ca h thp phn. Cc m u vo t 9-14 ng vi cc k hiu c bit nh nu, cn m 15 s tt tt c cc vch.

on sng th 8 ca n hin th l du chm thp phn (dp). Cc thit b hin th loi ny c nhiu kiu vi mu sc, kch thc khc nhau v c c tnh pht sng rt tt.

V mt in, cc LED hot ng nh diode chun, ch khc l khi phn cc thun i hi in p gia anode v Cathode cao hn. c cng sng khng i, thit b hin th phi c cp dng.

Cc thit b hin th 7 vch c th c cc tnh:

+ vi kiu cathode chung, iu khin bi mc logic dng. + vi kiu anode chung, iu khin bi mc logic m.

Vi mch TTL 74LS47 l mt b iu khin- hin th c dng ph bin. Vi mch ny c cc u ra o do s dng vi LED anode chung. Hnh di y ch ra s chn v s mch bn trong ca vi mch.

Chn LT (Lamp Test) c dng kim tra tnh trng hot ng (sng hay cht) ca cc vch; trong khi chn RB (Ripper Blanking) c dng tt tt c cc vch khi yu cu trng thi khng hin th s.

BomonKTDT-HGTVT

69

4. Thit k mch chuyn m

Cc bc thit k mch chuyn m hon ton tng t nh thit k mch t hp, ngha l qua cc bc sau:

+ Lp bng chn l ca mch

+ Ti thiu ho cc hm ra

+ Xy dng s mch vi loi cng cho trc 5. Mt s vi mch chuyn m thng dng

Tn vi mch Hnh dng thc t

74147

B m ho s h thp phn thnh m BCD 8421

74147I9I8I7I6I5I4I3I2I1

A0A1A2A3

U14

7447 / 74LS47

Chuyn i m BCD thnh m 7 vch hin th theo h 10 bng led 7 thanh c anode chung

74LS47A3A2A1A0

testRBI

gfedcba

RBO

U7

PTH-DTT

70

7442

Gii m BCD 8421 thnh s h thp phn 74LS42

A3A2A1A0

9876543210

U11

7448

Chuyn i m BCD thnh m 7 vch hin th theo h 10 bng led 7 thanh c cathode chung

74LS48A3A2A1A0

testRBI

gfedcba

RBO

U8

74LS138

B gii m / phn knh 1 - 8

74LS138A2A1A0

E3E2E1

Q7Q6Q5Q4Q3Q2Q1Q0

74154

B gii m / phn knh 4 ng thnh 16 ng

74LS154

E1E0

A3A2A1A0

1514131211109876543210

U10

4028

Gii m BCD 8421 h 10 4028

A3A2A1A0 Q0

Q1Q2Q3Q4Q5Q6Q7Q8Q9

U12

BomonKTDT-HGTVT

71

Phn III

Mch dy

PTH-DTT

72

CHng 7:

Cc phn t nh c bn

I.Khi nim chung Nh ni, mch dy l mch c tn hiu ra khng ch ph thuc vo tn hiu

vo m cn ph thuc vo trng thi trong ca mch, ngha l mch c kh nng lu tr nh trng thi.

Cc phn t nh c bn to thnh mch dy c gi l cc flip-flop (mch bp bnh), chng l cc phn t nh n bit v ch c kh nng nh c 1 ch s nh phn. II. nh ngha v phn loi 1. nh ngha

Flip flop / FF l phn t c kh nng lu tr 1 trong 2 trng thi l 0 hoc 1. FF thng c nhiu u vo v 2 u ra c tnh lin hp (u ra ny l o

ca u ra kia), k hiu l Q v Q . Tn gi ca cc u vo tu thuc vo tng loi FF, s ni c th sau.

K hiu v tnh tch cc trong mch FF:

FLIP - FLOP

Cc u vo kh

Q

Q

mc +

mc -

sn + sn -xung tch cc sn

xung tch cc mc +

xung tch cc sn +

xung tch cc mc -

BomonKTDT-HGTVT

73

2. Phn loi FF

C th phn loi FF theo 2 cch nh sau:

3. Biu din FF

m t mt FF ngi ta c th dng 1 trong 3 cch sau: + Dng bng chn l

+ hnh chuyn i trng thi

+ Phng trnh c trng III. cc loi FF v iu kin ng b 1. Flip-Flop kiu RS

RS FF l mch Flip-Flop n gin nht ch c 2 u vo iu khin R (reset xo) v S (set thit lp), RS-FF c th c xy dng t 2 cng NAND hay 2 cng NOR. Hnh di y ch ra bng trng thi rt gn v s ca mch vi cc cng NAND v k hiu ca RS - FF

R, S l cc u vo iu khin

Qn l trng thi ca FF ti thi im hin ti t

Q l trng thi s chuyn ti ca FF sau thi gian qu , tc trng thi ca FF thi im tip theo

Gi thit, ti thi im bt u, S=1 v R= 0. Mc u ra ca cng 1 l thp (0) v iu ny to nn trng thi cao trn u ra ca cng 3 (Q=1). Tuy nhin, u ra ca cng 2 mc cao, bi th cng 4 c c hai u vo u mc cao (t cng 2

Flip-flop

D - FF

SYNC

Theo chc nng

ASYNC

JK - FF

RS - FF

T - FF

Normal M / S

Theo cch lm vic

PTH-DTT

74

v 3) nn u ra ca n s mc thp (Q =0). Flip-Flop trng thi SET v u ra Q =1 bt k Qn trc l 0 hay 1.

Khi S=0 v R=1, Flip-Flop s chuyn trng thi v u ra: Q=0; Q =1. Trng hp ny, Flip-Flop c RESET hay xo v 0, trng thi logic 0 trn Q d trc Qn l 0 hay 1.

Trng thi m trong , c hai u vo u mc R = S = 0 c gi l trng thi nh, v u vo s duy tr trng thi trc , Qn.

Nu u vo SET v RESET ng thi mc cao (S = R = 1), ta s c trng thi sau:

Q = Q = 1.

c coi l trng thi khng xc nh (khng s dng hay cm) R-S Flip-Flop khng c thit k hot ng trong trng thi R=S=1. Nhn xt:

+ Phng trnh c trng ca RS FF l RQnSQ .+= + S lun a Q v ga tr 1 + R lun a Q v gi tr 0

+ FF tt, tc chuyn trng thi t 1 sang 0 vi phng trnh Toff = RQnS

+ FF bt, tc chuyn trng thi t 0 sang 1 vi phng trnh Ton = QnRS

RS Flip-Flop vi u vo xung nhp

Cc h thng tun t thng yu cu cc Flip-Flop thay i trng thi ng b vi xung nhp. Khi ngi ta coi FF nh mt mch cht hay RS FF ng b hay RST FF hay RS FF nhp. iu ny c th thc hin c bi vic thay i mch nh sau:

Khi cha c xung nhp, Flip-Flop s gi nguyn trng thi khng ph thuc vo R v S (trng thi nh), ngha l trng thi ca FF b cht li .

Khi c xung nhp:

nu R = S = 0, u ra ca Flip-Flop s khng i;

nu R = 0, S = 1, Flip-Flop s c trng thi u ra: Q = 1, Q = 0;

BomonKTDT-HGTVT

75

nu R = 1, S = 0 ta s c trng thi u ra: Q = 0 v Q = 1.

Tm li: Khi khng c xung nhp FF khng thay i trng thi (khng ph thuc vo tn hiu u vo iu khin) v ch khi c xung nhp Ck mch mi lm vic theo bng chc nng (ph thuc vo tn hiu u vo iu khin)

Cc bin th ca RS FF

s dng c c t hp cm R = S = 1 ngi ta ch to cc bin th ca RS FF nh FF R, FF S v FF E. Cc FF ny c s dng kh rng ri trong cc khu iu khin ca h thng s.

Flip Flop R: ng vi t hp cm u ra Q = 0

Flip flop S : ng vi t hp cm u ra Q = 1

Flip flop E: ng vi t hp cm FF khng chuyn trng thi 2. JK Flip-Flop.

JK FF l mt loi FF vn nng v c nhiu ng dng

JK Flip-Flop cng tng t nh mt R-S kho v c cc u ra hi tip v u vo nh hnh di y

Mt u im ca J-K Flip-Flop l n khng c trng thi khng xc nh nh ca R-S khi c hai u vo mc 1.

V d, nu J = K = 1; Q = 1 v Q = 0; khi c xung nhp n, ch c cng 2 cho php truyn d liu vo, cn cng 1 s ngn li. Mc 0 ti u ra ca cng 2 s khin cho phn t nh chuyn trng thi. Nh vy, khi cc u vo u mc cao, u ra s o hay lt (toggle) trng thi ti mi xung nhp vo.

Nhn xt:

+ Phng trnh c trng ca JK FF c dng: QKQJQ += . + C s tng ng gia JK v RS, J tng ng vi S, K tng ng vi R

nhng t hp 11 trong JK vn c s dng m khng b cm nh trong RS + JK = 00 FF lun gi nguyn trng thi

JK = 01 FF lun chuyn n trng thi 0

JK = 10 FF lun chuyn n trng thi 1

JK = 11 FF lun lt trng thi

JK Flip-Flop ch c mt kh nng cho trng thi khng xc nh, l khi di xung nhp ln hn thi gian truyn t. Gi thit, Flip-Flop ang trong trng

PTH-DTT

76

thi: Q = 0 , Q =1 v J = K = 1;

Khi c xung nhp n, u ra s o trng thi sau mt khong thi gian truyn t t :

Q = 1 v Q =0;

Tuy nhin, do vn c xung nhp kch thch, u ra s hi tip tr li u vo khin mch c xu hng dao ng gia 0 v 1. Bi th, ti thi im cui ca xung nhp, trng thi ca Flip-Flop s khng c xc nh. Hin tng ny gi l hin tng ua vng quanh v c th gy nn chuyn bin sai nhm ca mch. Ngi ta khc phc hin tng ny bng cch s dng mch JK FF kiu ch t. JK Flip-Flop kiu ch t.

JK FF kiu ch t c s cu trc nh sau:

Mch bao gm 2 na ging nhau, mi na l mt RS Flip-Flop, FF th nht gi l FF master (ch) v FF th 2 gi l FF slave (t). u vo ca FF ch l u vo ca mch v u ra ca FF t l u ra ca mch. Tn hiu hi tip t u ra ca FF t v u vo ca FF ch. Cc xung a ti phn t l o vi xung a ti phn ch.

Cc u vo Preset v Clear s c chc nng ging nh ca u vo Set v Reset. Chng tc ng n u ra mt cch khng ng b, tc chng s thay i trng thi u ra m khng ph thuc vo s c mt ca xung nhp; v ch yu a u ra v mt trng thi bit no . (ngi ta cn gi y l cc u vo iu khin trc tip)

Gi thit cc u vo ny l khng tch cc (khi Pr = Cl = 1), khi c xung nhp n, Flip-Flop s thay i trng thi nh trong bng chn l sau:

CK J K Qn+1

0

x

0

0

1

1

x

0

1

0

1

Qn

Qn

0

0

nQ

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Vi Qn+1: trng thi k tip;

Qn : trng thi trc . x: trng thi khng xc nh.

Trong khong thi gian xung nhp l cao, phn T kho, bi th cc u ra Q v Q s khng thay i. Khi xung nhp chuyn t 1 v 0, khi T s chuyn trng thi trong khi khi Ch s kho. Ni cch khc, d liu trn J v K trc tin c truyn n khi Ch ti sn tng ca ca xung nhp v truyn ti khi T ti sn xung; nh vy, trng thi khng xc nh ca u ra nh trng hp J-K Flip-Flop s c loi b. 3. D Flip-Flop

D FF l loi FF ch c mt u vo iu khin D

Phng trnh c trng ca D l Q = D

Thc cht D FF chnh l mt khu tr c thi gian t l thi gian qu ca mch. u ra Q chnh l tr ca u vo sau khong thi gian t, v vy FF ny c tn l D FF (delay FF)

Ch to D FF t JK FF

Nu t mt JK Flip-Flop thm vo mt b o nh hnh di th u vo K lun l b ca J v s to nn mch D Flip-Flop. Hot ng ca n rt n gin, khi c xung ng h n, d liu ti u vo s c truyn v gi nguyn ti u ra.

Ngoi ra cng c th ch to D FF t RST FF bng cch thm cng NOT gia

hai u vo S v R tng ng vi J v K nh hnh trn. Bin th ca D FF

Trn thc t ngi ta s dng bin th ca D l DV FF. Loi FF ny c bng trng thi v s xy dng t cc cng NOR nh sau:

D Q Q

0 0 1

1 1 0

SDCP

RQ_Q

U3A

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T bng trng thi ta thy:

+ Khi V = 1 FF DV hot ng nh mt FF D thng thng + Khi V = 0 FF khng i trng thi vi bt k mc logic no ca D

4. Flip-Flop kiu T.

FF T l mt FF c 2 u ra v 1 u vo T. T FF c bng trng thi nh sau:

T Qn+1

0 Qn

1 Qn

Khi T = 0 FF gi nguyn trng thi

Khi T = 1 FF lt trng thi (toggle)

Phng trnh c trng ca T FF: QnTQ = Nh vy mch T FF thay i trng thi tun t theo mi ln c xung kch

thch

Ch : Khi u vo T c thi gian tn ti mc logic cao trong mt khong di hn so vi thi gian chuyn trng thi (thi gian tr) ca mch th mch s tip tc lt trng thi ti khi ht thi gian tn ti mc logic cao ca T, qu trnh lm cho vic xc nh chnh xc mch ang trng thi no l khng th, do T ch c th lm vic ch ng b (v thc t thi gian tn ti mc logic cao ca T lun ln hn rt nhiu thi gian tr ca mch)

Ch to T FF t JK FF

R rng T FF n gin l mt JK Flip-Flop vi c J v K u mc logic 1.

V J = K = 1 nn Flip-Flop ny s lt (Toggle) trng thi mi khi xung nhp chuyn t 1 v 0.

Hnh bn l s mch v k hiu ca T Flip-Flop .

V D Qn+1

1 0 0

1 1 1

0 0 Qn

0 1 Qn

Q

Q

V

D U1D

U1C

U1B

U1A

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Bin th ca T FF

Trn thc t ngi ta s dng bin th ca T l TV FF. Loi FF ny c bng trng thi nh sau:

T bng trng thi ta thy:

+ Khi V = 1 FF TV hot ng nh mt FF T thng thng

+ Khi V = 0 FF khng i trng thi vi bt k mc logic no ca T

Nhn xt chung v ch lm vic ca cc loi FF:

+ Cc D FF v RS FF c th lm vic ch ng b hoc khng ng b v vi mi tp tn hiu vo iu khin lun tn ti t nht 1 trong cc trng thi n nh (Q = Qn)

+ Cc T FF v Jk FF khng th lm vic ch khng ng b v mch s ri vo trng thi dao ng (chuyn trng thi lin tc gia 0 v 1). Khi JK = 11 hoc T = 1 hai loi FF s dao ng, do chng lun phi lm vic ch ng b.

IV. Chuyn i gia cc loi FF

4 loi FF va xt trn c th chuyn i ln cho nhau

Phng php chuyn i gia loi FF i thnh FF j c m hnh ho theo s sau:

Cc bc thc hin:

+ Xc nh h hm i = f(j, Q) theo bng cc u vo kch ca cc FF

+ Ti thiu ho cc hm ny v xy dng s

Bng u vo kch ca cc FF

Mch logic

FF

loi i

Q

Q

V T Qn+1

1 0 Qn

1 1 Qn

0 0 Qn

0 1 Qn

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Qn Qn+1 JK RS D T

0 0 0X X0 0 0

0 1 1X 01 1 1

1 0 X1 10 0 1

1 1 X0 0X 1 0

v d: thit k mch t hp chuyn i RS FF thnh JK FF

Trc ht ra s thit k mch logic ca hm R = f(Q, J, K)

S = g(Q, J, K)

Lp bng Karnaugh ca R theo Q, J, K ta c:

Q / JK 00 01 11 10

0 X X

1 1 1

Vy: R = QK

Lp bng Karnaugh ca S theo Q, J, K ta c:

Q / JK 00 01 11 10

0 1 1

1 X X

Vy: S = J.Q

Nh vy mch thc hin chuyn i t RS FF sang JK FF s c dng nh sau:

U2BJ

KU2A

SR Q

_QU1

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Chng 8

B m I. nh ngha v phn loi 1. nh ngha: B m l mt mch dy tun hon c cc c im sau: + Mt u vo( m) v mt u ra(kt qu) + S trng thi trong bng h s m Di tc dng ca tn hiu vo, b m s chuyn trng thi ny n trng thi khc, theo mt trnh t nht nh. C sau K tn hiu vo m, mch li tr v trng thi ban u: 2. Phn loi a, Phn loi theo cch lm vic: + B m ng b(Synchronous Counter) l b m m cc FF dng( m ho cho cc trng thi trong ca b m) cng mt lc khi c tn hiu vo m m khng qua cc trng thi trung gian. Cc tn hiu xung nhp(Ck) c a ng thi vo cc FF + B m khng ng b(Asynchronous Counter), trong b m tn ti t nht mt cp chuyn trng thi(Si ->Sj) m trong cc FF khng thay i trng thi cng mt lc, tn hiu xung nhp Ck khng c a ng thi vo cc FF. Gi s ban u b m khng ng b ang trng thi cn bng n nh Si, khi c tn hiu vo m ch c mt s FF b tc ng, sau s thay i trng thi ca FF

B m (h s m K)

Tn hiu m: X Tn hiu ra: Y

S khi ca b m

Si 0001

Si 0010

Si 0001

Si+1 0010

Si+m

Sj 0100

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ny s lm thay i trng thi ca FF khc, cho n khi b m trng thi cn bng n nh b. Phn loi theo h s m: - B m c h s m K = 2

n (trong n l s t nhin), v d K=2, 4, 6, 8, 10,... - B m c h s m K # 2

n (trong n l s t nhin), v d K=3, 5, 7, 9, 11,... c. Phn loi theo hng m - B m thun, cn gi l b m tng: mi khi c tn hiu vo m gi tr ca b m tng ln 1 - B m nghch, cn gi l b m gim: mi khi c tn hiu vo m gi tr ca b m gim i 1.

Khi nim b m thun nghch trong k thut s ch mang tnh tng i, tu theo vic m ho(quan im) m b m s l thun hay nghch, c ngha l mt b m s l m thun khi m ho kiu ny, nhng s l m nghch khi m ho kiu khc. C th c loi th 3 l va thun va nghch tu theo tn hiu iu khin. d. Phn loi theo kh nng lp trnh

- B m c kh nng chng trnh ho - B m khng c kh nng chng trnh ho.

Ch : mt b m c th thng bao gm tt c cc c tnh trn: v d B m c K=9, m tng, ng b, khng th chng trnh... 3. hnh trng thi ca b m:

- nh: l cc trng thi bn trong ca b m, gm c K trng thi, tc l c K nh.

- Cung: L tn hiu vo m/ kt qu ra m; khi khng c tn hiu vo m(tn hiu vo m khng tch cc)- X = 0(o), trng thi ca b m gi nguyn; khi c tn hiu vo m(tn hiu vo m tch cc), b m thay i trng thi. Tn hiu ra ca b m ch xut hin(Y=1) khi b m trng thi K-1 (SK-1) v c tn hiu vo X

Tc l Y=(SK-1)X V khi c tn hiu ra mch s tr v trng thi ban u, v qu trnh m tip tc.

0X/0

X/0

1X/0

X/0

K-2 X/0

X/0

K-1

X/1

X/0

hnh trng thi ca b m

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VD: Gi s m s xe trn mt on ng, theo nguyn tc c xe i qua mt vch ngang th tin hnh m Khi , tn hiu vo m tch cc X chnh l xe i qua vch, gi s khi tin hnh m cha c xe no i qua, khi c 1 xe i qua, th trng thi b m thay i: s xe tng ln 0->1, c nh th cho n 9, khi n 9 th nu c mt xe i qua ta s c 1_0 xe, vit 0 nh 1(tn hiu ra l 1- hng chc), b m quay v trng thi ban u 0(ta hay ni l vit 0 nh 1). Ch : mun quan st c trng thi ca b m, th cn c mch di m, nh v d trn th ting Vit quy c c l: mt, hai, ba, bn... th ngi bit ting Vit s hiu l c 1, 2, 3, 4,... xe i qua vch. II. M ca b m. Ging nh mch dy ni chung cc trng thi ca b m c m ho bi mt m c th. Cng mt b m, c th c nhiu cch m ho cc trng thi trong. Cc m thng dng thit k b m: a. M nh phn: M nh phn l loi m m cc bit ca n c trng s l 1-2-4-8-16-32-...., bt tr nht(LSB) l 1=20, bit tip theo tng ng l 2=21, ... , v bt gi nht MSB l 2n-1. Dng n bit nh phn m ho c 2n trng thi. b. M Gray: L loi m khng c trng s, khong cch Hamming gia 2 t m k nhau l 1(2 t m k nhau ch khc nhau mt bin). Dng n bit nh phn m ho c 2n trng thi. c. M BCD L m nh phn m ho s thp phn, m ny dng 4 ch s nh phn m ho mt ch s thp phn, chng c gi l decard. Ch dng 10 t hp m ho s t 0(0000) n 9(1001), cc s ln hn th s dng t hp ca chng.

V d: 12=0001.0010 Ch : cc loi m trn xem bng m d. M Johnson:

Dng n bit nh phn, s m ho c 2n trng thi, theo nguyn tc: + Hai t m k nhau ch khc nhau mt bin. + Trong bng m cc bt bng 1 c y dn ln t bt tr nht n bt gi

nht, v khi y ht th n li vi dn i t bt tr nht(s bt 1 tng dn t tri sang phi, khi y th li gim dn tri sang phi):

VD: n=2 t hp th nht : 0 0 t hp th hai : 1 0

t hp th ba : 1 1 t hp th t : 0 1 n=3 t hp th nht : 0 0 0 t hp th hai : 1 0 0

t hp th ba : 1 1 0 t hp th t : 1 1 1

t hp th nm : 0 1 1 t hp th su : 0 0 1

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n=4 0000 1000 1100 1110 1111 0111 0011 0001

...... e. M vng: M vng c nguyn tc lp m: + Dng n bt nh phn th m ho c n trng thi + 2 t m k nhau lun khc nhau 2 bin + Trong mt t m ch c duy nht mt bt l 1 v c dch t bt tr nht

n bt gi nht, cc bt khc l 0. VD: n=3 1 0 0 0 1 0 0 0 1 n =4 1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1 .......

III.Cc bc thit k b m Cng ging nh cc bc thit k mch dy, gm c 4 bc c bn nh

sau: - Bc 1: T bi ton cho, xc nh tn hiu vo m, h s m K, t

v

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