ình Phần 1: Thiết kế hệ logic tổ...

Bi ging Vi mch s Bin son Ng Vn Bnh

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Phn 1: Thit k h logic t hp I. Gii thiu chung

V c bn h iu khin logic c chia lm 2 loi ln: - H t hp. - H tun t.

Trong h t hp u ra ti mt thi im bt k ch ph thuc vo trng thi cc u vo ti thi im , ngha l khng c phn t nh trong mch, i vi h tun t th khc: Trng thi ng ra ti thi im ang xt khng nhng ph thuc trng thi vo ti cng thi im m cn ph thuc vo trng thi vo trong qu kh c ngha l phi c phn t nh trong mch.

Mt h logic tun t c th cha cc h logic t hp con, nhng d liu thit k h t hp c th c cho di dng:

- Mt tp hp cc mnh . - Biu thc Boole. - Bng s tht.

Nu bit mt cch biu din c th suy ra cch biu din khc, nh trong cc lnh vc thit k k thut khc s lng thit b x dng cn phi nh nht gim chi ph, kch thc, tit kim nng lng v tng tin cy. Cc phng php t c s thc hin hm Boole mt cch n gin nht cn ph thuc vo nhiu yu t.

Mt cch o phc tp ca hm Boole l m s lng "literal" tc s lng ch c trong biu thc Boole, literal s xc nh lng dy ni v s lng u vo ca mch v vy cn phi gim s lng literal.

Mt vn khc l s lng cng cn thit chnh iu ny quyt nh kch thc ca mch, mt thit k n gin nht l dng t cng nht ch khng phi t literal. Yu t th ba l s mc logic, gim s mc logic s lm gim thi gian tr v tn hiu i qua t cng hn nhng nu ch ch n thi gian th c th li lm cho s lng cng tng ln.

Trong phn ny trnh by cch thit k mt h t hp thc hin mch logic hai mc, cch dng cc vi mch c trong thc t cho n cch tng hp logic nhiu mc t c s lng cng cn dng l t nht. II. Quy trnh thit k Quy trnh thit k thng c thc hin theo mt s bc sau y:

- Phn tch yu cu v xc nh tn hiu vo - ra. - Xc nh bng trng thi v bng s tht. - n gin ha. - Vit phng trnh tng ca tch hoc tch ca tng. - V s AND - OR hoc OR - AND. - Bin i sang s vi mch thng dng. - Chn linh kin v rp mch.

V d: Thit k mch logic iu khin mt bn cha nc vi s cng ngh v yu cu

nh sau:


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- Bm P ch chy khi ging y nc ( cm bin D tc ng) v bn cha y (cm bin A khng tc ng).

- Van V bnh thng lun lun m sn sng cung cp nc v ch ng li khi ging ht nc v ng thi bn cng cn nc (cm bin C khng tc ng) gi li mt lng nc an ton.

- Ci hoc n bo ng sng khi h thng b s c (A tc ng nhng C li khng tc ng). Gii 1. Phn tch yu cu v xc nh tn hiu vo - ra T s cng ngh v yu cu iu khin suy ra tn hiu vo l: cm bin A, C v D ( n gin xem nh khng c cm bin B) v tn hiu ra gm: bm P, van V v n BD 2. Xc nh bng trng thi v bng s tht. Bng trng thi l mt dng din t khc ca yu cu iu khin ni dung ca bng s cho bit quan h gia tn hiu ra vi tn hiu vo.


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A C D P V BDK.tac ong K.tac ong K.tac ong Dng ong ToiK.tac ong K.tac ong Tac ong Chay M ToiK.tac ong Tac ong K.tac ong Dng M ToiK.tac ong Tac ong Tac ong Chay M ToiTac ong K.tac ong K.tac ong Dng M SangTac ong K.tac ong Tac ong Dng M SangTac ong Tac ong K.tac ong Dng M ToiTac ong Tac ong Tac ong Dng M Toi

Bng s tht l bng trng thi khi thay vo cc gi tr logic tng ng 0 v 1. Do

t mt bng trng thi c th dn n nhiu bng s tht khc nhau ty theo h thng thc t. Trong trng hp ny c th quy nh nh sau:

Vi cc cm bin: Tc ng tng ng vi 1 v khng tc ng l 0 Bm P chy l 1 v dng l 0. Van ng l 1 v m l 0. n BD sng l 1 v ti l. 0. Suy ra bng s tht nh sau.

Hnh 1.3 Bng s tht

3. n gin ha Qu trnh n gin c th thc hin bng bng Karnaugh hoc i s logic, bng Karnaugh l mt phng php th, trc quan , d hiu v thng c p dng trong trng hp s lng ng vo khng nhiu lm. Hnh 1.4 Bng Karnaugh ca hm P


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4/ Vit phng trnh tng ca tch hoc tch ca tng Phng trnh n gin ca hm P: P = D A ABC ABC ABC ABC Phng trnh n gin ca hm BD: BD = AC Ring hm V ch c mt trng hp bng 1 nn khng cn n gin V = D AC 5/ V s AND - OR hoc OR - AND

Hnh 1.6 S mch dng AND - OR

S vi mch cn dng thc hin s AND - OR 7404 - 6 o 7408 - 4 AND 2 input 7411 - 3 AND 3 input

6/ Bin i sang s vi mch thng dng Trong cng ngh vi mch vic thc hin bng cc cng NAND v NOR l hiu qu hn. Trong thc t, cc cng AND v OR c thay th bng cc cng NAND v


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NOR vi mt b o c t thm pha sau. Ngoi ra mt hm logic bt k gm c AND, OR v NOT c th c thc hin bng cch ch dng cc cng NAND hoc NOR. S chuyn i ny da trn nh l De Morgan

AB A B .A B AB

Hnh 1.7 Kt qu ca nh l De Morgan

Mt cng AND tng ng mt cng NOR vi tt c cc ng vo o, v mt

cng OR tng ng mt cng NAND vi tt c cc ng vo o. T nhng nguyn tc va trnh by s hai mc logic AND - OR c chuyn thnh dng NOR - NOR nh sau:

Hnh 1.8 S mch dng NOR S vi mch cn dng 7402 - 4 NOR 2 input 7427 - 3 NOR 3 input

III. Vi mch phc hp Nh bit, mch logic nhiu mc c th c thc hin bng cch ghp cc cng

ri, trong cng ngh TTL cng nh CMOS c ch to sn mt s vi mch nhiu mc logic n gin , l cc cng phc hp AOI (AND - OR - INVERT) hoc OAI (OR - AND - INVERT) nhiu u vo

Mt khi AOI l mt mch logic ba mc bao gm cc cng AND mc th nht, mt cng OR mc th hai v mt cng o u ra. Tng t mt khi OAI gm cc cng OR mc th nht, mt cng AND mc th hai v mt cng o u ra. Tm li


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nhiu cng ri rc c tch hp li thnh mt cng phc hp c ni dy sn bn trong

Hnh 1.9 Khi AOI 2 ngn 2 u vo Trong hnh 1.9 trnh by mt cng AOI hai ngn (mt cng AND gi l mt ngn) hai u vo (mi cng AND c 2 ng vo). S lng nn bng vi s lng u vo ca cng OR V d: Thc hin hm XOR bng khi AOI

Phng trnh ca hm XOR: A B AB AB Nu dng cc cng ri th phi cn 2 cng AND 2 u vo, mt cng OR 2 u vo v 2 cng o. thc hin bng cng AOI ch cn tm b ca hm XOR di dng tng cc tch

A B AB AB Sau a A v B vo mt ngn ca AOI, A, B vo ngn th hai, cng o ng ra ca AOI s tr v gi tr ng ca hm XOR

Tn hiu ra - Z: Tn hiu iu khin thang chy ln - Y: Tn hiu iu khin thanh chy xung Tn hiu trung gian


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Vi mc ch gim bt s bin s trong khi xy dng v bin i cc hm logic cn thit phi to ra thm cc bin trung gian, cc bin ny l kt qu t hp t cc tn hiu vo ban u tha mn c yu cu hot ng ca thang my.

- c: Tn hiu cho php thang my hot ng c = b1.b2.b3...b6.b7.b8.b9

- e: Lnh cho thang my i ln. e = a3 + a2.d1

- f: Lnh cho thang my i xung. f = a1 + a2.d3 - g: Tn hiu bo hon thnh mnh lnh. g = a1,d1 + a2.d2 + a3.d3 Nh vy t 15 bin ban u nay thu gn li ch cn 4 bin mi to thnh 16 t hp

bin (2n), ngoi ra thang my cn c 3 trng thi: Dng, ln v xung. Bc tip theo l lp bng quan h bin s - trng thi. Nhn xt: - Thang my lun dng khi khng c c v c g. - Thang my khng thay i trng thi khi c c, khng g v ng thi c e v f hoc

khng c c e v f.

Trong cc thit b logic nhiu u ra i khi khng cho php nhiu u ra cng tn ti song song m ch mt m thi. Do , cn phi dng tn hiu ra hi tip tr v ng vo, trong trng hp thang my trnh trng hp ny c th chn 2 ng ra z v y lm tn hiu hi tip hnh v sau trnh by quan h gia cc tn hiu ny vi trng thi ca thang my.


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IV. S dng hm XOR v XNOR

Trong nhiu tnh hung thit k c th x dng phng php n gin ha cc hm logic bng ton t XOR v XNOR v cc hm ny c x dng rng ri nn chng c ch to sn trong cng ngh vi mch, phn di y s trnh by phng php nhn dng hm XOR v XNOR trn ba Karnaugh.

Trong phng php ti thiu ha AND-OR v OR-AND, cc s 0 hoc 1 cnh nhau theo chiu dc v chiu ngang c dn li vi nhau, trong trng hp ti thiu XOR v XNOR cc i din nhau theo phng cho hoc cc nhm i xng nhau qua mt giao im ca cc ng k trong bng c gi l Diagonal v cc hay nhm i xng nhau qua mt hng hoc ct c gi l Offset nh trnh by hnh sau y:


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Hnh 1.16 Bng Karnaugh vi diagonal v offset

Hnh 1.16a l mt bng Karnaugh 3 bin A, B, C trong diagonal 1 c im i xng l M, diagonal 2 c im i xng l N, offset 1 c ct i xng l AB, suy ra cc biu thc rt gn XOR v XNOR nh sau:

offset 1 = ( ) ( )CAB CAB C AB AB C A B offset 2 = ( ) ( )CAB CAB C AB AB C A B

diagonal 1 = ( ) ( )CAB CAB A CB CB A C B diagonal 2 = ( ) ( )CAB CAB A CB CB A C B


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Hnh 1.16b l bng Karnaugh 4 bin A, B, C, D trong diagonal 1 to nn bi 2 nhm i xng nhau qua im I v offset 1 to bi hai nhm i xng nhau qua hng CD, kt qu n gin nh sau:

Diagonal 1 = ( ) ( )CAB CAB A CB CB A C B Offset 1 = ( ) ( )CDA CDA A CD CD A C D V d: Dng hm XOR v XNOR n gin ha hm Y cho trong bng Karnaugh sau y

Theo cch n gin thng thng ta c kt qu

Y1 = C AB CDA CDA CAB CDAB CDAB Theo phng php XOR v XNOR ta c

Diagonal 1 = ( ) ( )CAB CAB B CA CA B C A Diagonal 2 = ( ) ( )CDA CDA D CA CA D C A

Offset 1 = ( ) ( )CDAB CDAB DB CA CA DB C A Suy ra: Y2 = Diagonal 1+ Diagonal 2 + Offset

Hnh 1.18 Hai cch thc hin hm Y V. Hazard - Glitch trong mch t hp Glitch l mt xung khng mong mun xut hin ti ng ra ca mch t hp, mch c th to ra glitch gi l mch c hin tng hazard (may ri), cng c trng hp mch c kh nng hazard nhng li khng to ra glitch, trong phn di y s trnh by phng php thit k mch khng c hazard (hazard-free circuit) 1. Khi nim v Hazard Cc Hazard gy rc ri cho h thng trong 2 trng hp


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- Trng thi kugic ng ra khng n nh, c th khc phc bng cch ko di thi gian p ng ca h V.D: Gim tn s xung ng h

- Ng ra ca mch c hazard c ni n cc ng vo khng ng b ca mch tip theo sau v s lm cc mch ny hot ng khng chnh xc

Cc phng php loi b kh nng hazard u gi thit rng cc xung glitch ng ra l do s thay i ca 1 bt u vo

Cc loi hazard khc nhau c trnh bi trong hnh. Mt hazard tnh xut hin khi u ra tri qua mt s chuyn tip nht thi mc ng l ra n khng c thay i: Hazard tnh loi 1 lm u ra gim xung 0 trong mt thi gian ngn (xung m) v ngc li hazard tnh loi 0 lm xut hin xung dng ng ra.

Hazard ng lm ng ra thay i nhiu ln trong khi ng ra n ch c php chuyn mt ln t 0 ln 1 hoc t 1 xung 0, v s to ra cc xung glitch trong mch logic a mc l loi mch c nhiu ng dn t u vo n u ra vi cc thi gian truyn khc nhau, rt kh loi tr hazard ng, phng php gii quyt tt nht l chuyn mch a mc c hazard ng thnh mch logic 2 mc khng c hazard.


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Hnh 1.19 Cc loi Hazard

2. Pht hin v loi tr hazard trong mch logic 2 mc Kho st hm sau y F (A,B,C,D) = (1,3,5,7,8,9,12,13)

Hnh 1.20 Bng Karnaugh ca hm F

Xt trng hp khi u vo ABCD = 1100 chuyn sang 1101. T s cho thy khi u vo bng 1100, u ra ca cng G1 bng 1 trong khi u ra ca cng G2 bng 0. Do , u ra ca G3 bng 1. Khi u vo chuyn sang 1101 th u ra ca cc cng vn gi nguyn. By gi xem s thay i ca ng vo t 1101 sang 0101 lc ny A chuyn xung 0 v A chuyn


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ln 1 (nhng phi qua mt thi gian tr ca cng o) do s c mt khong thi gian ngn c A v A cng bng 0, ng ra ca G1 v G3 cng bng 0 v F cng bng 0, sau khi A ln n mc 1 th F bng 1 tr li, mt xung glitch xut hin ng ra.

T bng Karnaugh cho thy khi cc gi tr u vo thay i nhng cng nm trong mt biu thc n gin th khng th xy ra glitch, nhng nu chuyn t biu thc n gin ny sang biu thc n gin khc th c th xy ra glitch. Tuy nhin, tc G1 chm hn G3 nhiu th s khng xy ra glitch ng ra F nhng hazard th lun lun tn ti.

Phng php khc phc l thm vo cc biu thc n gin sao cho bao ph mi thay i n bt u vo, trong v d trn nu thm vo phng trnh hm F biu thc CDB th F s lun bng 1 bt chp s thay i ca A nh ni trn.

T bng Karnaugh hnh 1.20 vit li phng trnh ca hm F di dng tch cc tng. F = ( )( )D A C A

D dng nhn thy rng trong mch tn ti hazard tnh loi 0 khi u vo thay i t 1110 sang 0110, loi tr hazard ny phi thm vo s hng (C + D)B v phng trnh F tr thnh. F = ( )( )D A C A (C + D)B

Nh vy, s tn ti Hazard 1 ca hm F s tng ng vi s tn ti hazard 0 ca hm F. 3. Hazard ng L cc bin thin xy ra nhiu ln ng ra trong mt qu trnh thay i u vo, nguyn nhn l do tn ti nhiu ng tn hiu trong mng a mc m mi ng c tr khc nhau

Hnh 1.22 Mch c Hazard ng

T mch in k bn cho thy c 3 ng khc nhau t B v B n u ra. Gi s

ABC = 000 th F =1 v sau ABC = 010 tng ng F = 0, khi ABC = 000 lc ny G1 = 0, G2 = 1v G3 = 1, G4 = 1. By gi B chuyn t 0 ln 1 vi tc cc cng nh trong hnh v, qu trnh bin i trong mch c m t nh sau:

G2 chuyn t 1 sang 0, tip theo G3 xung 0 v G5 cng xung 0.


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Tip theo G1 ln 1 lm G3 ln 1 v G5 cng ln 1, lc ny u ra chuyn t 1 xung 0 ri ln 1. Cui cng G4 chuyn t 1 xung 0 lm G5 xung gi tr cui cng l 0. Nh vy u ra bin i t 1 xung 0, ln 1 ri li xung 0.

VI. Cc vi mch thng dng trong h t hp

1. B dn knh (multiplex) Cn gi l b a hp hoc chn tn hiu (selector) c nhim v chn mt trong s cc

tn hiu u vo a ti mt u ra duy nht di s iu khin ca cc ng vo iu khin. Ngc li b phn knh (distributor) hoc gii a hp (demultiplex) s dn tn hiu t mt u vo duy nht n mt trong cc ng ra di s iu khin ca cc ng vo iu khin.

Nh vy, b dn knh l mt mch logic t hp bao gm 2n u vo d liu, n ng vo iu khin v mt u ra d liu, gi tr nh phn ca u vo iu khin l a ch ca u vo d liu.

T bng s tht suy ra phng trnh Boole tng ng:

Z = 1 2AI AI Nu A = 0 th I0 c chn v A = 1 th I1 c chn, b dn knh nh trn gi l b dn knh 2:1


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cc b dn knh c m t bng s lng u vo d liu v t c th suy ra s

lng tn hiu iu khin, t b dn knh 2 u vo suy ra phng trnh cc b dn 4:1 v 8:1 nh sau:

Z = 1 2 3 4BAI BAI BAI BAI Z =

1 2 3 4 5 6 7 8CBAI CBAI CBAI CBAI CBAI CBAI CBAI CBAI Dng tng qut minterm ca b dn knh

Thit k b dn knh S b dn knh dng cc cng logic c suy ra t cc phng trnh phn trn,

sau y l s b dn knh 4 u vo


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Hnh 1.25 B dn knh dng cng logic

Mch dng 4 cng AND 3 u vo, mt cng OR 2 u vo v 2 cng o, tng cng l 7 cng.

C th xy dng mt b dn N:1 t nhiu b dn vi s u vo t hn N, hnh v sau y trnh by b dn 8:1 t hai b 4:1 v mt b 2:1. Lp th nht chn mt u vo trong s cc u vo t I0 ...I3 v m u t I4...I7 dng 2 tn hiu iu khin B, C, lp th hai chn mt trong hai nhm da trn tn hiu A. Nhng cng c th thc hin b dn 8:1 t 4 b dn 2:1 v m b dn 4:1

X dng b dn knh nh mt khi logic Ngoi chc nng chn tn hiu nh bit, thc t mt b dn l mt kiu thc hin

bng s tht trc tip bng phn cng. Xt hm F (A, B, C) = m0 + m2 + m6 + m7 c th thc hin hm ny bng b dn

8:1 nh hnh v sau y. Cc bin vo A, B, C c ni n cc u vo iu khin ca b chn, ng vo Ii

c t 1 tng ng vi minterm ca hm F, tt c cc u vo d liu khc c gn mc 0. Trong v d ny I0, I2, I6 v I7 c gn mc 1 cn I1, I3, I4, I5 c gn mc 0.

Tuy nhin, c th x dng b dn n gin hn thc hin hm ny, hai bin A, B a vo cc u vo iu khin v cc u vo d liu ca b dn ni vi 0, 1, C v C

Xem bng s tht sau y:


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Hnh 1.26 Biu din F nh hm ca 0, 1 v C

F = ABC ABC ABC ABC V cui cng l s dng b dn 4:1

Hnh 1. 28 Cc gi tr c th ca hm F

Trong trng hp tng qut, mt hm bt k c th c thc hin bng cch chn (n

- 1) bin vo lm tn hiu iu khin cho mt b dn knh c 2 n-1 u vo d liu, mi gi tr t hp ca (n - 1) bin ng vi hai hng trong bng s tht, gi tr ca hm trong hai hng ny c th l cc tr s cho trong hnh 1...Nu gi tr hm trong hai hng u bng 0


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hoc 1 th u vo tng ng s l 0 hoc 1 nu khng th u vo s c ni vi bin n hoc o ca n.

Cc b dn knh h TTL Cu to cc b dn knh trong thc t c thm mt hay nhiu tn hiu iu khin khc mc ch cho php thit b hot ng. Trong hnh v sau y trnh by b dn TTL 74157 gm 4 b 2:1 ng ra tc ng mc cao v 74158 gm 4 b 2:1 ng ra tc ng mc thp.

Hnh 1.29 S v bng s tht b dn knh TTL 157 v 158

2.. B phn knh (demultiplex) B phn knh cho php tn hiu u vo ca n dn n mt trong cc ng ra, mi ng ra c phn bit bi mt m nh phn duy nht, do b phn knh cn c chc nng gii m. VD: Mt b gii m 1:2 gm 2 u vo: u vo d ABC D ABC D ABCD liu G v u vo iu khin S v 2 u ra Q0 v Q1, phng trnh cc u ra nh sau; Q0 = GS Q1 = G.S Nu G = 0 th 2 u ra bng 0, nu G = 1 th th 1 trong 2 u ra s bng 1 ty thuc vo gi tr ca S, mt b phn knh hay gii m thng c gi theo s lng u vo iu khin v s lng u ra V.D: 1:2; 2:4; hoc 3:8 Cch thc hin Hnh v sau trnh by mch gii m 1:2 bng cc cng logic, hai tn hiu vo l tn hiu cho php G v tn hiu chn. Khi tn hiu chn l 0 th tn hiu ti G s c dn n ng ra 1v lc ny ng ra 2 mc 0, ngc li. khi ng chn mc 1 th G c dn n ng ra 2 v lc ny ng ra 1 mc 0, nu G = 0 th c 2 ng ra u bng 0, b gii m ny c tn hiu cho php tc ng mc 1, hnh 1.30 trnh by mch gii m c tn hiu cho php G tc ng mc 0.

Hnh 1.30 Mch gii m 1:2


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Hnh 1.31 Mch gii m 2:4 dng cng AND v NOR

X dng b phn knh nh mt khi logic Mt b phn knh c th xem nh mt b to cc tch y (minterm), hnh 1.32

v s khi ca mt b gii m 3:8 c 3 tn hiu iu khin l A, B, C, khi mi u ra ca b gii m s tng ng vi mt minterm ca 3 bin A, B, C. Do , mt b gii m c th thc hin mt hm logic bt k di dng tng cc tch, v d hm sau y:

F1 = ABC D ABC D ABCD F2 = ABCD ABC

F3 = ABC D Chuyn cc hm v dng tch cc tng chun

F1 = ( )( )( )ABCD ABCD ABCD A B C D A B C D A B C D

F2 = ( )( )ABCD ABC A B C D A B C

F3 = ( )ABCD A B C D Hnh 1.33 l s thc hin dng b gii m

Hnh 1.32 B gii m 3:8


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Hnh 1.33 Thc hin hm logic 4 bin

Cc b phn knh h TTL Hnh 1.34 gii thiu hai b gii m h TTL l 74139 gm 2 b gii m 2:4 v

74138 l b gii m 3:8, trong 74139 ng vo G v cc ng ra tc ng mc thp. 74138 c 3 u vo cho php G1, G2A v G2B v 3 ng chn tn hiu A, B, C, bng

s tht ca 2 vi mch ny c trnh by trong hnh 1.34. Trong thc t c th kt hp cc b phn knh chun to nn cc b phn knh ln

hn v k c b dn knh nh trong hnh 1.35 cho thy hai cch to nn b dn knh 32:1 v trong hnh 1.36 l b gii m 5:32 c to nn t mt b gii m 2:4 v 4 b gii m 3:8:


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Hnh 1.34 B gii m TTL 138 v 139

VII. Vi mch s lp trnh - PLD (programmable logic device)

gim c n mc ti thiu s lng linh kin cn dng, mt yu cu t ra l phi tm c mt khi logic c th x dng cho nhiu thit k khc nhau, v d cc khi AOI nhng cng khng gim ng k s lng thit b tr khi c c mt khi logic cha hng trm cng nhng cu to ca khi ny phi nh th no ?

Mt gii php hp l l sp xp cc cng AND v OR theo mt cu trc mng c tng qut ha m cc im ni gia chng vi nhau c th lp trnh thc hin mt hm xc nh. Nhng khi logic a nng nh vy l mt trong cc h vi mch s lp trnh c tn gi ring l PLA (programmable logic array) - mng logic lp trnh.

1. PLA Hnh 1.37 trnh by s khi ca mt mng PLA gm c nhiu u vo v nhiu

u ra, c t chc thnh cc mng con AND v mng con OR. Mng con AND s chuyn cc u vo thnh cc tch s tng ng vi cc im ni c lp trnh, mng con OR s cng cc tch s ny vi nhau to ra biu thc dng tng cc tch ca hm.


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Mt khi PLA c th thc hin mt tp hp cc hm kh phc tp. phc tp ph

thuc vo s lng u vo, s lng cc tch s (s lng cc cng AND) v s lng u ra (s lng cc cng OR) m PLA c th cung cp.

V d mt FPLA cng ngh TTL c th c 16 u vo, 48 tch s v 8 u ra c ch to gn trong mt v 24 chn d liu. N tng ng vi 48 cng AND 16 u vo v 8 cng OR 48 u vo.

V d thc hin hm Boole sau y: ,C A CA F0 = A BC F2 = AB BC F1 = A.C + A.B F3 = A BC

C th m t bng tp hp cc bin A, B, C nhng ch ny tng ng vi s u vo ca mng AND, cc tch s , , , ,AC A BC BC AB v cc hm F0, F1, F2 v F3. Cc u ra ca mng AND l u vo ca mng OR v cc u ra ca mng OR k cc hm, thc hin cc hm ny phi cn mt PLA c t nht 3 u vo, 5 tch s v 4 u ra.

Mt cch tin li m t cc hm l dng ma trn c tnh, mt bin ha ca bng s tht c v trong hnh 1.38, bng m t cc u vo no phi ni vi cng AND to nn tch s tng ng, (1 = bin khng b; 0 = bin b; - = khng ni) v nhng tch s no phi c OR vi nhau to thnh cc u ra cui cng (1 = ni; 0 = khng ni). Cc hng xc nh cc tch s, cc ct biu din cc u vo v u ra. Mt tch s c th tham gia trong nhiu hm nu trong hng ca n c nhiu s 1. Trong hnh 1.38 d dng nhn thy cc tch s , ,A BC AB c dng chung cho nhiu hm.


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Hnh 1.38 Ma trn c tnh PLA

Hnh 1.39 Mng PLA trc khi lp trnh

V mt cu to phn cng ca mng PLA th gia cng v cc u vo c t cc

cu ch c th lm t bng cch cho dng in ln chy qua, phn mm lp trnh s phn tch hm Boole xc nh cu ch no phi ph hng v cu ch no cn li.


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Hnh 1.40 Mng PLA sau khi lp trnh

Hnh 1.41 trnh by k hiu thng dng biu din mng PLA, cc dy ni n i vo AND v OR tng trng cho cc u vo, cc du X cho bit cc cu ch ang trng thi ni

Hnh 1.41 K hiu PLA 4 u vo, 4 tch s v 4 u ra

2.. PAL (programmable array logic)

Trong hnh 1.41 cho thy c 2 mng AND v OR u c th thay i theo yu cu ca ngi x dng. Nhng khng phi tt c cc mng logic lp trnh u cho php lp trnh y nh PLA v d h PAL gm cc mng AND lp trnh cn cc im ni gia cc tch s vi ng vo cng OR c xc nh trc, s lng cc tch s vo mt cng OR thng l 2, 4, 8, 16...


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C mt s b tr trong cc mng PAL gia phc tp ca cc hm v s lng tchs trn mt cng OR vi s lng cc hm c lp m mng c th thc hin> Cng OR c fan - in cng cao th PAL cng t u ra.

V d mt h PAL c th c 3 dng: Mi dng u c 16 u vo, 16 tch s nhng khc nhau v t chc ca mng OR: 4 cng OR mi cng 4 u vo, 2 cng OR mi cng 8 u vo v 1 cng OR 16 u vo cn mng AND th vn hon ton c th lp trnh. Hnh 1.42 trnh by s mng PAL c 4 u vo, 4 tch s v 2 u ra v mng OR c nh v 2 tch s mi cng.

Hnh 1.42 Mng PAL 4 u vo, 4 tch s v 2 u ra

im khc nhau c bn gia PLA vi PAL l PLA c th chia x cc tch s chung

cho nhiu u ra cn PAL th khng th c. V vy nn dng PLA khi cn thc hin cc hm c x dng chung tch s.

Mt khc v mt tc PLA chm hn PAL do c nhiu tip gp lp trnh, cc tip gip ny c cu to t cc cu ch c tr khng cao hn cc tip gip c nh v Do , tn hiu i qua hai mng tip gip lp trnh s b tr nhiu hn so vi khi i qua mt tip gip lp trnh nh trong PAL.

3. V d Thit k b chuyn i m BCD bt sang m Gray 4 bt Mch c 4 bt BCD ng vo l ABCD v 4 bt gray ng ra WXYZ


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Hnh 1.43 Bng s tht b chuyn m BCD Gray

Hnh 1.44 bng Karnaugh cc hm W, X, Y, Z


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Hnh 1.45 Thc hin b chuyn m bng PAL

V tp hp cc hm khng c tch s no chung chia x nn thch hp thc hin bng mng PAL nh trnh by trong hnh 1.24 gm 4 cng OR 4 u vo, trong hnh cn cho thy c nhiu cng AND khng dng nu dng PLA th s n gin hn nhng chm hn.

Vi cng ngh PLD cho php thc hin cc hm ch cn mt vi mch duy nht, vi v d trn nu dng cc cng vi mch ri th kt qu nh trnh by hnh 1.46 sau y

Hnh 1.46 B chuyn m dng vi mch ri

Trong hnh 1.46 cho thy phi cn n 15 cng ly t 5 IC khc nhau. Nh vy l PLD rt hiu qu trong yu cu gim s lng linh kin. V d 2: Thit k b so snh 2 s nh phn 2 bt

u vo l 2 s nh phn 2 bt AB v CD, u ra l 4 hm logic c nh ngha nh sau: AB = CD (EQ =1) AB = CD (NE = 1) AB < CD (LT = 1) AB > CD (GT = 1)


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Hnh 1.47 Bng Karnaugh ca b so snh 2 bt

T bng Karnaugh suy ra cc hm c ti thiu ha EQ = ABC D ABCD ABC D ABCD NE = CA DB DB AC LT = C A ABD DC B GT = CA CDB DB DAB

Cc hm x dng 14 tch s trong c 2 tch s ,C A CA c dng 2 ln nn thc hin s ny bng mng PLA l thch hp. Mt PLA hay PAL c th thay th n hng vi chc cng logic ri l rt bnh thng, y l cch gim linh kin rt hiu qu. Do , cc nh thit k hin nay rt thng p dng khi thit k cc h t hp.


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Hnh 1.48 S b so snh 2 bt dng PLA


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8. Bi tp 1. Thit k mch cnh bo nhit cho bn X l ha cht. Nhit ca bn c theo

di v kim tra bng 3 cm bin t 3 v tr khc nhau trn bn. Nhit ca bn c cnh bo bng 2 n led: led xanh v led .

- Khi khng c qu nhit hoc ch c 1 v tr qu nhit th led xanh sng, led tt. - Khi c 2 v tr tr ln qu nhit th led xanh tt, led sng v nhp nhy 10Hz. 2. Thit k mch logic iu khin mt bn cha nc vi s cng ngh v yu cu

nh sau: - Bm P ch chy khi ging y nc ( cm bin D tc ng) v bn cha y (cm

bin A khng tc ng). - Van V bnh thng lun lun m sn sng cung cp nc v ch ng li khi

ging ht nc v ng thi bn cng cn nc (cm bin C khng tc ng) gi li mt lng nc an ton.

- Ci hoc n bo ng sng khi h thng b s c (A tc ng nhng C li khng tc ng).

3. X dng hm XOR v XNOR n gin biu thc sau:

a. F = (2,3,5,6,8,9) b. F = (2,3,5,6,8,9,12,15)

4. X dng hm XOR v XNOR n gin biu thc sau: a. F = (0,3,5,6,11,12,15) N = 4,7,8

b. F = (1,7,9,10,12,13,15) N = 3,5,8,11 5. Th no l hin tng chu k v chy ua trong mch tun t khng ng b? hy nu

cc trng hp chy ua trong mch tun t khng ng b? 6. Th no l Hazard v Glitch trong mch t hp? C my loi Hazard? Khi trong mch

c Hazard s xy ra hin tng g? 7. FPGA l g? Hy nu cu trc chung ca mt FPGA? 8. FPGA c bao nhiu loi? Cc loi FPGA c trn th trng thng dng hin nay? 9. Hy nu cu trc v cc c tnh k thut ca h vi mch lp trnh GAL16V8? 10. Hy nu cu trc v cc c tnh k thut ca h vi mch lp trnh GAL20V8? 11. Vi mch lp trnh PLA l g? Hy v mt m hnh ca PLA vi 3 ng vo, 3 ng ra v

3 tch s? 12. Cho bit s khi ca vi mch lp trnh PLA ? Hy v mt m hnh ca PLA vi 2

ng vo, 3 ng ra v 4 tch s? 13. Cho bit s khi ca vi mch lp trnh PLA ? Hy v mt m hnh ca PLA vi 2

ng vo, 2 ng ra v 5 tch s? 14. So snh s ging v khc nhau ca 2 vi mch lp trnh PLA v PAL? Ging nhau: PLA v PAL u c cu to t 2 mng AND v OR.

ình Phần 1: Thiết kế hệ logic tổ...

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