Bai giang Ky thuat so- tkhao.ppt

Trng Cao ng in t in lnh HNTn hc phn : K Thut sS n v hc trnh : 4L thuyt : 60 TitGio vin : Nguyn Phng Anh

Ni dung

Ni dungL thuytChng 1: Gii thiu chung v K thut s10Chng 2: Cc cng logic c bn15Chng 3: Mch Flip Flop15Chng 4 : Mch dy10Chng 5: Thit k mch s dng MSI, LSI10Tng s60

Ti liu tham khoL thuyt mch logic v k thut s Nguyn Xun Qunh, NXB i hc2. K thut s tp I, II Nguyn Vn Tiu, NXB i hc3. K thut s Nguyn Thy Vn, NXB Khoa hc & K thut, 19954. K thut s Nguyn Ph TinNXB Gio dc

Tiu chun nh gi sinh vin- im chuyn cn 10%- im thi gia hc phn 30 %- im thi cui hc phn 60%

Chng 1 : gii thiu chung v k thut s v i s logic1.1 Cc h thng tng t v h thng s1.1.1. Cc i lng tng t v sa. Tn hiu tng tL tn hiu lin tc theo gi tr v theo thi gianN/xt: th biu din ln ca t/hiu theo t/gian l 1 ng lin L tn hiu lin tc theo ln, do n nhn tt c cc gi tr trong phm vi no

b. Tn hiu sL tn hiu ri rc theo thi gian v theo lnN/xt :Tn hiu s ch nhn xut hin ticc thi im nht nh v ti cc thi im xut hin tn hiu s li chnhn cc gi tr nht nh

Mc Logic (Logic Level)H thng s nh phn ch c 2 s: 0 v 1Trong mch s c 2 mc in p i din cho 2 gi tr 0 v 11: Mc in p cao (High)0: Mc in p thp (Low)HIGH=1UncertainLOW=05.0 Volts2.0 Volts0.8 Volts0.0 VoltsVH(max)VH(min)VL(max)VL(min)

- Cc tn hiu s c dng sng c chu k hoc khng c chu k1010tWTWTWTW

- Gin nh th (Timing Diagram)Trong nhiu h thng s, cc tn hiu s cn c ng b ho theo 1 dng sng nh th c bn gi l xung nhp (Clock)Clock10Bit Time1010101010

1.1.2. Cc h thng tng t v h thng sa. H thng tng t L h thng m cc thit b trong h thng x l tn hiu tng tb. H thng s L h thng m cc thit b trong h thng x l tn hiu s

c. u nhc im ca K thut s D thit k Lu tr d dng Chnh xc, tin cy* u im Cc thao tc c th lp tnh Chu t nhiu hn Kh nng t hp cao

* Nhc ima s cc tn hiu cn x l l tn hiu tng t ng dng KTS C b chuyn i t tn hiu tng t sang tn hiu s

Chuyn i ADCX l tn hiu sChuyn i DACT/hiu tng tT/hiu tng t

1.2. Cc h thng m v m1.2.1. Cc h thng ma. m khng theo v tr L h thng m m gi tr ca cc ch s trong 1 s khng ph thuc vo v trVD: Ch s la m.I ; II; III; IV; V; VI; VII

b. m theo v tr L h thng m m gi tr ca cc ch s trong 1 s ph thuc vo v tr ca chng trong s VD: S thp phn1 2 3 4Hng nghnHng trmHng chcHng n v

Mt s khi nim- C s (r- radix): S lng k t ch s s dng biu din trong h thng s m- Trng s (Weight): i lng biu din cho v tr ca 1 con s trong chui s.- Gi tr ca 1 s: Tnh bng tng theo trng sGi tr = Tng (K s x Trng s )

a. S thp phn (Decimal): C s r = 10Kt qu = 4000+1000+0+70+3+0.6+0.02+0+0.0005 = 41073.6205

b. S nh phn (Binary): C s r = 2Kt qu = 16+8+0+2+1+0.5+0+0.125+0.0625= 27.6875

c. S thp lc (Hecxa - Decimal): C s r = 16

Kt qu = 12288+3072+112+10+0.375+0.0546675+0.000076293 = 15462.42976

1.2.2. Chuyn i gia cc h m Quy tcMun chuyn i phn nguyn ca s A sang c s bt k R, ta ch vic chia ln lt gi tr ca A cho R. Cc s d nhn c trong cc ln chia l cc ch s A khi biu din trong h c s R, tnh t ch s c trng s thp nht a. Xt bin i phn nguynLu : Khi khng vit c s bn cnh mc nh hiu s biu din h thp phn

VD: Chuyn i s (345) 8 sang c s 9 ?(345) 8 = ( ?) 92299254927902274(345) 8 = ( 274) 9

y l gi tr Ca (345)8 h 10

VD: Chuyn i s 11 sang h nh phn ?11 = ( ?) 211251221210101111 = ( 1011) 2

201

VD: Chuyn i s 700 sang h thp lc(Hecxa) ?700 = ( ?) 167001643121621116022BC700 = ( 2BC) 16

Quy tcMun chuyn i phn phn ca s A sang c s bt k R, ta ch vic nhn ln lt gi tr phn phn ca A cho R. Cc phn nguyn nhn c trong cc ln nhn l cc ch s A trong phn phn khi n biu din trong h c s R, tnh t ch s c trng s cao nht a. Xt bin i phn phnLu : Trong cc ln nhn, nu khng xut hin phn nguyn th coi nh phn nguyn tng ng bng 0. Cn nu trong ln nhn no xut hin phn nhn khc 0, th trc khi nhn phI b phn nguyn ny i

VD: (0,0001) 2 = (?) 8y l gi tr ca(0,0001)0,0625 x 8 = 0.5Phn phn = 0Phn phn = 0.50.5 x 8 = 4.0 Phn nguyn = 4Phn nguyn = 0Vy (0.0001)2 = (0.04)8

1.2.3. S nh phnCc tnh cht ca s nh phn S nh phn n bt c 2n gi tr t 0 n 2n-1S nh phn c gi tr 2n c biu din 100 (n bit 0) v gi tr 2n-1 l s 1.1 (n bit 1)Bit c trng s nh nht l LSB (Least Singificant Bit) v bit c trng s ln nht MSB (Most Singificant Bit)S nh phn c gi tr l l s c LSB =1,Ngc li gi tr chn l s c LSB =0

Cc tnh cht ca s nh phn Cc bi s ca bit1B (Byte) = 8bit1MB = 210KB = 220B1KB = 210B = 1024B1GB = 210MB

* Cc php tnh vi s nh phna. Php cng0000011111101

ABA+B

b. Php tr0000011111101

ABA-B

c. Php nhn000001111001

ABA*B

d. Php chia011011

ABA:B

1.2.5. M ha M ha l gn mt k hiu cho mt i tng thun tin cho vic thc hin mt yu cu c th no Nhm k hiu sau khi m ha gi l cc m. M BCD 8421 (Binary Coded Decimal) M BCD dng s nh phn 4 bt c gi tr tng ng thay th cho tng s hng trong s thp phnVD: S 625 10 c m BCD l 0110 0010 0101

M Gray M Gray l m hai s lin tip ch khc nhau 1 phn t nh phn (1 bt) M ASCII M Parity pht hin li M LED 7 on M d 3 M d 3 c to thnh bng cch cng thm 3 n v vo m BCD 8421

i Binary sang m GrayCh s u tin ca m Gray ging ch s u tin ca m nh phn. - Cng khng nh tng cp bit lin k ta s thu c ch s tip theo trong m Gray.M Gray

VD.(1100)binary = M Gray??????? Bc 1 Ch s u tin ca m Gray ging ch s u tin ca m nh phn. 1 1 0 0 binary 1 Gray Bc 2 Cng khng nh hai bit u tin ca s nh phn. Kt qu thu c l s Gray tip theo. 1 1 0 0 binary 1 0 Gray Bc 3 Cng hai bit k tip ca s nh phn ta nhn c ch s Gray tip theo. 1 1 0 0 binary 1 0 1 Gray Bc 4 Cng hai bit cui cng ca s nh phn ta nhn c bit cui cng ca m Gray 1 1 0 0 binary 1 0 1 0 Gray

Chuyn i t m Gray sang m nh phn: - S dng phng php tng t trn, tuy nhin c mt s khc bit. Ex.(1010)Gray = binary Bc 1 Ch s u tin ca m Gray ging ch s u tin ca m nh phn. 1 0 1 0 Gray 1 binary Bc 2 Cng theo ng cho nh di nhn c t m nh phn tip theo 1 0 1 0 Gray 1 1 binary Bc 3 Tip tc cng theo ng cho nhn c cc t m nh phn tip theo 1 0 1 0 Gray 1 1 0 binary 1 0 1 0 Gray 1 1 0 0 binary

M LED 7 onabdcefg

1.3 i s logic (Boole)1.3.1. Cu trc i s BOOLEi s BOOLE l cu trc i s c nh ngha trn 1 tp phn t nh phn B ={0, 1} v cc php ton nh phn: AND; OR; NOT

XYX.Y(X AND Y)001101010001

XYX+Y(X OR Y)001101010111

X (NOT X)0110

1.3.2. Cc tin a. Phn t ng nhtVi php ton OR, phn t ng nht l 0x + 0 = 0 + x = xVi php ton AND, phn t ng nht l 1x.1 = 1.x = xb. Tnh giao honx + y = y + xx . y = y. x

1.3.2. Cc tin c. Tnh phn bx + (y . z) = (x + y) . (x + z)x . (y + z) = x . y + x. zd. Phn t b

1.3.3. Cc nh l c bna. nh l 1:b. nh l 2:x + x = xx . x = xc. nh l 3:x + 1 = 1x . 0 = 0d. nh l 4 (L hp thu):x + x . y = xx. (x + y) = xe. nh l 5 (L kt hp):x + (y + z) =(x + y) + zx. (y . z) = ( x . y) . zf. nh l 6 (L DeMorgan):

Mt s CT thng dng

1.3 i s logic (Boole)1.3.2. Hm BOOLEHm BOOLE l 1 biu thc c to bi cc bin nh phn v cc php ton nh phn NOT, AND, OR. Vi cc gi tr cho trc ca cc bin, hm BOOLE s c gi tr l 0 hoc 1

Hm F = 1 nu:x = y = 1 (bt chp z)hoc x = y = 0, z = 1Ngc li: F = 0Ta c th biu din hm BOOLE bng bng gi tr

1.3.3 Cc nh lut c bn ca i s logica. Hm ORb. Hm AND

A BX = A+B0 00 11 01 1 0111

A BX = A.B0 00 11 01 1 0001

c. Hm NOTd. Hm EX-OR

1.3.4. Cc phng php biu din hm logica. Phng php dng bng gi tr L bng lit k t hp cc gi tr ca bin s (u vo ) v cc gi tr tng ng ca hm (u ra)

b. Phng php i s/ l: Mt hm n bin bt k F(X) =F(X1XiXn) c th biu din dng CTT (Chun Tc Tuyn) hoc CTH (Chun Tc Hi) Dng CTT: L tng ca nhiu thnh phn, mi th/phn l tch gm y cc bin

Dng CTH: L tch ca nhiu th/phn, mi th/phn l tng gm y cc bin

Cch vit Hm s di dng CTT Ch quan tm n t hp bin m hm c gi tr = 1. S ln hm = 1 s chnh l s tch ca biu thcTrong mi tch, cc bin c g/ tr = 1 c gi nguyn, cc bin c g/tr = 0 ly ph nh. Hm F bng tng ca cc tch K hiu

Cch vit Hm s di dng CTH Ch quan tm n t hp bin m hm c gi tr = 0. S ln hm = 0 s chnh l s tch ca biu thcTrong mi tng, cc bin c g/ tr = 0 c gi nguyn, cc bin c g/tr = 1 ly ph nh. Hm F bng tch ca cc tng K hiu

Trng hp tu nh (Dont care)Hm BOOLE c th khng c nh ngha cho ht tt c cc t hp cc bin ph thuc. Khi ti cc t hp khng s dng ny, hm BOOLE s nhn gi tr tu nh, ngha l hm BOOLE c th nhn gi tr 0 hoc 1K hiu: d (i vi CTT)D (i vi CTH)

Ta c th biu din hm BOOLE theo dng chnh tc:F(A,B,C) = (2,3,5) + d(0,7)

VD: F = (1,2,4,7) vi d = 0,5,6Trong : 1,2,4,7 l gi tr thp phn ca cc t hp bin m ti hm nhn g/tr = 10,5,6 l cc gi tr thp phn ca t hp bin m ti gi tr hm khng xc nh

Dng CTT: H/s F(X) = 1 ti cc t Hp gi tr bin tng ng Vi gi tr thp phn l 1,2,4,7

VD: F = (0,3,5,6) vi D= 1,2Trong : 0,3,5,6 l gi tr thp phn ca cc t hp bin m ti hm nhn g/tr = 01,2 l cc gi tr thp phn ca t hp bin m ti gi tr hm khng xc nh

Dng CTH: H/s F(X) = 0 ti cc t Hp gi tr bin tng ng vi gi tr thp phn l 0, 3, 5 6

V d1 : F(x,y,z) = xy + zTa c th chuyn v dng CTT

V d 2: Ta c th chuyn v dng CTT

c. Biu din bng Karnaugh (ccn)t g/tr 1 vo cc tng ng ti hm = 1 . t k hiu x vo cc ti hm khng xc nh. Cc cn li t gi tr 0 hoc b trng.Hai c gi l k nhau khi t hp bin m chng biu din ch khc nhau 1 bin

Ba 2 binVD: F(A,B)=0011BAF0011BAF

11x

Ba 3 binVD: F(A,B,C)=000101BCAF1110000101BCAF1110

xx111

Ba 4 binVD: F(A,B,C,D)=0001CDABF1110000111100001CDABF111000011110

x11x1111x11

1.4. Ti thiu ha hm logicPhng php i sy l phng php rt gn hm Boole bng cch s dng cc nh lut bin i trong i s Boole.VD: Hy rt gn hm sau: F =

VD3: Ti thiu ha biu thc:

Bi Tp: Rt gn biu thc

CMR: Nu A = BC + BC thA + AB = 0b. AB = AC

Rt gn Ba Karnaugh* Nguyn tcLin kt i: Khi lin kt OR (2 1) k cn vinhau trn ba K, ta s c 1 tch s mt i 1 bin so vi tch chun (Bin mt i l bin khc nhau gia 2 lin kt). Hoc khi lin kt (AND) (2 0) k cn vi nhau trn ba K, ta s c 1 tngmt i 1 bin so vi tng chun (Bin mt i l bin khc nhau gia 2 lin kt

000101BCAF1110F = AC: A=0; C=1000101BCAF1110F = B+C: A=1; C=0

11

00

Rt gn Ba Karnaugh* Nguyn tcLin kt 4: Tng t nh lin kt i khi ta lin kt 4 k cn nhau ta s loi c 2 bin (2 bin b loi l 2 bin khc nhau gia 4 t hp)

0001BCAF111001F = C0001BCAF111001F = A

0000

Rt gn Ba Karnaugh* Nguyn tcLin kt 8: Tng t nh lin kt i khi ta lin kt 8 k cn nhau ta s loi c 3 bin (3 bin b loi l 3 bin khc nhau gia 8 t hp)

0001CDABF111000011110F = B0001CDABF111000011110F = C

11111111

00000000

VD: Ti thiu hm sau bng bng Karnaugh

CDABF00011110000111101111111F= Nhm 1 + Nhm 2 + Nhm 3


CDABF000111100001111011111F= Nhm 1 + Nhm 2111


CDABF000111100001111011F= Nhm 1 + Nhm 211

VD: Biu din hm logic 2 bin theo bng sau bng bng Kanaugh ?X1X2F00110011Hm F = 0 tng ng t hp gi tr 0 0Hm F = 1 tng ng t hp gi tr 0 1Hm F= 1 tng ng t hp gi tr 1 0Hm F = 0 tng ng t hp gi tr 1 1

VD: Biu din hm logic 3 bin theo bng sau bng bng Kanaugh ?X1X2X3F100100x11101110101

S chuyn i gia cc cch biu dinT bng chn l xy dng biu thc logic ca hmT bng chn l c th vit biu thc logic di dng CTTNguyn tc: Tng ng vi mi gi tr 1 u ra ca hm trong bng chn l l 1 tch cc bin u vo trong biu thc, ti dng tng ng trong bng chn l nu cc bin u vo no l 0 th bin c vit o trong tch, nu bin no c gi tr 1 th gi nguyn.

Theo nguyn tc, chng ta ch quan tm n cc g/tr 1 u ra ca FTi chng c cc tch u vo trong biu thc

T biu thc logic ca hm xy dng bng chn l A. B tng ng t hp g/tr 11

Chuyn i sang bng KarnaughABCF100100101001110010

Chuyn t bng Karnaugh sang biu thc logicABCF100100110001110001Hm ra F = 1 tng ng t hp 101Hm ra F = 1 tng ng t hp 111Hm ra F = 1 tng ng t hp 110

Bi tp:Hy rt gn hm 3 bin ,bng phng php i s v bng bng Karnaugh ?

Bi tp:Hy vit hm logic c cho dng ba Karnaugh trong cc trng hp sau, di dng rt gn.

CDABF00011110000111101111CDABF0001111000011110111CDABF00011110000111101111CDABF0001111000011110111111111111111

Bi tp:Hy vit hm logic c cho dng ba Karnaugh trong cc trng hp sau, di dng rt gn.

CDABF00011110000111101111CDABF00011110000111101111CDABF000111100001111011CDABF000111100001111011111111111111111111111

Rt gn cc hm sau:

Chng 2 : Cc cng logic c bn2.1 Cc cng logic c bn

Cc cng logic c bn1. Cng NOT

Cc cng logic c bn2. Cng ANDxyzKL: Vi cng AND c nhiu ng vo, ng ra s l 1 nu tt c cc ng vo u l 1

Cc cng logic c bn3. Cng ORxyzKL: Vi cng OR c nhiu ng vo, ng ra s l 1 nu c t nht 1 ng vo l 1; hoc ng ra s l 0 nu tt c cc ng vo u l 0xyz=x+y

Cc cng logic c bn4. Cng NANDxyzKL: Vi cng NAND c nhiu ng vo, ng ra s l 0 nu tt c cc ng vo u l 1

Cc cng logic c bn5. Cng NORxyzKL: Vi cng NOR c nhiu ng vo, ng ra s l 1 nu tt c cc ng vo u l 0

Cc cng logic c bn6. Cng XORxyzKL: Vi cng XOR c 2 ng vo, ng ra s l 1 nu 2 ng vo l khc nhau.Vi cng XOR c nhiu ng vo, ng ra s l 1 nu tng s bit 1 ng vo l s l

Cc cng logic c bn7. Cng XNORxyzKL: Vi cng XNOR c 2 ng vo, ng ra s l 1 nu 2 ng vo l ging nhau.Vi cng XNOR c nhiu ng vo, ng ra s l 1 nu tng s bit 1 ng vo l s chn

Thc hin hm BOOLE bng cng logic1. Cu trc cng AND - ORCu trc AND OR l s logic thc hin cho hm BOOLE biu din theo dng tng cc tch.VD: F(A,B,C,D)=ABD + CDABCDF

Thc hin hm BOOLE bng cng logic2. Cu trc cng OR - ANDCu trc OR - AND l s logic thc hin cho hm BOOLE biu din theo dng tch cc tng.VD: F(A,B,C,D)=(A + D) (B + C + D)ABCDF

Thc hin hm BOOLE bng cng logic3. Cu trc ton cng NANDCu trc NAND l s logic thc hin cho hm BOOLE m biu thc c dng b ca 1 s hng tch.- Dng nh l De-Morgan bin i s hng tng thnh tch - Cng NOT cng c thay th bng cng NAND ni chung 2 ng vo.

Thc hin hm BOOLE bng cng logic4. Cu trc ton cng NORCu trc NAND l s logic thc hin cho hm BOOLE m biu thc c dng b ca 1 s hng tng.

Bi tp:Cho cc tn hiu A, B a vo mch OR nh sauHy xc nh tn hiu ra X ca mch OR?AB

Bi tp:Cho cc tn hiu A, B a vo mch AND nh sauHy xc nh tn hiu ra X ca mch AND ?AB

Dnh cho Sinh vin luyn tp V s mch logic thc hin hm f(X1, X2) = X1 . X2 + X1 + X2 V s mch logic thc hin hm f(X1, X2, , X3, , X4) = (X1 + X2 ). (X3 + X4)

Hy CM cc ng thc sau:A + A B = A + B2. Cho 2 mch logic sau

CMR F1 = F2

b. X1 F2 = X2 X2 F1 = X1

2.2 Thit k v phn tch mch t hp1. Ni dung bi tonGi thit: Cho s ca h logic t hpb. Kt lun: Tm hiu nguyn l hot ng ca h logic cho2. Cc bc thc hinNhn dng cc phn t c trong s b. Vit biu thc quan h gia hm ra vi bin vo ca tng phn tc. Tm hiu nguyn l hot ng ca hm ra & bin voLp bng Gi tr ca s V th thi gian- Kt lun nhim v ca h logic cho2.2.1. Phn tch mch t hp

1. Ni dung bi tonGi thit: Cho h logic t hp 1 trong cc dng sauBng Gi trBiu thc i sBng Karnaugh- M t thng qua cc mnh .b. Kt lun: a ra s logic 2. Cc bc thc hin2.2.2. Thit k mch t hpTi thiu hm, cc hm choChn phn t v s

VD1: Xc nh hm ra ca mch logic sau:X

VD 2: Hy thit k mch logic c 3 u vo A, B, C u ra F. Hot ng ca mch ny nh sau:u ra F = 1 khi c 2 hoc 3 u vo bng 1u ra F = 0 trong cc trng hp cn li.

Cc bc thit k:B1: t nhim v ca mch chng ta thit lp c bng chn l ca mch yu cu nh sau

B3: Ti thiu ha hm ra F bng phng php bng KarnaughABCF00011110011111F= AB + BC + AC

B4: V s logic ca mch

VD 3: Hy thit k mch logic c 3 u vo P, Q, R u ra S. Hot ng ca mch ny nh sau:u ra S = 1 khi P=0 hoc khi Q = R = 1u ra S = 0 trong cc trng hp cn li.

Cc bc thit k:B1: t nhim v ca mch chng ta thit lp c bng chn l ca mch yu cu nh sau

B2: T bng chn l ta xc nh c hm ra F nh sau : B3: Ti thiu ha hm ra F bng phng php bng KarnaughPQRF000111100111111

B4: V s logic ca mch

2.3. Cc mch t hp thng gp2.3.1. B cng nh phn

HY QUAN ST

1. Hy thc hin chuyn i s sau t h thp phn sang h nh phn. (11)10 = ( )2 (7)10 = ( )2

2. Nu cc quy tc cng 2 s nh phn 1 bt

(11)10 = (1011)2 (7)10 = (0111)2

Quy tc cng 2 s nh phn 1 bt 0 + 0 = 1 1+ 0 = 1 0 + 1 = 0 1+ 1 = 0 nh 1 CU TR LI

VD: A = 1010 ; B = 1111. Hy tnh C= A+BC = A + B = 1 0 1 01 1 1 11011011101525

Chc nng chung ca cc b cng l thc hin cc php cng nh phn1) Xy dng s khi b na tnga. Khi nim b cng

b. Bng chn lc. Hm gi trd. S logic

Thc cht vic cng cc s nh phn nhiu bt chnh l vic cng cc cp bt tng ng trong cc ct thuc cc s nh phn. Ni cch khc vic cng cc s nh phn nhiu bt c th quy v vic cng cc s nh phn 1 bt. Tuy nhin khi cng mt cp bt chng ta phi cng thm c bt nh (nu c) t vic cng cp bt trong ct trc .2) B cng nh phn 1 bta. Khi nim

Ai; Bi : L cc bt nh phn cn cng vi nhau (chng tng ng thuc ct th i ca 2 s nh phn A, B)Ci-1: L bt nh t php cng cc bt trong ct i-1 ca 2 s A, BCi: L bt nh ca chnh php cng cc bt trong ct i, nh sang ct tip theo i+1Si: L bt kt qub. Xy dng mch logic thc hin php cng cc s nh phn 1 bt

c. Bng chn l ca mch cng nh phn 1 bt

Rt gn Si v CiAiBiCi-1Si00011110011111d. Hm gi tr

AiBiCi-1Ci00011110011111

S mch logic cn thit kAi. Bi????????????

Ai. Bi S

B na tng1

C

SB na tng2

C

e. S mch cn thit k

ng dng b tng y trong n v ALU ca CPU my vi tnh.

ng dng b tng y 1 bt xy dng B tng nhiu bt trong my tnh.

S dng b tng y xy dng b bin i m BCD_nh phn.3) ng dng

B chuyni ADB chuyni ADMch tngy 4 bt Song songB chuyni DA11+7=18

TNG KT BIB tng y : Thc hin cng 2 s nh phn mt bt c nh u vo.

CiSiAiBiCi-1Bn tng 2Bn tng 1B tng y

c. B cng song song 4 btDa trn b cng 1 bt chng ta c th xy dng cc b cng 4 bt, tng ng cng cc s nh phn 4 bitS dng 4 b cng nh phn 1 bt cng cccp bt tng ng cc ct ca cc s nh phn 4 bt cn cng. Tt nhin u ra nh t ct ny phi c a vo u vo nh ct tip theo.

S mch cng nh phn 4 btCng 1 bt (0)

Cng 1 bt (3)

Cng 1 bt (2)

Cng 1 bt (1)

B3B1B2B0Nhn xt: Ta thy b cng trn khi thc hin cng phi tin hnh ln lt t phi qua tri, ging nh thc hin php ton bng tay. Chng ta mt 4 xung nhp tng ng cho vic cng ln lt cc ct. Nn tc cng s chm. Khc phc nhc im ny ngi ta s dng b cng nh nhanh

IC thc hin php cng 4 bt nh phn song song l IC 74LS83B cng 4 bt 74LS83Ai,Bi: Cc bt cn cngSi: Cc bt tngCo, C4: Bt nh u vo v u ra ca IC

IC thc hin php cng 8 bt nh phn song song l 2 IC 74LS83 ghp vi nhauB cng 4 bt 74LS83C0 C4

Kim tra nhn thcS mch logic sau y c tng ng vi b bn tng khng? Chng minh

Hnh 1ABCS

Cu tr li

Ta c: ABA.BC=A.BA+B

2.3.2. B Tra .B trChc nng chung ca cc b tr l thc hin cc php tr nh phnVD: A = 1111 ; B = 1010. Hy tnh C= A - BC = A - B = 1 1 1 1

01101510051 0 1 0

VD: A = 1111 ; B = 1010. Hy tnh C= A - BC = A - B = 1 1 0 001111205070 1 0 1111

B na hiua. Bng chn lb. Hm gi trc. S logic

* B Tr nh phn 1 btThc cht vic tr cc s nh phn nhiu bt chnh l vic tr cc cp bt tng ng trong cc ct thuc cc s nh phn. Ni cch khc vic Tr cc s nh phn nhiu bt c th quy v vic tr cc s nh phn 1 bt. Tuy nhin khi tr mt cp bt chng ta phi tr thm c bt nh (nu c) t vic tr cp bt trong ct trc .

Mch logic thc hin php tr cc s nh phn 1 btAi; Bi : L cc bt nh phn cn tr vi nhau (chng tng ng thuc ct th i ca 2 s nh phn A, B)Ci-1: L bt nh t php cng cc bt trong ct i-1 ca 2 s A, BCi: L bt nh ca chnh php cng cc bt trong ct i , nh sang ct tip theo i+1Hi: L bt kt qu

Bng chn l ca mch tr nh phn 1 bt

Rt gn Hi v CiAiBiCi-1Si0001111001AiBiCi-1Ci000111100111111111

S mch logic cn thit k

B na hiu 1

B na hiu 2

kt st in t

B kim tra so snh l mt mch t hp c kh nng thc hin chc nng so snh hai s nh phn ch ra mi quan h gia chng.Gi: F1=1 L hm tn hiu ra khi a = b F2=1 L hm tn hiu ra khi a < b F3=1 L hm tn hiu ra khi a > b Ta lp bng chn l nh sau:Mi quan h gia hai s nh phn A v B m mch c th ch ra c l A = B hoc A < B hay A > B.Gi thit c hai s nh phn 1 bit l a v b2.3.3. B so snh

a. B so snh 2 s nh phn 1 btF1: Bo hiu A = BF2 : Bo hiu A< BF3: Bo hiu A>B

ABF1F2F3S logic

b. B so snh 2 s nh phn nhiu btB so snh83(A < B) = F1(A =B) = F2(A > B) = F3

b. B so snh 2 s nh phn n btC 2 s nh phn n bt (AnAn-1A1)(BnBn-1B1), chng ta phi thc hin cc php so snh trn tng cp bt ly t 2 s t phi qua tri.u tin so snh An vi Bn. Nu An > Bn hoc (An < Bn) th quyt nh ngay A > B hoc (A < B). Nhng nu An= Bn th cha KL c g v phi tip tc so snh ti cp bt An-1 v Bn-1. Qu trnh tin hnh tng t cho n khi gp 1 cp bt Ai&Bi sao cho Ai > Bi hoc Ai < Bi lc ta KL A>B hoc A

So snh trc tip

F1 = (A

F1= ABa3 = b3a1 = b1a0 > b0a3 = b3a2 = b2a1 > b1a3 = b3a3 > b3a2 > b2

- F1 = 1 khi A = B- F2 = 1 khi A < B- F3 = 1 khi A > B Ta gi F1, F2, F3: L cc hm tn hiu u ra ca b so snhF11 =1 L hm tn hiu ra khi a1 = b1 F21 =1 L hm tn hiu ra khi a2 = b2F31 =1 L hm tn hiu ra khi a3 = b3F12 =1 L hm tn hiu ra khi a1 < b1 F22 =1 L hm tn hiu ra khi a2 < b2F32 =1 L hm tn hiu ra khi a3 < b3F13 =1 L hm tn hiu ra khi a1 > b1 F23 =1 L hm tn hiu ra khi a2 > b2F33 =1 L hm tn hiu ra khi a3 > b3 Ta c cc tn hiu vo t cc b so snh 1 bit nh sau:

Cn c trn bin lun ta c:F1 = F31 F21 F11 F2 = F32 + F31 F22 + F31 F21 F12F3 = F33 + F31 F23 + F31 F21 F13

Nh v d minh ho u tin, cc em cho bit thit k c mch kho in t trn kt vi m kho l 4 ch s thp phn th:1. Ta cn s dng b kim tra so snh bao nhiu bit? Ti sao? B kim tra so snh 3 bit B kim tra so snh 4 bit B kim tra so snh 5 bit2. Ta cn s dng bao nhiu b kim tra so snh? Ti sao? 3 b kim tra so snh 4 b kim tra so snh 5 b kim tra so snh B kim tra so snh 4 bit 4 b kim tra so snh

F1x: tn hiu so snh bng ca b kin tra so snhF11, F12, F13, F14: Tn hiu ra ti cc b so snh 1, 2, 3, 4

Fm kho

Bi ton t raXc nh s bit Bin lun cc trng hpa ra hmn tpc trc bi B dn knh (chun b cho bui sau)

Bi 3. B dn knh (Multiplexer - MUX)1. Xy dng s khia) Khi nim

- Vi mch hot ng khi u vo cho php E mc

tch cc. Gi s mc tch cc l mc cao th E=1

mch hot ng, E=0 mch khng hot ng.

Khi MUX lm vic: Nu cc u vo la chn

Sn-1 S1 S0 biu din s thp phn i no th

knh Xi s c ara u ra Y.c) Nguyn tc hot ng

S tng qutMUX...XoX1X2SoS1YEX: Cc tn hiu voE: Tn hiu chnS: Cc tn hiu vo la chn

Y: u ra

Bng chn l tng qut

VD: Thit k mch MUX c 2 u vo d liu Io,I1 mt u vo chn S, u ra Z.Ta c bng chn l nh sau

S mch MUX 2 u vo

VD: Thit k mch MUX c 4 u vo d liu X0, X1, X2 , X3 hai u vo chn S0, S1, u ra Y.

XOX1X2X3YS1S0E00* CHO E=1+ S1=0,S2=0+ S1=0,S2=11+ S1=1,S2=001+ S1=1,S2=111b. Nguyn tc hot ng

c) Bng chn lQuy c chn c tng u vo ring bit dng 2 ng a ch iu khin chn l S1, S2Gi s b dn knh c lm vic khi E=1 v b kha khi E=0

d) Phng trnh logic

e) S mch logic

3. ng dng ca MUX* nh tuyn lung d liu

* S dng chuyn i tn hiu song song thnh ni tip* To cc hm logicGi s chng ta thit lp cc gi tr c nh cho cc u vo d liu th chng ta c th thit lp c cc hm u rang dng ca MUX

DEMUXMUXng truyn dnCc Knh raCc Knh voTng kt

Kim tra nhn thcCu hi kim tra Cu 1: B dn knh 8 u vo d liu cn bao nhiu u vo la chn A.2 B.3 C.4 D.8 Cu2: Thit k b dn knh 8 u vo d liu cn bao nhiu cng NOT v ANDA.2-4 B.3-4 C.2-8 D.3-8

p n Cu 1: B dn knh 8 u vo d liu cn bao nhiu u vo la chnB.3 Cu2: Thit k b dn knh 8 u vo d liu cn bao nhiu cng NOT v AND D.3-8Kim tra nhn thc

Bi tp v nhHy thit k b dn knh 8 u vo d liu?

IC 74HC151 l IC MUX 8 u vo d liu

Bng chn l IC 74HC151

ng dng ca cc MUXnh tuyn lung d liuMUXCc knh voCc knh raCc tn hiu chn knh

To cc hm logicGi s chng ta thit lp cc gi tr c nh cho cc u vo d liu th chng ta c th thit lp c cc hm u ra

Th I1=1; I2 = 1; I7 = 1; I0 = I3 = I4 = I5 = I6 = 0

S thc hin hm Z mong mun da vo IC 74HC151

2.3.5. B phn knh ( DE MUX)

L mch logic c 1 u vo v c u ra, ti 1

thi im u vo d liu c a ra mt u ra

no ph thuc vo n tn hiu u vo chn

S tng qutDEMUX...XSoS1Y0EX: u voE: Tn hiu chnS: Cc tn hiu vo la chnYi: u raY1

Bng chn l tng qut

Vi mch s c hot ng khi tn hiu chn mch E mc tch ccKhi DEMUX lm vic, cc u vo nh phn Sn-1S1S0 biu din mt s thp phn i no , lc u vo X ca DEMUX s c a ra ti u ra d liu duy nht YiBiu thc logic ca cc u ra Yi, theo u vo d liu X v cc u vo chn Si nh sau

VD: Thit k mch DEMUX c 2 u ra d liu Y0, Y1 v mt u vo d liu X, mt u chn S* Ta c bng Chn l ca mch cn thit k nh sau:* Biu thc u ra Y0, Y1 nh sau

* S mch DEMUX 2 u ra nh sau

ng dng b phn knhDng phn knh tn hiu ng hTrong trng hp ny cc mch phn knh c s dng a mt tn hiu ng h ln lt ra cc knh u ra ng b thng tin cc knhDng phn knh trong h thng truyn d liu ng bTrong H/thng truyn tin ng b thng pha pht cc knh tn hiu c dn li sau c pht ra knh truyn dn, nhim v pha thu l phi tch cc knh thng tin ring l khi nhn c 1 lung d liu duy nht t knh truyn dn. lm c vic ny pha thu phi c b phn knh

DEMUXMUXng truyn dnCc Knh raCc Knh vo

2.3.7.Cc mch chuyn m2.3.7.1 Cc b m haB chuyn i m l mch logic nhn mt loi m u vo, bin i a ra loi m theo yu cu u ra. Cc b chuyn i m li c th c chia lm 2 loi l b m ha v b gii mM ha l qu trnh dng vn hay k hiu biu th 1 i tng.Nu dng m nh phn biu th i tng Qu trnh m ha nh phn. 1 k t nh phn c 2 gtr 0&1, tng ng vi vic biu din 2 tn hiu. Hay nu dng n k t nh phn(n s nguyn)s biu din c tn hiu khc nhau

S tng qut b m ha

VD: B m ha nh phnGi s: N=8= 8 u vo ca b m ha xut hin k t Y0, Y1, Y2, Y3, Y4, Y5, Y6, Y7 cn 3 bt biu din 8 tn hiu vo ny u ra tng ng A, B, C.Vy b m ha thc hin qu trnh bin i 1 trong 8 ng thnh 3 ng.

Bng chn l

Cu trc b m ha

VD: B m ha Thp phn nh phn

L mch in chuyn 1 k t thp phn tiu vo 0, 1, 29 thnh 1 t m biu din s nh phn tng ng trong h nh phn ti u ra.S dng 4 bt nh phn A, B, C, D biu din cc k t 0 9

Bng chn l

chng minh:Nu chn cc phng php m ha khc nhau Cu trc ca s logic s rt khc nhauTrong tt c Cc phng php m ha c th c 1 phng php cho cu trc n gin nht Phng php m ha ti u- Vic chn ra phng php ti u l phc tp

2.3.7.2 Cc b gii mMch ny c nhim v bin i m nhn c thnh m ban u.B gii m(CODER)S tng qut b m ha

1. Thit k b gii m BCD sang thp phna) S khi

CDAB0000111100001111010CDAB10001111000011110CDAB20001111000011110CDAB3000111100001111000000000101000000000000000001000000000xxxxxxxxxxxxxxxxxx

CDAB40001111000011110CDAB50001111000011110CDAB60001111000011110CDAB700011110000111100100000000xxxxxx0100000000xxxxxx

CDAB80001111000011110CDAB90001111000011110

2. Thit k b gii m BCD sang 7 thanha) S khi

b) Bng gi tr

CDABa0001111000011110CDABc0001111000011110CDABb0001111000011110CDABd0001111000011110

CDABe000111100001111010CDABg0001111000011110CDABf000111100001111001100010010011101111001011111xxxxxxxxxxxxxxxxxx

Cc em t v s logic

2.3.6. Mch to v kim tra chn lTrong cc h thng thng tin, pha thu i khi nhn c cc thng tin b sai (so vi pha pht) do nhiu nguyn nhn. Do vy h thng phi c h thng pht hin v sa li. Mt trong c ch n gin nht l phng php s dng cc m chn l (m PARITY).Phng php: pha pht mi khi pht i 1 nhm bt no s pht km theo 1 bt PARITY (bt chn l)c th l bt 0 hoc 1 m bo tnh chn hay l ca s bt 1 trong nhm (tnh c bt PARITY)S bt 1 trong nhm l chn hay l l do quy c to m PARITY chn hay l

VD: pha pht s pht km bt PARITY theo cc nhm 7 bt (gi s s dng m PARITY chn)Mun pht 7 bt 1111100 th bt PARITY km theo l phi l 1 v nhm 8 bt pht i l 11111100Nh vy s bt 1 l 6 (tha mn PARITY) chn Hay mun pht 7 bt 1110001 th bt PARITY km theo l bt 0 v nhm 8 bt pht i l 01110001 Tng t trong trng hp m PARITY l th bt PARITY c km theo nhm bt cn m ha l 0, 1 m bo s bt 1 l l

VD: Khi pht nhm bt 01110001 th s bt 1 l 4 (chn) nhng khi mt bt no b bin i VD bt cui cng th nhm bt ny tr thnh 01110000 S bt 1 l 3 (l) v vi bt bt k no cng vy s lm thay i tnh chn l ca m PARITY. Nh vy nu pha thu kim tra tnh chn l ca m nhn c, nu pht hin tnh chn l b sai th h thng hon ton pht hin c ra li.Ch : Pha pht v thu phi ng nht cng mt loi m PARITYPhng php ny ch pht hin ra li n, cn cc li kp th h thng khng nhn ra li.

Chng 3. Mch Flip - Flop3.1. Phn t nh c bn a) Khi nim Mch t hp l mch tn hiu ra ch ph thuc vo tn hiu vo. Cc phn t c bn xy dng nn h t hp l cc mch AND, OR, NAND, NOR. Mch ny khng c nh. Thc t t ra nhiu nhim v m ch cc mch t hp th khng thc hin c.

Mch dy(mch tun t) l mch tn hiu ra ph thuc khng nhng vo tn hiu vo m cn ph thuc vo trng thi trong ca mch, ngha l mch c lu tr, nh cc trng thi. xy dng mchdy, ngoi cc mch t hp c bn nh AND, OR, NOTcn phi c cc phn t nh. Cc mch ny c gi l Flip Flop (FF) Chng l cc phn t nh n bt v c kh nng nh c 1 ch s nh phn

b) nh nghaFF l phn t c kh nng lu tr (nh) 1 trong 2 trng thi 0 hoc 1

c) Phn loiPhn loi theo tn hiu iu khin- Phn loi theo chc nng* Khng c tn hiu iu khin (Khng ng b)* C tn hiu iu khin (ng b)

Cc k hiu v tnh tch ccTch cc l mc thp LTch cc l mc cao HTch cc sn dng ca xung nhpTch cc sn m ca xung nhp

3.2. Cc loi FF3.2.1. SR-FFSR-FF l loi FF n gin nht, ch c 2 u vo iu khin trc tip R (RESET) v S (SET). Mch khng c u vo iu khin ng b v xung nhpa. Khi nim v mch cht

* Mch cht SR dng NANDCc u vo S, R12S mch c ch to t cc cng NAND

* Nguyn l hot ngNAND1Q=0NAND2R=1Q=0Q=1y l trng thi n nh ca mch, m khng c s thay i cc u vo

Cc em t xt s n nh ca mch tng t nh TH1

Kt Lun: Khi S=1; R=1 ta lun c trng thi u ra mch khng i so vi u vo.(trng thi n nh)

Chng ta xt s chuyn trng thi ca mch trn khi c s thay i u vo.

Gi nguyn u vo R=1Chuyn mc logic ca u vo S t 1 v 0 (S=0)Q=1NAND2: R=1 Q=1

Gi nguyn u vo R=1Chuyn mc logic ca u vo S t 1 v 0 (S=0)Q=1NAND2 : R=1 Q=1

Gi nguyn u vo S=1Chuyn mc logic ca u vo R t 0 v 1 (R=1)Q=0NAND2: R=1 Q= 0

Gi nguyn u vo S=1Chuyn mc logic ca u vo R t 1 v 0 (R=0)NAND2: R=0 Q=0 NAND1: S=1 Q= 1

y l trng thi khng s dng (cm) v lm cho c Q v chuyn ln mc logic1y l trng thi khng mong mun v Q v theo thit k phi c gi tr o nhau u vo R=0, S=0 l t hp cm i vi cc mch SR

Bng chn l hot ng mch SR cu trc t NAND.Nhn xt: Khi S=0 th Q=1Khi R=0 th Q=0Nn thng gi S: u vo thit lpR: u vo xa

K hiu mch SR cu trc t cc NAND

* Mch cht SR dng NORCc u vo S, R

Cc em t phn tch nguyn l hot ngCa mch tng t nh mch trn

Bng chn l hot ng mch SR cu trc t NOR.- (Cm)

K hiu mch SR cu trc t cc NOR

b) Cc FF ng bPhn trn chng ta xt mch SR nhng vi phng din mch cht. V trng thi mc nh cc u vo R=S khng i lc u ra lun c gi(cht) ti mt gi tr khng i. u ra s thay i khi c u vo no b bin i. y chnh l vn ny sinh trong 1 h thng s s dng nhiu phn t nh, bi chng ta kh qun l thi im chuyn trng thi ca tt c chng. Gii quyt vn ny ta phi i n thit k cc FF ng b.

Cc FF phi thit k sao cho phi c thm 1 tn hiu ng h u vo. Tn hiu ny thng c dng xung vung, khi s chuyn trng thi ca FF ch c th c thc hin ti thi im xut hin ca sn trc hay sn sau ca xung ng h.

K hiu FF ng b s dng cc xung ng h hay ng b kch bng sn trc(sn dng) hay sn sau(sn m)

(Kch bng sn dng)u vo iu khin(Kch bng sn m)u vo iu khin

1. Cc SR - FF ng ba. K hiub. Bng chn lq: Trng thi ca u ra Q trc khi c xung kch thch

2. JK - FFa. K hiub. Bng chn l

Lc miu t s thay i trng thi u ra Q ca JKFF khi c cc u vo J,K v tn hiu ng h tng ngQ

3. D - FFa. K hiub. Bng chn l(DFF lt bng sn dng)

Lc miu t s thay i trng thi u ra Q ca DFF khi c cc u vo D v tn hiu ng h tng ngQ

4. T - FFa. K hiuQQCLKTb. Bng chn l(DFF lt bng sn dng)

Lc miu t s thay i trng thi u ra Q ca TFF khi c cc u vo T v tn hiu ng h tng ngTCLKQ

3.3. S chuyn i gia cc FFt vn : phn trn chng ta xt bng chn l ca cc FF. Qua ta hon ton c th xc nh u ra ca cc FF khi bit cc u vo v trng thi trc ca chng.Khi thit k mch logic c nh th ta phi lm ngc li, Ngha l bit trng thi trc ca FF phi xc nh cc u vo nh th no u ra l trng thi mong mun.

I. Cc bng ng dnga) DFFNhn xt: u vo kch D lun c gi tr ging gi tr Q. (ni cch khc mun Q c gi tr bao nhiu th u vo D phi c thit lp trc gi tr by nhiu)q: Trng thi trc ca mch Q:Trng thi mi cn chuyn nD: u vo

Cc bng ng dng (tip)b) TFFNhn xt: Mun trng thi tip theo ca mch o trng thi trc th phi c u vo kch T=1, cn ngc li mun gi nguyn trng thi FF th u vo kch T=0q: Trng thi trc ca mch Q:Trng thi mi cn chuyn nT: u vo

Cc bng ng dng (tip)c) JK-FFq: Trng thi trc ca mch Q:Trng thi mi cn chuyn nJ, K: u voX : L cc c gi tr ty (0 hoc 1)

Cc bng ng dng (tip)d) SR-FFq: Trng thi trc ca mch Q:Trng thi mi cn chuyn nS, R: u voX: L cc c gi tr ty (0 hoc 1)

II. S chuyn i gia cc FF1. Khi nimNguyn tc chung xy dng 1 mch logic c chc nng ging mt FF X no , da vo mt FF X c. Th hin lc sauMch logicCn thit ku vo XMch logic ca FF Xu ra

Nhn xt:Vic chuyn i cc FF thc cht l phi xy dng c mch logic cn thit. Mch ny bao gm cc u vo l cc u vo FF cn thit k. Cc u ra chnh l u ra ca FF c. xy dng phn mch logic ta da trn nguyn tc ng nht bng kch ca FF X v FF X.Ngha l: Khi cc u vo ca FF-X cn thit k lm thay i trng thi u ra ca n, th ng thi cc u vo ca trng thi ny cng to ra cc tn hiu vo tng ng cho FF X, sao cho X cng chuyn trng thi ging ht nh X. V ta thy u ra ca X& X l nh nhau

Mch logicCn thit k2. S chuyn i gia SR-FF sang JK-FFJKRSMch logic ca JK-FF

a) Bng kchq: Trng thi trc Q: Trng thi cn chuyn n

b) Bng chn l ca SSqJK00011110010011x00xBng Karnaugh S

qJ KS00 x00 x11x 101x 0x

c) Bng chn l ca RRqJK0001111001xx000110Bng Karnaugh R

qJ KR00 xx0 x0 1x 111x 00

Mch chuyn i

Mch logicCn thit k3. S chuyn i gia SR-FF sang D-FFDRSMch logic ca SK-FF

b) Bng chn l ca SSqD0101010xBng Karnaugh S

qDS00001 10011x

c) Bng chn l ca RRqD0101x010Bng Karnaugh R

qDR00x010 101110

Mch chuyn i

Mch logicCn thit k4. S chuyn i gia SR-FF sang T-FFMch logic ca T-FF

b) Bng chn l ca SSqT010101x0Bng Karnaugh S

qTS00001 11010x

c) Bng chn l ca RRqD0101x001Bng Karnaugh RRqD0101

qTR00x010 111100

TMch chuyn i

5. S chuyn i gia JK-FF sang D-FFDMch logic ca D-FF

b) Bng chn l ca JJqD010101xxBng Karnaugh S

qDJ00001 10x11x

c) Bng chn l ca KKqD0101xx10Bng Karnaugh K

qDK00x011 10x110

Mch chuyn i

6. S chuyn i gia JK-FF sang T-FFMch logic ca T-FF

b) Bng chn l ca JJqT010101xxBng Karnaugh S

qTJ00001 11x10x

c) Bng chn l ca KKqT0101xx10Bng Karnaugh K

qTK00x01x 111100

Mch chuyn i

3.4. Cc ng dng ca FF3.4.1. B mKhi nimB m l 1 mch dy tun hon c 1 u vo m v 1 u ra, mch c s trng thi trong bng chnh h s m (k hiu K). Di tc dng ca tn hiu vo m mch s chuyn t trng thi ny n trng thi khc theo 1 th t nht nh.C sau K tn hiu vo m, mch li tr v trng thi xut pht ban u3.4.1. B m

S b m B m l mch logic c nh, v trng thi m ca cc b m ng nhin phi ph thuc vo trng thi trc .- B m gi l c c s k khi m n chuyn qua k trng thi khc nhau, sau quay tr li trng thi ban u.

VD: B m s thp phn c s 10. u tin c gi tr m l 0, c sau khi c 1 xung kch n tng ln 1, 2, 3 n gi tr 9. Nu chng ta tip tc kch xung b m li quay v gi tr 0 v lc xut hin 1 xung u ra y. C b m tng, gim, Thm ch c th m theo 1 bng m cho trc.- Trong qu trnh m chng ta mun hin th gi tr hin hnh m b m ang m c. lm c iu ny, trng thi hin hnh ca b m (thng l u ra ca cc FF) c qua mch gii m v hin th bng n LED

B m c s k

B gii m

Hin th

u vo mu ra m

II. Phn loi b mB mP/loi theo cch lm vicP/loi theo KP/loi theo hng mP/loi theo kh nng lp trnhng bKo ng bm thunm nghchC th LTko th LT

1. Phn loi theo s ng bCc b m ng b: Trong cc b m loi ny cc FF c kch bi cng 1 xung ng hb. Cc b m khng ng b: Trong cc b m loi ny cc FF c th khng c kch cng bi 1 xung ng h, nn s chuyn trng thi ca cc FF trong cc b m loi ny c th khng ng thi.

2. Phn loi theo s hng mCc b m tng (m thun): L b m m mi khi c tn hiu vo m X th trng thi trong ca b m tng ln 1Vd: B m ang trng thi S5: 110 khi c tn hiu vo m X n s chuyn sang trng thi S6: 110b. Cc b m gim (m li): L b m m mi khi c tn hiu vo m X th trng thi trong ca b m gim i 1.Vd: Ban u b m ang trng thi S9: 1001 khi c tn hiu X n s chuyn sang trng thi S8: 1000

c. B m thun nghch: B m ny va c kh nng m thun, va c kh nng m nghch. Trong trng hp ny phi a thm tn hiu iu khin m thun hay m nghchCTN: Tn hiu iu khin vic m thun hay nghch

3. Phn loi theo c s mCc b m c s y l cc b m m s trng thi chuyn qua ca n l ly tha ca 2. D thit k.b. Cc b m c s khc y l cc b m m s trng thi chuyn qua ca n khng l ly tha ca 2. Kh thit k hn.

Ni dungGi thit: Cho s nguyn l h logic c nhb. Kt lun: Tm hiu s nguyn l ca h logic c nh Bng Trng thi hnh trng thi- V th thi gian

2. Cc bc thc hinNhn xtb. Vit phng trnh hm kchc. Lp bng kchd. Vit biu thc hm rae. Lp bng trng thi

Bi tp minh ho

Bi tp 1: Phn tch b m tin Khng ng b (KB) 4 bt (c s 16)

Li gii:Nhn xt: - y l h logic c nh - Khng ng b - Cc FF lt bng sn m ca xung nhp. Nn s chuyn trng thi ca cc FF ch c thc hin khi c s kch thch ca xung nhp t 1 0 Bng chn l

11010000000010011000010101011011100001100100110100111011011111114678910111213141501235A, B, C, D l cc u ra tng ng ca cc FF

Bi tp 2: Phn tch b m 8, tin, KB sau

Li gii:Nhn xt: - y l h logic c nh - Khng ng b - Cc FF lt bng sn m ca xung nhp. Nn s chuyn trng thi ca cc FF ch c thc hin khi c s kch thch ca xung nhp t 1 0 Bng chn l

Bi tp 3: Phn tch b m 8, li, KB sau

Li gii:Nhn xt: - y l h logic c nh - Khng ng b - Cc FF lt bng sn m ca xung nhp. V ta ni vo T cho nn khi Q chuyn trng thI t 1 v 0 th s o li t 0 v 1 Bng chn l

Mch m KB c nhc im l tch lu s ch tr ca cc FF

Bi tp 2: Phn tch h logic sau m xung, h m 8, thun, B11

Li gii:Nhn xt: - y l h logic c nh - ng b (chu tc ng 1 xung H)b. Vit phng trnh hm kchJ1 = 1J2 = q1J3 = q1.q2K1 = 1K2 = q1K3 = q1.q2c. Lp bng kch

J1 = 1K1 = 1 J2 = q1J3 = q1.q2K2 = q1K3 = q1.q2

d. Vit biu thc hm raKhng c e. Lp bng trng thi* Bng Trng thi l thuyt * Bng Trng thi thc t

q1 q2 q3Q1 Q2 Q30 0 01 0 0 1 0 00 1 00 1 01 1 01 1 00 0 10 0 11 0 11 0 10 1 10 1 11 1 11 1 10 0 0

f. hnh trng thi01234567Xung CXung CXung CXung CXung CXung CXung CXung C

Bi tp 3: Phn tch h logic sau H m 10, thun, BJ1Q1K1J2K2Q2J3K3Q3J4K4Q411

Li gii:Nhn xt: - y l h logic c nh - ng b (chu tc ng 1 xung H)b. Vit phng trnh hm kchJ1 = 1J2 = q1.q4J3 = q1.q2 J4 = q1.q2.q3K1 = 1K2 = q1K3 = q1.q2 K3 = q1c. Lp bng kch

J1 = 1K1 = 1J2 = q1.q4K2 = q1 J3 = q1.q2 K3 = q1.q2 J4 = q1.q2.q3K3 = q1

d. Bng trng thi l thuyt

d. Bng trng thi thc t

f. hnh trng thi01236789Xung CXung CXung CXung CXung CXung CXung CXung C45Xung CXung C

Bi 1: Phn tch h m xung, m 8, thun, KBDCBQ1Q2Q3Xg111111

Bi 2: Phn tch h m xung, m 10, thun, KBJ1Q1K1J2K2Q2J3K3Q3J4K4Q4111111

Bi 3: Phn tch h m xung, m 5, thun, KBC s dng u vo xaT1CLRT2CLRT3CLRQ1Q2Q3Xg111111

Ni dungGi thit: Cho h logic c nh 1 trong nhng dng sau- Bng Trng thi- hnh trng thi- M t thng qua cc mnh b. Kt lun: V s ca h logic

2. Cc bc thc hina ra bng trng thi m hab. a ra bng kchc. Ti thiu hm kch v hm rad. V s

Bi tp minh ho

Bi tp 1: Thit k b m 8, thun, BLi gii:

a.Bng trng thi Tng qutb.Bng m ha trng thic.Bng trng thi m ho

sq1q2q300001001201030114100510161107111

q1q2q3Q1Q2Q3000001001010010011011100100101101110110111111000

d.Bng kch (chn phn t JK-FF)

e.Ti thiu hm kchq1q2q3J1100100xxx01111xq1q2q3K11001001xx0111xx

q1q2q3J2100100x1x01111xxq1q2q3K21001001x10111x0xe.Ti thiu hm kch

q1q2q3J3100100xx10111xx111q1q2q3K310010011x011111xxxe.Ti thiu hm kch

f. V s J1J2Q1Q2Q3Xg CK1K2J3K3

Bi tp 2: Thit k b ghi dch 3 btLi gii:xXung CNguyn l: q3 Ra ngoiq2 a sang nhp vo q3q1 a sang nhp vo q2x a sang nhp vo q1q1q2q3xq1q2

a. Bng trng thi m ha

b. Bng kch (chn phn t TFF)

c. Ti thiu hm3 hm kch T1, T2, T3- 3 bin q1, q2, q3 v 1 bin vo x 4 bin Bng Karnaugh c 16

q3xq1q2T100011110000111101111q3xq1q2T20001111000011110q3xq1q2T3000111100001111011111111111111111111

Kt qu:

T1T2T3Q1Q2Q3XgXx

Bi tp 3: Thit k h m 10, thun, BLi gii:

a.Bng trng thi Tng qutb.Bng m ha trng thic.Bng trng thi m ha

sq4q3q2q100000100012001030011401005010160110701118100091001

q4q3q2q1Q4Q3Q2Q100000001000100100010001100110100010001010101011001100111011110001000100110010000

d. Lp bng kch (chn phn t JKFF)

Kt qu:

Cc em t v s

Bi 1: Thit k h logic iu khin bng n qung co gm 3 ch S E LTrnh t sng tt cc ch:s0: t0 t1: - - - (cc ch tt)s1: t1 t2: Ss2: t2 t3: S Es3: t3 t4: S E Ls4: t4 t5: - - -s5: t5 t6: S E Ls6: t6 t7: S E L t7 t8: - - - (v trng thi ban u)

Bi 2: Thit k h logic iu khin bng n giao thng 1 ng t. Bng n gm 3 n Xanh, , Vng. Trnh t tt sng cc n : X V - - X

3.4. Cc ng dng ca FF3.4.2. B ghi dchI. Khi nimThanh ghi dch c ng dng trong lu tr d liu. Tuy nhin thanh ghi dch do kh nng lu tr c hn nn ch s dng nh b nh tm thi. (Lu Kq php tnh)II.Thanh ghi dch chuynThanh ghi c xy dng trn c s cc DFF (hoc cc FF khc thc hin chc nng ca DFF) v trong mi DFF s lu tr 1 bt d liu

to TG nhiu bt, ngi ta ghp nhiu DFF li vi nhau theo quy lut sau:Ng ra ca DFF ng trc c ni vi ng vo ca DATA ca DFF ng sau (Di+1 = Qi) TG c kh nng dch phiHoc ng ra ca DFF ng sau c ni vi ng vo DATA ca DFF ng trc (Di = Qi+1) TG c kh nng dch tri

III.Phn loiPhn loi theo s bt d liu lu tr: 4 bit, 5bit, 8 bit16 bit, 32 biti vi TG ln hn 8 bt Khng dng h TTL m dng h CMOS* Phn loi theo hng dch chuyn d liu trong TGTG dch triTG dch phiTG va dch tri va dch phi

* Phn loi theo ng vo d liuNg vo d liu ni tipNg vo d liu song song: Song song KB, song song B* Phn loi theo ng raNg ra ni tipNg ra song song- Ng ra va ni tip va song song

IV .Nhp d liu vo FFNhp d liu vo FF bng chn Preset (Pr) Khi Load = 0Cng NAND 3 & 2 b kha Ng vo Pr & Clr = 1FF t do D liu A ko nhp c vo FF

PrClrLoadA231 Khi Load = 1Cng NAND 3 & 2 m Ng vo Pr = A; Clr = ANu A = 0 Pr =1; Clr = 0 Q = A = 0Nu A = 1 Pr =o; Clr = 1 Q = A = 1Vy Q = A D liu A c nhp vo FFTuy nhin vi cch ny phi dng nhiu cng logic ko kinh t v phi dng chn Clr xa nn phi thit k ng b

Khc phc nhc im ny ta dng mch sau, chn Clr trng tng ng mc 1Khi Load = 0Cng NAND khaPr = Clr = 1 FF t do d liu khng c nhpKhi Load = 1 Cng NAND mPr = ANu A = 0 Pr = 1, Clr = 1 Q = A = 0Nu A = 1 Pr = 0, Clr = 0 Q = A = 1Vy Q = A D liu A c nhp vo FF

VD: 1 thanh ghi 4 bt c kh nng di phiDSR: (Data Shift Right): Ng vo Data ni tip(ng vo dch phi)Q1, Q2, Q3, Q4: Cc ng ra song song

Chng 4:Mch dy4.1. Cc khi nim c bn v mch dy4.1.1. M hnh TQ

4.1.2. Phng php m t mch dya. M hnh ton hcDng 1 h phng trnh ton hc biu th mi quan h vo ra ca h tun t. Gi:V : L tp tn hiu voR : tp tn hiu raX : tp hm kchS : tp trng thi trongC th m t bi phng trnh sau:

A1. Otomat (H pt) MEALYR = f1 (V, S)(1)

S = f2 (X, S)(2)

X = f3 (V, S)(3)Gi l hm ra: Mi quan h gia u ra vi tc ng vo v bin trng thi

(2) Gi l hm chuyn i trng thi ca cc phn t nh

(3) Gi l hm kch thch cho cc phn t nh

a2. Otomat (H pt) MOORER = f1 (S)(1)S = f2 (X, S)(2)X = f3 (V, S)(3)Hm ra ch ph thuc vo bin trng thi khng ph thuc vo bin vo

a3. Mch dy ng bL mch dy c iu khin theo tn hiu B bn ngoi (xung nhp Ck). S chuyn trng thi ca mch t trng thi Si Sj ch xy ra khi c xung nhp tc ng. Khong thi gian gia 2 xung nhp phi ln h lun lun trng thi n nh trc khi chu tc ng bn ngoi.a4. Mch dy khng ng bL mch dy m s hot ng ca mch khng c tn hiu B .S thay i ca tn hiu l ngu nhin. i vi h ny qu trnh chuyn t trng thi vng Si Sj c th lt qua 1 s cc trng thi khng vng trung gian

3. Cc phng php m t mch dy1. Bnga. Bng chuyn i trng thiHng : Ghi trng thi trongCt : Ghi tn hiu voGiao im ca hng v ct: l ghi trng thi trong tip theo m mch s chuyn n ng vi tnhiu vo v trng thi hin ti nh ghi trn ct v hng tng ng.

VSTrng thi trongTn hiu voTrng thi s chuyn bin n SBng chuyn i trng thiS

b. Bng tn hiu raHng : Ghi trng thi trongCt : Ghi tn hiu voGiao im ca hng v ct: Tn hiu ra tng ng

VSTrng thi trongTn hiu voTn hiu raBng tn hiu raR

c. Bng chuyn i trng thi / RaC th gp 2 bng chuyn i trng thi v bng tn hiu ra thnh 1 bng chung, gi l bng chuyn i trng thi / ra ca bng ta ghi trng thi m mch s chuyn bin n v tn hiu ra S / R tng ng vi trng thi hin ti v tn hiu vo.

VSTrng thi trongTn hiu voTrng thi s chuyn bin ti v tn hiu raBng chuyn i trng thi / RaS/ R

V d: Mt mch dy cTp tn hiu vo V = {V1, V2, V3 }Tp tn hiu trong S ={ S1, S2, S3, S4, S5 }Tp tn hiu ra R = { 0, 1 }Chc nng ca mch c xc nh bng bng chuyn i trng thi v bng ra

VSBng chuyn i trng thiSBng tn hiu raVSRVSS/RBng chuyn i trng thi v tn hiu ra

V1V2V3S1S2S3S4S5111100010001100

V1V2V3S1S2S3S4S5S2/1S4/1S1/1S3/1S5/0S4/0S2/0S1/1S4/0S3/0S3/0S4/1S1/1S2/0S4/0

2. hnh trng thiM hnh MealyTp cc trng thi trong tp cc nh MTp cc tn hiu Vo / Ra Tp cc cung KCc cung c hng I t trng thi Si Sj, ghi Vo / Ra tng ng.

Bng chuyn i trng thi v tn hiu raS1S2S3S4S5V1/1V1/1V3/1V3/0V2/1V3/1V1/0V2/0V3/0V1/1V2/1V3/0V3/0V2/0VSS/R(V1+V2+V3)/1(V1+V3)/1

V1V2V3S1S2S3S4S5S2/1S4/1S1/1S3/1S5/0S4/0S2/0S1/1S4/0S3/0S3/0S4/1S1/1S2/0S4/0

Chuyn t m hnh Mealy sang MooreBc 1: ng vi mi cp (S,/R) ca Mealy ta qui nh 1 trng thi tng ng q ca MooreBc 2: Thnh lp bng chuyn i trang thi cho Moore. Ghi mi trng thi q (ng vi cp S/R ca Mealy 1 tn hiu ra tng ng R)

V d: Mch dy c m t bi m hnh Mealy c hnh trng thi v bng chuyn i trng thi, tn hiu ra nh trn.S0S1S21/00/00/01/00/11/0VSS/R

01S0S1S2S1/0S1/0S1/1S0/0S2/0S0/1

Cc bc chuyn iBc 1: T bng chuyn i trng thi /Tn hiu ra gn cc trng thi trong Qj ca m hnh Moore nh sau:S1/0 : Q0 S2/0: Q2S0/0 : Q1S1/1: Q3Bc 2: Thnh lp bng chuyn i trng thi cho MooreVSS/R

01S0S1S2S1/0S1/0S1/1S0/0S2/0S0/1

Bng chuyn i trng thi

MealyMooreTn hiu raTrng thi hin tiTrng thi s chuyn ti V = 0 V = 1S1/0S0/0S2/0S1/0Q0Q1 Q2Q3Q0Q0Q3Q0Q2Q1Q1Q20001

hnh trng thi (Moore)Q0/0V=0Q1/0Q2/0Q3/11011001

Nhn xt: M hnh Mealy v m hnh Moore l 2 m hnh dng biu din h dy, chng tng ng vi nhau M hnh Mealy: C hm ra ph thuc c tn hiu vo v trng thi trong ca mchM hnh Moore: C hm ra ch ph thuc vo trng thi trong ca mch m thi

4.2. Cc bc thit k mch dyBi ton (cha hnh thc ho)Hnh thc hoOtomat nh phnH hm ca mchS

Bc 1: Bi ton cha hnh thc hoNhim v thit k c m t bng ngn ng hoc lu thut ton. Ni chung cha c hnh thc hoBc 2: Hnh thc hoPhin dch cc d liu thnh 1 hnh thc m t hot ng ca mch bng cch hnh thc ho d kin ban u dng bng trng thi hay hnh trng thiRt gn cc trng thi ca mchBc ny tin hnh trn Otomat cha phi l nh phn c V: tp vo; R: tp ra; S: tp trng thi

Bc 3: Otomat nh phnM ho tn hiu vo, ra, trng thi trong c Otomat nh phnBc 4: H hm ca mchXc nh phng trnh logic ca mch v ti thiu ho cc phng trnh nyBc 5: Xy dng s mchT h phng trnh ca mch ta xy dng s mch thc hin

Thit k mch dy t hnh trng thiGi thit: Cho hnh trng thI ca mch c tp tn hiu vo V, tp tn hiu ra R, tp trng thi trong S (cha m ho nh phn)Kt lun: H phng trnh nh phn ca mch ( ti thiu ho) Sau v s mch

Cc bc thit kB1: M ho tn hiu vo V, tn hiu ra R, trng thI trong S chuyn Otomat ban u thnh Otomat nh phn c cc tp tn hiu vo X, tn hiu ra Y, trng thi trong QB2: Xc nh h pt tn hiu ra: Y= fi (X,Q). Pt ny c xc nh trn cung vi m hnh kiu Mealy v trn nh vi m hnh MooreTi thiu cc hm ny

B3: Xc nh hpt hm kch cho cc FF v ti thiu cc FF ny.Thut ton x pt u vo kch cho cc FF t hnh trng thii vi FF Qi bt k, s thay i trng thi Qi n cc Qi ch c th c 4 kh nng. Cc cung biu din s thay i trng thi t Qi Qi c k hiu nh sau00 l (0) loi 01 1 l (1) loi 10 1 l (2) loi 21 0 l (3) loi 3

Thut ton xc nh pt u vo kch cho FF loi DTrn c s: Di = QiDi = Qi = cc loi cung i ti nh c Qi = 1 = cc cung loi (1) + (2)Ti thiu ho hm Di va tm c rt ra c pt u vo kch cho FF Qi loi D

Thut ton xc nh pt u vo kch cho FF loi TTrn c s: Ti = Qi Qi = Qi Trong : Qi =1 khi Qi thay i t 0 =>1 hoc1 => 0 Tin hnh: - in s thay i gi tr Qi vo cc cung- Ti = Qi = cc cung c Qi thay i = cung loi (2) + (3)Ti thiu ha hm Ti va tm c, rt ra pt u vo kch cho T FF th i

Thut ton xc nh pt u vo kch cho FF loi JK

Xc nh:Ton = cc cung m Qi c bt (thay i t 0 1 cung loi (2)) = cung loi (2)Ton = Toff = cc cung m Qi tt (thay i t 10 cung loi (3)) = (3)Toff =

Thut ton xc nh pt u vo kch cho FF loi RS

Pt u vo S ca RS FF c xc nhS = Ton + [ Cc cung loi (1) ]Pt u vo R ca RS FF c xc nhR = Toff + [ Cc cung loi (0) ][ ] c ly gi tr khng xc nh, s li dng nhng gi tr ny ti thiu ha.

VD: Thit k mch m B K = 5 c hnh trng thi v ha trng thi cho hnh sau: dngD FF v cc mch NANDb. T FF v cc mch NANDc. JK FF v cc mch NANDd. RS FF v cc mch NAND00000Q1Q2Q30000010100111000001111001Q1Q2Q3Y01234xxxX: ng vi c trng thi ko s dng

Li gii- B m c K = 5 c 5 trng thi 0,1,2,3,4. Tn hiu ra ca b m s xut hin khi ang trng thi 4 lc n quay v trng thi ban u - Mch c 5 trng thi c m ha t nht bng 3 bin nh phn, tng ng vi 3 FF Q1, Q2, Q3 - Pt tn hiu ra YY = Ti thiu ha hm ny da vo bng trng thi Y = Q1.X

Xc nh pt u vo kch cho cc FF Q1, Q2, Q3 cc FF ny l:D-FFCc cung i ti nh c Qi =1+ Q1=1 Khi cung i ti nh (4), n xut pht t nh (3) i n nh (4)D1=Cc cung i ti nh (4) = (3)=+ Q2=1 Khi cung i ti nh (2) & (3)D2=Cc cung i ti nh (2) = (3) = (1)+ (2) =

+ Q3=1 Khi cung i ti nh (1) & (3)D3=Cc cung i ti nh (1) = (3) = (0)+ (2) =

0001111001Q1Q2Q3D11xx0001111001Q1Q2Q3D21x1xxx0001111001Q1Q2Q3D31x1xxD1 = Q2.Q3Cc em t v s

Xc nh pt u vo kch cho cc FF Q1, Q2, Q3 cc FF ny l:T-FFin s thay i gi tr Qi vo cc cungMch m t trng thi (0)=>(1) tc l 000 => 001Q3 thay i t 01 => Ghi Q3 ln cung (1)=>(2) tc l 001 => 010Q2 thay i t 0=>1. Q3 thay i t 1=>0(2)=>(3) tc l 010 => 011Q3 thay i t 0 =>1(3)=>(4) tc l 011 =>100Q1 thay i t 0=>1, Q2 thay i 1=>1, Q3 thay i t 1=>0(4)=>(0) tc l 100 =>000 Q1 thay i 1=>0

00000Q1Q2Q3000001010011100

0001111001Q2Q3Q1T11x1xx0001111001Q2Q3Q1T21x1xx0001111001Q2Q3Q1T31x1xx11T1 = Q1 + Q2Q3T2 = Q3Cc em t v s

Mch dy ng bI. Phn tch mch dy ng b1. Cc bc phn tch mch dy Ba. S mchT s cho trc cn xc nh chc nng tng phn t c bn ca s , mi lin h gia cc phn t .b. X cc u vo, u ra, s trng thi trong ca mchCoi nh mch l hp en cn phi x s u vo, u ra ca mch. c im ca cc u vo, u ra . x c s trng thi trong ca mch, chng ta cn phi x xem mch c xy dng t bao nhiu p/t nh, t x c s trng thi trong c th c ca mch

c. X pt hm ra, hm kch ca cc FFDa vo s cho trc, ch ra h pt hm ra, hm kch cc FFd. Lp bng trng thi, bng ra nh phnBng trng thi, bng ra nh phn l bng biu din quan h gia trng thi chuyn bin ti, tn hiu ra nh phn vi trng thi hin ti v tn hiu vo tng ngDa vo pt hm kch, hm ra x c trn v da vo pt c tnh ca FF x c trng thI chuyn bin ti v tn hiu ra tng ng vi tn hiu vo v trng thi hin ti ca mch

T bng trng thi v bng ra lp trn, xy dng c hnh trng thi v tn hiu ra ca mchd. Lp bng trng thi, bng ra nh phne. Chc nng ca mchSau khi lp c hnh trng thi, da vo hnh x c chc nng ca mch

Cc bc phn tch mch dy BS mchX cc u vo, ra. S trng thi trong ca mchX h pt hm ra, hm kch ca cc FF hnh trng thiChc nng ca mch

VD: phn tch mch dy B c s sauJAKAABBX1CkXClockZAB

Bc 1: S mch (Hnh trn)Mch c 2 u vo: - Tn hiu vo X- Xung tc ng CkC 1 u ra : ZC 2 p/t nh JKFF l A & BBc 2: X u ra, u vo, s trng thi trong ca mch- Mch c th c biu din bi mt hp en c 2 tn hiu vo v X & Ck c 1 u ra Z- C 2 FF S trng thI trong ca mch l AB = 00, 01, 11, 10

Bc 3: X pt hm ra & hm kch cho cc FF- Pt hm ra : Z = A.B.Ck Pt hm kch cho A FF; B- FFJA = BJB = AKA = 1KB = Bc 4: Bng trng thi, bng ra nh phn

Q: Trng thi hin tiq: Trng thi tip theo

JKQ00q01010111q

0001111001KQJq1100100Bng Karnaugh1Thay JA, KA, JB, KB bc 3 vo pt (*)(*)A, B: Trng thi hin tiA, b: trng thi tip theo

Da vo h pt chuyn i trng thi v pt hm raLp bng Chuyn i trng thi v bng Ra nh phn theo trng thi hin ti (A, B) v tn hiu vo X

Nhn xt: Z = 1 A & B ng thi =1 v c xung nhp tc ng. Cho nn khong thi gian Z = 1 bao gi cng bng khong thi gian tn ti xung nhp Ck

Bc 5: hnh trng thiA.B 00A.B 01A.B 11A.B 10S0S1S2S3S0 S1 & S2 S0 mc cho X = 0, X = 1 Ko ph thuc vo X, ch ph thuc vo xung kch

Cc bc thit k mch dy BBc 1: Xc nh bi tony l bc u tin v quan trng nht thc hin tt vic thit k. Trc ht phi xc nh yu cu t ra cho mch thc hin v phi xc nh mch c thit k t nhng yu t no?Bc 2: Xc nh tn hiu vo raCoi mch nh 1 hp enc cc u vo, ra. Cn xc nh nhng c im c bn ca cc u vo u ra

Bc 3: hnh trng thi, bng trng thi, Bng raVic xy dng hnh trng thi ko da trn quy tc no Da vo kinh nghimBc 4: Rt gn trng thiS trng thi trong ca mch gn nh t l vi s FF c dng trong mch. Vic ti thiu trng thI ch yu da vo khi nim trng thi tng ng. Cc trng thi tng ng c thay bng 1 trng thi chung, i din cho chng.

Bc 5: M ha trng thinu s lng trng trong l N, s lng bin nh phn cn dng l n, th phI tha mn k: Bc 6: Xc nh h pt ca mchC 2 cch x h pt:Lp bng Chuyn i trng thi v tn hiu ra nh phn t x c pt u vo kch cho cc FF v pt ca tn hiu ra, sau chng ta tin hnh ti thiu ha cc pt Da trc tip vo hnh trng thI, vit h pt Ton, Toff v pt ca tn hiu ra, sau ti thiu ha h pt

VD: Thit k mch dy B thc hin nhim v kim tra dy tn hiu vo dng nh phn c di bng 3 c a vo lin tip trn u vo X. Nu dy tn hiu vo c dng l 010 hoc 011, 110, 111, Z=1 Trong trng hp khc Z = 0Li gii:

Bc 1: Xc nh bi tonMch thit k c nhim v pht hin dy tn hiu vo. Nu dy tn hiu vo c dng l 010, 011, 110, 111 th Z = 1 bo hiu l mch nhn c mt trong dy tn hiu vo .

Bc 2: Xc nh tn hiu vo raMch thit k l mch B nn c u vo X v u vo xung nhp Ck, 1 tn hiu ra Z.

Mch logic

XCkZ

Bc 3: hnh trng thi, bng trng thi, Bng raS0S1S2S3S4 S5 S6 X111110011010

Bc 4: Rt gn trng thi

Gp cc trng thi tng ngSi tng ng Sj khi* Si, Sj l 2 trng thI ban u th vi mi dy tn hiu vo c th chng lun lun cho dy tn hiu ra ging nhau.* Nu c nhiu trng thi tng ng vi nhau tng i mt th chng tng ng vi nhau Ktra cc nhm c tng ng ko, ta Ktra:Nhm cc trng thi tng ng phi C nhng hng trong bng tn hiu ra ging nhau- Nhm cc trng thi tng ng phi C nhng hng trong bng trng thi cng 1 ct l tng ng (ng vi 1 t hp tn hiu vo)

Quy tc CadwellCc hng tng ng trong ma trn ra ging nhauTrong ma trn ra, cc hng tng ng phi tha mn 1 trong 3 iu sau:+ Cc hng trong ma trn trng thi ging nhau+ cc trng thi trong cng 1 ct nm trong nhm trng thi c xt+ Cc trng thi trong cng 1 ct l cc trng thi tng ng.

Bng trng thi sau khi gp S3 & S5, S4 & S601S0S1S2S35S46XS01S0S1S2S35S46XS

S2Z=0S1Z=0S46Z=0S35Z=0S46Z=0S35Z=0S0Z=1S0Z=1S0Z=1S0Z=1

01S0S1S2S35S46XS* Gp tip S1, S2* hnh trng thiA.B 00A.B 01A.B 10A.B 11S0S12S35S46

S2Z=0S1Z=0S46Z=0S35Z=0S46Z=0S35Z=0S0Z=1S0Z=1S0Z=1S0Z=1

Bc 5: M ha trng thi

Bc 6: Xc nh h pt ca mchBng trng thi

Bng Karnaugh0001111001ABXJA11xxx0x00001111001ABXKAxx111x1x0001111001ABXKB1011xxxx0001111001ABXJBxxxxx101

Bc 7: V s mchJAKAABBX1CkABZCK

Chng 5: Thit k mch s dng MSI, LSIM u M t cc bc thit k mch s dng SSI v MSI, LSISSI MSI, LSI Bi tonX p/trnhTi thiu haS Bi tonChn vi mch MSI, LSINi cc MSI, LSI cho trc dng trn SSI nu cnS

B chn d liua. S TQ b chn knhMUX...n u vo iu khinYETrong thc t thng gp cc loi MUX sau;MUX 2 1MUX 4 1MUX 8 1MUX 16 1

VD: MUX 4 1MUXfED0D1D2D3ABD0, D1, D2, D3: 4 u vo d liuA, B: 2 u vo iu khinf: u d liu raE: Tn hiu chn mch lm vicVD: Khi u vo AB c g/tr 00 u ra f s cho ra tn hiu c u vo d liu D0 (f=D0)a. S khi

b. S chn 1 trong 4 u voc. Xy dng MUX 4 u vo t mch t hpPt tn hiu ra

S mchEABfD0D1D2D3

Cc ng dng ca MUX* c lm b chn d liu hay chuyn mch in tD liu vo c chn a ti u ra ph thuc vo tn hiu iu khin hay tn hiu chnS cc ng d liu vo c th tng ln nh s dng MUX c nhiu u vo hoc bng cch mc t hp nhiu MUX c s u vo nh

* Bin i dng thng tin song song thnh ni tip u raBin i thng tin a n u vo dng song song 8 bit thnh dng ni tip u ra. Vic iu khin tn hiu ra c thc hin nh 1 b m nh phn 3 u rafD0D1D7AB.CmCk

* To hm logicTQ: mt b MUX 1 c th dng to hm logic bt k c n+1 bin vo, trong n bin s a vo n u iu khin cn 1 bin cng vi cc hng s 0,1 c a vo u vo cn li, ty thuc vo gi tr hm s.

MUX 41 c: 2 bin u voTa s a 2 bin u vo n cc u vo iu khin ca MUX 41. C 3 kh nng chn 2 bin ny trong 3 bin ca hm F. Cc kh nnga) 2 bin iu khin l A & Bb) 2 bin iu khin l B & Cc) 2 bin iu khin l A& C

Sau khi chn bin iu khin, cn xc nh cc g/tr a vo u vo MUX thc hin hm f ban u. Dng bng ccn c tin hnh nh sau: n bin /khin c phn thnh vng khc nhau trn bng Cacno, nh du l D0,D1..D2n-1. (Di ng vi gi tr thp phn ca n bin /khin l i ) in gi tr hm s cho trc vo bng Cacno Ti thiu hm cho trong tng vng Di, gi hm s ny l Di, Di chnh l gi tr u vo ti Di ca MUX. Nu Di l cc hm 1 bin hoc hng 0, 1 bi ton gii xong.- Trong TH ngc li tip tc dng MUX hoc cc cng logic thc hin

TH1: 2 bin iu khin l A & B- 2 bin /khin Bng Cacno c 4 ca DABf0101D0D1D2D3- Xd bng Cacno c 2 bin /khin AB v 1 bin vo CD0D1D0D2D3D0D2D1D3Vng D tng ng vi gi tr thp phnA=0, B=0 l vng D0A=0, B=1 l vng D1A=1, B=0 l vng D2A=1, B=1 l vng D30123A=0, B=0 l vng D0A=0, B=1 l vng D1A=1, B=0 l vng D2A=1, B=1 l vng D3

- in cc g/tr hm s cho trc vo bng Cacno1111- Ti thiu ha hm cho, cho tng vng Di

- S to hm f dng MUX 41 vi 2 bin /khin ABfCCAB01D3D2D1D0

TH2: 2 bin iu khin l A & C- 2 bin /khin Bng Cacno c 4 ca DACf0101D0D1D2D3- Xd bng Cacno c 2 bin /khin AC v 1 bin vo BD0D1D0D2D3D0D2D1D3Vng D tng ng vi gi tr thp phnA=0, C=0 l vng D0A=0, C=1 l vng D1A=1, C=0 l vng D2A=1, C=1 l vng D30123A=0, C=0 l vng D0A=0, C=1 l vng D1A=1, C=0 l vng D2A=1, C=1 l vng D3

- S to hm f dng MUX 41 vi 2 bin /khin ABfBBAC01D3D2D1D0

TH3: 2 bin iu khin l B & C- 2 bin /khin Bng Cacno c 4 ca DBCf0101D0D1D2D3- Xd bng Cacno c 2 bin /khin BC v 1 bin vo AD0D1D0D2D3D0D2D1D3Vng D tng ng vi gi tr thp phnB=0, C=0 l vng D0B=0, C=1 l vng D1B=1, C=0 l vng D2B=1, C=1 l vng D30123B=0, C=0 l vng D0B=0, C=1 l vng D1B=1, C=0 l vng D2B=1, C=1 l vng D3

- S to hm f dng MUX 41 vi 2 bin /khin BCf10BCD3D2D1D0

a) Dng MUX 4 1 & cc mch NANDC 2 bin iu khin, chn 2 bin iu khin l A, BXd bng Cacno c 2 bin /khin AB CDABF0001111000011110D0D1D2D3D0D0D0D1D1D1D2D2D2D3D3D3

- in cc g/tr hm s cho trc vo bng CacnoABF0001111000011110111111111- Ti thiu ha cc hm cho, cho cc vng Di- Vng D0 ng vi A=0; B=0. v tng t ta c kt qu:

- S to hm f dng MUX 41 vi 2 bin A, B

b) Dng MUX 8 1 & cc mch NANDC 3 bin iu khin, chn 3 bin iu khin l A, B,CCDABF0001111000011110D0D1D2D4D0D1D6D6D2D7D7D4D3D3D0D1D3D2D4D5D6D7Xd bng CacnoD5D5

Vng D0 ng vi A=0; B=0; C=0 D0 = 1Vng D1 ng vi A=0; B=0; C=1 D1 = 0Vng D2 ng vi A=0; B=1; C=0 D2 = DVng D3 ng vi A=0; B=1; C=1 D3 = 1Vng D4 ng vi A=1; B=0; C=0 D4 = DVng D5 ng vi A=1; B=0; C=1 D5 = DVng D6 ng vi A=1; B=1; C=0 D6 = 0Vng D7 ng vi A=1; B=1; C=1 D7 = D

S to hm f dng MUX 81 vi 3 bin A, B, Cf01BCD3D2D1D00D1A

Bai giang Ky thuat so- tkhao.ppt

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