Dien Tu So - DHBK Ha Noi

IN T S

Digital Electronics

B mn K thut my tnhKhoa Cng ngh thng tin Trng H Bch Khoa H Ni

a ch lin h ca tc giVn phng:B mn K thut my tnh Khoa Cng ngh thng tinTrng i hc Bch Khoa H NiP322 C1 S 1, i C Vit, H NiT : 04 8696125Ging vin: Nguyn Thnh Kin Mobile: +84983588135Email:[email protected]

Mc ch mn hc Cung cp cc kin thc c bn v:Cu toNguyn l hot ngng dngca cc mch s (mch logic, IC, chip)Trang b nguyn lPhn tchThit kcc mch s c bnTo c s cho tip thu cc kin thc chuyn ngnh

Ti liu tham kho chnhIntroductory Digital Electronics - Nigel P. Cook - Prentice Hall, 1998Digital Systems - Principles and Applications - Tocci & Widmer - Prentice Hall, 1998http://ktmt.shorturl.com

Thi lng mn hcTng thi lng: 60 titL thuyt: 45 tit, ti ging ngThc hnh: 15 tit.M phng mt s mch in t s trong gio trnh s dng phn mm Multisim v8.0Hng dn thc hnh ti phng myC1-325, C Nguyt B mn KTMT lin hNp bo co thc hnh km bi thiKhng c bo co thc hnh => 0 im.

Ni dung ca mn hcChng 1. Gii thiu v in t sChng 2. Cc hm logicChng 3. Cc phn t logic c bnChng 4. H t hpChng 5. H dy

in t sChng 1GII THIU V IN T S

B mn K thut My tnh, Khoa Cng ngh Thng tinTrng i hc Bch Khoa H Ni

Gii thiu v in t sin t s

Gii thiu v in t s (tip)H thng in t, thit b in t

Cc linh kinin, in t(component)

Ccmchin t(circuit)

Ccthit b,h thngin t(equipment, system)

Gii thiu v in t s (tip)S v tng t:Trong khoa hc, cng ngh hay cuc sng i thng, ta thng xuyn phi tip xc vi s lngS lng c th o, qun l, ghi chp, tnh ton nhm gip cho cc x l, c on phc tp hnC 2 cch biu din s lng:Dng tng t (Analog)Dng s (Digital)Dng tng t:VD: Nhit , tc , in th ca u ra microL dng biu din vi s bin i lin tc ca cc gi tr (continuous)Dng s:VD: Thi gian hin trn ng h in tL dng biu din trong cc gi tr thay i tng nc ri rc (discrete)

Gii thiu v in t s (tip)H thng s v tng t:H thng s (Digital system)L t hp cc thit b c thit k x l cc thng tin logic hoc cc s lng vt l di dng sVD: My vi tnh, my tnh, cc thit b hnh nh m thanh s, h thng in thoing dng: lnh vc in t, c kh, tH thng tng t (Analog system)Cha cc thit b cho php x l cc s lng vt l dng tng tVD: H thng m-ly, ghi bng t

Gii thiu v in t s (tip)Cng ngh s - u, nhc im so vi tng tDng cng ngh s thc hin cc thao tc ca gii php tng tu im ca cng ngh s:Cc h thng s d thit k hn: Khng cn gi tr chnh xc U, I, ch cn khong cch mc cao thpLu tr thng tin dC cc mch cht c th gi thng tin lu ty chnh xc cao hnVic nng t chnh xc 3 ch s ln 4 ch s n gin ch cn lp thm mch h tng t, lp thm mch s nh hng U, I v thm nhiuCc x l c th lp trnh ct b nh hng bi nhiuC th ch to nhiu mch s trong cc chip

Gii thiu v in t s (tip)Cng ngh s - u, nhc im so vi tng tHn ch: Th gii thc ch yu l tng tCc s lng vt l trong thc t, t nhin ch yu l dng tng t. VD: nhit , p sut, v tr, vn tc, rn, tc dng chy

Chuyn icc u vothc t dngtng tthnhdng s

X lthng tinS

Chuyn icc u ra sv dngtng t thc t

Gii thiu v in t s (tip)S kt hp ca cng ngh s v tng t!

in t sChng 2CC HM LOGIC


Ni dung chng 22.1. Gii thiu2.2. i s Boole2.2. Biu din cc hm logic di dng chnh quy2.3. Ti thiu ha cc hm logic

2.1. Gii thiuMch logic (mch s) hot ng da trn ch nh phn:in th u vo, u vo hoc bng 0, hoc bng 1Vi 0 hay 1 tng trng cho cc khong in th c nh ngha snVD: 0 0.8V : 0 2.5 5V : 1Cho php ta s dng i s Boole nh l mt cng c phn tch v thit k cc h thng s

Gii thiu (tip)i s Boole:Do George Boole sng lp vo th k 19Cc hng, bin v hm ch nhn 1 trong 2 gi tr: 0 v 1L cng c ton hc kh n gin cho php m t mi lin h gia cc u ra ca mch logic vi cc u vo ca n di dng biu thc logicL c s l thuyt, l cng c cho php nghin cu, m t, phn tch, thit k v xy dng cc h thng s, h thng logic, mch s ngy nay.

Gii thiu (tip)Cc phn t logic c bn:Cn gi l cc cng logic, mch logic c bnL cc khi c bn cu thnh nn cc mch logic v h thng s khc

Gii thiu (tip)Mc tiu ca chng: sinh vin c thTm hiu v i s BooleCc phn t logic c bn v hot ng ca chngDng i s Boole m t v phn tch cch cu thnh cc mch logic phc tp t cc phn t logic c bn


1. Cc nh nghaBin logic: l 1 i lng c th biu din bng 1 k hiu no , v mt gi tr ch ly gi tr 0 hoc 1.Hm logic: l biu din ca nhm cc bin logic, lin h vi nhau thng qua cc php ton logic, v mt gi tr cng ly gi tr 0 hoc 1.Php ton logic: c 3 php ton logic c bn:Php V - "AND"Php Hoc - "OR"Php o - "NOT"

Cc nh ngha (tip)Cc gi tr 0, 1 khng tng trng cho cc con s thc m tng trng cho trng thi gi tr in th hay cn gi l mc logic (logic level)Mt s cch gi khc ca 2 mc logic:Mc logic 0Mc logic 1Sai (False)ng (True)Tt (Off)Bt (On)Thp (Low)Cao (High)Khng (No)C (Yes)(Ngt) Open switch(ng) Closed switch

2. Biu din bin v hm logicDng biu Venn (le):Mi bin logic chia khng gian thnh 2 khng gian con.Khng gian con th nht, bin nhn gi tr ng (=1), khng gian con th cn li, bin nhn gi tr sai (=0).VD: F = A AND B

Biu din bin v hm logic (tip)Dng biu thc i s:K hiu php V AND: .K hiu php Hoc OR: +K hiu php o NOT: VD: F = A AND B hay F = A.B

Biu din bin v hm logic (tip)Dng bng tht:Dng m t s ph thuc u ra vo cc mc in th u vo ca cc mch logicBng tht biu din 1 hm logic n bin c:(n+1) ct:n ct u tng ng vi n binct cn li tng ng vi gi tr ca hm2n hng:tng ng vi 2n gi tr ca t hp bin

Biu din bin v hm logic (tip)Dng ba Cc-n:y l cch biu din tng ng ca bng tht.Trong , mi trn ba tng ng vi 1 dng ca bng tht.Ta ca xc nh gi tr ca t hp bin.Gi tr ca hm c ghi vo tng ng.

Biu din bin v hm logic (tip)Dng biu thi gian:L th biu din s bin i theo thi gian ca bin v hm logicVD: vi F = A . B

3. Cc php ton logic c bn

4. Tnh cht ca php ton logic c bnTn ti phn t trung tnh duy nht trong php ton AND v ORCa php AND l 1: A . 1 = ACa php OR l 0: A + 0 = ATnh cht giao honA.B = B.AA + B = B + ATnh cht kt hp(A.B).C = A.(B.C) = A.B.C(A + B) + C = A + (B + C) = A + B + C

Cc tnh cht (tip)Tnh cht phn phi(A + B).C = A.C + B.C(A.B) + C = (A + C).(B + C)Tnh cht khng s m, khng h sA.A.A. .A = AA+A+A+ +A = APhp b

5. nh l DeMorgano ca mt tng bng tch cc o thnh phn

o ca mt tch bng tng cc o thnh phn

Tng qut:

6. Nguyn l i ngui ngu:+ i ngu vi .0 i ngu vi 1V d:(A + B).C = A.C + B.C (A.B) + C = (A + C).(B + C)


2.2. Biu din cc hm logic di dng chnh quy

1. Tuyn chnh quynh l Shannon: mt hm logic bt k c th c trin khai theo 1 trong cc bin di dng tng ca 2 tch logic nh sau:

V d:

Mt hm logic bt k u c th chuyn v dng tuyn chnh quy nh p dng nh l Shannon cho dng tuyn

p dng nhanh nh l Shannon

2. Hi chnh quynh l Shannon: mt hm logic bt k c th c trin khai theo 1 trong cc bin di dng tch ca 2 tng logic nh sau:

V d:

Mt hm logic bt k u c th chuyn v dng hi chnh quy nh p dng nh l Shannon cho dng hi

p dng nhanh nh l Shannon

3. Biu din hm logic di dng s


2.3. Ti thiu ha cc hm logicMt hm logic c gi l ti thiu ho nu nh n c s lng s hng t nht v s lng bin t nht.Mc ch ca vic ti thiu ho: Mi hm logic c th c biu din bng cc biu thc logic khc nhau. Mi 1 biu thc logic c mt mch thc hin tng ng vi n. Biu thc logic cng n gin th mch thc hin cng n gin.C hai phng php ti thiu ho hm logic: Phng php i sPhng php ba Cc-n

1. Phng php i s

Phng php nhm s hng

Thm s hng c vo biu thc

Loi b s hng thaTrong v d sau, AC l s hng tha:

Ti thiu ha?

Bi tp p dngVD1: Ti thiu ha cc hm sau bng phng php i s:a.b.

2. Phng php ba Cc-nQuy tc lp ba Cc-n:2 lin k nhau ch sai khc nhau 1 gi tr ca 1 bin (tng ng vi t hp bin khc nhau 1 gi tr) Ba Cc-n c tnh khng gian

Ba Cc-n cho hm 2, 3, 4 bin

Quy tc nhm (dng tuyn chnh quy)Nhm cc lin k m gi tr ca hm cng bng 1 li vi nhau sao cho:S lng cc trong nhm l ln nht c th c,ng thi s lng trong nhm phi l ly tha ca 2,V hnh dng ca nhm phi l hnh ch nht hoc hnh vungNhm c 2n loi b c n binBin no nhn c gi tr ngc nhau trong nhm th s b loiCc nhm c th trng nhau mt vi phn t nhng khng c trng hon ton v phi nhm ht cc bng 1S lng nhm chnh bng s lng s hng sau khi ti thiu ha (mi nhm tng ng vi 1 s hng)

V d

Trng hp c bitNu gi tr hm khng xc nh ti mt vi t hp bin no :K hiu cc khng xc nh bng du Nhm cc vi cc 1Khng nht thit phi nhm ht cc

Bi tp p dngTi thiu ha cc hm sau bng phng php ba Ccn:a. F(A,B,C,D) = R(0,2,5,6,9,11,13,14) b. F(A,B,C,D) = R(1,3,5,8,9,13,14,15)c. F(A,B,C,D) = R(2,4,5,6,7,9,12,13)d. F(A,B,C,D) = R(1,5,6,7,11,13) v F khng xc nh vi t hp bin 12,15.

in t sChng 3CC PHN T LOGIC C BN


Ni dung chng 33.1. Khi nim3.2. Thc hin phn t AND, OR dng Diode3.3. Thc hin phn t NOT dng Transistor3.4. Cc mch tch hp s

3.1. Khi nimC 3 php ton logic c bn:V (AND)HOC (OR)O (NOT)Phn t logic c bn (mch logic c bn, cng logic) thc hin php ton logic c bn:Cng V (AND gate)Cng HOC (OR gate)Cng O (NOT inverter)Cc mch s c bit khc: cc cng NAND, NOR, XOR, XNOR

1. Cng V (AND gate)Chc nng:Thc hin php ton logic V (AND)u ra ch bng 1 khi tt c cc u vo bng 1Cng V 2 u vo:K hiu:

Bng tht:Biu thc: out = A . BABout000010100111

2. Cng HOC (OR gate)Chc nng:Thc hin php ton logic HOC (OR)u ra ch bng 0 khi tt c cc u vo bng 0Cng HOC 2 u vo:K hiu:

Bng tht:Biu thc: out = A + B

ABout000011101111

3. Cng O (NOT inverter)Chc nng:Thc hin php ton logic O (NOT)Cng O ch c 1 u vo:K hiu:

Bng tht:Biu thc: out = AAout0110

4. Cng V O (NAND gate)Chc nng:Thc hin php O ca php ton logic Vu ra ch bng 0 khi tt c cc u vo bng 1Cng V O 2 u vo:K hiu:

Bng tht:Biu thc: out = A . BABout001011101110

5. Cng HOC O (NOR gate)Chc nng:Thc hin php O ca php ton logic HOCu ra ch bng 1 khi tt c cc u vo bng 0Cng HOC O 2 u vo:K hiu:

Bng tht:Biu thc: out = A + BABout001010100110

6. Cng XOR (XOR gate)Chc nng:Exclusive-ORThc hin biu thc logic HOC C LOI TR (php ton XOR - hay cn l php cng module 2)u ra ch bng 0 khi tt c cc u vo ging nhauCng XOR 2 u vo:K hiu:

Bng tht:Biu thc:ABout000011101110

7. Cng XNOR (XNOR gate)Chc nng:Exclusive-NORThc hin php O ca php ton XORu ra ch bng 1 khi tt c cc u vo ging nhauCng XNOR 2 u vo:K hiu:

Bng tht:Biu thc:ABout001010100111

8. Bi tp Cho cc biu thi gian sau, hy cho bit tng biu thi gian biu din hot ng ca cng no?E0 (EA, EB) = ?

Bi tp (tip)E0 (EA, EB) = ?

3.2. Thc hin phn t AND, ORDiode:K hiu:

Chc nng: cho dng in i qua theo 1 chiu t A n KHot ng:Nu UA > UK th IAK > 0, Diode lm vic ch Thng

Nu UA UK th IAK = 0, Diode lm vic ch Tt

Phn t AND 2 u vo dng DiodeXt mch hnh bn.Gi s ly TTL lm chun cho hot ng ca mch.Ln lt t in p 0V v 5V vo 2 u vo A v B, sau o in p ti u ra S, ta c:

S = A.B

Phn t OR 2 u vo dng DiodeXt mch hnh bn.Gi s ly TTL lm chun cho hot ng ca mch.Ln lt t in p 0V v 5V vo 2 u vo A v B, sau o in p ti u ra S, ta c:S = A+B

3.3. Thc hin phn t NOTTransistor lng cc:C 2 loi: NPN v PNPTransistor c 3 cc:B: Base cc gcC: Collector cc gpE: Emitter cc phtChc nng: Dng khuch i (thng) dng IC bng vic iu khin dng IBHot ng:IB = 0, Transistor lm vic ch khng khuch i (tt), IC = 0IB > 0, Transistor lm vic ch khuch i (thng), IC = .IB, trong l h s khuch i.

Phn t NOT dng TransistorXt mch hnh sau.

Gi s ly TTL lm chun cho hot ng ca mch.Ln lt t in p 0V v 5V vo u vo A v chn Rb nh sao cho Transistor thng bo ha, sau o in p ti u ra S, ta c:

3.4. Cc mch tch hp sCc phn t logic c cu thnh t cc linh kin in tCc linh kin in t ny khi kt hp vi nhau thng dng cc mch tch hp hay cn gi l IC (Integrated Circuit).Mch tch hp hay cn gi l IC, chip, vi mch, bo c c im:u im: mt linh kin, lm gim th tch, gim trng lng v kch thc mch.Nhc im: hng mt linh kin th hng c mch.C 2 loi mch tch hp:Mch tch hp tng t: lm vic vi cc tn hiu tng tMch tch hp s: lm vic vi cc tn hiu s

Phn loi mch tch hp sTheo mt linh kin:Tnh theo s lng cng (gate).Mt cng c khong 210 transistorVD: cng NAND 2 u vo c cu to t 4 transistorC cc loi sau:SSI - Small Scale Integration: cc vi mch c mt tch hp c nh: < 10 cng/chipMSI - Medium Scale Integration: cc vi mch c mt tch hp c trung bnh: 10 100 cng/chipLSI - Large Scale Integration: cc vi mch c mt tch hp c ln: 100 1000 cng/chipVLSI - Very Large Scale Integration: cc vi mch c mt tch hp c rt ln: 103106 cng/chipULSI - Ultra Large Scale Integration: cc vi mch c mt tch hp c cc k ln: > 106 cng/chip

Phn loi mch tch hp s (tip)Theo bn cht linh kin c s dng:IC s dng Transistor lng cc:RTL Resistor Transistor Logic (u vo mc in tr, u ra l Transistor) DTL Diode Transistor Logic (u vo mc Diode, u ra l Transistor)TTL Transistor Transistor Logic (u vo mc Transistor, u ra l Transistor)ECL Emitter Coupled Logic (Transistor ghp nhiu cc emitter)IC s dng Transistor trng - FET (Field Effect Transistor) MOS Metal Oxide SemiconductorCMOS Complementary MOS

c tnh in ca ICDi in p quy nh mc logicVD: vi chun TTL ta c:

Di in p khng xc nh5V2V0.8V0VVo5V3,5V0,5V0VRaDi in p khng xc nh

c tnh in ca IC (tip)Thi gian truyn: tn hiu truyn t u vo ti u ra ca mch tch hp phi mt mt khong thi gian no . Thi gian c nh gi qua 2 thng s:Thi gian tr: l thi gian tr thng tin ca u ra so vi u voThi gian chuyn bin: l thi gian cn thit chuyn bin t mc 0 ln mc 1 v ngc li.Thi gian chuyn bin t 0 n 1 cn gi l thi gian thit lp sn dngThi gian chuyn bin t 1 n 0 cn gi l thi gian thit lp sn mTrong l thuyt: thi gian chuyn bin bng 0Trong thc t, thi gian chuyn bin c o bng thi gian chuyn bin t 10% n 90% gi tr bin cc i.

c tnh in ca IC (tip)Cng sut tiu th ch ng:Ch ng l ch lm vic c tn hiuL cng sut tn hao trn cc phn t trong vi mch, nn cn cng nh cng tt.Cng sut tiu th ch ng ph thucTn s lm vic.Cng ngh ch to: cng ngh CMOS c cng sut tiu th thp nht.

c tnh c ca ICL c tnh ca kt cu v bc bn ngoi.C 2 loi thng dng:V trn bng kim loi, s chn < 10V dt bng gm, cht do, c 3 loiIC mt hng chn SIP (Single Inline Package) hay SIPP (Single In-line Pin Package) IC c 2 hng chn DIP (Dual Inline Package) IC chn dng li PGA (Pin Grid Array): v vung, chn xung quanh

c tnh c ca IC (tip)Mt s dng IC:

c tnh nhit ca ICMi mt loi IC c ch to s dng mt iu kin mi trng khc nhau ty theo mc ch s dng n.IC dng trong cng nghip: 0C70CIC dng trong qun s: -55C 125C

VD: Phn t AND dng IC

VD: Phn t AND dng IC (tip)

VD: Phn t OR dng IC

VD: Phn t NAND dng IC

VD: Phn t NOR dng IC

VD: Phn t XOR v XNOR dng IC

Cc phn t logic c bnAND: 74LS08OR: 74LS32NOT: 74LS04/05NAND: 74LS00NOR: 74LS02XOR: 74LS136NXOR: 74LS266

Bi tp p dngBiu din cc phn t logic hai u vo AND, OR v phn t logic mt u vo NOT ch dng phn t NAND.

in t sChng 4H T HP


Ni dung chng 44.1. Khi nim4.2. Mt s h t hp c bn

4.1. Khi nimH t hp l h m tn hiu ra ch ph thuc vo tn hiu vo ti thi im hin tiH t hp cn c gi l h khng c nhH t hp ch cn thc hin bng nhng phn t logic c bn

Ni dung chng 44.1. Khi nim4.2. Mt s h t hp c bn

4.2. Mt s h t hp c bnB m haB gii mB chn knhB phn knhCc mch s hc

1. B m haM ha l vic s dng k hiu biu din c trng cho mt i tng no .K hiu tng ng vi mt i tng c gi l t m.Th d:

B m ha (tip)Chc nng: thc hin vic m ha cc tn hiu tng ng vi cc i tng thnh cc t m nh phn.

Th d:

i tngT mB m ha

tn hiu

tn hiu

B m ha

A

B

C

D

S0S1

V d - B m ha bn phmM ha bn phm:Mi phm c gn mt t m khc nhau.Khi mt phm c nhn, b m ha s cho ra u ra l t m tng ng gn cho phm .Hy thit k b m ha cho mt bn phm gm c 9 phm vi gi thit trong mt thi im ch c duy nht 1 phm c nhn.

B m ha bn phm (tip)S khi:Mt b 9 phm, phi s dng 4 bit m ha.Vy c 9 u vo, 4 u ra.M ha u tin:Nu 2 hoc nhiu phm ng thi c nhn, th b m ha ch coi nh 1 phm c nhn, v phm c m cao nht.

P1

P2

P9

BMHbnphm9 phmVccA

B

C

D

B m ha bn phm (tip)Bng m ha:

B m ha bn phm (tip)Lp biu thc u ra ph thuc u vo:A = 1 khi P8 hoc P9 c nhn, tc l khi P8 = 1 hoc P9 = 1Vy A = P8 + P9B = 1 khi P4 hoc P5 hoc P6 hoc P7 c nhn, tc l khi P4 = 1 hoc P5 = 1 hoc P6 = 1 hoc P7 = 1Vy B = P4 + P5 + P6 + P7C = 1 khi P2 hoc P3 hoc P6 hoc P7 c nhn, tc l khi P2 = 1 hoc P3 = 1 hoc P6 = 1 hoc P7 = 1Vy C = P2 + P3 + P6 + P7D = 1 khi P1 hoc P3 hoc P5 hoc P7 hoc P9 c nhn, tc l khi P1 = 1 hoc P3 = 1 hoc P5 = 1 hoc P7 = 1 hoc P9 = 1Vy D = P1 + P3 + P5 + P7 + P9V mch:

Bi tp v nhTm hiu hot ng ca bn phm my tnh n gin TLTK: www.wikipedia.org

2. B gii mChc nng: B gii m thc hin chc nng ngc vi b m ha.Cung cp thng tin u ra khi u vo xut hin t hp cc bin nh phn ng vi 1 hay nhiu t m c chn.T t m xc nh c tn hiu tng ng vi i tng m ha.

Hai trng hp gii mGii m cho 1 t m:Nguyn l: ng vi mt t hp cn gii m u vo th u ra bng 1, cc t hp u vo cn li, u ra bng 0.VD: S = 1 nu (AB) = (10), S = 0 nu (AB) (10)

Gii m cho ton b m: Nguyn l: ng vi mt t hp no u vo th 1 trong cc u ra bng 1, cc u ra cn li bng 0.

A

BSBGM

A

BS0S1S2S3

BGM

V d - B gii m BCDBCD: m ha s nguyn thp phn bng nh phn

B gii m BCD (tip)Xc nh u vo v u ra:Vo: t m nh phn 4 bit ( c 16 t hp)Ra: cc tn hiu tng ng vi cc s nh phn m t m m haTa ch s dng 10 t hp, cn 6 t hp khng s dng n c coi l khng xc nh.

BCD Binary Coding Decimal

B gii m BCD Bng tht

Tm biu thc ca tng u ra

Tm biu thc ca tng u ra (tip)




V mch

3. B chn knhMultiPlexor MUXC nhiu u vo tn hiu v 1 u raChc nng: chn 1 tn hiu trong nhiu tn hiu u vo a ra u ra

MUX 2-1S khi:

Tn hiu chn:

Tn hiu ra:

E1

E0

C0

S

MUX 4-1S khi:

Tn hiu chn:

Tn hiu ra:

E3E2E1E0

C1

C0

S

V d - Thit k MUX 2-1Bng tht:

V d - Thit k MUX 2-1 (tip)Biu thc u ra S:

V d - Thit k MUX 2-1 (tip)S mch:

Minh ha

4. B phn knhDeMultiPlexor DeMUXC 1 u vo tn hiu v nhiu u raChc nng: a tn hiu t u vo ti 1 trong nhng u ra

DeMUX 1-2S khi:

Tn hiu chn:

E

C0

S0

S1

DeMUX 1-4S khi:

Tn hiu chn:

E

C1

C0

S0S1S2S3

V d - Thit k DeMUX 1-2Bng tht:

Biu thc u ra:

5. Cc mch s hcB cngB trB so snh

a. B cngChc nng: thc hin php cng gia 2 s nh phn.Bn tng (Half-Adder):Thc hin php cng gia 2 bit thp nht ca php cng 2 s nh phn.S khi:

Bn tng (tip)Bng tht:

Biu thc u ra ph thuc u vo:

S mch:

Minh ha

Mch test

B cng y (Full-Adder)Chc nng: thc hin php cng gia 2 bit bt k ca php cng 2 s nh phn.S khi:ri: bit nh u vori+1: bit nh u ra

B cng y (tip)Bng tht:


B cng y (tip)S mch:

Minh ha

Mch test

B cng nhiu bity l b cng 2 s nh phn n bit, kt qu nhn c l 1 s nguyn n+1 bit.S :

Minh ha

Mch test

b. B trChc nng: thc hin php tr gia 2 s nh phn.Bn hiu (Half-Subtractor):Dng thc hin php tr gia 2 bit thp nht trong php tr gia 2 s nh phnS khi:Di: hiuBi+1: bit mn

Bn hiu (tip)Bng tht:


S mch:

Minh ha

Mch test

B tr y (Full-Subtractor)Chc nng: dng thc hin php tr gia 2 bit bt k trong php tr 2 s nh phn.S khi:

B tr y (tip)Bng tht:


B tr y (tip)S mch:

Minh ha

Mch test

c. B so snhDng so snh 2 s nh phnC 2 kiu so snh:So snh n gin:Kt qu so snh: bng nhau, khc nhauSo snh y :Kt qu so snh: ln hn, nh hn, bng nhauC 2 loi b so snh:B so snh n ginB so snh y

B so snh n ginGi s cn xy dng b so snh n gin 2 s A v B:A a3 a2 a1 a0B b3 b2 b1 b0u ra SS = 1 A = BS = 0 A B

B so snh n gin (tip)Ta c:

Suy ra:

B so snh n gin (tip)S mch:

B so snh y B so snh 2 bit y :u vo: 2 bit cn so snh ai v biu ra: 3 tn hiu bo kt qu ln hn, nh hn, bng nhau ca 2 bitai > bi Gi = 1 cn Ei, Li = 0ai < bi Li = 1 cn Ei, Gi = 0ai = bi Ei = 1 cn Gi, Li = 0S khi:

B so snh 2 bit y (tip)Bng tht:

Biu din u ra theo u vo:

S mch:

Minh ha

B so snh y 2 s nh phnCu to: gm cc b so snh 2 bitC tn hiu CS (Chip Select)CS = 0, tt c cc u ra = 0 (khng so snh)CS = 1, hot ng bnh thngBiu din cc u ra ca b so snh 2 bit theo u vo:

Minh ha

Mch test

VD: B so snh 2 s nh phn 3 bitS mch b so snh 2 s nh phn 3 bit:A = a2a1a0B = b2b1b0

Mch test

Minh ha

Bi tp chng 4Bi 1: Tng hp b chn knh 4-1. Bi 2: Thit k b tr/nhn 2 s 2 bit.Bi 3: Tng hp b chn knh 2-1 ch dng NAND. Bi 4: Tng hp mch t hp thc hin php ton sau : M = N + 3, bit rng N l s 4 bit m BCD cn M l s 4 bit.

in t sChng 5H DY


Ni dung chng 55.1. Khi nim5.2. M hnh ca h dy5.3. Cc Trigger5.4. Mt s ng dng ca h dy

5.1. Khi nimH dy l h m tn hiu ra khng ch ph thuc vo tn hiu vo ti thi im hin ti m cn ph thuc vo qu kh ca tn hiu vo.H dy cn c gi l h c nh. thc hin c h dy, nht thit phi c phn t nh. Ngoi ra cn c th c cc phn t logic c bn.

Phn loi h dyH dy ng b: khi lm vic cn c 1 tn hiu ng b gi nhp cho ton b h hot ng.H dy khng ng b: khng cn tn hiu ny gi nhp chung cho ton b h hot ng. H dy ng b nhanh hn h dy khng ng b tuy nhin li c thit k phc tp hn.


M hnh ca h dyM hnh ca h dy c dng m t h dy thng qua tn hiu vo, tn hiu ra v trng thi ca h m khng quan tm n cu trc bn trong ca h.

M hnh ca h dy (tip)C 2 loi m hnh:MealyMooreHai loi m hnh trn c th chuyn i qua li cho nhau.

a. M hnh MealyM hnh Mealy m t h dy thng qua 5 tham s:X = {x1, x2, ..., xn}Y = {y1, y2, ..., yl}S = {s1, s2, ..., sm}FS(S, X)FY(S, X)

M hnh Mealy (tip)Gii thch cc k hiu:X l tp hp hu hn n tn hiu u voY l tp hp hu hn l tn hiu u raS tp hp hu hn m trng thi trong ca hFS l hm bin i trng thi. i vi m hnh kiu Mealy th FS ph thuc vo S v X FS = FS(S, X)FY l hm tnh trng thi u ra: FY = FY(S, X)

b. M hnh MooreM hnh Moore ging nh m hnh Mealy, nhng khc ch l FY ch ph thuc vo S:FY = FY(S)

Bng chuyn trng thiM hnh Mealy:

Bng chuyn trng thi (tip)M hnh Moore:

V d v m hnh h dyS dng m hnh Mealy v Moore m t h dy thc hin php cng.V d:

V d: M hnh MealyX = {00, 01, 10, 11} - do c 2 u voY = {0, 1} - do c 1 u raS = {s0, s1} - s0: trng thi khng nh- s1: trng thi c nhHm trng thi FS(S, X):FS(s0, 00) = s0 FS(s0, 01) = s0FS(s0, 11) = s1 FS(s0, 10) = s0FS(s1, 00) = s0 FS(s1, 10) = s1FS(s1, 01) = s1 FS(s1, 11) = s1

V d: M hnh Mealy (tip)Hm ra FY(S, X):FY(s0, 00) = 0 FY(s0, 11) = 0FY(s0, 01) = 1 FY(s0, 10) = 1FY(s1, 00) = 1 FY(s1, 10) = 0FY(s1, 11) = 1 FY(s1, 01) = 0

Bng chuyn trng thi

hnh chuyn trng thi

V d: M hnh MooreX = {00, 01, 10, 11} - do c 2 u voY = {0, 1} - do c 1 u raS = {s00, s01, s10, s11} - sij: i = 0 l khng nhi = 1 l c nhj = tn hiu ra

V d: M hnh Moore (tip)Hm trng thi FS(S, X):FS(s00, 00) = s00 FS(s00, 10) = s01FS(s00, 01) = s01 FS(s00, 11) = s10FS(s01, 00) = s00 FS(s01, 10) = s01FS(s01, 01) = s01 FS(s01, 11) = s10FS(s10, 00) = s01 FS(s10, 10) = s10FS(s10, 01) = s10 FS(s10, 11) = s11FS(s11, 00) = s01 FS(s11, 01) = s10FS(s11, 11) = s11 FS(s11, 10) = s10Hm ra FY(S):FY(s00) = 0 FY(s01) = 1FY(s10) = 0 FY(s11) = 1

Bng chuyn trng thi

hnh chuyn trng thi


TriggerPhn t c bn ca h dy chnh l cc phn t nh hay cn gi l cc triggeru ra ca trigger chnh l trng thi ca nMt trigger c th lm vic theo 2 kiu:Trigger khng ng b: u ra ca trigger thay i ch ph thuc vo tn hiu u voTrigger ng b: u ra ca trigger thay i ph thuc vo tn hiu vo v tn hiu ng b

Cc kiu ng bng b theo mc:Mc cao:Khi tn hiu ng b c gi tr logic bng 0 th h ngh (gi nguyn trng thi)Khi tn hiu ng b c gi tr logic bng 1 th h lm vic bnh thng.Mc thp:Khi tn hiu ng b c gi tr logic bng 1 th h ngh (gi nguyn trng thi)Khi tn hiu ng b c gi tr logic bng 0 th h lm vic bnh thng.

Cc kiu ng b (tip)ng b theo sn:Sn dng:Khi tn hiu ng b xut hin sn dng (sn i ln, t 0 1) th h lm vic bnh thngTrong cc trng hp cn li, h ngh (gi nguyn trng thi).Sn m:Khi tn hiu ng b xut hin sn m (sn i xung, t 1 0), h lm vic bnh thngTrong cc trng hp cn li, h ngh (gi nguyn trng thi).

Cc kiu ng b (tip)ng b kiu xung: Khi c xung th h lm vic bnh thngKhi khng c xung th h ngh (gi nguyn trng thi).

Cc loi TriggerC 4 loi Trigger:RS Reset - Set Xa - Thit lp D Delay TrJK Jordan v Kelly Tn 2 nh pht minhT Toggle Bp bnh, bt tt

a. Trigger RSS khi:

Trigger RS hot ng c c 2 ch ng b v khng ng b

Bng chuyn trng thi ca RS

V dCho Trigger RS ng b mc cao v th cc tn hiu R, S nh hnh v. Hy v th tn hiu ra Q.

V d (tip)

b. Trigger DTrigger D c 1 u vo l D v hot ng 2 ch ng b v khng ng b.Ta ch xt trigger D hot ng ch ng b.

Trigger D ng bTrigger D ng b theo mc gi l cht D (Latch)

Trigger D ng b theo sn c gi l xc pht sn (Edge trigged)

Bng chuyn trng thi ca D

V d 1Cho cht D kch hot mc cao. Hy v tn hiu ra Q dng trn cng trc thi gian vi tn hiu vo D.

V d 1 (tip)

V d 2Cho trigger D xc pht sn dng. Hy v tn hiu ra Q dng trn cng trc thi gian vi tn hiu vo D.

V d 2 (tip)

c. Trigger JKTrigger JK ch hot ng ch ng bS khi:

Bng chuyn trng thi ca JK

J ~ SK ~ R

d. Trigger TTrigger T ch hot ng ch ng bS khi:

Bng chuyn trng thi ca T


1. B m v chia tn sB m c dng m xungB m c gi l module n nu n c th m c n xung: t 0 n n-1C 2 loi b m:B m khng ng b: khng ng thi a tn hiu m vo cc u vo ca cc triggerB m ng b: c xung m ng thi l xung ng h clock a vo tt c cc trigger ca b m

B m khng ng b module 16 m t 0 n 15 v c 16 trng thiM ha thnh 4 bit A,B,C,D tng ng vi q4,q3,q2,q1Cn dng 4 trigger (gi s dng trigger JK)

B m khng ng b module 16Bng m xung:

Biu thi gian:

NX: B m ny ng thi cng l b chia tn sB m khng ng b module 16

C 10 trng thi cn dng 4 TriggerGi s dng Trigger JK c u vo CLR (CLEAR: xa) tch cc mc thpNu CLR = 0 th q = 0C mi khi m n xung th 10 th tt c cc q b xa v 0S : (cc J=K=1)B m khng ng b module 10

B m ng b module 8C 8 trng thi cn dng 3 TriggerGi s dng cc Trigger JKBng m xung:

B m ng b module 8 (tip)

B m li khng ng b module 8Gi s dng Trigger JK c u vo PR (PRESET: thit lp trc) tch cc mc thpNu PR = 0 th q = 1u tin cho PR = 0 th q1q2q3 = 111Sau cho PR = 1, h hot ng bnh thng

xungq3 q2 q1012345678111100001110011001101010101S m765432107

B m li khng ng b module 8

2. Thanh ghiThanh ghi c cu to gm cc trigger ni vi nhauChc nng: lu tr tm thi thng tinDch chuyn thng tinLu : c thanh ghi v b nh u dng lu tr thng tin, nhng thanh ghi c chc nng dch chuyn thng tin. Do , thanh ghi c th s dng lm b nh, nhng b nh khng th lm c thanh ghi.

Phn loiVo ni tip ra ni tip

Vo ni tip ra song song

Vo song song ra ni tip

Vo song song ra song song

01010011

01010011

01010011

01010011

V dThanh ghi 4 bit vo ni tip ra song song dng Trigger D

V d (tip)Bng s liu kho st:

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