chuong 01 mot so khai niem mo dau 1716

7/29/2019 Chuong 01 Mot So Khai Niem Mo Dau 1716

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K THUT S(Digital Electronics)

Th.S ng Ngc KhoaKhoa in - in T

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Ni dung mn hc

Chng 1: Mt skhi nim m u Chng 2: H thng s Chng 3: Cc cng logic v i sBoolean Chng 4: Mch logic Chng 5: Flip-Flop Chng 6: Mch shc Chng 7: B m v thanh ghi Chng 8: c im ca cc IC s Chng 9: Cc mch sthng gp Chng 10: Kt ni vi mch tng t Chng 11: Thit b nh


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Gio trnh v ti liu tham kho

Bi ging Th.S ng Ngc Khoa K thut s- Nguyn Thy Vn, NXB.KHKT K thut s1 - Nguyn NhAnh, NXB. HQG Digital Systems: Principles and Applications

Ronald J.Tocci, Prentice-Hall

4

Chng 1Mt skhi nim m u

Th.S ng Ngc Khoa

Khoa in - in T


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M t shc

Tn hiu analog (tng t) l tn hiu cgi tr thay i mt cch lin tc

Tn hiu digital (s) l tn hiu c gi trthay i theo nhng bc ri rc.

Analog == Tng t. Digital == Ri rc (step by step)

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Tn hiu analog v digital

Tn hiu analog Tn hiu digital


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Nhng i lng sau y l analog haydigital? Cng tc 10 trng thi Dng ng ra ca mt thit b in. Nhit phng. Tc ca mt mt in.

Nt iu chnh m thanh ca radio.

V d 1.1

Digital

Analog

Analog

Analog

Analog

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H thng s

H thng s l mt kt hp ca cc thitb c thit k lm vic vi cc ilng vt l c miu t di dng s.

V d: my vi tnh, my tnh tay, cc thitb audio/video s, in thoi s, truynhnh k thut s


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u im ca k thut s

Nhn chung, h thng sd thit k. Cc thng tin c lu trd dng. chnh xc cao. C th lp trnh hot ng ca h thng. Cc mch s t b nh hng bi nhiu. Nhiu mch sc th c tch hp vo

trong mt IC.

10

Hn chca k thut s

Trong thc tphn ln cc i lng lanalog.

xl tn hiu analog, h thng cnthc hin theo ba bc sau:

Bin i tn hiu analog ng vo thnh tnhiu s(analog-to-digital converter, ADC) X l thng tin s Bin i tn hiu digital ng ra thnh tn

hiu analog (digital-to-analog converter,DAC)


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Mt sv d v h thng s

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H thng iu khin nhit


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a CD (Compact Disk)

m thanh ca cc nhc c v ting ht sto ra mt tn hiu in p analog trongmicrophone.

Tn hiu analog ny s c bin ithnh dng s.

Thng tin ss c lu trtrong a CD Trong qu trnh playback, my CD nhn

thng tin st a CD v bin i thnhtn hiu analog, sau khuch i v ara loa.

14

La chn gia digital & analog

H thng sphi thm vo 2 b ADC vDAC (phc tp, tn km)

H thng syu cu thm thi gian chocc qu trnh bin i (hn chtc )

Trong phn ln cc ng dng, h thng

sthng c u tin ng dng do ccu im ca n. Mch analog c sdng d dng cho

qu trnh khuch i tn hiu.Kt hp gia analog v digital


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Gi tr in p trong Digital

Binary 1: in p t2V n 5V

Binary 0: in p t0V n 0.8V Not used: in p t0.8V n 2V, vng

ny c thgy ra li trong mch s.

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Mch s

Mch sphi c thit k in png ra nm trong khong logic 0 hoclogic 1

Mt mch s lm vic vi cc gi tr ng

vo l logic 0 hoc 1 m khng quan tmn gi tr in p thc t. Mi mt mch stun theo mt tp hp

cc quy lut logic nht nh.


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Mch stch hp

Phn ln cc mch s c tch hp trongIC.

Mt sk thut chto IC TTL CMOS

Nhng h IC ny s c cp trong

chng 8.

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Truyn song song

Truyn ni tip

Truyn song song & ni tip


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B nh

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B nh

Trng thi ca mch c th c lu trsau khi chm dt tn hiu ng vo.

Thuc tnh lu gi tr ca n tng ngvi thit b nh nn c gi l b nh

(memory) B nh thng c lm tcc mch

Latches (cht) hoc Flip-Flop.


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Cc phn chnh ca my tnh

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Cu hi?


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1

Chng 11Thit b nh


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Thit b nh

Mt h thng thng sdng B nh trong (lm vic) tc cao B nh ngoi (lu tr) tc thp hn


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Thut ngthng sdng

Memory Cell: mt thit b hay mt mch ckhn nng lu trmt bit d liu

Memory Word: mt nhm cc bit, thngthng mt tc 8 64 bit.

Byte: mt nhm 8 bit. Dung lng: m t kh nng lu trca b

nh. Dung lng m t sword c trong bnh. 1K = 2

10

word 1M = 220 word

1G = 230 word

2K x 8 = 2.210 x 8 = 2.1024.8 word

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Thut ngthng sdng

Address: l sxcnh v tr ca t(word) trong b nh.

Lnh c: thc hinvic c d liu r a tb nh.

Lnh ghi: thc hinlnh ghi d liu vob nh.


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Thut ngthng sdng

RAM: Random-Access Memory. SAM: Sequential-Access Memory ROM: Read Only Memory RWM: Read/Write Memory Static Memory Devices: d liu c lu mi

mi khi cn ngun cung cp. Dynamic Memory Devices: d liu khng

c lu mi mi, lu d liu c lu trta

cn rewritten d liu. Main Memory: b nh lm vic Auxiliary Memory: b nh thcp dng

lu tr.

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Hot ng ca b nh

1. Xc nh a ch trong b nh c truy cpbi lnh ghi hoc c.

2. Xc nh lnh (ghi hoc c) cn thc hin.3. Cung cp d liu lu vo b nh trong qu

trnh ghi.

4. Nhn d liu ng ra trong qu trnh c.5. Enable hay Disable sao cho b nh p ng

n a ch v lnh thc thi.


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Cu trc ca b nh

Cu trc b nh 32x4

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Cu trc ca b nh

a) Ghi d liu 0100 vo b nh ti a ch 00011.b) c d liu 1101 tb nh c ch 11110


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Bi tp 01

Xc nh gi tr ca cc ng vo v ng rakhi c d liu t a ch 00100.

Xc nh gi tr ca cc ng vo v ra khighi dliu 1110 vo 01101.

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Bi tp 02

Cho mt b nh c dung lng 4Kx8 Cn bao nhiu ng d liu ng vo v ng

ra.

Cn bao nhiu ng a ch. Tnh tng sbyte c trong b nh.


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Kt ni gia CPU v b nh

Kt ni vi thit b iu khin (CPU)

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Kt ni vi thit b iu khin (CPU)

Vit d liu1. CPU cung cp a ch nh phn.2. CPU a dliu vo data bus3. CPU kch hot tn hiu iu khin ph hp.4. B nh s gii m a ch nh phn5. Data c a n a ch c chn.

c d liu1. CPU cung cp a ch nh phn.2. CPU kch hot tn hiu iu khin ph hp.3. B nh s gii m a ch nh phn4. B nh a dliu ph hp ln data bus


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ROM (Read Only Memories)

ROM l b nh bn dn c thit k lu dliu lu di.

Trong qu trnh hot ng, d liu khng thghi vo ROM nhng c th c ra tROM.

ROM c th c np d liu bi nh sn xuthoc ngi sdng.

D liu trong ROM khng b mt bi khi h thngb mt in

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ROM (Read Only Memories)


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ROM(tt)

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Cu trc ca ROM


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Cu trc ca ROM

Cu trc bn trong ca ROM rt phc tpnhng, bao gm nhng phn chnh sau: Ma trn thanh ghi: gm nhng thanh ghi lu

trd liu trong ROM. Mi thanh ghi chac mt tv c mt a ch tng ng.

Gii m a ch: bao gm gii m a ch hngv gii m a ch ct.

B m ng ra: d liu c chn s c

n b m ng ra khi CS mc thp. Khi CS mc cao, cc ng ra ca b m s trngthi tng tr cao.

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Gin thi gian


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MROM (Mask-programmed ROM)

MROM l ROM m dliu c nhp binh sn xut theo yu cu ca khch hng.

Phim m bn (mask) c sdng ktni trong ROM.

C hiu qu kinh tkhi sn xut vi slng ln

Cu trc ca mt MROM 16 bit nh nhsau

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Khi CE disabled, tt c ccchc nng ca chip s disabled.

Khi OE disabled, chnhngng ra 3 trng thi l disabled

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PROMs (Programmable ROMs )

PROM l cc loi ROM c th c lp trnh(np dliu) bi ngi sdng.

PROM c cu trc da vo cc kt ni nuchy(cu ch).

Khi np d liu cho ROM th chng trnhs nu chy cc kt ni tng ng.

PROM l loi ROM sdng mt ln. Kinh ttrong trng hp sdng vi s

lng nh


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Bipolar PROM ph bin l 74186, ROMny c cu trc gm 64 t8 bit.

TBP28S166 cng l mt bipolar PROM cdung lng 2K x 8.

MOS PROM c dung lng ln hn bipolarPROM. TMS27PC256 l mt MOS PROMc dung lng 32K x 8.


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EPROM (Erasable Programmable ROM)

EPROM c th c lp trnh bi ngi sdng v n cng c th c xa v lptrnh li.

Phi c mch np dliu chuyn dngdnh ring cho tng ROM.

Sdng tia UV xa dliu Tt c d liu trong EPROM s c xa S ca mt EPROM tiu biu (27C64)

nhsau:

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EPROM (Erasable Programmable ROM)


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EEPROM (Electrically Erasable PROM)

EPROM c hai nhc im chnh l: Chng ta phi tho chng ta khi mch

xa v lp trnh li. Mi ln xa v lp trnh li phi lm thc hin

cho ton b ROM Thi gian xa lu (khong 30 pht)

EEPROM c thkhc phc c nhng

nhc im trn. Sdng in p xa d liu C thxa d liu cho tng byte

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EEPROM (Electrically Erasable PROM)

EEPROM 2864 8K x 8


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CD ROM

CD ROM l nhng a c sn xut vimt b mt phn x.

Dliu s c lu trn a bng ccht chy hay khng mt khe trn b mta.

L phi php lu dmt cch kinh t

dung lng ln dliu

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Flash Memory


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28F256A flash memory

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S chc nng ca 28F256A


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ROM c th c sdng trong bt k ngdng no cn lu trd liu t hay khngthay i. Firmware: dng lu trdliu firmware cho cc h

thng microcomputer. Bng dliu: lu trdata cho nhng ng dng tra d

liu. Chuyn i dliu: lu trdata cho cc php bin i. To ra nhng hm bin i dliu.

ng dng ca ROM

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ng dng ca ROM

V d ng dng ca ROM trong mch tao ra sngdng hnh sin.


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Bi tp 03

Sdng ROM thit kmt mch c 3bit ng vo v ng ra thhin bnhphng gi tr ng vo.

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Bi tp 04

Sdng mch cho trang 21 thit lpmch thhin hm

y = 3x + 5


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Ngc li vi ROM, RAM bn dn l bnh c thghi v c c.

Nhc im chnh: dliu d b thay i. u im chnh: c thghi v c mt

cch nhanh chng v d dng.

RAM bn dn

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Cu trc ca RAM

Tng tnhROM, RAM cng bao gmmt sthanh ghi. Mi thanh ghi cha mttv c mt a ch duy nht.

Thng thng dung lng ca RAM l 1K,

4K, 8K, 16K, 64K, 128K, 256K. Kch thc ca mt ttrong RAM c th

l 1, 4 hay 8 bit.


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Cu trc ca RAM 64x4

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Cu trc ca RAM 64x4

RAM ny cha 64 t, mi t4 bit. Mi thanh ghi c mt a ch tng ng 0-

6310. Do vy cn tt c 6 ng a ch. 6 ng a ch c a qua mt b gii

m 6->64. Ng ra no mc cao th thanh ghi tng

ng c chn.


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Bi tp 05

Hy tnh dunglng ca nhngRAM sau y

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SRAM (Static RAM)

RAM tnh l RAM m dliu c lu trtrong RAM trong sut thi gian RAM ccp ngun.

Mi cell ca SRAM cha 1 bit v c cu

to tFlip-Flop. Gin thi gian truy cp SRAM nhsau:


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SRAM (Static RAM)

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SRAM (Static RAM)

Mt SRAM thc t6264C


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DRAM c chto tcng ngh MOS. So viSROM chng c dung lng cao hn v yu cucng sut cung cp tht hn.

Gi tr ca DRAM c lu trong nhng t in. Do sr r in tch ca t in nn DRAM yu

cu phi c np li in sau mt khong thigian nht nh.

Thng thng SRAM yu cu np li d liu sau2-10ms.

DRAM (Dynamic RAM)

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Cu trc ca DRAM


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Hot ng ca DRAM

Trong qu trnh ghi cng tc SW1 v SW2 ngcn SW3 v SW4 m.

Trong qu trnh c SW2, SW3, SW4 ng cn

SW1 m.

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M rng tv dung lng

M rng t Kt hp hai RAM 16x4

thng RAM 16x8


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M rng tv dung lng

M rng dung lng Kt hp hai chip 16x4

thnh b nh 32x4

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Bi tp 06

Tnh dung lng ca mch sau y


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Bi tp 07

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Bi tp 08


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Bi tp 09

TRAM 6206 (slide 41) hy thit kbnh 4K x 8.

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Cu hi?


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1

Chng 10Kt ni vi mch tng t


2

Kt ni vi mch tng t

Transducer: bin i i lng vt l thnh tn hiu in Analog-to-digital converter (ADC)

Digial system: x l tn hiu

Digital-to-analog converter (DAC)

Thc thi k t qu


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Bin i D/A

Nhiu phng php ADC s dng DACVrefc s dng xc nh ng ra full-

scale.

Trong trng hp tng qut, ng ra analog= K x gi tr s ng vo

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Bin i D/A

DAC 4 bit, ng ra in p tng t


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Ng ra tng t

Ng ra ca b bin i DAC khng honton l tn hiu analog bi v n ch xcnh mt sgi tr nht nh.

Vi mch trn, ng ra ch c thc nhnggi tr, 0, 1, 2, , 15 volt.

Khi sng vo tng ln th tn hiu ng ra

cng ging vi tn hiu tng t.

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Bc nhy

Bc nhy ca b bin i D/A c nhngha l khong thay i nh nht ca ng rakhi c s thay i gi tr ng vo.

B bin i D/A N bit: smc ng ra khc

nhau =2^N, s bc nhy =2^N-1 Bc nhy = K = Vref/(2^N-1)


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Bc nhy

Bc nhy = 1 volt

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V d 10-1

Cn sdng b DAC bao nhiu bit c thiu khin motor thay i tc mi 2 vng.

1000rpm/2rpm(per step) = 500 steps

2N - 1 > 500 steps. Suy raN= 9?


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Ng vo BCD

Trng s ca nhng ng vo khc nhau

Ng vo 2 s BCD

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Mch o Mch khng o

Vi

Mch bin i D/A

Tnh cht ca Opamp

Vo/Vi = 1+R2/R1Rin = infinity

Vo /Vi = - R2/R1Rin = R1


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V1

V2

V3

R1

R2

R3

Rf

Mch bin i D/A

Trng sca nhng ng vo khc nhau

Vo = -Rf(V1/R1 + V2/R2 + V3/R3)

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Mch bin i D/A

Bc nhy= |5V(1K/8K)| = .625VMax out = 5V(1K/8K + 1K/4K + 1K/2K + 1K/1K) = -9.375V


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B DAC 4 bit

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DAC vi ng ra dng in

Bin i dng sang p


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Mch bin i D/A

Vi nhng mch bin D/A trn, trngscc bit c xc nh da vo gi trca cc in tr.

Trong mt mch phi sdng nhiu intr vi nhng gi tr khc nhau B DAC 12 bit

in tr MSB = 1K in tr LSB = 1x212 = 2M

Mch sau ch sdng 2 gi tr in tr

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Mch bin i D/A

DAC R/2R


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DAC Thng sk thut

Nhiu b DAC c tch hp vo trongnhng IC, mt sthng stiu biu ca n Resolution: bc nhy ca b DACAccuracy: sai ssai sca b DAC Offset error: ng ra ca DAC khi tt c ng vo

bng 0

Settling time: thi gian yu cu DAC thchin bin i khi ng vo chuyn i ttrngthi all 0 n trng thi all 1

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IC DAC

AD7524 (Figure 11-9) CMOS IC

8 bit D/A

Sdng R/2R

Max settling time: 100 ns Full range accuracy: +/- 0.2% F.S.


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IC DAC

Khi ng vo CS v WR mc thp, OUT1 l ng ra analog.Khi c 2 mc cao, OUT1 c cht v gi tr nh phn ng vokhng c bin i ng ra.

OUT2 thng thng c ni t

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ng dng DAC

Control Sdng ng ra sca my tnh iu

chnh tc ca motor hay nhit .Automatic testing

To tn hiu tmy tnh kim tra mchannalog

Signal reconstruction Ti to tn hiu analog ttn hiu s. V d h

thng audio CDA/D conversion


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V d 10-2

Sdng DAC iu chnh bin catn hiu analog

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Bin i A/D

ADC miu t gi tr analog ng vo bnggi tr snh phn.

ADC phc tp v tn nhiu thi gian bin ihn DAC

Mt sADC sdng b DAC l mt phnca n Mt opamp c sdng lm b so snh

trong ADC


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Bin i A/D

B m nh phn c sdng nh l mtthanh ghi v cho php xung clock tng gi tr bm cho n khi VAXVA

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Hot ng ca bADC

Lng START bt u qu trnh bin i Control unit thay i gi tr nh phn trong thanh

ghi

Gi tr nh phn trong thanh ghi c bin ithnh gi tr nh phn VAX

B so snh so snh VAX vi VA. Khi VAX < VA, ngra b so snh mc cao. When VAX > VA, ng rac mc thp, qu trnh bin i kt thc, gi tr nhphn nm trong thanh ghi.

B phn iu khin s pht ra tn hiu end-of-conversion signal, EOC.


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Bin i A/D

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Bin i A/D

Dng sng thhin qu trnh mytnh thit lp mt chu trnh bin il lu gi tr nh phn vo b nh.


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Sai slng t

C thgim sai s lng tbng cch tng sbit nhng khng th loi b hon ton

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Khi phc tn hiu

Sau khi kt thc mt qu trnh ADC ta sc gi tr nh phn ca mt mu.

Qu tr khi phc tn hiu analog nhsau


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Khi phc tn hiu

Aliasing Nguyn nhn l do tn s ly mu khng ng Gii hn Nyquist

Tn s ly mu phi t nht ln hn 2 ln tn scaonht ca tn hiu ng vo.

Ly mu tn snh hn 2 ln tn sng vo sto nn k t qu sai khi khi phc tn hiu.

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Qu trnh ly mu khng ng


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ADC xp x lin tc (SDC)

Sdng rng ri hn ADC Phc tp hn nhng c thi gian bit i

ngn hn Thi gian bin i c nh, khng ph

thuc vo gi tr analog ng vo Nhiu SAC c tch hp trong nhng IC

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Successive-approximation ADC


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Successive-approximation ADC

SAC 4 bit sdng DAC c bc nhy 1 V

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ADC0804 SAC 8 bit


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ADC0804 SAC 8 bit

C hai ng vo analog cho php hai ng vo vi sai.

Ngng xc nh ti 1/2LSB. V d, bc nhy l10mV, bit LSB s trng thi 1 ti 5mV.

IC c thanh ghi xung clock bn trong to r a tn sf = 1/(1.1RC). Hoc c thsdng xung clock bnngoi.

Nu sxung clock c tn s606kHz, thi gian bini xp x 100us.

Sdng ni t ring bi v t ca thit b stnti nhiu do qu trnh thay i dng t ngt khithay i trng thi.

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ng dng ca IC ADC0804


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Flash ADC

Tc bin i cao Mch phc tp hn nhiu

Flash ADC 6 bit yu cu 63 b so snh tng t Flash ADC 8 bit yu cu 255 b so snh tng t Flash ADC 10 bit yu cu 1023 b so snh tng t

Thi gian bin i khng sdng xung clockdo vy qu trnh bin i l lin tc. Thi gian

bin i rt ngn ch khong 17 ns. B bin i flash 3 bit c miu t nhhnh sau

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Flash ADC 3 bit


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Mch ly mu v gi

40

Cu hi?


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1

Chng 9Cc mch sthng gp


2

Ni dung

Mch gii m/Mch m ha Mch ghp knh Mch phn knh Mch so snh Chuyn m Data Bus


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Gii m BCD - Decimal

Mt smch gii m khng sdng ttc 2N ng vo ca n

Mch gii m BCD Decimal c 4 ng vov 10 ng ra

Ng ra trng thi tch cc (mc thp)ch khi m BCD tng ng vi n ca n ng vo

Khi ng vo khng phi l gi tr BCD thkhng c ng ra no tch cc IC 74LS42 gii m BCD - Decimal

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Gii m BCD - Decimal


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Gii m BCD LED 7 on

LED 7 on c chto t7 LED thngthng

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IC hin th LED 7 on

LED


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LED v LCD

n LED (Light Emitting Diode) pht sngkhi c dng in chy qua n

LCD (Liquid Crystal Display) hin th tinhth lng

LCD hot ng vi tn hiu xoay chiuin p thp, tn sthp

n LED to ra nh sng mnh hn, LCDsdng t cng sut hn

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Nguyn l hot ng ca LCD

Control = low, ng ra ca EX-OR s ging ving vo. in p trn LCD = 0, LCD = off

Control = high, ng ra ca EX-OR s ngcvi sng ng vo. in p trn LCD l sngvung 5 v -5V, LCD trng thi on


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u tin trong m ha

Trong trng hp c nhiu ng vo trng thi tch cc th ng ra s tngng vi ng vo c trng scao nht

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V d mch m ha


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V d mch m ha

Sdng IC 74LS147 Cc cng tc tng ng vi cc nt nht

t0 n 9 Bnh thng tt c cc cng tc m, cc

ng vo trng thi cao, BCD ng ra l0000

Khi c mt phm nhn, mch s to ramt m BCD tng ng

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Mch ghp knh


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Mch phn knh 1 - 8

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Mch phn knh

IC gii m 74LS138 c th c sdng lmb phn knh vi ng vo E1 lm ng vo data


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Thng sdng v p

Mc in pVIH(min): in p ng vo mc cao: gi tr

in p thp nht cho mc logic 1 ng vo.VIL(max): in p ng vo mc thp: gi tr

in p cao nht cho mc logic 0 ng vo.VOH(min): in p ng ra mc cao: gi tr

in p thp nht cho mc logic 0 ng ra.VOL(max): ip p ng ra mc thp: gi tr

in p cao nht cho mc logic 0 ng ra.

4

Thng sdng v p

Mc dng in IIH: Dng in ng vo mc cao: dng in

chy vo mc logic 1 IIL: Dng in ng vo mc thp: dng in

chy vo mc logic 0 IOH: Dng in ng ra mc cao: dng in

chy ra mc logic 1 IOL: Dng in ng ra mc thp: dng in

chy ra mc logic thp


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H IC TTL

S mch ca cng NAND TTL

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IC TTL cng NAND

Ng ra trng thi thp


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IC TTL cng NAND

Ng ra trng thi cao

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Kt ni gia cc mch logic


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H IC TTL

Mch TTL c cu trc tng tnhtrn Ng vo l cathode ca tip gip PN

Ng vo mc cao s turn off mi ni v chc dng r r chy qua.

Ng vo mc thp s turns on mi ni v cdng tng i ln chy qua.

Phn ln mch TTL c cng cu trc ngra, tng tnhtrn.

16

Mch TTL cng NOR


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27

Cng ngh MOS

MOSFETs - Metal Oxide SemiconductorField Effect Transistors

Sn xut n gin v r hn Tiu tn t nng lng C ththc hin c nhiu mch hn D b tc ng bi tnh in

28

Cng ngh MOS

K hiu MOSFET knh N v knh P


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29

Cng ngh MOS

Trng thi ca MOSFET

30

Logic MOSFET knh N


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39

Cng Tristate

B m Tristate c sdng khi nhiutn hiu sdng chung bus

40

Giao tip gia cc IC

Li (driver) cung cp tn hiu ng ra. Ti (load) nhn tn hiu. Mch giao tip (interface circuit) kt ni

gia thit b li v ti. Kt ni gia nhng h IC khc nhau trong

cng mt mch.


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1

Chng 7B m v thanh ghi


2

B m khng ng b

Xt b m 4 bit hnh 7-1 Xung clock ch c a n FF A, ng vo J,

K ca tt c cc FF u mc logic 1. Ng ra ca FF sau c ni n ng vo CLK

ca FF trc n. Ng ra D, C, B, A l mt snh phn 4 bit viD l bit c trng scao nht.

y l b m khng ng b v trngthi ca cc FF khng thay i cng vixung clock.


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3

Hnh 7-1 B m 4 bit

4

Quy c v trng s

Trong phn ln cc mch, dng tn hiuthng chy ttri sang phi.

Trong chng ny, nhiu mch in cdng tn hiu chy tphi sang tri.

V d, trong hnh 7-1: Flip-Flop A: LSB

Flip-Flop D: MSB


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9

Tr trong b m khng ng b

Cu trc ca b m khng ng b khn gin nhng vn tr khi truyn tnhiu qua mi FF s lm hn chtn sca b m.

Vi b m khng ng b ta phi c TclockNx tpd

Fmax=1/(Nx tpd)

10

Tr trong b m khng ng b

B m 3 bitvi nhng tnsxung clockkhc nhau


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11

Cu hi?

Trong b m bt ng b, tt c cc FFthay i trng thi cng mt lc? Sai

Gi sb m trong hnh 7-1 ang trng thi 0101. Sau 27 xung clock, trngthi ca b m s l bao nhiu? 0000

SMOD ca b m c 5 Flip-Flop?

12

B m ng b

Trng thi ca tt c cc FF s c thayi cng mt lc vi xung clock.

Hnh sau m t hot ng ca mt bm ng b

Mi FF c ng vo J, Kc kt ni sao chochng trng thi cao ch khi ng ra ca ttc cc FF sau n u trng thi cao.

B m ng b c thhot ng vi tn scao hn b m khng ng b.


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13

B m ng b MOD-16

14

Hot ng ca mch

B thay i trng thi theo xung clock chkhi A = 1.

C thay i trng thi theo xung clock chkhi A = B = 1.

D thay i trng thi theo xung clock chkhi A = B = C = 1.


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15

Bng chn tr

16

B m c s MOD < 2N

Sdng ng vo khng ng b (clear,set) buc b m b i mt strngthi.

Trong hnh 7-2, ng ra cng NAND cni n ng vo khng ng b CLEARca mi Flip-Flop.

Khi A=0, B=C=1, (CBA = 1102= 610) ngra cng NAND s tch cc v cc FF s bCLEAR v trng thi 0.


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17

Hnh 7-2 B m MOD-6

18

Trng thi tm

Lu rng trong hnh 7-2, 110 l mttrng thi tm thi. Mch ch tn ti trng thi ny trong thi gian rt ngnsau s chuyn sang trng thi 000.

000

001

010

011

100

101

000 Ng ra ca FF C c tn sbng 1/6 tn s

ng vo.


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19

S trng thi

20

Thit kb m MOD-X

Bc 1: Tm sFF nh nht sao cho 2N X. Kt ni cc FF li vi nhau. Nu 2N = Xth khng lm bc 2 v 3.

Bc 2: Ni mt cng NAND n ng vo

CLEAR ca tt c cc FF. Bc 3: Xc nh FF s mc cao ng vi

trng thi b m = X. Ni ng ra ca ccFF n ng vo ca cng NAND.


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21

B m MOD-14 v MOD-10

B m khng ng b

22

B m MOD-14 v MOD-10

B m ng b


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23

B m thp phn

B m thp phn L bt k b m no c mi trng thi

phn bit.

B m BCD L mt b m thp phn m cc trng thi

trong b m tng ng t0000 (zero) n

1001 (9)

24

B m MOD-60 khng ng b


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25

V d 7-2

Xc nh mch b m ng b MOD-60

26

Cu hi?

Trong b m MOD-13, ng ra ca FF noc ni n ng vo cng NAND camch clear?

Tt c cc b m BCD l b m thp

phn? Cho mt b m thp phn, tn sng

vo l 50KHz. Tn sng ra l bao nhiu?


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27

B m xung khng ng b

111110101100011010001000

B m ln c thchuyn thnh b mxung bng cch sdng nhng ng rao li cc ng vo xung clock.

28

B m xung MOD-8


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29

B m xung ng b

B m xung ng b c cu to honton tng tnhb m ln ng b.

Ch khc l sdng cc ng ra o iu khin.

30

B m ln/xung ng b


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31

IC b m khng ng b

IC 74LS293 h TTL C 4 J-K Flip-Flop, Q3Q2Q1Q0 Mi FF c mt ng vo CP (clock pulse) tng

tnhng vo CLK. Ng vo clock ca Q1 vQ0 c ni n chn 11 v chn 10.

Ng vo clear ca mi FF c ni n ng raca mt cng NAND hai ng vo MR1 v MR2.

Q3Q2Q1 c ni vi nhau hnh thnh nn mtb m 3 bit.

Q0 khng c ni n cc phn khc.

32

IC b m khng ng b


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33

V d 7-3

Sdng IC 74LS293 thit lp b mMOD-16

34

V d 7-4



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35

V d 7-5

Sdng IC 74LS293 thit lp b m MOD-14 Trng hp ny cn phi sdng thm mt

cng AND

36

V d 7-6


Sdng 2 IC mc ni tip nhau.


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37

IC b m

Gii thiu IC b m 7 bit h CMOS

38

IC b m ng b

TTL 74ALS160 B m c 4 FF FF thay i trng thi theo cng dng ca

xung clock

IC c mt ng vo CLEAR khng ng b. B m c thpreset n bt k gi tr no

(theo cc ng vo A, B, C, v D) bng cchtch cc ng vo LOAD.

B m c iu khin bi cc ng vo khcnhau, thhin trong bng Function table.


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39

74ALS160

40

74ALS160

V d


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41

74ALS160

V d

42

74ALS160

C thkt hp nhiu IC to ra mt bm ln hn


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43

IC b m ng b

74ALS193/HC193 B m ng b MOD-16

Presettable up/down

Preset khng ng b Master reset khng ng b

44

IC 74ALS193


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45

IC 74ALS193

Mch to ra TCU v TCD

46

74ALS193 - b m ln


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47

74ALS193 - b m xung

48

V d 7-7

B m xung MOD-5


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49

IC 74ALS193

Kt hp nhiu IC m rng b m

50

Gii m b m

Gii m l bit i gi tr nh phn ngra thnh gi tr thp phn.

B gii m sau tch cc mc cao, c thsdng cc n LED thhin cc s

thp phn t0 7.


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51

Gii m b m MOD-8

52

Thit kb m ng b

Bng chuyn i trng thi ca JK-FF

0X1111

1X0110X1100

1

X00000

KJK tipHin tiChuyn i


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53

Trnh tthit k

Bc 1: Xc nh sbit (sFF cn) Bc 2: v s chuyn i ca tt c

cc trng thi, bao gm c nhng trngthi khng xut hin trong chu trnh.

Bc 3: da vo s chuyn i trngthi thit lp mt bng, trong lit

k tt c cc trng thi hin ti v ktip.

54

Trnh tthit k

Bc 4: trong bng va to, thm ct gitr J, K ca cc FF. Vi mi trng thi hinti, xc nh gi tr ca J v Kb mchuyn n trng thi ktip.

Bc 5: thit kmch logic to ra ccmc logic cho mi ng vo J v K. Bc 6: xc nh s mch.


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55

V d, thit kb m MOD-5

S chuyn i trng thi 000001010011100000

56

Trng thi hin ti v ktip

0001118

0000117

0001016

0000015

0011104

1100103

0101002

100000Line 1

ABCABC

NEXTPRESENT


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57

Bng trng thi ca mch

1x1x1x0001118

x01x1x0000117

1xx01x0001016

x0x01x0000015

1x1xx10011104

x10xx01100103

1xx1x00101002

x1x0x0100000Line 1

KAJAKBJBKCJCABCABC

NEXTPRESENT

58

Xc nh cc gi tr J v K

Tnh gi tr ca JA


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59

Tng tta c

60

S mch


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61

V d 7-8

Sdng D-FF thit kb m MOD-5

xxx000111

0xx000011

x0x000101

00x000001

xx1001110

1x0110010

x10010100

100100000

DADBDCABCABC

NEXTPRESENT

62

V d 7-8

Xc nh gi tr cc ng vo D


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63

V d 7-8

S mch

64

Mch thanh ghi tch hp

Thanh ghi c th c phn loi da vocch d liu c a vo v cch m dliu c ly ra: Parallel in/parallel out (PIPO)

Serial in/serial out (SISO) Parallel in/serial out (PISO)

Serial in/parallel out (SIPO)


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65


Vo song song, ra song song :

IC 74174 v 74178

1 0 1 0 1 1 0 1

1 0 1 0 1 1 0 1

Parallel out

Parallel inMSB LSB

66

PIPO 74ALS174/74HC174

74ALS174/74HC174 Thanh ghi 6 bit

D5, D0: ng vo song song

Q5, Q0: ng ra song song

D liu c a vo thanh ghi theo cnhdng ca xung clock Master reset c threset tt c cc FFs

mt cch c lp vi xung clock


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67


68


74ALS174 c sdng lm b thnhghi dch


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69


Vo ni tip, ra ni tip:

IC 4731B

1 0 1 0 1 1 0 1

Serial In Serial out

0 1 1 0

70

SISO - 4731B


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71


Vo song song, ra ni tip:

IC 74165,74LS165,74HC165

1 0 1 0 1 1 0 1

1 0 1 0 1 1 0 1

Parallel InMSB LSB

Serial Out

1 0

72

PISO - 74HC165

74HC165 Thanh ghi 8 bit

D liu ni tip c a vo DS D liu song song khng ng b c a

vo qua P0 - P7 Ch c ng ra Q7 c sdng

CP l ng vo xung clock

CP INH ng vo ngn xung clock SH/LD ng vo load d liu


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73

PISO - 74HC165

74


Vo ni tip, ra song song:

IC 74164,74LS164,74HC164

Serial In

1 0 1 0 1 1 0 1

1 0 1 0 1 1 0 1

0 1

Parallel outMSB LSB


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75

SIPO 74ALS164/74HC164

74ALS164 Cha thanh ghi dch 8 bit A v B l hai ng vo ca mt cng AND, ng

ra ca cnf AND l u vo ni tip.

Qu trnh dch xy ra khi c cnh m caxung clock

76

IC 74ALS164


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77

V d IC 74ALS164

78

Thanh ghi dch

3684368363

V d v thanhghi dch trongmy tnh


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79

B m thanh ghi dch

B m vng (FF cui ni n FF u tin) FF cui cng s dch gi tr ca n n FF u

tin

D-FF c sdng (JK-FF cng c th csdng)

Phi bt u vi trng thi ch c mt FF cgi tr 1 v nhng ci cn li trng thi 0.

80

Dch vng MOD-4


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81

Dch vng 4 bit

B m MOD-4

82

Cu hi?


148/278

1

Chng 6Mch shc

Th.S ng Ngc Khoa

Khoa in - in T

2

Mch shc

ALU (arithmetic/logic unit) s ly data ttrongb nh thc thi nhng lnh theo control unit


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3

Mch shc

V d qu trnh mt lnh c thc thi: n v iu khin ra lnh cng mt s c ch nh

trong b nh vi sc trong thanh ghi accumulator.

Scng c truyn tb nh n thanh ghi B.

D liu trong thanh ghi B v thanh ghi accumulators c cng li vi nhau.

Kt qu s c lu vo trong thanh ghi accumulator

Gi tr trong thanh ghi accumulator s c gichon khi c lnh mi.

4

B cng nh phn song song

A, B l gi tr cn cng. C l gi tr nh. Sl kt qu ca php cng


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5

Qu trnh x l php cng

6

V d 6-1

Hy thit kmt b cng y : B cng c 3 ng vo

2 ng vo thhin scn cng

1 ng vo cha snh ng vo

C 2 ng ra 1 ng ra l kt qu ca php cng

1 ng ra l snh ng ra


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7

V d 6-1 Gii

Bng chn tr

8

V d 6-1 Gii

S mch kt qu


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9

V d 6-2

Gii thch hot ngca mch sau

10

IC b cng

IC 74HC283 l IC b cng song song 4 bit

A v B l hai s4 bit

C0 l snh ng vo, C4 l snh ng ra


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11

IC b cng

Ta c thni tip hai b cng 4 bit tora mt b cng 8 bit

12

B cng BCD

C thm phn mch x l trng hptng ln hn 9

(18)01001

(17)10001

(16)00001

(15)11110

(14)01110

(13)10110

(12)00110

(11)11010

(10)01010

S0S1S2S3S4


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13

B cng BCD

X=S4+S3(S2+S1)

14

B cng BCD ni tip


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15

IC ALU

ALU c ththc thi nhiu ton tv hmlogic khc nhau, cc ton tv hm nyc xc nh bi mt m ng vo.

74LS382 (TTL) v HC382 (CMOS) l thitb ALU tiu biu c ththc hin 8 hmkhc nhau.

16

IC ALU


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17

V d 6-3

Hy sdng 2 IC 74LS382 to thnhb cng 8 bit

18

Cu hi?


157/278

1

Chng 5Flip Flops


2

Gii thiu

S h thng stng qut bao gmthnh phn nh v cc cng logic


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3

Flip-Flops

Thnh phn nh ph bin nht l cc Flip-flop, flip-flop c cu thnh tnhngcng logic n gin.

K hiu tng qut ca mt flip-flop

4

Mch cht cng NAND

Mch cht cng NAND l mt flip-flop n gin. Mch cht c hai ng vo l set v clear (preset). Ng vo tch cc mc thp, ng ra s thay i

trng thi khi c xung thp ng vo. Khi mch trng thi set

Khi mch trng thi clear (preset)

1Qv0Q ==

0Qv1Q ==


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5

Mch cht cng NAND

Mch cht cng NAND c hai trng thi n

nh (trng thi cht) ng vi trng hpSET = CLEAR = 1.

6

Trng thi SET mch cht

Khi ng vo SET chuyn ttrng thi caoxung trng thi thp, trong c hai trnghp ng ra Q s trng thi cao


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7

Trng thi clear mch cht

Khi ng vo CLEAR chuyn ttrng thi caoxung trng thi thp, trong c hai trnghp ng ra Q s trng thi thp

8

Mch cht cng NAND

SET = RESET = 1. Trng thi n nh, ng ra vn gitrng thi trc .

SET = 0, RESET = 1. Q mc cao. SET = 1, RESET = 0. Q mc thp. SET = RESET = 0. Ng ra khng c xc nh chnh

xc do c hai trng thi set v clear cng tc ng.


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9

M t tng ng mch cht

Ng ra mch cht nh trng thi trc v

ng ra ch c ththay i khi mt trong haing vo trng thi tch cc

10

V d 5-1

Kha chng ny


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11

Mch cht cng NOR

Tng tnhmch cht cng NAND ch khc vtr hai ng ra Q v Q c thay i cho nhau.

Ng vo tch cc mc cao

12

Dng sng mch cht cng NOR


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13

V d 5-2

Khi mt ngun nh sng h thng s bo ng. Cng tc SW1 dng reset h thng

14

ng b v bt ng b

H thng sc thhot ng trng thi: Bt ng b (Asynchronously): Trng thi ng

ra s thay i khi c bt k sthay i no ng vo.

ng b (Synchronously): Ng ra ch thay iti nhng thi im c cnh xung clock (ngb vi cch xung clock)


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15

Xung clock

Vi h thng ng b, ng ra thay i trngthi ti nhng thi im c cnh xung clock. Cnh xung dng Positive-going transitions (PGT) Cnh xung m: Negative-going transitions (NGT)

16

Flip-Flops v xung clock

Trong cc FF c ng vo xung clock (CLK) (a) Xung clock tch cc cnh dng (b) Xung clock tch cc cnh m


165/278

17

Dng sng ca SC-FF

18

SC-FF tch cc cnh m


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19

Cu trc bn trong SC-FF

Bao gm: Mch pht hin cnh xung Mch thit lp trng thi Mch cht cng NAND

20

Mch pht hin cnh xung

Pht hin cnh dng Pht hin cnh m


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21

JK-FF

Hot ng ging SC-FF. J l ng set, K lng clear

Khi c J v Ku mc cao, ng ra s otrng thi so vi trng thi trc .

C thtch cc cnh dng hay cnh mxung clock.

22

JK-FF


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23

JK-FF tch cc cnh m

24

Cu trc bn trong ca JK-FF

Khc nhau duy nht gia JK v SC-FF lJK c phn hi tip tn hiu.


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25

D Flip-Flop

Ch c mt ng vo D, tng ng vi ngvo data.

Ng ra Q s c cng gi tr vi ng vo Dkhi c tc ng ca cnh xung clock.

Trong nhng thi im khc, D-FF s lugi tr trc ca n.

c sdng trong ng dng truyn dliu song song

26

D Flip-Flop


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27

D-FF v JK-KK

C thto ra D-FF tJK-FF

28

Truyn dliu song song


171/278

29

Mch cht D

Khng c mch pht hin cnh xung Ng vo xung clockc thay bng ng

vo enable

Ng ra c xc nh theo ng vo ch khienable mc cao

30

Mch cht D


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31

V d 5-3

32

Ng vo khng ng b

S, C, J, K v D c gi l nhng ng vong b bi v nh hng ca chng ngb vi xung clock.

Ng vo khng ng b hot ng c

lp vi nhng ng vo ng b, chng cthset (1) hoc clear (0) Flip-Flop vo btk thi im no.


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33

JK-FF vi ng vo khng ng b

34

V d 5-4


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35

ng dng ca Flip-Flop

36

ng dng ca FF

Mt s ng dng ca flip-flop B m Lu d liu nh phn Truyn d liu nh phn gia cc thit b


175/278

37

ng b tn hiu

a sh thng hot ng ch ngb.

Cc tn hiu tnhin l nhng tn hiukhng ng b.

Chng ta phi ng b nhng tn hiu nyvi xung clock.

38

ng b tn hiu

Tn hiu khng ng b A c thto ranhng mu xung khng ng.


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39

ng b tn hiu

40

Lu v truyn dliu

FF thng c sdng lu v truyndliu dng nh phn.

Nhm FF sdng lu data l thanh ghi Dliu c truyn khi data chuyn i

gia nhng FF hoc thanh ghi. Trong trng hp truyn ng b, cn

phi c xung ng b


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41

Truyn dliu ng b

42

Truyn dliu song song


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43

Thanh ghi dch

Trong trng hp ny dliu s ctruyn ni tip.

44

Truyn data gia hai thanh ghi


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45

Chia tn s

46

B m


180/278

47

Cu hi?


181/278

1

Chng 4Mch logic


2

Biu din bng biu thc i s

Mt hm logic n bin bt k lun c thbiu din di dng: Tng ca cc tch (Chun tc tuyn - CTT):

l dng tng ca nhiu thnh phn m mithnh phn l tch ca y n bin.

Tch ca cc tng (Chun tc hi CTH): ldng tch ca nhiu thnh phn m mithnh phn l tng ca y n bin.


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3

Dng chun tc tuyn

F=ABC+ ABC + ABC + ABC

Dng chun tc hi

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


= )6,5,2,1(FA B C F

0

0

00

00

000 0

0

0

0

0

1

1

11

1

11

1

1

0

111

11

0

1

1

0

3

2

1

4

7

6

5

V tr

= )7,4,3,0(F

4


X = 1 ghi XX = 1 ghi X

X = 0 ghi XX = 0 ghi X

Lu cc gi tr 0Lu cc gi tr 1

Tch ca cc tngTng ca cc tch

Chun tc hiChun tt tuyn


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5

Rt gn mch logic

Lm cho biu thc logic n gin nht v dovy mch logic sdng t cng logic nht.

Hai mch sau y l tng ng nhau

6

Phng php rt gn

C hai phng php chnh rt gnmt biu thc logic. Phng php bin i i s: sdng

cc nh l v cc php bin i Boolean rt gn biu thc.

Phong php ba Karnaugh: sdng baKarnuagh rt gn biu thc logic


184/278

7

Phng php bin i i s

Sdng cc nh l v cc php bin iBoolean rt gn biu thc.

V d:

BD(A+B)(A+B+D)DBC+AD(B+C)AC(ABD)+ABCD+ABC

A(B+C)ABC+ABC+ABC

A(B+C)ABC+AB(AC)

Rt gnBiu thc ban u

?

8

V d 4-1

Hy rt gn mch logic sau


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9

Bi ton thit k

Hy thit kmt mch logic c: Ba ng vo

Mt ng ra Ng ra mc cao ch khi a sng vo

mc cao

10

Trnh tthit k

Bc 1: Thit lp bng chn tr.

1111

1011

1101

0001

1110

0010

0100

0000

xCBA

Mchlogic

A

BC

x


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11

Trnh tthit k

Bc 2: Thit lp phng trnh tbngchn tr.

1111

1011

1101

0001

1110

0010

0100

0000

xCBA

A.B.C

A.B.C

A.B.C

A.B.C

ABCCABCBABCAx +++=

12

Trnh tthit k

Bc 3: Rt gn biu thc logic

ABACBCxABCCABABCCBAABCBCAx

ABCCABCBABCAx

++=+++++=

+++=


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13

Trnh tthit k

Bc 4: V mch logic ng vi biu thclogic va rt gn

ABACBCx ++=

14

V d 4-1

Hy thit kmt mch logic c 4 ng voA, B, C, D v mt ng ra. Ng ra ch mc cao khi in p (c miu t bi 4bit nh phn ABCD) ln hn 6.


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15

Kt qu

16

V d 4-3

Thit kmch logic iu khin mch phunnhin liu trong mch t nhsau:

Cm bin c kh cn t

Cm bin ngn la gia A v B


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17

Ba Karnaugh

18

Phng php ba Karnaugh

Ging nhbng chn tr, ba Karnaugh l mt cchthhin mi quan h gia cc mc logic ngvo v ng ra.

Ba Karnaugh l mt phng php c sdng n gin biu thc logic.

Phng php ny d thc hin hn phng phpi s. Ba Karnaugh c ththc hin vi bt k sng

vo no, nhng trong chng trnh ch kho st sng vo nh hn 6.


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19

nh dng ba Karnaugh

Mi mt trng hp trong bng chn trtng ng vi 1 trong ba Karnaugh

Cc trong ba Karnaugh c nh ssaocho 2 k nhau ch khc nhau 1 gi tr.

Do cc k nhau ch khc nhau 1 gi trnn chng ta c thnhm chng li to

mt thnh phn n gin hn dng tngcc tch.

20

Bng chn tr K-map

Y

0

10

1

Z

1

01

1

X

0

01

1

Gi tr 0

Gi tr 1

Gi tr 2

Gi tr 3

1

1

0

1

Mt v d tng ng gia bng chn tr vba Karnaugh

0

1

2

3

Y

Y

X XZ


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21

Y

Y

X X

0

0

1

0

X Y

Y

Y

X X

0

0

0

1 X Y

Y

Y

X X

1

0

0

0

X Y

Y

Y

X X

0

1

0

0 X Y

Xc nh gi tr cc

22

Nhm cc k nhauX Y

Y

Y

X X

1

0

1

0

X Y

YY

Y

X X

1

0

1

0

Z = X Y + X Y = Y ( X + X ) = Y


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23

Nhm cc li vi nhau

Nhm 2 1 k nhau, loi ra bin xuthin c hai trng thi b v khng b.

Nhm 4 1 k nhau, loi ra 2 bin xuthin c hai trng thi b v khng b.

Nhm 8 1 k nhau, loi ra 3 bin xuthin c hai trng thi b v khng b.

24

Y

Y

X X

1

1

0

0

XY

Y

X X

1

0

1

0

Y

K-map 2 bin: nhm 2

Y

Y

X X

0

1

0

1 Y

Y

Y

X X

0

0

1

1X


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Y

Y

X X

1

1

1

1

1

K-map 2 bin: nhm 4

26

V d K-map 2 bin

S

0

1

0

1

T

1

0

1

0

R

0

0

1

1

1

0

1

0

T = F(R,S) = S

S

S

S

R R

0

1

2

3

T


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27

K-map 3 bin

0

1

2

3

4

5

6

7

C

0

1

0

1

0

1

01

Y

1

0

1

1

0

0

10

B

0

0

1

1

0

0

11

A

0

0

0

0

1

1

11

1

0

1

1

1

0

0

0

0

1

2

3

6

7

4

5

C

C

A B A B A BA BY

28

K-map 3 bin: nhm 2

C

C

A B A B A BA B

0

0

0

0

0

1

0

1

A C

0

0

0

1

0

1

0

0

B C

0

0

0

0

1

1

0

0

A B

0

0

0

0

0

0

1

1

A B

0

0

1

1

0

0

0

0

A B

1

1

0

0

0

0

0

0

A B

0

1

0

0

0

0

0

1

B C

1

0

0

0

0

0

1

0

B C0

0

1

0

1

0

0

0

B C

0

0

0

0

1

0

1

0

A C

0

1

0

1

0

0

0

0

A C

1

0

1

0

0

0

0

0

A C


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29

K-map 3 bin: nhm 4

C

C

A B A B A BA B

0

0

0

0

1

1

1

1

A

0

0

1

1

1

1

0

0

B

0

1

0

1

0

1

0

1

C

1

1

1

1

0

0

0

0

A

1

0

1

0

1

0

1

0

C

1

1

0

0

0

0

1

1

B

30

K-map 3 bin: nhm 8

C

C

A B A B A BA B

1

1

1

1

1

1

1

1

1


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31

Ba Karnaugh 4 bin

01111

0011111011

00011

11101

00101

01001

10001

01110

10110

01010

10010

01100

00100

01000

10000

FDCBA

ABCD

00 01 11 10

00

01

11

10

F

32

Ba Karnaugh 4 binAB

CD00 01 11 10

00

01

11

10

F

Lu cc khiu trong

ba Karnaugh


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33

Ba Karnaugh 4 bin

0

0

0

1

1

0

0

1

0

0

1

0

0

1

0

1

01111

0011111011

00011

11101

00101

01001

10001

01110

10110

01010

10010

01100

00100

01000

10000

FDCBA

ABCD

00 01 11 10

00

01

11

10

F

34

K-map 4 bin: nhm 2

1

0

0

0

0

0

0

0

0

0

0

1

1

0

0

1

AB

CD00 01 11 10

00

01

11

10

FACD

BCD


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35

K-map 4 bin: nhm 4

0

0

1

0

0

0

1

0

0

0

1

0

0

0

1

0

ABCD

00 01 11 10

00

01

11

10

F

CD

36

K-map 4 bin: nhm 4

0

0

0

0

0

1

1

0

0

1

1

0

0

0

0

0

AB

CD00 01 11 10

00

01

11

10

F

BD


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37

K-map 4 bin: nhm 4

1

1

0

0

0

0

0

0

0

0

0

0

1

1

0

0

ABCD

00 01 11 10

00

01

11

10

F

BC

38

K-map 4 bin: nhm 4

1

0

0

1

0

0

0

0

0

0

0

0

1

0

0

1

AB

CD00 01 11 10

00

01

11

10

F

BD


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K-map 4 bin: nhm 8

0

0

0

0

1

1

1

1

1

1

1

1

0

0

0

0

ABCD

00 01 11 10

00

01

11

10

F

B

40

K-map 4 bin: nhm 8

1

1

1

1

1

1

1

1

0

0

0

0

0

0

0

0

AB

CD00 01 11 10

00

01

11

10

F

A


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K-map 4 bin: nhm 8

1

1

1

1

0

0

0

0

0

0

0

0

1

1

1

1

ABCD

00 01 11 10

00

01

11

10

F

B

42

Rt gn bng ba Karnaugh

Bc 1: Biu din hm cho trn baKarnaugh.

Bc 2: Nhm cc c gi tr bng 1 theocc quy tc:

Tng cc l ln nht. Tng cc phi l 2n (n nguyn). Cc ny phi nm k nhau.


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Rt gn bng ba Karnaugh

Bc 3: Lm li bc 2 cho n khi tt ccc logic 1 u c sdng.

Bc 4: Xc nh kt qu theo cc quy tc: Mi nhm s l mt tch ca cc bin. Kt qu l tng ca cc tch trn.

44

V d 4-4

J L

J K JKL

M = F(J,K,L) = J L + J K + J K L

1

0

1

1

0

0

0

1

L

L

J K J KJ K

0

1

2

3

6

7

4

5

J KM


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V d 4-5

A C

Z = F(A,B,C) = A C + B C

1

0

0

0

0

1

1

1

C

C

A B A B A BA B

0

1

2

3

6

7

4

5

B C

46

V d 4-6

1

1

0

1

1

1

0

0

C

C

A B A B A BA B

A B A B

A C

B C

F1 = F(A,B,C) = A B + A B + A C

F2 = F(A,B,C) = A B + A B + B C


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V d 4-7

0

1

4

5

12

13

8

9

3

2

7

6

15

14

11

10

W X W X W XW X

Y Z

Y Z

Y Z

Y Z

0

1

0

1

0

0

0

1

1

0

1

0

1

1

0

0

W X Y

X Y ZW Z

F1 = F(w,x,y,z) = W X Y + W Z + X Y Z

48

V d 4-8

Rt gn biu thc sau y:

f(A,B,C,D) =(2,3,4,5,7,8,10,13,15)

11

111

11

11

AB

CD00 01 11 10

00

01

11

10

F


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V d 4-8

11

111

11

11

AB

CD00 01 11 10

00

01

11

10

F

BD

ABCABD

ABC

f(A,B,C,D) =BD + ABC + ABD + ABC

50

Trng thi Dont Care

Mt smch logic c c im: vi mtsgi tr ng vo xc nh, gi tr ng rakhng c xc nh c th.

Trng thi khng xc nh ca ng rac gi l trng thi Dont Care.

Vi trng thi ny, gi tr ca n c th l0 hoc 1.

Trng thi Dont Care rt tin li trongqu trnh rt gn ba Karnaugh.


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V d trng thi Dont Care

52

V d 4-9

Z01010101

01010101

F21x100x0x

x101x111

Y00110011

00110011

X00001111

00001111

W00000000

11111111

Y Z

F2 = F(w,x,y,z) = X Y Z + Y Z + X Y

X Y Z

X Y

0

1

4

5

12

13

8

9

3

2

7

6

15

14

11

10

W X W X W XW X

Y Z

Y Z

Y Z

Y Z

X

X

1

1

1

1

1

0

1

0

X

X

0

X

1

0


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V d 4-10

Xc nh biu thc cho bngchn tr sau y

D0101010101010101

F01011x010001xxxx

C0011001100110011

B0000111100001111

A0000000011111111

Dng chun tc tuynf(A,B,C,D)= (1,3,4,7,11) + d(5,12,13,14,15)

Dng chun tc hif(A,B,C,D)= (0,2,6,8,9,10)D(5,12,13,14,15)

54

V d 4-10

f(A,B,C,D)= (1,3,4,7,11) + d(5,12,13,14,15)

f(A,B,C,D)= (0,2,6,8,9,10)D(5,12,13,14,15)

x

1x11

xx1

x1

AB

CD00 01 11 10

00

01

11

10

F

0x00

x

0xx

0x0

AB

CD00 01 11 10

00

01

11

10

F

CTT CTH


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V d 4-10

x

1x11

xx1

x1

AB

CD00 01 11 10

00

01

11

10

F

0x00

x

0xx

0x0

AB

CD00 01 11 10

00

01

11

10

F

f(A,B,C,D) =CD +f(A,B,C,D) =CD + BC +f(A,B,C,D) =CD + BC + ADf(A,B,C,D) =(B+D)f(A,B,C,D) =(B+D)(A+C)f(A,B,C,D) =(B+D)(A+C)(C+D)

56

K-map 5 bin

f(A,B,C,D,E) = (0,2,4,7,10,12,13,18,23,26,28,29)BC

DE00 01 11 10

00

01

11

10

F BC

DE00 01 11 10

00

01

11

10

FA=0 A=1

0

1

3

2

4

5

7

6

12

13

15

14

8

9

11

10

16

17

19

18

20

21

23

22

28

29

31

30

24

25

27

26


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K-map 5 bin

f(A,B,C,D,E) = (0,2,4,7,10,12,13,18,23,26,28,29)BC

DE00 01 11 10

00

01

11

10

F BC

DE00 01 11 10

00

01

11

10

FA=0 A=1

1

1

1

1

1

1

1 1

1

1

1

1

58

K-map 5 bin

f(A,B,C,D,E) = (0,2,4,7,10,12,13,18,23,26,28,29)

11

1

1

111

BC

DE00 01 11 10

00

01

11

10

F

11

1

1

1

BC

DE00 01 11 10

00

01

11

10

FA=0 A=1

ABDEBCD

BCDE

CDE

f(A,B,C,D) =ABDE+BCD+BCDE+CDE


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Cng EX-OR

Cng EX-OR c hai ng vo. Ng ra ca cng EX-OR mc cao ch khi

hai ng vo c gi tr khc nhau.

60

Cng EX-OR


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IC EX-OR 74LS86

62

Cng EX-NOR

Cng EX-NOR c hai ng vo. Ng ra ca cng EX-NOR mc cao ch

khi hai ng vo c gi tr ging nhau.


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63

Cng EX-NOR

64

V d 4-11

Sdng cng EX-NOR n gin mchlogic sau


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65

Mch to v kim tra parity

66

Mch Enable/Disable


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c im ca IC s

IC c cu to tcc in tr, diode,transistor, cc linh kin ny c t trnmt lp bn dn lm nn.

trnh cc tc ng c hc, ha hc, ICc ng trong nhng v silicon hocplastic.

Chip thc tnh hn hnh dng ca nrt nhiu

68

Dng c bn ca IC s

Dng hai hng chn song song


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Dng c bn ca IC s

Dng hai hng chn song song

70

Dng c bn ca IC s

Dng ng v hp (flat pack)


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gn IC

thun li trong qu trnh lp rp vthay i, IC thng c gn trn cc .

72

Mch stch hp (IC)

1,000,000 or moreGiga-scale integration (GSI)

100,000 to 999,999Ultra large-scale integration (ULSI)

10,000 to 99,999Very large-scale integration (VLSI)

100 to 9999Large-scale integration (LSI)

12 to 99Medium-scale integration (MSI)


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IC sBipolar v Unipolar

IC s c phn thnh IC bipolar v ICunipolar.

IC bipolar l nhng IC c to thnh tnhng transistor BJT (PNP hoc NPN)

IC unipolar c to thnh tnhngtransistor hiu ng trng (MOSFET)

74

Cng NOT bipolar v unipolar


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H IC

IC s c phn thnh hai loi chnh lTTL v CMOS.

H TTL l nhng IC bipolar (bng 4-1) H CMOS l nhng IC unipolar (bng 4-2)

76

H TTL (Bng 4-1)

74ALS0474ALSAdvanced low-powerSchottky TTL

74AS0474ASAdvanced Schottky TTL

74LS0474LSLow-power Schottky TTL

74S0474SSchottky TTL

7404 (NOT)74Standard TTL

V d ICK hiuPhn loi TTL


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H CMOS (Bng 4-2)

74ACT0274ACTAdvanced-performance CMOS, not pin butelectrically compatible with TTL

74AC0274ACAdvanced-performance CMOS, not pin orelectrically compatible with TTL

74HCT0274HCTSilicon-gate, high-speed, pin-compatible and electrically compatiblewith TTL

74HC0274HCSilicon-gate, pin-compatible with TTL,high-speed

74C0274CMetal-gate, pin-compatible with TTL

4001 (NOR)40Metal-gate CMOS

V d ICK hiuPhn loi CMOS

78

Ngun cung cp v ni t

c thsdng c nhng IC stacn phi cung cp ngun cho n. Chn ngun (power) k hiu l VCC cho h

TTL v VDD cho h CMOS. Chn t (ground)


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Mc in p TTL

Maximum 5.0 V

4.0 V

3.0 V

2.0 V

1.0 V

Minimum 0.0 V

Input

Voltage5.0 V Maximum

4.0 V

3.0 V

2.0 V

1.0 V

0.0 V Minimum

Output

Voltage

UndefinedRegion

Typical 3.5 V

Typical 0.1 V

UndefinedRegion

2.0 V

HIGH HIGH

2.4 V

0.8 V

LOW0.4 V

LOW

80

Mc nhiu TTL

Mc nhiu (0.4 V)

Mc nhiu (0.4 V)

Maximum 5.0 V

4.0 V

3.0 V

2.0 V

1.0 V

Minimum 0.0 V

InputVoltage

5.0 V Maximum

4.0 V

3.0 V

2.0 V

1.0 V

0.0 V Minimum

OutputVoltage

UndefinedRegion

Typical 3.5 V

Typical 0.1 V

UndefinedRegion

2.0 V

HIGH HIGH

2.4 V

0.8 V

LOW0.4 V

LOW


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Mc in p CMOS

Maximum 5.0 V

4.0 V

3.0 V

2.0 V

1.0 V

Minimum 0.0 V

InputVoltage

1.0 V

LOW

3.5 V

HIGH

UndefinedRegion

5.0 V Maximum

4.0 V

3.0 V

2.0 V

1.0 V

0.0 V Minimum

OutputVoltage

HIGH4.9 V

LOW

0.1 V

UndefinedRegion

82

Mc nhiu CMOS

Mc nhiu (0.9 V)

Mc nhiu (1.4 V)

Maximum 5.0 V

4.0 V

3.0 V

2.0 V

1.0 V

Minimum 0.0 V

InputVoltage

1.0 V

LOW

3.5 V

HIGH

Undefined

Region

5.0 V Maximum

4.0 V

3.0 V

2.0 V

1.0 V

0.0 V Minimum

OutputVoltage

HIGH4.9 V

LOW

0.1 V

Undefined

Region


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Ng vo khng kt ni

Vi h TTL, ng vo khng kt ni lmvic ging nhmc logic 1, tuy nhin khio th in p DC ti chn nm trongkhong 1,4 1,8V.

Vi hc CMOS tt c cc ng vo phic kt ni.

84

Nhng li bn trong IC

Ng vo hoc rab ni n thoc ngun VCC


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85

Nhng li bn trong IC

Ng vo hoc ra b h mch

86

Nhng li bn trong IC

Ngn mch gia hai chn


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87

Nhng li bn ngoi IC

ng dy tn hiu b h mch: dy t,mi hn khng tt, chn IC gy, chn IC gy.

ng dy tn hiu b ngn mch: dong dy, mi hn, board mch b t.

Ngun cung cp khng ng. Output loading: khi ng ra kt ni vi qu

nhiu ng vo khc.

88

Cu hi?


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1

Chng 3Cc cng logic &i sBoolean


2

Hng sBoolean v bin

Khc vi cc i skhc, cc hng vbin trong i sBoolean ch c hai gitr: 0 v 1

Trong i sBoolean khng c: phn s,sm, ly tha, cn s,

i sBoolean ch c 3 ton t: Cng logic, hay cn gi ton tOR Nhn logic, hay cn gi ton tAND B logic, hay cn gi ton tNOT


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3

Closed switchOpen switchYesNo

HighLowOnOff

TrueFalseLogic 1Logic 0

Hng sBoolean v bin (tt)

Gi tr 0 v 1 trong i sBoolean mang ngha miu t cc trng thi hay mc logic

4

Bng chn tr

Bng chn tr miu t mi quan h giagi tr cc ng vo v ng ra. V d:


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5

Biu thc Boolean ca cng ORx = A + B

Cng OR

6

Cng OR (tt)

Ng ra trngthi tch cckhi t nht mtng vo trng thi tchcc.


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7

IC cng OR 74LS32

8

IC cng OR 74LS32


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9

Cng OR (tt)

Cng OR c thc nhiu hn 2 ng vo.

10

V d 3-1

Cng ORc sdng trong mt hthng bo ng.


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11

V d 3-2

Biu thi gian cho cng OR.

12

V d 3-3

Biu thi gian cho cng OR.


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13

Biu thc Boolean ca cng ANDx = A * B

Cng AND

14

Cng AND (tt)

Ng ra trngthi tch cckhi tt c ccng vo trng thi tchcc.


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15

IC cng AND 74LS08

16

Cng AND (tt)

Cng AND c thc nhiu hn 2 ng vo.


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17

V d 3-4

Biu thi gian cho cng AND.

18

Mch Enable/Disable

Cng AND c sdng lm mt mchkha n gin


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19

Cng NOT lun lun ch c mt ng vo Biu thc Boolean ca cng NOT

x = A

Cng NOT

20

IC cng NOT 74LS04


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21

IC cng NOT 74LS04

22

Ng ra ca cng NOT xc nh trng thica nt nhn.

V d 3-5


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23

Miu t i smch logic

Bt k mch logic no cng c th cxy dng t3 cng logic c bn: AND,OR v NOT.

V d: x = AB + C x = (A+B)C x = (A+B) x = ABC(A+D)

24

V d 3-6


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25

V d 3-7

26

V d 3-8


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27

Xc nh gi tr ng ra

Cho mch c biu thcx =ABC(A+D) Xc nh gi tr ng ra x khi A=0, B=1,

C=1, D=1 Gi tr ng ra c th c xc nh

28

Thit lp bng chn tr

V d hy thit lp bng chn tr ts mch logic sau y

2INPUTS = Strng thi ng vo23 = 8 trng thi

111011101001110010100

000xCBA


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29

Thit lp bng chn tr

111011101 001

110010100000

xCBA

0

0

0

0

10

0

0

101

0011

30

Thit lp mch tbiu thc

Hy thit kmt mch logic c xc nhbi biu thc:y = AC + BC + ABC

Khi mt mch c nh ngha bi biuthc logic, ta c ththit kmch logictrc tip tbiu thc .

Biu thc gm 3 thnh phn OR vi nhau. Ng vo ca cng OR l ng ra ca cc

cng AND


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31


32


V d hy thit lp mch logic cho biuthcx = (A + B)(B + C)


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33

Biu thc Boolean ca cng NORx = A + B

Cng NOR

34

IC cng NOR 74LS02


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35

V d 3-9

Biu thi gian cho cng NOR.

36

Biu thc Boolean ca cng NANDx = A * B

Cng NAND


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IC cng NAND 74LS00

38

V d 3-10

Biu thi gian cho cng NAND.


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Cc nh l c bn trongi sBoolean

40

Cc nh l n binx * 0 = 0

x * 1 = x

x * x = x

x * x = 0

x + 0 = x

x + 1 = 1

x + x = x

x + x = 1


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Cc nh l nhiu bin

Lut giao honx * y = y * x

x + y = y + x

Lut kt hp(x * y) * z = x * (y * z)

(x + y) + z = x + (y + z)

42

Cc nh l nhiu bin (tt)

Lut phn phix * (y + z) = xy + xz

(x + y)(w + z) = xw + xz +yw + yz

Lut hon nguynx = x


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Mt scng thc thng dng

a) x.y + x.y = x

b) x + x.y = x

c) x + x.y = x + y

44

nh l DeMORGAN

nh l DeMORGAN 2 binx.y = x + y

x + y = x.y

nh l DeMorGAN nhiu bin

x.y.z.w = x + y + z + w x + y + z + = x.y.z


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45

p dng nh l DeMORGAN

46

p dng nh l DeMORGAN


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47

S a nhim ca cng NAND

48

S a nhim ca cng NOR


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49

Miu t cng logic

50

Miu t cng logic (tt)

Khi mt ng vo hay ng ra trn cnglogic c k hiu vng trn th ng vo hayng ra c gi l tch cc mc thp.

Trng hp ngc li, khng c vngtrn, th gi l tch cc mc cao.


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52



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Cu hi?


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1

Chng 2H thng s


2

nh ngha

Mt h thng sbao gm cc k ttrong nh ngha cc php ton cng, tr,nhn, chia.

H c sca mt h thng s l tng ktc trong h thng s .

Trong k thut sc cc h thng ssauy: Binary, Octal, Decimal, Hexa-decimal.


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3

nh ngha (tt)

0, 1, 2, 3, 4, 5, 6, 7, 8 , 9

A, B, C, D, E, F16Hexa-decimal

0, 1, 2, 3, 4, 5, 6, 78Octal

0, 12Binary

0, 1, 2, 3, 4, 5, 6, 7, 8 , 910Decimal

Cc k tc trong h thngC sH thng s

4

H thng sthp phn

H thng sthp phn c phn bcctrng snhsau:

. 10-210-1100101102103104

Trng s102Trng s101

Trng s100Trng s10-1

Trng s10-2

Du thp phn


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5

V d: phn tch sthp phn 2745.21410

2745.21410 =

(2 x 103) + (7 x 102) + (4 x 101) +

(5 x 100) + (2 x 10-1) + (1 x 10-2) +(4 x 10-3)

H thng sthp phn (tt)

103 102 101 100 10-1 10-2 10-3

Du thp phn

412.5472

Most significant digit (MSL) Least significant digit (LSD)

6

H thng snh phn

H thng snh phn c phn bcctrng snhsau:

. 2-22-12021222324

Trng s22Trng s21

Trng s20Trng s2-1

Trng s2-2

Du phn s


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7

V d: phn tch snh phn 1011.1012

1011.1012 =

(1 x 23) + (0 x 22) + (1 x 21) +

(1 x 20) + (1 x 2-1) + (0 x 2-2) +(1 x 2-3) =

H thng snh phn (tt)

23 22 21 20 2-1 2-2 2-3

Du phn s

101.1101

Most significant bit (MSB) Least significant bit (LSB)

11.62510

8

Cng hai bit nh phn

Php cng nh phn

1011101

110

000

A + BBA


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9

Cng hai snh phn khng du

Php cng nh phn (tt)

(9)1001

(6)+110

(3)11a)

(6.125)110.001

(2.750)+10.110

(3.375)11.011b)

10

Nhn 2 bit nh phn

Php nhn nh phn

111001

010

000

A x BBA


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11

Nhn 2 snh phn

Php nhn nh phn

1110

x 1011

1110

1110

0000

1110

10011010

12

Trong trng hp cn thhin du, snh phn sdng 1 bit xc nh du.

Bit ny thng v tru tin

Bit du bng 0 xc nh sdng.

Bit du bng 1 xc nh sm.

Snh phn c du


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Snh phn 6 bit c du

Snh phn c du

0 1 1 0 1 0 0

A6 A5 A4 A3 A2 A1 A0

Bit du (+) Gi tr = 5210

1 1 1 0 1 0 0

A6 A5 A4 A3 A2 A1 A0

Bit du (-) Gi tr = -5210

14

Bi trong h nh phn

o lng dung lng ca b nh, nv Kilo, Mega, Giga c sdng

GGiga230

MMega220

KKilo210

K hiun vBi

1073741824

1048576

1024

Gi tr


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15

Bi trong h nh phn

V d

/ 230 =

16

H thng sbt phn

H thng sbt phn c phn bcctrng snhsau:

V d: phn tch sbt phn 37283728 = (3 x 82) + (7 x 81) + (2 x 80)= (3 x 64) + (7 x 8) + (2 x 1)

=

. 8-28-18081828384

25010


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17

H thng sthp lc phn

H thng sthp lc phn c phn bcctrng snhsau:

V d: phn tch sthp lc phn 3BA163BA16 = (3 x 16

2) + (11 x 161) + (10 x 160)

= (3 x 256) + (11 x 16) + (10 x 1)=

. 16-216-1160161162163164

95410

18

M BCD (Binary coded decimal)

Mi chstrong mt sthp phn cmiu t bng gi tr nh phn tng ng.

Mu chsthp phn s c miu tbng 4 bit nh phn.

0111

7

1000

8

1001

9

0110010101000011001000010000

6543210


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M BCD

V d hai sthp phn 847 v 943 c miut bi m BCD nhsau:

011101001000

748

001101001001

349

20

So snh BCD v Binary

M BCD sdng nhiu bit hn nhng qutrnh binn i n gin hn

(BCD)0001 0011 011113710=

(Binary)10001001213710=


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Bng chuyn i

1000 0101F17111115

1000 0100E16111014

1000 0011D15110113

1000 0010C14110012

1000 0001B13101111

1000 0000A12101010

100191110019

100081010008

0111771117

0110661106

0101551015

0100441004

001133113

001022102

000111011

00000000BCDHexadecimalOctalBinaryDecimal

22

Sdng bit Parity pht hin li

Trong qu trnh truyn d liu nh phn,nhiu c thgy nn nhng li trnng truyn.

Phng php n gin pht hin li lsdng bit Parity


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Trong phng php ny, mt bit m rngs c thm vo, bit m rng c gil bit Parity

24


Gi tr ca bit Parity ph thuc vo phngphp sdng v sbit 1 trong khung dliu. Phng php Parity chn: tng sbit 1 trong

khung d liu (kc bit parity) phi l schn. D liu 1 0 1 1, bit parity thm vo 1 1 0 1 1

Phng php Parity l: tng sbit 1 trongkhung d liu (kc bit parity) phi l s l. D liu 1 1 1 1, bit parity thm vo 1 1 1 1 1


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Bin i gia cc h c s

Hexadecimal

Decimal Octal

Binary

26

Binary Decimal

Cch thc hin: Nhn mi bit vi trng s2n ca n Cng cc k t qu li vi nhau

Binary Decimal


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Binary Decimal (tt)

V d: bin i (10101101)2 sang thp phn

Binary 1 0 1 0 1 1 0 1

Gi tr

xxxxxxxx

2021222324252627

128 + 32 + 8 + 4 + 1Kt qu

17310

28

Decimal Binary

Cch thc hin: Chia 2 ly phn d Sd u tin l bit LSB (least significant bit) Sdcui cng l bit MLB (most significant bit)

Decimal Binary


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29

Decimal Binary

V d: bin i 6710 sang nh phnBc 1: 67 / 2 = 33 d1Bc 2: 33 / 2 = 16 d1Bc 3: 16 / 2 = 8 d0Bc 4: 8 / 2 = 4 d0Bc 5: 4 / 2 = 2 d0Bc 6: 2 / 2 = 1 d0Bc 7: 1 / 2 = 0 d1

1 0 0 0 0 1 12

30

Octal Binary

Cch thc hin: Bin mi k tstrong Octal thnh 3 bit nh

phn tng ng.

Octal Binary

111110101100011010001000Binary

76543210Octal


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31

Octal Binary (tt)

Bin i 4728 sang h nh phn

Bin i 54318 sang h nh phn010111100

274

001011100101

1345

1011000110012

1001110102

32

Hexa Binary

Cch thc hin: Bin mi k ts

trong Hexa thnh 4 bitnh phn tng ng.

111115F

111014E

110113D

110012C

101111B

101010A

100199

100088

011177

011066

010155

010044

001133

001022

000111

000000

BinaryDecimalHexa

Hexa

Binary


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Hexa Binary (tt)

Bin i 47C16 sang h nh phn

Bin i 10AF16 sang h nh phn110001110100

C74

1111101000000001

FA01

10000101011112

100011111002

34

Decimal Octal

Cch thc hin: Chia 8 ly phn d Sd u tin l LSD (least significant digit) Sdcui cng l MLD (most significant digit)

Decimal Octal


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Decimal Octal (tt)

V d: bin i 123410 sang bt phnBc 1: 1234 / 8 = 154 d2Bc 2: 154 / 8 = 19 d2Bc 3: 19 / 8 = 2 d3Bc 4: 2 / 8 = 0 d2

2 3 2 28

36

Decimal Hexa

Cch thc hin: Chia 16 ly phn d Sd u tin l LSD (least significant digit) Sdcui cng l MLD (most significant digit)

Decimal Hexa


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Decimal Hexa (tt)

V d: bin i 466010 sang thp lc phnBc 1: 4660 / 16 = 291 d4Bc 2: 291 / 16 = 18 d3Bc 3: 18 / 16 = 1 d2Bc 4: 1 / 16 = 0 d1

1 2 3 416

38

Binary Octal

Cch thc hin: Bt u tbn tri, nhm snh phn thnh

cc nhm 3 bit

Bin i mi nhm 3 bit thnh mt sOctal

Binary Octal


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39

Binary Octal (tt)

V d: bin i 10110101112 sang Octal

13278

1 011 010 111

10110101112 =

7231

40

Binary Hexa

Cch thc hin: Bt u tbn tri, nhm snh phn thnh

cc nhm 4 bit

Bin i mi nhm 4 bit thnh mt sHexa

Binary Hexa


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41

Binary Hexa (tt)

V d: bin i 101011010101110011010102sang Hexa

56AE6A16

101 0110 1010 1110 0110 1010

101011010101110011010102 =

A6EA65

42

Octal Hexa

Cch thc hin: Bin i sOctal thnh sBinary Bin i sBinary thnh sHexa

Octal Hexa

7/29/2019 Chuong 01 Mot So Khai

chuong 01 mot so khai niem mo dau 1716

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