Giáo trình điện tử số cơ bản

8/2/2019 Gio trnh in t s c bn

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LI GII THIU

Cng vi s tin b ca khoa hc v cng ngh, cc thit bin tang v s tip tc cng dng ngy cng rng ri v mang li hiu qu cao trong hu ht cc lnh vc kinh t k thutcng nhi sng x hi.

Vic x l tn hiu trong cc thit bin t hin i u da trn csnguyn l s. Bivy vic hiu su sc vin t s l iu khng th thiu c i vi k sin t hin nay.

Nhu cu hiu bit v k thut s khng phi ch ring i vi cc k sin t m cn i vinhiu cn b k thut chuyn ngnh khc c s dng cc thit bin t.

Ti liu ny gii thiu mt cch h thng cc phn t cbn trong cc mch in t s kthp vi cc mch in hnh, gii thch cc khi nim cbn v cng in t s, cc phng php

phn tch v thit k mch logic cbn.Ti liu bao gm cc kin thc cbn v mch cng logic, csi s logic, mch logic t

hp, cc trig, mch logic tun t, cc mch pht xung v to dng xung, cc b nhthng dng.c bit l trong ti liu ny c b xung thm phn logic lp trnh v ngn ng m t phn cngVHDL. y l ngn ng ph bin hin nay dng to m hnh cho cc h thng k thut s. Ttc gm 9 chng. Trc v sau mi chng u c phn gii thiu v phn tm tt gip ngihc d nm bt kin thc hn. Cc cu hi n tp ngi hc kim tra mc nm kin thcsau khi hc mi chng. Trn cscc kin thc cn bn, ti liu c gng tip cn cc vn hin i, ng thi lin h vi thc t k thut.

Ti liu gm c 9 chng c b cc nh sau:Chng 1: Hm

Chng 2: i s Boole v cc phng php biu din hm

Chng 3: Cng logic TTL v CMOS

Chng 4: Mch logic t hp.

Chng 5: Mch logic tun t.

Chng 6: Mch pht xung v to dng xung.

Chng 7: B nhbn dn.Chng 8: Logic lp trnh.

Chng 9 : Ngn ng m t phn cng VHDL.

Do thi gian c hn nn ti liu ny khng trnh khi thiu st, rt mong ngi c gp .Cc kin xin gi v Khoa K thut in t 1- Hc vin Cng ngh Bu chnh vin thng.

Xin trn trng cm n.


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Chng 1: Hm

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CHNG 1: HM

GII THIU

Khi ni n sm, ngi ta thng nghngay n h thp phn vi 10 ch sc khiu t 0 n 9. My tnh hin i khng s dng s thp phn, thay vo l s nh phn vi haik hiu l 0 v 1. Khi biu din cc s nh phn rt ln, ngi ta thay n bng cc s bt phn(Octal) v thp lc phn (HexaDecimal).

m s lng ca cc i lng l mt nhu cu ca lao ng, sn xut. Ngng mt qutrnh m, ta c mt biu din s. Cc phng php m v biu din sc gi l hm.

Hm khng chc dng biu din s m cn l cng c x l.C rt nhiu hm, chng hn nh h La M, La Tinh ... Hm va c tnh a dng va

c tnh ng nht v ph bin. Mi hm c u im ring ca n nn trong k thut s s sdng mt s h b khuyt cho nhau.

Trong chng ny khng ch trnh by cc h thp phn, h nh phn, h bt phn, h thplc phn v cn nghin cu cch chuyn i gia cc hm. Chng ny cng cp n s nh

phn c du v khi nim v du phy ng.

NI DUNG

1.1. BIU DIN S

Nguyn tc chung ca biu din l dng mt s hu hn cc k hiu ghp vi nhau theo quic v v tr. Cc k hiu ny thng c gi l ch s. Do , ngi ta cn gi hm l hthng s. S k hiu c dng l cs ca h k hiu l r. Gi tr biu din ca cc ch khcnhau c phn bit thng qua trng s ca h. Trng s ca mt hm bt k s bng ri, vi il mt s nguyn dng hoc m.

Bng 1.1 l lit k tn gi, s k hiu v cs ca mt vi hm thng dng.

Tn hm S k hiu Cs (r)

H nh phn (Binary)

H bt phn (Octal)

H thp phn (Decimal)

H thp lc phn (Hexadecimal)

0, 1

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

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

0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C, D, E, F

2

8

10

16

Bng 1.1

Ngi ta cng c th gi hm theo cs ca chng. V d: H nh phn = H cs 2, Hthp phn = H cs 10...


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Chng 1: Hm

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Di y, ta s trnh by tm tt mt s hm thng dng.

1.1.1 H thp phn

Cc k hiu ca h nh nu bng 1.1. Khi ghp cc k hiu vi nhau ta sc mtbiu din. V d: 1265,34 l biu din s trong h thp phn:

3 2 1 0 1 21265.34 1 10 2 10 6 10 5 10 3 10 4 10 = + + + + +

Trong phn tch trn, n10 l trng s ca h; cc h s nhn chnh l k hiu ca h. Nhvy,gi tr biu din ca mt strong h thp phn sbng tng cc tch ca k hiu (c trongbiu din) vi trng stngng. Mt cch tng qut:

n 1 1 0 1 m10 n 1 1 0 1 m

mi

in 1

N d 10 ... d 10 d 10 d 10 ... d 10

d 10

= + + + + + +

=

trong , 10N : biu din bt k theo h 10,

d : cc h s nhn (k hiu bt k ca h),

n : s ch sphn nguyn,

m : s ch sphn phn s.

u im ca h thp phn l tnh truyn thng i vi con ngi. y l h m con ngid nhn bit nht. Ngoi ra, nhc nhiu k hiu nn kh nng biu din ca h rt ln, cch biu

din gn, tn t thi gian vit v c.Nhc im chnh ca h l do c nhiu k hiu nn vic th hin bng thit b k thut s

kh khn v phc tp.

Biu din s tng qut:

Vi cs bt k r v d bng h s a tu ta s c cng thc biu din s chung cho tt ccc hm:

n 1 1 0 1 mn 1 1 0 1 m

m

iin 1

N a r ... a r a r a r ... a r

a r

= + + + + + +

=

Trong mt s trng hp, ta phi thm ch s trnh nhm ln gia biu din ca cc h.

V d: 10 8 1636 , 36 , 36 .

1.1.2 H nh phn

1.1.2.1. Tchc hnhphn

H nh phn (Binary number system) cn gi l h cs hai, gm ch hai k hiu 0 v 1, c

s ca h l 2, trng s ca h l 2

n

. Cch m trong h nh phn cng tng t nh h thp phn.Khi u t gi tr 0, sau ta cng lin tip thm 1 vo kt qum ln trc. Nguyn tc cngnh phn l : 0 + 0 = 0, 1 + 0 = 1, 1 + 1 = 10 (102 = 210).


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Trong h nh phn, mi ch s ch ly 2 gi tr hoc 0 hoc 1 v c gi tt l "bit". Nhvy, bit l s nh phn 1 ch s. S bit to thnh di biu din ca mt s nh phn. Mt s nh

phn c di 8 bit c gi 1 byte. S nh phn hai byte gi l mt t (word). Bit tn cng bnphi gi l bit b nht (LSB Least Significant Bit) v bit tn cng bn tri gi l bit ln nht(MSB - Most Significant Bit).

Biu din nh phn dng tng qut :

2 n 1 n 2 1 0 1 2 mN b b ....b b .b b ....b =

Trong , b l h s nhn ca h. Cc ch s ca h sng thi cng bng ly tha catrng s tng ng. V d :

1 1 0. 0 0 s nh phn phn s

2 1 0 1 22 2 2 2 2 trng s tng ng.

Cc gi tr 210 = 1024 c gi l 1Kbit, 220 = 1048576 - Mga Bit ...

Ta c dng tng qut ca biu din nh phn nh sau:

n 1 1 0 1 m2 n 1 1 0 1 m

mi

in 1

N b 2 ... b 2 b 2 b 2 ... b 2

b 2

= + + + + + +

=

Trong , b l h s nhn ly cc gi tr 0 hoc 1.

1.1.2.2. Cc php tnh trong hnhphn

a. Php cng

Qui tc cng hai s nh phn 1 bit nu trn.

b. Php tr

Qui tc tr hai bit nh phn cho nhau nh sau :

0 - 0 = 0 ; 1 - 1 = 0 ; 1 - 0 = 1 ; 10 - 1 = 1 (mn 1)

Khi tr nhiu bit nh phn, nu cn thit ta mn bit k tip c trng s cao hn. Ln tr ktip li phi tr thm 1.

c. Php nhn

Qui tc nhn hai bit nh phn nh sau:

0 x 0 = 0 , 0 x 1 = 0 , 1 x 0 = 0 , 1 x 1 = 1

Php nhn hai s nh phn cng c thc hin ging nh trong h thp phn.

Ch : Php nhn c th thay bng php dch v cng lin tip.

d. Php chia

Php chia nh phn cng tng t nh php chia hai s thp phn.

u im chnh ca h nh phn l ch c hai k hiu nn rt d th hin bng cc thit b c,in. Cc my vi tnh v cc h thng su da trn cshot ng nh phn (2 trng thi). Do


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, h nh phn c xem l ngn ng ca cc mch logic, cc thit b tnh ton hin i - ngnng my.

Nhc im ca h l biu din di, mt nhiu thi gian vit, c.

1.1.3 H bt phn v thp lc phn

1.1.3.1 Hbt phn

1. T chc ca h :Nhm khc phc nhc im ca h nh phn, ngi ta thit lp cc hm c nhiu k hiu hn, nhng li c quan h chuyn i c vi h nh phn. Mt trong s l h bt phn (hay h Octal, h cs 8).

H ny gm 8 k hiu : 0, 1, 2, 3, 4, 5, 6 v 7. Cs ca h l 8. Vic la chn cs 8 lxut pht t ch 8 = 23. Do , mi ch s bt phn c th thay th cho 3 bit nh phn.

Dng biu din tng qut ca h bt phn nh sau:n 1 0 1 m

8 n 1 0 1 m

mi

in 1

N O 8 ... O 8 O 8 ... O 8

O 8

= + + + + +

=

Lu rng, h thp phn cng m tng t v c gii rng hn h bt phn, nhng khng

th tm c quan h n10 2= (vi n nguyn).

2. Cc php tnh trong h bt phn

a. Php cng

Php cng trong h bt phn c thc hin tng t nh trong h thp phn. Tuy nhin,khi kt qu ca vic cng hai hoc nhiu ch s cng trng s ln hn hoc bng 8 phi nhlnch s c trng s ln hn k tip.

b. Php tr

Php tr cng c tin hnh nh trong h thp phn. Ch rng khi mn 1 ch s ctrng s ln hn th ch cn cng thm 8 ch khng phi cng thm 10.

Cc php tnh trong h bt phn t c s dng. Do , php nhn v php chia dnh linh mt bi tp cho ngi hc.

1.1.3.2 Hthp lc phn

1.Tchc ca h

H thp lc phn (hay h Hexadecimal, h cs 16). H gm 16 k hiu l 0, 1, 2, 3, 4, 5,6, 7, 8, 9, A, B, C, D, E, F.

Trong , A = 1010 , B = 1110 , C = 1210 , D = 1310 , E = 1410 , F = 1510 .

Cs ca h l 16, xut pht t yu t 16 = 24. Vy, ta c th dng mt t nh phn 4 bit(t 0000 n 1111) biu th cc k hiu thp lc phn. Dng biu din tng qut:


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n 1 0 1 m16 n 1 0 1 m

mi

i

n 1

N H 16 .... H 16 H 16 .... H 16

H 16

= + + + + +

=

2. Cc php tnh trong h cs 16

a. Php cng

Khi tng hai ch s ln hn 15, ta ly tng chia cho 16. S dc vit xung ch s tngv s thng c nhln ch s k tip. Nu cc ch s l A, B, C, D, E, F th trc ht, ta phii chng v gi tr thp phn tng ng ri mi cng.

b. Php tr

Khi tr mt s b hn cho mt s ln hn ta cng mn 1 ct k tip bn tri, ngha l

cng thm 16 ri mi tr.c. Php nhn

Mun thc hin php nhn trong h 16 ta phi i cc s trong mi tha s v thp phn,nhn hai s vi nhau. Sau , i kt qu v h 16.

1.2. CHUYN I CS GIA CC HM

1.2.1. Chuyn i th cs 10 sang cc h khc

thc hin vic i mt s thp phn y sang cc h khc ta phi chia ra hai phn:

phn nguyn v phn s.i vi phn nguyn: ta chia lin tip phn nguyn ca s thp phn cho cs ca h cn

chuyn n, s d sau mi ln chia vit o ngc trt t l kt qu cn tm. Php chia dng likhi kt qu ln chia cui cng bng 0.

V d: i s 5710 sang s nh phn.

Bc chia c d

1

2

3

4

5

6

57/2

28/2

14/2

7/2

3/2

1/2

28

14

7

3

1

0

1

0

0

1

1

1

LSB

MSB

Vit o ngc trt t, ta c : 5710 = 1110012

i vi phn phn s:ta nhn lin tip phn phn s ca s thp phn vi cs ca h cnchuyn n, phn nguyn thu c sau mi ln nhn, vit tun t l kt qu cn tm. Php nhndng li khi phn phn s trit tiu.

V d: i s 57,3437510 sang s nh phn.


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Phn nguyn ta va thc hin v d a), do ch cn i phn phn s 0,375.

Bc Nhn Kt qu Phn nguyn

1

2

3

4

0,375 x 2

0,75 x 2

0,5 x 2

0,0 x 2

0.75

1.5

1.0

0

0

1

1

0

Kt qu : 0,37510 = 0,01102

S dng phn nguyn c v d 1) ta c : 57,37510 = 111001.01102

1.2.2. i mt biu din trong h bt k sang h thp phn

Mun thc hin php bin i, ta dng cng thc :

n 1 0 1 m10 n 1 0 1 mN a r .... a r a r .... a r

= + + + + +

Thc hin ly tng v phi s c kt qu cn tm. Trong biu thc trn, ai v r l h s v cs h c biu din.

1.2.3. i cc s th nh phn sang h cs 8 v 16

V 8 = 23v 16 = 24 nn ta ch cn dng mt s nh phn 3 bit l ghi 8 k hiu ca h c

s 8 v t nh phn 4 bit cho h cs 16.

Do , mun i mt s nh phn sang h cs 8 v 16 ta chia s nh phn cn i, k tdu phn s sang tri v phi thnh tng nhm 3 bit hoc 4 bit. Sau thay cc nhm bit phn

bng k hiu tng ng ca h cn i ti.

V d:

a.i s110111,01112 sang shcs8

Tnh t du phn s, ta chia s ny thnh cc nhm 3 bit nh sau :

110 111 , 011 100

6 7 3 4

Kt qu: 110111,01112 = 67,348. ( Ta thm 2 s 0 tin bin i).

b.i snhphn 111110110,011012 sang shcs16

Ta phn nhm v thay th nh sau :

0001 1111 0110 0110 1000

1 F 6 6 8Kt qu: 111110110,011012 = 1F6,6816


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1.3 S NH PHN C DU

1.3.1 Biu din s nh phn c du

C ba phng php th hin s nh phn c du sau y.1. Sdng mt bit du. Trong phng php ny ta dng mt bit ph, ng trc cc bit tr

s biu din du, 0 ch du dng (+), 1 ch du m (-).

2. Sdng php b 1. Gi nguyn bit du v ly b 1 cc bit tr s (b 1 bng o ca ccbit cn c ly b).

3. Sdng php b 2

L phng php ph bin nht. S dng th hin bng s nh phn khng b (bit du bng0), cn s m c biu din qua b 2 (bit du bng 1). B 2 bng b 1 cng 1.

C th biu din s m theo phng php b 2 xen k: bt u t bit LSB, dch v bn tri,gi nguyn cc bit cho n gp bit 1 u tin v ly b cc bit cn li. Bit du gi nguyn.

1.3.2 Cc php cng v trs nh phn c du

Nh ni trn, php b 1 v b 2 thng c p dng thc hin cc php tnh nhphn vi s c du.

1. Biu din theo bit du

a. Php cng

Hai s cng du: cng hai phn tr s vi nhau, cn du l du chung.

Hai s khc du v s m c tr snh hn: cng tr s ca s dng vi b 1 ca s m.Bit trn c cng thm vo kt qu trung gian. Du l du dng.

Hai s khc du v s m c tr sln hn: cng tr s ca s dng vi b 1 ca s m.Ly b 1 ca tng trung gian. Du l du m.

b. Php tr. Nu lu rng, - (-) = + th trnh t thc hin php tr trong trng hp nycng ging php cng.

2. Cng v trcc stheo biu din b 1

a. CngHai s dng: cng nh cng nh phn thng thng, k c bit du.

Hai s m: biu din chng dng b 1 v cng nh cng nh phn, k c bit du. Bit trncng vo kt qu. Ch , kt quc vit di dng b 1.

Hai s khc du v s dng ln hn: cng s dng vi b 1 ca s m. Bit trn ccng vo kt qu.

Hai s khc du v s m ln hn: cng s dng vi b 1 ca s m. Kt qu khng c bittrn v dng b 1.

b. Tr thc hin php tr, ta ly b 1 ca s tr, sau thc hin cc bc nh php cng.


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3. Cng v trnhphn theo biu din b 2

a. Cng

Hai s dng: cng nh cng nh phn thng thng. Kt qu l dng.

Hai s m: ly b 2 c hai s hng v cng, kt qudng b 2.

Hai s khc du v s dng ln hn: ly s dng cng vi b 2 ca s m. Kt qu baogm c bit du, bit trn bi.

Hai s khc du v s m ln hn: s dng c cng vi b 2 ca s m, kt qudngb 2 ca s dng tng ng. Bit du l 1.

b. Php tr

Php tr hai s c du l cc trng hp ring ca php cng. V d, khi ly +9 tri +6 ltng ng vi +9 cng vi -6.

1.4. DU PHY NG

1.4.1 Biu din theo du phy ng

Gm hai phn: s m E (phn c tnh) v phn nh tr M (trng phn s). E c th c di t 5 n 20 bit, M t 8 n 200 bit ph thuc vo tng ng dng v di t my tnh. Thngthng dng 1 s bit biu din E v cc bit cn li cho M vi iu kin:

1/ 2 M 1

E v M c thc biu din dng b 2. Gi tr ca chng c hiu chnh m bomi quan h trn y c gi l chun ha.

1.4.2 Cc php tnh vi biu din du phy ng

Ging nh cc php tnh ca hm m. Gi s c hai s theo du phy ng chun ha:

( )xE xX 2 M= v ( )yE

yY 2 M= th:

Tch: ( )x y ZE E Ex y zZ X.Y 2 M .M 2 M+

= = =

Thng: ( )x y wE E

Ex y wW X / Y 2 M / M 2 M

= = =

Mun ly tng v hiu, cn a cc s hng v cng s m, sau s m ca tng v hius ly s m chung, cn nh tr ca tng v hiu s bng tng v hiu cc nh tr.

TM TT

Trong chng ny chng ta gii thiu v mt s h m thng c s dng trong hthng s: h nh phn, h bt phn, h thp lc phn. V phng php chuyn i gia cc hm.

Ngoi ra cn gii thiu cc php tnh s hc trong cc h.


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CU HI N TP

1. nh ngha th no l bit, byte?

2. i s nh phn sau sang dng bt phn: 0101 1111 0100 1110a. 57514

b. 57515

c. 57516

d. 57517

3. Thc hin php tnh hai s thp lc phn sau: 132,4416 + 215,0216.

a. 347,46

b. 357,46

c. 347,56

d. 357,67

4. Thc hin php cng hai s c du sau theo phng php b 1:

0000 11012 + 1000 10112

a. 0000 0101

b. 0000 0100

c. 0000 0011

d. 0000 0010

5. Thc hin php cng hai s c du sau theo phng php b 2:

0000 11012 1001 10002

a. 1000 1110

b. 1000 1011

c. 1000 1100

d. 1000 1110

6. Hai byte c bao nhiu bit?

a. 16

b. 8

c. 32

d. 64


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Chng 2:i sBoole v cc phng php biu din hm

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CHNG 2: I S BOOLE V CC PHNG PHP BIU

DIN HM

GII THIU CHUNG

Trong mch s, cc tn hiu thng cho hai mc in p, v d 0 V v 5 V. Nhng linhkin in t dng trong mch s lm vic mt trong hai trng thi, v d transistor lng cclm vic ch kha (tt), hoc thng..

Do vy, m t hot ng ca cc mch s, ngi ta dng h nh phn (Binary), haitrng thi ca cc linh kin trong mch c m ha tng ng thnh 1 v 0.

Mt b mn i sc pht trin t cui th k 19 mang tn chnh ngi sng lp ra n,i s Boole, cn c gi l i s logic rt thch hp cho vic m t mch s. i s Boole lcng c ton hc quan trng thit k v phn tch mch s. Cc k s, cc nh chuyn mntrong lnh vc in t, tin hc, thng tin, iu khin.. u cn phi nm vng cng c ny cthi su vo mi lnh vc lin quan n k thut s.

84 nm sau, i s Boole c Shannon pht trin thnh l thuyt chuyn mch. Nhcc cng trnh ca Shannon, v sau ny, cc nh k thut dng i s Boole phn tch vthit k cc mch vi tnh. Trng thi "ng", "sai"trong bi ton logic c thay th bng trngthi "ng", "ngt"ca mt chuyn mch (CM). Mi quan h nhn qu trong bi ton logic c

thay bi mi quan h gia dng in trong mch vi trng thi cc CM gn trn on mch y.Mi quan h ny sc th hin bng mt hm ton hc, c tn l hm chuyn mch. Khi ,cc trng thi ca CM : "ng" = 1 v "ngt" = 0. Hnh 2-1 m tiu va ni. y, trng thica CM c k hiu bng ch ci A.

V thc cht, hm chuyn mch l mt trng hp cth ca hm logic. Do , i s Boole ng vi trng hpny cng c gi l i s chuyn mch. Mc d vy, trongmt s ti liu ngi ta vn thng gi n l i s logic hayi s Boole.

Ngy nay, i s Boole khng ch gii hn trong lnhvc k thut chuyn mch m cn l cng c phn tch vthit k cc mch s, c bit l lnh vc my tnh. Cu kinlm chuyn mch c thay bng Diode, Transistor, cc mchtch hp, bng t... Hot ng ca cc cu kin ny cng cc trng bng hai trng thi: thng hay tt, dn in haykhng dn in... Do , hai gi tr h nh phn vn cdng m t trng thi ca chng.

i s logic ch c 3 hm cbn nht, l hm "V",

hm "Hoc" v hm "o". c im ni bt ca i s logicl c hm ln bin ch ly hai gi tr hoc 1 hoc 0.

CM trngthi Ngt:A= 0

CM trngthi ng:A=1


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Trong chng ny, ta s cp n cc tin , nh l, cc cch biu bin hm Boole vmt s phng php rt gn hm. Ngoi ra, chng ny cng xt cc loi cng logic v cc thams chnh ca chng.

NI DUNG

2.1 I S BOOLE

2.1.1. Cc nh l cbn:

STT Tn gi Dng tch Dng tng

1 ng nht X.1 = X X + 0 = X

2 Phn t 0, 1 X.0 = 0 X + 1 = 1

3 B X.X 0= X X 1+ =

4 Bt bin X.X = X X + X = X

5 Hp th X + X.Y = X X.(X + Y) = X

6 Phnh p X X=

7 nh lDeMorgan

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

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

Bng 2.1. Mt snh l thng dng trong i s chuyn mch2.1.2 Cc nh lut cbn:

+ Hon v: X.Y Y.X= , X Y Y X+ = +

+ Kt hp: ( ) ( )X. Y.Z X.Y .Z= , ( ) ( )X Y Z X Y Z+ + = + +

+ Phn phi: ( )X. Y Z X.Y X.Z+ = + , ( ) ( )X Y . X Z X Y.Z+ + = +

2.2 CC PHNG PHP BIU DIN HM BOOLE

Nh ni trn, hm logic c th hin bng nhng biu thc i s nh cc mn tonhc khc. y l phng php tng qut nht biu din hm logic. Ngoi ra, mt s phng

php khc cng c dng biu din loi hm ny. Mi phng php u c u im v ngdng ring ca n. Di y l ni dung ca mt s phng php thng dng.

2.2.1 Bng trng thi

Lit k gi tr (trng thi) mi bin theo tng ct v gi tr hm theo mt ct ring (thngl bn phi bng). Bng trng thi cn c gi l bng sththay bng chn l.


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i vi hm n bin s c 2n t hp c lp. Cc t hp ny c k hiu bng ch mi, vi i= 0 n 2n -1 (xem bng 2-2) v c tn gi l cc hng tch hay cn gi l mintex.

V mi hng tch c th ly 2 gi tr l 0 hoc 1, nn nu c n bin th s hm m bng

trng thi c th thit lp c s l:n2N 2=

2.2.2 Phng php bng Cc n (Karnaugh)

T chc ca bng Cc n:Cc t hp bin c vit theo mt dng (thng l pha trn) vmt ct (thng l bn tri). Nh vy, mt hm logic c n bin s c 2n . Mi th hin mt

hng tch hay mt hng tng, cc hng tch trong hai kcn ch khc nhau mt bin.Tnh tun hon ca bng Cc n:Khng nhng cc kcn khc nhau mt bin m cc

u dngv cui dng, u ctv cui ctcng chkhc nhau mt bin (k c 4 gc vung cabng). Bi vy cc ny cng gi l kcn.

Mun thit lp bng Cc n ca mt hm cho di dng chun tng cc tch, ta ch vicghi gi tr 1 vo cc ng vi hng tch c mt trong biu din, cc cn li s ly gi tr 0 (theonh l DeMorgan). Nu hm cho di dng tch cc tng, cch lm cng tng t, nhng cc ng vi hng tng c trong biu din li ly gi tr 0 v cc khc ly gi tr 1.

2.2.3 Phng php i sC 2 dng biu din l dngtuyn (tng cc tch)v dng hi (tch cc tng).

+ Dng tuyn: Mi s hng l mt hng tchhay mintex, thng k hiu bng ch"mi".

+ Dng hi: Mi tha s l hng tnghay maxtex, thng c k hiu bng ch"Mi".Nu trong tt c mi hng tch hay hng tng c mt cc bin, th dng tng cc tch hay tchcc tng tng ng c gi l dng chun. Dng chun l duy nht.

Tng qut, hm logic n bin c th biu din ch bng mt dng tng cc tch:

( )

n2 1

n 1 0 i ii 0f X ,...,X a m

==

m A B C f

m0

m1m2

m3

m4

m5

m6

m7

0

00

0

1

1

1

1

0

01

1

0

0

1

1

0

10

1

0

1

0

1

0

00

0

0

0

0

1

Bng 2.2. Bng trng thi hm 3 bin


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hoc bng ch mt dng tch cc tng:

( ) ( )

n2 1

n 1 0 i i

i 0

f X ,...,X a m

=

= +

y, ai ch ly hai gi tr 0 hoc 1. i vi mt hm th mintex v maxtex l b ca nhau.

2.3 CC PHNG PHP RT GN HM

2.3.1. Phng php i s

Da vo cc nh l hc a biu thc v dng ti gin.

V d: Hy a hm logic v dng ti gin:

f AB AC BC= + +

p dng nh l, A A 1+ = , X XY X+ = ta c:

( )f AB AC BC A A

AB ABC AC ABC

AB AC

= + + +

= + + +

= +

Vy nu trong tng cc tch, xut hin mt bin v o ca bin trong hai s hng khcnhau, cc tha s cn li trong hai s hng to thnh tha s ca mt s hng th ba th shng th ba l tha v c th bi.

2.3.2 Phng php bng Cc n

Phng php ny thng c dng rt gn cc hm c s bin khng vt qu 5.

Cc bc ti thiu ha:

1. Gp cc k cn c gi tr 1 (hoc 0) li thnh tng nhm 2, 4, ...., 2i . S trongmi nhm cng ln kt qu thu c cng ti gin. Mt c thc gp nhiu ln trong ccnhm khc nhau. Nu gp theo cc c gi tr 0 ta s thu c biu thc b ca hm.

2. Thay mi nhm bng mt hng tch mi, trong gi li cc bin ging nhau theo dngv ct.

3. Cng cc hng tch mi li, ta c hm ti gin.

V d: Hy dng bng Cc n gin c hm :

( ) ( )f A,B,C 1, 2, 3, 4, 5= Li gii:

00 01 11 10

0 1 1 1 0

1 1 1 0 0

Hnh 2-2

ABC

1f B= 2f AC=


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+ Xy dng bng KN tng ng vi hm cho.

+ Gp cc c gi tr 1 k cn li vi nhau thnh hai nhm (hnh 2-2)

Li gii phi tm :

1 2f f f B AC= + = +

Nu gp cc c gi tr 0 li theo hai nhm, ta thu c biu thc hm b f:

f AB BC= +

2.3.3. Phng php Quine Mc. Cluskey

Phng php ny c th ti thiu ha c hm nhiu bin v c th tin hnh cng vicnhmy tnh.

Cc bc ti thiu ha:1. Lp bng lit k cc hng tch di dng nh phn theo tng nhm vi s bit 1 ging

nhau v xp chng theo s bit 1 tng dn.

2. Gp 2 hng tch ca mi cp nhm ch khc nhau 1 bit to cc nhm mi. Trong minhm mi, gi li cc bin ging nhau, bin bi thay bng mt du ngang (-).

Lp li cho n khi trong cc nhm to thnh khng cn kh nng gp na. Mi ln rt gn,ta nh du # vo cc hng ghp cp c. Cc hng khng nh du trong mi ln rt gn sc tp hp li la chn biu thc ti gin.

V d.Hy tm biu thc ti gin cho hm:( ) ( )f A,B,C,D 10, 11, 12, 13, 14, 15=

Gii: Bc 1: Lp bng (bng 2.3a):

Bng a Bng b

Hng tch

sp xp

Nh phn

A B C D

Rt gn ln u.

A B C D

Rt gn ln th 2.

A B C D

10

12

11

13

14

15

1 0 1 0

1 1 0 0

1 0 1 1

1 1 0 1

1 1 1 0

1 1 1 1

1 0 1 - # (10,11)

1 - 1 0 # (10,14)

1 1 0 - # (12,13)

1 1 - 0 # (12,14)

1 - 1 1 # (11,15)

1 1 - 1 # (13,15)

1 1 1 - # (14,15)

1 1 - - (12,13,14,15)

1 - 1 - (10,11,14,15)

Bng 2.3

Bc 2: Thc hin nhm cc hng tch (bng 2.3b).


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Tip tc lp bng la chn tm hm ti gin (Bng 2.4):

A BCD10 11 12 13 14 15

1 1 - -

1 - 1 - x x

x x x

x

x

x

Bng 2.4

T bng 2-4, ta nhn thy rng 4 ct c duy nht mt du "x" ng vi hai hng 11-- v 1-1-.Do , biu thc ti gin l :

( )f A,B,C,D AB AC= +

2.4 CNG LOGIC V CC THAM S CHNH

Cng logic csl mch in thc hin ba php tnh cbn trong i s logic, vy ta sc ba loi cng logic csl AND, OR v NOT.

2.4.1 Cng logic cbn

2.4.1.1 Cng AND

Cng AND thc hin hm logic

( )f f A,B A.B= = hoc nhiu bin:

( )f A,B,C,D, ... A.B.C.D...=

a) Theo tiu chun ANSI b) Theo tiu chun IEEE

Hnh 2-4a,b. K hiu ca cng AND.

Nguyn l hot ng ca cng AND:

Bng trng thi 2.5a,b l nguyn l hot ng ca cng AND (2 li vo).

A

Bf

f

A

B

CD

E

A&

Bf

f&

A

BC

D

E


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A B f A B f

0 0 0 L L L

0 1 0 L H L1 0 0 H L L

1 1 1 H H H

a) Ghi theo gi tr logic b) Ghi theo mc logic

Bng 2.5a,b. Bng trng thi m t hot ng ca cng AND 2 li vo.

Theo qui c, logic 1 c thay bng mc in th cao, vit tt l H (High) cn logic 0

c thay bng mc in th thp, vit tt l L (Low) (bng 2-5b). Cng AND c n li vo s c2n hng tch (dng) trong bng trng thi.

Khi tc ng ti li vo cc chui xung s xc nh, u ra cng s xut hin mt chuixung nh ch hnh 2-4. th ny thng c gi l th dng xung, th dng sng hay th thi gian.

T th, ta nhn thy rng, ch ti cc thi im t2n t3 v t7n t8 trn c hai li vou c logic 1 nn li ra cng ly logic 1. ng vi cc khong thi gian cn li v hoc c hai livo bng 0, hoc mt trong hai li vo bng 0 nn li ra ly logic 0. Hot ng ca cng AND

nhiu li vo cng xy ra tng t.

C th gii thch d dng mt vi ng dng ca cng AND qua th dng xung.

V d : Dng cng AND to "ca" thi gian. Trong ng dng ny, trn hai li vo cacng AND c a ti 2 chui tn hiu s X, Y c tn s khc nhau. Gi s tn s ca X ln hntn s ca Y. Trn u ra cng AND ch tn ti tn hiu X, gin on theo tng chu k ca Y. Nhvy, chui s Y ch gi vai tr ng, ngt cng AND v thng c gi l tn hiu "ca". Hotng ca mch c m t bng hnh 2-5.

1 1Li vo A

Li ra f

tt0 t1 t2 t3 t4 t5 t6 t7 t8 t9 t10

Li vo B

11

1 1

00000000

0 0 0 0 0 01 1

0 1 1 10 0 0 0

Hnh 2-4. th dng xung vo, ra ca cng AND


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Ty theo iu kin cho trc, c thng dng mch theo cc mc ch khc nhau. Nu bit rng xung ca Y ( thng ly bng 1s ) th s xung xut hin u ra chnh bng tn sca X. Ngc li, nu tn s ca X cho, chng hn bng 1 Hz ( Tx = 1s ) th ch cn m sxung trn u ra ta c th tnh c rng xung ca Y. y chnh l phng php o tn sv thi gian c ng dng trong kthut hin nay.

2.4.1.2 Cng OR

Cng OR thc hin hm logic: ( )f A,B A B= +

hoc vi hm nhiu bin: ( )f A,B,C,D... A B C D ...= + + + +

K hiu ca cng ORc biu din hnh 2-6a, b.


Hnh 2-6 a, b. K hiu ca cng OR.

Tng t nh cng AND, nguyn l hot ng ca cng OR c thc gii thch thngqua bng trng thi (Bng 2.6a,b) v th dng xung - hnh 2-7.

A B f A B f

0

0

1

1

0

1

0

1

0

1

1

1

L

L

H

H

L

H

L

H

L

H

H

H

a) Theo gi tr logic b) Theo mc in th

Bng 2.6 a, b. Bng trng thi ca cng OR.

X

1s 1s Y

f

Hnh 2-5. M hnh dng cng AND to ca thi gian

A1

BF

A

BF

E

F

A

CD

BA

1B

FCDE


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Mt cng OR c n li vo s c 2n hng tch trong bng trng thi ca n.

2.4.1.3. Cng NOT

Cng NOT thc hin hm logic:f A=

K hiu ca cng NOT c ch ra trn hnh 2-8 a, b.

a) Theo tiu chun ANSI. b) Theo tiu chun IEEE.Hnh 2-8a,b. K hiu ca cng NOT

Hot ngca cng NOT kh n gin, nu li vo:

A 0= th A 1= ,

nu A 1= th A 0=

Nguyn l ny c minh ho bng th dng xung hnh 2-9.

Hot ng ca cng NOT c tm tt bng 2.7a,b.

A f A f

0

1

1

0

L

H

H

L

a) Theo gi tr logic b) Theo mc logic

Bng 2.7a, b. Bng trng thi ca cng NOT.

2.4.2 Logic dng v logic m

Logic dngl logic c in th mc H lun ln hn in th mc L (Hnh 2-10).

fB

tt0 t1 t2 t3 t4 t5 t6 t7 t8 t9 t10

0 1 1 0 0 0 1 1 1 0 A

0 0 1 1 1 0 0 1 0 0

0 1 1 1 1 0 1 1 1 0

Hnh 2-7. th dng xung ca cng OR.

A A

Hnh 2-9

A A

A

A1

A1A

A

A


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Hnh 2-10a,b. th dng xung ca logic dng

Logic m th ngc li, logic 1 c in th thp hn mc 0. Khi nim logic m thngc dng biu din tr cc bin. Logic m v mc m ca logic l hon ton khc nhau.

2.4.3 Mt s cng ghp thng dng

Khi ghp ba loi cng logic cbn nht s thu c cc mch logic tn gin n phctp. y ta ch xt mt vi mch ghp n gin nhng rt thng dng.

2.4.3.1 Cng NAND

Ghp ni tip mt cng AND vi mt cng NOT ta c cng NAND (Hnh 2-11).

Hnh 2-11. S cu to cng NAND

Hm ra ca cng NAND 2 v nhiu bin vo nh sau:

f AB

f ABCD...

=

=

K hiu cng NAND (hnh 2-12a,b) v bng trng thi (bng 2-8).

a) Theo tiu chun ANSI b) Theo tiu chun IEEEHnh 2-12a,b. K hiu ca cng NAND

AB

f

A

B f

C

A &B

f

f&

A

B

CD

E

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

0t

V

H

L

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

t

V

H

L

0

a) Logic dng vi mc dng.

b) Logic dng vi mc m.

A

B

AB f AB=


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Bng 2.8a,b. Bng trng thi ca cng NAND

2.4.3.2 Cng NOR

Cng NORc thit lp bng cch ni tip mt cng OR vi mt cng NOT.

T hnh 2-13 ta c th vit c hm ra ca cng NOR 2 v nhiu li vo nh sau:

f A B hay f A B C ...= + = + + +

Hnh 2-13. S cu to cng NOR

K hiu ca cng NOR 2 li vo nh chhnh 2-14a,b.

a) Theo tiu chun ANSI. b) Theo tiu chun IEEE.

Hnh 2-14a, b. K hiu cng NOR 2 li vo

Hot ng ca cng NORc gii thch bng bng trng thi nh chbng 2.9a,b.

A B f A B f

0

0

1

1

0

1

0

1

1

0

0

0

L

L

H

H

L

H

L

H

H

L

L

L

Bng 2.9a, b. Bng trng thi ca cng NOR 2 li vo.

2.4.3.3 Cng khc du

Cng khc du cn c mt s tn gi khc: cng Cng Modul-2, cng XOR.

A

Bf

A 1B

f

A B f

0

01

1

0

10

1

1

11

0

A B f

L

LH

H

L

HL

H

H

HH

L

AB

A B+ A B+


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Hnh 2-15. S ca cng XOR 2 li vo

T hnh 2-15, ta c biu thc ca hm khc du 2 li vo l:

f AB AB= + hay theo qui c f A B=

K hiu ca cng XOR 2 li vo nh hnh 2-16a, b.


Hnh 2-16a, b. K hiu ca cng XOR 2 li vo

Bng trng thi ca cng XOR hai li vo c trnh by bng 2.10a,b.

A B F A B F

0

0

1

1

0

1

0

1

0

1

1

0

L

L

H

H

L

H

L

H

L

H

H

L

Bng 2-10a,b. Bng trng thi ca cng XOR 2 li vo

Hot ng cng XOR nhiu li vo cng tng t nh cng 2 li vo, ngha l nu s bit 1

trn tt cc cc li vo l mt s l, th hm ra ly logic 1; ngc li nu tng s bit 1 trn cc livo l mt s chn, th hm ra ly logic 0. C th dng cng XOR 2 li vo thc hin hmXOR nhiu bin.

2.4.3.4 Cngng du (XNOR)

Cng XNOR thc hin biu thc logic sau:

f AB AB hay f A B A ~ B= + = =

K hiu ca cng XNOR hai li vo c trnh by hnh 2-17.

A =1B

fAB

f

A

B

f AB AB= + B

AB

AB

A


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Hnh 2-17. K hiu ca cng XNOR 2 li vo

Nu tng s bit 0 trn tt c cc li vo l mt s l, th hm ra ca XNOR s ly logic 1.Nu tng s bit 0 trn tt c cc li vo l mt s chn, th hm ra li ly logic 0.

XOR v XNOR l hai loi cng c rt nhiu ng dng trong k thut s. Chng l phn tchnh hp thnh b cng, tr , so snh hai s nh phn v.v...

2.4.4 Cc tham s chnh

2.4.4.1 Mc logic

Vo Ra Vo Ra

a) i vi h TTL b) i vi h CMOS

Hnh 2-19a, b. Mc logic ca cc h cng TTL v CMOS

Mc logic l mc in th trn u vo v u ra ca cng tng ng vi logic "1" v logic

"0", n ph thuc in th ngun nui ca cng (VCC i vi h TTL (Transistor TransistorLogic) v VDD i vi h MOS (Metal Oxide Semiconductor)). Lu rng, nu mc logic vovt qu in th ngun nui c th gy h hng cho cng.

Mc TTL

Mc TTL l mt chun quc t, trong qui nh:

- in th ngun nui VCC , VDD bng + 5 vn hoc bng - 5,2 vn;

- Mc in th tng ng vi logic H v L trn u vo, u ra ca cng nh chhnh 2-18a,b.

Nhn xt: + Mc vo ra i vi cng TTL v CMOS (Complementary Metal OxideSemiconductor) khc nhau rt nhiu;

5v

4v

3v

2v

1v

0v

VVHmax

VVHmin

VVLma0,8v

VRHmax VVHmax VRHmax

VRHmin

VRLmax

VVHmin

VVLma

VRHmin

VRLmax

2,4v

0,4v

3,5v

1,5v

4,9v

0,1vNL

NH

NL

NH

A

Bf

A =1B

f


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+ Mc vo ra snh hng n phng v nhiu ca cng.

2.4.4.2 chng nhiu

chng nhiu (hay phng v nhiu) l mc nhiu ln nht tc ng ti li vo hoc lira ca cng m cha lm thay i trng thi vn c ca n.

a) Tc ng nhiu khi mc ra cao b) Tc ng nhiu khi mc ra thp

Hnh 2-20a, b, M t tc ng nhiu n cc cng logic

nh hng ca nhiu c th phn ra hai trng hp :

+ Nhiu mc cao: u ra cng I ly logic H (hnh 2-20a), tt nhin, u ra cng II l logicL, nu cc cng vn hot ng bnh thng. Khi tnh ti tc ng ca nhiu, ta c:

RHmin NH VHmin NH VHmin RHminV V V V V V+

Vi cng TTL: NLV 2V 2,4V 0,4V = Vi cng CMOS: NLV 3,5V 4,9V 1, 4V =

+ Nhiu mc thp: u ra cng I ly logic L (hnh 2-20b), tng t ta c:

RLmax NL VLmax NL VLmax RLmaxV V V V V V+

Vi cng TTL: NLV 0,8V 0,4V 0,4V =

Vi cng CMOS: NLV 1,5V 0,1V 1,4V =

2.4.4.3 Hsghp ti K

Cho bit kh nng ni c bao nhiu li vo ti u ra ca mt cng cho.

H s ghp ti ph thuc dng ra (hay dng phun) ca cng chu ti v dng vo (hay dnght) ca cc cng ti c hai trng thi H, L.

VTT TT

Cng I Cng II

VRH

VVL

VVH

VNH

TT TT

Cng I Cng II

VVH VRH

VRLVVL

VNL


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a) Mc ra ca cng chu ti l H b) Mc ra ca cng chu ti l L

Hnh 2-21a,b. M t v h s ghp ti.

2.4.4.4. Cng sut tiu th

Hnh 2-22. Hai trng thi tiu th dng ca cng logic

ICCH - L dng tiu th khi u ra ly mc H,

ICCL - L dng tiu th khi u ra ly mc L.

Theo thng k, tn hiu s c t l bit H / bit L khong 50%. Do , dng tiu th trungbnh ICCc tnh theo cng thc :

ICC = (ICCH + ICCL)/ 2

Cng sut tiu th trung bnh ca mi cng s l :

P0 = ICC . VCC

2.4.4.5. Trtruyn lan

Tn hiu i qua mt cng phi mt mt khong thi gian, c gi l tr truyn lan.

Hnh 2-23. Minh ho tr truyn lan ca tn hiu

Vo Ra

Vo

Ra

tTHL tTLH

H

+VccICCH

LH L

+Vcc

ICCLHH

AB

Cng chu ti

AB

Cc cng tiH L

IRH IRL

Cng chu ti Cc cng ti


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Tr truyn lan xy ra ti c hai sn ca xung ra. Nu k hiu tr truyn lan ng vi sntrc l tTHL v sn sau l tTLH th tr truyn lan trung bnh l:

tTtb = ( t THL + t TLH )/2

Thi gian tr truyn lan hn ch tn s cng tc ca cng. Tr cng ln th tn s cng tccc i cng thp.

TM TT

Trong chng 2 chng ta gii thiu v cc phng php biu din v rt gn hm Boole.Ngoi ra cn gii thiu mt s cng logic thng dng v cc tham s chnh ca chng.

CU HI N TP

Bi 2.1 Rt gn hm sau theo phng php dng bng Karnaugh:

1. F (A, B, C) = (0, 2, 4, 6,7).

a. AB C+

b. AB C+

c. AB C+

d. AB C+

2. F (A, B, C, D) = (0, 1, 8, 9, 10)

a. BC D+

b. BC ABD+

c. BC ABD+

d. BC ABD+

2.2 Rt gn hm sau theo phng php i s

1. CD CD . A C D+ +

a. CD

b. CD

c. CD

d. CD

2. A BC . A B BC C A+ +

a. AB AC+

b. AB AC BC+ +

c. AC BC+

d. AB BC+

2.3 Rt gn hm sau theo phng php Quine-Mc.CLUSKEY:

F (A, B, C, D) = (2, 3, 6, 7, 12, 13, 14, 15).a. AC AB+


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b. AC AD+

c. AC AB+

d. AC AB+

2.4 Hai mch in hnh di y l tng ng

a. Do u bng A+Bb. Do u bng Bc. Do u bng ABd. Do u bng A+AB

Bi 2.5 Phn tch ngha cc tham s chnh ca cc h cng logic.

Bi 2.6 Trnh by v phng v nhiu ca cc h cng logic? Tnh phng v nhiu ca

mt cng logic h TTL, bit VVL = 0 V 0,8 V, VVH = 2,0 V 5,0 V, VRL = 0 V 0,4 V, VRH =

2,4 V 5,0 V?

a. NH NLV 0.4V, V 0.4= =

b. NH NLV 0.4V, V 0.4= =

c. NH NLV 0.4V, V 0.4= =

d. NH NLV 0.4V, V 0.4= =

Bi 2.7 Cho mch in nh hnh 1. Biu thc hm ra l:

Hnh 1

a. AB AB+ b. AB AB+ c. AB AB+ d. AB AB+

Bi 2.8 Phn tch ngha ca vic ti u ho mch in ca cc h cng logic? Cho v d

minh ho?

Bi 2.9 Chng minh cc ng thc:a. A B A B AB = +

A

B

F

AB B

A


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b. AB (A B C) = ABC

c. A B C = A B C

Bi 2.10 Lit k 3 phn t logic cbn trong k thut s?

a. AND, OR v NOTb. NAND, AND v NOTc. AND, NOR v NANDd. AND, OR v XNOR

Bi 2.11 Phn t logic AND 2 li vo cho u ra bng 1 khi cc u vo l bao nhiu?

a. 0 v 0b. 0 v 1c. 1 v 0d. 1 v 1

Bi 2.12 c biu thc A+B nh th no?

a. A AND Bb. A XOR Bc. A OR Bd. A NAND B


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CHNG 3: CNG LOGIC TTL V CMOS

GII THIU

Xt v mt cbn th c hai loi linh kin bn dn l lng cc v n cc. Da trn cclinh kin ny, cc mch tch hp c hnh thnh v c sn trn th trng. Cc chc nng kthut s khc nhau cng c ch to trong nhiu dng khc nhau bng cch s dng cng nghlng cc v n cc. Mt nhm cc IC tng thch vi cc mc logic ging nhau v cc in pngun thc hin cc chc nng logic a dng phi c ch to bng cch s dng cu hnhmch chuyn bit c gi l h mch logic.

Cc yu t chnh ca mt IC lng cc l in tr, it v cc transistor. C hai loi hotng cbn trong cc mch IC lng cc:

Bo ho.

Khng bo ho.

Trong mch logic bo ho, cc transistorc vn hnh trong vng bo ho, cn trong ccmch logic khng bo ho th cc transistor khng lm vic ti vng bo ho.

Cc h mch logic lng cc c bo ho l:

Mch logic in tr- Transistor (RTL).

Mch logic it Transistor (DTL).

Mch logic Transistor Transistor (TTL).

Cc h mch logic lng cc khng bo ha l:

Schottky TTL.

Mch logic ghp cc pht (ECL).

Cc linh kin MOS l cc linh kin n cc v ch c cc MOSFET c vn hnh trongcc mch logic MOS. Cc h mch logic MOS l:

PMOS.

NMOS.

CMOS

Trong chng 3 s trnh by cc h cng logic ch yu v c dng ph bin hin nay.Phn cui ca chng trnh by mt s mch cho php giao tip gia cc h logic TTL v CMOS.


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NI DUNG

3.1. CC H CNG LOGIC

3.1.1. H DDL

DDL (Diode Diode Logic) l h cng logic do cc diode bn dn to thnh. Hnh 3-1a,b ls cng AND, OR 2 li vo h DDL.

Hnh 3-1. Mch in cng AND v OR h DDL.Bng trng thi sau th hin nguyn l hot ng ca mch thng qua mc in p vo/ra

ca cc cng AND v OR h DDL

AND OR

A (V) B (V) F (V) A (V) B (V) F (V)

0 0 0,7 0 0 0

0 3 0,7 0 5 4,3

3 0 0,7 5 0 4,3

3 3 4,7 5 5 4,3

Bng 3-1. Bng trng thi ca cng AND v OR h DDL

u im ca h DDL:

Mch in n gin, d to ra cc cng AND, OR nhiu li vo. u im ny chophp xy dng cc ma trn diode vi nhiu ng dng khc nhau;

Tn s cng tc c tht cao bng cch chn cc diode chuyn mch nhanh;

Cng sut tiu th nh.

Nhc im :

fD2

B

D1A

R1

+5V

fA

B

a) Cng AND

R1

f

D2B

D1A

fA

B

b) Cng OR


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phng v nhiu thp (VRL ln) ;

H s ghp ti nh.

ci thin phng v nhiu ta c th ghp ni tip mch ra mt diode. Tuy nhin,khi VRH cng b st i 0,6V.

3.1.2. H DTL

thc hin chc nng o, ta c thu ni tip vi cc cng DDL mt transistor cngtc ch kho. Mch cng nh thc gi l h DTL (Diode Transistor Logic). V d, hnh3-2a, b l cc cng NOT, NAND thuc h ny.

Hnh 3-2. S mch in ca h cng TDL.

Trong hai trng hp trn, nhcc diode D2, D3 chng nhiu trn li vo ca Q1cci thin. Mc logic thp ti li ra f gim xung khong 0,2 V ( bng th bo ho UCE ca Q1). DoIRHmax v IRLmax ca bn dn c th ln hn nhiu so vi diode nn h s ghp ti ca cng cngtng ln.

Bng cch tng t, ta c th thit lp cng NOR hoc cc cng lin hp phc tp hn. Vti ca cc cng l in trnn h s ghp ti (c bit i vi NH) cn b hn ch, mt khc trtruyn lan ca h cng ny cn ln. Nhng tn ti trn sc khc phc tng phn cc hcng sau.

3.1.3. H RTL

H RTL (Resistor Transistor Logic) l cc cng logic c cu to bi cc in tr vtransistor. Hnh 3-3 l s ca mt mch NOT h RTL.

Khi in p li vo l 0 V, in p trn base ca transistor s m nn transistor cm nh vyli ra trn collector ca transistor smc cao. Do li ra ny c ni ln ngun +5 V thng quadiode D nn gi trin p li ra lc ny khong 5,7 V, nhn mc logic cao. Khi in p li vo l5 V do hai in trli vo c gi tr ln lt l 1 k v 10 k, nn in p ti base s ln lmtransistor thng lm cho in p li ra l 0 V. Nh vy logic li ra s l o ca logic ca tn hiuli vo.

Tng t nh mch hnh 3-3, nu mt in trc ni thm li vo nh hnh 3-4 saumch s trthnh mch NOR h RTL.

5k

Q1

2kf

+5V

D3D1

4k

+5V

D2A

D4B

5k

Q1

2kf

+5V

D3D1

4k

+5V

D2A

a) b)


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Hnh 3-3. Cng NOT h RTL

Bng 3-2 th hin quan hin p ca cng NOR h RTL, ch khi c hai li vo A v Bcng gi tr 0 V th transistor mi cm v li ra nhn logic cao. Cc trng hp khc u dn

n transistor thng v lm gi tr logic li ra mc thp.A (V) B (V) F (V)

0 0 5,7

0 5 0

5 0 0

5 5 0

Bng 3-2. Bng trng thi ca cng NOR h RTL

Hnh 3.4. Cng NOR h RTL

3.1.4. H TTL

Do hn ch v tc , h DTL trnn lc hu v b thay th hon ton bi h mchTTL. Hn ch tc ca DTL c gii quyt bng cch thay cc it u vo thnh transistoralp tip gip BE.

a. Cng NAND TTL


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Hnh 3-5. S mch in mt cng NAND 2 li vo.

Hnh 3-5 l s nguyn l ca mch NAND TTL. N c thc chia ra thnh 3 phn.Transistor Q1, trR1 v cc diode D1, D2 to thnh mch u vo, mch ny thc hin chc nng

NAND. Transistor Q2, cc trR2, R4 to thnh mch gia Q3, Q4, R3 v diode D3 to thnh mchli ra nh phn tch trn.

Khi bt k mt li vo mc thp th Q1u tr thnh thng bo ho, do , Q2 v Q4ng, cn Q3 thng nn u ra ca mch smc cao. Li ra s ch xung mc thp khi tt c

cc li vo u mc logic cao v lm transistor Q1 cm. Diode D3c s dng nh mch dchmc in p, n c tc dng lm cho Q3 cm hon ton khi Q2 v Q4 thng. Diode ny nhiu khicn c mc vo mch gia collector Q2 v base ca Q3.

fA

B

+Vcc

R14k

D2D1

A Q1

B

Q4

f

D3

300R3

Q3

R21,6k

Q2

R41k

Q2

R14k

R24k

f

+Vcc

R51,6k

Q6

R7130

Q7

R61 k

D4

Q8

D3

R41 k

Q5

Q4

R31,6k

B

A

D2

Q3

Q1

D1

Hnh 3-6. S mch in ca mt cng OR 2 li vo.


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b. Cng OR TTL

Hnh 3-6 l s ca mt cng OR h TTL tiu chun hai li vo. Trong trng hp ny,mch vo s dng cc bn dn n. Tuy nhin, nguyn l hot ng ca mch vo ny cng

ging vi cng NAND hnh 3-5.c. Cng collector h

Nhc im ca h cng TTL c mch ra khp kn l h s ti u ra khng th thay i,nn nhiu khi gy kh khn trong vic kt ni vi u vo ca cc mch in t tng sau. Cnglogic collector hkhc phc c nhc im ny. Hnh 3-7 l s ca mt cng TTL ocollector h tiu chun. Mun a cng vo hot ng, cn u thm tr gnh ngoi, t cccollectorn +Vcc.

Hnh 3-7. Mch in ca mt cng NOT collector h.

Mt nhc im ca cng logic collector h l tn s hot ng ca mch s gim xungdo phi s dng in trgnh ngoi.

d. Cng TTL 3 trng thi

Mt cng logic, ngoi hai trng thi cao v thp ti u ra ca n cn c mt trng thitrung gian c gi l cng ba trng thi. Trng thi trung gian ny cn c tn l trng thi u rac trkhng Z cao hay trng thi treo. Cng c k hiu nh chhnh 3-8.

Tng t nh cng collector h, cc h cng logic u c cng 3 trng thi. Hnh 3-8 lmt v d v mch in ca cng NAND ba trng thi h TTL tiu chun .

Hnh 3-8. K hiu ca cng ba trng thi : (a) cng NOT; (b) cng AND.

Hot ng ca cng NAND 3 trng thi c gii thch bng bng trng thi 3-3. Khi trn

li vo E c mc logic thp, cng hot ng nh mt cng NAND. Trn li ra fs tn ti haitrng thi cao v thp nh thng l.

Q3 f

D1

R1

4k

Q1A

+5V

Q2

R21,6k

R3

1,6k

A f

A

E

F

(a)

A

E

FB

(b)


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Hnh 3-9. Mch in cng NAND 3 trng thi v s tng ng ca n.

Ngc li, khi trn li vo E mc cao thbt lun trn hai li vo A, B c gi tr logic no(du x trong bng trng thi mang ngha tuchn) li ra f lun trng thi treo, hay th ni.Trng thi ny tng ng vi trng thi u rakhng c ni ti mt im no trong mch. ng

vi trng thi ny, trkhng Z trn u ra ca cng,nhn t pha ti vo s rt ln. Theo s tngng, lc ny c Q4, Q5u kho. Li ra f dngnh b treo trong mch. Do , trng thi ny cnc gi l trng thi treo.

Trong k thut s, cng ba trng thi thngc dng lm cc bm u ra, kho iu khinhng d liu ...

e. H TTL c diode Schottky ( TTL + S )

Cng TTL tiu chun c nhc im chung l thi gian tr truyn lan ln. Nguyn nhnca nhc im ny l do tt c bn dn trong mch u cng tc ch bo ho. Mt trongnhng bin php gim nh tr truyn lan l s dng diode Schottky chng hin tng bo hony.

Diode v bn dn Schottky

Cu to ca diode Schottky cng ging nh diode Silic. Nhvic chn thm mt lp oxitkim loi vo gia tip gip p-n m in th phn cc ca n l 0,4 Vdc (thp hn 0,6 vn i vidiode Silic v cao hn 0,2 vi diode Ge).

K hiu ca diode v bn dn Schottky cho hnh 3-10.

+5V

Q3

R3

1,6k

Q5

D2

f

Q4

R5

130

R41k

D1

A

R1

4k

Q1

R2

4k

2E

+Vcc

R5

4

Q5

Li ra Z caoB

E A B f

L L L H

L L H H

L H L H

L H H L

H x x -

H x x -

H x x -

H x x -

Bng 3-3. Bng trng thi cacng 3 trng thi.


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Hnh 3-10. Cu to ca diode Schottky

Mch in dng diode Schottky chng bo ho cho cc bn dn nh hnh 3-10b. ngin, ngi ta gi mch ny l bn dn Schottky v k hiu nh hnh 3-10c.

Mch in h cng TTL + S

Nu thay tt c diode v bn dn trong mch in ca h TTL tiu chun bng cc diodev bn dn Schottky, ta s c mch in h cng TTL+S. Hnh 3-11 l mt v d v cng NANDdng diode Schottky.

Nhs dng diode v bn dn Schottky m tn s cng tc ca h cng ny tng ng k.Thi gian tr truyn lan ca cng TTL+S khong 3 ns, cng sut tiu th khong 19 mW.

Khi ch tiu thi gian tr khng cn cao th gi tr cc in trphn cc c tng ln gim dng tiu th ca mi bn dn xung. H cng nh th c tn gi l TTL+LS (TransistorTransistor Logic + Lowpower Schottky Diode). Cng sut tiu th ca h cng ny ch khong 2mW v thi gian tr truyn lan vn t khong 9,5 ns.

Nu cn nng cao tn s cng tc, ngoi vic gim tr s cc in trphn cc, ngi ta cndng cc cch ni mch ci tin. H cng thu c c tn l TTL+AS.

C

B

E

C

B

E

a) K hiu DiodeSchottky

b) Cu to bn dnSchottky

c) K hiu bndn Schottky

R350

R2900

Q3

+Vcc

500R6

250R5

Q4

B

A

R18,2k

Q1

D1 D2

Q2

Q6

R43,5k

Q5

f

Hnh 3-11. Mch in ca cng NAND 2 li vo h TTL+S


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3.1.5. H MOS FET

Bn dn trng (MOS FET) cng c dng rt ph bin xy dng mch in cc loicng logic. c im chung v ni bt ca h ny l:

Mch in ch bao gm cc MOS FET m khng c in tr

Di in th cng tc rng, c th t +3 n +15 V

tr thi gian ln, nhng cng sut tiu th rt b

Tu theo loi MOS FET c s dng, h ny c chia ra cc tiu h sau.

1. Loi PMOS

Mch in ca h cng ny ch dng MOSFET c knh dn loi P. Cng ngh PMOS chophp sn xut cc mch tch hp vi mt cao nht.

Hnh 3-12 l s cng NOT v cng NOR loi PMOS. y MOSFET Q2, Q5 ngchc nng cc in tr.

Hnh 3-12. Mch in ca cng NOT v NOR theo cng ngh PMOS.

2. Loi NMOS

Hnh 3-13. Mch in cng NAND v NOR theo cng ngh NMOS.

VSS

S

G

D

Q2

VDD

S

G

D

Q1A

f = A

VSS

S

GD

Q5

A

B

VDD

S

GD

Q4

S

GD

Q3

f= A+B

a) Cng NOT b) Cng NOR

VSS

VDD

Q1

Q2

Q3

A

B

f

A B

VSS

VDD

Q1

f

Q3Q2

a) Cng NAND b) Cng NOR


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Hnh 3-13 l s cng NAND v NOR dng NMOS. Du + trn cc li vo mun ch cctnh ca tn hiu kch thch. Trong trng hp ny, Q1 cng ng chc nng l mt in tr. ivi cng NAND, ta nhn thy rng ch khi trn c hai li vo A v B u ly mc cao th u rami c mc thp. ng vi 3 t hp bin vo cn li, li ra fu c logic thp. Hot ng cacng NOR cng c gii thch tng t.

3. Cng CMOS

CMOS l vit tt cc t ting Anh Complementary MOS. Mch in ca h cng logicny s dng c hai loi MOS FET knh dn P v knh dn N. Bi vy c hin tng b dng introng mch. Chnh v th m cng sut tiu th ca h cng, c bit trong trng thi tnh l rt b.

Hnh 3-14 l mch in ca cng NOT v NAND thuc h CMOS. im ni bt trongmch in ca h cng ny l khng tn ti vai tr cc in tr. Chc nng logic c thc hin

bng cch thay i trng thi cc chuyn mch c cc tnh ngc nhau. Du tr v du cng trn

cc ca cc MOSFET ch ra cc tnh iu khin chuyn mch. Nhc im cu trc mch, mcVRL, VRHt c gn nh l tng.

minh ho, ta c th tm hiu hot ng ca cng NOT. T hnh 3-14a, d thy rng, nutc ng ti li vo A logic thp th Q1 s thng, Q2 kho. Li ra f gn nhc ni tt ti VDD vcch ly hn vi t, ngha l VRH VDD. Ngc li, khi A ly mc cao, Q1 mv Q2ng. Do ,li ra f gn nh ni t v cch ly vi VDD. Ni khc i, VRL 0.

Hnh 3-14. Mch in ca h cng CMOS.4. Cng truyn dn

Da trn cng ngh CMOS, ngi ta sn xut loi cng c th cho qua c tn hiu s lntn hiu tng t. Bi vy cng c gi l cng truyn dn. S nguyn l v k hiu cngtruyn dn nh hnh 3-15.

S

G

DD

G

S

VDD

Q1

Q2

fA

S

G

D

S

G

D

Q4

A

B

VDD

Q2

Q3

Q1f

D

G

S

a) Cng NOT b) Cng NAND


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- 1,75 V

- 0,9 V

- 1,4 V - 1,2 V Vo

Ra

a) Mch in nguyn l b) th mc vo/ra

Livo

-Vcc = - 5V

A

Li ra NOR

Li ra OR

R5

Q3

R3

Q4

RE

R6

Q5

R4

Q2

R2

Q1

R1

-1,29

R9

D1D2

Q6

R7

R8

+Vcc

Q8

Q7B

D

C

Hnh 3-15. Cng truyn dn.

Mch nguyn l ca cng truyn dn cng s dng hai MOSFET c knh dn ngc nhau.Tuy nhin cch iu khin trng thi cc chuyn mch li khc vi cng logic thng thng.

Trong trng hp ny, ngi ta phn cc sao cho khi c tn hiu iu khin th c hai chuynmch Q1 v Q2 cng dn in. Khi , mch tng ng nh mt dy dn. Cc cng o (trongs k hiu) m bo cc tnh iu khin ph hp cho c hai cc G ca mi MOSFET.

Tnh dn in ca cng truyn dn ph thuc mnh vo tn s cng tc v gi tr ti. V sdng cng ngh CMOS nn tn s cng tc ca cng ch gii hn 6 MHz.

H CMOS cng c cng D hv cng ba trng thi nh h TTL.

3.1.6- H ECL

ECL (Emitter Coupled Logic) l h cng logic c cc E ca mt s bn dn ni chung vi

nhau. H mch ny cng s dng cng ngh TTL, nhng cu trc mch c nhng im khc hnvi h TTL. Ngoi vic s dng hi tip m trn in trRE chng bo ho, mch in ca hECL cn tn dng c u im ca mch khuch i vi sai, nn tn s cng tc h ny l caonht trong cc h. Ngoi tr thi gian tr, tt c cc tham s cn li u km hn cc h khc.

Hnh 3-16. Cng OR/NOR thuc h ECL.

Hnh 3-16 l mch in v th mc vo ra ca mt cng OR/NOR thuc h ECL. V in

thtrn hai cc collector ca Q4, Q5 l b nhau nn c th ly ra cc E ca Q7 chc nng ORv cc E caQ8 chc nng NOR. mch hot ng theo logic mc m, +Vcc c ni t, -

Ra/Vo

S

G

D+5V

Q1

Q2

Vo/RaS D

G

Ra/VoVo/Rao

iu khin

a) Mch in b) K hiu


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Vcc c ni ti m ngun. Mc logic trong mch c bin i t gi tr thp l -1,75 V n gitr cao l - 0,9 V so vi in tht. Khi mun c mc logic ra dng cc cc E ni ti t.

3.2. GIAO TIP GIA CC CNG LOGIC CBN TTL-CMOS V CMOS-TTL

Trong nhiu ng dng, yu cu chuyn i cc tn hiu gia cc mc logic khc nhau nht TTL sang CMOS hoc ngc li. Cc cng logic collector h hoc cc mch khuch itransistorn gin thng c s dng trong cc mch chuyn i ny.

3.2.1. Giao tip gia TTL v CMOS.

to c giao tip gia TTL v CMOS th ta phi n ngun cung cp ca 2 h. HTTL cn in p cung cp l + 5V, h CMOS c th dng in p cung cp t +3V n +15V.

a. Cng in p cung cp +5V.

Trong trng hp ny in p ra ca TTL nh hn so vi in p vo ca CMOS. Do vy taphi dng mch b sung tng hp hai loi IC khc nhau.

Gii php tiu chun l dng in trko ln gia iu khin TTL v ti CMOS nh hnh3-17.

Hnh 3-17. iu khin TTL v ti CMOS

b. Khc in p cung cp.

in p cung cp dng cho IC CMOS thch hp nht l t +9V n +12V. Mt cch dngin p cung cp ln l s dng IC TTL hmch Collector nh hnh 3-18, v tng ra caTTL hcc C ch gm transistor nhn dng vi cc C th ni. hnh ny cc C hc ni

vi ngun cung cp +12V qua in trko ln 6,8k. Khi li ra ca h TTL mc L th dng

ca n l:

Inhn dng =12V

1,76mA6,8k

=

Khi li ra ca TTL mc H th li ra ca cc C h tng ln mt cch thng n+12V. Trong trng no th cc li ra ca TTL cng u tng hp vi cc trng thi li voca CMOS.

+ 5V

Ti CMOS

iu khin

TTL

Rp


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Hnh 3-18. iu khin TTL hmch Collector v ti CMOS

c. B chuyn mc ngun dng CMOS.

Hnh 3-19 l b chuyn mc CMOS 40109. Tng li vo ca IC dng in p cung cp+5V trong khi tng li ra dng +12V.

Trong hnh 3-19, IC TTL tiu chun iu khin b chuyn mc ngun, n ko IC TTL ln tnht l +2,4V. in trko ln tip tc a in p ln cao n mc +5V, mc ny m bo chcchn li vo mc H. Li ra ca b chuyn mc ni vi ngun +12V.

Hnh 3-19. B chuyn mc CMOS cho php s dng hai loi ngun +5V v +12V.

3.2.2. Giao tip gia CMOS v TTL

to ra c giao tip gia h CMOS v TTL th ta phi quan tm n vn chuynmc in p cho ti khi trng thi li ra ca CMOS ph hp vi li vo ca TTL. Ta phi m

bo chc chn li ra trng thi L ca CMOS lun lun nh hn 0,8 V(y l in p li vo lnnht trng thi L ca h TTL). in p li ra trng thi H ca CMOS lun lun ln hn 2

V(y l in p li vo nh nht trng thi H ca h TTL).a. Cng in p cung cp +5V.

Theo s liu k thut ca IC 74Cxx th trng hp xu nht dng li ra ca CMOS iukhin TTL l:

IOL MAX = 360A ; IOH MAX = - 360A

iu ny c ngha l iu khin CMOS c th cho nhn dng l 360 A khi trng thi L, l dng vo i vi IC TTL loi Schottky cng sut thp. Mt khc, iu khin CMOS c thcho dng ngun 360 A, n ln hn mc cn thit iu khin dng vo trng thi H. Nh

vy h s ghp ti gia CMOS v 74LS l bng 1.

+ 12V

Ti CMOSTTL hmch

Collector

6,8k

+ 5V

+ 5V

B chuyn mc40109 Ti CMOSiu khin

TTL

+ 12V

3,3k


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i vi loi IC TTL cng sut thp th c dng li vo l 180 A th h s ghp ti giaCMOS v 74L l bng 2.

IC CMOS khng thiu khin trc tip IC TTL tiu chun, v dng li vo trng thi L

yu cu l 1,6 mA, m transistor nhn dng ca IC CMOS c in trxp x 1,11k (trng hpxu nht). Nn in p li ra ca IC CMOS bng 1,6 mA x 1,11k = 1,78 V. in p ny qu lni vi li vo trng thi L ca IC TTL.

- Dng tng m bng CMOS.

Hnh 3-20. Tng m CMOS c thiu khin ti TTL tiu chun

Hnh 3-20 l mch iu khin IC CMOS vi h s ti qua tng m. Tng m c dng raln. V d IC 74C902 c 6 tng m CMOS, mi tng m c dng li ra trong trng hp xunht l:

IOL MAX = 3.60mA

IOH MAX = 800A

V ti TTL tiu chun c dng li vo trng thi L bng 1,6mA v dng li vo trngthi H l 48 A, IC 74C902 c thiu khin hai ti TTL tiu chun.

Cc IC khc c dng lm tng m nh hnh 5-19 l IC CD4049A, 4050: o;CD405CA: khng o, 74C901: o

b. Khc in p cung cp.

Cc tng m CMOS nh 74C902 c th dng in p cung cp t +3V n +15V v inp li vo t -0,3 V n +15V> in p li vo c th ln hn in p cung cp m khng lmhng loi IC dng lm tng m ny. V d ta c th dng in p li vo trng thi H l +12Vngay khi in p cung cp ch bng 5V.

Hnh 5-23 l mch iu khin CMOS dng in p cung c p +12V, trong khi tng mCMOS c in p cung cp l +5V.

Hnh 3-21.iu khin CMOS hot ng thch hp nht vi ngun cung cp +12V.

+ 5VTng m

CMOS

Ti TTLiu khin

CMOS

+ 5VTng m

CMOS

Ti TTLiu khin

CMOS

+ 12V


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c. Giao din ca hcc mng.

Ta bit IC TTL hmch Collector, tng li ra ca transistor nhn dng vi cc C thni. Tng t nh vy i vi IC CMOS cng c hcc mng. V d: IC 74C906 c 6 tng m

hcc mng.

Hnh 3-22. Tng m CMOS hcc mng lm tng dng nhn.Hnh 3-22 l mch dng tng m CMOS hcc mng lm giao din iu khin CMOS v

ti TTL. in p cung cp cho hu ht cc tng m l +12V. Tuy vy c th ni tng m hccmng vi ngun cung cp +5V qua mt in trko ln (pull up) c gi tr 3,3k. Cch ni nyc u im l ciu khin CMOS v tng m CMOS u c cung cp ngun +12V, khngk li ra hcc mng giao din vi TTL

TM TT

Chng 3 trnh by cu trc, nguyn l v c im ca cng thng dng. Xut pht t

thc t mch in vi mch ho, nn trng tm ch nghin cu ca chng ta l cc cng cvi mch ho.

C 2 loi vi mch s ph bin nht : TTL v MOS. TTL l cng nghin hnh trong nhmcng ngh transistor bao gm TTL, HTL, ECL, MOS l cng ngh vi mch s dng MOSFET,trong in hnh l MOS

ng thi trong chng 3 cng a ra vn giao tip gia cc h cng vi nhau.

CU HI N TP

1. Chc nng ca mch logic RTL c s nh hnh v sau:

a. NOR

+ 5V

Tng mCMOS hcc mng

Ti TTLiu khinCMOS

+ 12V

3,3k


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b. OR

c. AND

d. NAND2. Vi mch c s nh trong cu hi 1, nhng in p logic li vo tng ng vi cc

mc logic cao v thp ln lt l 10 V v 0 V th chc nng ca mch l g?

a. NOR

b. OR

c. AND

d. NAND

3.

Cho mch c s nh s sau, in p logic li v tng ng vi cc mc logic cao vthp ln lt l 1 V v 0 V, nu chc nng ca mch?

a. NOR

b. OR

c. AND

d. NAND

4. Chc nng ca diode D3 trong s sau l g?

a. Cch ly transistor Q3 v Q4

b. Dch mc in p lm cho Q3 v Q4 khng bao gicng ng hoc cng m


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c. Chng nhiu li ra

d. Cch ly Q4 khi mch ngoi ni vo u ra f

5. Chc nng ca mch biu din trong s nh cu hi 4 s thay i th no nu diode D3chuyn ti chn base ca transistor Q3 (cathode D3 ni vi base Q3 cn anode ni vicollector Q2)?

a. Q3 lun cm

b. Q3 lun m

c. Chc nng ca mch khng thay i

d. Li ra lun trng thi treo

6. Cng collector hs hot ng bnh thng nh cc cng logic bnh thng nu :

a. Li ra c ni ln ngun thng qua mt trgnh

b. Li ra c ni ln ngun thng qua mt t gnh

c. Li ra ni xung t thng qua mt tr

d. Li ra ni xung t thng qua mt t

7. Tc dng ca trng thi trkhng li ra cao trong cng ba trng thi l :

a. a ra mc logic th 3 l trung bnh ca hai mc cao v thp

b. Cch ly gia cc li ra ca cc cng logic khi chng cng c ni vo mt livo

c. C mc logic thp nhng trkhng cao

d. C mc logic cao nhng trkhng cao

8. Mch in c biu din trong s sau c cn hot ng nh bnh thng khng nunh diode D1 b ni tt ?

a. Mch trthnh cng NAND vi hai trng thi li ra nh cc cng NAND thngb. Mch trthnh cng NOR


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c. Trng thi li ra khng theo logic cbn no

d. Vn hot ng bnh thng l cng NAND 3 trng thi

9. Mch in nh trong cu hi 8 c cn hot ng nh bnh thng khng nu nhin trR4 c gi tr bng 10 k?

a. N s hot ng nh mch NOR

b. N s hot ng nh mch XOR

c. Vn hot ng bnh thng

d. C ba cch tr li trn u sai

10.Vi mch in TTL nh s trong cu hi 4, hin tng g s xy ra khi mt trong haili vo lng?

a. Li vo ny c tnh logic 0

b. Li vo ny c tnh logic 1

c. Mch khng hot ng


11.So snh cng NOT h MOS v CMOS ta thy :

a. Cng sut tiu th ca MOS cao hn CMOS

b. Cng sut tiu th ca CMOS cao hn MOSc. Cng sut tiu th ca hai h nh nhau


12.C cho php u vo ca mch CMOS l lng khng? C th ni u vo lngtng ng vi mc cao khng?

a. c- C th coi l mc 1

b. c- Phi coi l mc 0

c. Khng c- mch hot ng bnh thng th u vo khng dng phi nivi mc logic 0

d. Khng c- mch hot ng bnh thng th u vo khng dng phi nivi mc logic 1

13.Cng truyn dn l cng

a. Ch cho php tn hiu si qua theo mt chiu nht nh

b. Ch cho php tn hiu si qua theo hai chiu

c. Ch cho php tn hiu tng ti qua theo mt chiu nht nh

d. Cho php tn hiu tng ti qua theo hai chiu


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14.u im ca cc cng logic h ECL l

a. Tn s cng tc nhanh

b. in p ngun nui thp

c. Cng sut tiu th thp

d. chng nhiu cao


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Chng 4: Mch logic thp

48

CHNG 4: MCH LOGIC T HP

GII THIU CHUNG

Cc hm logic c thc hin nhcc h vt l gi l cc h logic hay l cc mch logic.Trong chng 4 chng ta cp n cc mch logic t hp, tc l cc mch m tn hiu u rach ph thuc vo tn hiu u vo ca mch ti thi im ang xt. Ni cch khc, cc tn hiura khng ph thuc vo "lch s " ca tn hiu vo trc , ngha l cc h ny lm vic theonguyn tc khng c nh. Hot ng ca cc mch t hp c m t bng cc bng trng thihoc bng cc hm chuyn mch logic c trng cho quan h gia cc i lng vo v ra ca hthng. V mt cu trc, cc mch t hp khng cha mt thit b hoc mt phn t nhthng tinno c.

Trong chng ny cp n cc mch in c th thc hin cc chc nng khc nhau cah thng s. Cc mch in ny c thit k da trn cc cng logic t hp. Cc cng logic nyc tch hp trong mt IC cva (MSI) c cha khong vi chc ti vi trm cc cc cng logiccs c xt n chng 4. Nhng linh kin ny c ch to nhm thc hin mt s cchot ng thu nhn, truyn ti, bin i cc d liu thng qua tn hiu nh phn, x l chng theomt phng thc no .

Phn u ca chng gii thiu cch phn tch v thit k cc mch logic t hp n gin.

Phn tip theo gii thiu v Hazard trong mch logic t hp. y l phn rt quan trng khithit k mch. Nu khng n hin tng ny c th dn n s lm vic sai lch ca c hthng. Phn tch v nhn dng Hazard c ngha rt quan trng khng nhng trong tng hp cch logic m c trong tng chn on trng thi lm vic ca chng.

Phn tip theo gii thiu mt s mch t hp thng dng trong cc h thng s:

- M ho v gii m cc lung d liu nh phn.

- Hp knh v phn knh chn hoc chia tch cc lung s nh phn theo nhng yu cunht nh nh tuyn cho chng trong vic truyn dn thng tin,

- Cc mch cng, tr.- Cc php so snh snh gi nh tnh v nh lng trng s ca cc s nh phn.

- Mch to v kim tra tnh chn l.

- n v s hc v logic (ALU).


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49

NI DUNG

4.1 KHI NIM CHUNG

Cn c vo c im v chc nng logic, cc mch sc chia thnh 2 loi chnh: mch thp v mch tun t (mch tun tc trnh by chng sau).

1)c im cbn ca mch thp

Trong mch s, mch t hp l mch m tr sn nh ca tn hiu u ra thi im angxt ch ph thuc vo t hp cc gi tr tn hiu u vo. c im cu trc mch t hp l ccu trc nn t cc cng logic. Vy cc mch in cng chng 2 v cc mch logic chng 3u l cc mch t hp.

2) Phng php biu din chc nng logic

Cc phng php thng dng biu din chc nng logic ca mch t hp l hm slogic, bng trng thi, s dng logic, bng Cac n (Karnaugh), cng c khi biu th bng ththi gian dng xung.

i vi vi mch cnh (SSI) thng biu din bng hm logic. i vi vi mch c va(MSI) thng biu din bng bng trng thi.

S khi tng qut ca mch logic t hp c trnh by hnh 4-1.

Nh vy, mch logic t hp c th c n li vo v m li ra. Mi li ra l mt hm ca ccbin vo. Quan h vo, ra ny c th hin bng h phng trnh tng qut sau:

Y0 = f1(x0,x1,...,xn-1);

Y1 = f2(x0,x1,...,xn-1);

Ym-1 = fm-1(x0,x1,...,xn-1).

T, ta thy rng c im ni bt ca mch logic t hp l hm ra ch ph thuc cc binvo m khng ph thuc vo trng thi ca mch. Cng chnh v th, trng thi ra ch tn ti trongthi gian c tc ng vo.

Th loi ca mch logic t hp rt phong ph. Phm vi ng dng ca chng cng rt rng.

Mch logic thp

x0

x1

xn-1

Y0

Y1

Ym-1

Hnh 4-1 S khi tng qut ca mchlogic t hp.


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VAC

A B

1

0

1

0

Hnh 4-2 Mch in ca h thngchiu sng

4.2 PHN TCH MCH LOGIC T HP

Phn tch mch logic t hp l nh gi, ph phn mt mch . Trn cs, c th rtgn, chuyn i dng thc hin ca mch in c c li gii ti u theo mt ngha no y.

Mch t hp c th bao gm hai hay nhiu tng, mc phc tp ca ca mch cng rtkhc nhau.

Nu mch n gin th ta tin hnh lp bng trng thi, vit biu thc, rt gn, ti u (nucn) v cui cng v li mch in.

Nu mch phc tp th ta tin hnh phn on mch vit biu thc, sau rt gn, ti u(nu cn) v cui cng v li mch in.

4.3 THIT K MCH LOGIC T HP

Thit k l bi ton ngc vi bi ton phn tch. Ni dung thit kc th hin theo tunt sau:

1- Phn tch bi ton cho gn hm v bin, xc lp mi quan h logic gia hm v ccbin ;

2- Lp bng trng thi tng ng;

4- T bng trng thi c th vit trc tip biu thc u ra hoc thit lp bng Cac n tngng;

4- Dng phng php thch hp rt gn, a hm v dng ti gin hoc ti u theo

mong mun;5- V mch in th hin.

V d : Mt ngi nh hai tng. Ngi ta lp hai chuyn mch hai chiu ti hai tng, sao chotng no cng c th bt hoc tt n. Hy thit k mt mch logic m phng h thng ?

Li gii:

+ Nu k hiu hai cng tc l hai bin A, B. Khi tng 1 ta bt n v ln tng 2 th tt ni v ngc li. Nh vy n ch c th sng ng vi hai t hp chuyn mch v tr ngc nhau.Cn n tt khi v tr ging nhau. H thng chiu sng trong c s nh hnh 4-2.

Bng trng thi m t hot ng ca h nh chbng 4-1.

Biu thc ca hm l: f A B A B = A B= +

hoc

f AB A AB B=

y l hm cng XOR quen thuc ccchng trc. Hm ny c thc th hin bng nhiu kiu mch khc nhau. Hnh 4-3 l mtdng s th hin hm f.


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4.4 HAZARD TRONG MCH T HP

4.4.1. Khi nim.

Vic thit k cc mch logic nhn chung khng phc tp, v cn c biu thc ton l ta c

th v ra c mch in v lp rp thnh h thng iu khin. Trn thc t, khng phi mch nocng c th hot ng tt c, nguyn nhn l do cu trc ca mch t hp gy ra, hin tnghot ng khng n nh xy ra trong mch t hp c gi l hazard.

Hazard cn c gi l s "sai nhm", hot ng lc c lc khng ca mch logic. S"sai nhm" ny c th xy ra trong mt mch in hon ton khng c hng hc linh kin. Tc ltrong mch, cc linh kin hon ton tt nhng iu khin chc nng lc c lc khng. Nichung l mch hot ng khng c s tin cy. Hin tng ca Hazard trong mch t hp c thgp l:

- Hazard ch xut hin mt ln v khng bao gigp li na.

- Hazard c th xut hin nhiu ln (theo mt chu k no hoc khng theo mt chu kno).

- Hazard c th do chnh chc nng ca mch in gy ra. y l trng hp kh gii quytnht khi thit k.

Nh ta bit, mt trong cc c tnh quan trng nht ca mch in khi hot ng l quntnh, linh ng hay s chm tr ca mch. Chnh s chm tr ny lm cho tn hiu tu vokhng th truyn ngay tc khc ti u ra ca mch in, iu ny lm cho cc thit biu khin

pha sau khng th c phn ng tc khc i vi tn hiu a vo. Do tt c cc mch in u cthi gian tr nht nh, ngay ccc mch vi in t cng c thi gian tr. S thay i nhit

mi trng cng lm cho thi gian tr thay i, dn n s sai lch khi iu khin ca mch logic, chnh l hazard.

4.4.2. Bn cht ca Hazard

hiu c nguyn nhn xut hin hazard trong mch logic t hp, hazard ch xut hintrong mch t hp m khng xut hin bt k h thng in t no khc. Ta xt v d sau:

Gi s tn hiu vo l X = (x1,x2, x3, x4) thay i gi tr t (0 0 0 1) n (1 1 1 1), tc l (X)thay i t QP. Nhn vo bng Cac n (hnh 4-4) ta thy p ng ra ca mch logic t hp khitn hiu vo b thay i c gi tr:

A B f0 0 00 1 11 0 1

1 1 0

A

Bf

Bng 4-1. Bng trng thi mt hot ng ca h chiu sng

Hnh 4-3. S logic th hin hm f

f(Q) = f(0001) = 1 f(P) = f(1111)= 1


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Nh vy tn hiu vo (X) thay i gi tr t Q = (0001) n P = (1111) lm cho p ng raca mch b thay i gi tr t , s thay i iu khin u ra ca mch theo sthay i tn hiu vo (X) iu ny hon ton chnh xc, khi hazard khng xut hin v

khng xy ra iu khin b sai nhm.Nhng thc t c th khng c nh vy v khi tn hiu vo thay i t Q = (0001) n P

= (1111), ta thy tn hiu x1, x2, x3 b thay i cn gi tr x4 khng b thay i. Mch in no cngxut hin thi gian tr l () v s thay i gi tr (01 hay 10) ca tn hiu u c thi gian trnht nh.

Trong trng hp ny, cc tn hiu vo (x1, x2, x3) c gi tr logic b thay i khi ta thay ib tn hiu vo, v chng s c mt thi gian tr nht nh (c th rt nh, cs hay ns). Mtkhc, thi gian tr ca mi ng tn hiu vo (xi) li khc nhau, d cng mt chng loi IC. Nhvy nu (x1, x2, x3) c thay i ng thi v chng c thi gian tr khc nhau th vn xy ra hin

tng "chy ua" ca tn hiu vo ti u ra ca mch in.

V c s "chy ua" gia ba tn hiu vo (x1, x2, x3) (x4 khng thay i nn khng ua), gis x2 chy nhanh hn (c thi gian tr nh hn) x1, x2 (gi s thi gian tr ca hai tn hiu ny

bng nhau). Mi quan h ny ta c th biu din nh sau:

(X) (x1 x2 x3 x4) p ng ra

t0 0 0 0 1 f(Q) = 1

t'0 0 1 0 1 f(0101) = 0

t1 1 1 1 1 f(P) = 1

Do x2 "chy" nhanh hn x1 v x3 nn gi tr ca x2 chuyn t 0 sang 1 trc gi tr ca x1 vx3. Sau mt thi gian th (x1, x3) mi chuyn t 0 sang 1.

Quan h "chy ua" gia ba tn hiu vo c minh ho bng biu sau:

Do x2 "chy nhanh" hn (x1, x3) nn trong khong thi gian t xut hin mt xung zrnht thi. Nh vy trong thi gian tr ca mch tn hiu ra thay i t 101 (ng ra lkhng c thay i), to ra mt xung kim nht thi. Hin tng xut hin mt xung zr u

ra ca mch c gi l hin tng hazard v y l hazard nht thi, n ch xut hin trong thigian tr sau li mt ngay. Nh vy ta c th ni rng s "chy ua" ca tn hiu vo gy ra

1 sang 1

x1x2 00 01 11 10

00 1 1 1 0

01 0 0 0 1

11 1 1 1 0

10 0 0 1 1

Hnh 4-4. Mch chc nn lo ic

x3x4

Mch

logic

x1x2x3x4

f(x)

t0

t1t'0


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hazard, hay thi gian tr ca mch s lm xut hinhazard, l tn hiu iu khin khng mong munu ra.

Xung Hazard l mt xung kim xut hin ura ca mch logic t hp, v thi gian xut hin (t)nh hn thi gian tr ca mch () nn xung hazardc th xut hin nhng khng gy nguy him, khnggy ra siu khin sai nhm. V xung hazard quhp nn nng lng ca n khng ln c thkch nhm hay kch c cc mch in tip theo,do d c xung hazard nhng mch in vn hotng tt. Xung hazard ch tht s nguy him khi

rng t ln th n c nng lng lt chuynmch in ti p theo gy ra hin tng iu khinnhm.

Nh vy c th thy vi b tn hiu vo thayi kiu khc vi t hp trn th c th khng xut hin xung hazard. Hay vi mt chc nng khcd c hin tng "chy ua" tn hiu vo gia (x1,x3 v x2) nh v d trn nhng c f(0101) = 1th hazard cng khng th xut hin do xung zr nht thi khng c.Do vy ta thy hin tnghazard xut hin rt ngu nhin cho d mch in cha ton cc linh kin tt.

4.4.3. Phn loi.

u tin ta cp n mt snh ngha tn gi khi ni v hazard nh sau:Q = (q1, q2, ....qk, qk+1,...qn )

P = ( n1kk21 q,...q,q...q,q + )

y P v Q l tp tn hiu vo ca mch, nhng yu cu gia P v Q cn c s lng v trthay i gi tr logic 2, v ch khi tp tn hiu vo thay i gi tr logic ng thi vi t nht 2 vtr (2 bin s) th mi xut hin hin tng "chy ua" tn hiu vo, v khi hazard mi c khnng xut hin. Cn nu tn hiu vo ch thay i gi tr ln lt trn tng u vo mt th skhng c hin tng chy ua tn hiu v hazard khng th xut hin c.

nh ngha 1: Nu tp tn hiu vo (X) thay i t Q sang P th c gi l c s chuyni t Q sang P (Q P).

nh ngha 2: Hazard nht thi xut hin trong mch logic t hp l hin tng tn hiu ramt hoc nhiu u ra ca mch xut hin khc vi cc gi tr quy nh cho chng theo hmBoole trong thi gian chuyn i t Q P.

nh ngha 3: Hazard nht thi xut hin trong mch logic t hp trong thi gian chuyni t Q P gi l hazard tnh nu v ch nu f(Q) = f(P). yf(X) l hm logic c thchin bi cc mch cho.

nh ngha 4: Hazard nht thi xut hin trong mch logic t hp trong thi gian chuyni t Q P gi l hazardngnu v ch nu f(P)f(Q)= . Nh vy khi c hazard nht thi

thi gian tr

x1, x3

0x2

0f(x)

0

t

t

t

Q P

Hnh 4-5. Hin tng hazard

t0 t'0 t1

t

1 0 1


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x1, x4

0x2

0f(x)

0

t

t

t

Q P

Hnh 4-6. Chy ua hazard tnh

t0 t'0 t1

t

ng th tn hiu u ra thay i t nht ba ln, v d 1010, ngha l c t nht hai xungnhiu xut hin. Loi hazard ny thng xy ra trong cc mch t hp.

nh ngha 5: Hazard nht thi gi l hazard hm s trong thi gian chuyn i t QP

nu:-f(Q)=f(P)

- Hmf(X) ly c hai gi tr 1 v 0 trong thi gian chuyn i t QP

nh ngha 6: Hazard nht thi gi l hazard logic trong thi gian chuyn i t QP nu:

-f(Q)=f(P)

- Hmf(X) ch nhn mt gi tr nh nhau (hoc 0 hoc 1)

- Trong thi gian chuyn i t QP xut hin mt xung hazard u ra.

4.4.3.1. Hazard tnh trong mch logic.

Do c hin tng "chy ua" gia cc tn hiu vo vi nhau trong thi gian chuyn t QPm xut hin hazard. Nu f(Q) = f(P) tc l c s thay i ca tn hiu vo nhng siu khin u ra ca mch logic vn khng i d l 0 hay 1, nhng xut hin hazard, khi s lng tn hiuchy ua khng nhiu, chnh l hazard tnh.

Hazard nht thi cng chnh l hazard tnh, tc l loi hazard ch xut hin nh mt xungkhng theo quy nh ca hm logic. Hin tng ny khng nguy him, v rng ca xunghazard tnh t lun nh hn thi gian tr ca mch, nn mch logic vn hot ng bnh thng

d c xut hin hazard.Nhng hazard tnh nguy him ch: n c th gy ra "sai nhm" cho iu khin ca h

thng logic khi gi tr rng hazard (t) ln, iu ny s xy ra khi s "chy ua" ca tnhiu vo qu chnh lch, ngha l c tn hiu vo "chy" qu nhanh cn tn hiu khc li "chy"qu chm, hin tng ny c minh hohnh 4-6.


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Ta thy x2 trong qu trnh "chy ua" (thay i gi tr logic) "chy" nhanh hn so vi tn

hiu x1, x4, th hin hnh v dc xung x2 ln hn, iu lm cho t ca xung hazard tngtheo, khi xung hazard trnn "nguy him" hn v n c th kch lt chuyn mt mch in tip

sau h thng mch logic, gy hin tng iu khin "sai nhm" trong mch logic.4.4.3.2. Hazardng trong mch logic.

Trong thc t khi thay i tn hiu vo ca mch logic ng vi qu trnh chuyn i (QP)c th c rt nhiu tn hiu vo cng thay i khi c s chy ua ca cc tn hiu vo ti u raca mch. V d trng hp Q = (0000); P = (1101), d dng nhn thy c s chy ua (X)

(X) (x1 x2 x3 x4)

t0 0 0 0 0 f(Q) = 1

t'0 0 1 0 0 f(X') = 0

t"0 1 1 0 0 f(X") = 1

t1 1 1 0 1 f(P) = 0

Do c nhiu tn hiu vo ng thi thay i gi tr logic t 0 sang 1 v t 1 v 0 m mi tnhiu vo c tc "chy" khc nhau nn v tnh lm cho gi tr hm f(X) u ra thay i nhhnh bn. Hin tng tn hiu ra f(X) thay i gi tr t 1010 c gi l hazard ng, tc

l xut hin nhiu xung khng cn thit trong khong thi gian tr ca mch (). Nh vy trongthi gian rt nh xut hin rt nhiu xung hazard nh hn th ta c th hiu l xung hazardng khng c g nguy him c, v mt xung b chia ra nhiu xung con th nng lng cn rt nhv rng xung qu b nn khng kch mch khc c. Hin tng ny ta c th hiu l khin dang sng ta cho tn hiu thay i n tt nhng do c hin tng chy ua nn sau khi ntt th li hi sng ln ri mi tt hn.

Hazard ng t c kh nng gy ra iu khin "sai nhm" trong mch logic t hp.

(X)

0f(x)

0

t

t

Q P

Hnh 4-7. Hazard ng

t0 t'0 t

"0 t1


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4.4.3.3. Hazard hm strong mch logic.

Hazard c th xut hin do chc nng ca mch trong c hai trng hp l hm f(X) ly gitr logic l 0 hoc 1.

Hazard nht thi gi l hazard hm strong thi gian chuyn i t QP nu:

-f(Q)=f(P)

- Hmf(X) ly c hai gi tr 1 v 0 trong thi gian chuyn i t QP

iu ny c ngha l trong thi gian chuyn i QP th hm logic khng thay i gi tr(f(Q)=f(P)), nhng nu ly f(Q)=f(P) = 0 th th hazard vn xut hin hoc ly f(Q)=f(P)=1 thhazard vn xy ra. Hin tng ny c gi l hazard hm s. Trn thc t c nhng hm shazard nht thi ch xut hin khi iu khin logic l 1 (f(X) = 1) cn iu khin logic u ra l0 th khng c hazard nht thi xut hin v ngc li c thiu khin ra khng b hazard.

nguy him ca hazard hm s cng ging nh hazard tnh, nhng n nguy him hnmt mc na v bt k qu trnh iu khin no (0 hay 1) u c kh nng xut hin hazard, tc liu c kh nng gy ra "sai nhm" khi iu khin mch.

4.4.3.4. Hazard logic trong mch logic.

y l loi hazard nguy him nht, hay gy ra iu khin "sai nhm" nhiu nht trong cch thng mch t hp iu khin.

Bn cht ca loi hazard ny nh sau:

Khi tp tn hiu vo ca hm logic thay i ng thi nhiu bin trong thi gian chuyn iQ P, m mi mt ln tn hiu vo c thi gian tr khc nhau, trong qu trnh "chy ua" nygp phi trng hp Q = (00000), P = (11101)

(X) (x1 x2 x3 x4 x5)

t0 0 0 0 0 0 f(Q) = 1

t'0 0 0 1 0 0 f(X') = 0

t"

0 0 1 1 0 0 f(X") = 0

t"'0 0 1 1 0 1 f(X"') = 0

t1 1 1 1 0 1 f(P) = 1


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x1

x2

x3

1

1 2

2

3

Hnh 4-10. Phng php khcphc Hazard

Hin tng hazard logic c m t trnhnh 4-8:

Hazard nht thi gi l hazard logic

trong thi gian chuyn i t QP nu:-f(Q)=f(P)

- Hmf(X) ch nhn mt gi tr nh nhau(hoc 0 hoc 1)

- Trong thi gian chuyn i t QPxut hin mt xung hazard c rng t ln u ra, khi qu trnh chy ua ngu nhin cacc tn hiu vo to ra hm f(X) c cng mtgi tr logic.

Nh vy trong qu trnh chuyn i t Q P ca tp tn hiu vo, c nhiu tn hiu cngthay i gi tr v hm logic v tnh hay ngu nhin xy ra trng hp c cng mt gi tr logichazard u ra f(X) ca mch. iu to nn mt xung hazard u ra ca ca mch rngt ln ln rt nhiu, khi t ln lm cho xunghazard c nng lng ln kh nng kchchuyn mt mch ti p theo sau mch iukhin, iu gy ra hin tng iu khin"sai nhm" trong h thng logic t hp. yl iu v cng nguy him i vi cc h

thng t hp cln c nhiu u vo.

Trn thc t qu trnh chuyn i t QP trong mch logic t hp rt phc tp, rtt khi gp tng loi hazard ring bit m gps t hp hn lon cc loi hazard trn. Hintng ny c minh ho bng hnh 4-9.

Tm li, mi mt mch iu khin c th xut hin nhiu loi hazard, c mch logic c slng bin s "chy ua" rt ln nhng hazard li khng xut hin, nhng c mch rt n ginth hazard li xut hin v gy ra iu khin "sai nhm". V vy mun khc phc c hazard th

phi cn c vo mch in c th ca n, ri dng k thut phn tch pht hin kh nng xut hinhazard, sau tm cch khc phc hazard. Sau y l mt vi bin php khc phc v hn ch sxut hin hazard trong h thng logic diu khin.

4.4.4. Cc bin php khc phc Hazard.

Nh phn tch trn, hazard xut hin doc s chy ua tn hiu vo trong h logic t hp,ni cch khc hazard xut hin l do s khc nhauv thi gian tr truyn lan t u vo n u ra

ca mch, t ta c nhng bin php khc phchazard nh sau:

(X)

0f(x)

0

t

t

Q P

Hnh 4-8. Hazard lo ic

t0 t'0 t

"0 t

"'0 t1

t

(X)

0

f(x)

0

t

t

Q P

Hnh 4-9. Hin tng tng qut xut hin Hazard

t0 t1


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- Bin php n gin nht lm bin mt hazard l khng xut hin qu trnh chy ua cacc tn hiu vo trong mch logic, ngha l ch thay i gi tr logic trn mt u vo tn hiu. Khich c mt tn hiu vo "chy" trong mch logic th s khng cn "ua" tn hiu na v chc chnhazard khng th xut hin. Nhng nh vy cng c ngha l tng tn hiu vo thay i gi trlogic s lm cho mch hot ng chm chp, v khng phi qu trnh iu khin no cng cho

php lm nh vy, thng thng c s thay i nhiu tn hiu vo cng mt lc.

- Tip theo khi phi chp nhn qu trnh chuyn i t QP c nhiu tn hiu thay i hayc nhiu bin (X) chy ua. Cch khc phc l chn gi tr linh kin hay IC c thi gian tr nh.V ta bit hazard ch xut hin trong thi gian tr ca mch, cng nh ngha l xung hazard c rng t nh, v nh vy n khng c nng lng kch chuyn mch tip theo.Nhng khichn linh kin lp rp h thng hay chon IC c nh tc l phi chn linh kin, IC c cht lngcao, ngha l gi thnh ca hiu hnh tng, y cng l vn cn quan tm khi thit mch.

- Khi ta chp nhn c s chy ua tn hiu vo (X) trong qu trnh chuyn i t QP,ng thi khng dng linh kin c cht lng cao gim gi thnh v mch vn hot ng ttng thi khng c hazard xut hin, th ta c th dng phng php khc phc hazard bng cchthm cc mch tr trn ng truyn tn hiu, m bo cho thi gian chy ua ca cc tn hiul tng ng nhau. Phng php ny c minh hohnh 4-10:

Ta bit tn hiu x2 chy nhanh ti u ra, nn trn ng truyn ca x2 ta cho thm hai cngo c thi gian tr l 1 v 2 cho tn hiu trn x2 xut hin ng thi vi x1 v x3, khi hazard s khng xut hin hoc s lm gim bt hazard . Phng php ny c gy ra hazard nung tr thm vo li lm cho x2 chy qu chm v li pht sinh hin tng chy ua tn hiu

vo. trnh xy ra hin tng chy ua tn hiu vo, cn bit chnh xc thi gian tr1 v 2,

sau phi to ra c cng o c thi gian tr bng ng gi tr1 v 2.

- mc cao hn khi ta phi chp nhn c s chy ua tn hiu vo trong qu trnh chuyni Q P, khng mun dng linh kin c cht lng cao, ng thi thm cc mch tr (khngnh hng ti chc nng ca mch logic) nhng vn khng th khc phc ht hazard th khi tadng xung ng b, tc l ta bt chp c s chy ua ca tn hiu vo, v gia cc ng truyntn hiu tu vo ti u ra c thi gian tr khc nhau. Nhng tn hiu truyn lan trong h logicd nhanh, d chm, n trc hay n sau th chng chc lan truyn khi c s cho php ca

xung ng b. Xung ng b thng thng "ch" theo ng tn hiu chy chm nht, khi ccxung n sm phi "ch" cho y cc tn hiu khc khi xung ng b mi cho php truyntip. Nu cho thm vo mch iu khin xung ng b th cng c th gim ng knh hngca hazard.

- Trong trng hp cc phng php nu trn u c p dng nhng hin tng hazardvn xut hin th ta buc phi thay i chc nng iu khin, tc l thay i chc nng ca hmlogic ca h thng iu khin tc l phi xy dng mch in khc.

Nh vy c c mt mch iu khin tt, cht lng cao th phn cng xy dng nnmch in mang tnh quyt nh. Ngi thit k phi hiu rt k v su sc h thng k thut m

mnh thit k th mi c th khc phc c hazard trong mch in, cng nh phi bit thm haybt cc mch in ph nh th no m khng lm thay i chc nng ca h thng. T lm


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cho mch c cht lng cao hn, gi tr kinh t cng cao hn. iu ny cung d hiu l cc mchin c cng chc nng iu khin nhng mi hng sn xut li a ra mt mch khc nhau vgi tr kinh t ca chng cng khc nhau, tu thuc vo trnh v s quan tm n vic tng tin cy, tng cht lng iu khin mch ca hng. Nhng bn cht vn ch l lm gim ti a khnng xut hin hazard trong mch.

4.5. MCH M HO V GII M

4.5.1. Mt s loi m thng dng.

4.5.1.1. M BCD v m d3.

M BCD (Binary Coded Decimal) l m c cu to bng cch dng t nh phn 4 bit m ha 10 k hiu thp phn, nhng cch biu din vn theo thp phn. V di vi m NBCD,cc ch s thp phn c nh phn ho theo trng s nh nhau 23, 22, 21, 20 nn c 6 t hp d,

ng vi cc s thp phn 10,11,12,13,14 v 15. S xut hin cc t hp ny trong bn tin cgi l li d.

Do trng s nh phn ca mi v tr biu din thp phn l t nhin nn my c th thchin trc tip cc php tnh cng, tr, nhn, chia theo m NBCD. Tuy nhin nhc im chnhca m l tn ti t hp ton Zero, gy kh khn trong vic ng b khi truyn dn tn hiu.

V vy, ngi ta s dng m D-3 c hnh thnh t m NBCD bng cch cng thm 3vo mi t hp m. Nh vy, m khng bao gm t hp ton Zero. M D-3 ch yu c dng truyn dn tn hiu m khng dng cho vic tnh ton trc tip.

Th p phn BCD 8421 M d 3012345678

9

000000010010001101000101011001111000

1001

001101000101011001111000100110101011

1100Bng 4-2. M BCD 8421 v m d 3

4.5.1.2. M Gray.

M Gray cn c gi l m cch 1, l loi m m cc t hp m k nhau ch khc nhauduy nht 1 bit. Loi m ny khng c tnh trng s. Do , gi tr thp phn c m ha chc gii m thng qua bng m m khng th tnh theo tng trng s nhi vi m BCD.

M Gray c thc t chc theo nhiu bit. Bi vy, c thm theo m Gray.

Cng tng t nh m BCD, ngoi m Gray chnh cn c m Gray d-3.


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Th p phn Gray Gray d 3

0

123456789

1011

12131415

0000

00010011001001100111010101001100110111111110

1010101110011000

0010

01100111010101001100110111111110101010111001

1000000000010011

Bng 4-3. M Gray v Gray d 3

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