Cong Tru Nhi Phan

8/6/2019 Cong Tru Nhi Phan

1/23

_________________________________________________________________Chng 6Mch lm ton VI - 1

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CHNG 6: MCH LM TON S B

PHP TRS NH PHN DNG S B 1PHP TRS NH PHN DNG S B 2

PHP TON VI S C DU MCH CNG

Bn phn Ton phn

Cng hai s nhiu bt MCH TR

Bn phn Ton phn

Tr hai s nhiu bit Cng & tr hai s nhiu bit trong mt mch

MCH NHNMch nhn cbn

Mch nhn ni tip - song song n gin MCH CHIA

Mch chia phc hi s b chiaMch chia khng phc hi s b chia

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6.1 S b

Cho s dng N, n bit, cc s b ca N c nh ngha:S b 2: (N)2 = 2

n - N (s 2n gm bit 1 v n bit 0 theo sau)S b 1: (N)1 = (N)2 -1 = 2

n - N - 1Th d 1: N = 1010

S b 2 ca N l (N)2 = l 10000 - 1010 = 0110V s b 1 ca N l (N)1 = 0110 - 1 = 0101

Th d 2: N = 110010101100 (N)2 = 001101010100 v (N)1 = 001101010011Nhn xt:- c s b 2 ca mt s, bt u t bit LSB (tn cng bn phi) i ngc v bn

tri, cc bit s gi nguyn cho n lc gp bit 1 u tin, sau o tt c cc bit cn li.

- c s b 1 ca mt s, ta o tt c cc bit ca s.T cc nhn xt trn ta c th thc hin mt mch to s b 1 v 2 sau y:

(H 6.1)

- Khi C=1, B l s b 1 ca b (B1 v b1 l bit LSB)

Nguyn Trung Lp K THUT S


2/23

_________________________________________________________________Chng 6Mch lm ton VI - 2

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- Khi C=0, B l s b 2 ca b.Tht vy, cc biu thc logic ca B theo b v C l:

CbB 11 = )bCbB 122 += (

)bbCbB 2133 ++= (

- Khi C=1 , cc ng ra cng OR lun bng 1, cc cng EX - OR lun c mt ng vobng 1 nn ng ra l o ca ng vo cn li, ta c:

111 b1bB ==

22122 b1b)b1bB ==+= (

332133 b1b)bb1bB ==++= (

- Khi C=0

111 b0bB ==

12122 bb)b0bB =+= (

= b2 nu b1=0 v 2b nu b1 = 1

)b(bb)bb(0bB 2132133 +=++=

= b3 nu b1 v b2u =0= 3b nu (b1 v/hoc b2 = 1)

Nh vy tt c cc bit sau bit 1 th nht tnh t bit LSB u bo v B chnh l s b2 ca b

Chng ta cng c th thit k mch to s b hai bng cch dng FF RS, c ng voR, S tc ng mc cao, kt hp vi cc cng logic nh (H 6.2). Mch ny dng kh tin likhi cn thc hin bi ton cng v tr nhiu bit kiu ni tip.

(H 6.2)Bt u, Preset mch ng ra Q = 1, cng G3ng, G2 m, cho s B i qua m

khng bo cho n khi c bit 1 u tin n, cng G1 mcho xung ng hi qua, FF RSc reset, Q = 0, Q = 1, G2ng, G3 m, s B i qua cng G2 v bo. ng ra c s

b 2 ca B.

6.2 Php trs nh phn dng s b 1:

Cho hai s dng A v B c n bit (nu s bit khc nhau, ta thm s 0 vo , m khng

lm thay i tr, c hai c cng s bit)a/ - AB



3/23

_________________________________________________________________Chng 6Mch lm ton VI - 3

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Kt qu A-B l s 0 hoc m, php tnh c thc hin nh sau:Tnh A - B:A - B = A-B+2

n-1-2n+1= A+(2n -B-1 ) - 2

n+1= A+(B)1 - 2

n+1= - {2n - [A+(B

)

1] -1}

= - [A+(B)1]1Vy A-Bc c bng cch cng s b 1 ca B vo A ri ly b 1 ca tng v thm

du tr . Nh vy thc hin php tnh tr ta ch cn dng php cng v php o

Th d 3 : Tnh 1001 - 11010 dng s b 1Ta c A = 01001 (thm s 0 vo c 5 bit nh s B

B = 11010 (B)1 = 00101A-B = - [A+(B)1]1 = - (01001+00101) =- (01110)1

= - (10001)Trong h thp phn, y l bi ton 910 - 2610 = -1710 thy du trc nhn ra nh th no, ta vit li php ton:

Khng c s trn (hay s trn =0) l du hiu ca kt qu m (hoc =0) v ta phi lyb 1, thm du trc kt qu cui cng: (01110)1 = - 10001

Th d 4: Tnh 10110 - 10110A = 10110 v B = 10110 (B)1 = 01001

Trong php cng u tin khng c s trn, kt qu xem nh s m (hoc =0) ly b 1ca kt qu ta c A-B=00000.

b/ - A >BKt qu A-B l s dng, php tnh c thc hin theo qui tc sau:Cng A vi (B)1 ri thm 1 v khng quan tm ti s nhcui cng

Th d 5: Tnh 110101 - 100110 dng s b 1A = 110101 v B = 100110 (B)1 = 011001

B qua s nhcui cng, ta c kt qu A-B=001111.

Trong h thp phn l bi ton 5310 - 3810 = 1510.



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_________________________________________________________________Chng 6Mch lm ton VI - 5

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Th d 8 : Tnh 10110 - 10110A = 10110 v B = 10110 (B)2 = 01010

B qua s trn ta c A-B=00000.

6.4 Php ton vi s c du

Cho ti gichng ta thc hin cc php ton vi s khng du v i khi xut hindu tr trong kt qu. Trong my tnh, iu ny c th khc phc c bng cch dng s cdu.

Vi qui c s dng c bit du l 0 v s m c du l 1

Th d 9: +1010 = 01010 +1510 = 01111 +2310 = 010111-1010 = 10110 - 1510 = 10001 - 2310 = 101001

C th thy rng s m ca mt s l s b 2 ca n k c bit du.Vi cch biu din s c du, php ton tr trthnh php ton cng:

A-B= A+(-B)

Th d 10: Tnh A-B=01110 - 01001; B= 01001 = +910 - 910 = 10111

Bit du =0 ch kt qu dng, b bit trn C2.Vy A-B= 00101 [(+1410 )-(+910)] = +510

Nu Ahoc B u dng hoc m , kt qu c th cn thm mt bit do trn s. Trongtrng hp ny bit trn u tin thuc kt qu v C2 l bit du

Th d 11: Tnh A+B vi A= 01110 (+1410) v B= 01001 (+910)Kt qu l 010111 = +2310 vi C2 = 0 l bit du



6/23


7/23

_________________________________________________________________Chng 6Mch lm ton VI - 7

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6.5.2 Mch cng ton phn(Full adder,FA) :

L mch cng hai bit cng v tr trong hai s nh phn nhiu bit, ni cch khc, yl mch cng hai bit , gi s th n, v bit nhc c t php cng hai bit th n-1 ca hai snh phn . Ta c bng s tht

Cn-1 BB

n An Sn Cn00001111

00110011

01010101

01101001

00010111

Dng bng Karnaugh ta xc nh c Sn v Cn nh sau:

)B(ACBAC

)B(ACS

nn1nnnn

nn1nn

+=

=

(H 6.4)C th thy mt mch cng ton phn gm hai mch cng bn phn v mt cng OR

6.6 Cng hai s nh phn nhiu bit:

6.6.1 Cng ni tip

Trong cch cng ni tip, ngi ta dng cc ghi dch chuyn cc bit vo mt mchcng ton phn duy nht, s nht ng ra Cnc lm tr mt bit nhFF D v a vo ngvo Cn-1. Nh vy tc ca php cng ty thuc vo tn s xung CKv s bit phi thc hin.

(H 6.5)



8/23

_________________________________________________________________Chng 6Mch lm ton VI - 8

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6.6.2 Cng song song

Trong cch cng song song, cc bit c a ng thi vo cc mch cng ton phnv s nhca kt qubit thp c a ln bit cao hn (H 6.6).

(H 6.6)Chnh v phi ch s nhm tc cng cn hn ch. Mun nng tc cng ln,

ngi ta thc hin php cng song song nh trc s nh.

6.6.3 Mch cng song song nh trc s nh

tng tc ca mch cng song song, ngi ta to trc cc s nha ngthi vo mch cng

T biu thc xc nh s nh

)B(ACBAC nn1nnnn +=

t Pn = AnBBn v Gn = An Bn

Ta xc nh c C1, C2, C3 .... nh sau:



9/23

_________________________________________________________________Chng 6Mch lm ton VI - 9

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(H 6.7)

Nhn thy thi gian tnh s nhging nhau cc tng v bng t1+t2 . t1 l thi giantruyn ng thi qua cc cng AND v t2 l thi gian truyn qua cng OR.

S khi mch cng song song nh trc s nh:

(H 6.8)

Trn th trng hin c IC 7483 (tng ng 4008 ca CMOS) l IC cng 4 bit theokiu nh trc s nh.

6.6.4 Cng hai s BCD

Trn th trng c cc IC cng s nh phn, trong lc trn thc t nhiu khi chng ta

cn cng cc s BCD cho kt qu l s BCD.Chng ta tm cch dng IC 7483 (4008) cng hai s BCDHai s BCD c tr t 010n 910 khi cng li cho kt qu t 010n 1810. c c

kt qu dng BCD ta phi hiu chnh kt qu c c t mch cng nh phn.Di y l kt qu tng ng gia 3 loi m: thp phn, nh phn v BCD

NhPhn B C

DBCD c

TP S=C4 S4 S3 S2 S1 S =C4 S4 S3 S2 S1 theo NP01

234567891011121314

151617

00

0000000000000

011

00

0000001111111

100

00

0011110000111

100

00

1100110011001

100

01

0101010101010

101

00

0000000011111

111

00

0000001100000

000

00

0011110000001

111

00

1100110000110

011

01

0101010101010

101

01

234567891617181920

212223



10/23

_________________________________________________________________Chng 6Mch lm ton VI - 10

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18 1 0 0 1 0 1 1 0 0 0 24

Nhn thy:- Khi kt qu


11/23

_________________________________________________________________Chng 6Mch lm ton VI - 11

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cng mt chui s, nhiu mch cng ton phn sc s dng, s nhc luli a vo mch cng bit cao hn.

Th d 14 : Vi 3 s 3-bit X (X3X2X1), Y(Y3Y2Y1), Z (Z3Z2Z1) mch cng c dng

(H 6.10)Ngi ta dng mch cng loi ny thc hin bi ton nhn. c kt qu nhanh hn, c th dng mch (H 6.11)

(H 6.11)

6.7 Mch trnh phn:

6.7.1 Mch trbn phn

L mch tr hai s 1 bit (H 6.12)

(H 6.12)



12/23

_________________________________________________________________Chng 6Mch lm ton VI - 12

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6.7.2 Mch trc s nh(mch tr ton phn)

L mch tr 2 bit c quan tm ti s nhmang t bit trc

Rn-1 An BBn Dn Rn

00001111

00110011

01010101

01101001

01001101

Bng s tht

Dng bng Karnaugh xc nh c cc hm Dn v Rn)B(ARD nn1nn =

)BA(RBAR nn1nnnn +=

V mch (H 6.13)

(H 6.13)

Nhn thy cu to mch tr ging nh mch cng, ch khc mch to s nh

6.7.3 Tr s nhiu bit

Ta c mch tr s nhiu bit bng cch mc song song cc mch tr 1 bit (H 6.14)

(H 6.14)

6.7.4 Cng v trs nhiu bit trong mt mch

Nhc li thc hin php ton tr, ngi ta cng vi s b 1 v cng thm 1 (hoccng vi s b 2), nh vy thc hin php tr A - B ta tnh A + (B)1 + 1 . Mch (H 6.6)c sa i c thc hin php cng v tr ty vo ng iu khin C (H 6.15)

- Khi C=0, ta c mch cng- Khi C =1, ta c mch tr



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14/23

_________________________________________________________________Chng 6Mch lm ton VI - 14

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(H 6.17)

Dng IC cng 4 bit (7483 hoc 4008) mch nhn hai s 4 bit c dng (H 6.18)

(H 6.18)

6.8.2. Mch nhn ni tip - song song n gin (H 6.19)

(H 6.19)

Trong mch ny, mt trong hai sc a ni tip vo mch (trong trng hp nyl s b nhn) v s cn li a song song vo mch.

- S nhn (b4b3b2b1) a song song vo mch qua cc cng AND ng thi kim sotcc cng ny: ng vi bit 1 s b nhn qua mch ti mch cng (cng 2 v 4); ng vi bit 0ng ra cng AND bng khng (cng 1 v 3)



15/23

_________________________________________________________________Chng 6Mch lm ton VI - 15

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- S b nhn a ni tip vo mch theo th t t bit LSB. Cc FF D c tc dng dchkt qu ca php nhn (l cc tch tng phn) trc khi a vo mch cng cng cc tchtng phn ny.

Th d 15 : Xem bi ton nhn 10x14. S nhn l 1010 (1010) v s b nhn l 1110(1410). Qu trnh nhn gii thch nh sau:

P8 P7 P6 P5 P4 P3 P2 P1ABCD

0000

0001

0001

0101

0100

0100

0000

0000

Output 1 0 0 0 1 1 0 0100011002 = 14010C th thy rng ng ra A lun lun bng 0 v bit LSB ca s nhn = 0. Ng ra B c

gi tr ca s b nhn c lm tr 1 bit (1 xung ng h). Ng ra C c lm tr 2 bit v lunbng 0 (Ging nh A). Ng ra D ging nh B nhng tr 3 bit. iu ny c th so snh vi bi

ton trn giy

S b nhnS nhn

11

10

11

00

ABCD 1

01

101

0100

0100

0000

0000

Tch 1 0 0 0 1 1 0 0

Mun khng s dng mch cng s nhiu bit, ngi ta dng mch (H 6.20)

(H 6.20)Mch (H 6.20) cn (n-1) mch cng v mch tr (FF D) cho s nhn n bit. Cc cng

AND cho php cc bit ca s b nhn i qua khi s nhn l 1, s b nhn (vi s bit bt k)c cho vo mch ni tip vi bit LSB vo u tin.

Ng ra cng 4 sau 4 xung Clock l 1110. Ng ra cng 3 lun lun bng 0.Mch cng A cng s ng ra 3 v ng ra 4 b tr 1 bit:

101

01

00

00

1 1 1 0 0

Tng t mch cng B cng s b nhn vi kt quA c lm tr 1 bit

1 1

1

1

1

0

1

0

0

01 0 0 0 1 1 0



16/23

_________________________________________________________________Chng 6Mch lm ton VI - 16

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v mch cng C

1 0 0 001

01

00

00

1 0 0 0 1 1 0 0

Lu l mch (H 6.20) kt qu cho ng ra mch cng C vi bit LSB ra u tin,

tuy nhin mch ny cha quan tm ti s nh.Mch (H 6.21) cho kt qu vi s nh.

(H 6.21)V (H 6.22) l mt mch thc t dng ghi dch 4 bit c ng vo/ra song song, mt

mch cng 4 bit v mt chip 4 cng AND 2 ng vo thc hin bi ton nhn.

(H 6.22)

6.9 Mch chia

Nguyn tc ca php chia s nh phn l thc hin php so snh mt phn ca s bchia (s bit u tin bng vi s bit ca s chia) vi s chia, nu s b chia ln hn s chia ththng s =1, thc hin php tr, ngc li th thng s =0, sau dch tri phn cn li cas b chia mt bit (hoc dch phi s chia 1 bit) ri tip tc thc hin bi ton so snh gingnh trn. Cng vic c lp li cho n khi chm dt.

S (H 6.23) tm tt gii thut thc hin bi ton chia



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18/23

_________________________________________________________________Chng 6Mch lm ton VI - 18

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Dch

s b chia - s chia

Thng s = 0

Kt qudng ?

Thng s = 1

s b chia + s chia

Chia xong?

Kt thc

NoYes

No

Yes

(H 6.24)

thc hin php chia theo s trn, ngoi cc thanh ghi cha cc s b chia, schia , s thng ngi ta phi dng thanh ghi cha s b chia c phc hi.

6.9.2 Php chia khng phc hi s b chia

H thng sn gin hn nu chng ta dng php chia khng cn phc hi s b chia

theo nguyn tc nh di y.Quan st gin (H 6.24) ta thy c 2 trng hp:



19/23

_________________________________________________________________Chng 6Mch lm ton VI - 19

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S chia ln hn s b chia (nhnh bn phi)Lu l dch s chia v bn phi 1 bit tng ng vi chia s cho 2

Nhnh bn phi ca s trn gm 2 bi ton:- Cng s b chia vi s chia.

- Tr s b chia cho 1/2 s chia (tr b chia cho s chia dch phi)Hai bc ny c th gom li thnh mt bc duy nht nh sau:- Cng s b chia vi s chia dch phi.

S chia nh hn s b chia (nhnh bn tri)Sau khi ly kt qu =1, lnh k tip thc hin l tr s chia dch phi.T cc kt qu nhn xt trn c th thay s (H 6.24) bi s gii thut thc hin

php chia khng cn phc hi s b chia (H 6.25)

Dch

Thng s = 0

Kt qudng ?

Thng s = 1

s b chia + s chia

Kt thc

Chia xong?

s b chia - s chia

Dch

No

Ye s No

s b chia - s chia

(H 6.25)

Da vo s (H 6.25), cc bc thc hin bi ton chia nh sau:



20/23

_________________________________________________________________Chng 6Mch lm ton VI - 20

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- S chia (SC) ln hn s b chia (SBC) (SBC - SC < 0), thng s l 0, dch phi schia 1 bit (thc t ta mang thm 1 bit ca s b chia xung), thc hin bi ton cng s chia vs b chia

- S chia nh hn s b chia (SBC - SC > 0), thng s l 1, dch phi s chia 1 bit,thc hin bi ton tr (cng s b 2) s b chia cho s chia

n gin, gi s s chia v b chia u dng (MSB = 0), s b chia gm 6 bit vs chia gm 4 bit.

Th d 1: Thc hin bi ton chia 2110 = 0101012 cho 710 = 01112.S b 2 ca 0111 l (0111)2 = 1001

Ghi ch:

(1) S 1 trn mi tn ch rng kt qu php ton tr l s m, bc k tip l di vcng s chia

(2) S 0 trn mi tn ch rng kt qu php ton tr l s dng, bc k tip l di vtr s chia (cng s b 2)

Thng s c c t cc s trn m trn php tnh ta ghi trong vng trn.Kt qu: thng l 011(=3) v s d l 0000(=0)Bi ton trn cho kt qu vi 3 bc cng/tr. Tuy nhin nu ta chia 21 cho 1 th cn

ti 6 bc cng tr c thng s 6 bit. Mt cch tng qut s bc ca bi ton bng vis bt ca s b chia.

Ta c th lm li bi ton vi 6 bc cng/tr ((thm 3 bit 0 cho s b chia)



21/23

_________________________________________________________________Chng 6Mch lm ton VI - 21

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Th d 2 v 3 di y l bi ton 6 bcTh d 2 : Chia 21 cho 6 c kt qu 3 v s d l 3

Th d 3 : Chia 21 cho 5, c kt qu 4 v s d l 1. Tuy nhin trn php ton tathy php cng vi s chia cui cng cho kt qu m (s 1100) nn iu chnh s d ta

phi cng s chia vo v b qua s trn.



22/23

_________________________________________________________________Chng 6Mch lm ton VI - 22

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(1) Cng s chia vo iu chnh s d

Mch thc hin cc bi ton ny cho (H 6.26).Trong (H 6.26) bc u tin c thc hin bi cc cng EX-OR trn cng c ng

iu khin = 1 thc hin bi ton tr. Sau bc th nht, bit th t ca mch cng (S4) squyt nh php ton sau l cng (S4=1) hay tr (S4=0) s b chia vi s chia. S nhca

bi ton cui cng (bc 6) l bit LSB ca thng s. V mch cng cui cng c thit kkt hp vi cc cng AND x l kt qu ca s d nh trong hai th d 2 v 3. Nu kt quca bi ton bc 6 c S4 = 1 th cng AND m thc hin bi ton cng vi s chia iu chnh s d.



23/23

_________________________________________________________________Chng 6Mch lm ton VI - 23

(H 6.26)

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