Kỹ thuật số - Lý thuyết và ứng dụng: Phần 2

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CHUONG 7

GHI CHUYEN

Hoc xong chuong nay hoc vien co kha nang:

□ Hieu duoc cac nguyen ly cau tao va hoat dong
cua cac loai bo ghi chuyen.

□ Xay dung duoc cac loai bo ghi chuyen.

□ Ap dung cac loai bo ghi chuyen trong cac mach
thuc te.

TONG QUAN

Bo ghi chuyen la mot nhom cac flip flop du’Oc dung de luu trCr so nhi phan.
Phai can co mot flip flop cho m6i bit cua so nhi phan. Vi du, mot bo ghi dung de
lifu trir 8 bit nhi phan can phai co 8 flip flop. Cac flip flop du’Oc noi vdi nhau sao cho
so nhi phan co the di vao bo ghi chuyen va co the chuyen du’Oc cac bit ra ngoai flip
flop cuoi cung. Mot nhom cac flip flop thu’c hien chirc nang do du’Oc goi la bo ghi
chuyen.

Cac bit cua so nhj phan co the dich chuyen tir vi tri nay sang vi tri khac
trong bo ghi chuyen theo mot hirdng hoac theo ca hai hirdng. Co hai cach dira so
lieu vao bo ghi chuyen la so lieu vao noi tiep hoac so lieu vao song song. Tirong
tir, cung co hai cach dira so lieu ra khoi bo ghi chuyen: so lieu ra noi tiep va so
lieu ra song song. Tir do, co 4 loai cau true bo ghi chuyen co ban nhu” mo ta tren
Hinh 7.1.

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Ngo vao so lieu

noi tiep

Ngo vao so lieu
noi tiep

Ngo ra so lieu
a'. . . V

noi tiep
a)l3p ghi chuyen ra/vao noi tiep
Ngo vao so lieu

song song

Ngo ra so lieu
/^. . • ft
noi tiep

; ra so lieu

b) Bo ghi chuyen vao noi tiep /ra song song
song song

Ngo vao so lieu

it.

song song

ra so lieu
---^ N«5

song song

7 K

c) EJo ghi chuyen vao song song/ra noi tiep d) E3o ghi chuyen ra/ vao song song

Hinh 7.1: Cac loai bo ghi chuyen

7.1. BO GHI CHUYEN VAO NOI TIEP- RA NOI TIEP

Cac flip flop JK hoac D du’Oc sir dung de tao nen cac bo ghi chuyen.
De chuyen bit 0 vao flip flop thi J=0 va K=1

De chuyen bit 1 vao flip flop thi J=1 va K=0.

Dieu quan trong la J va K phai du’Oc dieu khien de cung cap cho so lieu vao
chuan xac. Cac mu’c logic cua J va K co the du’Oc thay doi trong khi xung nhip
dang 0 mu’c H (hoac L), can nhd rang cac mu’c nay phai that on dinh tCr tri/dc khi
cho den tan sau khi xung nhip chuyen trang thai. O day, ta dung flip flop JK
Master-Slave tac dong khi xung nhip 6 sirdn xuong. Hinh 7.2 minh hoa hoat dong
cua flip flop JK Master-Slave tirong Crng vdi sirdn xuong cua xung nhip.

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Ngo vao
/{ i • A

eo lieu

Clock
(Xung nhip)

•J Q

>
—

K Q

Then gian

Clock O—

J 1

0.

a

1

•

o

-o o o o

Hinh 7.2

b)

Trudc thdi diem a, Q=1.

- Tai thdi diem a, Q Reset tro thanh mu’c 0 (so 0 du’Oc chuyen vao flip flop ).
- Tai thdi diem b, Q khong thay doi, vi flip flop dang chi/a so 0 va so 0 khac
du’Oc chuyen vao flip flop .

- Tai thdi diem c, Q Set trd thanh mu’c 1 (so 1 du’Oc chuyen vao flip flop ).
- Tai thdi diem d, so 0 khac du’Oc chuyen vao flip flop .

Ta da dira 4 bit so lieu vao flip flop nay trong day thdi gian sau: bit 0 tai a, bit

0

khac tai b, bit

1

tai c va bit

0

tai d.

Bay gid hay them 3 flip flop nira va du’Oc ket noi nhu

1

Hinh 7.3, dong thdi
Reset tat ca cac flip flop va dung cac tin hieu nhir tren dira vao flip flop .

Tai thdi diem a: tat ca cac flip flop Reset, tat ca cac ngo vao J d mu’c 0, tat ca
cac ngo vao K d mu’c 1.

Khi Q

4

Reset (so 0 d Q

3

du’Oc chuyen vao Q4). Tirong tir, so 0 d Q

2

du’Oc dira
vao Q3, so 0 d Q, du’Oc dira vao Q

2

va so 0 d ngo vao so lieu diroc dira vao Q v
Cac ngo ra cua flip flop sau thdi diem a la

0 ^ 2 0 3 0 4

= 0000.

Tai thdi diem b: tat ca cac flip flop deu chCra so 0. Nhir vay, so 0

0

3

chuyen
vao Q4, s o 0 d Q

2

chuyen vao Q3, sd 0 d Q, chuyen vao Q

2

va so 0 d ngo vao so
lieu chuyen vao Qv Cac ngo ra sd lieu la: Q.,Q

2

3

4

= 0000.

Tai thdi diem c: cac flip flop van con chi/a cac so 0. Sd 0 d Q

3

chuyen den Q4,
so 0 d Q

2

chuyen den Q3, sd 0 d Q, chuyen den Q

2

nhirng luc nay so 1 d ngo vao
sd lieu chuyen vao Q,. Cac ngo ra sd lieu bay gid la

0 ^ 2 0 3 0 4

= 1000.

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Ngo vao
eo lieu

Clock

Thai gian

Clock

So lieu
vao
noi tiep

l
k 0

1
J 0

1
K

u
Q1 1
0
1

W

u

1
u
1
Q4 0

J

Hinh 7.3: Bo ghi chuyen 4 bit vao noi tiep

Tom lai, ta da chuyen 4 bit so lieu theo kieu noi tiep vao cac flip flop . Tai thdi
diem d, bon bit so lieu nay bieu thi so nhi phan 0100. Gia du rang ta bat dau
chuyen so LSB trudc, nhu vay, LSB nam 6 Q4 con MSB nam 6 Q v Bon flip flop nay
du’Oc coi nhu1 mot bo ghi chuyen 4 bit. Day chi'nh la ky thuat thudng dung de tao
nen bo ghi chuyen ngo vao noi tiep.

Mach dien Hinh 7.3 yeu cau co hai tin hieu vao J va K, ro rang la J = K (hoac
K = J). Co the noi rang, J luon luon la phan bu cua K va ngupc lai. Neu ta ket noi
mot cong Invecto giua J va K trong flip flop Q v6i ngo vao J, luc do, ta chi can mot
ngo vao so lieu, chi can mot ngo vao J. Nhu vay, ta co the dung flip flop D nhu

trong Hinh 7.4a de thuc hien dung yeu cau tren. Mot bo ghi chuyen ngo vao noi

tiep 4 bit dupe tao bang flip flop D d Hinh 7.4b sir dung flip flop D lam bo ghi
chuyen co Uu diem la chi can mot tin hieu vao va cach ket noi mach dupe don gian
hon.

Vi du:

Ve cac dang xung de chuyen so 1010 ra khoi bo ghi chuyen nhu 6 Hinh 7.4d

Ldi giai:

Cac dang xung cua bo ghi chuyen dung nhu tren Hinh 7.4d. Dang xung ky

hieu K dupe loai bo va dang xung J dupe ky hieu D. Ngo vao so lieu noi tiep duoc
xep loai hoac la JK hoac la D tuy thuoc loai flip flop dupe sir dung. Bay gid hay xet
so lieu chuyen ra ngoai bo ghi chuyen nhu the nao?

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Hay xem xet bo ghi chuyen

0

Hinh 7.4. Gia du co so nhi phan 4 bit

0,020304

= 1010 du’Oc lull tri/ trong bo ghi chuyen. Neu tin hieu xung nhip co dang nhir

0

Hinh 7.4d, ta hay xem xet tai cac thdi diem a, b, c, d.

- Trudc thdi diem a: bo ghi chuyen lull tru” so nhi phan

0,020304

= 1010. So
LSB (so 0) xuat hien

0

0 4.

- Tai thdi diem a: toan bo so nhi phan du’Oc di chuyen sang flip flop ben phai.
So 0 du’Oc chuyen vao Q

3

va so LSB (so 0) du’Oc chuyen sang phai, ra khoi flip flop

cuoi cung va bien mat. Bo ghi chuyen giufcac bit

0,620304

= 0101 va so LSB thCr

hai (so 1) xuat hien tai Q4.

- Tai thdi diem b: tat ca cac bit du’Oc chuyen sang flip flop ben phai, so 0 du’Oc
chuyen vao Q, va so LSB thir 3 xuat hien tai Q4. Bo ghi chuyen giCr cac bit

0,020 304

=0010.

- Tai thdi diem c: tat ca cac bit du’Oc chuyen sang flip flop ben phai, so 0 du’Oc
chuyen vao Q, va so LSB (so 1) xuat hien tai Q4. Bo ghi chuyen giuf cac bit

0,020304

= 0001.

- Tai thdi diem d: so MSB du’Oc chuyen ra ngoai flip flop cuoi cung va bien

mat, so 0 chuyen vao Q,. Bo ghi chuyen giu’ so Q,Q

2

3

4

= 0000.

Nqo vao So lieu

Clock

Ngo vao eo lieu

Clock

D 02 D 03 D 0 4
--->

— C>

Q2 0 3 0 4

---1

»---c)
Then' gi an

Clock

02.

03

0 4
1

O .
1 ■

0

1
O.

1 '
0

1
o

-1

0
1

o

d)

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Tom lai, so lieu du’Oc IlTu tru* trong bo ghi chuyen da du’Oc chuyen ra khoi bo

ghi chuyen 6 flip flop cuoi cung Q4 dirdi dang noi tiep. Nhir vay, khong chi ngo vao
la noi tiep ma ngo ra bo ghi chuyen cung la noi tiep. Oieu quan trong khi thirc hien
la so nhi phan du’Oc liru trCr trong bo ghi chuyen du’Oc chuyen ra ngoai 0 dau cuoi
ben phai bo ghi chuyen va bien mat sau 4 xung nhip.

So do chan IC 54/74L91 du’Oc trinh bay tren Hinh 7.5. IC nay la bo ghi

chuyen 8 bit ho TTL MSI. IC co 8 flip flop RS ket noi theo kieu noi tiep ngo vao va
noi tiep ngo ra. Flip flop hoat dong Crng v6i sirdn am cua xung nhip. Tuy nhien, vi
dung mach dao trirdc cac flip flop nen so lieu du’Oc di chuyen tirong Crng vdi sirdn
dirong cua xung nhip.

Cac flip flop RS du’Oc noi hoan toan giong flip flop JK nhir tren Hinh 7.3.

Mach dao du’Oc noi giufa R va S 6 flip flop dau, co nghTa la mach hoat dong nhir flip
flop D. Vi vay, ngo vao bo ghi chuyen co mot dirong duy nhat de chuyen so lieu
vao bo ghi chuyen theo kieu noi tiep. Ngo vao so lieu du’Oc dira vao hoac chan 12
(A) hoac chan 11 (B). Chu y rang mu’c so lieu A (hoac B) du’Oc dao bang cong
NAND de dira vao ngo vao R cua flip flop dau tien, mCrc tin hieu sau khi ra khoi

cong NAND lai du’Oc cong dao thirc hien dao mot Ian nu’a de den ngo vao S. Do
do, so 1 Set flip flop dau, noi cach khac, so 1 du’Oc dira vao flip flop dau tien khi
sirdn dirong cua xung nhip xuat hien.

14 13 12 11 10 .9 8

a Q A B GND CLK

IC 54/74L91

Vcc

1 2 3 4 5 6 7

| 5a do chan IC 54/74L91: Bo ghi chuyen 8 bit ngo vao noi tiep. ngo ra noi tiep

i—O

Clock

rC>

5 Q

r-C >

s c

R Q

-S Q

R Q

s Q

-d>

R Q

S Q

R a

S Q

R Q

S Q

R Q

b) So do logic

Hinh 7.5

Neu co 2 tin hieu vao bo ghi chuyen, ta dira vao cong NAND (chan 11 va 12),
neu chi cd mot tin hieu, ta cd the noi hai chan A va B lai vdi nhau de d in tin hieu
vao.

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So lieu

(DATA

O' I

Clock-—q>
R

a) Cac mile logic diroc chf dan bang cac mui ten
se set Flip Flop

b) Cac mtic logic duoc ch’ dan bang cac mui ten
se reset Flip Flop

Hinh 7.6:

Vi du:

Kiem tra xem cac mu’c logic dira den ngo vao cua bo ghi chuyen dung IC

54/74L91 nhu1 the nao? Cac bit 1 va 0 du’Oc dira vao bo ghi chuyen ra sao?

Ldi giai:

Ngo vao logic va flip flop dau tien du’Oc ve 0 Hinh 7.6a, so lieu 1 du’Oc dira
vao ngo vao chan A. Ngo vao R co mu’c 0, ngo vao S co mu’c 1. Vi flip flop du’Oc
Set khi sirdn dirong cua CLK xuat hien, so lieu 1 du’Oc dira vao flip flop.

6 Hinh 7.6b so lieu 0 du’Oc dira vao chan A. Ngd vao R cd mu’c 1, ngd vao S
cd mire 0. Flip flop du’Oc Reset khi sirdn dirong xung nhip CLK xuat hien. Nhir vay,
so lieu 0 du'Oc dira vao bo ghi chuyen.

7.2. BO GHI CHUYEN VAO NOI TIEP-RA 30N G SONG

Bo ghi chuyen vao noi tiep-ra song song cd so lieu du’Oc dira vao noi tiep
nhirng cac so lieu du’Oc dira ra song song, cac sd lieu xuat hien d ngd ra cung mot
luc. Thirc hien dieu nay bang cach noi ngd ra cua mdi flip flop den mot chan ra cua
IC. Hinh 7.7b la so do chan IC 54/74164. Bo ghi chuyen 8 bit cd 8 ngd ra, mdi ngd
ra la mot flip flop trong bo ghi chuyen. Mach logic cua loai bo ghi chuyen nay du’Oc
gidi thieu d Hinh 7.7a.

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CLEAR

CLOCK. - ^ o > • — »
A
B
-H

-QA
>

■s QA
a ) So d o logic

b) S o d o chan

n
QB

S OB

QA
K)
T
QB
K
QC
>
5
QC
-A.
P _
QD
>
6 QD
A .

K QE
-o>

5 QE

_d.

R QF
•—0(>

S QF

QC QD QE

T

QF

R QO R X

> >

S QO

T 5 QH

QO

n

Cac ngo ra song song
Ngo ra so lieu

Vcc | QH QO QF QE 1 CLR CLK

14 13 12 11 10 I » I

IC54/74164

TU

QB QC QP | GND

Ngo vao
a' |. a

eo lieu

Ngo ra so lieu

Hinh 7.7

Hoat dong cua B nhir sau:

- Khi B 6 mu’c H: cong NAND hoat dong va so lieu ngo vao noi tiep qua cong
NAND va cong Invecto. So lieu vao du’Oc chuyen noi tiep vao bo ghi chuyen.

- Khi B d mu'c L: ngo ra cong NAND 6 mu’c H, qua cong Invecto, tin hieu 6
mu’c L, dong so lieu bi can lai va khi xuat hien sirdn di/ong xung nhip, so 0 du’Oc
dira vao bo ghi chuyen. Sau 8 xung nhip, bo ghi chuyen chCra day so 0.

- Cung nhu1 IC 54/7491, neu chi can mot dirong tin hieu, co the noi hai chan A

va B lai vdi nhau va dira so lieu vao hai chan do.

7.3. BO GHI CHUYEN VAO SONG SONG-RA NOI TIEP

Ta can nghien ciru bo ghi chuyen vao song song-ra noi tiep co s in tren thi
trirdng, do la IC 54/74166, bo ghi chuyen 8 bit va ta tap trung nghien cull mot trong
8 mach flip flop trong IC de hieu ro nguyen ly hoat dong cua bo ghi chuyen nay.

So do chan cua IC 54/74166 du’Oc trinh bay tren Hinh 7.8a. Day la bo ghi
chuyen 8 bit. So lieu vao co the hoac song song hoac noi tiep, so lieu ra noi tiep.
IC nay co 8 flip flop RS. Hay phan ti'ch mot trong cac mach flip flop nay, sau do,
them vao cac khoi logic.

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a) So do chan b) Mach logic

Hinh 7.8: IC 54/74166

Nhan thay rang flip flop RS chuan hoa theo thdi gian co mach dao ngo vao
chinh la flip flop D. Neu mot bit so lieu x du’Oc xung nhip chuyen vao flip flop thi
phan bu cua x phai du’Oc dira vao ngo vao so lieu. Vi du: neu x = 0, thi R = 0 va
S = 1 va so 1 du’Oc xung nhjp chuyen vao flip flop khi xung nhip xuat hien.

Bay gid them cong NOR (Hinh 7.9b). Neu mot chan cua cong NOR 6 mu’c 0,

mot bit so lieu x du’Oc dira vao chan con lai, gia tri x du’Oc cong NOR dao. Vi du:
neulc = 1 , 6 cong NOR ngo ra co x = 0, nhir vay, so 1 du’Oc dira vao flip flop khi
xung nhip xuat hien. Cong NOR hay du’Oc dung de dira so lieu tir hai nguon so lieu
khac nhau, hoac x, hoac x2. Khi cho x2 tiep dat thi so lieu x, du’Oc chuyen vao flip
flop, ngirpc lai, neu x, tiep dat thi x2 chuyen vao flip flop.

Viec dira them 2 cong AND va 2 cong dao nhir trong Hinh 7.9c du’Oc phep

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cua cong NOR trong khi chan du”6i cua cong NOR co mu'c L. Mat khac, khi day
dieu khien 6 mu’c L, cong AND tren ngirng hoat dong, trong khi cong AND dirdi
hoat dong. Dieu do cho phep x2 xuat hien d chan dirdi cua cong NOR trong khi
chan tren cua cong NOR co mu’c L.

Tom lai, day dieu khien d mu’c H: bit so lieu d x1 du’Oc chuyen vao flip flop 6
xung nhip ngay sau khi x1 xuat hien; day dieu khien d mu’c L: bit so lieu 6 x2 di/pc
chuyen vao flip flop 6 xung nhip ngay sau khi x2 xuat hien.

Vi du:

Cho mach nhu" 0 Hinh 7.9c, hay viet cac mu’c logic co mat d chan moi cong
neu day dieu khien co mu’c 1, x, = 1 va x2 = 1.

Ldi giai:

Cac mu’c logic du’Oc ghi d cac chan cua mdi cong logic tren Hinh 7.9d. Gia tri

so lieu 1 d x, du’Oc chuyen vao flip flop khi xung nhip xuat hien.

Khi xem xet can than thay 8 mach cho tren Hinh 7.9a ghep lai vdi nhau tao

thanh bo ghi chuyen 54/74166 cho tren Hinh 7.8b. Chung dirpc ket noi de co hai

cach hoat dong khac nhau:

- So lieu di/a vao song song.

- Khai thac so lieu di chuyen noi tiep qua bo ghi chuyen tir flip flop QA dau
tien den flip flop QH cuoi cung.

Neu ngo vao so lieu ky hieu la x2 trong Hinh 7.9c du’Oc mang den cho tirng

flip flop, 8 ngo vao so lieu nay ddng vai trd cac ngd di/a so lieu vao song song cho
mot so 8 bit ABCDEFGH. Tam ngd vao nay du’Oc ghi nhan la A, B, C, D, E, F, G,

H nhi/trong Hinh 7.8. Day kiem tra cd ten Shift/Load.

- Khi duy tri day dieu khien (Shift/Load) nay d mu’c L, cong AND di/di se hoat
dong cho moi flip flop va so 8 bit nay se du’Oc nap vao cac flip flop bang mot xung
nhip, mach tren la bo ghi chuyen cd ngd vao song song.

- Khi duy tri day dieu khien (Shift/Load) d mu’c H, cong AND tren se hoat
dong cho moi flip flop, mdi xung nhip xuat hien se chuyen mot bit so lieu tCr mot flip
flop vao flip flop lien sau, tien hanh theo hirdng tir QA den QH. Noi cach khac, so
lieu se du’Oc dich chuyen noi tiep qua bo ghi chuyen. O mach flip flop dau tien
trong bo ghi chuyen, ngd vao cong AND tren cd ten la Serial Input (ngd vao noi
tiep). Nhir vay, so lieu cGng cd the nap vao bo ghi chuyen theo kieu noi tiep.

Tom lai, khi Shift/Load d mu’c L: mot xung nhip nap 8 bit so lieu ABCDEFGH
vao bo ghi chuyen theo kieu song song. Khi Shift/Load d mu’c H: cac xung nhip se
chuyen so lieu qua bo ghi chuyen theo kieu noi tiep.

Can chu y rang xung nhip du’Oc nap vao bang cong NOR hai ngd vao. Khi
Clock Inhibit d mu’c L, tin hieu xung nhip qua cong NOR dirpc dao lai va bo ghi
chuyen dirpc dap Crng theo sirdn xung dirong. Khi Clock Inhibit d mu’c H, ngd ra
co’ng NOR d mu’c L, xung nhip khong tdi dirpc cac flip flop. Luc nay, bo ghi chuyen
lull giCr noi dung trong bo ghi chuyen.

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Mu’c L d ngo vao Clear co the du’Oc nap vao tai thdi diem bat ky, khong can
quan tam den thdi diem xung nhip xuat hien va se reset ngay lap tCrc toan bo flip
flop tro ve trang thai "0". Khi khong dung tdi chan Clear, no phai luon luon du’Oc
duy tri 0 mu’c H.

7.4. BO GHI CHUYEN VAO SONG SONG-RA SONG SONG

De dap ting yeu cau co bo ghi chuyen vao song song-ra song song ta co the
thu’c hien mot cach don gian bang cach them vao dirong ra cho moi flip flop 0 IC
54/74166 da de cap tren day.

IC 54/74198 la loai IC TTL-MSI 8 bit co kha nang cho tin hieu vao song song
va cho tin hieu ra song song. IC co 24 chan, chCrc nang cua cac chan du’Oc ghi

trong Hinh 7.10. Ngoai chu t nang cho tin hieu vao song song va lay tin hieu ra

song song, bo ghi chuyen nay con co the chuyen so lieu tir trai qua phai va ngirpc
lai. Ta hay nghien cuu cach chuyen sang trai cac so lieu trong bo ghi chuyen.

IC 54/7495A la bo ghi chuyen 4 bit, ngo vao song song, ngo ra song song. IC
nay cung co ngo vao noi tiep, co the dung IC nay chuyen so lieu tir trai sang phai
(tir Qa sang QB) va theo hirdng ngirpc lai, chuyen tir phai sang trai. So do chan IC
va do thi logic cho tren Hinh 7.11. Cac flip flop va logic dieu khien du’Oc dung 6 IC
nay hoan toan phu hpp vdi flip flop va logic dieu khien cua IC 54/74164 nhu1 da
trinh bay 0 Hinh 7.9c.

Cac ngo ra song song la QA, QB, Qc, QD cua 4 flip flop trong bo ghi chuyen.

Khi chan Mode Control (kieu dieu khien) 6 mu’c H, cong AND d ngo vao ben

phai cua moi cong NOR hoat dong, con cong AND ben trai ngirng hoat dong. Cac
so lieu 6 chan A, B, C va D du’Oc nap vao bo ghi chuyen moi khi sirdn am cua
xung nhip xuat hien. Nhir vay la ngo vao so lieu song song.

Khi chan Mode Control (kieu dieu khien) d mu’c L, cong AND 6 ngo vao ben
trai cua cong NOR hoat dong, con cong AND ben phai cua cong NOR ngirng hoat
dong. Ngo vao so lieu doi vdi flip flop QA bay gid la ngo vao noi tiep. Ngo vao so
lieu cua flip flop QB lay tir QA va cis nhir the tiep tuc cho den flip flop Q0. 6 moi
sirdn am sung nhip xuat hien mot bit so lieu du’Oc dira noi tiep nhau vao bo ghi
chuyen ma flip flop QA la flip flop dau tien. Moi bit so lieu luu trCr du’Oc dich sang flip
flop lien ke ben phai, cho den flip flop cuoi cung QD Bo ghi chuyen hoat dong nhir
vay la dich chuyen so lieu tCr trai sang phai va co ngo vao so lieu noi tiep.

</div>
(12)<div class='page_container' data-page=12>

Co hai ngo vao xung nhip: Clock 1 va Clock 2, phu hpp vdi yeu cau can phan biet
xung nhip dung de dich sang phai vdi xung nhip dung de dich sang trai. Neu
khong can tach xung nhip dich sang phai va xung nhip dich sang trai co the ket noi
Clock 1 va Clock 2 lai v6i nhau.

56 lieu vao

S Q

— 0>
R

Clock-00

a) Flip Flop D

—•£>>•7

Day 3ieu khien

c) Bo sung day dieu khien Clock

"J5T >

x2___ *Z____
00

0 ° * — 5 Q w

Clock

-c>

b) Bo sung them cong NOR

d) Vi du 5 clock_________________

Hinh 7.9

Ngo vao
noi tiep
dich trai
Vcc 51

24 23 22

Input H QH Input G QG Input F QF Input E QE Clear

51 L H QH G QG F QF E QE

SO IC 54/74193 Clear

; /V QA B QB : QC QD CLK

LiJ

l

A

j

L±J UJ L®J CD [

j

EI [a] IioI l l DU

1 Input A QA Input B QB Input C QC Input D QD Clock GND

5 0

Ngo vao

A'. . • a'

noi tiep
dich phai

Hinh 7.10

</div>
(13)<div class='page_container' data-page=13>

7.5. BO OEM VONG

Cac bo ghi chuyen da trinh bay 0 phan tri/6c phai co mach bo trp de dieu
khien ngo vao D cua flip flop D (hoac cac ngo vao RS, cac ngo vao JK) de dam
bao thao tac chuan xac khi dich chuyen. Mach logic tao nen cac dang xung dieu
khien du’Oc lay tir phan dieu khien cua he thong. Phan dieu khien la nguon cung

cap xung nhip va cac tin hieu dieu khien khac rat quan trong. Mach phan hoi co
the di/a vao mach ghi chuyen co ban.

O u t o u li C'oc*. 2

— > C I<xk I L

Oj o, 0o n notei

S«rn> A

,n p g ( 0 GNO

R fT51 pn [io] 'TI

I i ( i

X

( O« o. Qc o*

V

CKt CK?V t

1 r

S f U i n o u l

A B c 0 M o d e 1

) i | t |

l i i i l l LU i±J liJ LU LiJ

if-OuH
fjJ Ptn ovji

O l l t >nput|

im« ( - 9 5 A . w S 3 s ) l L S 5 i, r r '. o e t r t ^ e i y V'jtsai:

i? l ld<} C -J

</div>
(14)<div class='page_container' data-page=14>

Hay bat dau tCr bo ghi chuyen noi tiep nhir IC 54/74164. Mot trong nhCrng ap
dung logic la ket noi ngo ra cua flip flop cuoi cung QH quay tro lai ngo vao D cua
flip flop A dau tien (Hinh 7.13a). Chu y rang cac ngo vao so lieu dirpc ket noi vdi
nhau. Gia du, tat ca cac flip flop deu du’Oc Reset va xung nhip hoat dong. Oieu gi
se xay ra? Cau tra Idi se la: khong cd gi xay ra vi ngo vao D cua flip flop dau tien
(ngo vao cua A va B) dang 6 mu’c 0. Do do, moi khi xung nhip len mu'c 1, so 0 cl
moi flip flop se du’Oc dich sang flip flop ben canh trong khi so 0 d flip flop cuoi cung
H theo mach phan hoi chuyen dich trd lai mach flip flop A dau tien. Noi cach khac,
tat ca cac flip flop dang d trang thai Reset, mdi mot xung nhip lai Reset cac flip flop
mot Ian nu’a va moi ngd ra flip flop duy tri mu’c 0. Hay xem bo ghi nhir mot ong
dirng day so 0 (giong nhirbong ping-pong trong ong), no dich vong quanh trong bo
ghi chuyen va dich dan sang flip flop ben moi khi xung nhip xuat hien. Gia du rang,
Qa d mu’c 1, con tat ca cac flip flop khac d mu’c 0, va xung nhip hoat dong. Chinh d
thdi gian dau khi xung nhip len mu’c 1, so 1 d A se dich chuyen sang B va A du'Oc

Reset vi so 0 d H se dich chuyen vao A. Tat ca cac flip flop khac van duy tri mu’c 0.
Xung nhip thir hai se dich chuyen tir C sang D, v.v... Nhir vay, so 1 nay se dich
chuyen tCr flip flop nay sang Flip flop lien ben moi khi cd xung nhip tac dong
(chuyen tir 0 len 1). Khi so 1 den flip flop H, xung nhip tiep theo se chuyen so 1
vao flip flop A theo day phan hoi. Mot Ian nCra, lai coi bo ghi nhu" ong dirng day
bong ping-pong, cd 7 qua trang (so 0) va mot qua den (sd 1) chuyen dong tuan
hoan vong quanh bo ghi chuyen theo chieu kim dong ho, chuyen sang flip flop
lien ke ben moi khi cd xung nhip xuat hien. Dang xung cua bo dem vong cho tren

Hinh 7.13b.

S e ria l d a ta in p u t

Hinh 7.12: Day ket noi cua IC 54/7495A de thirc hien chuyen dich trai

</div>
(15)<div class='page_container' data-page=15>

cioc*

°- j

1

____________________ i i____________________ r

°> 1 1— |_________________ L_J— 1__________________

! 1— 1 : j— i

l— 1 ; i i

_J !

■---
--

-b | C.>c d in q it iin q k h i -b o qJvi c h u y e n c « n v l m u c 1 v.» u a y m in 0

Hinh 7.13: Bo dem vona

Dang xung cua bo dem vong thirdng du’Oc dung de dieu khien he thong
digital. Bo dem vong du’Oc dung de dieu khien cac bien co phai xay ra theo mot
trinh tir thdi gian nghiem ngat, vi du, bien co A xay ra, tiep den la bien co B, roi den
bien co C v.v... So do logic 6 Hinh 7.14 chi ra cach tao ra Reset, Read, Bu va
Write dung lam cac xung dieu khien, cac xung do xuat hien tuan tii, xung no sau
xung kia. Cac tin hieu dieu khien nay xuat hien 6 cac ngo ra cua cac flip flop A. B,
D va du’Oc chi ra tren Hinh 7.13.

kE SEt ~

CONTROL
LOGIC

(LOGIC
D\e\J KHIeN)

COMPLEMENT
WRITE

tin
nieu
ra

(QD) COMPLEMENT

(QE) WRITE 1 1

n

i i

_ n _

</div>
(16)<div class='page_container' data-page=16>

Tuy vay, bo dem vong cung con nhieu van de can giai quyet. Oe tao ra cac

dang xung nhuf tren Hinh 7.13, bo dem can co mot va chi co mot so 1 duy nhat.

Thdi co de co xung nay khi bat dau dong nguon vao he thong rat hiem hoi. Neu tat
ca cac flip flop deu Reset khi dong nguon thi bo dem vong khong lam viec nhi/
chung ta thay trirdc day. Mat khac, neu mot vai flip flop 6 trang thai Set khi dong
nguon, chuoi cac dang xung phoi hop la ket qua cua cac flip flop du’Oc Set. Do do
can Preset bo dem de co cac trang thai mong muon trudc khi sir dung chung.

Ta hay xem xet mot phi/ong phap de xoa flip flop, di/a chung ve mu’c 0 va
xac lap mot mu’c 1 duy nhat.

Vi du, bo ghi chuyen dung IC 54/74164, 8 bit co chan Clear. Hay chi ra
phi/ong phap xac lap mu’c 1 va duy tri cac mu’c 0.

Nhu” Hinh 7.15a chi ra phi/ong phap don gian :"Reset khi dong mach".

Phirong phap nay du’Oc sir dung rong rai de tao ra mot xung am co do rong nho khi
dong mach nguon cung cap. Tri/dc khi dira nguon cung cap vao bo ghi chuyen,
dien ap tren tu bang 0 khi di/a dien ap nguon vao tu, tu du’Oc nap len den dien ap
+Vcc vdi hang so thdi gian RC va duy tri mu’c dien ap + Vcc trong suot thdi gian

duy tri dien ap nguon. Dang xung du’Oc chi ra tren Hinh 7.15a. Neu diem A du’Oc

noi vdi ngo vao Clear cua IC 54/74164, tat ca cac flip flop se tir dong du’Oc Reset
ve 0 khi dien ap + Vcc du’Oc dira vao mach IC.

Mach logic bo sung vao di/dng phan hoi nhi/ ve 6 Hinh 7.15b se tao ra mu’c

1 duy nhat de Set bo ghi. Ta xem mach hoat dong nhir the nao?
^ r r

i

c c T o C le a r

(O en m a c h x o a )

P o w e r o n
(D o n g m a c h n g u o n )

Hinh 7.15a: Mach Reset khi dong nguon

1. Xung :"dong mach nguon Reset" du’Oc dao va de Set ban dau cho flip flop
X. mCrc 1 nay du’Oc dira vao cong OR de ngo ra cong OR co mu’c 1.

Khi xung nhip dau tien den no se chuyen dich mu’c 1 nay vao QA.

2. Khi Qa len mu’c 1, no se Reset flip flop X. Tai thdi diem nay QA co mi/c 1,
cac Qb, Qc .... Qh co mu’c 0. X se duy tri mu’c 0 khi nao con nguon di/a vao flip flop
X. So lieu ti/ 0 H se di/a qua cong OR tri/c tiep di/a den ngo vao so lieu AB. Mu’c 1
duy nhat va bay mu'c 0 se dich chuyen vong quanh bo ghi chuyen, dich ti/ flip flop
nay sang flip flop lien ben, roi lai tii flip flop QH chuyen len flip flop QA moi khi xung
nhip xuat hien.

</div>
(17)<div class='page_container' data-page=17>

(Xung nhip)

< 1 : : ; 1 ■ '

1

. . i i • i

Pang xung dieu

khien can co ________

Hinh 7.16

Vi bo dem vong d Hinh 7.13 co the hoat dong vdi nhieu mu’c 1. Vi du, no co
the du’Oc dung de tao ra dang xung dieu khien phirc hpp. Gia du, can co dang

xung dieu khien nhir ve tren Hinh 7.16. Dang xung nay co the de dang du’Oc tao

nen bang cach Preset bo dem 6 Hinh 7.13 v6i mu’c 1 o A va C con cac flip flop

</div>
(18)<div class='page_container' data-page=18>

CAU HOI VA BAI TAP

7.1. Xac dinh so flip flop can thiet de xay diing bo ghi chuyen co kha nang
lull tru” du’Oc:

a) So nhi phan 6 bit,

b) Cac so thap phan len den 32,

c) Cac so thap luc phan len den F.

7.2. Mot bo ghi dich co 8 flip flop. So nhi phan co do I6n bao nhieu co the liru
trir vao bo ghi chuyen nay?

7.3. Ke ten 4 loai bo chuyen va ve so do khoi cho moi loai.

7.4. Ve dang xung dich chuyen so nhi phan 1010 vao trong bo

nhir ve tren Hinh 7.3.

7.5. Ve dang xung dich chuyen so nhi phan 1001 vao trong bo

nhir ve tren Hinh 7.4.

7.6. Bo ghi chuyen ve tren Hinh 7.3 I iru tru1 so nhi phan 0100. Ve dang xung
doi vdi chuyen tiep 4 xung nhip, gia du rang ca J va K deu 6 mu’c L.

7.7*. Can bao lau de chuyen so nhi phan 8 bit vao trong IC 54/74164 nhirve

tren Hinh 7.7. Neu xung nhip co tan so:

a) 1 MHz,
b) 5 MHz.

7.8*.. Tren co so bai tap so 7.5. tan so xung nhip ci/c dai la bao nhieu neu
thdi gian truyen dan so lieu la 30 ns?

7.9*. Doi vdi mach tren Hinh 7.9 viet mu’c logic tren moi chan cua cong, cho
biet:

a) Day dieu khien CONTROL = 1, X, = 0, X2 = 1,

b) Day dieu khien CONTROL = 0, = 0, X2 = 1.

7.10*. Ve tat ca cac dang xung ngo vao va ngo ra doi vdi IC 54/74166 nhi/

tren Hinh 7. 8 gia du rang so thap phan 190 du’Oc chuyen dich vao bo ghi chuyen

theo kieu:

a) song song,
b) noi tiep.

7.11*. Ve cac dang xung can dira vao va dich chuyen sang trai mot bit don 1
thong qua bo ghi chuyen tren Hinh 7.12.

7.12. Cd mach nhu' tren Hinh 7.12, giai thi'ch hoat dong cua mach khi dich

chuyen cac bit tir phai sang trai.

ghi chuyen
ghi chuyen

</div>
(19)<div class='page_container' data-page=19>

7.13*. Bo ghi chuyen Hinh 7.13 co the de dang xoa tat ca cac bit 0 bang
cach dung ngo vao Clear. Hay xem xet neu muon thiet ke mach logic de xac lap
bo ghi chuyen vdi sir chuyen doi cac bit 1 va bit 0.

7.14. Giai thi'ch hoat dong cua IC 4017. SCr dung IC nay de xay dirng mach
nhay dudi.

7.15. Xay dirng mach tao tre 8 |ns bang cach sir dung bo ghi chuyen. Giai
thi'ch hoat dong cua mach. Muon tang thdi gian tre len gap ddi (16 |is) thi can tien
hanh ra sao?

7.16. Sir khac nhau co ban giCra mach logic to hop va mach logic tuan tir la
gi?

7.17. Phan biet cac tir viet tat dirdi day thirdng du’Oc sir dung trong cac ky
hieu logic cua flip flop truyen thong:

a) CLK
b) CLR
c) D
d) R
e) S

7.18. Khi mot flip flop du’Oc dung de duy tri tam thdi so lieu thirdng du’Oc goi
bang tir nao?

7.19. Mot flip flop du’Oc kich hoat bang sirdn am xung nhip. Hoi flip flop thay
doi trang thai khi xung nhip chuyen tir mu’c nao den mu’c nao?

</div>
(20)<div class='page_container' data-page=20>

CHUONG 8

BO DEM

Hoc xong chuong nay hoc vien co kha nang:

□ Hieu duoc nguyen ly cau tao va hoat dong cua cac

loai bo dem va cac kieu dem.

□ Phan tich duoc cac dang xung cua cac loai bo dem

□ Ve du'Oc cac dang xung khi cho biet so do nguyen
ly cua cac loai bo dem.

□ Xay dung duoc cac loai bo dem va cac kieu dem.

□ Ap dung cac cach xay dung cac loai bo dem va
cac kieu dem vao cac Ung dung thuc te.

TONG QUAN

Trong he thong digital, bo dem la bo kien du’Oc su” dung nhieu va rat linh
hoat. Xung nhip dieu khien bo dem, nen co the dung bo dem de dem so chu ky
xung nhip. Vi da biet chu ky xung nhip nen bo dem du’Oc dung lam cong cu do thdi
gian, chu ky hoac tan so. Cd hai loai bo dem co ban la bo dem dong bo va bo dem
khong dong bo. Bo dem kieu song Ian truyen cd hoat dong va cau tao don gian, it
linh kien nhirng han che ve toe do dem. Mdi flip flop do flip flop phi'a trirdc lam cho
thay doi trang thai, do vay, tre truyen dan bi don ti'ch lai. Bo dem nhir vay du’Oc goi
la bo dem noi tiep hoac bo dem khong dong bo.

Bo dem song song hoac bo dem dong bo cd toe do hoat dong nhanh. 6 day,
xung nhip lam cho cac flip flop dong thdi thay doi trang thai. Tre truyen dan cua bo
dem bang tre truyen dan cua mot flip flop. Bo dem loai nay cd toe do truyen d in
nhanh, nhirng lai ddi hoi nhieu linh kien.

Cac bo dem noi tiep va song song du’Oc dung phoi hop de dung hoa giCra toe
do hoat dong va phan cirng. Cd bo dem thirc hien dem tien, cd bo dem thirc hien

dem lui. Bo dem cd the xoa de mdi flip flop deu chCra so 0 hoac dat trirdc so lieu
ngd vao de cho moi flip flop chCra mot bit du’Oc dir kien trirdc.

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(21)<div class='page_container' data-page=21>

8.1. BO DEM KHONG DONG BO

Bo dem nhi phan khong dong bo co the du’Oc cau tao bang cac flip flop JK.
Hinh 8.1 la 3 flip flop JK master/slave ket noi thanh 3 tang. Xung nhip la dang
xung vuong dieu khien flip flop A. Ngo ra cua flip flop dieu khien flip flop B va ngo
ra cua flip flop B lai dieu khien flip flop C. Tat ca cac ngo vao JK du’Oc ket noi vdi +
Vcc. Oieu do co nghTa la mdi mot flip flop se chuyen trang thai bu (toggle) moi khi
sirdn am xung nhip den ngd vao cua xung nhip CLK.

Khi ngd ra cua mdi flip flop du’Oc dung nhir ngd vao xung nhip cua flip flop
lien ke sau, bo dem nhu” vay goi la bo dem khong dong bo. Flip flop A phai chuyen
trang thai trirdc khi no tac dong den flip flop B, flip flop B phai thay doi trang thai
trirdc khi no tac dong den flip flop C. Viec lat trang thai xay ra tir flip flop nay sang
flip flop lien ke ben giong nhu song Ian truyen tren mat nirdc. Vi ly do do ma tre
truyen dan cua ca bo dem bang td’ng tre truyen dan cua rieng tirng flip flop. Vi du,
trong bo dem gom 3 flip flop, mdi flip flop cd thdi gian tre truyen dan la 10 ns, vay
tre truyen dan cua bo dem la 30 ns.

Hinh 8.1b la dang xung cua bo dem. Gia du rang, luc bat dau, cac flip flop
duoc reset (phuc hoi) de cd cac ngd ra 6 mu’c 0. Ta xem flip flop A chira LSB (so
cd trong so nho nhat) va flip flop C chira MSB (so cd trong so Idn nhat), nhir vay,
luc dau bo dem cd CBA = 000.

J1TL

f

Clock

— J A

-c>

— K A

J B

c>

AC B

~ J C

— AC C
B

Cac ngo ra
a) 3 o d e m nhi phan khong dong bo 3 b it

Thai gian h

-1_

Clock

1___ I

L

B

-C.

b) D ang xung

C

J t

-i---Thuf t j

d e m C B A

O o o o

1 O O 1

2 O 1 O

3 O 1 1

4 1 O O

5 1 O 1

6 1 1 O

7 1 1 1

O o o o

c ) B ang siJ t h a t

</div>
(22)<div class='page_container' data-page=22>

Tai moi thdi diem cua sirdn am xung nhip xuat hien thi flip flop A chuyen
trang thai. Nhir vay, tren true thdi gian,

- tai thdi diem a, flip flop A len mu’c 1,
- tai thdi diem b, flip flop A xuong mu’c 0,
- tai thdi diem c, flip flop A tro lai mu’c 1 v.v...

Nhan thay rang, tan so xung 6 ngd ra cua flip flop A bang 1/2 tan so xung nhip.
Vi flip flop A cd ngd ra noi vdi ngd vao xung nhip cua flip flop B, nen tai mdi
thdi diem dang xung cua flip flop A ve mu’c 0, flip flop B thay doi trang thai. Nhir
vay, tai thdi diem b, flip flop B cd mu’c 1, roi flip flop B xuong mu’c 0 tai thdi diem d
va lai len mu’c 1 tai thdi diem f. Nhan thay rang tan so xung 6 ngd ra cua flip flop B
bang 1/2 tan so cua flip flop A va bang 1/4 tan so xung nhip.

Vi flip flop B cd ngd ra ket noi vdi ngd vao xung nhip cua flip flop C, tai thdi
diem d dang xung cua flip flop B xuong mu’c 0, flip flop C chuyen trang thai. Nhu
vay, flip flop C cd ngd ra len mu’c 1 tai thdi diem d va lai trd ve mire 0 tai thdi diem
h. Tan so xung 6 ngd ra flip flop C bang 1/2 tan so d ngd ra cua flip flop B va bang
1/8 tan so xung nhip.

Nhan thay rang, trang thai ngd ra cua cac flip flop la mot so nhi phan tirong
dirong vdi so sirdn am xung nhip xuat hien. Trirdc thdi diem a, tren true thdi gian,
trang thai ngd ra la CBA = 001; tai thdi diem b chuyen thanh CBA = 010, v.v...
Thirc te, khi kiem tra cac dang xung, nhan thay rang, bo dem thirc hien dem tirng

xung nhip va dem theo he nhi phan (xem bang tren Hinh 8.1c).

Vi moi trang thai ngd ra bieu thi tren bang sir that la mot so nhi phan tirong
dirong vdi so xung nhip, ba flip flop trong Hinh 8.1 la bo dem nhi phan 3 bit khong
dong bo. Bo dem nay cd the dung de dem so xung nhip len den circ dai la 7 xung.
Bo dem bat dau dem tir 000 len den 111. Tai thdi diem nay, bo dem reset trd lai
000 va bat dau chu ky dem lai. Ta noi rang bo dem thirc hien cach dem tien.

De dang nhan thay rang, mot bo dem cd n flip flop se cd 2n trang thai ngd ra.
Vi du: mot bo dem gom 3 flip flop vira trinh bay d tren cd 23 = 8 trang thai ngd ra
(tir 000 den 111). Mot bo dem gom 5 flip flop cd 25 = 32 trang thai ngd ra (tir 00000
den 11111) v.v... Mot bo dem nhi phan Idn nhat cd the du’Oc bieu thi bang n tang
flip flop cd sd thap phan tirong dirong la 2n - 1. Vi du, bo dem 3 flip flop dem duoc
nhieu nhat 23 - 1 = 7 , so thap phan nhieu nhat ma bo dem 5 flip flop dem du’Oc la
25 - 1 = 31. Bo dem 6 flip flop dem du’Oc 63.

Bo dem 3 flip flop du’Oc goi la bo dem mo-dun 8 (hoac Mod-8) vi cd 8 trang
thai. Bo dem 4 flip flop la bo dem Mod-16. Mo-dun cua bo dem la tong trang thai
ma bo dem thu’c hien.

Hinh 8.2 la so do chan va bang sir that cua IC 54/74L93. Day la bo dem 4 bit
nhi phan thuoc loai mach TTL MSI, cd the lam bo dem Mod-8 hoac bo dem Mod-

16. Neu xung nhip dira den ngd vao B, luc do cac ngd ra la QB, Qc va Qz va bo

dem nay la bo dem Mod-8. Trong trirdng hop nay flip flop A khong dung den.

Mat khac, neu xung nhip du’Oc dira den ngd vao A va QA cua flip flop nay
du’Oc ket noi vdi ngd vao B, ta cd bo dem Mod-16, 4 bit, kieu khong dong bo.
Cac ngd ra la QA, QB, Qc va QD. Bang sir that cua bo dem 4 bit loai nay cho tren

Hinh 8.2c

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(23)<div class='page_container' data-page=23>

Tat ca cac flip flop cua IC 54/74L93 co cac ngo vao reset true tiep va hoat
dong d mu'c L tich c l / c . Nhi/ vay, can co cac ngo vao reset R0(1) va R0(2) cua 2
ngo vao cong NAND d cung mu'c H de reset tat ca cac flip flop cung mot luc. Bo
dem du’Oc reset, khong lien quan den xung nhip.

Dang xung cua IC 74L93 dem theo Mod-16 du’Oc bieu thi tren Hinh 8.3. Tai
thdi diem a, tr§n true thdi gian, bo dem co cac gia tri 6 ngo ra la 0000. Moi khi
si/on am xung nhip xuat hien, bo dem lai dem mot so, bat dau tir 0000 tai thdi
diem a den 1111 tai thdi diem b. Sang den thdi diem c, bo dem reset thanh 0000
va viec dem lap lai nhi/ trudc. Ro rang, day la bo dem Mod-16, vi cd 16 trang thai
(tir 000 den 1111) va so thap phan Idn nhat du’Oc liru trir trong flip flop la so 15

(1111).

enput A ■
r u i t u ;

. <1118!

^11

■

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Hinh 8.3: Cac dang xung cua Mod-16 trong trirdng hop su1 dung IC 54/74L93

Mot sir thay doi ly thu va co loi cua bo dem khong dong bo 3 bit 6 Hinh 8.1

du’Oc chi ra tren Hinh 8.4. He thong xung nhip van dira den ngo vao xung nhip cua
flip flop A, nhirng nay ngo ra A dieu khien flip flop B, tirong tir nhir vay, ngo ra B
dieu khien flip flop C. Hay chu y dang xung cua bo dem theo kieu ket noi nay.

Vcc

Thai gian.
Clock

Trang

3

thai C A

7 1 1 1

6 1 1 0

5 1 0 1

4 1 0 0

3 0 1 1

2 0 1 0

1 0 0 1

0 0 0 0

7 1 1 1

c)
Hinh 8.4: Bo dem lui

Flip flop A chuyen trang thai doi lap khi sirdn am xung nhip xuat hien nhir da
trinh bay 6 tren. Nhirng flip J o p B se chuyen trang thai doi lap khi flip flop A len
mu’c 1, chu y rang moi khi A len mu’c 1 thi A tir mu’c 1 chuyen xuong mure 0 va

sirdn am xung nhip tac dong vao flip flop B lam no chuyen trang thai. Tren true thdi
gian, B chuyen trang thai doi lap tai cac thdi diem a, c, e, g va i.

Tirong tii, flip flop C du’Oc B tac dong va flip flop C chuyen trang thai doi lap
khi flip flop B len mu’c 1. Flip flop C chuyen mu’c 1 tai a, xuong mu’c 0 tai thdi diem
e va len mu'c 1 tai thdi diem i.

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Nhu” vay tai thdi diem a, noi dung bo dem la ABC = 111, thay doi thanh 110
tai thdi diem b va thanh 101 tai thdi diem c. Nhan thay rang tri so cac ngo ra cua
bo dem cirgiam 1 moi khi cd xung nhip xuat hien. isloi mot cach khac, bo dem hoat
ddng theo kieu dem lui, cac ket qua du’Oc tom tat trong bang sir that d Hinh 8.4c

va day van la bo dem Mod-8 vi bo dem cd 8 trang thai va la bo dem lui.

Bo dem tien-lui, khong dong bo, 3 bit, dem tuan tir so nhi phan du’Oc gidi
thieu tren Hinh 8.5. Day la siJ ket hpp 2 bo dem da trinh bay tren day. Doi vdi bo
dem loai nay, de dem tien, mdi flip flop phai lat trang thai d phan thu’c (Q) cua flip
flop lien trudc (doi lap vdi phan bu Q ). Neu dem tien, day dieu khien dem lui dat d
mu’c 0, day dieu khien dem tien dat d mu’c 1, so do dang xung trong trirdng hpp
nay du’Oc trinh bay nhir tren Hinh 8.1.

Hinh 8.5

Mat khac, neu day dieu khien dem lui dat d mu’c 1 va day dieu khien dem tien
dat d mCrc 0, mdi flip flop phai lat trang thai d phan bu cua flip flop lien trirdc. Nhir

vay bo dem trong trirdng hpp nay la kieu dem lui, cd dang xung nhir tren Hinh 8.4.

Qua trinh dem nay cd the du’Oc tiep tuc doi vdi cac flip flop tiep sau de tao
nen bo dem tien-lui cd mo-dun Idn hon. Tuy vay, de xac dinh toe do circ dai cua bo

dem khi hoat dong, ta can tinh den cac do tre cua cac cong.

8.2. CONG GIAI MA

Co’ng giai ma can du’Oc ket noi vdi ngd ra cua bo dem, bang cach nhir vay,
ngd ra cua co’ ng se d mu’c 1 chi khi nao cac noi dung cua bo dem bang vdi trang
thai da cho. Vi du, cong giai ma du’Oc ket noi vdi bo dem khong dong bo 3 bit nhir

a Hinh 8.6a, se giai ma trang thai 7 (CBA = 111). Nhir vay, ngd ra cua co'ng se 6
mire 1 chi khi nao A = 1, B = 1 va C = 1 va dang xung xuat hien 6 ngd ra cua cong
mang so hieu 7. He thu’c logic doi vdi co’ng nay dirpc viet la 7 = CBA. So sanh vdi

bang sir that cua bo dem nay (nhu’ d Hinh 8.1) cho thay dieu kien CBA = 111 la

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Hinh 8.6

Bay trang thai khac cua bo dem co the du’Oc giai ma theo cach tirong tir tren
day. Vi du, de giai ma trang thai 5, bang sir that cho thay rang CBA =101 la trang
thai duy nhat. Be ngo ra cua cong o mu’c 1 trong thdi gian nay, ta phai dung C, B
va A d cac ngo vao cua cong. Can chu y rang neu B = 0, thi B = 1. He thu’c logic

chuan xac la 5 = CBA, cong ioai nay cho tren Hinh 8.6c. Bang xung cho tren Hinh

8.6b ma so 5 va ma so 7.

Tat ca 8 cong can giai ma 8 trang thai cua bo dem 3 bit nhir tren Hinh 8.1

du’Oc chi ra tren Hinh 8.7a. Cac ngo ra cua cong du’Oc bieu thi tren Hinh 8.7b. Cac
dang xung du’Oc giai ma la mot day cac xung dirong xay ra theo chuoi thdi gian
nghiem ngat va rat thuan loi khi dung lam tin hieu dieu khien he thong digital. Neu

ta xem trang thai 0 nhir la sir kien thCr nhat, roi den trang thai 1 la sir kien thu1 hai,
trang thai 2 la sir kien thCr 3 v.v... cho den khi trang thai 7. Ro rang la bo dem dang
dem tien theo ma so thap phan tir 0 den 7 roi lai dem tir 0.

Neu 8 cong nay du’Oc noi vdi bo dem tien-lui nhu” 6 Hinh 8.5, thi cac dang

xung du’Oc giai ma se xuat hien nhir tren Hinh 8.7b, day la bo dem theo kieu dem

tien. Neu bo dem theo kieu dem lui, co dang xung du’Oc giai ma xuat hien nhir tren

Hinh 8.7c, trong trirdng hop nay, neu trang thai 0 duoc xem nhir la siJ kien dau
tien, tiep theo la trang thai 7 la sir kien thir hai, v.v... xuong den trang thai 1. Ro
rang la bo dem dang dem lui theo ma so thap phan tir 7 den 0 roi lai dem b it dau
tir 7.

Vi du, dung IC 54LS11 co 3 cong ANB 3 ngo vao de giai ma cac trang thai 1,
4, 6 cua bo dem tren Hinh 8.5.

a biet rang so do logic va so do chan cua IC 54LS11_du;oc gidi th ie u jre n

Hinh 8.8. Cac he thu’c logic doi vdi cac trang thai tren la: 1 = C B A. 4 = C B A va
6 = C B A. Cach noi day tir cac ngo ra flip flop cua bo dem IC 54LS11 diroc chi ra
tren Hinh 8.8.

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Hay xem xet cac dang xung do bo dem Hinh 8.5 tao ra, day la hoat dong
dem tien. Xung nhip va moi ngo ra cua flip flop du’Oc ve lai tren Hinh 8.9, thdi gian
tre truyen dan cua mdi flip flop co ke den khi ve cac dang xung A. B va C. Chu y
rang xung nhip lam flip flop A lat trang thai va dang xung A nhu” vay bi tre di mot
khoang thdi gian b in g tp ke tu1 khi siren am xung nhip truyen dan. Tren Hinh 8.9

cung ve phan bu cua A la A de tham khao.

a) Dem tien b) Dem lui

Hinh 8.7: Cac cong giai ma cua bo dem nhi phan khong dong bo 3 bit

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Nhung hay quan sat xung tap nhieu xay ra khi bo dem tien hanh tCr trang thai
7 den trang thai 0. Tren true thdi gian, A xuong mu’c 0 (A len mu'c 1) tai thdi diem
a. Vi thdi gian tre cua flip flop B mai tan thdi diem b mdi xuong mu’c 0. Nhu1 vay,
giura 2 diem a va b tren true thdi gian, ta cd dieu kien C = 1, B = 1 va A = 1, do do,
ngd ra cua cong d mu’c 1 va mot xung tap nhieu xuat hien. Hay xem dang xung 6 =
C B A . Phu thuoc vao cac ngd ra cua cong giai ma nhir the nao de cac xung tap
nhieu (cac xung khong mong muon) cd the hoac khong cd the anh hirdng nghiem
trong den bo dem. Cac xung tap nhieu cd do rong vai nano giay va ngay ca khi
quan sat tren may hien sang cGng con kho khan. Nhirng IC TTL chuyen trang thai
rat nhanh va mach TTL se dap Crng ngay ca cac xung tap nhieu nho nhat xuat
hien, ma cac xung tap nhieu thirdng xuat hien vao cac thdi diem khong mong
muon. Do do, phai coi chi/ng tranh dieu kien nay. It nhat la cd hai giai phap doi vdi

xung tap nhieu. Phuang phap thCr nhat la lay mau cac co’ ng, ta se xem xet phirong

phap nay ngay 6 phan dirdi day. Phwong phap thu" hai la dung bo dem dong bo,

mot loai bo dem kha quan trong va du’Oc trinh bay d phan tiep theo.

A-B -

C

C i o c k

-X u n g t a p n h i e u

J

a) D a n g x u n g b) C o n g g ia i m a c o lay m au

Hinh 8.9

Hay xem xet viec sir dung cong AND 4 ngd vao de giai ma trang thai 6 d

Hinh 8.9b, 6 day, xung nhip du’Oc dung nhir la xung chon. Khao sat dang xung
tren Hinh 8.9b thay ro rang xung nhip 6 mu’c 0 giCra cac diem a va b tren true thdi
gian. Vi xung nhip phai 6 mu’c 1 de cho ngd ra cua co’ ng d mu’c 1, vi vay, xung tap
nhieu khong the xuat hien. Noi mot cach khac, xung nhip d mu’c 1 khi C = 1, B = 1
va A = 1 va xung doi vdi trang thai 6 xuat hien dung luc no can du’Oc xuat hien.
Chu y rang do rong xung dirong d trang thai 6 dung bang do rong phan duong cua
xung nhip. Hay quan sat dang xung 6 = C B A x xung nhip CLK. Ky thuat nay cd
the du’Oc ap dung doi vdi cac cong giai ma trang thai 7 khac cua bo dem nay (hoac
cho bo dem bat ky khac) va cac dang xung ngd ra du’Oc giai ma se khong cd cac
xung tap nhieu.

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8.3. BO OEM DONG BO

Bo dem khong dong bo co cau tao don gian nhat, nhung co han che khi lam
viec 6 tan so cao. Nhir da trinh bay 6 phan tren, moi flip flop co mot thdi gian tre. O
bo dem khong dong bo, cac thdi gian tre nay la cong tinh, to'ng thdi gian tre cua bo
dem xap xi bang tong so thdi gian tre cua cac flip flop. Kha nang xay ra xung tap

nhieu d ngo ra cua cac cong giai ma co the co d bo dem khong dong bo. Ca hai
van de tren co the khac phuc nhd bo dem dong bo. Oieu khac nhau chinh la moi
flip flop du’Oc chuyen trang thai dong bo vdi xung nhip.

Hinh 8.10 la cau true cua be dem nhi phan song song, cung vdi bang sir that
va cac dang xung doi vdi trudng hdp dem tu nhien theo kieu tuan tu. Vi moi trang
thai ting v6i mot so nhi phan tuong duong, nen ta gan cho moi trang thai la mot so
dem. 6 day, tu tudng cd ban la giu cho ngd vao J va K cua mdi flip flop d mUc 1,
nhu vay, flip flop se chuyen trang thai ifng vdi mdi sudn am cua ngd vao xung
nhip. Tiep do, ta dung cac co’ ng AND de chon mdi xung nhip thU tu den flip flop C
v.v...Cau hinh logic nay thudng mang y nghTa dieu khien tUng mdi flip flop mot.

Xung nhip dupe true tiep dua den flip flop A. Vi flip flop A chuyen trang thai
vdi mdi sudn am xung nhip 6 ngd vao CLK va flip flop A se chuyen trang thai doi
lap khi ca hai ngd vao J va K deu 6 mu’c 1, nen flip flop A se chuyen trang thai Ung
vdi moi sudn am xung nhip.

Mdi khi A 6 mu’c 1, co’ ng AND X dUdc hoat dong va mot xung nhip duoc
chuyen qua co’ng de den ngd vao cua CLK cua flip flop B. Sau do B chuyen trang
thai vdi mdi sudn am khac tai cac thdi diem b, d, f va h tren true thdi gian.

Vi co’ng AND Y dupe hoat dong va se chuyen xung nhip den flip flop C chi khi
ca hai A va B d mu’c 1, flip flop C chuyen trang thai Crng vdi moi sudn am xung nhip
thU tu tai cac thdi diem d va h tren true thdi gian.

Xem xet cac dang xung va bang su that, thay rang bo dem nay thuc hien
dem tien chuoi so nhi phan tu nhien tU 000 den 111, dem tang tUng so moi khi
sudn am xung nhip xuat hien. Day la bd dem nhi phan dong bo Mo-dun 8, theo
kieu dem tien.

Vdi cau hinh bo dem loai nay, ta khac phuc dupe hien tupng xung tap nhi<§u
da gidi thieu d phan tren day. Cac dang xung cua bo dem nay dupe ve lai d Hinh
8.11 va ta cd tinh den thdi gian tre truyen d in cua tUng flip flop. Nghien cuti ky cac
dang xung nay, ta rut ra ket luan sau:

- Sudn am xung nhip la tac nhan lam cho mdi flip flop chuyen trang thai doi
lap.

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- Ket qua cua chuyen trang thai dong bo khong lam nay sinh ra cac xung tap
nhieu 6 ngo ra cua cong giai ma, nhu' cong doi vdi so 6 du’Oc chi ra tren Hinh 8.11

Do do, cac cong giai ma tren Hinh 8.7 du’Oc sir dung cho bo dem nay ma khong so

xay ra xung tap nhieu.

n

U r

Hinh 8.10

— l _

Hinh 8.11

Ta hay so sanh cac dang xung nay voi cac dang xung cua bo dem khong

dong bo 6 Hinh 8.9.

Bo dem tien, mac song song (dong bo) co the du’Oc xay di/ng theo cach
tirong tii va du’Oc chi ra 0 Hinh 8.12. O mot he thong dong bo. thdi gian tai do mot
flip flop bat ky chuyen trang thai du’Oc xac dinh bang cac trang thai cua tat ca cac

flip flop d phia trirdc cua flip flop vtia chuyen trang thai. Theo kieu dem tien. mot
flip flop phai chuyen trang thai doi lap moi khi tat ca cac flip flop dting phia trirdc no
d trang thai 1 va xung nhip lam chuyen trang thai. Theo kieu dem lui. flip flop
chuyen trang thai doi lap chi xay ra khi tat ca cac flip flop phia trirdc 0 trang thai 0.

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Bo dem tren Hinh 8.12 la bo dem dong bo, 4 bit, dem tien-lui. De dem tien,
xung nhip cua bo dem du’Oc dua den ngo vao dem tien trong khi ngo vao dem lui
dupe giu 6 mu’c 0. De dem lui, xung nhip cua bo dem du’Oc dua vao ngo vao dem
lui trong khi duy tri ngo vao dem tien 6 mu’c 0.

Khi duy tri ngo vao dem lui 6 mu'c 0 (dem tien) se lam cho cac cong AND Y,,
Y2 va Y3 khong hoat dong. Xung nhip du’Oc dira vao ngo vao dem tien se dua
th in g vao flip flop A va se dieu khien cac flip flop khac bang cac cong X v X2 va X3.
Bo dem nay se van hanh dung nhu 6 bo dem dong bo da thao luan trudc day tren

Hinh 8.10. Dieu khac nhau duy nhat la 6 cho day la bo dem Mod-16, dem tien
tUng so nhi phan va tac dong voi sudn am xung nhip, bat dau tU 0000 va dem len

den 1111. Dang xung chuan xac dupe chi ra tren Hinh 8.12b.

Day dem tien

Day dem lui

a) 5a do logic

C L K

T h a i

g ia n j b

I

b) C a c d a n g x u n g d e m t i e n

k i m

--- s---1--- r n--- r
C L K

J 1 _ J !_ l L _ r i _ j L i l_ J L_i L _ r T _ i L

c

o

c ) C a c d a n g x u n g d e m lu i

Hinh 8.12

Neu day dem tien duy tri 6 mUc 0, cac cong AND phia tren cua Hinh 8.12 la

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se di th in g den flip flop A va dieu khien cac flip flop sau nhd cac cong AND Y,, Y2
va Y3.

Flip flop A se chuyen trang thai doijap moi khi co sudn am xung nhip chuyen
den nhu chi ra tren Hinh 8.12c. Mdi khi A o mu’c 1, cong AND Y, se hoat dong va
xung nhip chuyen trang thai se lam cho flip flop B chuyen trang thai doi lap tai cac
thdi diem a, c, e, g v.v... moi khi ca hai ngo ra A va B deu 6 mu’c 1, cong AND Y2

dupe hoat dong va nhu vay xung nhip se dieu khien flip flop tai_a,_e, i, m va q.
Tuong tu, cong AND Y3 se hudng xung nhip vao flip flop D chi khi A, B va C deu d
mu’c 1. Nhu vay, flip flop D se chuyen trang thai doi lap tai cac thdi diem a va i tren

true thdi gian. Cac dang xung tren Hinh 8.12c chifng to rang bo dem dang hoat

dong theo kieu dem lui, thu’c hien dem moi Ian mot soi nhi phan, bat dau tir 1111
xuong den 0000.

Neu nghien cifu so do logic cua IC TTL 54/74193 nhu gidi thieu tren

Hinh 8.13, ta thay rang IC nay dieu khien logic giong nhu bo dem gidi thieu tren

Hinh 8.12. IC MSI nay la mot bo dem dong bo, 4 bit, dem tien va dem lui. IC nay
co chan xoa (Clear) va chan xac lap trudc (Preset), cac ti'nh chat nay dupe trinh
bay d phan sau. Bay gid ta can xem xet logic dieu khien doi vdi moi flip flop va
cong OR cung hai cong AND, tai cong OR dung de cung cap xung nhip cho moi
flip flop.

Dang xung cua IC 54/74193 dupe gidi thieu tren Hinh 8.12, chi cd dieu can

chu y la dang xung 6 ngd ra A, B, C va D thay doi trang thai tuong ting vdi sudn

am xung nhip. Trong mach logic 6 Hinh 8.12, xung nhip dua den mot trong hai

ngd vao dem tien hoac dem lui deu qua mot co’ ng dao trudc khi dua den co’ ng logic
AND-OR cua mdi ngd vao xung nhip cua flip flop.

Bo dem tien-lui dong bo co the dupe xay dung theo so do logic hoi khac mot
chut, nhu chi ra tren Hinh 8.14. 6 bo dem loai nay, tai thdi diem mot flip flop nao

do thay doi trang thai duoc xac dinh bdi cac trang thai cua cac flip flop dirng phia
trudc trong bo dem. 6 kieu dem tren, flip flop phai chuyen trang thai doi lap moi khi
tat ca cac flip flop phia trudc dang 6 trang thai 1 va khi xung nhip den ngd vao
xung nhip CLK lam cho flip flop nay chuyen trang thai. 6 kieu dem lui, flip flop
chuyen trang thai doi lap khi tat ca cac flip flop d phia trudc dang 6 trang thai 0.

Bo dem dac biet nay lam viec theo kieu kiem che vi moi flip flop thay doi
trang thai theo sudn am xung nhip khi cac ngd vao J va K deu d mu’c 1 va khong
thay doi trang thai khi cac ngd vao J va K deu 6 mu’c 0.

Tom lai, kieu dem cua bo dem loai tien-lui nay thuc hien theo trinh tu thdi
gian nghiem ngat nhUsau:

1. Thiet lap mu’c 1 hoac mu’c 0 6 cac ngd vao J va K.
2. Cho xung nhip chuyen tir mu’c 1 xuong mu’c 0.

3. Xem ngo ra cua flip flop de xac dinh flip flop da chuyen trang thai doi lap
chua.

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Hinh 8.14b gidi thieu bang su1 that cua bo dem tren Hinh 8.14a. De dap ting
yeu cau can phai chuyen trang thai moi khi xung nhip chuyen trang thai, cac ngo
vao J va K cua flip flop A phai du’Oc giCr 6 mu’c 1. Dieu nay thi'ch hpp cho ca kieu
dem tien Ian dem lui.

Trong kieu dem tien, flip flop B doi hoi dieu kien flip flop A 6 mu'c 1 va xung
nhip chuyen tu” mu’c 1 xuong mu’c 0 de no chuyen trang thai. Moi khi day dem tien
va flip flop A, ca hai deu 6 mu’c 1, ngo ra cua cong X! 6 mu’c 1. Moi khi mot trong
hai ngo vao Z, 6 mu’c 1 thi ngo ra cua no cung 6 mu’c 1, do do, cac ngo vao J va K
cua flip flop B cung 6 mu’c 1. Khi 6 kieu dem tien, sirdn am xung nhip den ngo vao
xung nhip CLK cua flip flop lam cho no chuyen trang thai bu, cti nhir the dem tien

len, bat dau tu" 1 len 2 hoac 3 len 4.

6 kieu dem lui, flip flop B phai thay doi trang thai moi khi A 6 mu’c 1 va sirdn
am xung nhip xuat hien. Ngo ra cong Y, 6 mu’c 1, do do, cac ngo vao J va K cua
flip flop B d mu’c 1 moi-khi A va day dem lui cung 6 mu’c 1. Nhir vay, flip flop B thay
doi trang thai moi khi A 6 mu’c 1 va xung nhip chuyen tir 1 xuong 0. Viec dem lui
thu’c hien tir 0 den 15 hoac tir 14 den 13...

6 kieu dem tien, xung nhip dich chuyen xuong mu’c 0, lam cho C chuyen
trang thai doi lap moi khi ca A va B deu 0 mu’c 1 (dich chuyen tir 3 len 4; 7 len 8; 11
len 12 va 15 den 0) Ngo ra cua cong X2 0 mu’c 1 moi khi ca hai A va B deu 6 mu’c
1 va day dem tien cGng 6 mu’c 1. Do do, cac ngo vao J va K cua flip flop C 6 mu’c 1
trong khoang thdi gian nay va C chuyen trang thai.

6 kieu dem lui C chuyen trang thai moi khi ca A va B deu 6 mu’c 1. Ngo ra
cua cong Y2 6 mu’c 1 vao luc ca hai A va B deu a mu’c 1 va day dem lui 6 mu’c 1.
Do do, cac ngo vao J va K cua flip flop C 6 mu’c 1 trong khoang thdi gian nay va C
chuyen trang thai, do la viec chuyen tir 15 xuong 0; 12 xuong 11; 8 xuong 7 va 4
xuong 3.

Trong kieu dem tien, D phai chuyen trang thai doi lap moi khi A, B va C tat
ca deu d mire 1. Ngo ra cua cong X3 6 mire 1. Nhu” vay, cac ngo vao J va K cua flip
flop D d mire 1 moi khi tat ca A, B, C va day dem tien deu 6 mu’c 1. Flip flop D
chuyen trang thai khi dem tir 7 len 8 va tir 15 den 0.

9 ?

</div>
(34)<div class='page_container' data-page=34>

D ata l" p ’Jt A 115(

‘i'rn U it

C ti.!'»? Up

;C*rr;i I -is.

input B i l l

Data input C iioj

Data Input D'91
C l e a r

.c a d : 1''

.— . j i r y C » ■.

O u ' . p j t O i

Outtt/i 0,

Oulffu*. CL

II Hi l l

i ( M i l l !
r r n

r~T

~ —

T J n ;! 11 t——

m r F >

.T ■* ■■ ~N f/i

„

• I

:j j So d o l o g . c

Hinh 8.13

Hay xem so do logic cua IC TTL MSI 54/74191 gi6i thieu tren Hinh 8.15. Day

la bo dem tien-lui dong bo, IC nay co chan Preset va chan Clear.Ta so sanh so do

logic cua IC 54/74191 voi bo dem 6 Hinh 8.14 de hieu ro hoat dong cua bo dem

dung IC 54/74191.

</div>
(35)<div class='page_container' data-page=35>

Can chu y rang ngo vao xung nhip qua mot cong dao trirdc khi xung nhip di
den tiing flip flop. Nhi/ vay, cac ngo ra cua cac flip flop Master-Slave se chuyen
trang thai chi khi nao xung nhip d ngo vao cua flip flop chuyen trang thai tCr mu’c 1
xuong mu’c 0 va do do, xung nhip 6 ngo vao chung chuyen tCr 0 len 1. Cac dang
xung du’Oc trinh bay tren Hinh 8.15b.

Dem tie n

Dem lui

D c 5 A Bern

0 0 0 0 0

0 0 0 1 1

0 0 1 0 2

0 0 1 1 3

0 1 0 0 4

0 1 o 1 5

0 1 1 0 6

0 1 1 1 7

1 0 0 0 &

1 0 0 1 9

1 0 1 0 10

1 0 1 1 11

1 1 0 0 12

1 1 0 1 13

1 1 1 0 14

1 1 1 1 15

0 0 0 0 0

b ) d a n g 5U th a t

Hinh 8.14

8.4. BO OEM MOD-3

Cac bo dem da trinh bay tren day la cac bo dem dong bo va khong dong bo,
dem tien, dem lui. Tat ca deu la cac bo dem nhi phan, co Mo-dun la 2n, trong do, n
la so flip flop. Cac bo dem nhir vay goi la "dem tii nhien" cua 2n.

Mot bo dem Mod-2 du’Oc xay diing boi mot flip flop duy nhat; mot bo dem

Mod-4 can 2 flip flop, no dem tirng trang thai rieng biet.

</div>
(36)<div class='page_container' data-page=36>

Ou'cuo

i C w l I / ---*\

M i x / r !Ol «kil*

Ycc %1« A OOC* Cl&Cfc m in Loao C j'. j C O j j D

71 P . R . F I L r T u ' I t T —

<ff W II>t jSJKJfcWJ#

. — j. _ 6 ---- 1 JT

p
rd? » rw !

c * ^:t o r v or..* ‘*t* <Wbt)%( ^_0*d c

_ o

I 1

J

0 , G *

Oo<wV
go Of
OW i TrpU t %0

*411 0 * • <2* - S.

. . , — 1

t r . - i - -i 1 , j , I
0 - - Cr ar*J C?» _ D ! J

I I I L i i L u Id L i ) Llj J j

011 * 0 u , 0 * £/u£tl« Q o v irv .1

up ^’-’c

CfiO

int Cuipju --- ^ OutCUII

'npurt

*9t '!»Si9S cocrir*

RHx*« c'ock

M i Kirr\>t\ p r j i

Output O,

Oukxji Q,

•6>

f ---O ulout Qr

» v

Cxi'SUC Oc

Hinh 8.15 a) Sd do chan cac IC 74191 va 74LS191 (bo dem nhi phan)
b) Cau true ben trong cua IC

</div>
(37)<div class='page_container' data-page=37>

'19 ?. 'LS191 Binary coumef

Typ>c«l to*d, count, end inhibit •*o<nrvc*i

l l i u n r e t ^ d t x l o w is i n * f o l l o w i n j M Q u m c t
l o a d I f x e v e t l t o b i n a r y t h ir t e e n .

2 C o u n t u p t o f o u r t e e n , f i f t w n ( m * x n n u m l , l e r c . o n e . a n d t w o
3 . i n h t D i i .

A. Count oown to one, zero (minimum], <ihe«n, fourteen, end thirteen

D a ta in p u t s

ftippie dock

Hinh 8.15c

Thuting ngirdi ta yeu cau xay dirng cac bo dem co Mo-dun khac vdi 2, 4, 8,
16 v.v... Vi du, can co bo dem Mod-3 hoac Mod-5. Mot bo dem co Mo-dun nho hon
luon luon co the du’Oc xay dirng tir bo dem co Mo-dun I6n hon bang cach bo qua
mot so trang thai. Nhu” vay, bo dem da du’Oc cach dieu. Trirdc het, can xac dinh so
flip flop can co de xay dung loai bo dem cach dieu. So flip flop du’Oc xac dinh bang
cach chon so dem tir nhien thap nhat, nhirng Idn hon so dem cua bo dem du’Oc
cach dieu. Vi du, bo dem Mod-7 doi hoi co 3 flip flop, vi 8 la so tu nhien thap nhat
Idn hon so dem cua bo dem cach dieu co Mod-7.

</div>
(38)<div class='page_container' data-page=38>

TUdng tu". mot bo dem co 3 flip flop co so dem tu* nhien la 8, nhi/ng do bo Qua
so dem, nen co the xay dung bo dem cach dieu co mod la 7, 6 hoac 5.

Co nhieu phirong phap xay dirng cac bo dem co kieu dem cach dieu. Bo dem
loai nay co the la loai bo dem dong bo, bo dem khong dong bo hoac phoi hdp ca
hai loai bo dem dong bo va khong dong bo va co giai phap dem nhay. Vi du, neu
bo dem Mod-6 dung 3 flip flop se co 2 trong so 8 trang thai can du’Oc bo qua. Bo
dem Mod-3 di/dc xem xet ngay trong phan nay, con d phan sau ta xem xet bo dem
Mod-5.

Thai g ia n

Clock

Vcc
J

Ot>

K

+ Vcc
J 3

—
K &■

A
) S o d o lo g ic

i

C lo c k

_n_

A

---b ) 5 a d d da ng xung

& A Bern

O O O

O 1 1 C lo c k_____ w M o d -3

1 0 2 Xung n h ip

O O O a\ t>\

c ) B ang s u th a t d ) K h o i lo g ic

Hinh 8.16

Hinh 8.16 la bo dem Mod-3 dung 2 flip flop. Vi 2 flip flop co so dem tir nhien
la 4, nen bo dem cach dieu Mod-3 da bo qua mot trang thai. Cac dang xung va
bang sir that d Hinh 8.16 chi ra rang bo dem thu’c hien dem cac chuoi nhi phan 00,
01, 10, roi quay lai so 00. Bo dem bo qua so nhi phan 11. Bo dem thirc hien phep

dem nhir sau:

1. Trudc thdi diem a tren true thdi gian, A = 0, B = 0; tai thdi diem a, sirdn am
xung nhjp den gay ra:

- A chuyen trang thai doi lap vi cac ngo vao J va K deu d mu’c 1.

- B reset ve 0 (B vcn da d 0), vi ngo vao J 6 mCrc 0 va ngo vao K o mu’c 1.
2. Trudc thdi diem b tren true thdi gian, A = 1 va B = 0; tai thdi diem b, sirdn
am xung nhip den, gay ra:

- A chuyen trang thai bu, (tir 1 ve 0) vi cac ngo vao J va K deu o mu’c 1
- B chuyen trang thai bu (tir 0 len 1) vi cac ngo vao J va K deu d mire 1

3. Trudc thdi diem c tren true thdi gian, A = 0 va B = 1; tai thdi diem c, sirdn
am xung nhip den, gay ra:

- A reset ve 0 ( A von da d mu’c 0), vi ngo vao B 6 mu’c 0 va ngo vao K 6 mu’c 1.

</div>
(39)<div class='page_container' data-page=39>

- B reset ve 0 vi ngo vao J d mu’c 0 va ngo vao K 0 mu’c 1

4. Bo dem tien hanh qua 3 trang thai, tang len theo birdc dem la 1 khi co
sirdn am xunhg nhip xuat hien d ngo vao xung nhip CLK.

Bo dem Mod-3 dung 2 flip flop du’Oc coi nhir la mot khoi logic nhi/ chi ra tren

Hinh 8.16d. Khoi logic nay co ngo vao xung nhip, cac ngd ra A va B. Bo dem
Mod-3 du’Oc coi nhir la khoi chia 3, vi dang xung 6 ngo ra B (hoac A) co chu ky
bang 3 Ian chu ky xung nhip. Noi cach khac, bo dem Mod-3 chia tan so xung nhip
cho 3. Chu y rang, day la bo dem dong bo vi ca hai flip flop cung chuyen trang thai

dong bo vdi xung nhip.

Neu ta xem mot flip flop la mot bo dem Mod-2, ta thay rang mot bo dem Mod-
4 (2 flip flop mac noi tiep) la hai bo dem Mod-2 mac noi tiep. Tirong tir, mot bo dem
Mod-8 la ket noi mot cach don gian 2x2x2 v.v... Nhu” vay, cac bo dem co Mod cao
hdn co the du’Oc xay dirng tir cac bo dem co Mod nho hon. Vi du, khi noi mot flip

flop vdi ngo ra B cua mot bo dem Mod-3 nhir d Hinh 8.16, ta co mot bo dem

Mod-6 (3 x 2 = Mod-6) nhir chi ra tren Hinh 8.17. Ngd ra cua flip flop du’Oc ky hieu C. Chu y
rang, xung cua bo dem Mod-6 co dang doi xCrng va no co tan so bang 1/6 tan so
xung nhip.

+ Vcc

Clock
(Xung n h ip )

a ) 3 3 d e m M o d 6 = 3 x 2 h ) P a n g xung

+ Vcc

Hinh 8.17

Clock
(xu n g n h ip )

a ) 3 3 d e m M o d - 6 = 2 x 3

Clock

Q

A

3

J---L

1___ T

t>) D ang xung

</div>
(40)<div class='page_container' data-page=40>

Ta cung co the xay dirng bo dem Mod-6 bang cach ket noi flip flop 6 phia

trirdc bo dem Mod-3 (Hinh 8.18), luc do, ta co bo dem Mod-6 = 2 x 3 . Dang xung

cua bo dem nay du’Oc bieu thi tren Hinh 8.18b. IC TTL MSI 54/7492A la bo dem

chia 12. Khi xem xet ky, ta thay rang 0 so do logic, cac flip flop QB, Qc va QD dung

la bo dem 3 x 2 nhir tren Hinh 8.17. Nhi/ vay, neu xung nhip di/a den ngo vao B

cua IC '92A va cac ngo ra lay tu” QB, Qc va QD, thi day la bo dem Mod-6 ('92A la ky
hieu pho bien va thu gon cua IC 54/7492A).

Mat khac, neu xung nhip du’Oc dira den ngo vao A va QA du’Oc ket noi vdi ngo
vao B, ta co bo dem Mod-12 = 2 x 3 x 2 . Bang sir that cua cau hinh bo dem Mod-
12 cho tren Hinh 8.19b. Day la bo dem khong dong bo vi tat ca cac flip flop deu
khong dong thdi chuyen trang thai. Nhir vay, co kha nang xay ra hien tirpng xung

tap nhieu xuat hien 6 ngo ra cua cac cong giai ma khi du’Oc dung vdi bo dem nay.

Ti/ cac phan tren, ta thay rang co the xay di/ng cac bo dem co so dem tir
nhien (2, 4, 8, 16 v.v...) va bo dem Mod-3. Ta cung co the ket noi cac bo dem nay
de tao nen cac bo dem co cac Mod 2, 3, 4, 6, 8, 9, 12 v.v...

Ngo vao A (14)

Ngo vao 3 0)

(6)
RO(1) ---,

RO( 2) IZ I* .

a) 5 a do logic

-O

J Q

(12)

> CLK.
K

j a

>CLK

Z?“

□ J Q

- £ > cLK

K Q

Y

01) QB

(9)

J Q (6 )

> CLK
K

Dem QD QC QB QA

0 L L L L

1 L L L H

2 L L H L

3 L L H H

4 L H L L

5 L H L H

6 L H H L

7 L H H H

6 H L L L

9 H L L H

10 H L H L

11 H L H H

t>) Bang su that

Input A NC QA QB GND QC QD

fwl Ib] [tFI I~n1 [iol in 171

LJ LJ Lll LJ LJ L_l l_J

Input B NC NC NC Vcc ROd) RO(2)

Logic d ifcn g : Xem bang c h itc nang
G h i chu: C h in QA k e t n o i vcri ngo vao B

c) 5 a do chan IC '92A, 'L592

Hinh 8.19

8.5. BO DEM MOD-5

Bo dem dung 3 flip flop nhir tren Hinh 8.20 co so dem tir nhien la 8. nhi/ng
no du’Oc ket noi de bo qua 3 trang thai. Thirc chat, bo dem nay la bo dem tien,

</div>
(41)<div class='page_container' data-page=41>

birdc dem la 1, dem chuoi nhi phan tu1000 den 100, do do, day la bo dem Mod-5.
Hay xem xet hoat dong cua bo dem nay.

Cac dang xung cho thay rang flip flop A chuyen trang thai moi khi xung nhip
chuyen tir 1 xuong 0, trir trirdng hop bo dem dem tir 4 den 0. Nhir vay, flip flop A
can phai chuyen trang thai_mdi khi xung nhip den ngo vao CLK va khong lat trang
thai khi dem 4. Chu y rang C d mu’c 1 trong moi birdc dem, tru" budc dem 4. Neu C
du’Oc ket noi den ngo vao J cua flip flop A, ta se co tin hieu ngirng dem nhir mong
muon. Oieu do dung vi cac ngo vao J va K 6 flip flop A bang 1, trir khi dem 4. Nhir
vay, flip flop chuyen trang thai moi khi sirdn am xung nhip xuat hien. Tuy nhien, khi
dem 4, J = 0 va thdi gian tiep sau, sirdn am xung nhip den ngo vao CLK, flip flop A
se khong xac lap. Viec ket noi nhu” vay lam cho flip flop A thu’c hien dem chuoi so

nhu1 mong muon, ket qua chi ra 6 Hinh 8.20.

Dang xung tren Hinh 8.20b chi ra rang flip flop B phai chuyen trang thai mdi
khi A chuyen tir 1 xuong 0. Nhir vay, ngo vao xung nhjp cua flip flop B do flip flop A

dieu khien, dieu nay du’Oc the hien tren Hinh 8.20c.

Neu xung nhip lam cho flip flop C chuyen trang thai khi J = 0 va K = 1, thi moi
xung nhip se phuc hoi flip flop nay. Bay gid, ngo vao J 6 mu’c 1 chi khi nao dem 3
thi C se 6 mu’c 1 khi dem 4 va C d mire 0 trong tat ca cac birdc dem khac. Cac
mu’c can thiet doi vdi ngo vao J do co’ng AND co 2 ngo vao la A va B qui dinh. Vi A
va B ca hai deu d mu’c 1 chi khi dem 3, nhu” vay, ngo vao J cua flip flop C se d mu’c
1 chi khi nao dem 3. Do do, khi sirdn am xung nhip lam cho chuyen trang thai cac
flip flop tir dem 3 len dem 4 thi flip flop C se du’Oc xac lap. Tai tat ca cac thdi diem
khac, ngo vao J cua flip flop C deu d mire 0 va flip flop C du’Oc duy tri d trang thai

phuc hoi. Bo dem Mod-5 du’Oc gidi thieu tren Hinh 8.20.

Vca—

C lock'

J1TL

j a

>

K A

J B

-£>

k b

— j c

-C>
— K C

Clock

c ) 3 o dem nhi phan M od-5

A B C

d ) Khoi logic

c B A Bern

o O O O

o O 1 1

o 1 O 2

o 7 / 3

I O o 4

o o o O

Clock

-A

B

C

a ) r a n g e u t h a t i?) Dana xune

</div>
(42)<div class='page_container' data-page=42>

Trong khi xay di/ng bo dem loai nay, can luon luon kiem tra cac trang thai
du’Oc bo qua de dam bao chac chan bo dem khong thao tac sai. Bo dem nay bo
qua cac trang thai 5, 6 va 7 trong khi no thao tac dem chuoi so. Tuy nhien, co the
xay ra trudng hpp bo dem du’Oc xac lap 6 mot trong cac trang thai dirpc bo qua, khi
nguon du’Oc di/a vao bo dem. Can kiem tra thao tac cua bo dem khi b it dau tCr mot
trong ba trang thai cam de dam bao rang bo dem khong thao tac sai va tien hanh
dem chuoi so dem mong muon.

Gia du rang, bo dem dang a trang thai 5 (CBA = 101). Khi xung nhip lien ke
sau chuyen tir mu’c 1 xuong mu’c 0, cac tri/dng hpp sau day cc the xay ra:

1. Vi C = 0, flip flop A phuc hoi. Do do, A chuyen trang thai t i / 1 xuong 0.
2. Khi A chuyen tir 1 xuong 0, flip flop B lat trang thai va B chuyen tir mu’c 0
len mu’c 1.

3.Vi ngo vao J cua flip flop C 6 mu’c 0, flip flop C phuc hoi va C chuyen tCr
mu’c 1 xuong mu’c 0.

4. Do do, bo dem chuyen tCr trang thai cam 5 den trang thai dung qui dinh,

trang thai 2 (C B A = 010) sau mot xung nhip.

Bay gid, gia du rang bo dem bat dau tir trang thai cam 6 (C B A = 110). Sirdn
am xung nhip lien ke sau xuat hien, cac hien ti/dng sau day se xay ra:

1. Vi C d mire 0, flip flop A phuc hoi. Vi A da 6 mu’c 0, no van duy tri mire 0.
2. Vi A khong thay doi, flip flop B khong doi va B van duy tri mu’c 1.

3. Vi ngo vao J cua flip flop C O mu’c 0, flip flop C phuc hoi va C thay doi tir
mu’c 1 xuong mu’c 0.

4. Do do, bo dem nhir vay tien hanh ti/ trang thai cam (so 6) den trang thai
qui dinh 2 sau mot xung nhip.

1 j

e - J i J ^ L j ^ ' i j n j - u n _ m - L n _ r L r L

,n • c < fr 4 S u t*

i i i.' i ^ i ;; i a

</div>
(43)<div class='page_container' data-page=43>

Cuoi cung, gia du rin g bo dem bat dau 6 trang thai cam 7 (CBA = 111). Khi
sudn am xung nhip lien ke sau xuat hien, cac sir kien sau day xay ra.

1. Vi C 6 mu'c 0, flip flop A phuc hoi va A chuyen tti mu’c 1 xuong mire 0.

2. Vi A chuyen tu’ mu’c 1 xuong mu’c 0, flip flop B lat trang thai va B chuyen tis

mire 1 xuong mu’c 0.

3. Ngo vao J cua flip flop C d mu’c 1, do do, flip flop C chuyen trang thai doi
lap tu” mu’c 1 xuong mu’c 0.

4. Nhu vay, bo dem tien hanh tu1 trang thai cam 7 sang trang thai qui dinh 0
sau mot xung nhip.

Khong co mot trong ba trang thai cam nao gay ra cho bo dem thao tac sai.
Bo dem tir dong thoat ra ngoai trang thai cam nao do chi sau mot xung nhip.

Cau true cua bo dem Mod-5 nay co the du’Oc coi nhu la mot khoi logic nhu
trinh bay tren Hinh 8.20d va co the dung cach ghep 5 x 2 hoac 2 x 5 de tao nen
bo dem Mod-10 hoac con goi la bo dem thap phan.

Bo dem Mod-10 dupe cau tao bang cach dung bo dem Mod-5 6 Hinh 8.20 va

ket noi vdi mot flip flop nhir tren Hinh 8.21. Hinh 8.21 cGng ve dang xung va bang
sir that cua bo dem Mod-10. Chu y rang bo dem Mod-10 tren day tien hanh chuoi
dem co 2 va co 5 ma khong dem true tiep du’Oc chudi so nhi phan.

Bo dem thap phan co the dupe tao nen mot cach de dang bang cach ket noi
bo dem Mod-5 6 Hinh 8.20 va mot flip flop, nhirng cau hinh la 2 x 5, nhir tren Hinh
8.22. Bang sir that va dang xung tirong Crng vdi bo dem Mod-10 = 2 x 5 du’Oc gidi
thieu tren Hinh 8.22. Chu y rang, bo dem thap phan nay dem tri/c tiep chudi so nhi
phan, dem tir 0000 len den 1001 va trd ve 0000.

</div>
(44)<div class='page_container' data-page=44>

IC TTL MSI 54/7490A la mot bo dem thap phan. Sd do logic, bang sir that va
chan IC dirpc gi6i thieu tren Hinh 8.23. Quan sat ky ta thay rin g cac flip flop QB,

Qc, Qd tao nen bo dem Mod-5 dung nhir bo dem Mod-5 d Hinh 8.20 Tuy vay,

nhan thay rang flip flop QD 6 IC '90A la mot flip flop RS, flip flop nay co ngo ra Q0
ket noi ngirpc tro lai ng5 vao A, ta co bo dem cd 2 va cd 5, nhir trinh bay d phan
tren. Hay nghien ciru sd do logic va bang sir that cua IC '90A vi IC nay dirpc dung
rong rai trong thuc te. Mot ap dung kha hap dan ve viec sir dung IC 54/7490A lam
bo dem den 999. Ba IC '90A mac noi tiep nhau. IC dau (phia phai) dem cac xung
ngd vao d ngo vao CLK. Ta goi IC nay la bo dem ddn vi.

IC '90A d giCra co birdc dem 1, mdi Ian dem bang 10 xung ddn vi, vi D cua he
dem ddn vi mang sang khoi 6 giCra mdi khi dem tir 9 den 0. Khoi 6 giCra dupe goi la
bo dem hang chuc.

IC '90A d ben trai co birdc dem 1, mdi Ian dem bang 100 xung ddn vi va bang
10 Ian birdc dem hang chuc. Do do, khoi nay dupe goi la bo dem hang tram.

Hoat dong cua bo dem da dirpc giai thich ro. Mach logic cd kha nang dem
cac xung ngd vao tir 1 len den 999. Trudc het, can Reset tat ca cac flip flop trong
IC '90A, sau do, tien hanh dem so xung 6 ngd vac cua bo dem ddn vi. Cach s§p
xep cac IC '90A theo tirng tang nhir vay dude dung rong rai trong cac Von ke
digital, cac bo dem tan so va trong nhieu irng dung can den dem thap phan.

Can chi ra rang IC 54/7490A chi la mot trong so cac IC loai TTL MSI dung
lam bo dem thap phan. Dac biet, IC 54/74176 la bo dem thap phan khong dong bo
kha pho’ bien khac. Ngoai ra con cd cac IC 54/74160, 54/74162, 54/74190 va
54/74192 la cac bo dem thap phan dong bo kha pho’ bien. Mdi loai IC ke tren co
nhung dac diem rieng can nam vung cac dac diem do dirpc gidi thieu trong cac
thong bao ky thuat ve IC ke tren.

</div>
(45)<div class='page_container' data-page=45>

\ L9C LS90

a ) S<? <1d lo -g .c

?.:a ; £•!■ Li'^o

CHw6» il**n' HCO

|Xu”n Qii cf^u Uf

9 t - A , X S 9 9

CfJ •Jem V) “*.VM (5 <?|
I X o n q l i f cf"* . f$)

[ _

N j o i .» N ao r.l

d 0

___ Oc <■># a . G «in 0 . ° 9

£

<>,

0 f T L L 0 L L L L

> i £ H 1 i L L H

2 L H L 2 I L H L

■» l L */ *

3 <■ L H V

- *■ h L L •» t H L i

5 L h W H

5 * L L L

6 L H >y L to_ >/ L L H

7 H / / H 1 M I

8 H t u L 8 L N

0 H I f H g H / / i t

b ) B u n g S t / t l i . i t

IC * 9 0 A 0 0 . ’ I S S O ( n h t n t u - i f i - n x t t o i v i t
r v p u l A ,\*C

j—0> B ^tu.j —|

* j . „ I j

i ~ i i r 1

L! I ’J 'T ’ f l i J l?J !»J L’ i

I^DUl ft fT.. NC

L o g ic d i/tfn rj, x e m C i r i g chifC
C»hi c h u A . Ng<> r j Q A k e t i w i v<ii m jd v i o B

B. Nlji* ra <jD K'-t n o i vt*i n r jo v.it» A

Hinh 8.23

-TLTLR

D e m h a n g tra m B e rn h a n g c h u c D e m d o n v i

</div>
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S N 5 4 1 6 3 , S N 7 4 1 6 3 S y n c h r o n o u s b in a r y c o u n t a r t

S N 5 4 1 6 1 . S N 7 4 1 6 1 S y n c h ro n o u i b in a ry c o u n te rs a ra s im ilar;
ho w e ve r, th e C L E A R is a sy n c h ro n o u s as s h o w n fo r tha
S N 5 4 1 60 . S N 7 4 1 6 0 decade c o u n te rs a t le ft.

J o r N d u a l- in - lin e o r W fla t p a ck a g e ( T o o v ie w !

Outputs

D a ta inputs

P o sitiv e -logic: See d e s c rip tio n .

Hinh 8.25

8.6. BO DEM XAC LAP TR UD C

Ta da nghien cifu hoat dong cua cac bo dem cac chuoi so nhi phan theo kieu
dem tien, dem lui, ta cGng da nghien cu'u hai loai bo dem co kieu dem cach dieu,

</div>
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do la bo dem Mod-3 va Mod-5. Tu1 cac loai bo dem co ban va ket noi phoi hpp cac
bo dem co ban nay ta co the tao nen cac bo dem co mo-dun 2, 3, 4, 5, 6, 7, 8, 9,
10 v.v... Kha nang xay dung nhanh chong va de dang mot bo dem co Mo-dun nhu
mong muon quan trong den mu’c cong nghe ban dan cung cap cho chung ta nhieu
IC TTL MSI de thu’c hien muc dich tren. Bo dem xac lap trudc (Preset) la khoi logic
co ban du’Oc dung de thuc hien bo dem co Mo-dun can co.

Hau nhu tat ca cac bo dem xac lap trudc la loai IC TTL MSI dupe cau tao
bang 4 flip flop (thudng la loai flip flop JK Master-Sslave) va loai bo dem nay
thudng du’Oc goi la bo dem 4 bit, cd the la dong bo hoac khong dong bo. Khi bo
dem tien chudi so nhi phan tii nhien tu” 0000 len den 1111, bo dem do du’Oc goi la
bo dem nhi phan. Vi du, IC 54/74161 va 54/74163, ca hai deu la bo dem nhi phan
dong bo hoat dong theo kieu dem tien. Cac IC 54/74191 va 54/74193 cOng la cac
bo dem nhi phan dong bo, chung hoat dong hoac la theo kieu dem tien hoac la
theo kieu dem lui.

Vi bo dem thap phan rat quan trong, nhieu bo dem 4 bit co ban du’Oc ket noi
ben trong IC de tao ra bo dem cach dieu Mod-10 hoac con goi la bo dem thap
phan. Vi du, IC 54/74160 va 54/74162 la cac bo dem thap phan dong bo theo kieu
dem tien. Cac IC 54/74190 va 54/74192 cung la cac bo dem thap phan dong bo,
nhung cd the dem tien hoac dem lui.

* * i ' 9

Cac bo dem de cap tren day, tat ca deu la mach TTL MSI. O' cac IC nay, ta
chi can quan tam den cac ngo vao, ngo ra va tin hieu dieu khien.

So do chan va so do logic cua bo dem 4 bit dong bo cua IC 54/744163 dupe

gidi thieu tren Hinh 8.25. Nguon cung cap +Vcc va GND (dat) Ian lupt la cac chan

16 va 8. Xung nhip dupe dua den chan 2. Ta can chu y rang cac ngo ra chuyen
trang thai tuong Ung vdi sudn duong xung nhip.

Cac ngo ra cua 4 flip flop la QA, QB, Qc va QD trong khi ngd ra "CO NHO"
(Carry) d chan 15 du’Oc dung de tao nen bo dem cd nhieu tang lien tiep de dem
hang don vi, hang chuc va hang tram...

Hai ngo vao Enable (P d chan 7 va T d chan 10) du’Oc dung de dieu khien bo
dem. Neu mot trong hai chan Enable d mu’c 0, bo dem se ngUng dem, con neu ca
hai ngo vao deu d mu’c 1 thi bo dem hoat dong.

Chan Clear d mu’c 0 se reset tat ca cac flip flop, cac ngo ra cua cac flip flop d
mu’c 0 ngay khi xung nhip tiep sau xuat hien, luc do khong can quan tam den cac
mu’c logic d cac ngo vao Enable. Cach tien hanh nay dpc goi ia reset ddng bo vi
no ddng thdi xay ra doi vdi tat ca cac flip flop luc sudn duong xung nhip xuat hien.
Mat khac, nhan thay rang IC 54/74161 cd chan Clear khong ddng bo, vi no xay ra
ngay lap tUc khi chan Clear chuyen trang thai tu’ 1 xuong 0, khong lien quan den
mu’c cua xung nhip, cua Enable va Load.^

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Doi voi bo dem thirc hien dem tien chuoi so nhi phan, can duy tri cac ngo vao
Enable, ngo vao Load va Clear 6 mu’c 1. Voi dieu kien nay, bo dem se dem tien co

birdc dem la 1 mdi khi sirdn dirong xung nhip xuat hien va dem tu 0000 len den
1111, roi lap lai chudi dem tren. Vi xung nhip tac dong dong bo den tat ca cac flip
flop, cac ngo ra thay doi trang thai cung mot luc va khong co hien tirpng xung tap
nhieu xuat hien. So do trang thai cho tren Hinh 8.26a chi ra chudi dem tu1 nhien, a

do, moi hinh vuong tirong Crng vdi mot birdc dem (hoac trang thai) va mui ten chi ro
bo dem tien hanh dem tir trang thai nay sang trang thai tiep sau nhir the nao.

0
13

| 12 10

Den ngd vao xoa
(chan 1)

a) Sd do trang thai bo dem mod-16 b) Cong giai ma so dem 9

c) SO do trang thai d i/d c cach dieu doi vdi bo dem mod-10

Hinh 8.26

Cd the de dang cach dieu de thay doi do dai so dem bang cach sir dung ngd
vao Clear dong bo. Oieu nay thirc hien mot cach de dang bang cach dung cong
NAND de gia ma so dem Idn nhat can cd, va dung ngd ra cua cong NAND nay de
xoa bo dem mot cach dong bo de tro ve day so nhi phan 0000. Sau do, bo dem
bat dau dem tir 0000 len den so dem Idn nhat can cd, va roi lai xoa de trd ve so
0000. Day la giai phap ky thuat dirpc sir dung de xay dirng mot bo dem cd mo-dun

can cd.

Vi du, can cd bo dem cd so dem Idn nhat la 9, ta noi cac ngd vao cua cong

NAND de giai ma sd dem 9 = DCBA = 1001.

Tir

do, ta cd bo dem Mod-10 vi chuoi

so dem nhi phan

tir

0000 len den 1001. Cong NAND du’Oc dung de giai ma so 9

theo so do trang thai cach dieu chi ra tren Hinh 8.26b va 8.26c. Tir Hinh 8.26b

nhan thay rang can

sir

dung hai cong dao de cd QB va Qc. So do trang thai cach

dieu cd nhung hinh vuong lien net 6 bo dem cach dieu, Mod-10 va cac hinh vuong
dirt net bieu thi cac trang thai du'pc bo qua.

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Bo dem Mod-12 co so dem Idn nhat la 1110 = 10112. Bo dem se dem tCr 0000
len den 1011 roi quay trd lai so 0000. Nhu vay, cac ngo vao cong NAND phai la

Q0, Q

>

Q

-Xac lap cac dang xung voi cac ngo vao Clear, Preset, dem, ngUng hoat dong

(Inhibit) cua cac IC 54/74163 (va 54/74161) du 'O c gidi thieu tren Hinh 8.27. Can

nghien cun ky de hieu sau sac cach dieu khien cac ngo vao nay.

So do logic va xac lap cac dang xung doi voi cac IC 54/74160 va 54/74162
d u ’Oc cho tren Hinh 8.28 (cac chan cua IC nay giong voi chan IC 54/74163 cho

trudc day). Hai bo dem nay d u ’O c cach dieu va chung la cac bo dem thap phan.

Hon nu’a, cac ngo vao, ngo ra va cac day dieu khien cua hai bo dem tren day cung
hoan toan giong vdi cac IC 54/74163 va 54/74161. Cac bo dem nay dem tien moi
khi si/on duong xung nhip xuat hien, tien hanh dem tu” 0000 len den 1001 roi quay
trd lai 0000. So do trang thai chi ra tren Hinh 8.26c. do la so do trang thai cua bo
dem Mod-10 hoac bo dem thap phan.

Mot vi du k h a c , muon x a y di/ng bo dem co Mo-dun 12, ta tien hanh nhu1 sau:

5N54161, SN&4163. SN74161, SN74163 Synchronous binary counters
Typtc*J ckaf, pretet. count. and inhibit iaqu#nc«i

Illustrated below is the following lequence.
1 Clear ootouts to zero.

2. Preset to binary twelve.

3. Count to thirteen, fourreen, fifteen, 2ero, one. and two

4. Inhibit.

Clear (SN54161, SN74161)

Clear (SN54163. SN74163!

Load

A

Data inputs

B

V

U _ r

Clock (SN 54161, SN74161)

Clock (SN54163, SN74163i

L J (Asynchronous'

Outputs

-( S y n c h r o n o u s )

"LJ---m j i J i J ^ n _ r L r i r L n _ r L r L ri i

112 *13 14 15 0

| h --- C o u n

t-Clear Prmet

- I n h i b i t

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IC 54/74193 la bo dem nhi phan dong bo theo kieu dem tien-lui. IC co chan

reset chung va co the reset de dem so can co voi cac ngo vao nap so lieu mac

song song. Ky hieu logic doi vdi IC TTL MSI nay dupe chi ra tren Hinh 8.29a.

Chan PL la ngo vao dieu khien de nap so lieu vao cac chan PA, PB, Pc va PD. Khi
IC dung lam bo dem, 4 chan nay duoc bo ngo va PL phai duy tri 6 mu’c 1. Chan
MR la chan reset chu, binh thudng chan nay dupe duy tri 6 mu’c 0 (mu’c 1 6 chan
MR se reset tat ca cac flip flop).

Cac ngo ra TCuVa TCD dupe dung de dieu khien cac bo kien noi tiep phia

sau. Cac ngo vao xung nhip la CPu va CPD. Dua xung nhip vao CPy, bo dem se
dem tien, con neu dua xung nhip vao chan CPD, bo dem se dem lui. Chu y rang,

xung nhip can dupe noi vdi mot trong hai chan CPy hoac CPD, nhung khong dua

vao hai chan nay cung mot luc. Cac ngo vao khong dung tdi can dupe duy tri 6

mu’c 1. Cac ngo ra cua bo dem la QA, QB, Qc va QD.

So do trang thai chi ra cac trang thai on dinh cua bo dem, dong thdi cung chi
ra qua trinh dem tu” budc nay sang budc tiep theo. So do trang thai cua IC

54/74193 dupe gidi thieu tren Hinh 8.29b. Moi hinh vuong la mot trang thai b’n

dinh, cac mGi ten chi ra chuoi so dem theo ca hai cach dem tien va dem lui. Day la
bo dem 4 bit, cd 16 trang thai o’n dinh, dupe danh so 0, 1, 2, ....15.

C a c b o d e m t h a p p h a n d o n g b o s i r d u n g c a c IC 5 4 /7 4 1 6 0

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> .V 5 4 I * i 0 . .>.‘ « 5 4 1 v > 2 . S N 7 4 1 G 0 . S N 7 4 1 6 2 S y n c n r o n o y i c o w m e n

T y p * C 4 i c l* - * r . c o u n t . a n d i n h i b i t m k j u •*><=••
I l l u s t r a t e d t > « lo w is t h a f o l l o w i n g » # a u t n c » .

1. C. ■c m s o u t p u n t o z a r o .
2 . P r * * « t t o 0 C O w n m .

3 . C o u n t t o ••Q h T , n ir > * , j a r o , o n * , t w o . a n d t h r e « .
4 . I n h i b i t

C l — r( S N 6 4 1 6 0 . S N 7 4 1 6 0 )
C l« » c ( S N 5 4 I 6 2 . S N 7 4 1 0 2 )

1_T

D a t a i n p u t s ■<

-> _ r

B S

C l o c k ( S N 5 4 1 6 0 . S N 7 4 1 6 0 )

C lo c k ( S N 5 4 1 6 2 . S N 7 4 1 G 2 !
ENA9LE P

E N A B L E T

f aA

! '

O u tp u ts <

i _
1QC
C a r r y

( A r y n c / i r o n o u i )

{ S y n c h r o n o u s )
"I__I

---.
T '

17

C l e a r P r e t a t

0 1

C o u n i

-Hinh 8.28

<*) (61

a --- |— y 0 ^ 7Oc°o

To P L

(e )

Hinh 8.29

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dich chuyen vao bo dem, dieu do co nghia la bo dem preset cac so lieu co mat 0
cac chan PD, Pc, PB va PA.

Ta cung co giai phap ky thuat khac de cach dieu bo dem. Dung cong NAND
de phat hien mot trang thai bat ky trong cac trang thai on dinh, vi du, trang thai 15

(1111), va dung ngo ra cua cong nay de dat PL a mire 0. Tai thdi diem nay, bo

dem da preset so lieu co 6 chan PD, Pc. Pb va PA.

0 1 2 3 4

Hinh 8.30

Vi du, gia su1 rang PD Pc PB PA = 100.1 (so 910). Khi xung nhip dupe dua vao,
bo dem thu’c hien dem tuan tu cho den 15 (11112). Tai thdi diem nay, PL se xuong

mu’c 0 va so 9 (1001) se dupe dua vao bo dem. Bo dem se thuc hien dem cac

trang thai 9. 10, 11, 12, 13 va 14 va tai so dem 15 bo dem lai se preset ve so 9.
Chuoi so dem de dang dupe chi ra tren so do trang thai 6 Hinh 8.30, Vi so 15

(1111) ton tai trong mot khoang thdi gian rat ngan de bo dem du dupe preset, nen
cac trang thai on dinh trong vi du tren day la 9. 10, 11, 12, 13 va 14. Do do, day la
bo dem Mod-6. Can chu y rang, giai phap ky thuat nay la khong dong bo, vi preset
hoat dong khong xay ra dong bo vdi xung nhip. Vi ly do nay, can de phong hien
tuong xung tap nhieu phoi hpp vdi cac ngo ra khi preset bo dem.

Ta xem xet mot trudng hpp khac, trudng hpp dem lui khi dung giai phap ky
thuat tren day. Gia du rang, bo dem tren van preset so 1001 (so 9), nhung xung
nhip dupe dua den chan dem lui CPD. Luc do, bo dem dem lui den 15, roi lai preset
den 9 va lap lai qua trinh dem. So do trang thai trong trudng hpp nay dupe chi ra

tren Hinh 8.31. Ro rang bo dem nay cd mo-dun la 10.

Hinh 8.31

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8.7 BO DEM CHUYEN (BO OEM JOHNSON)

6 Chirong 7, ngo ra cua flip flop cuoi cung d bo ghi chuyen dupe ket noi tro
lai ngo vao dieu khien cua flip flop dau tien. Cach ket cau nay cho ta bo dem vong
nhir ve tren Hinh 7.13 va giai phap ky thuat nay dupe goi la phan hoi true tiep. Neu
cac ngo ra cua flip flop cuoi cung dupe dao cheo nhau va ket noi trd ve cac ngo
vao dieu khien cua flip flop dau tien, giai phap ky thuat nay dupe goi la phan hoi
dao. Cach ket noi nay dan den bo dem co nhung tinh chat rat doc dao, bo dem
loai nay dupe goi la bo dem dich hoac la bo dem Johnson.

Trang thai 6 1 2 3 4 5 6 1 2 3 4

Hinh 8.32: Bo ghi chuyen 3 tang dung phan hoi dao de tao nen bo dem chuyen

Ba flip flop JK Master-Slave d Hinh 8.32 dupe ket noi thanh bo ghi chuyen

chuan. Them vao do, cac ngo ra cua flip flop cuoi cung dupe dao cheo nhau_va ket
noi trd ve cac ngo vao cua flip flop dau, ngo ra C ket noi v6i K cua flip flop A va C
dupe ket noi v6i J cua flip flop A.

Gia_du rang, tat ca cac flip flop deu dupe preset va xung nhip b it dau hoat
dong. Vi C d mu’c 1 con C d mire 0, nen mu’c 1 xac lap flip flop A khi xung nhip dau
tien xuat hien. Dong thdi, B va C van duy tri mu’c 0 vi cac ngo vao J cua cac flip
flop nay van con d mu’c 0 va cac ngo vao K cua chung d mu’c 1.

Trong chu ky xung nhip thir hai, A van con duy tri mu’c 1 vi C con_d mu’c 1 va
C con d mu’c 0. Dong thdi B dupe xac lap mu’c 1 vi bay gid A = 1 va A = 0; C van
khong thay doi vi B d mu’c 0 trong chu ky xung nhip nay.

Trong chu ky xung nhip thir ba, A va B van duy tri mu’c 1 va C dupe xac lap
mu’c 1 vi bay gid B d mu’c 1. Do do, sau ba chu ky xung nhip, tat ca 3 flip flop da
dupe thay doi tir trang thai 0 len trang thai 1.

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Trong chu ky xung nhip thir nam, A duy tri mu’c 0, B reset ve 0 (vi bay gid A 6
mu’c 0 va A 6 mu’c 1), va C duy tri mu’c 1.

Chu ky thu1 nam sau dua bo dem trd ve diem xuat phat ban dau, vi C dupe
reset ve 0 trong khi ca hai A va B dang duy tri mu’c 0. Nhir vay, bo ghi chuyen nay
v6i phan hoi dao da thuc hien mot chu ky dem tron ven trong 6 chu ky xung nhip.

Hinh 8.32b cho thay rang dang xung cua moi flip flop la xung vuong co chu
ky gap 6 Ian chu ky xung nhip. Hon nu’a, tat ca cac ngo ra cua 3 flip flop deu co
chu ky giong nhau, chi co dieu khac nhau la cac xung nay dich chuyen xung no
lech vdi xung kia dung mot chu ky xung nhip. Xung vuong nay chuyen dich sang
cac flip flop tiep sau, tien sang mot flip flop moi khi xung nhip chuyen ti/ trang thai

1 xuong trang thai 0. Vi hoat dong cua bo ghi chuyen nay la tuan hoan va cac
dang xung dich sang cac flip flop lien sau roi quay trd lai flip flop dau tien nen cau
hinh nay duoc goi la bo dem chuyen hoac bo dem Johnson. Bang sir that cua bo

dem loai nay dupe gidi thieu tren Hinh 8.33a. De de dang so sanh, Hinh 8.33b chi

ra bang su that nhi phan true tiep doi vdi 3 flip flop. Can chu y rang, bo dem
chuyen 3 flip flop dem 6 trang thai rdi rac. Qua 6 trang thai sap xep nhu vay, bo
dem chuyen dem tuan tu so nhi phan tuong Crng vdi 1-3-7-6-4-0. Nhu vay, sau
trang thai cua bo dem chuyen la 6 trang thai roi rac va co the giai ma dupe.

c B A

T rang
t h a i

Dem nhi phan

t\Jor\c\ diicm g C & A D em

0 O 1 1 1 0 0 0 O

0 1 1 2 3 0 0 1 1

1 1 1 3 7 0 1 0 2

1 1 O 4 6 0 1 1 3

1 0 O 5 4 1 0 0 4

0 0 0 6 O 1 0 1 5

1 1 0 6

0 0 1 1 1

1 1 1 7

0 0 0 O

a) b)

Hinh 8.33: Bang suth at cua bo dem chuyen 3 flip flop

Tuy nhien, can chu y rang bo dem chuyen bo qua cac so nhi phan 2 (010) va
5 (101), do do, phai kiem tra de xac dinh xem bo dem nay co dem cac so theo
dung qui dinh khong, vi co the bo dem chuyen hoat dong sai, dem mot trong hai
so cam tren day tai thdi diem ban dau dua nguon vao he thong. Mot trong hai
trang thai cam nay cung co the xay ra do nhieu hoac do mot vai hoat dong sai sot
khac. Ta hay xem xet cac trudng hpp bo dem thao tac sai nhu sau:

Gia du, luc dau, bo dem dang dem so nhi phan 2 (010). Thdi gian ke sau. xung

nhip chuyen xuong mu'c 0. A len mu’c 1, B chuyen xuong mu’c Ova C len mu’c 1.

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Nhir vay, bo dem se tien den dem so nhi phan 5 (101), day la trang thai cam
thu* hai.

Trong chu ky xung nhip thu1 hai, A xuong mu’c 0, B len mu’c 1 va C xuong
mu’c 0. Nhu vay, d xung nhip thu" hai, bo dem lai trd ve trang thai cam ban dau, so

nhi phan 2 (010). Do do, bo dem dao dong giua hai trang thai cam tren day va
khong hoat dong nhu mot bo dem Mod-6.

De tranh tinh trang nay, can phai dam bao c h ic chan rang bo dem khong the
duy tri mot trong hai trang thai cam ke tren. Mot phuong phap de_thuc hien dieu
tren day la dung cong NAND nhu chi ra tren Hinh 8.34. Khi A, B va C deu 0 mu’c 1
(tuong Crng v6i trang thai cam 010), ngo ra cua cong NAND d mu’c 0. Neu ngo ra
cong NAND dupe noi vdi chan PR (preset) cua flip flop A, thi A se set mu’c 1 moi
khi xay ra dieu kien nay. Nhu vay, bo dem ngay lap tire tien tir so 2 nhi phan (010)
sang so 3 nhi phan (011). Day la trang thai thu" hai trong chuoi dem binh thudng va
bo dem hoat dong dung nhu mong muon.

Vi du, thiet ke bo dem chuyen 4 flip flop. Hay dung so do dang xung de lap
bang suthat va chi ra chudi so dem can co.

A

-
5-c~

Ben PR

cua Flip Flop A

Hinh 8.34: Cong preset doi v6i bo dem tren Hinh 8.32

Lap danh sach cac trang thai cam va kiem tra cac trang thai nay de xac dinh
bo dem co hoat dong sai vi cac trang thai cam thi hay chi ra phuong phap giai
quyet.

Hinh 8.35 la bo dem chuyen va dang xung can thiet ke. Bang su that

dupe

xay dung tir so do dang xung. Vi bo dem dung

4

flip flop nen co

16

trang thai co
the xay ra. Vi bo dem dung

4

flip flop nen co

16

trang thai co the xay ra. Theo kieu
hoat dong cua bo dem can thiet ke, chuoi so dem co 8 trang thai. do do, 8 trang
thai cam phai

dupe

xac dinh. Tam trang thai cam tuong Crng la

2. 4, 5, 6, 9, 10, 11

13.

Cac so nay

dupe

chi ra trong bang tren Hinh 8.36. Hai cot ben phai cua

bang la trang thai mdi va so nhi phan mdi, tuong duong va bo dem se dem cac so
nay sau moi chu ky xung nhip. Vi du, sau mot chu ky xung nhip, bo dem se dem tu”

so nhi phan

2 (0010)

sang so nhi phan

5 (0101).

Bang chi ra rang neu bo dem mot

trang thai nao do trong so cac trang thai cam, thi no se dem chuoi nhi phan 2-5-

11-6-13-10-4-9-2.

Nhu vay, bo dem se sa vao kieu dem khong mong muon va cu”
lap lai mai. Chu y rang, bo dem van la bo chia 8. Tuy nhien, can phai dua bo dem
ve kieu dem mong muon.

Cac dang xung cua chuoi so cam dupe chi ra tren Hinh 8.36b. Mot phuong

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10). Dieu kien nay khong bao gid bang 1 trong chudi so dem cho phep, nen cong

NAND chi ra trong Hinh 8.34 co the dung de hieu chinh bo dem. Tuy nhien, dieu

can_thiet la noi cac ngo ra cua cong NAND chuyen tu* 1 xuong 0 (tire la moi khi
ABC =1) va do do, bo dem se tro ve chuoi so dem mong muon.

J U L

Clock

PR

L - jJ A

-Oy

r —IK A
CLR

T)--r-o>

PR

J p

K &

CLR

T>

PR
J 0

CLR
~Cr~
PR
J P
>
D
CLR
0

: f f (—e—

^_njn_n_run_nji_njnjn_njn_n

A
3
C
D
J

P c P A

T ran g
th a i

S o n h i p h a n 
T arrn n d ilrrn r\

O o O O 7 O

O o O 1 2 7

O o 7 7 3

O 7 7 7 4 7

1 7 7 7 5 15

1 7 7 O 6 14

1 7 O O 7 12

7 O O O & »

O O O O 7 O

b) c)

Hinh 8.35: Bo dem chuyen 4 tang

Chu y rang, n flip flop co the dupe dung de chia xung nhip cho 2n. Dieu do co
nghTa la 2 flip flop chia xung nhip cho 4, 3 flip flop chia xung nhip cho 6 v.v...Nhu
vay, co the xay dung bo dem co Mo-dun c h in bang each chon so thi'ch hpp cac
flip flop va noi chung thanh bo dem chuyen.

Pem
nhi phan
Trang
thai
Trang thai
mdi
Pem

nhi phan
mrri

2 0010 0101 5

4 0100 1001 9

5 0101 1011 11

6 OHO HOI 13

9 1001 0010 2

10 io io 0100 4

11 1011 OHO 6

13 1101 m o 10

a) t>)

Hinh 8.36

Xay dung cac bo dem co Mo-dun khac tir cau hinh bo dem chuyen co ban

that de dang. Cac dang xung tren Hinh 8.32 cho thay rang neu A reset ve 0 sdm

</div>
(57)<div class='page_container' data-page=57>

xung nhip, C cung nhu vay, dieu do co nghTa la A, B va C tat ca deu 6 mu’c 1 chi
trong hai chu ky xung nhip, con 3 chu ky xung nhip cac flip flop nay van duy tri
mu’c 0. Do do, bay gid bo dem chia 5, thay vi chung chia 6 va co 5 trang thai rdi

rac cua bo dem Mod-5.

Hinh 8.37 gidi thieu so do logic, cac dang xung va bang su that cua phiiong
phap thuc hien tren day. Vi can thiet lam cho A chuyen trang thai xuong 0 chi sdm
hon 1 chu ky xung nhip so vdi trudc day, nen ngo vao K cua flip flop A phai xuong
mu’c 0 sdm hon mot chu ky xung nhip. Do do, chi can ket noi ngo ra B vdi ngo vao
K cua flip flop A ma khong ket noi vdi ngo ra C. Nhu vay, bo dem cd mo-dun le bat
ky de dem so le m cd the duoc xay dung tu* bo dem cd mo-dun c h in m+1 bang
cach don gian la ket noi ngo vao K cua flip flop dau tien vdi ngo ra cd mu’c 1 cua
flip flop trudc cuoi.

Trang th a i
xung nhip

A
P
C

J IT L

o
D _
PP
A
K A
CLP
TT

3 4

- o >

. O

PP
J &

CLP

—c r

/ 2

H----

h-_r

pp

dp
*)

c p A Trang th a i

o o o O

o o 1 1

o 1 1 2

1m 1 o 3

1 o 4

o o O

c)
Hinh 8.37: Bo dem chuyen Mod-5

Bo dem nay van con cd hai so dem bi cam, so 2 (010) va so 5 (101). So 7
(111) cGng la so bi cam, nhung so 7 khong gay ra chuyen r lc roi nao vi bo dem se
chuyen mot cach tu nhien tir so 7 (111) sang so 6(110) va day la trang thai thu” tu
trong chudi so dem yeu cau. Hon nua, cd the thay rang, bo dem se chuyen tu” so 2
(010) sang so 5 (101) va sau do sang so 3 (011), so nay la trang thai thir hai trong
chudi so dem mong muon. Nhu vay, bo dem nay khong cd kieu dem bi cam duoc
duy tri mai nhu bo dem dich trudc day.

8.8. BO OEM CHUYEN MQD-10 CO GIAI MA

Bo dem chuyen 5 flip flop cd the dupe xay du’ng nhu tren Hinh 8.38. Day la

</div>
(58)<div class='page_container' data-page=58>

rC

JUL

Clock

Fk

J <4

> r °>

K A
CLR

K
CLR

J C

K C

CLR

— 0—
r—C>

1 ,0 1 2

I i r -4—M - 7 9 1.0T

/C P ‘

_ a ^ _

rc>

-a-CLK

~o~

Trang Si9 nhi phan

Hinh 8.38

Tu” do, ta thay neu bo dem dem 1 trong cac trang thai cam nay, no se thuc
hien mot trong hai tien trinh sau day: thif nhat, bo dem van chia xung nhip cho 10,
nhung no se di den mot trong hai chuoi so cam 2- 5-11-23-14-29-26-20-8-17-2

hoac 4-9-19-6-13-27-22-12-25-18-4; thCr hai, neu bo dem dem mot trong hai so 10

hoac 21, no se dem luan quan hai so nay. Be giai quyet tinh trang tren, lam cho bo

dem lam viec trong chudi so chuan xac, ta dung cong NAND nhu chi ra 6 Hinh

8.34. Ngo ra cua cong dupe ket noi vdi chan PR cua cac flip flop C, D va E.

Bo dem dich thap phan 6 Hinh 8.38 co 10 trang thai roi rac, do do, bo dem
co the duoc giai ma de tao thanh 10 dang xung thdi gian tuong tu nhu bo dem

Mod-8 6 Hinh 8.7. Trong bo dem khong dong bo Mod-8 co giai ma, ta can 8 cong

AND 3 ngo vao vi co 3 flip flop trong bo dem. Trong trudng hop nay, ta dung 10
cong AND 5 ngo vao de giai ma bo dem chuyen Mod-10 nay vi co 5 flip flop. Tuy

vay, xem xet cac dang xung tren Hinh 8.38 cho thay rang co the cau tao mach

don gian hon.

Nhan thay rang, trang thai 1 duoc xac dinh duy nhat khi A 6 mu'c 1 va B 6
mu’c 0. Nhu vay, chi can mot cong AND 2 ngo vao d_e giai ma trang thai 1. Cac ngo
vao cong AND nay la A va B, he thu’c logic la Xt = A B.

Tuong tu, trang thai 2 chi co q_thdi diem B 6 mu’c 1 va C 6 mu’c 0. Cac ngo
vao cua cong de giai ma 2 la B va C, va he thu’c logic chuan xac la x2 = BC. Cac
phuong trinh logic de duy tri trang thai duoc xay dung tuong tu cach lam tren day.
Cac phuong trinh nay cung vdi cac cong giai ma thi'ch hop duoc chi ra tren

Hinh 8.39

Nhu vay, ngoai nhung Uu diem khi sir dung bo dem chuyen, co mot iru diem
nu’a la chi can cong 2 ngo vao de giai ma mot trang thai nao do trong so cac trang
thai cua bo dem.

</div>
(59)<div class='page_container' data-page=59>

3 t > «

3 £ > - "

E >

\ x 7 = P C

-

_

J r -

O

~ ^ \ - X 9 = P E

“ ---^ E '

o

-X 2 = P

-XO = A P

i = CP

X 4

-
reX T

-X9_

XIO

7L

_ n _

J L

_ n _ n

J L

_n_

n _ r r

J L

r r r r

Hinh 8.39

Tren Hinh 8.40a. dang xung dupe dung nhu la mot tin hieu dieu khien. Van

de la can chi ra mach logic de tao ra tin hieu do.

Ta thay rang tin hieu dieu khien can co tuong ung vdi mu’c 1 6 cac trang thai
1, 3 va 5 va o tat ca cac thdi diem khac, no co mu’c 0. Giai phap trudc kia la dung 3

co’ng AND nhu chi ra tren Hinh 8.39 de giai ma. Neu cac ngo ra cua 3 cong nay

duoc dung nhu la ngd vao co’ng OR nhu trong Hinh 8.40b, tin hieu dieu khien x

dupe tao ra 6 ngo ra cua cong OR.

t>) c)

Hinh 8.40

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(60)<div class='page_container' data-page=60>

CAU HOI VA BAI TAP

8.1. Neu i/u diem cua bo dem khong dong bo.

8.2. Giai thi'ch hien tupng co xung tap nhieu trong bo dem khong dong bo.

8.3. Neu cach triet xung tap n h ilu trong bo dem khong dong bo.

8.4. Neu Uu diem cua bo dem dong bo.

8.5. Neu nhung Uu va nhupc diem cua bo dem phoi hpp dem khong dong bo

va dem dong bo.

8.6. Mo ta diem khac biet giua cac tin hieu ngo ra cua bo dem dong bo va bo
dem khong dong bo.

_ 8.7. 6 bo dem lui dau vao flip flop thu1 hai va thu” ba xuat phat tu1 dau nao cua
Q, Q?

8.8. Bo dem nao co the doi hudng dem?

89. Bo dem Mod-10 thudng dupe goi la bo dem gi?

8.10. Cum tu1 hoan chinh cua BCD la gi?

8.11. Trong bo dem nao tat ca cac dau ra cung thay doi dong thdi?

8.12. Flip flop cuoii cung cua bo dem tien 4 bit lat 6 Ian dem, sau khi ba flip
flop trudc co 6 ngo ra cac mu’c nao trong cac mu’c sau: 000, 101, 111, 110?

8.13*. Tan so cua xung nhip la bao nhieu neu chu ky cua B tren Hinh 8.1 la
1000 ns?

8.14. Xem xet neu ve cac dang xung doi vdi bo dem khong dong bo co 10 flip
flop thi co nhung kho khan chu yeu nao?

8.15*. Tan so xung nhip tren Hinh 8.1 la bao nhieu, neu chu ky dang xung tai
C la 24 ps?

8.16*. Can co bao nhieu flip flop de xay dung bo dem Mod-1287 So thap
phan Idn nhat co the lUu tru trong bo dem Mod-64 la bao nhieu?

8.17*. Ve chi'nh xac cac dang xung ngd ra doi vdi IC 74L93 dupe ket noi de
tao nen bo dem Mod-16.

8.18*. Ve cac cong can thiet de giai ma 16 trang thai cua IC 74L93 hoat dong

nnu tren Hinh 8.3.

8.19*. Sir dung cac dang xung co tren Hinh 8.9 va nghien cuti cac cong giai

ma nhu tren Hinh 8.7. Hay chi ra cac xung tap nhieu se xuat hien b£ng cach ve

cac dang xung dupe giai ma cua cac co’ ng.

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8.20*. Xac dinh so cac flip flop can thiet de xay dung cac bo dem sau day:

a) Mod-6

b) Mod-11

c) Mod-15

d) Mod-19

0) Mod-31

8.21*. Ve cac cong giai ma va tat ca cac dang sang doi vdi bo dem Mod-6

nhu tren Hinh 8.17

8.22*. Ve so do logic, lap bang su that va ve cac dang xung doi v6i bo dem

Mod-9 khi dung hai bo dem Mod-3 ket noi noi tiep.

8.23*. Ve cac cong giai ma va tat ca cac dang xung doi v6i bo dem nhu tren

Hinh 8.22.

8.24*. Co bao nhieu bo flip flop de xay du’ng cac bo dem co cac Mod dem
sau:

a) 5
b) 8
c) 9
d) 10
e) 21

8.25*. Ve mot bo dem chuyen gom 2 flip flop va ve dang xung cua bo dem
do. Lap bang suthat va kiem tra cac trang thai bat hop phap.

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(62)<div class='page_container' data-page=62>

CHUONG 9

BO NHO BAN DAN

Hoc xong chuong nay, hoc vien co kha nang:

□ Hieu duoc nguyen ly cau tao, cac tinh nang
co ban cua cac loai bo nhO ban din:ROM,
PROM, EPROM, RAM va DRAM.

□ Hieu duoc nguyen ly cau tao cac loai te bao
nho ban dan.

□ Phan tich duoc cac hoat dong cua cac loai
bo nhd ROM va RAM.

□ Thuc hien d'UOc cach md rong dung luong
bo nhO ban d§n.

□ ifng dung cac tinh nang cua cac loai bo
nho ban dan trong cac mach dien tit

TONG QUAN

Bo nhd ban dan thay the cac loai bo nho bang vat lieu tir. Cac tien bo m6i
cua cong nghe ban dan thdi gian gan day da cung cap nhieu mach nho loai MSI
va LSI co do tin cay cao va gia thanh ha. Vao dau nhung nam 60 cua the ky trudc,
gia thuong pham mot bit nhd vao khoang 2 USD (tuong duong 34000 VND). Den
nay, nhung nam dau cua the ky 21, gia thuong pham 128 Mega Byte (1024 000
x128 x 8 bit) vao khoang 20 USD (tuong duong 340000 VND). Nhu vay, gia
thuong pham cua mot bit nhd sau khoang 40 nam da giam di vao khoang 105.10‘
Ian. Bo nhd ban dan dien hinh co cac te bao nhd sap xep theo hinh chu nhat, gan
trong khoi hop nho bang nhua dang DIP (Dual in line package). Te bao nhd co ban
la mot mach flip flop tranzito hoac mach cd kha nang tich tru dien tich va duoc
dung de lu’u tru' mot.bit tin. Cac bo nhd thu'dng du'pc phan ra thanh cac loai bo nhd

</div>
(63)<div class='page_container' data-page=63>

Ii/C in g circ, MOS hoac CMOS tuy theo tranzito dung de cau tao cac te bao nhd.
Tong so cac te bao nhd trong mot bo nhd chinh la dung lupng cua bo nhd. Vi du,
mot chip nhd ludng cuc 1024 la mot bo nhd ban dan cd 1024 te bao nhd, moi te
bao gom mot flip flop do cac tranzito ludng c u t tao nen. Chip la mot thuat ngu' hay
dirpc dung hon la thuat ngu thiet bi nhd ban dan.

Noi chung, cac bo nhd MOS va CMOS tieu thu nang lirpng it hon, gia thanh
ha hon va kich thudc nho hon bo nhd ludng cuc, nhung bo nhd ludng cUc hoat
ddng nhanh hon. Mot cach tong quat, cac loai bo nhd cd the dupe chia ra thanh
cac loai bo nhd ROM va RAM.

RAM la loai bo nhd cd the nap cac tin vao hoac lay cac tin tu” bo nhd ra. Vi
mdi te bao la mot flip flop nen khi khong cd nguon cung cap, tin trong bo nhd cung
bi xoa, RAM cd loai te bao nhd nhu vay dupe goi la bo nhd xoa ngay.

ROM la loai bo nhd chi cd the lay (doc) so lieu ra. Vi du, ROM luu tru cac gia
tri cua ham luong giac, ham logarit thap phan, logarit tu nhien hoac cac chuong
trinh khong thay doi... de tim can bac hai cua mot so. Vi cac so lieu dupe lUu tru
vTnh vien d mdi te bao nen khi mat nguon cGng khong lam mat so lieu va nhu vay,
ROM la loai bo nhd khong xoa. Viec sir dung PROM cung giong nhu ROM, nhung
cd dieu khac vdi ROM la cac so lieu dupe ngUdi sir dung tu nap vao PROM.

EPROM la loai bo nhd so lieu cd the thay the trong tirng thdi gian nhd sir
dung loai PROM cd kha nang xoa duoc (EPROM). So lieu cd the dupe lap trinh
vao EPROM, sau do lai xoa di va lap trinh lai neu can.

Dudi day trinh bay cac loai bo nhd ban dan va cac dac diem hoat dong cua
chung. Cac bo nhd ban dan cd dung luong nho dupe thao luan trudc vi no cd cau
hinh don gian va de hieu. Cac cau true co ban nay dupe van dung vao cac bo nhd
cd dung lupng Idn hon.

9.1. OjA CHI HOA BO NHO BAN DAN

Lap dia chi la viec tien hanh lira chon mot trong cac te bao cua bo nhd de
nap tin vao hoac lay tin ra. De lua chon thuan tien, cac bo nhd thudng sap xep cac
vi tri te bao theo hang va theo cot nhu Hinh 9.1a. Trong trudng hpp dac biet nay,

cd m hang va n cot va bo nhd nay cd n x m te bao.

n c o t c o t 5 Te ha r nha A 3

Ft

1 ' /

m hang
n
C- %

I

1 n x m te bao

hang A

... I

1
1
1

a)

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(64)<div class='page_container' data-page=64>

Mach dieu khien phu hap v6i sU sap xep co ban cua bo nhd dupe thiet ke
sao cho chi co mot hang va chi co mot cot hoat dong va te bao nh6 6 vi tri giao

nhau cua cac hang va cot nay dupe chon. Vi du, 6 Hinh 9.1b, neu hang A va cot B

duoc kich hoat, thi te bao nh6 6 vi tri giao nhau cua hang va cot nay dupe chon,
tuc la tin co the dupe doc ra tu” te bao hoac viet vao te bao nay. Oe tien Ipi, te bao
duoc goi la te bao AB tuong ting vdi hang va cot dupe chon. Cach dia chi hoa nay
duoc goi la "dia chi" cua te bao. Su kich hoat cua mot hang hoac cot du’Oc thuc
hien bang cach dat mu’c logic 1 (hoac co the dat mu’c logic 0) len hang va cot do.
Van de nay dupe de cap den 6 phan sau. Bo nh6 co nhieu cau hinh khac nhau.
Nhung cau hinh cua bo nhd co 16 te bao sap xep thanh hinh chu nhat dupe trinh
bay 6 Hinh 9.2. Cac bo nhd 16 x 1 va 1 x 16 ve thuc chat tuong duong nhau, cGng
nhu vay, bo nhd 2 x 8 ve co ban tuong duong vdi vdi bo nhd 8 x 2. Do do, trong

nam bo nhd trinh bay tren Hinh 9.2 chi ro 3 loai cau hinh khac nhau, chung cung

chira so te bao nhu nhau.

^ Cl? t 16 c o t

t>) c )

Hinh 9.2

6 Hinh 9.2a viec chon ra mot te bao van doi hoi phai cd mot hang va mot
cot, dupe goi la dia chi duy nhat, 6 cau hinh nay 17 day dia chi phai dung tdi 16
hang va 1 cot hoac 1 hang va 16 cot. Mot trong hai trudng hpp nay, yeu cau toi
thieu that ra chi la 16 day. Tuy nhien, mot trong hai cach sap xep nhu o Hinh 9.2 b

chi yeu cau cd 10 day dia chi: 8 hang 2 cot hoac 2 hang 8 cot. Ro rang la su bo tri
tot nhat van la cau hinh tren Hinh 9.2c, vi no chi can cd 8 day dia chi: 4 hang va 4

cot.

Noi chung, su sap xep cd it day dia chi nhat v in la sap xep hinh vuong vdi n
hang va n cot va dung lupng bo nhd la n x n = n2 te bao. Vdi ly do nay, cau hinh
hinh vuong dupe dung rong rai trong cong nghiep che tao bo nhd ban dan. Su sap
xep n hang n cot thudng cd quan he vdi dia chi hoa dang ma tran. Ngupc lai, cau
hinh 1 cot n hang thudng dupe goi la dia chi hoa dudng th in g , vi viec lua chon ra

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(65)<div class='page_container' data-page=65>

mot te bao nh6 chinh la sir lira chon hang tirong Crng, con cot luon luon dupe
dung tdi.

Vi du, IC 54/747481A la mot RAM ludng cuc 16 bit, dupe sap xep theo cau
hjnh 4 x 4 nhu trinh bay tren Hinh 9.3. Co 4 day dia chi X (cac hang X,, X2, X3,
X4) va 4 day dia chi Y (cac cot Y 1( Y2, Y3, Y4). Viec lira chon mot te bao dupe thuc
hien bang cach dat mu’c logic 1 vao mot day X va vao mot day Y. Vi du, X, = 1 va
Y4 = 1, se lua chon te bao a hang duoi cung va cot ngoai cung ben phai (tat ca cac
day dia chi khac phai dat 6 mire logic 0). Dia chi cua te bao nay la X1Y4.

Hay lay bo nhd 4 x 4 tren Hinh 9.2c lam vi du. Be chon dupe mot te bao nhd
duy nhat ta chi cho mot hang va mot cot hoat dong. Nhu vay, ta phai dung bo giai
ma 1 cua 4 so nhi phan (Hinh 9.4). Vi du, lua chon te bao nhd a dia chi 43 (hang 4
cot 3) . Neu A4 = 1 va A3 = 1 bo giai ma se duy tri mu’c logic 1 cho hang thir tu,
trong khi cac hang khac 6 mu’c 0. Tuong tu, neu A2 = 1 va A, = 0, bo giai ma se
duy tri mu’c logic 1 cho cot thu” ba va tat ca cac cot khac 6 mu’c 0. Nhu vay, mot
ngo vao A4A3A2A., = 1110 se lira chon te bao nhd 43. Ta xem A4A3 la dia chi hang 2
bit va A?A, la dia chi cot 2 bit. Do do, mot te bao nhd dupe dac trung bang dia chi 4
bit A4A3A2A.|.

J o ’ A- - i i n c O '

'• A.*SN /
Smje
A<Jdi*»K W i l * , ---

-*•---: * C

i ■—

i o c i Se? ’C

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(66)<div class='page_container' data-page=66>

Mot vi du khac, dia chi AaA3A2A^ = 0 1 1 0 , lira chon te bao co hang la 2 va cot
la 3 (dia chi 23).

Cac bo giai ma dia chi trong Hinh 9.4 giam so day dia chi can thiet de dia chi
hoa mot te bao nh6 duy nhat. Cac bo giai ma nay co chira trong cung bo nho. Mot
bo giai ma nhi-thap phan co n ngo vao nhi phan se lira chon mot trong 2n day ra.
Vi du, mot bo giai ma co 3 ngo vao nhi phan se chon dupe 23 = 8 ngo ra, hoac bo
giai ma co 4 ngo vao se co 16 ngo ra.

Noi chung, mot dia chi B bit co the dupe dung cho bo nhd cd 2° te bao, trong

do, cd B/2 bit dung cho cac hang va B/2 bit dung cho cac cot (Hinh 9.5). Chu y

rang tong cac bit dia chi phai la so nguyen va chan (2, 4, 6, ...) vi ngd vao mdi bo
giai ma la B/2 bit, nen ngo ra mdi bo giai ma phai la B/2 day. Nhu vay, dung lirpng
bo nhd la:

2

8/2 n 2b

Vi du, mdi dia chi 12 bit cd the dung de tao nen bo nhd cd dung lupng la 2 12

= 4 0 9 6 bit theo cach thuc hien tren day.

Cac bo nhd thuong pham thudng cd dung lupng la 1024, 2048, 4096,
16384...Tat ca cac so nay deu la luy thCra 2 vdi so mu la so nguyen. Cung can noi
them la mot bo nhd cd dung lupng la 210 = 1024 bit dupe goi la 1 Kbit (1K), tuong
tu, cd bo nhd 16K(16384 bit), bo nhd 4K (4096 bit)....

Hinh 9.5

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(67)<div class='page_container' data-page=67>

Nhu" tren da trinh bay, ta mdi de cap den cac bo nh6 du’Oc truy cap den mot
te bao don. Viec truy cap den nhom bit - dac biet la nhom 4 bit (Nibble) va nhom 8
bit (Byte) la thuan Ipi hon ca. It nhat la ta co hai phuong phap thudng dung.
Phuong phap dau la truy cap nhom cua cac te bao tren cung mot bo nhd. Phuong
phap thu1 hai la ket noi song song cac bo nhd.

Hinh 9.6 gidi thieu so do logic 64 bit (16x4) cua bo nhd ludng cUc. Cd 16
hang te bao nhd va moi hang 4 te bao, nhu vay la 16x4 te bao, mdi te bao la mot
flip flop tranzito. Bo giai ma dia chi cd 4 bit dia chi va nhu vay, cd 16 day chon, mdi
day chon mot hang. Trong trudng hop nay, mdi mot day chon dupe noi vdi tat ca 4
te bao trong mot hang, nhu rhe, mdi day chon se lay ra 4 te bao cung mot luc. Do
do, moi day chon se lay ra mot tu” 4 bit (Nibble), thuan tien hon la lay ra tting te
bao don.

Cd the cho rang su sap xep nay giong nhu mot "ngan xep" cd 16 bo ghi
chuyen 4 bit. Day chinh la dang dia chi dudng thSng va 4 bit dia chi khi

dupe

giai

ma se chon ra mot trong 16 bo ghi chuyen 4 bit. Trong trudng hop nao do, khi so
lieu

dupe

doc tu” bo nhd nay, no xuat hien

6

4 day cua ngd ra so lieu: D1f D2, D3, D4
va

dupe

xem nhu mot tu” so lieu 4 bit. Tuong tu, so lieu

dupe

dua den bo nhd

de

nap vao

dupe

xem nhu la mot

tCr

so lieu 4 bit

0

cac day ngd vao so lieu: l^ l2. I3, l4.

Cac IC 54/74S89 va 54/74S189 deu la cac bo nhd ludng cue,

dupe

sap xep dung

nhu cau hinh

dupe

ve tren Hinh 9.14a.

Hinh 9.6: Bo nhd 64 bit (16x4)

9.2. ROM, PROM VA EPROM

- ROM sur dung cac diot

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Ngo vao G

dupe

dung de cho tat ca 32 cong giai ma ngo vao

dupe

ki'ch hoat
hay khong di/cfc kich hoat. Khi G 6 mu’c 1, tat ca cac cong giai ma dia chi bi ehcin
lai va bo nh6 khong hoat dong, do do, dat 8 day bit so lieu ngo ra d mu’c 1. Khi G d
mtic 0, so lieu 6 cac ngo ra se tirong Crng v6i tir 8 bit trong bo nhd va dupe chon
bang dia chi ngo vao. Tren hau het cac bo nhd co mot day d ngo vao thirc hien
cung chut nang nhir chan G, no thudng dupe chi thi la day chip dupe kich hoat
(CHIP ENABLE) hoac day lira chon chip (CHIP SELECT).

Cac mu’c logic 1 va 0 co the

dupe

liru trir trong 8 bit cua moi tir trong so 32 tir.
Cac mu’c logic nay phai du’Oc xac dinh trudc khi mua bo nhd va cac nha san xuat
phai cong bo trudc cac thong so ky thuat. Cac nha san xuat lap trinh cho bo nhd

trudc khi phan phoi cho khach hang. Mot khi bo nhd

dupe

lap trinh, noi dung cua

bo nhd khong the thay doi

dupe

nu’a. Cac nha san xuat deu phai ghi ro noi dung

cac thong so ky thuat cua cac bo nhd trong cac phieu ky thuat hoac thong bao
thong so ky thuat cua cac san pham. Phan lap trinh cho bo nhd la do nha san xuat
thu’c hien.

Viec sir dung ROM 54/7488A tuong doi don gian. Trirdc het, vi cac mach
logic la mach TTL nen IC phai dupe noi vdi nguon cung cap va tiep dat. Cac thong
so ky thuat cua ROM chi ro dien ap cung cap, binh thirdng Vcc = + 5V 6 chan 16
va ROM noi dat d chan 8. Cac ngo vao va ngo ra la cac chan con lai du’Oc chi
trong so’ tay chan linh kien (Hinh 9.9b).

Tam ngo ra so lieu la cac tranzito hd mach Colecto nen phai cd dien trd keo
len Rpu d moi ngo ra so lieu tir chan 1 den chan 7 va chan 9. Cu the la gia tri dien
trd keo len khoang tir 5100 den 5k£21 noi tir moi ngo ra den nguon cung cap +
Vcc.

Bay gid, yeu cau dat ra la dua dia chi ngo vao chi'nh xac de doc mot tir 8 bit
va dat day ngo vao G (day chon) d mu’c 0. Bang thong so ky thuat cua IC
54/7488A cho biet la thdi gian truy cap tp = 25 ns va circ dai la 45 ns.

Nhir vay, mot

tir

so lieu 8 bit se du’Oc lay ra d cac ngo ra Y 1f Y 2....Y8 trong

khoang 45 ns sau khi sudn sau cua xung G xuat hien, nhu chi ra tren Hinh 9.10a.

Cac day dia chi can phai on dinh trong thdi gian so lieu dang du’Oc doc ra cua bo

nhd. Hai dudng bieu dien nhu trong Hinh 9.10a la so lieu ra, dudng bieu di§n do

cd the chuyen

tir

mu’c logic 0 len mu’c logic 1 hoac tir mu’c logic 1 xuong

mire

logic

0. Trong Hinh 9.10b cach the hien dang xung don tirong duong vdi cach the hien

cac dang xung don d ngo ra tren Hinh 9.10a.

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(69)<div class='page_container' data-page=69>

1 .i« 1 SN > ằ ằ * * ô'ã% ô -! ãằ ô

Or /V Ooô'- .rv-i.n^ c>r >V f !* | p K K ^ i*

enable B i n a r y « > e c r

O u r p u t

fia] r a n r i n y i nTi rrn r^n m

C u t o u t s

P o s itiv e <ogic. S e « O « c r i o t i o n .

ib)

</div>
(70)<div class='page_container' data-page=70>

BANG CHON TU

T i r CAC NGO VAO

E D C B A*

0 L L L L L

1 L L L L H

2 L L L H L

3 L L L H H

4 L L H L L

5 L L H L H

6 L L H H L

7 L L H H H

8 L H L L L

9 L H L L H

10 L H L H L

11 L H L H H

12 L H H L L

13 L H H L H

14 L H H H L

15 L H H H H

16 H L L L L

17 H L L L H

18 H L L H L

19 H L L H H

20 H L H L L

21 H L H L H

22 H L H H L

23 H L H H H

24 H H L L L

25 H H L L H

26 H H L H L

27 H H L H H

28 H H H L L

29 H H H L H

30 H H H H L

31 H H H H H

Hinh 9.7c

— Chanomq---- ——

A O O H E S S

■ f . ■ I i i i i

SE L S C T
! m e G

I I

O^J !D u t
(ht<jr> to io w t

O u i o u t
! ! o w t o n iq f r l

i
i

O u t p u t

0---i a)

Hinh 9.10: Thdi gian truy nhap, tp cua ROM 54/7488A

So do logic doi v6i IC 54/74186 du’Oc chi ra tren Hinh 9.11. Chip nay la loai

IC TTL-LSI va no la PROM 512 bit, du’Oc cau tao gom 64 tir 8 bit. 6 bit dia chi
ABCDEF dupe giai ma de chon dupe mot trong so 64 tir, cac tir 8 bit trong so do
dia chi dirong th in g tuong tu so do dia chi cua IC 54/7488A. Y nghTa duy nhat cua
bo nho nay la ngUdi sir dung dupe quyen lap trinh cho bo nhd.

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(71)<div class='page_container' data-page=71>

Mot tCr 8 bit dupe nap day du co dang 0000 0000. Lap trinh cho bo nhd cd
nghia la xac dinh cac noi dung yeu cau cho moi tti trong bo nhd roi nap mu’c logic
1 vao cac vi tri can cd mu’c logic 1. Vi du, neu noi dung yeu cau cua tii thir 27 la
0101 1100, cac vi tri bit 3, 4, 5 va 7 se Ian lupt thay doi tU mu’c logic 0 vTnh cifu
sang mu’c logic 1 virih cCru. Mot khi lap trinh xong cac mu’c logic 1 hoac 0 duoc nap
vao PROM trd thanh vTnh ciru.

Ve co ban, viec lap trinh dupe thu’c hien bang cach dung mot xung ddng dua

den moi ngd ra, d do mot mu’c logic 1 phai xuat hien. Nhu vay, a Hinh 9.11b, bo

nhd dupe lap trinh bang cach:

1. Oat dia chi chi'nh xac (ABCDEF) cho mot tU de dupe lap trinh. Vi du, doi

vdi tu’ thu’ 27, dia chi la 00011011. Mu’c 1 la hd mach, mu’c 0 la ddng mach.

* G N D 2 D a ta o u t p u t s G N D 2

A D D R E S S

--- --- 1 G N I .•

! J G N D :

o '

---V D C _ J _

Hinh 9.11: a) So do logic va chan IC b) Lap trinh doi vdi 54/74186

2. Dat mot xung ddng den mdi bit can phai nhd mu’c 1. Be nhd noi dung 0101
1100, yeu cau cac xung dong, mdi xung tai mot thdi diem phai dupe dat vao cac vi
tri sd lieu Y3, Y4i Y5 va Y7.

3. Lap lai cac budc tren cho tat ca cac tti dupe lull trti trong bo nhd.

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(72)<div class='page_container' data-page=72>

Ve cd ban, IC 54/74186 lap trinh giong nhir IC 54/7488A. NhUng IC 54/74186
co hai day MEMORY/ENABLE, mot trong hai day nay neu d mu’c 0 s§ can trd viec
nap trinh, neu ca hai deu d mu’c 1, be nh6 mdi hdat dpng di/dc. Theo thong so ky
thuat, thdi gian truy cap la 50 ns, thi'ch ting vdi moi bo nhd TTL.

Mot nhifdc cua PROM la khi PROM dUdc lap trinh, cac noi dung duoc nap
vao bo nhd mot cach vTnh vien, khong the thay doi duoc; loi khi lap trinh khong the
stia lai cho dung dUdc. EPROM khac phuc dUdc nhi/dc diem nay.

EPROM co sd do cau true va dia chi tUdng tti nhu PROM nhi/ng EPROM
difdc cau tao bang cac tranzito MOS. Nhieu EPROM dung linh kien MOS deu
tuong hop vdi bo nhd TTL va ngay ca ky thuat lap trinh cho EPROM cung giong
nhu ky thuat lap trinh cho chip luong cue. Oieu khac duy nhat cua EPROM la cd
che nap vTnh vien cac mtic Icgic 0 va 1 vao te bao nhd.

Xung dong du’Oc dung de nap mtic Icgic 1 khi lap trinh chc PROM ludng cue
chi'nh la dung de dc't chay day noi trong bo nhd. Ky thuat nay cung du’Oc dung de
lap trinh cho EPROM loai MOS, nhung xung dong bay gid du’Oc dua vao thudng la

vai milisec (ms) de nap dien ti'ch co dinh vao te bao nhd dac biet nay. Dien ti'ch
duoc nap lam cho te bao luu trti mtic logic 1.

Oieu hap d in 6 day la co the lam phong dien va te bao nhd lai tro ve mtic 0.
Hon the ntia, viec lap trinh lai dupe lap lai nhu" qua trinh thu’c hien luc ban dau. Cac
te bao nhd dUdc "xda" bang cach chieu met tia cuc tim qua c ctia so’ bang thach
anh len mat cua chip. Tia cuc tim lam mat cac dien ti'ch da du’Oc nap va tat ca cac
te bac nhd trd lai mtic logic 0. Cac yeu cau ve lap trinh va xoa lenh da nap rat
khac nhau doi vdi tting loai EPROM. Can tham khao cac thong so ky thuat cua cac
nha san xuat EPROM.

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Hinh 9.12: EPROM Intel 8708,dung lUdng 8192 bit

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(73)<div class='page_container' data-page=73>

EPROM loai MOS du’Oc dung rong rai la loai Intel 8708. Bo nho nay la PROM
co the xoa duoc,' co dung lupng la 8192 bit (8192 bit dupe sap xep thanh 1024 tif,
mdi tu* 8 bit). Hinh 9.12 cho thay bo nhd co 8192 te bao nhd sap xep thanh 128
hang 64 cot. Bo giai ma X dung 7 bit dia chi de chon mot trong so 128 hang, trong
khi bo giai ma Y dung 3 bit dia chi de chon mot trong 8 nhom cua 8 bit tu’ 64 cot. 8
bit so lieu duoc dung d cac ngo ra qua cac bo khuech dai dem 3 trang thai.

t ACC: thdi gian truy cap la 450 ns va day la thdi gian tre de dia chi den du’Oc
ngo ra; tACC dupe chi ra tren hinh ve cac dang xung. Thdi gian tre Idn nhat cua
CHIP SELECT den ngo ra, ky hieu la t ^ cung duoc chi ra tren hinh ve, gia tri Idn
nhat cua t^o la 120 ns.

Nhu vay, de doc mot tti so lieu 6 chip, cac day dia chi phai o’n dinh va sau
khoang thdi gian 120 ns, day CHIP SELECT (CS) ve mtic 0, so lieu se xuat hien 6
cac ngd ra. Nhupc diem cua bo nhd nay la ddi hoi cung cap nguon + 12 VDC va
ca hai nguon + 5V va - 5V. Cac EPROM Intel 2716, 2732, 2764 khac phuc dupe
nhupc diem nay vi chi can cung cap nguon + 5 VDC.

INTEL 2716 la EPROM cd dung lirpng 16K (2K x 8). Bo nhd nay cd 2048 tti,
moi tti 8 bit, dung lupng nhd tong cong la 16384 bit, no hoan toan thich hpp vdi
TTL va cd thdi gian truy cap t ACC nho hon 450 ns. 11 bit dia chi se chon mot trong
2048 tti 8_bit (211 = 2048) va tti dupe chon se xuat hien d cac day ngo ra so lieu vdi

dieu kien CE va OE (OUTPUT ENABLE) d mtic 0. Hinh 9.11 gidi thieu so do logic

va chan cua EPROM 2716.

INTEL 2732 la EPROM 32 K(4K x 8), cac chan IC 2732 giong nhu chan IC
2716, nhung dung lupng bo nhd gap hai Ian.

Tuong tu, EPROM 2764 la EPROM cd dung li/png 64k (8k x 8).

Thu’c chat, so do logic a Hinh 9.13a khong chi rieng cua IC 2716 ma chung

cho moi ROM, PROM hoac EPROM. Ve co ban, no chi khac d tong so dia chi ngo
vao de chtia di/ng du so bit trong ma tran te bao nhd.

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</div>
(74)<div class='page_container' data-page=74>

PROM AMD (Advanced Micro Devices) la EPROM loai CMOS cung co sd d6
logic tUOng tu” voi 2716. EPROM nay co ky hieu AM 27512, no co 16 day ngo vao
dia chi va ma tran te bao nhd co dung lupng 524288 bit; dupe cau tao gom 65536
tir, m6i ti/ 8 bit. EPROM nay co hon nira trieu tin dupe liru truf trdng khoi hinh hop
co 28 chan, v6i be dai nho hon 4 cm, be rong khoang 1,25 cm. Dieu do noi len sire
chira that la tuyet dieu cua EPROM AM 27512 loai CMOS.

9.3. RAM

Dieu khac nhau co ban giCra RAM va ROM la d chd sc lieu co the

dupe

nap

vao RAM 6 dia chi mong muon. Le duong nhien, so lieu co the du’Oc dcc ra tir dia
chi nac d6 cua ca RAM lln ROM, cac chu ky dia chi hoa va chu ky doc doi v6i ca
RAM lln ROM la nhu nhau. Cac dac ti'nh cua RAM tinh thuoc ca hai loai MOS va

ludng cuc se

dupe

de cap den trong phan nay.

RAM tTnh dung mach flip flop lam te bao nhd co ban (hoac loai ludng cuc,
hoac loai MOS) va mot khi mot bit dupe nap vao flip flop, bit do du’Oc duy tri d flip
fldp chi/ng nao con nguon cung cap cho chip, nen co thuat ngCr "tTnh". Mat khac, te
bao nhd co ban 6 mot RAM "dong" dung dien ti'ch dupe nap co lien quan vdi loai
linh kien MOS de nhd mot bit tin. Vi dien ti'ch du’Oc nap nay se khong duy tri dupe
lau, dinh ky no dupe lam mdi lai, lam tuoi lai, vi vay co thuat ngu RAM "dong". Ca
hai loai RAM "tTnh" va RAM "dong" deu thuoc loai bo nhd xoa ngay, vi mot khi mat
nguon cung cap se mat luon ca sd lieu da du’Oc luu tru.

IC 7489 gidi thieu 6 Hinh 9.14 la RAM 64 bit loai TTL LSI sap xep theo 16 tir,
moi tir 4 bit. Khi giu ngo vao MEMORY ENABLE (ME) 6 mu’c logic 0, se lam cho
chip co kha nang hoat dong hoac doc ra hoac nap so lieu vao bo nhd va 4 day dia
chi so lieu se chon mot trong so 16 vi tri cua tir 4 bit de doc ra hoac nap vao. Roi
sau do, neu WRITE ENABLE (WE) du’Oc duy tri d mu’c 0, sU co mat 4 bit d cac ngo
vao so lieu (D,, D2, D3, D4) se dupe nap vao dia chi dupe lua chon. Ngirpc lai, neu
WE d mu’c 1, so lieu dang dupe nap d dia chi nhd se dupe dua den 4 day ngo ra so
lieu (Sn, S2, S3, S4). Phai noi them rang, cac ngo ra la cac tranzito hd mach
Colecto, nen can phai cd dien trd keo len Rpu cho moi ngd ra sd lieu. Dien trd Rpu
ket noi ngd ra vdi + Vcc. Bang trong Hinh 9.14c tom tat cac hoat dong cua loai bo
nhd nay.

</div>
(75)<div class='page_container' data-page=75>

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(76)<div class='page_container' data-page=76>

Hinh 9.14a: RAM 7489, 64 bit

Viec doc ra cua RAM khong khac vdi viec doc ra cua ROM. Doi vdi chip nay,
mot cach don gian la duy tri ME 6 mu’c logic 0 va WE 6 mu’c logic 1 va chon dia chi
yeu cau. TCr so lieu 4 bit xuat hien 6 cac ngo ra SENSE (doc ra). Dinh thdi cho

thao tac doc dupe chi ra bang cac dang xung 6 Hinh 9.14d. Thdi gian tre truyen

dan tpHL la khoang thdi gian tu' su'dn xuong cua xung ME den so lieu on dinh xuat
hien d ngo ra. Bang thong so ky thuat cho biet gia tri cuc dai la 50 ns. thong

</div>
(77)<div class='page_container' data-page=77>

thirdng la 35 ns. DiTcfng nhien la cac ngo vao cua dia chi phai on dinh khi toan bo
cac hoat dong doc ra bat dau vdi sudn xuong cua xung ME. Can chu y rang, sir
xuat hien so lieu d 4 ngo doc ra (SENSE) se la phan bu cua cua tir so lieu dirpc
lull tru.

n day
dia chi

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CE WE

Hinh 9.15: So do khoi cua RAM tTnh

TCr bang sir that, ta co nhan xet khi chip khong dupe chon la khi ME 6 mu’c
logic 1, tat ca cac ngo ra dat 6 mu'c 1, chung ta co mot kieu doc ]W E 6 mu’c 1). Vi
vay, trong cac dang xung cua bit dupe doc ra, tpLH la thdi gian tre tCr sudn len cua
xung ME den khi cac ngo ra cd trang thai logic 1, theo thong so ky thuat cho biet
thdi gian tre cuc dai la 50 ns, con thudng la 26 ns.

Khi viet vao, 4 bit dirpc dua vao ngd vao se du’Oc lull trir 6 dia chi bo nhd
dupe chon bang cach duy tri ngd vao ME 6 mu’c logic 0 (chon chip) va duy tri WE 6
mu'c 0. Dong thdi, phan bu cua so lieu dupe dua vao 4 ngd vao se xuat hien d 4

ngd ra. Hinh 9.14d gidi thieu cac dang xung dinh thdi cho thao tac viet vao.

Hay xem xet can than cac yeu cau ve dinh thdi cho mot chu ky viet vao.
Trudc het WE phai dupe duy tri d mu’c 0 trong mot khoang thdi gian tdi thieu de
nap tin vao cac te bao nhd, thdi gian nay du’Oc ky hieu la t^, tren dang xung, va
theo thong so ky thuat, ^ circ tieu bang 40 ns. ME (MEMORY ENABLE) chon chip

khi d mu’c 0 va dupe pnep dat mu’c 0 trung hpp vdi hoac trirdc khi thao tac viet vao
ddi hoi WE xuong mu’c 0.

Tiep theo, de nap vao bo nhd, cac so lieu d ngd vao phai on dinh trong mot
khoang thdi gian tdi thieu trudc WE va cGng phai on dinh sau WE. Khoang thdi
gian trudc khi WE xuat hien dupe ky hieu la t2 va la thdi gian xac lap sd lieu. Thdi
gian nay dupe do tCr cuoi cua sirdn sau tin hieu WE den diem d do so lieu phai
dupe on dinh. Theo thong sd ky thuat, thdi gian nay la 40 ns va trong trudng hpp
nay no cGng bang vdi t* Cac ngd vao sd lieu phai dupe giu on dinh doi vdi khoang
thdi gian sau khi xung WE tang len. Thdi gian do dupe goi la thdi gian duy tri so
lieu t3, va theo thong sd ky thuat, yeu cau tdi thieu t3 = 5 ns.

</div>
(78)<div class='page_container' data-page=78>

cua tin hieu WE den diem a do cac day dia chi ngo vao phai on dinh; yeu cau toi
thieu doi v6i t4 la 0,0 ns. Noi mot each khac la cac day dia chi ph§i on dinh trirdc
khi hoac dong thdi voi tin hieu WE xuong mu’c logic 0. Cac day dia chi cung phai
on dinh doi vdi khoang thdi gian sau khi tin hieu WE tang len den mu’c 1. Thdi gian
do duoc goi la thdi gian duy tri dia chi hoac thdi gian duy tri chon va ky hieu la t,,.
Yeu cau toi thieu doi vdi t5 la 5 ns.

Cuoi cung, sau thao tac viet vao, neu chip khong duoc chon (ME co mu’c
logic 1) cac ngo ra se trd lai trang thai 1. Thdi gian toi da de dieu nay xay ra la thdi
gian phuc hoi doc ra (SENSE) ky hieu la tSR va tSP = 70 ns la gia tri c u t dai theo
thong so ky thuat.

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Hinh 9.16: RAM 74S200, 256 bit

</div>
(79)<div class='page_container' data-page=79>

Hoat dong cua IC 7489 don gian va d l hieu, do do IC nay dupe nghien ciru
trong phan RAM. No co the dupe dung de tao nen cac bo nh6 co dung lupng Ion
hon bang cach noi song song, nhi/ng no khong thuc te khi chung ta muon co bo
nhd 16k, 32k, ...256k, 512k....Tuy nhien can nghien ctiu IC 7489 vi cac nguyen tac
co ban cua dia chi va cac hoat dong doc ra, nap vao la rat co ban va tuong tu nhu
doi vdi tat ca cac RAM tinh. Vi vay, ghi nhd cac nguyen tac co ban nay chung ta co
the xem xet cac chip khac co dung lupng bo nhd Idn hon.

So do khoi 6 Hinh 9.17 co the dung de mo ta hoat dong cua hau het cac

RAM tTnh. Cac RAM nay dupe cau tao vdi n day dia chi, cac day nay se chon duy
nhat chi mot trong so 2n te bao trong cach sap xep bo nhd, co nghTa la chon 1 bit
tai mot thdi diem, se cd dieu khien chip hoat dong (CE), kich hoat viet (WE) va
cung cap cho mot bit so lieu 0 ngo vao don D, va mot bit so lieu ngo ra don D0.

Vi du, IC 54/74200 0 Hinh 9.18 la RAM cd dung lupng 256 bit, dupe to’ chi/c
gom 256 tir, moi tu" 1 bit. 256 te bao dupe sap xep theo hinh vuong gom 16 hang
va 16 cot. 8 bit dia chi (28 = 256) dupe chia thanh mot nhom 4 bit, dupe giai ma de
chon mot trong 16 hang va mot nhom 4 bit, dupe giai ma de chon mot trong 16
cot. Cd mot bit so lieu ngo vao don, mot bit so lieu ngd ra don va mot day WE. Cd
3 ngd vao ME (M „ M2, M3), tat ca 3 ngd vao nay phai d mu’c 0 de chon chip hoac
lam cho chip du’Oc kich hoat.

Bang su that chi ra rang neu chip hoat dong, mot chu ky viet

dupe

bat dau

bang duy tri WE 0 mu’c 0 hoac mot chu ky doc du’Oc bat dau bang duy tri WE 6
mu’c 1. Neu mot vai hoac tat ca cac ngd vao MEMORY ENABLE 0 mu’c 1, chip bi
ngan can va ngd ra cd trang thai tro khang cao. Duong nhien la dinh thdi rieng

phai tuong tu vdi dang xung dinh thdi

dupe

xac dinh tren day doi vdi IC 7489.

Chung ta hieu hoat dong cua 54/74200 (viet gon la '200), rat don gian khi
dung nhieu chip '200 de tao nen bo nhd cd dung lupng Idn hon. Vi du, chung ta cd
the noi song song 4 chip nhu chi ra tren Hinh 9.19 de tao nen RAM cd 256 tu", moi
tir 4 bit. Khi noi song song 8 chip '200 se tao nen bo nhd cd tu” 8 bit v.v ... Khi noi
song song cac chip nhu vay thi viec dieu khien va dinh thdi hoan toan giong nhu
khi chi cd mot chip rieng le. Oieu khac duy nhat la 6 chd cd 4 bit so lieu vao va 4
bit so lieu ra (hoac 8 vao, 8 ra), tat ca cac bit so lieu deu song song vdi nhau.

</div>
(80)<div class='page_container' data-page=80>

Hinh 9.17: Bon '200 du’Oc sap xep thanh bo nhd 1024 bit co 256 tti 4 bit

Bo nhd MOS du’Oc dung rat pho bien la INTEL 2114. Day la bo nhd RAM tinh
cd 4096 bit, dupe sap xep thanh 1024 tCr, moi tCr 4 bit (cac thong so ky thuat d Hinh

9.17). Cau true cua IC nay hoan toan tuong tu’vdi IC 7489 nhUchi ra 0 Hinh 9.16,

nhung chu y rang IC 2114 cd dung lupng Idn hon cua IC 7489 la 16 Ian.

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</div>
(81)<div class='page_container' data-page=81>

Bo nhd co ban du'Oc sap xep thanh 64 hang va 64 cot, tong cong co 64 x 64
= 4096 bit. Sau bit dia chi (tCr A3 den As) dupe dung de chon mot trong 64 hang (26
= 64), 64 cot dupe chia ra thanh 16 nhom cac tti 4 bit va 4 bit dia chi (Aq, A 1f A2 va
Ag) dupe dung de chon mot trong so 16 nhom nay. Dia chi 10 bit se chon mot tu"
don 4 bit ti/ 64 hang va 16 cot, de cung cap mot bo nhd cd dung lupng b ln g 64 x
16 = 1024 tti 4 bit.

Hau het tat ca cac RAM tTnh cd dung lupng Idn hon 1024 bit deu thuoc loai

MOS, nhung so do khoi to’ng quat tren Hinh 9.15 van giu nguyen. Su thay doi duy

nhat la 6 cho cac bit dia chi ddi khi dupe ghep lai (ghep bo nhd thanh hai nhom) de
giu so chan yeu cau tren DIP la it nhat. Vi du, bo nhd cd 16 bit dia chi cd the dupe
thiet ke chi vdi 8 chan dia chi. Cac dia chi dupe ghep vao chip thanh 2 nhom 8,
trudc het cac bit tU 1 den 8 roi den cac bit tU 9 den 16. Cac bit dupe nap mot cach
don gian vao bo ghi chuyen d bo nhd, roi sau do dupe dua den ma tran te bao,
dong thdi de giai ma dia chi 16 bit can cd. Mot vi du bo ich cua loai ghep dia chi
nay la RAM "dong" 4116 dupe trinh bay 0 phan sau.

9.4. RAM DONG - DRAM

Nhu trinh bay tren day, su khac nhau co ban giua RAM tTnh va RAM dong la

6 cho te bao nhd. Te bao nhd dong su1 dung mot mach tranzito MOS de luu tru
dien ti'ch, va do do, nhd 1 bit. Te bao dong chiem dien ti'ch be mat tren chip silicon
nho hon nhieu so vdi te bao nhd tTnh; nhu vay, chip RAM dong cd dung lupng Idn
hon nhieu so vdi chip RAM tTnh tren cung mot ki'ch thudc be mat. Tuy nhien, vi
dien tich dupe luu tru se mat dan theo thdi gian, no phai dupe dinh ky nap lai (lam
tuoi lai), do do, RAM dong cd mot yeu cau hoat dong phu them, do la chu ky lam
tuoi lai cac te bao nhd.

</div>
(82)<div class='page_container' data-page=82>

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Hinh 9.19: a) RAM dong

</div>
(83)<div class='page_container' data-page=83>

RAM DONG TM S 4116, DUNG LlfONG 16384 BIT

THONG SO THAY DOI KY HIEU TMS4116-15 TMS4116-20 TMS4116-25 DON Vj

Min Max Min Max Min Max

CAC YEU CAU VE THCTI GIAN DIEN AP CUNG CAP VA DIEU KIEN HOAT DONG NGOAI TROT

t d p ) Thdi gian chu ky kieu trang t p c 170 225 275 ns

Thdi gian chu ky doc f R C 375 375 410 ns

t c m Thdi gian chu ky viet * w c 375 375 410 ns

1 cfn<vv) Thdi gian chu ky doc, thay doi viet t R W C 375 375 515 ns

< w ( c m Do rong xung, CAS d mu'c H (thdi gian trudc khi nap) t C P 60 80 100 ns

t w ( c u Oo rang xung, CAS d m tic L f CAS 100 10000 135 10000 165 10000 ns

t w tR H ) Do rang xung, RAS d mu'c H (thdi gian tri/dc khi nap) t R P 100 120 150 ns

t W(RL) Oo rang xung, RAS d mu'c L t R A S 150 10000 200 10000 250 10000 ns

t W(W) Oo rang xung viet t W P 45 55 75 ns

11 Cac thdi gian chuyen tiep (stidn trtidc, sifdn sau) doi

vdi RAS, CAS

t r 3 35 3 50 3 50 ns

t s u ic A ) Thdi gian xac lap dia chi cot t A S C -10 -10 -10 ns

t S U (R A ) Thdi gian xac lap dia chi hang t A S R 0 0 0 ns

t SUfD) Thdi gian xac lap so’ lieu I D S 0 0 0 ns

t s u , " , , Thdi gian xac lap lenh doc t R SC 0 0 0 ns

t s u i w c h) Thdi gian xac lap lenh viet trtfdc khi CAS d mite H f CtVL 60 80 100 ns

f s u i w R H i Thdi gian xac lap lenh vie t trtidc khi RAS d m tic H t RW L 60 80 100 ns

t h (c l c ai Thdi gian duy tri dia chi cdt sau khi CAS d m ilc L f C A H 45 55 75 ns

t h (R A i Thdi gian duy tri dia chi hang t R A H 20 20 36 ns

t h (R L C A ) Thdi gian duy tri dia chi cot sau khi RAS d m tic L t A R 95 120 160 ns

t h (C L D ) Thdi gian duy tri so lieu sau khi CAS 6 mu'c L t D H 45 55 75 ns

t h (r l d) Thdi gian duy tri so’ lieu sau khi RAS 6 mu'c L t D H R 95 120 160 ns

t h m D ) Thdi gian duy tri so lieu sau khi W d mire L f D H 45 55 75 ns

t h(ixi) Thdi gian duy tri lenh doc f R C H 0 0 0 ns

t h /c L W ) Thdi gian duy tri lenh viet sau khi CAS d mu'c L f W C H 45 55 75 ns

1 h iR L w i Thdi gian duy tri lenh viet sau khi RAS d m tic L t W C R 95 120 160 ns

1 r l c h Thdi gian t r i, RAS d mu'c Lden CAS d mtic H t C S H 150 200 250 ns

f c h r l Thdi gian tre, CAS d mu'c H den RAS d mure L f C R P -20 -20 -20 ns

t c l r h Thdi gian tre, CAS d mu'c L den RAS d mu'c H t R S H 100 135 165 ns

t c l w l Thdi gian t r i, CAS d m tic L den W d m tic L (chi chu
ky doc, chu ky tha y doi viet)

t CW D 70 95 125 ns

f r l c l Thdi gian tre, RAS d m tic L (gia tri c u t dai chi dSc
trting cho thdi gian dam bao truy cap)

t RC D 20 50 25 35 ns

t r l w l Thdi gian tre, RAS d m tic L den W d m tic L (chi chu
ky doc, chu ky thay doi viet)

• R W C 120 160 200 ns

1 w l c l Thdi gian tr i,"W d m tic L den CAS d m tic L (chu ky

viet sdm)

t1VCS -20 -20 -20 ns

t „ Khoang thdi gian lam mdi f R E F 2 2 2 ms

CAC DAC TINH CHUYEN MACH DOI VCTl DIEN AP CUNG CAP VA HOAT DONG TRONG DIEU KIEN NGOAI TRCfl
t a/i Thdi gian truy cap t t i CAS

Oieu kien thCr C L=100pF; tai:cac cong 74TTL

t CAC 100 135 165 ns

tad Thdi gian truy cap t ti RAS

Oieu kien thCr. CL=100pF; tai cac cong 74TTL

t RAC 100 200 250 ns

f ais tc H i Thdi gian ngo ra bi vo hieu hoa sau khi CAS d m tic H
Oieu kien thti: C L=100pF: tai:cac cong 74TTL

f o F F 0 40 0 50 0 60 ns

</div>
(84)<div class='page_container' data-page=84>

TMS 4116 dupe chi ra tren Hinh 9.19. Cc^mdt day vao so lieu don D, mot
day ra so lieu don Q va mot day dieu khien viet W. Co 16384 te bao nho d u ’O c sap
xep theo hinh vuong co 128 hang va 128 cot, vdi yeu cau co 14 day dia chi (214 =
16384). Chu y rang cac chan ra chi cung cap cho 7 day dia chi, dieu do co nghia
la 14 bit dia chi phai dupe ghep vao chip thanh 2 nhom, moi nhom 7 bit. Day la
chirc nang cua 2 ngo vao dieu khien khac nhau: RAS va CAS.

14 bit dia chi la A 13> A 12... A0 trong do 7 LSB la (A6, A5...Aq) la dia chi hang
va 7 MSB (A13, A12...A7) la dia chi cot. Viec chon mot bit don trong bo nho doi hoi
viec dia chi hoa cho chip theo 2 birdc: birdc thCf nhat, dja chi hang 7 bit

dupe

dua
den 7 chan dia chi tren chip, mot xung am tren day dieu khien xung chon dia chi
hang (RAS) se lay mau 7 bit nay dua vao cac chot hang tren chip. Bi/dc thir hai,
dia chi cot 7 bit

dupe

dua den 7 chan dia chi tren chip, mot xung am tren day dieu
khien xung chon, dia chi cot (CAS) se lay mau 7 bit nay dua vao cac chot cot tren
chip. 14 bit dia chi bay gid

dupe

liru trir tren chip va dupe giai ma de li/a chon ra
mot te bao nhd don. Neu ca hai RAS va CAS

dupe

duy tri 6 mu’c logic 0, dia chi

dupe

giai ma se duy tri trong mot khoang thdi gian va cac noi dung cua te bao nhd

dupe

chon cd the dupe doc ra, hoac mot bit so lieu mdi cd the d u ’O c nap vao te bao
d u ’O c chon.

S w i t c h i n g c h » r * c J * r n t i c j r e c o m m e n c e d l u p p i y v o iio ^ a 9 ru n g ?
[ ■ - ! . ]j A lt. ■ JO 15 j * h\T,A\i f ;d; t i.ãSôi * i f ? i

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Neu k i c h h o a t v i e t (W) dupe d u y tri o m u ’c l o g i c 1, c a c n o i d u n g c u a b i t d u ’Oc
c h o n se x u a t h i e n 6 n g d ra so l ie u , d o d o , Q t a o n e n m o t c h u ky d o c . Mat k h a c ,

</div>
(85)<div class='page_container' data-page=85>

neu W du’Oc giu16 mu’c logic 0, bit so lieu 0 ngo vao so lieu D se du’Oc nap vao vi tn

du’Oc chon, do do no tao ra mot chu ky viet.

Dirong nhien la can co nhCrng yeu cau dinh thdi nghiem ngat. Cac yeu cau ve
thdi gian nay du’Oc chi ro d cac dang xung djnh thdi gidi thieu tren Hinh 9.19c. Vi
du, cac ngd vao dia chi hang phai on dinh trong khoang thdi gian circ tieu tri/dc khi
xung RAS dat mire logic 0, day la thdi gian xac lap dia chi cho cac hang tASR hoac
ts RA . Hay xem xet cac dang xung dinh thdi chu ky doc cho cac khoang thdi gian
nay. Theo thong so ky thuat, thdi gian nay toi thieu bang 0,0 ns. Noi each khac,
cac dia chi hang phai o’n dinh thdi gian sau khi RAS dat mu’c logic 0. Can nghien
cCru ky cac dang xung nay. Vi hieu ro cac tinh chat cua IC nay ta se sir dung IC cd
hieu qua cao.

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Hinh 9.19d

</div>
(86)<div class='page_container' data-page=86>

can thiet de lam m6i mot hang 128 te bao. Tuy nhien, tat ca 128 hang phai du’Oc
lam m6i sau moi khoang thdi gian 2 ms.

Viec lam m6i co the du’Oc thu’c hien bang cach dung bo dem 7 bit dieu khien
de cung cap cac bit dia chi de chon mot hang trong mot thdi gian, sau do lai chon
tiep hang khac... RAS chi du’Oc dung tai moi vi tri dia chi de lam mdi lai hang do.
Gia du rang, mdi thdi gian chu ky tC(w) hoac tC(rd) la 375 ns, no can 0,375 x 128 =
48 ps de lam mdi tat ca 128 hang. Nhir vay, ngoai mdi khoang thdi gian 2 ms, thdi
gian 48 ps la thdi gian dung lam mdi lai, no chi chiem 48/2000 = 2,4% cua thdi
gian dung cho bo nhd. Viec lam mdi lai cd the du’Oc tien hanh cung mot luc cho tat
ca 128 hang hoac bang cach lam mdi lai Ian lirdt tting hang mot, vdi khoang thdi
gian bang nhau giCra cac hamg theo kieu "dinh ky".

Mot dac tinh hap dan va htiu ich cua TMS 4116 la no cd the du’Oc dia chi hoa
theo kieu dia chi trang. Theo kieu nay dia chi hang du’Oc luu trCr trong cac bo giai
ma ngd vao RAS va Ian lirot cac cot du’Oc chon trong cung mot hang bang cach
dung CAS . Uu diem Idn cua kieu dia chi trang la thdi gian chu ky doc hoac viet tc
(p) giam xuong con non mot nu’a, tCr 375 ns xuong con 170 ns.

Viec s i r dung RAM dong hoi phifc tap so vdi viec s i r dung RAM tTnh, vi cac bit

dia chi phai du’Oc ket hop lai; bo nhd phai du’Oc dinh ky lam mdi lai; viec lu ll tru* so
lieu vao bo nhd hoac doc so lieu tCr bo nhd ra phai du’Oc lam tach biet vdi cac chu
ky lam mdi. T a t ca cac chirc nang nay cd the dat du’Oc bang logic dieu khie n ngoai,

nhirng van de nay du’Oc don gian mot chut bang cach dung bo dieu khie n RAM

dong. IC TMS 4500A cua TEXAS INSTRUMENTS la mot trong cac mach dieu

khien RAM dong dung cho chip don. IC 4500A cd the du’Oc dung vdi IC 4116, cung

n h ir vdi cac RAM dong 8K, 32K va 64 K khac (Hinh 9.19).

IC TMS 450A thirdng du’Oc dung giao dien cac RAM dong vdi he thong pP va
IC TMS 4500A cd chtia cac bo ghep dia chi dieu khien lam mdi lai, dieu khien dinh
thdi de dap ting cac yeu cau truy cap bo nhd va cac yeu cau ve viec lam mdi. Cac
ngd ra tti MA0 den MA7 la 8 bit cua dia chi bo nhd du’Oc dung de dieu khien cac
RAM dong. Khuon dang (FORMAT) la 8 bit dia chi hang, 8 bit dia chi cot va 8 bit
dia chi lam mdi vao cac thdi gian yeu cau lam mdi lai, nhu1 chi ra tren Hinh 19d. Cd
2 ngd ra xung chon dia chi hang RAS0va RAS.,; CAS la ngd ra xung chon dia chi
cot cua cac RAM dong neu chan chon chip (CHIP SELECT:CS) 6 mtic 0.

Ngd vao REN, du’Oc dung de chon mot trong hai n hom cua RAM dong dang

du’Oc dieu khien bang 2 ngd ra RAS. K h i 6 m tic 1 thi RAS, du’Oc chon va khi 6 mtic

0 thi RAS0 du’Oc chon.

Mtic 0 6 mot trong hai dieu khien truy cap doc (ACR) hoac dieu khien truy
cap viet (ACW) se lam cho dia chi cot xuat hien 6 8 day dia chi ngd ra tti MAq den
MA7. Tuy nhien khi ca hai ngd vao nay 6 mtic 0, tat ca cac ngd ra deu 0 trang thai
trd khang cao._______

Ngd vao REFRED bat dau mot chu ky lam mdi. Ngd vao xung nhip ky hieu

CLK la ngd vao xung nhip cua he thong. FS0 , FS, va TWST du’Oc dung de x£c
dinh toe do lam mdi va dinh thdi.

Cd RAM dong khac du’Oc dung rong rai hon, do la IC 4164. Hoat dong cua IC
nay hoan toan giong vdi IC 4116, chi cd dieu khac duy nhat la ngudn cung cap la
mot nguon ddn + 5V. RAM dong 64 k (65536x1) tuy tirong thi'ch di/dc vdi nhau
nhirng cd so ky hieu khac nhau: AM 9064; F 4164, 2164, MK 4564. MCM 6665 va
TMS 4164.

</div>
(87)<div class='page_container' data-page=87>

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3C- I? K n -^ N Pt\K*. P C W .- TMS -*3TC *.

=T>

s-ss.

</div>
(88)<div class='page_container' data-page=88>

9.5. CAC TE BAO NHCJ

Te bao nh6 la don vi co ban de nh6 mot bit don cua tin trong bo nhd. Cac te
bao cau tao nen bo nhd ban d in thudng la cac flip flop du’Oc thiet ke bang cac
tranzito lirdng ci/c hoac MOS hoac mach nap dien tich dung tranzito MOS.

6 day ta chi nghien cull hoat dong cua te bao nhd ma khong nghien cull thiet

ke che tao. Sir hieu biet sau sac ve te bao nhd dan den viec nam vCrng cac yeu
cau cua mach, cung nhu1 danh gia du’Oc cac gidi han hoat dong cua bo nhd. Ta
nghien cull cac mach cd ban, minh hoa cho cac nguyen tac du’Oc dung cho hau
het cac te bao nhd. Doi vdi cac mach du’Oc nghien cUu, ta gia dinh mu’c dien ap
cao bang + Vcc va mu’c dien ap thap bang OV.

Te bao nhd dung trong RAM tTnh TTL cung vdi bo khuech dai doc ra va bo
khuech dai viet vao du’Oc gidi thieu tren Hinh 9.23. Hai tranzito lirdng circ Q 1 va Q2
ghep cheo nhau tao nen bo chot don va cd kha nang nhd mot bit tin, mach nay la
mot phan tir nhd. Moi tranzito cd 3 Emito. mach cua te bao nhd du’Oc noi vdi + Vcc
va dat (GND).

Neu day chon cot hoac hang 0 mu’c 0 (GND), te bao ngirng hoat dong (khong
chon). De chon mot te bao, ca 2 day cua hang va cot deu d mu’c 1 (+ Vcc). Khi
du’Oc chon, mot bit so lieu cd the du’Oc nhd trong te bao (viet vao) hoac cac noi
dung cua te bao cd the du’Oc doc ra. Ta xem xet si/ hoat dong cua te bao nhd.

Khi cac day chon hang va cot 6 mu’c 1, cac Emitd noi vdi cac day nay nhan

thien ap ngirpc, te bao tac dong nhu mot chot, cho nen mot trong hai tranzito dan,
tranzito con lai ngirng dan. Tranzito dan cho dong ra den Emito di xuong qua dien
trd Rs va di vao ci/c Badd tranzito cua bo khuech dai doc ra, lam no d in . Trong te
bao, d tranzitd ngirng dan, khong cd ddng den tu” Emito, nhi/ vay, tranzito cua
khuech dai doc d phi'a khong cd ddng Emitd bi ngirng dan. Ket qua la cac tranzito
cua bo khuech dai doc ra "bat chu’dc" cac tranzito trong te bao nhd. Do do, khi
du’Oc chon, cac noi dung cua te bao nhd du’Oc sir dung ngay lap tCrc d cac ngc ra
so lieu cua bo khuech dai doc ra. Vi du, neu bit 1 du’Oc nhd trong te bao, Q, dan
con Q2 ngirng dan. Khi du’Oc chon, ngo ra so lieu (DATA OUT) d mu’c 1, va phan
bu cua nd, DATA OUT se d mu’c 0.

Bo khuech dai viet (WRITE) du’Oc dung de liru trir tin vao te bao nhd khi ngo
vao W du’Oc duy tri d mu’c 1. Khi W d mu’c 0, bo khuech dai khong lam viec va cac
ngo ra cua nd khong ket noi vdi cac dien trd Rs. Cd nhieu cau hinh ve te bao nhd,
nhirng cac yeu cau co ban la WRITE 1 len mu’c logic 1 va WRITE 0 xuong mu’c
logic 0, mdi khi DATA IN d mu’c Icgic 1; ngirpc lai, WRITE 0 d mu’c logic 1 con
WRITE 1 xuong mu’c logic 0, moi khi DATA IN d mu’c logic 0. De lUu tru” mot bit vao
te bao, trirdc het te bao du’Oc chon bang cach cho hang va cot, ca hai deu d mu’c
logic 1. Sau do, mu’c logic d DATA IN cua bo khuech dai viet vao se lam cho
WRITE 1 len mu’c logic 1 va WRITE 0 xuong mu’c logic 0. Nhu” vay, ca 3 Emito cua
Q2 se d mu’c logic 1, lam cho Q2 ngirng dan va Qt dan va chot liru trCr bit 1. Ngirpc
lai, mCrc 0 d DATA IN se dat tat ca 3 Emitd cua Qt d mu’c Icgic 1, lam cho Q,
ngirng dan va Q2 dan, nhd bit 0 vao chot.

Trong bo nhd tTnh dung NMOS, te bao nhd du’Oc cau tao bang cach dung hai

bo dao NMOS, ghep cheo de tao nen mot chot (Hinh 9.23). Trong bo nhd tTnh

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dung CMOS, hai bo dao CMOS du'Oc ghep cheo de tao nen mot chot (Hinh 9.24).

Tir mot trong hai trudng hpp nay, mot bit so lieu du’Oc nap vao te bao hoac cac noi
dung du’Oc doc ra bang cac cong truyen d in NMOS. Trudc het, ta can xem xet
cong truyen dan NMOS hoat dong nhu the nao.

G a te

#1 #2

Hinh 9.22: Cong truyen dan NMOS

Hinh 9.22 gidi thieu mot tranzito NMOS loai giau dupe noi nhu mot cong
truyen dan. Tranzito dupe gia dinh truyen dan theo hai hudng: day cuc nguon va
day cuc mang, hai day nay co the thay doi cho nhau ma khong co dieu khac biet
nao xay ra. Ta goi ngan gon la day #1 va day #2. Oe day #1 la ngo vao thi day #2
la ngo ra. Day cong (GATE) la day dieu khien. Cac tin hieu dien ap hoac la 0 VDC
hoac la + Vcc.

Neu cuc GATE dupe duy tri 0 mUc 0 (0 V), tranzito ngung d in , cac ngo vao
va ngo ra bi phan cach nhau. Tin hieu khong truyen dan qua mach.

Neu cuc GATE dupe duy tri 6 mUc logic 1 (+ 5VDC), va dien ap 6 day #1
duoc dat d mUc 0, tranzito dan, dong se chay qua tranzito cho den khi
V0 = V, = 0,0 V.

Neu cu t GATE 6 mUc logic 1 va dien ap 6 day #1 dupe dat 0 mUc logic 1,
tranzito se dan neu VQ d mUc 0 va dong se chay qua tranzito cho den khi
V0 = V, = + Vcc. Hoac neu VQ cung 6 mUc logic 1, tranzito se duy tri ngung dan vi
V0 = V, = + Vcc.

Tom lai, cong truyen dan khong hoat dong khi nao cuc GATE d mUc logic 0

va ngo vao bi phan cach khoi ngo ra; nhu vay, cong truyen d in hoat dong nhu mot
cong-tac bi ho mach, cong truyen dan hoat dong khi nao c u t cong 0 mUc logic 1,
dien ap ngo ra bang dien ap ngo vao, VQ = V, . Nhu vay, cong truyen dan hoat
dong nhu cong-tac kin mach.

Mach trong Hinh 9.23 dien hinh cho te bao nh6 va khuech dai doc ra da

dung trong RAM tTnh loai MOS. Te bao nhd gom co hai bo dao loai NMOS, ghep
cheo de tao thanh chot. Co 4 cong truyen d in T,, T2, T3 va T4 dung de chon te
bao. Khi ca hang va cot deu 6 mUc 1, te bao

dupe

chon va cac noi dung cua no co
the

dupe

doc ra nhd bo khuech dai doc ra, hoac mot bit so lieu

dupe

nap vao te
bao nhd dung cong WRITE (W).

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so lieu 6 chan DATA IN (hoac 6 mu'c logic 1 hoac 6 mCrc logic 0) se di/pc noi tri/c
tiep den chot tai D. Neu DATA IN 6 mu’c 1, bit 1 du’Oc nap vao, vi chot se on dinh
vdi D co mu’c logic 1; ngirpc lai, DATA IN d mu’c logic 0, bit 0 du’Oc nap vao vi chot
on dinh v6i D co mu’c logic 0. Day la chu ky viet.

Ta lai gia dinh rang, te bao du’Oc chon nho duy tri ca hang lln cot 6 mu’c 1.
Hai cong truyen dan T3 va T4 cung hoat dong (chung hoat dong nha cong-tac kin
mach). Bat cir bit nao du’Oc nap vao chot D, phan bu cua bit do xuat hien tai D va
du’Oc ghep nhd cac cong truyen dan T3 va T4 ra den bo khuech dai doc ra. Neu
cong doc hoat dong (R d mu’c 1), bit so lieu du’Oc liru trCr d chot D se xuat hien 6
ngo ra so lieu (DATA OUT). Day la chu ky doc.

Hinh 9.23: Te bao nh6 dung hai bo dao NMOS ghep cheo de tao nen mot chot

Hai tranzito dau 6 bo khuech dai doc ra tao nen mot cong dao co ngo ra la D.
Hai tranzito 0 ngo ra bo khuiech dai doc ra Qt va Q2 co cac tin hieu vao~D va D v6i
ket qua la mot trong hai tranzito nay luon luon dan, tranzito con lai luon luon ngirng

dan. Vi du, neu bit so lieu du’Oc liru trir la 1, thi D = 1. Nen Q, se dan con Q2 ngirng
dan va ngo ra so lieu (DATA OUT) se 6 mu’c 1. Ngirpc lai, neu bit 0 du’Oc liru trir d
chot, D = 0 vi vay Q, ngi/ng dan, 0 2 dan va DATA OUT se d mu’c 0.

Mach trong Hinh 9.24 la dien hinh cua te bao nho va bo khuech da* doc ra

du’Oc dung trong RAM tTnh loai CMOS. Te bao nhd gom co hai cong dao CMOS,
ghep cheo de tao nen chot. Co 4 cong truyen dan T 1f T2, T3 va T4 du’Oc dung de
chon te bao. Khi ca hang Ian cot deu 6 mu’c logic 1, te bao du’Oc chon va noi dung

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cua no co the du’Oc doc nhd dung bo khuech dai doc ra, hoac bit so lieu co the
du’Oc nap vao te bao bang cach dung cong WRITE (W).

Chu ky viet cua mach nay hoan toan giong v6i chu ky viet cua mach NMOS
da du’Oc trinh bay tren day. Te bao du’Oc chon nhd duy tri ca hang Ian cot 6 mu’c 1
va sau do, bit cd mat 6 chan DATA IN se du’Oc nap vao chot neu W d mu’c 1 (cong
WRITE hoat dong).

Chu ky doc cGng hoan toan giong nhu” 0 mach NMOS da trinh bay tren day.
Te bao du’Oc chon bang cach cho ca hang Ian cot deu d mu’c 1 va cac noi dung
cua te bao nhd xuat hien d chan DATA OUT sau khi truyen qua bo dao khuech dai
doc ra.

c *

GATE
DdU o<i\

c---rA s c ;

Hinh 9.24: Te bao nhd dung hai bo dao CMOS ghep cheo de tao nen mot chot

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phai di/dc nap lai. Hdn nu’a, trong qua trinh doc ra, cac noi dung cua te bao dong
co the bi mat do tieu tan cac dien tich da du’Oc nap, ket qua la se bi mat cac bit da
du’Oc nhd. Day la hien tirpng doc xoa, hien tirpng nay khong xay ra d cac te bao
nhd tTnh. Ngay ca khi cac bo nhd dong co nhirpc diem nhu1 can phai co mach ho
trp va cac tin hieu dinh thdi de lam mdi lai cung nhir hien tirpng doc xoa thi iru
diem dung lirpng Idn cua chung van la dang ke.

Hau#* tu

S o v » c e p i j i c

I T

- L - / L doxid

a ____ ___________
• M O S F F T M0S C w » K H O f

f'VflCOH

T

r,s t a o f t le t b ao <Aho d ong voi m e t t x a x t z

Chen t e w
*o w«;f ie c7

9 0 » i? »r.

i f t . P M va.

\

I Gair

Fhfart ta

C a ; > * c 't o ' o ' a K

hia-JCT: I > c * !

it)) ~ 0 stt

Hinh 9.25

Hay xem xet mot te bao nhd dong co mot tranzito loai MOS nhir tren

Hinh 9.25. Te bao co mot tranzito NMOS va mot tu dien MOS du’Oc che tao tren
cung mot chip. Neu phien tu va de du’Oc noi vdi dat (GND) thi dien dung Cs thirdng
de chCra cac dien tich cho mot bit bang dien dung cua tu dien MOS (CMOS) mac
song song vdi tu dien tap tan cua tranzito C,). Bit 1 dirpc nhd vao te bao neu tu Cs
dirpc nap dien va bit 0 dirpc nhd neu Cs phong dien.

Ve cd ban, tranzito hoat dong nhir mot cong truyen d in . Chon hang se lam
cho cac tranzito trong ca hang cua bo nhd hoat dong. Sau do, d chu ky doc,

khuech dai doc ra phai tim cac noi dung cua te bao va dong thdi nap lai dien dung
nhd Cs . Chu y rang bo khuech dai doc ra giai quyet dirpc hien tirpng doc xoa ngay
do ddng rb va dong thdi dirpc dung de tao lai, 6 chu ky viet, cong WRITE dirpc
dung de nap bit 1 hoac bit 0 vao te bao bang cach Ian lirpt cho tu Cs nap hoac
phong dien./.

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CAU HOI VA BAI TAP

9.1. Trong bo nhd nao so lieu khong mat di khi mat nguon cung cap?

9.2. Bo nh6 nao van duy tri so lieu dirpc liru sau khi da ngat dien?

9.3. Bo nh6 co 10 dirong dia chi co the gan cho bao nhieu tu?

9.4. Bo nhd co dung lirpng 1024 x 8 du’Oc goi la bo nhd co bao nhieu tir 8 bit?

9.5. Cac bit 0 va 1 co the liru trCr trong phan tir nao cua bo nhd ban dan?

9.6. Bo nhd PROM co cac diot noi day chi cho phep lap trinh bao nhieu Ian?

9.7. Muon xoa cac vi tri nhd bang cach cho phong dien cac vi tri nhd trong
EPROM ngirdi ta can chieu loai anh sang nao? Trong bao lau?

9.8. Ram ngoai kha nang dung de doc cac so lieu ra con co kha nang nao

khac nu’a?

9.9. Cac mach dieu khien RAM dong cung cap 3 che do hoat dong: doc, ghi

va che do hoat dong nao nu’a?

9.10. RAM dong sir dung linh kien nao de liru trCr so lieu?

9.11. Theo dinh ky 2 ms, RAM dong phai du’Oc tac dong gi doi vdi cac te bao
nhd?

9.12. Tren thi trirdng cd loai chip dac biet nao dung de dieu khien va thu’c
hien tien trinh lam tiroi?

9.13. Giai thi'ch sir khac nhau giC/a PROM va EPROM?

9.14. Giai thi'ch the nao la "bo nhd xoa dirpc".

9.15. Giai thi'ch y nghTa RAS va CAS.
Doi vdi IC TMS 4116,

- Gia tri circ dai va circ tieu cua do rong xung cua RAS va CAS la bao

nhieu?

- Gia tri circ tieu cua thdi gian chu ky doc va viet la bao nhieu?

9.16. Hay chi ra cac trirdng hpp cd the sap xep theo hinh chCr nhat doi vdi bo
nhd cd 32 te bao nhd.

- Cd bao nhieu hang va bao nhieu cot doi vdi moi trirdng hpp?

- Can cd bao nhieu day dia chi trong moi trirdng hpp ke tren?

9.17*. Can cd dia chi nao cua cac day A3A2A 1A0 de chon te bao 21 trong

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9.18. Xac djnh can co bao nhieu bit dia chi doi v6i bo nhd co dung lupng sau:
a. 1024

b. 4096
c. 256
d. 16384

9.19*. Can dia chi EDCBA bang bao nhieu dudi dang nhi phan va dang thap
luc phan (hexadecimal) de lua chon tti 8 bit 6 hang thti 27 cua bo nh6 54/747488A

tren Hinh 9.9a?

9.20*. Viet bieu thtic dai so Boolean cho dia chi hang thti 15 doi v6i ROM
54/7488A tren Hinh 9.9a.

9.21*. Chi ra cach ket noi cac RAM 7489 de co bo nh6 co dung luong la 32 tti
4 bit.

9.22*. Ve so do logic cho bo nh6 co 256 tti 8 bit, v6i dieu kien sti dung IC

’200.

9.23*. Ve lai mach tren Hinh 9.23 va bieu thi cac mtic dien ap , gia du rang te
bao khong dupe chon va dang luu trti mtic 1.

9.24*. Ve lai mach tren Hinh 9.23, bieu thi mtic dien ap, neu mtic 1

dupe

luu
trti trong te bao va te bao

dupe

chon, gia du rang W

0

mUc

L.

9.25*. Ve cong truyen dan nhu tren Hinh 9.22, bieu thi mUc dien ap cho cac

trudng hpp sau day:

a. Cong (GATE) 6 mtic 1 va \ / : 6 mtic 1
b. Cong (GATE) d mtic 1 va V, 0 mtic 0
c. Cong (GATE) 0 mtic 0 va V, 6 mtic 1

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CHUONG 10

CAC BO CHUYEN DOI

■

A/D VA D/A

Sau khi hoc xong chuong nay, hoc vien co kha
nang:

□ Hieu duoc nguyen ly cau tao cua bo chuyen doi
A/D va bo chuyen doi D/A

□ Phan tich duoc hoat dong cua mot so mach cua
bo chuyen doi A/D co ban

□ Phan tich duoc hoat d'ong cua bo chuyen doi A/D
co bo so ap

□ Phan biet dupe mot so bo chuyen doi A/D loai
gan dung lien tiep va loai cd dien ap rang cua.

□ SCr dung duoc IC ADC 0804 thuong pham dung
lam bo chuyen doi A/D

TONG QUAN

Hau het cac dang tin hieu phan anh cac hien tirpng tu nhien deu la cac tin
hieu analog. Vi du nhir thdi gian, toe do, trong luong, ap suat, cirdng do anh sang...
tat ca deu la dang tin hieu analog trong tir nhien. He thong digital nhir chi ra tren

Hinh 10.1 dirdi day co ngo vao analog. Dien ap thay doi lien tuc tir OV den 3V. Ma
hoa la mot bo kien dac biet chuyen doi tin hieu analog thanh tin hieu digital. Bo ma
hoa trong Hinh 10.1 du’Oc goi la bo chuyen doi tin hieu analog sang tin hieu digital,
hoac ngan gon hon, du’Oc goi la bo chuyen doi A/D (ADC). Trong trirdng hpp nay,
ADC chuyen doi cac tin hieu analog thanh cac so lieu digital.

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Sd do he thong digital tren Hinh 10.1 co mot bo giai ma. Bo giai ma nay
thuoc loai dac biet chuyen doi tin hieu digital tir bo xir ly digital thanh ngo ra
analog. Vi du, ngo ra analog co dien ap thay doi lien tuc tir OV den 3 V. Ta goi bo
giai ma nay la bo chuyen doi D/A (DAC). Trong trirdng hpp nay, DAC giai ma cac
tin hieu digital thanh dang analog.

Toan bo he thong trinh bay tren Hinh 10.1 co the goi la he thong hon hpp vi
no co ca bo kien analog Ian bo kien digital. Cac bo ma hoa va bo giai ma nham
chuyen doi tin hieu tir analog sang digital va tir digital sang analog dirpc goi la cac
bo kien giao dien. Tir giao dien thirdng du’Oc dung de chi bo kien hoac mach
chuyen doi kieu hoat dong nay sang kieu hoat dong khac. Dien ap ngo vao cua tin

hieu analog cua he thong tren Hinh 10.1 tir 0 den 3V, dien ap nay co the do bo

bien nang (transducer) tao ra. Transducer la bo kien co chirc nang chuyen doi tin
hieu tir dang nang lirpng nay sang dang nang lirpng khac.

10.1. CHUYEN DOI TCr DIGITAL SANG A N A LO G (DAC)

Tir Hinh 10.1 gia du rang ta muon chuyen doi tin hieu nhi phan tir bo xir ly
digital sang ngo ra co dien ap tir 0 den 3y. Ta lap bang sir that cho bo giai ma.

Bang 10.1 trinh bay 4 ngo vao D, C, B va A dirdi dang nhi phan. Moi mu’c 1 nam
trong khoang tir +3V den +5V. Moi mu’c 0 vao khoang 0V. Cac ngo ra du’Oc the

hien bang dien ap 6 cot ngoai cung ben phai cua Bang 10.1. Theo bang nay, neu

so nhi phan 0000 xuat hien o ngo vao cua DAC thi ngo ra la 0V. Neu so nhi phan
0001 6 ngo vao thi ngo ra la 0,2V. Neu so nhi phan 0010 xuat hien o ngo vao, luc
do, ngo ra la 0,4V. Chu y rang, doi vdi m6i hang tir tren xuong dirdi cua Bang, ngo
ra analog lai tang len 0,2V.

Bang 10.1: Bang sir that cua DAC

CAC NGO VAO DIGITAL NGO RA
ANALOG

D c B A VON

Hang 1 0 0 0 0 0

Hang 2 0 0 0 1 0,2

Hang 3 0 0 1 0 0,4

Hang 4 0 0 1 1 0,6

Hang 5 0 1 0 0 0,8

Hang 6 0 1 0 1 1,0

Hang 7 0 1 1 0 1,2

Hang 8 0 1 1 1 1,4

Hang 9 1 0 0 0 1,6

Hang 10 1 0 0 1 1,8

Hang 11 1 0 1 0 2,0

Hang 12 1 0 1 1 2,2

Hang 13 1 1 0 0 2,4

Hang 14 1 1 0 1 2,6

Hang 15 1 1 1 0 2,8

Hang 16 1 1 1 1 3,0

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So do DAC du’Oc trinh bay tren Hinh 10.2. Cac ngo vao digital (D,C, B va A)
d ben trai. Bo giai ma co 2 phan: mang dien trd va bo khuech dai cong, ngo ra la
dien ap ben phai hinh ve.

Mang dien trd tren Hinh 10.2 du’Oc ti'nh toan sao cho mu’c 1 d ngo vao B co
gia tri gap 2 Ian mu’c 1 6 ngo vao A. Tirong tir nhir vay, mu’c 1 6 ngo vao C co gia
tri gap 4 Ian gia tri cua mu’c 1 0 ngo vao A. Mach nay du’Oc goi la mang dien tro

thang.

Van de co ban trong viec chuyen doi tin hieu digital thanh tin hieu analog
tirong dirong la chuyen n mCrc dien ap digital thanh dien ap analog ti/ong dirong.
Oieu do co the thirc hien bang cach thiet ke mang dien trd de chuyen moi mu’c
digital thanh dien ap (hoac dong) tirong dirong co trong so nhi phan.

Gia du ta can chuyen 8 tin hieu digital cho trong bang si/ that dirdi day thanh
cac dien ap analog tirong dirong. So nhd nhat la 000, hay lay so nay bang 0 V. Sd
Idn nhat la 111, hay lay sd nay tirong Crng bang +7 V. Xac dinh giai tin hieu analog
de trien khai. GiCra 000 va 111 co 7 mu’c roi rac. Do do chia tin hieu thanh 7 mu’c :
LSB la tin hieu digital cd mCrc chuyen doi nhd nhat. Nhir vay, si/ chuyen doi d ngo
ra analog LSB bang 1/7 dien ap ngo ra analog. Bo phan trd (mang dien trd) du’Oc
thiet ke sao cho sd 1 d vi tri nhi phan 2° tao ra dien ap bang: +7 V x 1/7 = +1 V d
ngo ra.

Ta thay ro rang rang 21 bieu thi sd co mCrc logic gao 2 Ian bit 2°. Do do, so 1
d vi tri nhi phan 21 phai tao nen si/ chuyen doi dien ap d ngo ra analog gap 2 Ian
cua so LSB. Dien ap ngo ra analog trong trirdng hop nay bang: +7 V x 2/7 = +2 V.

Tirong ti/, 22 = 4 = 2 x 2 1 = 4 x 2° va nhir vay 22 phai tao nen si/ chuyen doi
dien ap ngo ra bang 4 Ian dien ap LSB . Bit 22 phai tao nen dien ap ngo ra mot si/
chuyen doi bang +7 V x 4/7 = +4 V.

Cong viec du’Oc tiep tuc va mdi bit sau co gia tri gap 2 Ian bit trirdc. Nhir vay,
LSB cd sd nhi phan tirong dirong bang 1/7. Sd LSB lien sau cd sd bang 2/7, gap 2
Ian LSB trirdc. Sd MSB (trong trirdng hop nay he thong cd 3 bit) cd trong sd nhj
phan bang 4/7, gap 4 Ian LSB dau tien. Ta thay: 1/7+2/7=4/7=7/7=1 (tong cac
trong sd nhi phan phai bang 1).

Tom lai, trong sd nhi phan tirong dirong mang nhan LSB bang 1/(2n - 1), d do,
n la sd cac bit. Cac trong sd con lai du’Oc xac dinh bang cach nhan 2, 4, 8 v.v...

BANG SIf THAT CAC TRONG SO NHI PHAN TUONG DUKTNG

2* r 2*

0 0 0

0 0 1

0 1 0

0 1 1

1 0 0

1 0 1

1 1 0

1 1 1

BIT TRONG SO

2° 1/7

21 2/7

22 4/7

Tong: 7/7

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B o k h u e c h dai c on g tren Hinh 10.2 lay dien ap tCr m a n g dien trd v a k h u e c h 
dai len tha nh dien ap du’Oc ghi trong cot ngoai c un g ben phai c ua Bang 10.1. B o 
khuech dai c o n g thirdng d u n g IC khuech dai toan tir O P - A M P . B o k h u e c h dai c o n g 
con du’Oc goi la bo khuech dai ba c thang.

T o m lai, D A C co 2 phan: mot n h o m dien trd tao ne n m a n g dien trd t ha ng va 
mot bo k h u ec h dai d u ng de khu ec h dai cong.

(S s ) (4 s ) (2 s ) (1s)

0 C B A

Mang B o khuech

dien t r o d a i conq

► —

---Hinh 10.2. S o d o khoi D A C

10.2. BO KHUECH DAI TOAN T lf

B o khu e c h dai toan tir, O P - A M P , co d a c di em la trd k h an g v a o c a o va trd 
k hang ra thap. D o tang i'ch dien ap cua O P - A M P du’Oc x a c lap b a n g dien trd ben 
ngoai.

K y hieu cua O P - A M P du’Oc gidi thieu tren Hinh 10.3a. O P - A M P co 2 ngo

v ao: ng o v a o phia tren la ng o v a o dao, du’Oc ky hieu b a n g da u n g o v a o phi'a dirdi
la ngo v a o khong dao, du’Oc ky hieu b a n g dau +. N g o ra cua O P - A M P d phia ben 
phai cua ky hieu.

Ngo vao dao_

Ngo vao khong dao

OPAMP

Ngo ra (Output)

Rin

Vin

1

1

OPAMP

-Q

1

Vout

Hinh 10.3: O P - A M P a) K y hieu

b) N g o v a o va do tang ich du’Oc x a c lap b a n g dien trd ngoai

O P - A M P luon luon dirpc su” d u n g vdi linh kien k em theo. D i e n hinh la 2 dien 
trd trong Hinh 10. 3 du’Oc dira v a o de x ac lap d o tang i'ch dien ap c ua O P - A M P . R in 
la dien trd v a o v a R, la dien trd ph a n hoi. D o tang i'ch dien ap du’Oc x a c di nh b a n g 
c o n g thirc sau:

</div>
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A =

---Rin

Gia du, c a c gia tri dien trd ket noi vdi O P - A M P la R f = 10 k Q va R jn = 10 k£2. 
T h e o cong thu’c tren, do tang i'ch dien ap cua O P - A M P la:

R f 10000

A = ---= ---= 1

R in 10000

D o tang i'ch cua O P - A M P A = 1.

N eu trong Hinh 10.3b dien ap v a o V in = 5 V , thi dien ap ng o ra V out = 5V , dien 
ap ngo vao dao d a ng du’Oc sir d u n g va neu dien ap na y ba ng + 5 V , thi dien ap ngo 
ra bang - 5 V . B o tang ti'ch dien ap cua O P - A M P cGng co the du’Oc ti'nh b a ng c on g 
thu’c:

V0

A = 
---V in

Do do, do tang i'ch cua mach tren Hinh 10.3 du’Oc ti'nh nhir sau:

V0 ' 5

A = --- = --- = 1

V in 5

D o tang i'ch dien ap A = 1.

Gia du, dien trd ng o v a o va dien trd ph a n hoi Ian li/pt ba ng 1 k Q v a 1 0 k Q nhu” 
chi ra tren Hinh 10.4. Hoi do tang i'ch dien ap cua m a c h ba ng bao nhieu?

T a n g i'ch dien ap du’Oc ti'nh nhir sau:

R f 10000

A = --- = --- = 10

R in 10 00 0

D o tang i'ch dien ap A = 1 0. N e u dien ap ngo v a o b a n g + 0 , 5 V , thi luc do, dien 
ap tai ngo ra b a n g bao nhieu? N e u do tang i'ch b a ng 10, dien ap n g o v a o ba ng 
0, 5 V nhan vdi 10 Ian ba ng 5 V la dien ap n g o ra. D i e n ap ng o ra tai V0 b a ng - 5 V 
nhu" do bang v on ke tren hinh ve.

D o tang i'ch dien ap cua O P - A M P co the tha y doi neu ta thay doi ti le dien trd 
ngo va o R in va dien trd phan hoi R f. T a c u n g co the xac lap do tang i'ch c ua O P 
A M P ba ng c ach sir d u n g c a c gia tri khac nh au cua R in va R f.

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+ O—

1k
Vin 1/2 V

J L

Hinh 10.4: M a c h k h u e c h dai sir d u n g O P - A M P

10.3. DAC CO BAN

D A C dd n gian du’Oc trinh b a y tren Hinh 10.5. D A C c o 2 p h a n: m a n g dien trd 
d phan ben trai g o m c a c dien tro R 2, R3 va R 4. B o k h u e c h dai c o n g 6 ben phai 
g o m mot O P - A M P va mot dien trd p h a n hoi. D i e n ap n g o v a o (V,n) b a n g 3 V du’Oc 
dat v a o c ac c h u y e n m a c h D, C , B va A . D i e n ap ng o ra ( V Q) du’Oc d o tren von ke. 
O P - A M P doi hoi du’Oc c u n g c ap b a n g n g u o n kep + 1 0 V va - 1 0 V .

V6i tat ca c a c c h u y e n m a c h du’Oc ket noi v6i dat ( G N D ) b a n g 0 V , nhir chi ra 
tren Hinh 10.5, dien ap ng o v a o tai di em A b a n g 0 V va dien ap n g o ra c u n g bang
0 V. Khi dien ap 3 V du’Oc dat v a o n g o v a o c ua O P - A M P , ta ti'nh d o tang i'ch cua 
O P - A M P . D o tang i'ch phu t hu oc v a o dien trd p h a n hoi R f. N e u R f b a n g 10 kQ. va 
dien trd R in b a n g 150 kQ , do tang i'ch c ua O P - A M P ba ng :

R f 1 0 0 00

A = --- = --- = 0, 06 6 .

R,n 1 5 0 0 0 0

M u o n tinh dien ap ra, ta n h a n d o tang i'ch vdi dien a p v a o:
V0 = A x V in = 0 , 0 6 6 x 3 = 0,2 V

D ie n ap ng o ra b a ng 0,2 V khi n g o v a o la so nhi p h a n 0 0 0 1 . D i e u n a y thu’c 
hien d u ng yeu cau c ua han g thi/ 2 trong Bang 10.1.

B a y gid ta du a so nhi p h a n 0 0 1 0 v a o D A C trong Hinh 10.5. T h a n h d o n g cua 
c h u y e n m a c h B d U d c dat d vi tri 1, O P - A M P d U d c c a p n g u o n 3 V . D o tang i'ch luc 
n a y bang:

R f 1 00 0 0

A = --- = --- = 0 , 1 3 3

R, 7 5 0 0 0

N h a n do tang i'ch vdi dien ap v a o ta dUdc dien ap n g o ra b a n g 0 ,4 V . D i e n ap 
n a y thu’c hien d u n g y e u cau c ua h a n g thu” 3 trong Bang 10.1.

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Ngo vao nhj phan

8>$ 4s 2s 1s

Hinh 10.5: M a c h D A C

R f 10000

A = --- = --- = 0 , 5 3 5

R,n 18700

Nha n do tang i'ch vdi dien ap 3 V ta diroc dien ap ng o ra c ua O P - A M P ba ng
1,6 V. D ien ap n a y thuc hien d u n g yeu cau cua h ang thir 9 trong Bang 10.1.

Khi tat ca c a c c h u y e n m a c h de u d U d c dat d vi tri 1 trong Hinh 10.5, thi O P 
A M P nhan dUdc dien ap 3 V vi luc d o do tang ich tang len de n 1. D i e n ap dat va o 
O P - A M P co the tang den mu’c gidi han v a ba ng ±1 0 V ( b a n g vdi dien ap c un g cap 
cho O P - A M P ) . C d the bo’ s un g c a c vi tri so nhj pha n b a n g c ac h tang so lUdng 
chuyen mac h trong m a n g dien trd. N e u c h u y e n m a c h cd gia tri nhi pha n la 16s, luc 
do, can dien trd cd gia tri ba ng mot nira gia tri dien trd R4 v a ba ng 9 3 5 0 Q . G ia tri 
dien trd phan hoi cGng cd the thay doi de n 5 kQ. N g o va o, d o do, cd 5 so nhi phan 
va ngo ra dien ap bien doi tu" 0 V de n 3 V.

D A C tren Hinh 10.5 cd 2 nh Ud c diem: m a n g dien trd can cd nhieu loai tri so 
dien trd khac nhau va do chi'nh x ac c u a dien ap ng o ra thap.

10.4. CAC DAC LOAI THANG

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B o k hu e c h dai cong tren Hinh 10.6 c un g du ng loai O P - A M P nhi/ trong m a c h 
tren Hinh 10.5, O P - A M P c ung can den n gu o n c ung c ap kep. H o at d o n g c ua m a c h 
n a y tuong tu n h u m a c h tren Hinh 10.5. Bang 10.2 neu len chi tiet hoat d o n g cua 
D A C nay. C h u y r i n g ta d a ng sir d u n g dien ap ng o v a o la 3, 75 V c h o bo c h u y e n 
doi. Mdi b u d c d e m nhi phan tang len d ng o ra analog la 0, 25 V , nhir chi ra trong 
cot ngoai c un g ben phai cua Bang 10.2. N h6 lai la moi so 0 d n g o v a o co nghia la
0 V du pe dat v a o ng o va o do. Moi so 1 d ng o v ao co nghTa la dien ap 3 . 7 5 V dupe 
dat v a o ngo v a o do. S i r d u n g dien ap ng o v a o 3, 7 5 V vi dien ap n a y co lien quan 
mat thiet vdi ngo ra cua bo d e m T T L va c a c I C khac d u p e sir d u n g trong D A C . Do 
do, c a c ngo va o ( D. C , B. A ) trong Hinh 10.6 co the di/pc ket noi tri/c tiep de n cac

n g o ra cua IC T T L va hoat d o n g n h u da neu len trong Bang 10.2. T u y nhien, thuc 
te. c a c ngo ra cua IC T T L kho ng c ho c a c gia tri du mu'c chinh x ac c an thiet, c hung 
phai d up e hieu chinh de dat d u p e dien ap ng o ra chinh xac.

B AUG 1 0 .2 : B A N G S J T H A T C U A C H U Y E N o S l D / A

S»GOVAOM~. C A C N J& 3 v a on h i '’H A N CAC

KJ-*M*LQG

D 5 A

O 0 0 O 0

0 0 Q 0 2 5

0 0 0 0 5 0

0 0 } C.75

0 T 0 0 \DO

0 1 0 1 X25

0 1 O \5C

0 1 1 T .7 5

1 O o 0 2jOO

0 Q 1 Z 2 S

1 0 1 O 2 3 O

1 0 2 ,7 5

1 1 O 0 2 . 0 0

1 1 0 3 2 5

1 0 2 5C

1 1 2 .7 5

Hinh 10.6: M a c h cua D A C sir d u n g m a n g dien tro t ha ng R - 2 R

C o the dua them nhieu vi tri nhi ph a n ( 16s. 32s. 6 4 s. . . ) v a o m a n g dien trd cua 
D A C theo c ac mau dien trd n hi /t r ong Hinh 10.6.

D A C loai dien trd thang R - 2 R cd i/u di em la chi si/ d u n g 2 loai dien trd, con 
cau true khong khac gi D A C nhu" da trinh b a y tren Hinh 10.5, nd c u n g cd 2 phan: 
m a n g dien trd va bo khu ec h dai cong.

10.5. BO CHUYEN DOI TL/ANALOG SANG DIGITAL (ADC)

A D C la d a ng d a c biet c ua bo m a hoa. S o do khoi c u a A D C dirpc gidi thieu 
tren Hinh 10.7. N g o v a o la mot dien ap bien thien. D i e n ap trong tri/Ong h o p nay 
th a y doi trong kho an g tCr 0 V d e n 3 V . N g o ra c ua A D C cd gia tri nhi p h a n. A D C 
bien doi dien ap anal og ng o v a o thanh tti nhi p h a n 4 bit. C u n g n h u bo m a hoa 
khac. d ngo v a o va ng o ra cua A D C can cd c a c dien ap chinh xac. Bang 10.3 c ho

ta biet c ach hoat d on g cua A D C .

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Hang 1 la dien ap dat vao ngo va o cua A D C ba ng 0 V, ng o ra la so nhi pha n 
0000. Hang 2 la dien ap dat vao ngo v a o va ba ng 0,2 V, ng o ra cua A D C co so nhi 
phan la 0001. C h u y rang m6i khi dien ap ngo v ao tang len 0,2 V thi so nhi pha n 0
ngo ra d e m tang len 1. Cuoi c ung, hang 16 chi ra khi dien ap circ dai 3 V du’Oc dat 
vao ng o v ao cua A D C thi ngo ra co so nhi pha n la 1111. Bang sir that 10.3 cua 
A D C la dao ngirpc cua Bang sir that 10.1 cua D A C . C a c ng o v a o va ngo ra du’Oc 
dao ngirpc. B a n g sir that cua A D C don gian, tuy nhien c a c m a c h dien tir thirc hien 
bang sir that nay lai co phan phtic tap hon m a c h dien tir c ua D A C . S o do khoi cua 
mot loai A D C du’Oc trinh b a y tren Hinh 10.8. A D C co mot bo so ap, mot co’ ng A N D , 
mot bo d e m B C D va mot D A C .

NGO VAO ANALOG

NGO RA NHI PHAN

Hinh 10.7: S o do khoi A D C

Die n ap analog du’Oc dat v a o phan trai cua m a c h tren Hinh 10.8. B o so s anh 
kiem tra dien ap de n tir D A C . N e u dien ap ng o v a o a na l og tai A I6n hon dien ap 
ngo vao tai B cua bo so sanh, x ung nhip den bo d e m B C D lam bo d e m thirc hien 
de m tang len mot birdc d e m. B o d e m d e m tang c ho de n khi dien ap pha n hoi tir 
D A C lam c ho dien ap ng o v a o anal og I6n hon. L u c do, bo s o s a n h lam c ho bo d e m 
ngirng de m tang. G i a du, dien ap analog ng o v a o la 2 V . T h e o Bang 10.3 bo d e m 
nhi phan tang len de n 1010 trirdc khi no ngting d e m tang. B o d e m du’Oc p h u c hoi 
ve so nhi phan 0000 va bo d e m bat dau d e m lai.

T a hay x e m xet A D C tren Hinh 10.8. Gia du rang c o mtic logic 1 tai di em X d 
ngo ra cua bo so sanh. C u n g gia du rang bo d e m B C D c o so nhi p h a n 0 0 0 0 v a co

dien ap 0,55 V dat va o ng o v a o analog. Mtic logic 1 tai X c h o p h e p c o n g A N D hoat 
dong va x ung nhip dau tien xuat hien tai ngo v a o C L K c u a bo d e m B C D . Bo d e m 
de m tien len so nhi phan 00 01. S o 0001 du’Oc the hien tren ng o ra nhi p ha n va 
c ung du’Oc dira v a o D A C .

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NG<3 RA NHI PHAN B A N G 1 0 3 B A N G S i/ T *->T C b A A D C
Nq n v i c

Kqg r » nhj ?h£n

V6n 4* 2d

- . ft ■A

H A N G 1 OJO 0 0 0

H A N G 2 0 2 O 0 0 1

H A N G 3 0 .4 0 0 1 0

H A N G 4 0 .6 0 0 1 1

H A N G 5 0 1 0 0

H A N G 6 w 0 1 0 1

H A N G 7 12 0 1 1 0

H A N G 6> 1.4 0 1 1 1

H A N G 9 1.6 1 0 0 0

H A N G 10 1 0 0 1

H A N G 11 2 jO 1 0 1 0

H A N G 12 2 2 1 0 1 1

H A N G 13 2.4 1 1 0 0

H A N G 14 2 .6 1 1 0 1

H A N G 15 2 JS> 1 1 1 0

H A N G 16 3 .0 1 1 1 1

Hinh 10.8: S d d o khoi cua A D C loai co bo d e m B C D

T h e o Bang 10.1 ng o ra c u a so nhi phan 0 0 1 0 cho dien ap la 0, 4 V. D i e n ap 
0,4 V du’Oc phan hoi v e c ha n B c ua bo ao sanh. B o so s a n h lai kiem tra v a so sanh 
dien ap cha n B vOi chan A : dien ap c ha n A v a n Idn hon dien a p 6 c h a n B ( 0 , 55 V 
so v6i 0,4 V ) . B o so s an h lai c h o ra mot mu’c logic 1. M u ’c logic 1 lai kich hoat cong 
A N D , de dira mot x ung nhip tiep theo tdi bo d e m B C D . B o d e m thu’c hien d e m tang 
mot birdc d e m va ngo ra nhj p h a n co so nhj p h a n la 0 0 0 1 1 . S o nhj p h a n n a y cGng 
du’Oc dira den D A C .

T h e o Bang 10.1, n g o v a o 0011 c ho dien a p n g o ra la 0 ,6 V . D i e n ap 0, 6 V lai 
du’Oc phan hoi v e chan B c ua bo so sanh. B o so s a n h lai kiem tra v a so s a n h dien 
ap cua c han B vdi c h an A , Ian da u tien, ng o v a o c h a n B c d di en ap Idn h o n dien ap 
n g o va o cha n A . B o so s an h c h o mu’c logic 0 6 n g o ra, m u ’c logic n a y l am c h o cong

A N D khong hoat d on g , do do, x ung nhjp kho ng d e n du’Oc bo d e m . B o d e m B C D 
dirng lai 0 so nhj p ha n 0 0 1 1 . L u c nay, so nhj p ha n 0011 b a n g 0 , 5 5 V . Q u a n sat 
hang 4 Bang 10.3 ta thay 0, 6 V tirong Crng vdi n g o ra nhj p h a n la 0 0 11 . A D C da 
thirc hien viec c h u y e n doi theo d u n g Bang sir that.

N e u dien ap ng o v a o an al og la 1,2 V theo Bang 10.3 n g o ra nhi p h a n se la 
0 1 1 0. Bo d e m se d e m tir 0 0 0 0 len den 0 1 1 0 trirdc khi bi bo so s a n h lam dirng qua 
trinh d e m. C h u y rang c an mo t thdi gian de c h u y e n doi dien ap a n a lo g t ha nh so nhj 
phan. T u y nhien, hau het c a c trirdng hop, x u n g nhjp hoat d o n g du n h a n h de qua 
trinh c h u ye n doi dien ra k h o n g cd sir co nao.

10.6. BO SO AP

O phan trirdc ta cd d u n g bo so ap. T a da t ha y bo s o s a n h lam n h i e m vu so 
s a n h 2 dien ap va bao c ho ta biet mot trong hai dien ap, dien ap na o Idn hon. 
Hinh 10.9 la so d o khoi c o ba n c ua mot bo so sanh. N e u dien ap tai n g o v a o A Idn 
hon dien ap ng o v a o B bo so s a n h c ho mu’c logic 1 d ng o ra. N e u dien a p 6 n g o

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v a o B Idn hon 6 ngo vao A, ngo ra co mu’c logic 0. Ket qua co the tom tat la A > B, 
ngo ra co mu’c 1; B > A, ngo ra co mu’c 0. T h u ’c chat bo so s a n h la mot O P - A M P . 
Hinh 10.10a gidi thieu mach bo so sanh. C h u y rang, dien ap 1,5 V dat va o ng o 
vao A va dien ap 0 V dat vao ngo v a o B. N g o ra cua vo n met chi thi dien ap 3,5 V 
hoac mu’c logic 1. Hinh 10.10b chi thi dien ap ng o va o B da du’Oc tang len de n 2,0 
V. Dien ap 0 ngo vao A v i n giu’ gia tri 1,5 V . N hu ” vay, dien ap ngo v a o B I6n hon 
dien ap ngo vao A. N g o ra cua bo so s anh co mtic dien ap la 0 V (thirc chat dien 
ap vao khoang - 0, 6 V ) hoac mtic logic 0.

Hinh 10.9:Scf do khoi Hinh 10.10: Mach cua bo so ap a) Dien ap tai A Idn hon tai B

bo so ap b) Dien ap tai B Idn hdn tai A

Bo so sanh 0 A D C tren Hinh 10.8 lam viec hoan toan giong nhi/ bo so s anh 
vtia trinh bay 6 tren. D i o d Z e n e trong bo so s anh tren Hinh 10.10 giCr c ho dien ap 
ngo ra on dinh v ao kho an g +3 ,5 V . K h o n g co diod Z e n e , dien ap ng o ra v a o 
khoang +9 V v a - 9 V. D ie n ap +3 ,5 V rat thi'ch hpp vdi c a c I C T T L .

10.7. LOAI ADC KHAC

6 m u c 10.5 ta da x e m xet A D C co bo d e m x ung rang cira. Ngoai loai A D C ke 
tren, ngudi ta con sir d u ng mot so loai A D C khac. T r o n g m u c na y ta x e m xet hai 
loai A D C .

A D C s ud n d o c du’Oc gidi thieu tren Hinh 10.11. A D C n a y hoat d o n g giong 
nhir A D C co bo d e m x ung rang cua tren Hinh 10.8. B o tao x ung rang cua 6 p ha n 
trai cua Hinh 10.11 la phan m a c h phu mdi. B o tao x u n g tao ra c a c x ung rang cira, 
dupe the hien tren Hinh 10.12a.

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N06 RA CUA

S>6 ADC

Hinh 10.11: S d d o khoi cua A D C loai x u n g rang cira

Xung nhip den bo dem Q H D m m Xung nhip den bo de mn n n n n n n n n n n n

€>oc so nhi p h in 0011 0011 €>oc so nhi phan 0110 0110

a) b)

Hinh 10.12: C a c d a n g x ung cua A D C loai dien ap rang cira

a) V6i dien ap dat v a o la 3 V b) Vdi loai dien ap dat v a o la 6V

Hinh 10.12b la mot vi du khac. D i e n ap n g o v a o d e n A D C la loai x u n g rang 
cira co dien ap la 6V . D i e n ap rang cira bat da u tang tu* trai s a n g phai . N g o ra cua 
bo so sanh c ho mu’c logic 1 vi dien ap ng o ra tai d i e m ' A Idn hon dien ap c u a bo tao 
x ung rang cira tai di em B. B o d e m tiep tuc d e m tien. T a i di em Z dien ap tren sirdn 
d o c cua x ung rang cira cua bo tao x ung Idn hon dien ap n g o v a o V jn T a i d i e m nay 
ng o ra bo so s an h c ho mtic logic 0. MCrc logic n a y lam c h o c o n g A N D k h o n g hoat 
do ng , x ung nhip, d o do, kho ng den du’Oc bo d e m . B o d e m thirc hien d e m de n 01 10 
thi ngirng d e m. S o nhi p ha n 0 1 1 0 tirong ting vdi dien ap n g o v a o a n al o g la 6 V .

D ieu kho khan doi vdi A D C loai x ung rang cira la thdi gian doi hoi hoi dai de 
bo d e m thirc hien d e m tien de n c a c dien ap c a o hon. V i du, neu ng o ra nhi p h a n cd 
tam vi tri nhi pha n, bo d e m phai d e m tien len de n 255. D e k ha c p h u c nhi rp c di em 
nay, ta d u ng loai A D C g a n d u n g lien tuc.

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Hinh 10.13: S d do khoi cua A D C loai g an d u n g lien tuc

S o do khoi cua A D C loai g a n d u ng lien tuc du’Oc trinh ba y tren Hinh 10.13.
A D C g o m co bo so ap, bo D A C va mot khoi logic moi du’Oc goi la "khoi logic g an 
dung lien tuc".

Gia du, ta dat dien ap 7 V v ao ngo v a o analog. D a u tien, A D C loai g a n d u n g 
lien tuc doan dinh dien ap ng o va o analog, no g an c ho M S B mu’c logic 1. C o n g 
viec nay du’Oc trinh b a y 6 Butfc 1 tren Hinh 10.14, du’Oc "khoi logic g a n d u n g lien 
tuc" thirc hien. Ket qua g an so M S B nay (vi du la 10 00) du’Oc D A C pha n hoi ve bo 
so sanh. B o so s an h tra Idi cau hoi (6 Birdc 2, Hinh 10.14) so nhi p h a n vira g a n la 
1000 tirong Crng vdi dien ap Idn hdn h oac nho hdn dien ap ng o v a o ? T r o n g trirdng 
hdp nay bo so s an h tra Idi la dien ap Idn hdn dien ap ng o vao. S a u do, "khoi logic 
gan dung lien tuc" thirc hien nhiem vu 6 Birdc 3, Hinh 10.14, vi tri nhi p ha n 8s

(mtic logic 1) du’Oc xoa va thay the ba ng mu’c logic 0. Ket q ua x ac lap cho vi tri nhi 
phan 4s la 0 1 0 0 du’Oc c h u y e n tdi D A C va pha n hoi v e bo so sanh. B o so s a n h tra 
Idi tiep tuc cau hoi v e ket qua x ac lap c ho vi tri nhi p h a n 4 s mdi n a y du’Oc thirc hien 
d B ud c 4, Hinh 10.14: m U c logic 0 1 0 0 tuong Ung vdi dien ap Idn hon h ay nh o hon 
dien ap ngo v a o ? C a u tra Idi la nho hon. T i e p theo, "khoi logic g an d u n g lien tuc" 
thuc hien c on g vi ec 6 B u d c 5, Hinh 10.14. Vi tri nhi p h a n 2s d u p e x ac lap mtic 
logic 1, cho ket q u a x ac lap Ian na y la 0110. S o nhi p h a n n a y du p e c h u y e n tdi D A C 
va tuong ting vdi 0 1 1 0 la mtic dien ap d upe p h a n hoi ve bo so sanh. B o so s a n h 
tra Idi 6 B u d c 6: so nhj phan 0 1 1 0 tuong ting vdi dien ap Idn hon h ay nh o hon dien 
ap ngo v a o ? C a u tra Idi la nho hon. "Khoi logic g a n d u n g lien tuc" t hu c hien c o n g 
viec d B u d c 7, vi tri nhi phan 1s d upe x ac lap mtic logic 1. Ket qua cuoi c u n g du’Oc 
xac dinh bang so nhj p ha n 01 11. S o nhi pha n na y tuong ting vdi dien ap dat v a o 
ngo vao cua A D C la 7V.

C h u y rang Hinh 10.14 la so do khoi dupe "khoi logic g an d u n g lien tuc" t hu c 
hien. C a c cau hoi dupe bo so s anh tra Idi. C u n g can chu y the m la c a c c o n g vi ec 
do "khoi logic g a n d u n g lien tuc" thuc hien phu t hu oc v a o cau tra Idi trudc do: mtic 
dien ap Idn hon h a y nho hon mtic dien ap dat v a o n g o v a o A D C ( x e m b u d c 3 v a 
budc 5). ITu d i e m c u a A D C g an d u ng lien tuc la ton it thdi gian x ac lap mti c logic 
de cd dupe c au tra Idi. Q u a trinh digital hoa d i i n ra n h a n h hon. A D C g a n d u n g lien

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Bl/CTC 1

BI/OC 2

Bl/OC 3

Bl/OC 4

Hinh 10.14: Lull do hoat dong cua ADC loai gan dung lien tuc

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N g a y na y co tren 400 loai A D C khac nh au du’Oc su' d u ng rong rai. T u y nhien 
cac A D C co nh u n g yeu cau ky thuat c h u ng nhu1 sau:

- Cac loai ngo ra

Noi c hung, c a c loai A D C du’Oc phan loai 6 ngo ra h oa c la so nhi pha n h o a c la 
so thap phan. C a c A D C vdi ng o ra la so thap phan noi c h u n g du’Oc sir d u ng rong 
rai, ch§ng han nhir la von ke so. A D C vdi c a c ngo ra la so nhi p ha n cd tir 4 de n 16 
ngo ra. A D C vdi n gd ra nhi phan la c a c ngd v a o c ua c a c bo kien da n de n c a c he 
thong dira tren co sd vi xir ly, ta hay goi la A D C loai |jP.

- Do phan giai

D o phan giai cua mot A D C la so bit 6 n gd ra doi vdi khoi nhi phan. D oi vdi 
loai A D C ngd ra thap phan do phan giai 'a so con so d o c ra (vi du: 3 V2 h oa c 4 V2). 
Loai A D C vdi c a c ng d ra nhi phan cd do pha n giai la 4, 6, 8, 10, 12, 14 va 16 bit. 
C a c sai Idi cd the x a y ra la do sir d u ng c a c birdc nhi p ha n rdi rac. Vi du, mdi birdc

d A D C 4 bit bang 1/15 dien ap n gd v a o ( 24 - 1 = 15). D i e u na y c ho d o phan giai la 
6 , 7 % (1/15 x 100 = 6 , 7 % ) . T u y nhien, d A D C 8 bit c a c gia so nho hon. Mot bo A D C

8 bit cd 255 birdc rdi rac hoa, tirong Crng vdi d o phan giai la 0 , 3 9 % (1/255 x 100 = 
0 , 3 9 % ) . A D C 8 bit cd do phan giai tot hon h o a c "chinh x ac hon" A D C 4 bit.

- Do chinh xac

D o phan giai c ua A D C cd the cho la sai Idi digital von cd vi c a c IC sir d u n g 
cac birdc rdi rac 6 n gd ra. C a c ly do khac g a y ra sai Idi cua A D C cd the la c a c linh 
kien analog, c h S n g han nhir bo so sanh h o a c sir d u n g c a c dien trd cd tri so khong

chinh xac trong m a n g dien trd. Si r khong sai lech khi c h u y e n do7i tir tin hieu anal og 
sang tin hieu digital c ua A D C dirpc goi la do chinh x a c cua I C A D C . D o chinh x ac 
cua IC A D C cd ng d ra nhi phan trong p h a m vi tir ± V2 LSB de n ±2 LSB. D o chinh 
xac cua IC A D C cd n g d ra thap p ha n trong p h a m vi tir 0,01 de n 0, 05 % .

- Thdi gian chuyen doi

T h d i gian c h u y e n doi la mot trong c a c d a c tinh q u a n trong cua A D C . D o la 
thdi gian can thiet de A D C c h u y e n doi dien ap ngd v a o a n al og s an g c a c n g d ra so 
lieu nhi phan h o a c thap phan. C a c thdi gian c h u y e n doi dien hinh n a m trong p h a m 
vi tir 0,05 den 1 0 0 0 0 0 ps doi vdi cac IC cd c a c n g d ra nhi phan. C a c thdi gian 
chuyen doi doi vdi c a c A D C cd c a c ngd ra thap p ha n trong p h a m vi kho ng q ua tir 
200 den 400 ms.

- Cac dac tinh ky thuat khac

Bon d a c tinh ky thuat c h u ng cua A D C la dien ap c u n g cap, mu’c logic c ua n gd 
ra, dien ap ngd v a o v a c on g suat tieu tan circ dai. D i e n ap n g u o n c u n g c a p noi 
c hung la +5V. T u y nhien, mot vai IC A D C hoat d o n g vdi dien ap n g u o n c u n g cap tir 
+ 5 V den + 1 5 V . C a c mu’c logic n gd ra h oac la IC T T L h o a c la IC C M O S h o a c la 3 
trang thai. P h a m vi dien ap ngd va o noi c h u n g v a o k h o a n g 5V. C o n g suat tieu tan 
circ dai doi vdi c a c I C A D C trong p h a m vi tu” 15 de n 3 0 0 0 m W .

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10.9. IC A D C TH l/ O N G PHAM

I C A D C thirdng p h a m dtipc de c a p de n trong p h a n nay. Hinh 1015a liet ke 
do c ha n IC A D C 0804 loai 8 bit. B a n g trong Hinh 10.15 liet ke ten c a c c h an va cac 
ti'nh n a ng c ua c ac chan cua IC nay. A D C 0 8 0 4 du’Oc thiet ke de giao dien trtic tiep 
vdi c a c |iP 8080, 80 85 h o ac Z 8 0. Mot so c h an cua IC A D C 0 8 0 4 ti/dng ting vdi 
c h a n cua c a c juP thong d u ng . Vi du, A D C 0 8 0 4 sti d u n g R D , W R v a I N T R ti/dng

ting vdi R D , W R va I N T R tren jxP 8 0 85 . IC A D C 0 8 0 4 c u n g c o the giao dien vdi 
c a c |uP 8 bit khac c h i n g han nhir 65 02. N g o v a o c h a n dieu khien C S cua A D C 
0 8 0 4 nhan tin hieu c hon chip tti m a c h giai ma dia chi c ua j.iP.

A D C 0 8 0 4 la bo c h u y e n doi tti anal og s an g digital 8 bit, g a n d u n g lien tuc, 
loai C M O S . IC na y co c a c ng o ra 3 trang thai de s a o c ho no giao dien trtic tiep vdi 
B U S sd lieu cua he thong sti d u n g jaP. IC A D C 0 8 0 4 co c a c n g o ra nhi pha n va co 
d a c ti'nh la thdi gian c h u y e n doi n ga n, chi k h o an g 100 |l is. C a c n g o v a o va ngo ra 
c ua I C nay deu tirong hdp d u p e vdi c a c IC T T L va I C M O S . I C n a y co mot bo tao 
d a o do n g x ung nhip. IC A D C 0 8 0 4 k h o n g . c a n de n c a c linh kien R v a C be n ngoai 
phu trd de hoat dong. I C na y hoat d o n g vdi n g u o n +5 V tieu c h u a n va co the giai 
m a c a c dien ap anal og ng o v a o trong p h a m vi tti 0 V de n +5 V .

A D C 0 8 0 4 du’Oc sti d u n g nhir trong m a c h gidi thieu tren Hinh 10.16. Chtic 
n a n g cua m a c h la de ma hoa c a c dien ap khac nhau gitia V in( + ) v a V in( - ) so sanh 
vdi dien ap tha m khao ( 5 , 1 2 V trong vi du na y ) ttiOng ting vdi gia tri c u a so nhi 
pha n. Vi du, d o phan giai c ua I C A D C 0 8 0 4 la 8 bit h o a c 0 , 3 9 % . D i e u n a y co nghia 
la ( 5, 12 x 0 , 3 9 % = 0, 02 V ) khi tang moi 0 . 0 2 V dien ap tai c a c n g o v a o anal og thi 
bo d e m nhi phan d e m tang t he m 1. " C h u y e n m a c h khoi d o n g " tren Hinh 10.16 luc 
da u ddng , sau do m d de khdi d o n g c ho A D C c h a y tu1 do. G o i la " c h a y tti do" la vi 
no c h u y e n doi lien tuc ng o v a o a nal og s a n g c a c n g o ra digital. C h u y e n m a c h khdi 
d o n g can dtipc de ho mot khi A D C d a n g hoat d on g. N g o v a o W R co the dtipc xem 
la ng o v a o x ung nhip vdi n g o ra ngat I N T R phat x u n g de n n g o v a o W R tai cuoi moi 
c h u y e n doi analog-digital. S t i dich c h u y e n tti L len H c u a tin hieu tai n g o v a o W R 
khdi dau c ho mot q ua trinh c h u y e n doi. Khi viec c h u y e n doi ket thuc. S t i hien thi so 
nhi pha n dtipc hieu chinh kip thdi v a ng o ra cua c h a n ngat I N T R phat ra mot xung 
a m . X u n g a m n a y dtipc pha n hoi ve d o n g ho c ua n g o v a o W R va no b i t da u mot 
c h u y e n doi A/D khac. M a c h tren Hinh 10.16 thtic hien k h o a n g 5 0 0 0 de n 10000 
c h u y e n doi trong mot giay. T o e d o c h u y e n doi cua A D C 0 8 0 4 c a o vi no sti d u n g ky 
thuat g an d u n g lien tuc trong q ua trinh c h u y e n doi.

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DBo (LSB)

Vcc CLK R | 0 31 Df32 DB3 DE34 DF35 DB6 DB7 (MSB)

2 0 19 16 17 16 15 14 13 12 11

IC A D C 0 & 0 4

1 2 3

CS RD \NR

CLK IN

5 6 7 & 9 10

INTR Vin (+) Vin (-) A GND V ref/2 D GND

BANG TEN VA CHUfC N AN G CAC CHAN IC ADC 0804

S O CHAN K f HIEU NG O RA/NGO VAO

HO AC NG U O N C UNG C AP MOTA

1 CS Ngo vao Day chon chip til chan dieu khien cua pP

2 RD Ngo vao Day doc tC/ chan dieu khien cua pP

3 WR Ngo vao Day viet chan dieu khien cua pP

4 CLK IN Ngo vao Xung nhip

5 INTR Ngo vao Day ngat di den ngo vao ngat cua pP

6 v in(+) Ngo vao Dien ap analog (+)

7 Vin (-) Ngo vao Dien ap analog (-)

8 A GND Nguon cung cap Tiep dat cua analog

9 ref/2 Ngo vao Dien ap tham khao (+)

10 D GND Nguon cung cap Tiep dat cua digital

11 d b7 Ngo ra Ngo ra sd lieu MSB

12 d b6 Ngo ra Ngo ra so lieu

13 d b5 Ngo ra Ngo ra sd lieu

14 d b4 Ngo ra Ngo ra sd lieu

15 d b3 Ngo ra Ngo ra sd lieu

16 d b2 Ngo ra Ngo ra sd lieu

17 DB, Ngo ra Ngo ra sd lieu

18 DB0 Ngo ra Ngo ra sd lieu LSB

1 9 CLK R Ngo vao Dien trd ket noi ngoai cua dong ho
20 Vcc (hoac

tham khao)

Nguon cung cap Dien ap nguon cung cap + 5 V va dien ap

tham khao sd cap

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NGO VAO ANALOG

1 0 k

ôã-KHdl 'AO

7

2

i R1

10k

F

Cl

150 pF

n g5 r a d i g i t a l

129s 6 4 ^ 3 2 s 16- 8 s 4 s 2s 1s

O O O O O Q Q Q

+5V

Vcc

Vin (+) DB7

DB6
Vin (-)

ADC0604

DB5
DB4
36 CHUYEN & 6 \ ADC & BIT DB3

— DB2

WR
CLK R

DBO

CLK IN INTR

A GND DGND CS RD

Hinh 10.16: S o do noi d a y c ua m a c h sCr d u n g I C A D C 0 8 0 4 C M O S
cua Bo c h u y e n doi A/D

D i e n tro R: va tu dien C , du’Oc ket noi d e n c a c ng o v a o c ua C L K R va C L K IN 
cua IC A D C 0 8 0 4 tren Hinh 10.16 tao nen d o n g ho ben trong I C d e hoat dong. 
C a c ngo ra so lieu ( D B 7 - D B 0 ) dieu khien L E D hien thi so nhi pha n. C a c ng o ra so 
lieu la c a c ngo ra 3 trang thai vdi mu’c H tich circ.

N g o ra nhi phan tren Hinh 10.16 la ba o nhieu neu dien ap n g o v a o analog 
ba ng 1 , 0 V ? N h d lai rang cu1 mOi 0, 02 V b a n g d e m tang nhi pha n len 1 birdc. Chia
1 , 0 V cho 0 , 0 2 V b a ng 50 (thap p ha n) . C h u y e n doi so thap p h a n 50 s a n g so nhi 
pha n ba ng 0 0 1 1 0 0 1 0 . B o chi thi 6 n g o ra se hien thi so nhi p h a n 0 0 1 1 0 0 1 0 
( L L H H L L H L )./.

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CAU HOI VA BAI TAP

10.1*. C o the x e m bo ma hoa c h u ye n doi tin hieu anal og sang digital la bo 
chuyen doi gi?

10.2*. D A C g o m nhCrng thanh phan chu yeu na o?

10.3*. D o tang i'ch dien ap cua O P - A M P trong Hinh 10.3b du’Oc xac dinh
bang cach chia tri so dien tro nao cho tri so dien trd nao trong 2 dien trd: R ,n va 
R f? Ti'nh do tang ich dien ap cua O P - A M P neu co R in = 1 k£> va R f = 20 k£>.

10.4*. T h e o Hinh 10.3b, vOi dien ap ngo va o la 0,2 V, hoi dien ap ngo ra V0

bang bao nhieu?

10.5*. Ti'nh do tang i'ch cua O P - A M P tren Hinh 10.5, khi chi co c h u y e n mach
C (vi tri nhi phan 4s ) du’Oc dat tai mu’c logic 1?

10.6*. Vdi dCr lieu tren hinh 10.5 va ket qua cua bai tap 10.5, ti'nh dien ap ngo 
ra V 0 cua D A C ?

10.7*. C o ma c h tren hinh 10.6 va ba ng 10.2, khi chi co mot trong c a c c h u ye n 
mach A, B, C va D d vi tri logic 1 thi do tang i'ch cua O P - A M P la nh d nhat?

10.8. T r e n hinh 10.8, khi dien X co mu’c logic ba ng bao nhieu de bo d e m thu’c 
hien de m tien mot birdc d e m khi x ung nhip xuat hien?

10.9*. A D C thong dich:

a) T i n hieu d ngo v ao dirdi d a n g nao ?
b) T h a n h tin hieu d ngo ra dirdi d a n g nao?

10.10*. T h e o ba ng 10.3, neu dien ap analog d ng o v a o la 1 V , thi ng o ra nhi 
phan cd tri so la ba o nhieu?

10.11*. T h e o hinh 10.8, khi dien ap tai diem B nh d hdn dien ap tai di em A , thi 
luc do ngo ra cua bo so s anh tai di em X cd mu’c logic b a n g ba o nhieu? Vdi mu’c 
logic na y xung nhip bi co’ ng A N D ng an can h ay c ho q u a ?

10.12*. S d do khoi bo so s anh tren hinh 10.8 so s a n h hai dien ap, hai so nhi 
phan hay hai so thap pha n ?

10.13*. C d m a c h nhu” tren hinh 10.10, khi dien ap tai c o n g B tang da n va den

luc nao do trd nen Idn hdn dien ap tai di em A. Hoi ng o ra cua O P - A M P c h u y e n tir

mu’c logic nao s a n g mu’c logic nao?

10.14*. T r e n hinh 10.13, neu dien ap ngo v a o ( V in) la 2 V va dien ap rang 
ci/a la 0 V. Hoi ngo ra cua bo so ap d mCrc logic n a o ? va luc do, c o n g A N D nga n 
can hay cho x ung nhip c h u y e n q u a ?

10.15*. C d hai bo A D C 8 bit va 4 bit, hoi bo A D C nao c h o do p h a n giai Idn
hon?

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CLK IC 
7 4 9 0
A B C D

Vcc

3 C
I

GND

+ Vcc

4LS4 ;
A3 g
A2 f
Al e

AO d
c
b
a
test
RBI RBO

-a b c d e f q

i m i 11
LED
7 THANH

7 1 2 ___ 6

- c i

IC 7446

13

LE Da

12 11 10 9 1 5 " 14

tJC
l:

T f T T

■eg

E]Dc LflDeT7E Dg

LED 7 THANH

a)

T

CLK

Hien t h i b£ng LED
7 th a n h a n fit chung

1 I I I I

)(loai 150 Om)|

-L l

IC 7 4 4 7
Bo giai ma
LED 7 thanh

A B C D

Bo dem th a p ph3n

b)

Hinh 11.1a: I C 7 4 4 8 dieu khien L E D 7 thanh catot c h u n g
b) I C 7 4 4 7 dieu khien L E D 7 thanh anot c h u n g

Hinh 11.1a) la m a c h hien thi d u n g L E D 7 thanh co catot c h u n g , m a c h nay 
thirdng dirpc hien thi c a c con so thap phan don. B o giai m a B C D - 7 t ha nh d u n g IC 
7 4 4 8 tirong hpp vdi L E D catot c h u n g thirdng de dieu khien bo chi thi. I C 7 4 9 0 la bo 
d e m thap phan cd bon ng o ra flip flop la bon n gd v a o IC 7 4 4 8 . C a c n g d ra cua IC 
7 4 4 8 cd mu’c H tich circ, nd dieu khien bo kh u ec h dai d e m vi c a c tirong hpp ddng 
n g d ra qua nho khong du de dieu khien trifc tiep c a c L E D . B a y tranzito np n hoat 
d o n g nhir c h u y e n m a c h de ket noi + V c c vdi tha nh L E D . Khi n g d ra c ua 7 4 4 8 0 mu’c
H, tranzito dan, m a c h c u n g c a p n g u o n c ho thanh L E D . M a c h tirong difcJng de L E D 
phat sang du’Oc gidi thieu tren Hinh 11.2. Khi n gd ra c ua 7 4 4 8 0 mu’c L, tranzito 
ngirng dan, luc do khong cd tha nh L E D nao phat sang.

+ V cc

+ v

Hinh 11.2: Mach tirong duting LED thanh du’Oc phat sang

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M a c h trong Hinh 11.1b IC 7447 la m ac h anot c hu ng . Bo chi thi vdi L E D 7 
thanh thudng dtipc sir du ng de hien thi con so thap pha n don. Bo giai m a B C D - 
L E D 7 thanh thudng d u ng IC 7447 de dieu khien bo hien thi. Bon ngd v a o cua IC

74 4 7 la 4 ngd ra cua bo d e m thap phan 7490. C a n nhd rang IC 7 4 47 c ho c a c ngd 
ra L ti'ch ctic.

Neu ngd ra cua IC 7 447 6 m Uc L, thanh L E D se phat sang. Gia du, dien ap 
dat tren mot thanh L E D la 2 V, d d ng c h ay qua thanh L E D la: (5 - 2)/150 = 20 m A .

Ddng to’ng de hien thi so 8 la 20 m A x 7 = 140 mA . IC 7 44 7 can tieu thu d d n g la 
64 mA, n h u v a y de hien thi mot s d thap pha n don can d d n g toi da v ao khoang 
200 mA. Mot ma c h hien thi 4 con so thap pha n can 4 m a c h n h u trinh b a y tren 
Hinh 11.1, cd nghTa la can so d d ng to’ng c on g la 4 x 2 0 0 m A = 80 0 m A ; neu hien 
thi 6 so thap phan thi can d d n g la 1200 mA . R o rang rang d d n g tieu thu n h u v a y la 
qua Idn doi vdi c ac thiet bi nho. T u y nhien, ta cd the giam dd ng tieu thu hien thi 
bang cach sir du n g ky thuat da hpp.

V e co ban, ky thuat da hpp du pe thuc hien ba ng c ach dat d d ng v ao mdi con 
so hien thi cua c a c x ung lap ngan. N e u toe do lap lai cua x ung du Idn thi mat ngtidi 
nhin thay anh s ang o’ n dinh ma khong nhan ra L E D phat s ang nhap n h a y (vi du, 
kho ma nhan ra anh s a n g nhap n h a y cua bo hien thi khi nd phat sang vdi toe do 
50 Hz). C o n so thap p h a n don tren Hinh 11.3a d up e dien ap +5 V dat v ao thong 
qua mot tranzito pnp d u p e d u ng n h u mot c h u y e n m a c h . Khi ngd v ao ctic B a - d o 
cua tranzito d m U c H, tranzito ngting dan, bo hien thi cd d d n g ba ng 0. Khi n g d v ao 
chan B a - d o cua tranzito m d m Uc L, tranzito da n, va con so dtipc hien thi. N e u 
dang xung tren Hinh 11.3b dtipc dat v a o chan B a - d o c ua tranzito, tranzito se d i n 
va L E D se hien thi mot con so trong thdi gian 1 s trong mdi 4 ms. C h o du so khong 
dtipc hien thi trong thdi gian 3 ms, thi con so v i n dtipc Itiu trong m l t ta n h u khi cd 
ddng lien tuc c h a y qua L E D . Vi L E D dtipc phat s an g do x ung tac d o n g c U mdi 
4 ms, toe do lap lai la:

v lap = 1/0,004 = 2 5 0 H z .

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T h i r d n g ngirdi ta tiet kiem linh kien v a n a ng lupng c ua n g u o n c u n g c a p ne u 
4 ngo v a o A , B. C va D den bo giai m a L E D 7 thanh nhu1 tren H i n h 11 . 6 du’Oc p h a n 
kenh c u n g vdi dieu khien c a c c c n so. B o hien thi 4 con so nhir trong H i n h 1 1 . 8 sir 
d u ng 2 IC 54/74153 moi IC cd mot cap bo pha n kenh 4-1 d u n g cho 4 n g o ra cua 
moi I C 7 4 7 5 m a c noi tiep den moi bo giai ma 7 thanh don. T a x e m m a c h hoat 
do ng ra s a o ?

+ V c c .

O

1 ms

• 4 m s '

- 3 ms

b)

Hinh 11.5: L E D hien thi cd x u n g v a o c h an B a - d d c ua tranzito dieu khien

S o lieu ng o va o B C D du’Oc lull trir trong 4 chot loai D c ua I C 7 4 7 5 , c hung 
du’Oc ky hieu 1, 2 ,3 , 4. C h o t 1 liru trCr M S D v a chot 4 liru trir L S D . S o nhi p ha n 4 
bit cua M S D du’Oc ky hieu lA lB lc lD . Vi du, neu M S D = 7 , luc d o lA lB lc lD = 01 11 .

M6i I C 7 4 1 5 3 co hai bo phan kenh. B o n bo pha n kenh du’Oc ky hieu: A , B, C, 
D. C a c d a y c h on A va B cua 2 bo pha n kenh du’Oc ket noi s o n g s o n g va d o bo 
d e m da hop dieu khien ( d u n g n h u tren Hi nh 1 1 . 7 ) . Khi c a c ng o v a o d a y c h o n bang 
A B = 00, d a y so 1 cua mdi bo phan kenh se dtipc ket noi vdi n g o ra c u a bo phan 
kenh. N h u vay, c a c ng o ra cua bo p ha n kenh ket noi de n I C giai m a 7 4 4 7 se la cac 
d a y 1A 1B 1 C 1D, d a y la so nhi ph a n cua M S D . S o nhi p h a n d u p e IC 7 4 4 7 giai ma 
va dat v a o tat ca c a c L E D hien thi ket noi s o n g song. . T u y trong c u n g thdi gian 
nh u n g con so thU nhat d upe c h c n b a n g bo giai m a 7 4 1 5 5, nen M S D se d u p e hien 
thi tren L E D d ngoai c un g ben trai, con tat ca c a c L E D k ha c de u k h o n g phat sang.

D e n khi b o d e m da hpp d e m tien de n A B = 01, d a y so 2 c u a moi bo phan

kenh dupe c h on va s o nhi phan dupe ap dat de n 7 4 4 7 se la 2 A 2 B 2 C 2 D , d a y la 
M S D thU hai ( so ha n g tram). N g o ra dupe giai m a cua 7 7 4 4 7 lai dupe ap dat den 
tat ca c a c L E D hien thi dupe ket noi s o n g s on g, n h u n g c o n so thU 2 c o d a y d mUc 
L, nen c c n so "hang tram" b a y gid dupe phat s a n g ( c c n tat ca c a c s o k h a c khong 
phat s a n g tai thdi di em nay).

T u d n g tu, c c n so ha n g c h u c se d u p e hien thi khi c a c n g o v a o S e l e c t b a n g A B 
= 01, va so h a n g dd n vi d u p e hien thi khi c a c n g o v a o Se l ec t b a n g A B = 1 1. C h u y 
rang chi mot c on so d upe hien thi tai moi hdi di em va toe do lap V lap = 2 5 0 H z , nen 
k ho ng xuat hien n hap n h a y khi c on so d u p e hien thi. B o n con so d u p e ph a t s a n g 
n h u n g d d n g n g u o n c u n g cap de u n h u nh au c ho moi con so d on d u p e phat s a n g

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lien tuc. D o n g thdi, 6 mu’c do nao do m a c h da tiet kiem du’Oc chip. T i e t kiem linh 
kien lam tang con so thap phan du’Oc hien thi.

cic n qo v ac 3 C D t j

bo dem thap phjn

Hinh 11.6: L E D hien thi co x ung St robe dieu khien c a c chot de d u y tri hien thi

U llf
Oocv

.EO

a rc T

chong

y. Cor. e>o f

v, Oipn o o 3

. . . J--- r ~ i

. . I j ■ " i

n ___r i__r ~ " l

5 i i i j

i p : m - 7

</div>
(118)<div class='page_container' data-page=118>

K y thuat thirdng dirpc sir d u n g hien thi da hpp 4 con so thap p h a n nhu" tren 
Hinh 11.8 rat de m d rong cho hien thi nhieu hon 4 c on so thap pha n. D i e u can 
thiet la tang kich co cua bo d e m da hpp va thay the IC 7 4 1 5 3 b i n g I C co so ngo 
v a o nhieu hon.

Cac ngo vao &CD

Bo dsrr, da h e r

Hinh 11.8

T a t ca c a c m a c h trinh b a y 6 d a y dirpc pho du n g , nhirng c h u n g ta y thu’c dirpc 
rang tren thi trudng da co c a c chip LSI thuc hien da h p p rat tien Ipi tren c u n g mot 
chip, do la M M 7 4 C 9 2 5 , 926, 9 2 7 v a 928. IC M M 7 4 C 9 2 5 trinh b a y tren Hinh 11.9
la bo d e m 4 con so cd c a c bo dieu khien da hdp c a c n g o ra 7 thanh. C a c linh kien 
ngoai can co chi la c a c L E D 7 t hanh c ua bo hien thi va c a c dien trd ha n ddng.

T h u c te, bo d e m 4 con so c u n g n a m tren mot chip. Mot x u n g d u o n g d ng o vao 
R e s e t p h u c hoi bo d e m 4 bit va sau do bo d e m se d e m tien mot khi c o s ud n am 
x u n g nhip xuat hien. X u n g a m tren Latch E n a b l e luc d o se chot c a c noi d u n g cua 
bo d e m trong c a c chot 4 bit. B o n so da dupe luu tru s au d c dupe p h a n kenh, giai 
ma, va hien thi tren bo chi thi b a n g L E D 7 thanh la linh kien be n ngoai. S d d o dupe

gidi thieu tren Hinh 11.9b vdi bo hien thi catot chu ng .

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+Vcc

Hinh 11.9: B o d e m 4 con so co c a c bo dieu khien da hop c a c n g o ra 7 thanh

11.2. BO DEM TAN

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G ia du rang tin hieu n g o v a o can x ac dinh tan so tren Hinh 11.10 la mot x ung 
v u o n g co tan so 7, 50 k Hz . Bo chi thi se hien thi bao nhieu neu thdi gian kich hoat 
co’ng la t = 0,1 s? La bao nhieu neu thdi gian kich hoat c on g la t = 1 s v a t = 10 s?

Ldi giai:

Khi t = 0,1 s, bo d e m se d e m tien len de n 7 5 0 0 ( c a c c h u y e n mu’c logic / 1 s) x
0 , 1 s = 750.

Khi t = 1 s, bo d e m se d e m 7 5 0 0 ( c a c c h u y e n mu’c logic / 1 s ) x 1 s = 7500.
Khi t = 10 s, bo d e m se hien thi 7 5 0 0 ( ( c a c c h u y e n mu’c logic / 1 s) x 10 s = 
75 0 0 0 . Doi vdi tri/dng hdp cuoi c u n g can cd bo d e m chi thi 5 con so thap phan.

Vi du:

Xung c a n x a c dinh

JUUliL

a) S o do khoi Bo d e m tan

- i P

0 0 7 5 0 - 1 1

—

O 1 >

}

D au ph a y th a p phan
<»!

0 7 5 0 0

D a u c 'r a y

7 S 0 0 0

t 0 i*1

" 1 _

■ r - 10 i

_ r
D au p h a y th a p chan

Ic)

b) Dich c h u y e n da u p h a y thap pha n 
Hinh 11.10

N hi r trong Hinh 11.4, ket q ua d e m du’Oc cua bo d e m luon luon la con so ti le 
v6i tan so can xac dinh, ti le d o la 10, 1 h o a c 1/10... Vi v ay , d e v a n d e du’Oc don 
gian ta can dira dau p h a y thap p ha n v a o giCra c a c c on so hien thi s ao c h o bo d e m 
chi thi tri/c tiep tan so can x ac dinh.

Hinh 11.10b chi ra dau p h a y thap ph a n can di c h u y e n trong bo hien thi 5 con 
so thap phan vi do rong c ua bo hien thi can phai di/dc tha y doi. P h a n hien thi 6

- ► B o d e m

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phia tren cua Hinh 11.10b, tin hieu can d u o c xac dinh tan so co noi du ng hien th 
nhan vdi 10, vi v a y dau phay thap phan c h u y e n sang phai mot vi tri. P h an hien th

6 giCra c ho true tiep gia tri tan so cua tin hieu can xac dinh. P h an hien thi du*6i c un g 
cua Hinh 11.10b, noi dung hien thi phai chia cho 10 d e d u o c tan so cua tin hieu 
can xac dinh, vi v a y dau phay thap phan da dich sang trai mot vi tri.

S o do logic tren Hinh 11.11 gidi thieu mot c ach x ay d u ng bo d e m tan so vdi 4 
con so thap phan. Khoi khuech dai tao dieu kien cho tin hieu ngo va o can xac dinh 
tan so sao cho ngo vao la tin hieu tuong hop vdi T T L - loat x ung d u o n g tU m Uc 0 V 
den +5 V D C . Khi di qua cong d e m , ngo va o se hoat d on g n h u la x ung nhip doi vdi

bo dem. Bo d e m co the d u o c x ay d u n g tU bo d e m thap pha n 4 con so khi sir du ng 
cac IC 54/74160 va bo d e m d u o c ket noi vdi bo hien thi L E D da hop n h u da trinh 
bay tren Hinh 11.8. H o a c bo d e m v a bo hien thi co the d o c ket hop trong c ung mot 
chip, vi du n h u IC M M 7 4 C 9 2 5 d u o c trinh b a y tren Hinh 11.9.

Bo chia g o m 6 bo d e m thap phan (vi du n h u c a c IC 54/74160) ket noi noi 
tiep. N g o va o cua bo chia la x ung v u o n g tan so 100 k H z de n tU bo da o d on g tao 
xung nhip, x ung v u o n g d u o c chia thanh 1 0 H z , 1 H z va 0,1 H z c a c x ung nay d u o c 
dung de tao tin hieu cua cong Enabl e.

Khi xung v uo n g 1 H z d u o c sir d u n g de dieu khien flip flop c o n g , ngo ra cua 
flip flop Q co x ung v u o n g la 0,5 H z . N g o ra Q co m U c H trong thdi khoang chinh 
xac 1 s va m Uc L c un g trong thdi k h o an g 1 s, va n h u v a y tin hieu nay d u o c sir 
dung lam tin hieu c on g Enabl e. C h u y rang, tan so 10 H z se tao c on g 0,1 s va tan 
so 0,1 H z tao c on g 10 s T a hay d u n g c a c d a n g x ung tren Hinh 11.11 de x em xet 
cac chUc nang cua mach.

Mot chu ky do d u o c bat dau khi flip flop cong d u o c c h u y e n trang thai doi lap
H, du oc ky hieu khdi dau tren true thdi gian. L u c nay, ng o v a o ( I N P U T ) qua c on g 
de m di den bo d e m . (gia du rang bo d e m bat da u d e m tU 0 0 0 0 ) . V a o cuoi thdi 
gian t cua co’ ng E n ab le, flip flop c o n g c h u y e n trang thai bu tU H x u o n g L, bo d e m 
ngUng d e m tien v a Q c h u y e n m U c logic a m lam tac d o n g d e n IC 74 12. T u o n g tu, 
Q chu ye n m Uc H x ung n a y da c h u y e n ket qua d e m d u o c v a o chot hien thi. C o thdi 
gian tre truyen dan c u c tieu la 30 ns khi c h u y e n q ua IC 74121 va sau do x ung a m 
Reset xuat hien 6 ngo ra X. D o tre n a y d a m bao rang ket q ua d e m c ua bo d e m tac 
dong va o bo chi thi trudc khi bo d e m d u o c p h u c hoi. X u n g p h u c hoi R e s e t tU IC 
74121 cd do rong 1 ps du de x ac lap n hd c a c linh kien R ca C cua bo dinh thdi. 
Pha n cuoi cua x ung Res et la phan cuoi c ua chu ky d o ludng d u o c ky hieu la ket
thuc tren true thdi gian.

Doi vdi co’ ng 1,0-s dau p h a y thap pha n n a m 6 n g a y ben phai con so hang 
don vi, bo d e m cd the d e m len de n 9999, vdi do chinh x a c la ±1 b u d c d e m ( cd 
nghia la do chi'nh xac dat 1 phan 104). Vdi c on g 10-s, da u p h a y thap pha n n a m 
giua con so h an g c h u c va hang don vi, va vdi c on g 0,1-s da u p h a y thap phan na m

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tio d e m m i n i

10 : »0 J i o SO

__ L l . ,

I I I

T

i c r

1 1 n 1 1 n

l _ f / _ / u / _ /

f f ir # VT/StT: n a n g ?j~uc d o r vi

r Cb'ri^ ie n

ir

Ci?ri<3
F N / . P - :
;• '

J ~ L

: r

N^Ê3 .-i30
<tac
ãô &o

m>m

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O

74 76
< v .

6 K

---f C T n r ■ P P
3 o ch ia ta n

100 k U i

3 o ta o

. J U U L

- f(? s 10 ? r» TO f • - in

xu na nhip

1 0 M>

M.-0 i t 1.- C h u y e n

'n a c h c o n g
Khoi tao

c , C o n g E N A 3 L E

n s t n o n e L

K e t t h u c Kh oi t a o K e t t h u c

= L L Z 3 ^ u I

— J < i — J i

— ______m t m i A ________ j v m w w w L j ___

t i — ... “ t i r

-r t f 7 —

L .

I

Hinh 11.11: B o d e m tan so co 4 con so thap pha n

B o tao x ung nhip du’Oc xac lap tai tan so 100 k H z va tan so n a y c h o d o chinh 
x a c tren thdi gian c on g E n a b l e b a n g 1 phan 104 Ung v6i c o n g 0 , 1 -s. N h u v ay , do 
chinh xac na y co kha n a ng tuong hop v6i bo d e m .

11.3. DO THOI GIAN (DO DO RONG XUNG)

V6i s u thay doi nho, bo d e m tan tren Hinh 11.10 co the c h u y e n doi thanh 
d u n g cu do thdi gian. S o do logic tren Hinh 11.12 mi nh hoa y t uo ng co ban 
t hu dn g d u o c sir d u n g de cau tao nen d u n g cu do chu ky c ua mot d a n g x u n g co 
chu ky nao do. D i e n ap can x ac dinh chu ky d u o c c h u y e n q ua bo k h u e c h dai de 
tao ra d a n g x ung co chu ky tuong hop vdi c a c m a c h T T L , sau do d u p e d u a d e n flip 
flop J K . N g o ra cua flip flop nay d u o c d u ng lam tin hieu c o n g E n a b l e , vi no o m U c

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cao ting vdi do rong xung t bang vdi chu ky cua dien ap ngo va o can xac dinh do 
rong xung. B o dao dong va bo chia tan c u n g cap chuoi xung du’Oc dira qua cong 
dem va du’Oc d u ng lam xung nhip cho bo d e m. C a c ket qua d e m cua bo d e m va 
con so hien thi cua bo chi thi ti le vdi do rong x ung cua tin hieu ng o v a o can xac 
dinh do rong x ung t.

— ►I t \4—

Hinh 11.12: S o do logic d u ng cu do do rong x ung t

Vi du, neu tin hieu ngo vao can x ac dinh do rong x ung cd d a n g s ong hinh sin, 
tan so la 5 k H z va x ung nhip tti bo chia tan cd do rong la 1,0 ps va dan c ach cua 
mdi xung la 1,0 ps. B o d e m va bo chi thi se d o c la 200. R o rang dieu n a y cd nghTa 
la 200 ps vi 200 cua 0,1 c ac x ung na y se q ua co’ ng d e m trong thdi gian 200 ps luc 
ma tin hieu c ong E n ab l e d mu’c H. Bo d e m va bo chi thi cd do chi'nh x ac la cong 
hoac trir mot b u dc de m.

11.4. DONG HO DIGITAL

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50 Hz 1 chu ky/s 1 chu ky/phut 1 chu ky/gia

Pern giay Dem phut Pern gia

Hinh 11.13: S o do khoi d o n g ho digital

O e co tan so 1 Hz/s c an phai chia tan so 50 H z cho 50. T a lai chia 1 Hz/s cho 
60 c ho ket qua da ng x ung co tan so la 1- mi n' 1. C h i a d a n g x u n g n a y c ho 60 cho 
d a n g x ung mdi cd tan so la 1- h' 1. D a y la y tirong c o ba n de x a y dirng m a c h dong 
ho digital.

S o do khoi chi ro c ac chCrc n a n g thu’c hien cua d o n g ho tren Hinh 11.13. Bo 
d e m thir nhat chia 50 vdi m u c dich d on gian la chia tan s o c u a dien li/di 50 H z de 
du’Oc d a n g x ung v u o n g vdi tan so 1 Hz/s. B d d e m thir hai chia 60 n h a m tao ra sir 
thay doi mot trang thai trong mdi mot giay va cd 60 trang thai rdi rac, do do cd the 
giai ma de c ho tin hieu chi thi giay va bo d e m thirc hien dem giay.

Hang chuc Hang dan vj

Hinh 11.14: B o d e m 1 0 x 6 M o d - 6 0 cd giai m a h a n g d o n vi v a h a n g c h u c

B o d e m thir ba chia 60 de thay doi mot trang thai trong mdi phut va cd 60 
trang thai rdi rac cd the giai m a de c u n g c ap tin hieu hien thi phut. B o d e m luc na y 
thirc hien dem phut.

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B o d e m cuoi c ung cu" 60 phut ( mot gio) lai c un g cap mot trang t ha i^ n ht i vay, 
neu chia 12 se co 12 trang thai va co the giai ma de c un g c ap tin hieu hien thi gid. 
Luc nay, bo d e m thu’c hien dem gid.

Q A

E J

E CLR K

>

+Vcc
RESET

<3-7 4 1 6 0

CLK : 10
Q A Q & QC QD

J

H

-I

- i 1
7 4 4 7

J U L
1 chu kry/h

1 1 1 1

1 --- 1 LED 1 l

LED

Hang chuc H ang dan vi
Hinh 11.15: B o d e m gid M o d - 1 2

N h u da biet, cd nhieu c ach de thuc hien bo d e m . N h u n g d day, ta yeu cau 
thiet ke bo de m sao cho tdi uu hoa phan cUng.

f

Giai ma
hang chuc

Giai ma
hang dem vi

I

Giai ma Giai ma
hang chuc hang dem vi

r

Giai ma
hang chuc

Giai ma
hang dem vi

i i

Hang chuc Hang dem vi Hang chuc Hang dcrr\ vi Hang chuc Hang don vi

Hinh 11.16: D o n g ho Digital

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11.5. BO DjNH THOI LCD CO CANH BAO

H a u het c a c lo vi s o ng va lo dien deu co bo dinh thdi co c an g bao. Tri/6c kia, 
ngirdi ta d u n g bo dinh thdi co, n g a y n a y ngirdi ta sir d u n g c ac lo vi s o n g hien dai co 
c an h bao dien tir. Khai m e m v e bo dinh thdi du’Oc trinh b a y tren Hinh 11.17a)
T r o n g he thong nay, ban phi'm la ng o v a o con bo hien thi va loa bao hieu la c a c bo 
kien ngd ra. X i r ly va liru trCr c a c so lieu du’Oc c a c m a c h digital thirc hien trong khoi 
m a c h digital 6 Hinh 11.17a.

P h a n chi tiet m a c h digital du’Oc trinh b a y trong Hinh 11.17b). P h a n ma c h 
digital du’Oc chia thanh 4 m a c h bo phan: do la khoi tao quet, khoi d e m lui tir ngat, 
khoi dieu khien/giai ma/chot va khoi so s an h tri so. Khoi tao quet la bo d a o dong, 
trong trirdng hpp na y bo d a o d o n g c u n g c ap x u n g v u o n g co tan so 1 Hz . D o chinh 
xc a c cua bo dinh thdi co c a nh bao phu thu oc v a o d o chinh x ac c u a khoi dat thdi 
gian. Hoat d o n g cua bo dieu khien khdi d o n g n g o v a o lam c ho bo d e m lui thirc 
hien d e m lui tirng birdc d e m . Moi so nho hon d u p e chot va du’Oc giai ma. Khoi nay 
c u n g dieu khien c a c bo hien thi. Hinh 11.18 m o ta chi tiet m a c h dien tir co dinh 
thdi, hien thi ba ng hai so. M a c h sir d u n g L C D v a I C C M O S c o n g suat thap.

Canhbao
b)

Hinh 11.17: a) Khai quat v e bo dinh thdi co c a nh bao
b) S o d o khoi cua bo dinh thdi c o c a n h bao

Hinh 11.18 gidi thieu bo dinh thdi c d c a n h b a o d u n g L C D (hien thi d u n g tinh 
the long). B o dinh thdi hoat d o n g nhir sau:

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- X a c lap tai/dieu khien khdi tao den 0 (kieu tai)

- N a p tai c ho bo de m hang don vi nh d xac lap so B C D khi su1 d u n g 4 c h u y e n 
mach phia tren.

- N a p tai cho bo d e m hang c h u c nh d xac lap so B C D khi su" du ng 4 c h u y e n 
mach phia durcJi

- S o co hai con so du’Oc hien thi tren L C D

- Dich c h u y e n tai/dieu khien khdi tao den 1 (khoi tao kieu d e m lui)

Bo dinh thdi se bat dau d e m lui tirng giay mot. L C D hien thi thdi gian con lai 
trirdc khi coi canh bao keu. Khi ca hai bo d e m de u dat de n so 0, L C D d oc 00 va 
coi canh bao phat tieng keu. Birdc cuoi c un g la ngat m a c h ng uo n de m a c h ngat tin 
hieu am cua coi canh bao.

S o do noi d a y cua ma c h bo dinh thdi co ca nh bao du’Oc trinh b a y chi tiet tren 
Hinh 11.19. C h u y rang mdi IC du’Oc dat a vi tri tirong doi tren so do noi day. Ho at 
dong chi tiet cua bo dinh thdi co canh bao nhir sau:

(555) # 100 Hz

Hinh 11.18: S o do khoi cua bo dinh thdi L C D co c a n h bao

- Bo tao quet:

B o tao quet g o m bo tao x ung va bo chia tan. B o tao x u n g sir d u n g IC dinh thdi 
555 de tao x u n g co tan so 256. B o tao x ung n a y kh o ng that chinh x ac va on dinh, 
co the dieu chinh tri so cua R 1t thirdng R, du’Oc lay k h o a n g 20 kQ.

</div>
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dtipc ket noi vdi nhau. Hinh 11.19 trinh b a y haj_khoi 4 bit dtipc ket noi d e lam 
thanh c a c bo d e m chia 16. C h u y la cac ngo v a o C P la c a c ngo v a o d o n g ho v a chi 
co ngo ra Q D du ’Oc sir du ng . B o d e m chia 16 dau chia tan so 25 6 de c o 1 6 H z . Bo 
d e m thir hai chia tan so de dat dirpc 6 ng o ra tan so ye u cau la 1 H z .

- Bo dem lui tir ngirng dem

Hai bo d e m thap pha n sir d u ng IC 7 4 H C 1 9 2 . Khi ng o v a o tai d e n IC 
7 4 H C 1 9 2 , c a c bo d e m hoat do ng , c a c so lieu d n g o v a o ( A, B, C v a D ) n g a y lap 
tCrc dirpc c h u y e n v a o c a c flip flop cua bo d e m . S a u do, ng o ra c u a bo d e m tai cac 
ch a n Q a, Q b, Q c va Q D co c a c so lieu dirdi d a n g B C D . Khi tai/dieu khien khoi tao 6

mu’c H, tin hieu 1 H z kich hoat ngo va o bo d e m lui cua bo d e m h a n g d on vi. Bo 
d e m lui d e m g ia m di 1 tai moi birdc d e m khi x u n g nhip c h u y e n m u ’c tir L len H. N g o 
ra NHCf R A cua bo d e m lui h an g don vi c h u y e n tCr m u ’c L len H moi khi bo d e m lui 
hang don vi d e m tir 0 den 9. C a c bo d e m lui thirc chat du’Oc ket noi d e trd thanh 
bo d e m lui tir ngirng vi d a y ngirng d e m du’Oc ket noi vdi n g o v a o c h a n C l e a r c ua ca 
hai bo d e m d u n g I C 7 4 H C 1 9 2 . Khi d a y na y 6 mu’c H, ca hai bo d e m ngirng d e m va 
trd ve tri so 0 0 0 0.

- Bo so sanh trj so 8 bit

C a c bo so s an h 4 bit sir d u n g IC 7 4 H C 8 5 d e tao nen bo so s a n h tri so 8 bit

du’Oc gidi thieu tren Hinh 11.19. M u c dich c ua m a c h na y la phat hien khi na o cac 
ngo ra cua c a c bo d e m dat tdi tri so 0 0 0 0 0 0 0 0 BCD. Khi ca hai bo d e m dat tri so 0 
thi ngo ra cua bo so s an h tri so 8 bit ( A = B OUT) len mu’c H. D i e u n a y n h a m 2 mu c 
dich, thir nhat la tin hieu n a y lam ngirng hoat d o n g c ua ca hai bo d e m tai so 0 0 0 0; 
thu1 hai la mtic H tai n g o ra cua bo so s anh lam tranzito Q , d i n , d o n g q u a tranzito 
de n con ca nh ba o lam coi phat ra a m thanh. Di ot triet d o n g ngi rpc cd the do coi 
ca nh bao g a y ra.

- Bo dieu khien I bo giai ma

Hai IC 7 4 H C 4 5 4 3 d u n g trong bo dinh thdi c d 3 m u c dich. C a c chCrc n a n g cua 
IC 7 4 H C 4 5 4 3 dirpc trinh b a y tren Hinh 11.20. N g o v a o kich hoat chot ( L E ) dirpc 
g i n chat va o mu’c H trong m a c h dinh thdi. lam c h o chot kh o ng hoat d o n g . S o lieu 
B C D c h a y qua chot de n bo giai ma B C D - 7 thanh. B o giai m a t ho ng dich n g d vao 
B C D thanh m a 7 thanh. C u o i c u n g, m a c h c u a bo dieu khien d chip 7 4 H C 4 5 4 3 
c u n g cap na ng Itiong c ho c a c thanh c ua L E D .

D o n g ho hien thi d phia dirdi ben phai Hinh 11.9 tao x u n g v u o n g 100 Hz. 
X u n g na y dirpc giri tdi pha n ket noi c h u n g tren L C D v a n g d v a o c h a n P h c ua IC 
7 4 H C 4 5 4 3 . B o dieu khien L C D d IC 7 4 H C 4 5 4 3 dtia tin hieu d a o pha 180° d e n ca c 
thanh L C D de c a c thanh n a y hoat d on g . C a c t ha nh khong hoat d o n g n h a n dtipc 
mot tin hieu x u n g v u o n g c u n g pha tti ph a n bo dieu khien L C D cua I C 7 4 H C 4 5 4 3 .

</div>
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IC 7 4 H C 3 9 3

_ n

+5V
I
+Vcc

2 A QD

6 o dem

chia 16
CLEAR

GND

b ~

1 Hz

a) 5(7 do noi day cua bo chia 2 5 6 , dung 2 bo dem chia 16

NGO
VAO

b) Cau tr u e ben tro n g cua IC 7 4 H C 4 5 4 3 gom c h o t, bo giai ma va phan dieu khien

</div>
(130)<div class='page_container' data-page=130>

Tai/K.hoi ta o / P ie u khien
Ta i bo d e m h a n g d m vi

r , A T a i = 0

C h u y e n m a c h O j -

---D e m lui =
I s _____
C h u y e n m a c h O

-C h u y e n m a c h O

C h u y e n m a c h O —

NgQ v a o

Tai b3 d e m h iin g ch u c
C h u y e n m a c h O ■
C h u y e n m a c h O
C huyen m a c h u
---C h u y e n m a c h 0 — —

+Vcc QA
CLR QB
-|> P e m lui

tAi

A Dem xuong

f3 hang don vi
c (74H C 192)

NHCf RA
GNP^e

+ 5 V

13
15

n h<5r a

D % ngang dem

5 V
116

Dem tie n +Vcc
>Dem lui

CLR
T^' y v

Dem xuong QA
hiing chuc

OP
A ((74H C 192)

QC
&

C QD

D G ND

= & m Vcc
AO

A l E3o so sanh
A 2 t r i s6
A 3 4 b it

(7 4 H C 6 5 )
bO

31 A>f3o

A = f*o
A<f3o
b 2

P 3

A>f5 in A < [3 in
T^Ti
13
15
+5V
L
AO +Vcc
. . F3o so s in h
A l

t r i scf
A 2 4 b it
A 3

(7 4 H C 6 5 )
A > (3 in
A = f3 in
A < f3 in

A = f t OUT
PO

PI
A?

P 3 GND

!L

D iiy ngung dem

+5V

J U U l

# 1 0 0 Hz

3ra

</div>
(131)<div class='page_container' data-page=131>

CAU HOI VA BAI TAP

Hz.

11.2*. Mach trong Hinh 11.3 chi ra c ach hien thi L E D anot c h ung khi ap du ng 
ky thuat da hop. H a y chi ra cach hien thi L E D catot c h ung khi cung ap d u n g ky 
thuat nay.

11.3. Gia du rang bo d e m va bo chi thi tren Hinh 11.12 co the d e m de n 5 con 
so thap phan va bo chia tan c ung c ap x ung v u o n g co tan so 100 k H z va du’Oc 
dung lam xung nhip. D o c du’Oc tren bo chi thi la bao nhieu sau mot thdi gian t cua 
cong Enable, neu ngd va o cd tan so 200 H z d a n g xung v u o n g can xac dinh do 
rong xung?

11.4*. Giai thi'ch y nghTa do chi'nh xac c on g h oa c trti mot btidc d e m ap d u ng 
doi vdi Bai tap 11.4.

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TRA LOI CAC CAU HOI VA BAI TAP

■

CHlCONG 1: HE THONG SO VA MA

1.2 0*.
sfi t hApp hAn

B0I ra c A c so NHj PHAN BAT PHAN THAP LUC PHAN

3579 1101 1111 1011 6 7 7 3 DFB

246810 1100 0100 0001 1010 742032 3C41A

86420 0101 0001 1001 0100 250624 15194

45678910 0001 0011 1110 56200476 2B9013E

97531 0111 1100 1111 1011 276373 17CFB

1 . 2 1 * .

DOI RA CAC SO BAT PHAN THAP PHAN THAP LUC PHAN

SO NHI PHAN

1010 1100 254 172 AC

0001 0000 1010 412 266 10A

0111 10100110 3646 1958 7A6

0101 1111 0100 1110 57516 24398 5F4E

0010 10100011 1011 0100 1100 12435514 2767692 2A3B4C

CHL/ONG 2: CAC CONG LOGIC, BAI SO LOGIC, PHAN

TI'CH VA TONG HOP MACH LOGIC

■ ■

2.24.

a) Mu’c L b) Mu’c H
2.26.

A B C Y

0 0 0 0

0 0 1 1

0 1 0 1

0 1 1 1

1 0 0 1

1 0 1

1 1 0 1

1 1 1 1

D a y la ba ng sti that cua c o n g O R 3 n gd va o. D o do, ket noi t ha nh tang hai 
c o n g O R 2 n gd v a o tuong d u o n g vdi mot co’ ng O R 3 n g d vao.

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2.30*.

2.31*Gian tide cac bieu thu’c sau:

a) F = ( A + B ) + ( A + B ) = A + ( B + B ) = A + 1 = 1
b) Vi A C D + B C D = C ( A D + B D )

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

Dtia vao tinh chat A + A B = A + B v a vi C = 1 . C ta co:
F ( A B C D ) = A B + C.1 + C ( A D + B D )

= A B + C + A D + B D 
= C + A B + D ( A + B)

Ap dung dinh luat D e Mo rg an : A + B = A B, ta co:
F ( A B C D ) = C + A B + D ( A B )

Lai ap du ng ti'nh chat A B + A = A + B ta co:
F ( A B C D ) = C + A B + D

CHUONG 3: MACH XLf LY SO LIEU

■ ■

3.10*. Y = D9

3.12*. Ket noi c ac ng o v a o so lieu theo h u d n g da n sau:

</div>
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+ Ket noi dat v6i c a c ngo vao so lieu D, , D2 , D 3 , D 7 , D8 , D 9 , D10 V a D 13.
3.14*. Multiplexer Y 0 : Noi dat cac ngo vao so lieu D 8 , D 9 , D13 - c o n tat ca 
c a c ngo v ao so lieu khac 6 mu’c cao.

Multiplexer Y , : Noi dat c ac ngo vao so lieu D0 i D, , D14 , D 1S - c o n tat ca c a c 
ng o v a o so lieu khac 6 mu’c cao.

Multiplexer Y 2: Noi dat cac ngo vao so lieu D3 , D8 , D13 - c o n tat ca c a c ng o 
v a o so lieu khac d mu’c cao.

Multiplexer Y3 Noi dat c a c ngo vao so lieu D6 , D 7 , D14 - c o n tat ca c a c 
ngo v a o so lieu k hac d mu’c cao.

3.15*. a) khong b) Y5

3.17*. T r u d n g hop c

3.19*. a) C h i p ben phai b) Y6

3.21*.

a) 67 b) 8 1 3 c) 7 259 
3.23*. Y7

3.25*. T r u d n g hp p c 
3.27*. S a p si 3 m A

3.29*. a) C h a n 5 b) A B C D = 0111
3.30*. a) 0 b) 1 c) 0 d) 1

3.32*. C h a n 4 noi dat ( sau khi kh o ng ket noi vdi + 5 V ) v a ket noi c h a n 3 vdi 
+ 5 V (sau khi kho ng ket noi vdi dat)

CHlTONG 4: MACH SO HOC

■ ■

4.8*

T r u d c het ta c h u y e n doi n h u sau:

20 0 - ► C8H 1 1 00 1 0 00
125 - > 7 D H - ► 0111 1101 
L ap phep trU n h u sau:

1100 1000

- 0 1 1 1 1101

?

T h u c hien trU theo cot, ta co ket qua:

1100 1000
-0 1 1 1 1101
0100 1011

</div>
(135)<div class='page_container' data-page=135>

T h e o he thap luc phan, ta co:
C8H

- 7 D H 
4 B H
4,9.

T h u c hien phep ti'nh so hoc v6i do chi'nh x a c kep, dong nghTa vdi viec thuc 
hien phep ti'nh so hoc 16 bit. P h ep ti'nh so ho c sir d u ng c ac so 16 bit dti6i d a ng 
sau:

X15X14X13X12 X^X^XgXg x7x6x5x4 x3x2x1x0

C a c so giong n hu byte tren x15...x8 va byte dudi x7...x0. O e thuc hien phe p ti'nh 
so hoc 16 bit, m a y vi ti'nh phai hoat d o n g tach rdi tUng byte mot. M a y vi tinh cong 
byte dudi sau do cong byte cao. Dudi d a y la trinh tu lam viec cua m a y vi tinh:

18357 — * 4 7 B 5 H * 0 1 0 0 0111 1011 0101
12618 * 3 1 4 A H * 0011 0001 0 1 0 0 1010
Bu 2 cua 12618 bang:

- 1 2 6 1 8 — ► C E B6H — ► 1100 1110 1 0 1 1 0 1 1 0

Phep cong dtipc thtic hien theo hai btidc cua p he p ti'nh so hoc 8 bit. D a u tien, 
cac byte dtidi dtipc cong:

1011 0101

+ 1011 01 1 0

1 0 1 1 0 1011 — ► x8 x7x6x5x4 x3x2x,x0

Ma y vi ti'nh se Itiu trti byte dtidi x7x6x5x4 x3x2x1x 0. S O N H O x8 dtipc sir d u ng de 
cong vao byte tren. B a y gid, m a y vi ti'nh c o n g c a c byte tren va c ong ca S O N H O 
nhu sau:

1 < — x8

0100 0111 
+ 1100 1110

1 0001 0110 — ► 0001 0110

C a c Idi giai 8 bit dtipc ket hpp lai de co ket q u a cuoi cung:

0001 0110 0110 1011

C h u y rang M S B la bit 0, dieu do cd nghTa la Idi giai cd gia tri d u on g . Ket qua 
bang so thap phan la:

</div>
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4.17*.

- H e tha p luc phan: 0 2 C B H ;

- H e nhi phan: 0000 0 010 1100 1011

CHLfONG 5: IC TTL VA CMOS

5.3*. L o w - p o w e r Schottky 
5.5*. 100

5.7*. 20

5.9*. 31 2 Q

5.11*. Doi v6i bo ghi c huyen B, Disable can 6 mu’c L, c on tat ca c a c bo ghi 
c h u y e n k hac Disable can d mu’c H.

5.13*. 6, 03 m A
5.14*. 14,4 m A

CHUONG 6: FLIP FLOP

6.7*.

5 = 0.

<3=1

Q = 1

5 = 1

K = 1 K = 1

K = 0 (3 = 1

6.8* .

a) C; b) G

6.9*. D a u tien, flip flop dtidc p h u c hoi, ( Q = 0). T a i thdi di em t, C L K len mu’c

1, flip flop luc n a y dtidc ki'ch hoat va lap t u t dtidc xac lap ( Q = 1) vi R = o v a S = 1. 
T a i thdi di em t2 C L K x uo ng mu’c 0 va flip flop k ho n g hoat d o n g va chot a trang thai 
on dinh, Q = 1.

</div>
(137)<div class='page_container' data-page=137>

Gii/a t2va t3ca hai R va S thay doi trang thai, nhu ng vi C L K d mtic 0, flip flop 
v i n con khong du’Oc kich hoat (khong hoat d o n g ) va Q v i n du y tri mu’c 1.

Giua t3 va te, se dap ifng doi vdi bat ky s u thay doi nao cua R va S vi C L K d 
mu'c 1. N h u vay, tai t3 Q xuong mu’c 0 va tai t4 Q lai len mUc 1. T a i tg gia tri Q = 1

dupe chot va khong co sti thay doi nao x ay ra doi vdi Q giua t6 va t7 m a c du ca hai 
R va S thay doi.

Giua t7 va t8 khong co sti thay doi nao x a y ra doi vdi Q vi ca hai R va S deu a

mu’c 0.

X em Hinh 6.26

6.10*.

Vi J = K = 1 , flip flop c h u ye n trang thai doi lap moi khi cd xung nhip c huye n 
tdi. Dang xung tai Q cd chu ky Idn g a p hai Ian chu ky x ung nhip. Noi cach khac, 
tan so cua dang xung xuat hien d ng o ra Q b a n g 1/2 tan so x ung nhip C L K . M a c h 
hoat dong n hu la bo chia tan so: tan so n gd ra Q ba ng tan so ngd va o xung nhip 
C LK chia cho 2.

+Vcc

---k. a

---Hinh 6.28 a va b

6.13*. D o rong xung du oc x ac dinh n h u sau:
t = 1.1 ( R a. C ) = 1,1 ( 1 04 x 10-7) = 1,1 ms

6.14*. T U cong thUc ti'nh do rong x u n g ta cd the tinh gia tri cua dien d u ng C 
nhu sau:

t 10'2

C = --- = 
---1,1 R a 1,1 x 1 0 4

= 0, 909 p F

6.15*. f = 48 kHz, t, = 13 ps, t2 = 7,8 ps

CHl/ONG 7: BO GHI CHUYEN

7.7*.
a) 8p S
b) 6 ps

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(138)<div class='page_container' data-page=138>

7.9*. a) R = 1, S = 0, Q = 0

c) R = 0 , S = 1 , Q = 1
7.10*.

a) M S B thu1 nhat Shift/Load 6 mu’c L, A B C D E F G H = 1 011 1 1 10

b) M S B thir nhat, Shift/Load d mu’c H

7.20*. D = J Q + K Q . M a c h logic n h u sau:

t MSB thd n h a t, Shift/Load tymt/c L, A B C P E F G H = 1011 WO
Xung nhip

chan

Xung nhip

QA
QB
QC
QD
Q E
Q F
Q O
QH

!

i O

7

1
1
1

: o

► iSo di/oc !Uu trif
b) MSB thi/ nhat, Shift/Load a mtic H,

Xung nhip

J

Hinh BT 7.10

</div>
(139)<div class='page_container' data-page=139>

7.11*.

7.13*.

M O P E — I
CLOCK (L= t)

---P ---P A T A IN---PUT --- 1

i i

Q P
QC

Q 3
QA

I i 
-I---1

-t J--- hi

Hinh Bai tap 7.11

P e’chuyen d o i cac h it 1 va O m ach phan hoi duoc th a y th e hJmg mach sau:

Mach "dong nguon-xoa"

Vcc
+V

10k

dS^CLEAZ

luF

r

Hinh Bai tap 7.12

CHUONG 8: BO OEM

8.13*. 4 M H z

8.15*. Vi co 8 chu ky x ung nhip trong mot ch u ky c u a C , ch u ky x ung nhip 
phai la 24/8 = 3 ps. D o do, tan so x u n g nhip la 1 /(3x10 6) = 3 3 3 kHz.

</div>
(140)<div class='page_container' data-page=140>

S o thap phan Idn nhat co the luu tri/ trong bo d e m co 6 flip flop ( M o d -6) la

111111 = 6 3 . C a n liru y sti khac biet giCra M o d u lu s (tong so c a c trang thai) v a so 
thap p ha n Idn nhat.

8.17*. D a n g x ung chi'nh xac d u o c ve n h u sau:

i
Qf3 — [■

i
Q C - i

QD — 
|---HINH BT 8.17

Noi d u ng cua bo d e m la 0 0 0 0 tai di em a tren true thdi gian. Vdi moi c h u y e n 
dich sang s udn a m x u n g nhhip, bo d e m lai d e m len mot, c h o de n khi bo d e m co 
noi du ng cd noi d u n g la 1111 tai thdi di em b tren true thdi gian. T a i thdi d i e m c, bo 
d e m p hu c hoi ( R e se t ) ve 0 0 00 , va bo d e m lap lai q ua trinh d e m tuan tu. R o rang, 
d a y la bo d e m M o d - 1 6, vl co 16 trang thai (tU 0 0 0 0 de n 1111) v a s o t hap p h a n Idn 
nhat cd the d u o c Itiu tru trong c a c flip flop la so thap p h a n 1 5 ( 1 1 1 1 ) .

8.18*. C a n den 16 co’ ng N A N D 4 ng o v ao vdi c a c ng o v a o n h u sau:
A B C D , A B C D , A B C D , A B C D

A B C D , A B C D , A B C D , A B C D 
A B C D , A B C D , A B C D , A B C D 
A B C D , A B C D , A B C D , A B C D

A-5~

C~ '15'

Hinh Bai tap 8.18

</div>
(141)<div class='page_container' data-page=141>

C')

03

8.19*.

CLK
_A

A

C
C
A B C

ABC

A B C
A D C
A B C
A B C
A B C
A B C

HINH BT 8.19

8.2 0*.

a) 3 b)4 c)4 d) 5 e) 5

8.2 1*.

Aj

5 -

C ~

A

-/I

3

C
/I _

5

-c

-'2

' 5

"O_

"2 "
" 3 1 .

" 4 —

" 5

</div>
(142)<div class='page_container' data-page=142>

8.22*.

O 1 2 3 4 5 6 7 3 0 1

j ^ L T L J i J i r i T T r i J i - r L r

8.23*

A
§ -

c I

D

>

I
^
I
I
I
I

~ y

■

B I

C

D
/ i “

>

B
C I
D
j 6
* H
B I

C “ 1

D

>

Cac cong giai ma

8.24*.
a) 3 
8.25*.
CLK
B

C
D
A
B
C
P
j A
>
—
K A

HINH BT 8.22

_ >

>

Dem
xung nhip
0
1
2
3
4
5
6
7
&
9

Cac ngo ra cua bo dem th a p phan
HINH BT 8.23

b) 4 c ) 5 d ) 5 e ) 1 1

J 3

K 3

a / c j m n n n n n n

. _ n _ j

3--- —

HINH BT 8.25
8.26*. Nam chu ky xung nhip

</div>
(143)<div class='page_container' data-page=143>

CHl/ONG 9: BO NHC5 BAN DAN

9.17*. 0100

9.19*. E D C B A = 11011 = 1B 
9.20*. "15" = ( E D C B A ) G

9.21*. Hai chip g ong nhau dtidc ket noi lai vdi nh au n h u sau:

a. C a c ngo vao c h o n : ( A ket noi vdi A; B ket noi vdi B; Cke t noi vdi C ; D ket 
noi vdi D)

b. C a c ngo vao so lieu: ( D , ket noi vdi D, ; D2 ket noi vdi D 2; D3 ket noi vdi D 3; 
D4 ket noi vdi D 4)

c. C a c ngo ra doc: ( S . ket noi vdi S , ; S2 ket noi vdi S 2; S3 ket noi vdi S 3; S4 ket 
noi vdi S 14)

ME va W E du oc sir du ng de lua c h on mot chip h oa c c ac chip khac.

9.22*. V e hai m ac h giong n h u tren Hinh 9.17, m a c h no tren ma c h kia. H a y 
ket noi:

a. C a c d a y dia chi ket noi song song
b. Hai day M E ket noi lai vdi nhau.
c. Hai d a y W E ket noi lai vdi nhau.

Bon da y D A T A IN va D A T A . O U T c ua m a c h phi'a tren dtidc coi la 4 L S B va 4 
day D A T A IN va D A T A O U T cua m a c h dtidi dtidc coi la 4 M S B .

9.23*. Q , kin mach; Q2 hd m ac h; D A T A O U T = D A T A O U T = H
9.24*. Q , ngat mach; Q2 kin m a c h ; D A T A O U T = 1; D A T A O U T = 0
9.25*. V 0 :

a. o mUc H
b. 6 mUc L

c. C a c h ly khoi ngo vao Vdi mot tai dien trd tiep dat, V0 d mtic L.
Vdi mot tai dien trd ket noi vdi + V c c , ng o ra a m u c H./.

CHl/ONG 10: BO CHUYEN OOl

■

AID

VA D/A

10.1*. Bo c h u ye n doi tin hieu anal og s a n g digital ( A D C )
10.2*. M a n g dien trd va bo kh u ec h dai c o n g

10.3*. A = 20
10.4*. V 0 = - 4 V
10.5*. A = 0, 266
10.6*. V o = - 0 8 V
10.7*. A

10.9*. a) analog

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10.11*. M u ’c H, c on g A N D cho x ung nhip c h u y e n qua 
10.12*. Hai dien ap D C

10.13*. T u ’ m u’c H x uo ng mu’c L

10.14*. M u ’c logic H ; C o n g A N D c h o x u n g nhip c h u y e n qua. 
4

0 , 3 9 %

100 ps

10.10*.

0 1 0 1

10.15*
10.16*
10.17*

10.18* I N T R

CHl/ONG 11: MOT SO l/NG DUNG KY THUAT DIGITAL

11.1*. T o e do lap V|5P ba ng 125 H z co nghTa la tat ca c a c c o n s o phai du’Oc 
hien thi trong moi mot khoangt thdi gian la (1/12 5 ) s = 8 ms. C h i a k h o a n g thdi gian 
n a y c ho 6 c on so co nghTa la moi c on so se du’Oc phat s a n g trong k h o a n g thdi gian 
la 8 m s/6 = 1,33 m s ( cu n g co nghTa la L E D c o d o n g c h a y q u a trong thdi gian 1.33 
m s ) va ngirng phat s an g trong k h o an g thdi gian 6 , 6 7 m s ( c u n g co nghTa la L E D 
khong co d d n g c h a y qua trong thdi gian 6 , 6 7 m s ) . C a n c h u y ra ng d o rong x ung 
gi am x u ong , do s an g c ua L E D khi hien thi c u n g g ia m. D i e u c a n thiet la phai lam 
tang d d ng dinh c h a y qua mdi thanh c ua L E D b a n g c a c h g i a m tri so di en trd R da 
neu trong Hinh 11.1 v a Hinh 11.2.

11.2*. T r a n z i t o npn tren Hinh Bai tap 11.2 du’Oc dung lam ch u ye n m a c h giCra 
catot cua L E D hien thi va dat. Khi tranzito da n, d d n g c h a y q ua t h a n h L E D de phat 
sa n g. Khi tranzito n g u n g dan, khong cd d d n g c h a y q u a tha nh L E D , c a c th a nh L E D 
khong phat sang.

Ngo vao chan d a - d o

+Vcc-O

4^is

4 m s '

— 3 ms
b)
Hinh Bai tap 11.2

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D a ng x ung a ngo vao cua chan B a - d o du’Oc trinh b a y r r e n Hinh Bai tap 11.2
b). C h u y rang, phai co xung di/ong de tac d o n g ram tran^'to da n, d on g c h a y qua 
L ED va lam thanh L E D phat s an g trong k h oa n g thdi gian 1 m s trong cCr moi 4 ms.

11.3*. Gia du rang bo d e m va bo chi thj bat da u d e m tu 00000 . T i n hieu ngo 
vao co tan so la 200 H z se cho O c o n g E n a b l e la t = 1/200 = 5 0 0 0 ps X u n g vu o n q 
100 kHz du’Oc dung lam x ung nhip, d a y la chudi x ung d u o n g co da n cach giua 
xung trudc den xung sau la 10 ps ( T xungnhip = 10 ps). D o do, thgdi gian cong la t bo 
dem se dem tien 5000/10 = 5 0 0 b u d c d e m va con so n a y hien thi tren bo chi’ thi 
Vi m6i xung nhjp co thdi gian la 10 as, nen so hien thi tren bo chi thi d u o c d o c te 
500 x 10 = 5000 ps - d a y chinh la chu ky c ua tin hieu n g o v a o can d u o c x ac dinh.

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CAC

t l

T v i e t

t a t

TUTVIET TAT TU DAY DU

A/D Analoq/Diqital

A C Alternatinq C ur r e nt

A D C A n al o q to Diqital C o n v e r t e r
A L U Arithmetic and Loqical Unit

A N S I A m e r i c a n National S t a n da r d Institute

A S C I I A m e r i c a n St an d ar d C o d e for Information I n t e r c ha n ge
A S I C Application Specific Integrated Circuit

B C D B i n a r y - C o d e d Decimal

B C H Binar y C o d e d H e x ad e ci m a l

Bit B inar y Digit

C A S C o l u m n A d d r e s s St robe

C L K C l oc k

C L R C l e a r

C M O S C o m p l e m e n t a r y Met al -oxi de s e m i c o n d u c t o r
C P U Cent ral P r oc es si ng Unit

D A C Digital to A n a l o g C o n v e r t e r
D E M U X D e- mul ti pl ex

D T L D i o d e - T r a n s i s t o r Logic
E C L E m i t t e r - C o u p l e d Logi c
E P R O M E r a s a b l e P R O M

F A Full a d d e r

F E T Fi el d-Effect T r a ns i st or
F I F O First In First O u t

G n d G r o u n d
H H e x a d e c i m a l

H A Haft A d d e r

IC Integrated Circuit

L E D Light Emitting Di od e
L I F O L ast In First O u t
L S B L e as t Significant Bit
L S D L eas t Significant Digit
L SI L a r g e S c a l e Integration
M O S Metal O x i d e S e m i c o n d u c t o r
M S B Mo s t Significant Bit

M S D Mo s t Significant Digit
M S I M e d i u m S c a l e Integration

M U X Multiplex

P L A P r o g r a m m a b l e L o gi c A r r a y

</div>
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Pr Preset

P R O M P ro g ra mm a b le R e a d O n l y M e m o r y
R A M R a n d o m A c c e s s M e m o r y

R A S R o w A d d r e s s Strobe

R C T L Resistor C ap a ci t or T r a ns i st o r Logi c
R O M R ea d O n l y M e m o r y

R T L Resistor T r a n si st o r Logi c
T T L Transi st or T r a n s i s t o r Logi c

__________ Mi croproces s or

TA I LIEU TH A M KHAO

Arpad Barna, Dan I. Pcrat Integrated Circuit in Digital Electronic, J o h n Willey
& S o n s , 1997

James E. Palmer, Ph.D

Do Kim Bang

Ken Steiglitz

Introduction to digital s ys te ms , International 
Editions, S c h a u m ' s Outline Series, M c Gr a w- Hi l l , 
Inc., 1993

K y thuat - ly thuyet va ting d u ng , H o c vien B C V T , 
1993

A Digital signal P r o c e s si n g primer

with Applications to Digital A u d i o and C o m p u t e r 
Music, A d d i s o n - W e s l e y Publishing C o m p a n y , 
Inc., 1996

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MUC LUC

■ ■

TRANG

LOI NOI DAU ... 3 
CHITONG 1: HE THONG SO VA M A... 5

T o n g q u a n ...
1.1. S o nhi p h a n ...
1.2. C h u y e n doi nhi pha n - thap p h a n ...
1.3.C h u y e n doi thap p h a n - nhi p h a n ...
1.4.S o bat p h a n ...
1.5. S o thap luc p h a n ...
1. 6. Ma A S C I I ...
1. 7. Ma d t i 3 ...
1.8..M a G R A Y ...

Cau hoi va bai t a p...
CHlflJNG 2: CAC CONG LOGIC, DAI SO LO G IC , PH A N TICH VA TONG HOP M ACH LOGIC 23

T o n g q u a n ...
2.1. C o n g da o ( I nv ect o) ...
2.2. C o n g H o a c ( O R ) ...
2.3. C o n g V a ( A N D ) ... s...
2.4. C o n g H o a c - D a o ( N O R ) ...
2.5. C o n g V a - D a o ( N A N D ) ...
2.6. C o n g X - O R ...
2.7. Dai so l o g i c ...
2.8. Hai dinh ly D E M O R G A N ...
2.9. C a c dinh luat va dinh ly c ua dai so B O O L E ...
2.10. Phirong p ha p tong c u a tich ...
2.11. B a n g K A R N A U G H ...
2.12. C a c n h o m doi, n h o m bo n v a n h o m t a m ...
2.13. D o n gian phirong trinh logic b ang B a n g K A R N A U G H ...

2. 14. D i e u kien tuy n g h i ...
2. 15. P h u o n g p h a p ti'ch c u a tong ...
2. 16. G i a n Udc bieu thUc ti'ch c ua t o n g ...

Cau hoi va bai t a p...

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(149)<div class='page_container' data-page=149>

CHl/GTNG 3: M ACH XIJTLY SO LIEU

6 6

T o n g q u a n ...
3.1. C a c bo ghep kenh (Multiplexer) ...
3.2. C a c bo phan kenh ( D e m u l t i p l e x e r ) ...
3.3. Bo giai ma 1 - 1 6 ...
3.4. C a c bo giai ma B C D - t h a p p h a n ...
3.5. Bo giai ma du ng L E D 7 t h a n h ...
3.6. Ma h o a ...
3.7. Bo tao bit c h i n le - B o kiem tra bit ch a n l e ...

Cau hoi va bai t a p...
CHlfflNG 4: MACH SO HOC ... 93

T o n g q u a n ...
4.1. Phep cong nhi p h a n ...
4.2. Phep trCr nhi phan ...
4.3. S o nhi phan khong dau ...
4.4. C a c so co d a u ...
4.5. Phep bu 2 ...
4.6. Phep bu so h o c ...
4.7. C a c khoi so hoc ...
4.8. Bo cong - Bo tru"...
4.9. Phep nhan va phep chia nhi phan ...

Cau hoi va bai t a p...
CHUtfNG 5: IC TTL VA CMOS ... 119

T o n g q u a n ...
5.1. C a c IC ho 74xx ...
5.2. C a c thong so cua IC T T L ...
5.3. C a c cong ho m a c h C o l e c t o ...
5.4. C a c IC T T L ba trang t h a i ...
5.5. Dieu khien ngoai doi vdi c a c tai T T L ...
5.6. C a c IC T T L dieu khien tai n g o a i ...
5.7. Logic di/ong va logic a m ...
5.8. IC C M O S ...
5.9. IC ho 7 4 C x x ...
5.10. C a c dac ti'nh cua C M O S ...
5.11. G i ao dien cua T T L va C M O S ...
5.12. G i ao dien cua C M O S va T T L ...

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CHlfflNG 6: FLIP FLO P , XUNG NHIP VA DJNH THOTl 157

T o n g q u a n ...
6.1. Flip flop R S ...
6.2. Flip flop R S d on g bo tTnh...
6.3. Flip flop D ...
6.4. Flip flop D tac d o n g ba ng stidn x u n g n h i p ...
6.5. T h d i gian c h u y e n trang thai cua flip f l o p ...

6.6. Flip flop J K ...
6.7. Flip flop J K M a s t e r - S l a v e ...

6.8. T r i g o S c h m i t t ...
6.9. C a c d a ng xung n h i p ...
6.10. X u n g nhip T T L ...
6.11. B o dinh thdi d u ng IC 555 ... ,...
6.12. D a o d o n g ddn o’ n d u n g IC dinh thdi 5 5 5 ...

Cau hoi va bai ta p...

CHlfflNG 7: BO GHI CH U YEN ... 179

To’ng q u a n ...
7.1. Bo ghi c h u y e n v a o noi tiep-ra noi t i e p ...
7.2. Bo ghi c h u y e n v a o noi tiep-ra s o n g s o n g ...
7.3. B o ghi c h u y e n v a o s ong s on g- r a noi tiep ...
7.4. B o ghi c hu y e n v a o song s o ng - ra s o n g s o n g ...
7.5. B o d e m v o n g ...

Cau hoi va bai tap...

CHlfflNG 8: BO O EM ... 198

T o n g q u a n ...
8.1. B o d e m kho ng d o n g b o ...
8.2. Co’ ng giai m a ...
8.3. B o d e m d o n g bo ...
8.4. B o d e m M o d - 3 ...
8.5. B o d e m M o d - 5 ...

8.6. B o d e m x a c lap t r i r d c ...
8.7. B o d e m c h u y e n ...

8.8. B o d e m c h u y e n M o d - t O co giai m a ...

Cau hoi va bai ta p...

</div>
(151)<div class='page_container' data-page=151>

CHlfflNG 9: BO NHGT BAN DAN 240

T o n g q u a n ...

9.1. Dia chi hoa bo n h d ...
9.2. R O M , P R O M va E P R O M
9.3. R A M ...
9.4. R A M dong - D R A M ...
9.5. C a c te bao nhd ...

Cau hoi va bai ta p...

Tong q u a n ...
10.1. C h u y e n doi tu” digital sang anal og D A C ...
10.2. Bo khuech dai toan tCr...
10.3. D A C co ban ...
10.4. C a c D A C loai thang ...
10.5. Bo c huyen doi tti analog s ang digital ( A D C ) ...
10.6. Bo so a p ...
10.7. Loai A D C khac

10.8. C a c d ac tinh ky thuat cua A D C ...
10.9. IC A D C thuong p h a m ...

Cau hoi va bai ta p...

CHUttNG 1 1 : MOT SO LTNG DUNG KY THUAT DIGITAL ... 295

11.1. C a c bo hien thj da hop ...
11.2. Bo de m tan ...
11.3. D o thdi gian (do do rong x u n g ) ...
11.4. D o n g ho digital ...
11.5. Bo dinh thdi L C D cd canh bao ...
CHlfOfNG 10: CAC B0 CHUYEN DOI A/D VA D/A 275

TRA LCTI CAC CAU HOI VA BAI TAP
CACTI/VIET TAT

TAI LIEU THAM K H A O
MUC LUC

</div>
(152)<div class='page_container' data-page=152>

KY TH U A T SO

LY T^OYET

VK

&RG

Nha xuat ban Lao dong - X a hoi

Tang 6 - 41b - Ly Thai To - Ha Noi

DT: 04.8241706 - 9346024

Fax: (04).9348283

Chiu trach nhiem xuat ban:

Nguyen Dinh Thiem

Chiu Trach nhiem noi dung:

Nguyen Ba Ngoc

Bien tap va sua ban in:

Nguyen Quang Ha

Thiet ke bia:

Manh Hoa

In 700 cuon kh6 19x27cm tai xucmg in Cong ty Thanh Xuan
Giay chap nhan dang ky ke hoach xuat ban so 0 9 - 1 2 1 /L D X H
do Cue xuat ban cap ngay 27/8/2004

</div>
(153)<div class='page_container' data-page=153></div>

Kỹ thuật số - Lý thuyết và ứng dụng: Phần 2

<b>CHUONG 7</b>

<b>GHI CHUYEN</b>

<b>TONG QUAN</b>

<b>it. </b>

<b>7.1. BO GHI CHUYEN VAO NOI TIEP- RA NOI TIEP</b>

1

0

1

0

1

4

3

2

2

0 ^ 2 0 3 0 4

0

3

2

2

2

3

4

3

2

2

0 ^ 2 0 3 0 4

J

0

0,020304

0

0,020304

0

3

0,620304

0,020 304

0,020304

2

3

4

<b>7.2. BO GHI CHUYEN VAO NOI TIEP-RA 30N G SONG</b>

<b>T</b>

<b>n</b>

TU

<b>7.3. BO GHI CHUYEN VAO SONG SONG-RA NOI TIEP</b>

<b>7.4. BO GHI CHUYEN VAO SONG SONG-RA SONG SONG</b>

—•£>>•7

<b>"J5T ></b>

LiJ

l

A

j

L±J UJ L®J CD [

j

EI [a] IioI l l DU

<b>7.5. BO OEM VONG</b>

1

<b>n</b>

<b>i</b>

1

<b>CAU HOI VA BAI TAP</b>

<b>CHUONG 8</b>

<b>BO DEM</b>

<b>TONG QUAN</b>

<b>8.1. BO DEM KHONG DONG BO</b>

J1TL

<b>f</b>

-c>

c>

1___ I

L

^11

I IT

J

<b>o</b>

<b>8.2. CONG GIAI MA</b>

<b>8.3. BO OEM DONG BO</b>

<b>n</b>

U r

II Hi l l