De bi~u thi cae chi} s6 tu 0 + "9 trong M de'm 10 hay ter 0 + F (trong M d€m 16) (xem bang 3.1), ngucri ta thuemg su d\lDg t~p hqp 7 tbanh pMt sang.
ki hi¢u a, b. c, d, e. f, g nhu hloh 3.47. MOi chi} s6 dn bie:u thi lUang U'ng v6i
Cr '- I
" '-,,-
Hinh 1.47: Bilu Ihj cae chiJ sd di'm theo ma 7 thanh
mQt s6 [hanh ph8[ sang vA duqc dit!u khien nha cae bQ chuyen dOL X6t bQ chuyen dOi rna BCD - 8421 - 7 thanh c6 bang chan Ii nhu bang 3.17.
Bang 3.i7: Bdng chdn If gidi mo til BCD sang 7 thanh
Oau vilo a,u ra Chu.o
A, A, A, A, • b c d • I • hi •• thl
0 0 0 0 1 1 I, 1 1 1 0 0
0 0 0 1 0 1 1 0 0 0 0 1
0 0 1 0 1 1 0 1 1 0 1 2
0 0 1 1 1 1 1 1 0 0 1 3
0 1 0 0 0 1 1 0 0 1 1 4'
0 1 0 1 1 0 1 1 0 1 1 5
0 1 1 0 1 0 1 1 1 1 1 6
0 1 1 1 1 1 1 0 0 0 0 7
1 0 0 0 1 1 1 1 1 1 1 8
1 0 0 1 1 1 1 1 0 1 1 9
1 0 1 0 1 1 1 0 1 1 1 A
1 0 1 1 0 0 1 1 1 1 1 B
1 1 0 0 1 0 0 1 1 1 0 C
1 ' 1 0 1 0 1 1 1 1 0 1 D
1 1 1 0 1 0 0 1 1 1 1 E
1 1 1 1 1 0 0 0 1 1 1 F'
Trong bAng nAy, cae thanh dUQ'C phat sang-tuong Ung v6i mlfc lOgic 1. Hai phuong phap hien thi phO bie'n cua chil7 thanh Ia.:.
ai Phuung philp dung LED
M6i thanh dUQ'c phat sang nha mQt LED (xem m1,lc 2.2.Sb) ho~c vAi LED (ne'u lA thanh dAi, to). Bay LED trong mQt chO' s6 7 thanh c6 the n6i cat6t chung hay an6t chung nhu hlob 3.48. NgoAi rat v6i mOi ph4n til hien thi 7 thanh cc)n c6 diu chlm dp sau chil 56 M bieu thi ph4n Ie (nhll diu phay) ho~c them dlu (+). (-) vl m¢t.6 kl hi¢u khOe (hinh 3.49),
bi Phuung pkdp dung link Ihi lOng
Hien thj tinh the long LCD (Liquid:..Crystal Display) c6 nhi~u uu diem han hi~n thj bang LED nen hi~n nay phuang phOp nAy duge dftng phO bi~n, D6 Iii do:
- Tieu !hI} rlt ft di¢n nAng
- Ca'u truc gQn, d~p. dQ Mn co hQc cao - C6 tM ditu , khi~n true tie'p ti) eae IC
'
- C6 tM hitn thi nhOng dA'u hi~u phuc
t~p. ..
Nhuqc diem ella LCD lit d¢ sling y~u
nen trm t6i phiii dQC nho aoh sang ben ngoai.
Chi thi LCD duqc dung nhi~u a cae may tinh bifm (calculator), may tioh deh 'tay (computer), d6 chai tre em, cae d6ng h6 tren xe haL ..
M~t hien thi tioh the long tl1-o bbi 2 tlfm thuy tioh mong xt.p song song, each nhau khoang 10 !lm (hloh 3.50). M~t troog kinh sao c6 trang lOp di¢n . qrc S002 trong su6t mang hloh, dlfu c~n
the hi~n. Giiia hai tam kinh lit tinh the long duqc bao lXri m¢t khung kin. Tinh tht long 1. h<!p ell(t hllu
CO' d{ic bi~t mll. khi c6 tac d\log cua di~n
truOng thi di¢n C\IC se t6i di va mAt nhln
duqc hloh. dA'u cua di¢n C\I'C (mau den) tren n~n trong 5u6t cua pMn tioh the long kh()ng bi t6i. D6 .lit tcuCmg hqp hai mang
--{;., N v
,(' b b N
v v
,({
v , , I"
VI
N
v d d
,ô
VI
N , , ,ô
,ô f f VI ,({
-, v
,ô
VI g g ,({
v
,({
v dp dp I"
v
-, +C • b) . -c
Hlnh 3.48: each noi LED cat6r chung (II) va anb'r chung (Q) 1 0 9 8 7 6
.
f~b •
. . . . '@' .
1 2 3 4 5
~o
Hid 3.49: M~r Ydj pMn I~ hitn thi 7 rhanh
IQc ph<1n c1,I'c g6c 90°. Khi hai mang lQc phtln qrc song song thl c6 chit s6 trong tren nen t6i. De r6 chii' s6.1o .... i n~y c6 den chi~u sang phia sau qua tm. kinh tan x .... nhAm phan b6 deu anh sang (blnh 3.51).
•
1lnh "" ,M.
Hu6ng nhln
Hinl! 3.50: Cau Iqo man tinh thl lOng
. Ngu6n di¢n cAp cho cac di¢n Cl,I'C Ia ngu6n xung vu6ng, tan s6 30 Hz + 100 Hz, di¢n ap tu 3V + 8V ho~c c6 the ra't thAp (I,5V).
N4n Ihloljotlnh
HU'Ong nhln
LCD
N~u tinh the long c6 tr¢n cac h .... t mau, nguai ta
t~o duqc cac LCD mau lo .... i GHD-LCD (Guest- Host-Dichronic LCD). V6i kI thu~t nhi~u lOp, ch6ng nhi~u LCD m¢t mau Mn nhau, ta cung c6
m¢t LCD nhieu mau. Hinl! 3.51: LCD hi~1I ehil so
frollJi! suoi
3.3.4. BC;; chQn du Ii~u
Cac bQ chQn dO' li¢u duqc chia ra hai lo<,ti: b¢ chQn kenh hay d6n kenh c (MUX- Multiplexer) vi\ bQ pMn kenh (DEMUX- Demultiplexer).
a) B", ch{lR kinll MUX
BQ chQn kenh MUX I. h¢ lOgic t6 hqp co nhi¢m v~ chQn ra mQt dja chi trong cac dia chi di1li¢u dltu YaO va chuyen di1 li¢u
a d~a chi d6 ra dltu Ia.
M~t b¢ chQn kenh c6
DO
01
• •
• •
D&u vAo choph~
MUX
••••
Ak_1 Al AO
~ dis. chi
Hinh 3.52: Cdu true bi) ch9f1 kln.n
cAu truc nhu tren hlnh 3.52. Cac d~u vao baa g6m:
DlIumdAo
- Oic d.4u vao dO' li~u: de nh4p dO li~u (dOng thai) - Oic d.4u vao dja chi: de chQn dja chi dO' li~u
- D.4u vao cho phep (Enable): cho phep chuyen dO' li~u da chQn fa d4u Ta.
8() chQn kenh c6 tbe c6 rn()t d.4u ra ho~c hai d.4u ra (lien hqp vOi nhau).
N€u b() cl)Qn kenh c6 k d4u vao dja chi thl chi co the chQn dllqc rn()t trong 86 t6i da 2k d.4u V80 dO' li~u vll dllqc gQi la b() chQn kenh 2k bit.
Co thi noi, b() chQn kenh III bQ chuyen thOng tin du6'i dl,lng song song (vao dOng thai) a dliu V80 thanh n6i tilp (timg thOng tin) jJ dAu ra.
Hlnh 3.53 bieu thj sa dO logic cua ~ chQn kenh 4 bit. Nguyen H hOl,lt dQng cua bQ chQn kenh g4n gi6ng bQ gi8.i rna duai sl! di~u Mnh cua tin hi¢u dieu khien E. Kenh dugc truy~n dAn se duqc chQn dja chi theo cae dilu va.o d!a chi.
" ~
0, G,
D, 0,
- ~OR F
0, r- G,
I " F
- ~
0" \0,
A, "
" J
~ IV
Ao " v IV "
Hinh 3.53: S(1 tid Mgic ella b~ eht;Jn kenh 4 bit
Vi d1,l: CAn truy~n dAn dO li¢u dilu vao D2 = 1 tai dilu ra. Dia chi chQn Ao
= 0; AI = 1 va cho tin hi¢u dieu khien E = O. Khi d6, 4 dilu vao cong AND G2 deu c6 mUe lOgic 1 va dilu ra c6 muc lOgic 1, qua cong OR co F = 1 (F = 0),
nghla la D2 = 1 da dllqC chuyen ra F = 1. Nlu D2 = 0 cin chuyen thl trong 4 dAU vao cOng AND c6 mQt d4u jJ muc lOgic 0 va. dilu ra AND G2 co muc lOgic 0, d;:n Mn F = 0 (F = 1), nghia 1. D, = 0 dii dugc chuy~n Ta F = O.
mnh 3.54 IA sa d6IOgic cua b¢ chQn ktnh 8 bit vai Ie tuang Ihlg (3.54b).
~n dQC c6 the: tl! tim hie:u ffi.iJ.ch nhu mOt b~i t~p.
A2 AI,\ Ao 0.,
'" 0, D. 0, 0, D,
r r Yo: ~G' ~~:rr
.)