Bi Ging K Thut S
S tit :45 tit im tng kt = 20% im thng k+ 20% im gia k+ 60% im cui k iu kin thi kt thc mn : - im gia k >=4 -Vng mt F=?
CHNG 3 :MCH T HPB gii m (Decoder) Binary decoder BCD to Led 7 segment B m ha (Encoder) B dn knh (Multiplexer) B phn knh ( Demultiplexer)
B GiI M (DECODER)L mch logic t hp chp nhn mt tp hp u vo biu din mt s nh phn v ch kch hot u ra tng ng vi gi tr no . B gii m nh phn, xem xt u vo ca n, xc nh s nh phn hin din, v kch hot mt u ra tng ng vi s ; cn tt c u vo khc khng tch cc.
B gii m nh phnA0 A1 O0 O1N
AN 1
O2N 1
Decoder 2 to 4 ng ra tch cc mc cao.S logic0=
A.B
1=
A.B
2=
A.B
3=
A.B
Decoder 2 to 4 ng ra tch cc mc cao.Bng trng thi:
K hiu logic:
Decoder 2 to 4 ng ra tch cc mc thpS logic0
=A+B
1
=A+B =A+B =A+B
2
3
S logic
Decoder 2 to 4 ng ra tch cc mc thpBng trng thi:Ng vo B 0 0 1 1 A 0 1 0 1 Y 3 1 1 1 0 Ng ra Y 2 1 1 0 1 Y1 1 0 1 1 Y 0 0 1 1 1
K hiu logic:
Decoder 2 to 4 c ng vo cho php1
S logic0
3
Y0 = E + A1 + A 0Y1 = E + A 1 + A 0Y2 = E + A 1 + A 0
2
Y3 = E + A 1 + A 0E
1
0
Decoder 2 to 4 c ng vo cho php1
S logic0
3
2
1
E0
Decoder 2 to 4 c ng vo cho phpBng trng thi
K hiu logic
Y 3 Y 2 Y1 Y 0
Decoder 3 to 8 (74138)K hiu logic:
Y7 Y6 Y5 Y4 Y3
Bng trng thi:
G 2AG 2B
Y2Y1
Y0
G 2A G 2B
Y 7 Y 6 Y 5 Y 4 Y 3 Y 2 Y1 Y 0
L x x H
x H x L
x x H L
x x x x
x x x x
x x x x
H H H H H H H H H H H H H H H H H H H H H H H Hp ng vi m u vo nh phn A B C
Ghp cc Decoder tng s ng vo-ra Mi decoder ghp phi c ng vo cho php Ghp 2 decoder 2 to 4 1 decoder 3 to 8
A B(MSB)
Y Y Y Y
0 1 2 3
Y0Y1
2 to 4E
Y2Y3
A B(MSB)
Y Y Y
0 1 2 3
Y4 Y5 Y6
E
2 to 4
Y
Y7
GHI CHNg vo cho php E mc cao: E=1 : tch cc => cho php IC hot ng E=0 : khng tch cc => cm IC hot ng, tt c cc ng ra u khng tch cc. Ng vo cho php E mc thp: E=0 : tch cc => cho php IC hot ng E=1 : khng tch cc => cm IC hot ng, tt c cc ng ra u khng tch cc.
Cu hi n tp1.
Mi ln c th kch hot nhiu u ra ca b gii m hay khng?
2. Cc u vo cho php ca b gii m c chc nng g 3. Nu r trng thi ca cc u ra 74138 cho mi tp hp u vo sau y: a. G1 = G2 A = G2B = 1, C = B = 1, A = 0 b. G1 = 1, G 2 A = G 2 B = 0, C = 0, B = A = 1
B m ha (Encoder)Khi nim B m ha u tin
Khi nimi lp vi quy trnh gii m. B m ha c mt s u vo v ch mt trong s c kch hot ti thi im xc nh, v to m u ra n bt, ty thuc vo u vo no c kch hot. B m ha nh phn c 2n u vo s c n u ra
Encoder 4 to 2K hiu logic:Bng trng thi:E
x 3(MSB) x2 x1 x 0(LSB)EY1 Y0
Y1 Y0
42
x 3 x 2 x1 x 0
M ha u tinNu c 2 tn hiu cng mc tch cc th ch c ng vo c u tin cao hn mi cho tc ng ti ng ra V d Encoder c th t u tin gim dn t X0 n X3:
x 3 x 2 x1 x 0
Y1 Y0
V D :Cho b Encoder 4-2 c th t u tin gim dn t X3 n X0. Tm ng ra Y1Y0 tng ng vi cc gi tr ng vo sau: X3 X2 X1 X0 = 0110 X3 X2 X1 X0 = 1110 X3 X2 X1 X0 = 1100 X3 X2 X1 X0 = 0011
8 to 3 Encoder (74148)I0 I1 I2 I3 I4 I5 I6 I7 EI
A2 A1 A0GS EO
B dn knh (B chn d liu) Multiplexer/Data selector
{MUX 2N 1 T hp u vo iu khin quyt nh ng vo d liu no c truyn n u ra
Mux 2 to 1
0 1
Z = I 0 .S + I1.S
Mux 4 to 13 2 1 0
1 0
Mux 8 to 1
B phn knh/Demultiplexer
Ng vo d liu ch truyn n mt trong cc ng ra do m ng vo iu khin quyt nh
Demux 1-4
Led 7 on2 loi : Anode chung Cathode chung
Led Anode chungChn Com phi c mc logic 1, mun sng led th cc on a-f, dp s mc logic 0
Bng m cho led anode chung
Led Cathode chungChn Com phi c mc logic 0, mun sng led th cc on a-f, dp s mc logic 1
Bng m cho led Cathode chung
Chng 4 MCH TUN TMch cht Flip Flop B m
MCH CHTCht cng NOR
YU CUVit biu thc ng ra ca Q, Q Thay tng gi tr ca SET, CLEAR chng minh bng gi tr trn.
MCH CHTCht cng NAND
Flip-Flop(FF)Flip-Flop l mt mch tun t thng thng ly mu cc ng vo v thay i cc ng ra ca chng nhng thi im xc nh bi tn hiu ng h (xung clock). FF iu khin (tch cc) cnh ln v FF iu khin (tch cc) cnh xung.
Flip-Flop(FF)Xung clock (CK)
DFF
Biu din dng tn hiu ng ra theo tn hiu vo v xung Clock.Gi s ban u Q0=0. - FF tch cc cnh xung. - FF tch cc cnh ln.FF tch cc cnh xung
TFF
Biu din dng tn hiu ng ra theo tn hiu vo v xung Clock.Gi s ban u Q0=1.FF tch cc cnh ln.
JK FF
Biu din dng tn hiu ng ra theo tn hiu vo v xung Clock.Gi s ban u Q0=0. - FF tch cc cnh xung. - FF tch cc cnh ln.
Ngoi ra cc FF cn c cc ng vo iu khin l Preset(Pr) v Clear (Cl).
Preset (Pr) : ng lp oPr mc cao: Pr =1 :tch cc => ng ra Q =1. Pr =0 : khng tch cc oPr mc thp: Pr =0 :tch cc => ng ra Q =1. Pr =1 : khng tch cc
Clear (Cl) :ng xa oCl mc cao: Cl =1 :tch cc => ng ra Q =0. Cl =0 : khng tch cc oCl mc thp: Cl =0 :tch cc => ng ra Q =0. Cl =1 : khng tch cc
Lu khi gii bi ton FF:Bc 1: xt Pr, Cl :
Pr tch cc : Q = 1 Cl tch cc : Q = 0 Pr v Cl tch cc : ng ra Q khng xc nh c Pr v Cl khng tch cc : chuyn sang bc 2.
Bc 2: xt xung CK
Xung CK tch cc : ng ra ph thuc vo bng gi tr ca tng loi FF. Xung CK khng tch cc : ng ra khng i trng thi.
V D 1:Cho JKFF nh hnh vPR=0, CL=1, Q=? b) PR=1, CL=0, Q=? c) PR=0, CL=0, Q=? d) PR=1, CL=1, xung CK kch cnh ln, J=1, K=0, Q=? e) PR=1, CL=1, xung CK kch cnh xung, J=0, K=0, Q=?a)
V D 2 :Cho JKFF nh hnh vPR=0, CL=1, Q=? b) PR=1, CL=0, Q=? c) PR=1, CL=1, Q=? d) PR=0, CL=0, xung CK kch cnh ln, J=1, K=1, Q=? e) PR=0, CL=0, xung CK kch cnh xung, J=0, K=1, Q=?a)
B M (COUNTER)Modulo(MOD): l s trng thi khc nhau trong mt chu k m Phn loi :
Mch m ni tip (khng ng b): Mch m song song (ng b):
Mch m ni tipm MOD chn 2n Phng php thit k : MOD 2n s dng n JK FF m ni tip th xung CK ca FF sau c ly t ng ra Q ca FF trc Cc ng J&K ni chung v J=K=1.S m c ly t cc ng ra Q ca cc JKFF (dng BCD).
V d1: Thit k mch m c MOD=4 dng JKFF tch cc cnh xung.Gi s ban u cc ng ra bng 0.
Tn hiu ra c biu din nh sau
Mch m ni tipm MOD bt k 2n-1< MOD < 2n V d1: Thit k mch m ln dng JKFF tch cc cnh ln c Pr v Cl tch cc mc cao m chui s sau:0,1,2,3,4,5,0,1,2,.. v c th lp li. Mch m c MOD=6 tc l MOD l.Do 22