báo cáo mạch
TRANSCRIPT
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TRNG I HC IN LC
KHOA IN T VIN THNG
BO CO THC HNH XNG
MN K THUT SGing vin hng dn: HONG VN NG
Bo co nhm 3:
Thnh vin:
Nguyn Trng i
Phm Vn ng
inh Vn c
Nguyn Hu c
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BO CO THC HNH XNG
MN K THUT SGing vin hng dn: HONG VN NG
Bo co nhm 3:
Thnh vin:
Nguyn Trng i
Phm Vn ng
inh Vn c
Nguyn Hu c
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Bi 1: Cc phn t logic c bn
I, cc phn t logic
1.C bn: AND :
A AND B = A.B =>A
B0 1
0 0 0
1 0 1
OR
A OR B = A+B =>A
B0 1
0 0 1
1 1 1
NOT
Y = NOT A = A
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A Y
0 1
1 0
2.M rng NAND
NAND = NOT AND
AB
0 1
0 1 1
1 1 0
NORNOR = NOT OR
AB
0 1
0 0 0
1 0 1
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XOR
L s 1 => y =1Chn s 1 => y=0
AB
0 1
0 0 1
1 1 0
3. Kt lunTa c th dng cc phn t c bn hoc dng 1 trong ccphn t m rng biu din tt c cc php ton logic
II, h mCc h m hay gp nh 2, 8, 10, 16. H c s 10 (thp phn, Decimal system) H thp phn l
h thng s rt quen thuc, gm 10 s m nh ni trn.
Di y l vi v d s thp phn:N = 1998 = 1x103 + 9x102 + 9x101 + 8x100 = 1x1000 +9x100 + 9x10 + 8x1
N = 3,14 = 3x100 + 1x10-1 +4x10-2 = 3x1 + 1x1/10 +4x1/100 H 2 : cn gi l h nh phn (Binary system) , c biu
din bi 2 s 0 v 1.
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Mi s m trong mt s nh phn c gi l mt bit (vit ttca binary digit).
S N biu din trong nh phn (anan-1an-2.a1a0) c gi tr
l : N= an .2n +an-1 .2n-1 +an-2 .2n-2+.+a1 .21 +a0.20 H 8 : tng t. h 8 c biu din bi 8 ch s
(0,1,2,3,4,5,6,7)
S N biu din trong h 8 (anan-1an-2.a1a0) c gi tr l :N= an .8
n +an-1 .8n-1 +an-2 .8
n-2+.+a1 .81 +a0.8
0
H 16: h 16 c biu din bi 10 ch s t nhin v 6 ch
ci A,B,C,D,E,F.S N biu din trong nh phn (anan-1an-2.a1a0) c gi trl : N= an .16
n +an-1 .16n-1 +an-2 .16
n-2+.+a1 .161
+a0.160
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Bi 2: Thit k mch logic tng hp
I,cc bc thit k:
T yu cu thc t ca mch ta lp bng chn l hay bng stht. trong bng ghi r cc kt qu u ra vi tng trng hpu vo khc nhau.
VD: thit k mch mch c 2 u vo sao cho mch ch mkhi 2 uvo u l 1
Ta c bng chn l
A B Y
0 0 0
0 1 0
1 0 0
1 1 1
T bng chn l suy ra bng cacnoAB
0 1
0 0 0
1 0 1
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T bng cacno ta khoanhAB
0 1
0 0 0
1 0
T cc s khoanh ta suy ra php tonY= A.B
T php ton ta suy ra mch
II, thc hnh
Yu cu: thit k mch kha kt sao cho: A,B l bo v, C lch. Kt m khi A v B cng m hoc C m.Thit k:T yu cu ca bi ta suy ra bng chn l:
A B C kt
0 0 0 0
0 0 1 1
0 1 0 00 1 1 1
1 0 0 0
1 0 1 1
1 1 0 1
1 1 1 1
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T bng chn l ta suy ra ba cacno
Suy ra: Y = C + A.BSuy ra mch:
Cu m rng: thit k mch trn ch dung NAND:Thit k:T kt qu cu trc ta cY= C+ A.B
Y = C+ A.B
Y = C + A.B
Ta c mch nh sau
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Bi 3: MUX/ DEMUX
I, l thuyt
MUX cn gi l mch a hp hay mch chn d liu, gm 2nng vo d liu, n ng vo a ch (hay iu khin) v mt
ng ra. Khi c mt a ch c tc ng d liu ng votng ng vi a ch s c chn. y l mch MUXn gin vi 21 cng vo v 1 cng ra
S0= 0 ta c Y = C0S0= 1 ta c Y = C1
DEMUX l mch tch knh.
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S0= 0 ta c Y1 = YS0= 1 ta c Y2 = Y
II.thc hnhyu cu :
Thit k mch MUX/ DEMUX S dng cc phn t logic c bn V s dng ch dng NOR
THIT K :
Thit k mch MUX/ DEMUX S dng cc phn t logic cbnMUXT yu cu ca bi ta c bng chn l
C0 C1 S0 Y
0 0 0 0
0 0 1 0
0 1 0 0
0 1 1 1
1 0 0 1
1 0 1 0
1 1 0 1
1 1 1 1
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Suy ra ba cacnoC0S0
C100 01 11 10
0 0 0 0 1
1 0 1 1 1
Khoanh
Vy ta c pt :Y = C1S0 + C0 S0
SUY RA mch nh trn
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DEMUXT yu cu ca bi ta c bng chn lS0 Y Y1 Y2
0 0 0 0
0 1 1 0
1 0 0 0
1 1 0 1
Suy ra ba cacnoVi Y1 Vi Y2S0
Y0 1
0 0 0
1 1 0
Khoanh
S0Y
0 1
0 0 0
1 0 1
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Vi Y1
Vi Y2
V s dng NORSuy ra:
Y1 = S0 . YY2 = S0 . Y
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Suy ra mch
Ch dng NOR:MUXY = C1S0 + C0 S0
Y = C1S0 + C0 S0
Ta c tnh cht
A.B = A + B => A.B = A.B = A + B
Y = C1 + S0 + C0 +S0Ta li c tnh cht :
A = A + A
Y = C1 +C1 + S0 + S0 + C0 + C0 + S0
Vy:
Y = Y + Y
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Vy ta c mch MUX
DEMUX
Y1 = S0 . YY2 = S0 . Y
Ta c tnh cht
A.B = A + B => A.B = A.B = A + B
=> Y1 = S0 + Y=> Y2 = S0 + Y
Ta c mch DEMUX
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Bi 4: TRIGO
I, l thuyt:
Cht RS tc ng mc cao
Vi R,S l cc u vo, Q v Q l cc trng thi hin ti ca u ramch, Q+ l trng thi u ra tip theo ca mch
Ta c bng chn l di
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Vy ta c bng rt gn
+ Khi R=S=0 (c 2 ng vo u khng tc ng), ng ra
khng i trng thi.+Khi R=0 v S=1 (ng vo S tc ng), cht c Set (tct Q+=1)+Khi R=1 v S=0 (ng vo R tc ng), cht c Reset(tc t li Q+=0)+Khi R=S=1 (c 2 ng vo u tc ng), cht ri votrng thi cm
Cht RS tc ng mc thp:
Tng t nh cht RS mc cao ta s c bng rt gn
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+ Khi R=S=0(c 2 ng vo u khng c tc ng), chtri vo trng thi cm.+Khi R=0 v S=1 (ng vo R khng tc ng), cht c
Reset (tc t li Q+=0)+Khi R=1 v S=0 (ng vo S khng c tc ng), chtc Set (tc t Q+=1)+Khi R=S=1 (c 2 ng vo u c tc ng), ng rakhng i trng thi
II, thc hnh Lp bng hot ng, gii thch
Gii thch ng dng dng trigo chng ny xung cho cngtc c khGii thch:
Lp bng hot ng, gii thchCht RS tc ng mc cao
Tng t nh phn l thuyt trn ni. Mch trigo mc cao dng 2con NOR nn ta c th d dng suy ra kt qu u ra
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+ Khi R=S=0 (c 2 ng vo u khng tc ng), ng rakhng i trng thi.+Khi R=0 v S=1 (ng vo S tc ng), cht c Set (tct Q+=1)+Khi R=1 v S=0 (ng vo R tc ng), cht c Reset(tc t li Q+=0)+Khi R=S=1 (c 2 ng vo u tc ng), cht ri votrng thi cm
Gii thch ng dng dng trigo chng ny xung cho cngtc c kh
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Gii thch c ch chng rung lc:
Khi cng tc t 1 ta s c R = 1 , S = 0. Ta s c Q = 1 vQ+= 1. Khi ta bt cng tc ta s c R= 0, S=1 ta s c trngthi Q+ =0. Khi cng tc b rung lc ta s thy cng tc rikhi c 2 v tr 1, 2 nn c R, S u bng 1. Trigo roi votrng thi nh Q+ = Q =0. Cc trng thi khc cng th. Khicng tc b rung s lm cho c R & S ri vo trng thi 1 1khin trigo t trng thi nh. Ko di xung ra, xung ra s
vn l mt xung vung. Khng b mo.
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Bi 5: M BCD v hin th led 7 thanh
I, l thuyt:
M BCD:Ta dng 4 bit nh phn m ha cc s thp phn t 0 n9. Tha 6 s . ta c bng m di vi A, B , C, D l cc bitnh phn dng biu din.Vi led 7 thanh mc Vcc chung ta c
Cu to led 7 thanh gm 7 con led ring bit , ta gi l a, b, c,d, e, f, g. ta cn c 1 led l du chm gi l h.
ab
fg
ce
d h
S D C B A a b c d e f g h0 0 0 0 0 0 0 0 0 0 0 1 01 0 0 0 1 1 0 0 1 1 1 1 02 0 0 1 0 0 0 1 0 0 1 0 03 0 0 1 1 0 0 0 0 1 1 0 0
4 0 1 0 0 1 0 0 1 1 0 0 05 0 1 0 1 0 1 0 0 1 0 0 06 0 1 1 0 1 1 0 0 0 0 0 07 0 1 1 1 0 0 0 1 1 1 1 08 1 0 0 0 0 0 0 0 0 0 0 09 1 0 0 1 0 0 0 1 1 0 0 0
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II,Thc hnh. Thit k mch gii m BCD
S dng IC 7447
Thit k: Thit k mch gii m BCD
Ta c bnh chn l :
S D C B A a b c d e f g h0 0 0 0 0 0 0 0 0 0 0 1 01 0 0 0 1 1 0 0 1 1 1 1 0
2 0 0 1 0 0 0 1 0 0 1 0 03 0 0 1 1 0 0 0 0 1 1 0 04 0 1 0 0 1 0 0 1 1 0 0 05 0 1 0 1 0 1 0 0 1 0 0 06 0 1 1 0 1 1 0 0 0 0 0 07 0 1 1 1 0 0 0 1 1 1 1 08 1 0 0 0 0 0 0 0 0 0 0 09 1 0 0 1 0 0 0 1 1 0 0 0
Vi a
a = C . A + A .B .C .D
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Vi b
b = A . B .C+ A . B .C = C . ( A .B + A .B)Vi c
c = A.B.C
Vi d:
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. g = B .C .D + A .B .C
S dng IC 7447
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Bi 6: b mI, l thuyt:
Mc cao
Sn ln
Mc thp
Sn xung
C 4 kiu tc ng khc nhau ca CLKMc cao:
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1 0 1 0 1 0 1 0 1 0 1 0 1 0
1 1 0 0 1 1 0 0 1 1 0 0 1 1
1 1 1 1 0 0 0 0 1 1 1 1 0 0
1 1 1 1 0 0 0 01 1 0 0 1 1 0 0
3 2 1 0
IC m
R0 = MR0 = Master reset 1R1 = MR1 = Master reset 2CKA & CKB = CLKO v CLK1
CLK
Q0
Q1
S
m
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II, thc hnh
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BO CO MCH NG HI,THIT K
YU CU: thit k mch ng h bo gi, pht v 2 lednhy theo giy.PHN TCH :Mch cn thit k s c 4 led 7 thanh , 2 led bo gi v 2 ledbo pht. Nh vy cng s cn 4 IC 7447 gii m. V cnm c gi , c pht v giy nn ta cn tt c 6 IC m 7490hoc 7493. Trong mch ta dng 3 IC 7490 v 3 IC 7493.
Vi 2 IC dng m giy yu cu n 59 giy phi resetnn i vi IC th nht yu cu n khi m n 9 tc l ln10 th t th reset khi cc ga tr ca D-C-A-B l 1010 =>ta ni 2 chn reset vo chn 9 v 11 nh hnh di.
Vi IC m giy th 2 yu cu u vo l xung sn xungv mi ln kick xung trng vi mi ln reset ca IC th 1 U1.Ta xt IC U1khi reset l t 9 v 0 tc l t 1010 v 0000.Nh vy y c 2 xung c xn xung vo thi im reset
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l chn 9 v 11ca IC U1. Nhng trong 1 chu k m chn 9s c nhiu ln c xung sn xung, cn chn 11 th ch chota xung sn xung vo thi im reset nn ta chn chn 11
lm chn cp xung cho IC 7493 th 2 U2. Vi IC U2 ny tas c yu cu reset khi m n 5 tc l ln 6 th t th resetkhi cc gi tr ca D-C-A-B l 0110 => ta ni 2 chn resetvi chn 8 v chn 9. Nh hnh di.
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Tip n l 2 IC m pht ta s dng 2 con 7490 , tng tnh 2 IC m giy , ng h m pht yu cu m n 59th reset. Vi IC th nht do ta dng IC 7490 c th t reset
khi m n 9 nn ta ch cn ni mt cc chn reset. i viu vo, ta cn click xung sn di vo IC khi IC m giyhnh chc reset m hnh l U2 reset. Suy lun tng t nhvy ta phi ni chn 14 ca IC ny vi chn 8 ca U2
Vi IC th 2 ta c u vo ta suy ra tng t nh phn trns l chn 11 ca IC U3 v chn 11 cung cp xung snxung vo thi im U3 reset v ch cung cp xung snxung vo ng thi im . V chn reset yu cu m n5 tc l ln 6 th reset , ti thi im ta c D-C-A-B l0110 => ta ni 2 chn reset Ro vo chn 8 v 9 nh hnhdi.
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Cui cng l 2 IC m gi. Yu cu y l m n 23 tcl khi ti 24 th reset . vi IC m gi th nht U5 s c 3trng hp reset l
- ti chn gi ln mi gi ng h hin 09 ln 10, IC th1 m n 9 reset.
- Ti thi im 19 gi ln 20 gi , IC th 1 m n 9
reset.- Ti thi im 23 gi sang 0 gi , IC th 1 m n 3
reset.Do ta s dng IC 7490 lm IC m gi th nht do ICny c th t reset 9 p ng 2 trng hp reset u , tach phi lm n reset trng hp th 3 ti thi im 23gi sang 0 gi. Tc l khi IC th 1 m n 3 tc l
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chuyn ln 4 v IC th 2 m n 2 hay l ang 2. Nhvy cc gi tr ca A-B-C-D l :BCD
IC
D
Chn 11
C
Chn 8
B
Chn 9
A
Chn 12IC1 0 1 0 0
IC2 0 0 1 0
Nh vy U5 reset 23 sang 0 gi ta phi ni 2 chn Roca U5 vi chn 8 ca U5 v chn 9 ca IC m gi th 2 lU6.Tng t nh th U6 cng phi reset cng lc vi U5 nn tani cc chn reset ca U6 tong t nh 2 chn Ro ca U5, y ta c th dng U6 l 7490 hoc 7493 u c , tronghnh di ta dng 7493. Vi u vo ca U6 ta lm tng tnh cc trng hp trc ni vo chn 11 ca U5.
Vi xung ly vo ta c th dng thch anh hoc 555. Vi 2led nhy theo giy ta c th ni 2 led ni tip ri mc trctip vo chn xung ra ca b to xung, nu khng ta s dng
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xung qua tran khuyc i lm 2 led ny nhy theo nhpca xung. V mch nguyn l th nh hnh di.