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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.

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