basic logic gates with examples in bangla

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Information and communication technology Chapter 3 Part -2 By Arnab (cse department NSU)

Transcript of basic logic gates with examples in bangla

Information and communication technology

Information and communication technologyChapter 3 Part -2By Arnab(cse department NSU)

Digital logic gate + . + +

) +=)+=)+=)+=

).=).=).=).=

1)A+0=A 2)A.1=A3)A+A=1 4)A. A=05)A+A=A 6)A.A=A7)A+1=1 8)A.0=09)A+B=B+A 10)A.B=B.A11)A+(B+C)=(A+B)+C 12)A(BC)=(AB)C13)A(B+C)=AB+AC 14)A+BC=(A+B)(A+C)15) A+B= A . B 16) AB = A + B17) A+AB=A 18)A(A+B)=A19) A = A

eN -

)A1 + A2 + A3 ..AN = A1 .A2 . A3 . AN

)A1 .A2 .A3 AN = A1 + A2 + A3 + AN ) A + B + C = A .B .C ) A .B .C = A + B +C

ABCABCA+B+CA+B+C A + B + CABC ABC A B C000111011011001110101010010101101010011100101010100011101010101010101010110001101010111000100100

) (OR) (+)) (AND) (*)) (NOT) NOT(A)=A.

Y=A or B=A+B.ABY=A+B000011101111

Y=A AND B=A.B =AB

ABY=A.B000010100111

Y= not(A)= A .

AY= A.0110

, ) )

Y=not(A.B)= AB .

ABA.BA.B0001010110011110

Y=not(A+B)= A+B .

ABA+BY= A+B0001011010101110

Not gate prove through NAND:Y= A . A = A

AND gate prove through NAND:Y= A.B = A . B = AB

OR gate prove through NAND:

Y= A B = A + B = A+B

NOR NOT gate prove through nor gate:

Y= A+A = A

OR gate prove through nor gate:

Y= A+B = A + B =A+B

AND gate prove through NOR gate:

Y = A + B

= A . B = AB

Y= A . B Y= A + B=A + B =A . B= A+B = AB= A+B = AB

F(A,B,C)= A B + B C = A B + B C = (AB) (BC)

F=AB+BC+AC = AB+BC+AC = (AB) (BC) (AC)

( ) Y=A B=AB+AB

ABY=A B000011101110

( ) Y= A B = A B + A B

Y= A B= A B+ A B= A B + A B= (A B) (A B)

( ) Y= A B=AB +A B.

ABY = A B001010100111

( ) Y= A B= A B + A B= AB + A B

Y= A B=AB + A B= AB + A B= (AB) ( A B )

n n ,

10000000000100000000001000000100010000010000100010000001001000000010110000000111

Z0101010

1

D 0D 1D 2D 3D 4D 5D 6D 7 X Y

input

output

X=Q4+Q5+Q6+Q7Y=Q2+Q3+Q6+Q7Z=Q1+Q3+Q5+Q7

n n 1 BCD

000001010011100101110111

1000000001000000001000000001000000001000000001000000001000000001

XYZ

D 0D 1D 2D 3D 4D 5D 6D 7

input

output

D0= X Y ZD1= X Y ZD2= X Y ZD3= X Y ZD4= X Y ZD5= X Y ZD6= X Y ZD7= X Y Z

)- )-

-) ) ) 0+0 = 0 S= AB+AB 0+1 = 1 = A B 1+0 = 1 C= A.B=AB 1+1 = 0 1carry.

Half Adder Truth Table

S=A B C+ A B C+ A B C + ABC.C(out)= A B C + A B C + A B C + A B C

Full Adder Truth Table

SUM=P Q C CARRYOUT = C1 + C2 =S1 C = PQ+ C(P Q)

SUM=A B C CARRYOUT = C1 + C2 =S1 C = AB+ C(A B)

Thank you all