Analog & Digital Electronics: Course No: PH-218 Lec-29: Combinational Logic Modules
Analog & Digital Electronics: Course No: PH-218 Lec-29: Combinational Logic Modules
Analog & Digital Electronics: Course No: PH-218 Lec-29: Combinational Logic Modules
Course Instructor:
Dr. A. P. VAJPEYI
Department of Physics,
Indian Institute of Technology Guwahati, India 1
Combinational Logic Circuits
A Combinational logic is a circuits which employs two or more of the basic gates
to form a more useful, complex function.
Combinational logic
– has no memory
– the present output depends only on the present input
Sequential logic
– has memory
– the present output depends not only on the present input but also on the past
sequence of inputs (also called memory)
The objective is to use minimum number of components to ensure low cost, less
space, and power requirements.
Example of Combinational Logic Circuits
Let’s design a logic for an automobile warning buzzer using combinational logic.
The criteria for the activation of the warning buzzer is as follows –
– The buzzer activates if the headlights are ON and the driver’s door is opened,
or
– if the key is in the ignition and the door is opened.
K
D Y
D
Y
H Y = D(K+H)
D Y = KD+HD K
H
Combinational Logic Circuits
The OR function (X = A+B) is Boolean addition, and the AND function (X = AB) is
Boolean multiplication.
The operator precedence for evaluating Boolean expressions is (1) parentheses, (2)
NOT, (3) AND, and (4) OR
De Morgan’s Theorem
Theorem 1: AND gate with inverted output is equivalent to OR gate with inverted
inputs. In other words, NAND gate is equivalent to a OR gate with inverted inputs.
Theorem 2: OR gate with inverted output is equivalent to AND gate with inverted
inputs. In other words, NOR gate is equivalent to a AND gate with inverted inputs
Standard Forms
Most Boolean reductions result in an equation in one of the two forms –
Product of Sums (POS) form
Example of POS forms are: (1) X = (A+B’).(B+C) (2) X = (A+C’).(B’+E).(C+B)
mj is the symbol for each minterm, where j denotes the decimal equivalent of the
binary number.
The maxterm with subscript j is a complement of the minterm with the same
subscript j, and vice versa
Minterms and Maxterms
Any Boolean function can be written in Minterms and Maxterms in the below form
F(A,B,C) = Σ( 1,4,5,6,7) = m1+m4++m5+m6+m7
Solution1 : The function has three variables, A, B and C. The first term A is missing
two variables; therefore A = A(B+B’) = AB+AB’
The second term B´C is missing one variable: B’C = B’C (A+A’) = AB’C+A’B’C
Solution:
First, convert the function into OR terms using the distributive law:
F = xy + x´z = (xy+x’)(xy+z)
= (x+x’)(y+x’)(x+z)(y+z) = (x’+y)(x+z)(y+z)
The function has three variables: x, y and z. Each OR term is missing one
variable; therefore
(x’+y) = x’+y+zz’ = (x’+y+z)(x’+y+z’)
(x+z) = x+z+yy’ = (x+y+z)(x+y’+z)
(y+z) = y+z+xx’ = (x’+y+z) (x+y+z)
Combining all the terms and removing those that appear more than once
– Arranged with squares representing minterms which differ by only one variable
to be adjacent both vertically and horizontally.
– Squares on one edge of the map are regarded as adjacent to those on the
opposite edge.
Karnaugh map minimization method
The function is ‘mapped’ onto the Karnaugh map by marking a 1 in those squares
corresponding to the terms in the expression to be simplified.
If two or more pairs are also adjacent, these can also be combined using the
same theorem.
Terms C’ C
A’B’ 1
X = C’+A’B
A’B 1 1
AB 1
AB’ 1