NAND NOR Implementation

Baskı 2

NAND and NOR Implementations (Güncellenme Tarihi: 30.10.2012)

NAND Implementation Algebraic Method 1. Obtain the Boolean function F in the sum-of-products form. 2. Take its complement by applying DeMorgan’s theorem to obtain F in the form of product of complemented products. 3. Take the whole expression into a primed parenthesis to obtain F. Hence, the function becomes suitable for the NAND implementation. Example 1: Implement the Boolean function F = xy + x y + z with NAND gates. Step 1: The Boolean function F = xy + x y + z is given in the sum-of-products form. Step 2: F = (xy + x y + z) = (xy ) .(x y ) .z Step 3: F = [(xy ) .(x y ) .z ] x y

(xy )

x y

(xy )

F

z

x y

(xy )

x y

(xy )

[(xy ) .(x y ) ]

(xy ) .(x y ) F = [(xy ) .(x y ) .z ]

z

Note: A three-input NAND gate can be implemented with two-input NAND gates as follows: x y z

(x.y.z)

x y

(x.y)

(x.y)

(x.y.z)

z

Graphical Method 1. 2. 3. 4. 5.

Group 1’s on the Karnough map. Obtain the simplified Boolean function in the sum-of-products form. Draw the logic diagram with AND and OR gates. Insert a bubble at the output of each AND gate and at each input of the OR gates. Replace INVERT-OR gates with NAND gates.

NAND gate INVERT-OR gate 6. Check all bubbles in the logic diagram. For every bubble that is not compensated by another bubble along the same line, insert an inverter (a one-input NAND gate) or complement the input literal. Example 2: Implement the Boolean function F = xy + x y + z with NAND gates. x y x y

x y F

z

AND-OR implementation

F

x y z

x y

(xy )

x y

(x y )

F

z

INVERT-OR implementation 1/3

NAND implementation

Baskı 2

NOR Implementation Algebraic Method: 1. Obtain the Boolean function F in the product-of-sums form. 2. Take its complement by applying DeMorgan’s theorem to obtain F in the form of sum of complemented sums. 3. Take the whole expression into a primed parenthesis to obtain F. Hence, the function becomes suitable for the NOR implementation. Example 3: Implement the Boolean function F = (x + y).(y + z) with NOR gates. Step 1: The Boolean function F = (x + y).( y + z) is given in the product-of-sums form. Step 2: F =[(x + y).( y + z)] = (x + y) + (y + z) Step 3: F = [(x + y) + (y + z) ] x y y z

(x +y)

F (y+z)

Note: A three-input NOR gate can be implemented with two-input NOR gates as follows: x y z

(x+y+z)

(x+y)

(x +y)

x y

(x+y+z)

z

Graphical Method: 1. 2. 3. 4. 5.

Group 0’s on the Karnough map. Obtain the simplified Boolean function in the product-of-sums form. Draw the logic diagram with AND and OR gates. Insert a bubble at each input of the AND gates and at the output of each OR gate. Replace INVERT-AND gates with NOR gates.

INVERT-AND gate

NOR gate

6. Check all bubbles in the logic diagram. For every bubble that is not compensated by another bubble along the same line, insert an inverter (a one-input NOR gate) or complement the input literal. Example 4: Implement the Boolean function F = (x + y).( y + z) with NOR gates. x y

F

y z

AND-OR implementation

x y

F

y z

x y

F

y z

NOR-INVERT AND implementation

2/3

NOR implementation

Baskı 2

Example 5: The carrying capacity of a conveyor is 100 kg. Three objects are randomly loaded on the conveyor. The weights of the objects x, y and z are 25 kg, 70 kg, and 85 kg respectively. Design a logic circuit that can activate an alarm circuit when the total weight exceeds 100 kg. Implement the logic diagram with NAND gates. Solution: In order to consider all possibilities let’s construct a truth table satisfying the given conditions. The output of the logic circuit should be 1 for (x + y + z) > 100 kg x

0 0 0 0 1 1 1 1

y

0 0 1 1 0 0 1 1

z

0 1 0 1 0 1 0 1

F

0 0 0 1 0 1 0 1

F (x, y, z) =

Minterms

m0 m1 m2 m3 m4 m5 m6 m7

x

yz 0

x

1

(1, 3, 5, 7) y

00

mo m4

0 0

01

11

m1

m3

m5

m7

0 1

1 1

10

m2

0

m6

0

The simplified function: F = xz + yz

z

F = [(xz + yz)' ]'= [(xy)' .(yz)' ]' The output (F) takes the logical value of 1 which activates the alarm circuit. x y

(xy)

y z

(xy)

F

3/3