Logical gates subjects list
▲
Boolean algebra laws
Boolean algebra calculator
Karnaugh 3 & 4 variables tables
Logical gates symbols
Karnaugh maps
Seven Segment Decoder
Empty Karnaugh maps
Logical gates overview
▲
Logical gates are the basic of the computerized world and is based on the digital values of the binary numbers 0 and 1. The implementation of the logical gates are performed by the rules of the Boolean algebra, and based on the combinations of the operations OR, AND and NOT. The specific gate operation is attained by using diodes or transistors that acts like a switch 0 is off (0 Volt) and 1 is on (5 Volt).
Logical OR and AND gates description
Operation
Logical symbol
Switch operation
Electronics
OR
0
AND
Logical gates examples 1, 2 and 3
▲
F(X, Y,Z) = XY' + Z
Num
X Y Z
XY'
Output
1
0 0 0
0
0
2
0 0 1
0
1
3
0 1 0
0
0
4
0 1 1
0
1
5
1 0 0
1
1
6
1 0 1
1
1
7
1 1 0
0
0
8
1 1 1
0
1
Num
A B
Output
1
0 0
1
2
0 1
1
3
1 0
0
4
1 1
0
Output
From the output result we can see that the system can be simplified to the equivalent form.
F(A, B) = A XOR B
Num
A B
Output
1
0 0
0
2
0 1
1
3
1 0
1
4
1 1
0
Output
This system is the equivalent of the XOR gate.
Seven-segment Decoder
▲
Segment:
A seven-segment display is an electronic device for displaying decimal numbers. widely used for electronic clocks and counters. The display is designed by using logical gates.
Each segment of the number can be calculated by using the Karnaugh method (see above).
Logical gates example 4
▲
The truth table at points D, E, F, G, H and K of the system described above are.
A
B
C
D
E
F
G
H
K
0
0
0
0
0
0
0
0
0
0
0
0
1
0
0
1
0
0
0
0
1
0
0
0
0
0
0
0
0
1
1
0
0
1
0
0
0
1
0
0
0
1
1
0
0
0
1
0
1
0
0
1
0
0
0
1
1
0
0
1
1
0
0
0
1
1
1
0
0
1
0
0
1
0
0
0
0
0
0
0
0
1
0
0
1
0
0
1
1
1
1
0
1
0
0
0
0
0
0
1
0
1
1
0
0
1
1
1
1
1
0
0
1
1
1
1
1
1
1
0
1
1
0
1
1
1
1
1
1
0
0
1
1
1
1
1
1
1
1
0
0
1
1
1
Port
Value
Notes
E
F
G
D + F
H
A‧G
K
E + H
Logical gates example 5
▲
The truth table at points D, E, F, G, H, I, J, and K of the system described above are.
A
B
C
D
E
F
G
H
I
J
K
0
0
0
1
1
1
0
0
1
1
0
0
0
1
1
1
0
1
0
1
0
0
0
1
0
1
0
1
1
0
1
0
0
0
1
1
1
0
0
1
0
1
0
0
1
0
0
0
1
1
1
0
1
0
0
1
0
1
0
1
0
1
0
1
0
0
1
1
0
0
0
1
1
0
1
0
0
1
1
1
0
0
0
1
1
0
1
1
Port
Value
Notes
D
NOT Buffer
E
NOT Buffer
F
NOT Buffer
G
De Morgan's theorem
H
I
J
K
Logical gates example 6
▲
The truth table at points C, D, E, F and G of the system described above are.
A
B
C
D
E
F
G
0
0
0
1
0
0
0
0
1
0
1
1
0
1
1
0
0
1
1
0
1
1
1
1
1
0
0
0
Port
Value
C
D
E
F
(D equals 1 see above)
G
Logical gates example 7
▲
The truth table at points D, E, F, G, H and K of the system described above are.
A
B
C
D
E
F
G
H
K
0
0
0
0
0
0
0
0
0
0
0
1
0
0
0
1
1
1
0
1
0
1
1
0
0
1
1
0
1
1
1
1
1
1
0
1
1
0
0
1
0
0
0
1
0
1
0
1
1
0
0
0
0
0
1
1
0
0
0
0
0
0
0
1
1
1
0
0
1
0
1
1
Port
Value
Notes
D
E
F
G
H
K
Logical gates example 8 - 4 bit comparator
▲
If the 4 bits at the upper left side are the same as the 4 bits at the lower left side, then the output is 1 (green) else the output is 0 (red).
Logical gates 4 bits left shift example 9
▲
Left shift
is performed by moving all the bits one place to the left and filling 0 in the last place, most left bit is lost or moved to the next higher level.
For example, the byte
10011001
after left shift will be:
00110010
If we remember the most left bit, the left shift operation is equivalent to multiplying the number by 2. For example the byte 00001110 equals 14 decimal, after left shift we get 00011100 which is equal to 28.
Draw logical gate circuit from Boolean expression example 11
▲
If we have a logical expression such as
we can draw the equivalent
logical gate circuit by the following steps:
1
Divide the expression into 2 expressions separated by the OR
(+)
operator marked by X and Y.
this is equivalent to the gate:
2
The X expression can be divided into AND gate after simplifying:
this expression is the gate:
3
Perform the same process on the Y expression to get an AND of 2 expressions:
this expression is the gate:
4
Combine the gates of section 1, 2 and 3 to get the final scheme of the complete gate:
Full adder example 12
▲
The full adder adds the values of two bits A and B, if both bits are equal to 1 then it passed through the value of the carry to the next calculation level. For the first loop (rightmost bit) the input carry equals to 0. In order to add two bytes, we can add 8 full adders (see next example).
The truth table for full adder is:
Carry-in
A
B
D
E
F
Sum
Carry-out
0
0
0
0
0
0
0
0
0
0
1
1
0
0
1
0
0
1
0
1
0
0
1
0
0
1
1
0
0
1
0
1
1
0
0
0
0
0
1
0
1
0
1
1
1
0
0
1
1
1
0
1
1
0
0
1
1
1
1
0
0
1
1
1
For example 2 bits A and B
Port
Value
D
A'
·
B + A
·
B'
E
C
·
D = C·(A'·B + A·B') = C·A'·B + C·A·B'
F
A
·
B
Sum
Carry-out
Full adder draw with 7408, 7432 and 7486 chips
▲
The basic circuit for adding two bits contains the chips
IC 7486
- quad
XOR
gate,
IC 7408
- quad
AND
gate and
IC 7432
quad
OR
gate. We used only half of the gates of the
XOR
and
AND
gates and only 1 of the
OR
gate. All the chips have to be connected to the input voltage at V
cc
and to the ground at
GND
input.
The carry out values can be connected to the carry input of the next pair of bits and so on until we reach the number of bits that should be added, see next example for 4 bits addition, this case uses all the inputs of the three types of chips
7486, 7408
and
7432
.
4 bits adder example 13
▲
0
0
0
The 4 bits adder adds the values of two 4 bits values and gives the result at right, notice that the result contains additional carry that can be used for the next higher calculations. This process can be easily extended to more bits.
Example
- Add the 4 bits binary number
1011 (
decimal
11)
to the binary number
0110 (
decimal
6).
.
Notice that in binary addition
0 + 0 = 0
0 + 1 = 1
1 + 1 = 0
and carry 1.