A circuit outputs a digit in the form of 4 bits. 0 is represented by 0000, 1 by 0001, …, 9 by 1001. A combinational circuit is to be designed which takes these 4 bits as input and outputs 1 if the digit ≥ 5, and 0 otherwise. If only AND, OR and NOT gates may be used, what is the minimum number of gates required?
(A) 2
(B) 3
(C) 4
(D) 5
Answer: (B)
Explanation: We should get output 1 for values>=5
Making truth table for problem
| A |
B |
C |
D |
Op |
| 0 |
0 |
0 |
0 |
0 |
| 0 |
0 |
0 |
1 |
0 |
| 0 |
0 |
1 |
0 |
0 |
| 0 |
0 |
1 |
1 |
0 |
| 0 |
1 |
0 |
0 |
0 |
| 0 |
1 |
0 |
1 |
1 |
| 0 |
1 |
1 |
0 |
1 |
| 0 |
1 |
1 |
1 |
1 |
| 1 |
0 |
0 |
0 |
1 |
| 1 |
0 |
0 |
1 |
1 |
| 1 |
0 |
1 |
0 |
X |
| 1 |
0 |
1 |
1 |
X |
| 1 |
1 |
0 |
0 |
X |
| 1 |
1 |
0 |
1 |
X |
| 1 |
1 |
1 |
0 |
X |
| 1 |
1 |
1 |
1 |
X |
Putting this in kmap and solving
Â
Here crucial point is that we need to make pair of 8 elements using don’t cares also…so final expression is
A+BD+BC
Hence we’ll use two OR gate and one AND gate so total 3 gates.
Ans (B) part.
Quiz of this Question
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Last Updated :
19 Nov, 2018
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