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
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
- A+B(C+D)
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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