Topic: Digital Circuits
Given a single 3:8 active high output decoder. What is the minimum number of 3-input OR gates required to implement a one bit Full adder?
- A. 2
- B. 5
- C. 4
- D. 3
Answer: C. 4
Explanation: A 1-bit full adder produces Sum = $\Sigma(1, 2, 4, 7)$ and $C_{out} = \Sigma(3, 5, 6, 7)$. With a 3:8 decoder, these 8 minterms are directly available. Each output requires ORing 4 terms. A single 3-input OR gate can only accept 3 inputs. To sum 4 minterms for one output, you need at least two 3-input OR gates cascaded (e.g., ORing the first three minterms, then feeding that result and the fourth minterm into the second OR gate). Since there are two outputs (Sum and Cout), you need a minimum of $2 \times 2 = 4$ gates.