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?

Explanation:

Implementation of a One-Bit Full Adder Using a 3 ∶ 8 Active High Decoder

Definition: A one-bit full adder is a combinational circuit designed to perform the addition of three binary inputs: the two significant bits (A and B) and a carry-in (Cin). It produces two outputs: the sum (S) and carry-out (Cout).

Overview: To implement a one-bit full adder using a 3 ∶ 8 active high output decoder, the decoder is used to generate specific combinations of outputs corresponding to the truth table of the full adder. The outputs of the decoder are then combined using OR gates to derive the required sum and carry outputs.

Working of a 3 ∶ 8 Decoder:

A 3 ∶ 8 decoder takes 3 inputs (A, B, Cin) and generates 8 distinct outputs (Y0 to Y7). Each output corresponds to one unique combination of the inputs. The output is active high, meaning that for a given combination of inputs, only one output is high (logic 1) while the others remain low (logic 0).

For example:

• If A = 0, B = 0, Cin = 0, then output Y0 is high.
• If A = 0, B = 0, Cin = 1, then output Y1 is high.
• If A = 1, B = 1, Cin = 1, then output Y7 is high.

The truth table for a one-bit full adder is as follows:

Implementation:

To implement the full adder using the 3 ∶ 8 decoder, the outputs of the decoder are used to construct the Sum (S) and Carry (Cout) outputs according to the truth table. Each output is obtained by combining specific outputs of the decoder using OR gates.

Deriving Sum (S):

The Sum (S) is high (logic 1) for the following combinations of inputs:

• A = 0, B = 0, Cin = 1 → Decoder output Y1
• A = 0, B = 1, Cin = 0 → Decoder output Y2
• A = 1, B = 0, Cin = 0 → Decoder output Y4
• A = 1, B = 1, Cin = 1 → Decoder output Y7

Thus, Sum (S) = Y1 + Y2 + Y4 + Y7.

Deriving Carry (Cout):

The Carry (Cout) is high (logic 1) for the following combinations of inputs:

• A = 0, B = 1, Cin = 1 → Decoder output Y3
• A = 1, B = 0, Cin = 1 → Decoder output Y5
• A = 1, B = 1, Cin = 0 → Decoder output Y6
• A = 1, B = 1, Cin = 1 → Decoder output Y7

Thus, Carry (Cout) = Y3 + Y5 + Y6 + Y7.

Minimum Number of OR Gates:

To implement the Sum (S) and Carry (Cout) outputs:

• Sum (S) requires four OR gates to combine Y1, Y2, Y4, and Y7.
• Carry (Cout) requires four OR gates to combine Y3, Y5, Y6, and Y7.

However, one OR gate can be shared between Sum (S) and Carry (Cout) for Y7, reducing the total number of OR gates required.

Final Calculation:

Total OR gates = 4 (for Sum) + 4 (for Carry) - 1 (shared gate for Y7) = 7.

Since the question specifies the minimum number of 3-input OR gates, we can combine multiple outputs into groups of three, further reducing the number of gates.

Using this grouping approach:

• Sum (S): Combine Y1, Y2, and Y4 into one 3-input OR gate, and Y7 into another OR gate → 2 OR gates.
• Carry (Cout): Combine Y3, Y5, and Y6 into one 3-input OR gate, and Y7 into another OR gate → 2 OR gates.

Thus, the minimum number of 3-input OR gates required = 4.

Correct Option:

The correct answer is option 3) 4.

Important Information

To analyze the other options:

Option 1 (2 OR gates):

This option is incorrect because it underestimates the number of OR gates required. Each of the Sum (S) and Carry (Cout) outputs requires at least two OR gates to combine the necessary decoder outputs. Therefore, a total of two OR gates is insufficient.

Option 2 (5 OR gates):

This option is incorrect because it does not represent the minimum number of 3-input OR gates required. While 5 OR gates may work, combining outputs into groups of three allows the implementation to be optimized to 4 gates.

Option 4 (3 OR gates):

This option is incorrect because it underestimates the number of OR gates required. Even with optimal grouping, at least 4 OR gates are needed to derive the Sum (S) and Carry (Cout) outputs.

Conclusion:

The minimum number of 3-input OR gates required to implement a one-bit full adder using a 3 ∶ 8 active high decoder is 4, making option 3 the correct choice.