Semiconductors — JEE Main · 2023 · 30 Jan 2023 · Shift 2 Question with Solution

The output Y for the inputs A and B of circuit is given by

Truth table of the shown circuit is :

(A)\begin{tabular}{lll|l} & A & B & Y \\ \hline 0 & 0 & 1 \\ & 0 & 1 & 1 \\ & 1 & 0 & 1 \\ & 1 & 1 & 0 \end{tabular}
(B)\begin{tabular}{lll|l} & A & B & Y \\ \hline & 0 & 0 & 1 \\ & 0 & 1 & 0 \\ & 1 & 0 & 0 \\ & 1 & 1 & 1 \end{tabular}
(C)\begin{tabular}{lll|l} & A & B & Y \\ \hline & 0 & 0 & 0 \\ & 0 & 1 & 1 \\ & 1 & 0 & 1 \\ & 1 & 1 & 1 \end{tabular}
(D)\begin{tabular}{lll|l} & A & B & Y \\ \hline 0 & 0 & 0 \\ & 0 & 1 & 1 \\ & 1 & 0 & 1 \\ & 1 & 1 & 0 \end{tabular}✓ Correct

Answer: (D)

Solution

Call the output of the gate fed by both inputs

\overline{A \cdot B}

. This feeds, together with

A

, the upper gate to give

\overline{A \cdot \overline{A \cdot B}}

, and together with

B

the lower gate to give

\overline{B \cdot \overline{A \cdot B}}

. The final gate combines these two outputs as

Y = \overline{\overline{A \cdot \overline{A B}} \cdot \overline{B \cdot \overline{A B}}} = A \overline{B} + \overline{A} B

. This is high only when the two inputs differ, so the truth table reads

(0, 0) \to 0

(0, 1) \to 1

(1, 0) \to 1

(1, 1) \to 0

JEE Main · 2023 · 30 Jan 2023 · Shift 2 · Single correct option