Question 6: Semiconductors and Logic Gates

Question

$$ 	ext { Identify the correct truth table of the given logic circuit. } $$

Accepted Answer

Logic gate 1 is an AND gate with both inputs shorted together to $A$. So, the output $\left(Y_1ight)$ is just $A$.
$$ \begin{array}{|c|c|c|} \hline A & B & Y_1 \ \hline 0 & 0 & 0 \ \hline 0 & 1 & 0 \ \hline 1 & 0 & 1 \ \hline 1 & 1 & 1 \ \hline \end{array} $$
Logic Gate 2 is an NAND gate with inputs A and B and its output is $\mathrm{Y}_2$.
$$ \begin{array}{|c|c|c|c|} \hline A & B & A \cdot B & Y_2=\overline{A \cdot B} \ \hline 0 & 0 & 0 & 1 \ \hline 0 & 1 & 0 & 1 \ \hline 1 & 0 & 0 & 1 \ \hline 1 & 1 & 1 & 0 \ \hline \end{array} $$
Logic gate 3 is an AND gate with both inputs shorted to the output of the logic gate 2. So, the output ( $Y_3$ ) is just the output of logic gate 2.
$$ \begin{array}{|c|c|c|c|c|} \hline A & B & A \cdot B & Y_2=\overline{A \cdot B} & Y_3 \ \hline 0 & 0 & 0 & 1 & 1 \ \hline 0 & 1 & 0 & 1 & 1 \ \hline 1 & 0 & 0 & 1 & 1 \ \hline 1 & 1 & 1 & 0 & 0 \ \hline \end{array} $$
Logic gate 4 is an AND gate that takes the output from the outputs $(\mathrm{Y})$ of logic gate 1 and the logic gate 3.
$$ \begin{array}{|c|c|c|c|c|} \hline A & B & Y_1 & Y_3 & Y=Y_1 \cdot Y_3 \ \hline 0 & 0 & 0 & 1 & 0 \ \hline 0 & 1 & 0 & 1 & 0 \ \hline 1 & 0 & 1 & 1 & 1 \ \hline 1 & 1 & 1 & 0 & 0 \ \hline \end{array} $$
Therefore, the truth table of the given logic gate is,
$$ \begin{aligned} &\begin{array}{|c|c|c|} \hline A & B & Y \ \hline 0 & 0 & 0 \ \hline 0 & 1 & 0 \ \hline 1 & 0 & 1 \ \hline 1 & 1 & 0 \ \hline \end{array}\ &	ext { Hence, the correct option is (A) } \end{aligned} $$

Answer

$$ \begin{array}{|c|c|c|} \hline 	ext { A } & 	ext { B } & 	ext { Y } \ \hline \hline 0 & 0 & 0 \ \hline \hline 0 & 1 & 0 \ \hline 1 & 0 & 1 \ \hline 1 & 1 & 0 \ \hline \end{array} $$

Answer

$$ \begin{array}{|c|c|c|} \hline 	ext { A } & 	ext { B } & 	ext { Y } \ \hline \hline 0 & 0 & 0 \ \hline \hline 0 & 1 & 1 \ \hline 1 & 0 & 1 \ \hline 1 & 1 & 0 \ \hline \end{array} $$

Answer

$$ \begin{array}{|c|c|c|} \hline 	ext { A } & 	ext { B } & 	ext { Y } \ \hline \hline 0 & 0 & 1 \ \hline \hline 0 & 1 & 1 \ \hline \hline 1 & 0 & 1 \ \hline \hline 1 & 1 & 0 \ \hline \end{array} $$

Answer

$$ \begin{array}{|c|c|c|} \hline 	ext { A } & 	ext { B } & 	ext { Y } \ \hline \hline 0 & 0 & 1 \ \hline \hline 0 & 1 & 0 \ \hline 1 & 0 & 1 \ \hline 1 & 1 & 0 \ \hline \end{array} $$

Semiconductors and Logic Gates

\text { Identify the correct truth table of the given logic circuit. }

Choose the correct answer: