Question 10: Semiconductors and Logic Gates

Question

For the given circuit (shown in part (A)) the time dependent input voltage $v_{in}(t)$ and corresponding output $v_{o}(t)$ are shown in part (B) and part (C), respectively. Identify the components that are used in the circuit between points X and Y.

Accepted Answer

For the input voltages at A and B we assume a positive logic system where 0 V represents logic 0 (LOW) and 5 V represents logic 1 (HIGH).
The supply voltage $\mathrm{V}_{\mathrm{dc}}=5 \mathrm{~V}$ connected through a resistor R to the output C .
Diodes $\mathrm{D}_1$ and $\mathrm{D}_2$ are connected with their n -sides (cathodes) toward the inputs A and B . They act as switches.
$$ \mathrm{A}=0 \mathrm{~V}, \mathrm{~B}=0 \mathrm{~V} 	ext { (Both LOW) } $$
Both diodes are forward-biased because the p -side (connected to $\mathrm{V}_{\mathrm{dc}}$ ) is at a higher potential than the n -side (connected to 0 V ). Current flows through the diodes to the ground. This pulls the potential at C down to approximately 0 V , i.e., output $\mathrm{C}=0$ (LOW).
$$ \begin{array}{|l|l|l|} \hline 	ext { A } & 	ext { B } & 	ext { C } \ \hline 0 & 0 & 0 \ \hline \end{array} $$
$\mathrm{A}=0 \mathrm{~V}, \mathrm{~B}=5 \mathrm{~V}$ (One LOW, One HIGH)
Diode $\mathrm{D}_1$ is forward-biased and conducts current to the 0 V input. This shorts point C to ground through the conducting diode. Diode $\mathrm{D}_2$ is reverse-biased and does not conduct. Because $\mathrm{D}_1$ is conducting, point C remains at approximately 0 V , i.e., output $\mathrm{C}=0$ (LOW).
$$ \begin{array}{|l|l|l|} \hline 	ext { A } & 	ext { B } & 	ext { C } \ \hline 0 & 1 & 0 \ \hline \end{array} $$
$$ \mathrm{A}=5 \mathrm{~V}, \mathrm{~B}=0 \mathrm{~V} 	ext { (One HIGH, One LOW) } $$
Diode $\mathrm{D}_2$ is forward-biased and conducts current to the 0 V input at B . This pulls point C down to ground level, i.e., output $\mathrm{C}=0$ (LOW).
$$ \begin{array}{|l|l|l|} \hline 	ext { A } & 	ext { B } & 	ext { C } \ \hline 1 & 0 & 0 \ \hline \end{array} $$
$$ \mathrm{A}=5 \mathrm{~V}, \mathrm{~B}=5 \mathrm{~V} 	ext { (Both HIGH) } $$
Both n -sides and p -sides of the diodes are at 5 V (or the n -side is higher than the p -side). The diodes do not conduct, i.e., they are OFF. Since no current flows through the resistor R, there is no voltage drop across it. Therefore, point C stays at the supply voltage of 5 V , i.e., output $\mathrm{C}=1(\mathrm{HIGH})$.
$$ \begin{array}{|l|l|l|} \hline 	ext { A } & 	ext { B } & 	ext { C } \ \hline 1 & 1 & 1 \ \hline \end{array} $$
The resulting truth table for the given circuit is,
$$ \begin{array}{|l|l|l|} \hline A & B & C \ \hline 0 & 0 & 0 \ \hline 0 & 1 & 0 \ \hline 1 & 0 & 0 \ \hline 1 & 1 & 1 \ \hline \end{array} $$
Hence, the logic gate is an AND Gate, where the output is HIGH only when all inputs are HIGH.
Hence, the correct option is (A).

Answer

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Semiconductors and Logic Gates

For the given circuit (shown in part (A)) the time dependent input voltage v_{in}(t) and corresponding output v_{o}(t) are shown in part (B) and part (C), respectively. Identify the components that are used in the circuit between points X and Y.

Choose the correct answer: