Question 8: Semiconductors and Logic Gates

Question

Consider a circuit consisting of a capacitor $(20 \mu \mathrm{~F})$, resistor $(100 \Omega)$ and two identical diodes as shown in figure. The resistance of diode under forward biasing condition is $10 \Omega$. The time constant of the circuit is $\alpha 	imes 10^{-3} \mathrm{~s}$. The value of $\alpha$ is $\_\_\_\_$

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

2.2

Answer

2.0

Answer

2.1

Answer

2.4

Semiconductors and Logic Gates

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