Question

The truth table depicting the logic gate realized from the given arrangements of $p-n$ junction and resistor as shown in figure will be :

• A. $\begin{array}{ccc}\text{A}& \text{B}& \text{Y}\\ 0& 0& 0\\ 0& 1& 1\\ 1& 0& 1\\ 1& 1& 1\end{array}$
• B. $\begin{array}{ccc}\text{A}& \text{B}& \text{Y}\\ 0& 0& 1\\ 0& 1& 1\\ 1& 0& 1\\ 1& 1& 0\end{array}$
• C. $\begin{array}{ccc}\text{A}& \text{B}& \text{Y}\\ 0& 0& 0\\ 1& 0& 0\\ 0& 1& 0\\ 1& 1& 1\end{array}$
• D. $\begin{array}{ccc}\text{A}& \text{B}& \text{Y}\\ 0& 0& 1\\ 0& 1& 0\\ 1& 0& 0\\ 1& 1& 0\end{array}$

Answer 1

The correct option is A $\begin{array}{ccc}\text{A}& \text{B}& \text{Y}\\ 0& 0& 0\\ 0& 1& 1\\ 1& 0& 1\\ 1& 1& 1\end{array}$

In the given arrangement two diodes are connected in parallel which are further connected to a grounded resistor.

$\Rightarrow$ In the given circuit, the output will be obtained if current flows in the circuit.

Let us assign a high value $\left(1\right)$ for flow of current, similarly a low value $\left(0\right)$ for no current in circuit.

Now coming to diodes, if any of the diode is forward biased then it will allow current to flow through the resistor and the circuit gets completed.

Also, there will be no current through the resistor if only both the diodes are reverse biased.

For inputs,

Diode ( Forward bias ) $\to 1$ ( Current flow)

Diode ( Reverse bias ) $\to 0$ ( No Current)

Constructing the truth table:

$\begin{array}{ccc}\text{A}& \text{B}& \text{Y}\\ 0& 0& 0\\ 0& 1& 1\\ 1& 0& 1\\ 1& 1& 1\end{array}$

Hence, option $\left(A\right)$ is correct.

Why this Question? Tip : If we look at the circuit, we infer that current will pass through the resistor, if the inputs $A$, $B$ are such that either diode $D_1\text{or}{D}_2$ becomes forward biased. Thus, it is a realization of $\text{OR}$ gate.