Animated Solution for Physics - Semiconductors: In the logic circuit shown in the figure, if input $A$ and $B$ are 0 to 1 respectively, the output at $Y$ would be $x$. The value of $x$ is ......

Decoding the Logic Circuit: A Step-by-Step Signal Trace

Logic gates are the fundamental building blocks of digital electronics. When multiple gates are combined into a circuit, it might look like a complex maze of wires. However, the secret to solving these problems quickly and accurately lies in methodically tracing the signal from the inputs all the way to the output. Let's dive into this specific circuit and see how a simple substitution method can lead us straight to the answer.

We are presented with a logic circuit comprising two NOT gates, two NOR gates, and one final AND gate. The problem gives us specific input values: $A=0$ and $B=1$. Our goal is to find the final output $Y$, which is denoted as $x$.

The most efficient way to tackle this in an exam setting is direct substitution. Instead of writing out a long Boolean expression for the entire circuit, we can simply push the given 0s and 1s through the gates one by one.

Let's start at the very beginning with our inputs. We have $A=0$ and $B=1$.

The NOT Gates: The first components our signals encounter are the NOT gates. A NOT gate simply inverts the signal. For input $A$: The signal $0$ passes through the NOT gate, becoming $\bar{A}=1$. For input $B$: The signal $1$ passes through the NOT gate, becoming $\bar{B}=0$.

The Cross Connections: Here is where many students make a silly mistake. If you look closely at the circuit diagram, the wires for $A$ and $B$ branch off before they enter the NOT gates. This means the original signal $B=1$ is fed directly into the second input of the top NOR gate. Similarly, the original signal $A=0$ is fed directly into the first input of the bottom NOR gate.

Now we have all the necessary inputs to evaluate the two NOR gates. Remember, a NOR gate is an OR gate followed by a NOT gate. It outputs $1$ only if both inputs are $0$.

Top NOR Gate: The inputs are $\bar{A}=1$ and $B=1$. The OR operation gives $1+1=1$. Inverting this (the NOT part of NOR) gives us $\overline{1}=0$. So, the output of the top NOR gate is $0$.

Bottom NOR Gate: The inputs are $A=0$ and $\bar{B}=0$. The OR operation gives $0+0=0$. Inverting this gives us $\overline{0}=1$. So, the output of the bottom NOR gate is $1$.

We are almost there! The outputs of the two NOR gates now serve as the inputs for the final AND gate.

The inputs to the AND gate are $0$ (from the top NOR) and $1$ (from the bottom NOR). An AND gate performs logical multiplication and outputs $1$ only if both inputs are $1$. Therefore, $Y=0\cdot 1=0$.

The final output $Y$ is $0$, which means the value of $x$ is $0$.

By carefully tracing the path of each signal and applying the basic rules of logic gates step-by-step, we've effortlessly navigated the circuit. While you could also solve this by deriving the Boolean expression $Y=(\overline{\bar{A}+B})\cdot (\overline{A+\bar{B}})$, the direct substitution method is often much faster and less prone to algebraic errors during a timed exam.