Animated Solution for Physics - Semiconductors: In the following digital circuit, what will be the output at Z, when the input $(A,B)$ are $(1,0),(0,0),(1,1),(0,1)$?

The world of digital electronics is built upon the foundation of logic gates. These simple devices take binary inputs—zeros and ones—and perform logical operations to produce a binary output.

In this problem, we are presented with a network of four logic gates and asked to determine the final output for a series of input combinations. At first glance, the circuit might look like a tangled web of wires, but by breaking it down systematically, we can easily decode its behavior.

Let's start by identifying the components of our circuit. We have two inputs, $A$ and $B$, which feed into a two-layer network of gates.

In the first layer, we have a NAND gate at the top and an OR gate at the bottom. Both of these gates receive inputs directly from $A$ and $B$.

Moving to the second layer, we find an AND gate in the middle. This gate takes its inputs from the outputs of the first layer. Finally, the entire circuit culminates in a NOR gate, which produces our final output, $Z$.

To solve this systematically, let's assign intermediate variables to the outputs of each gate. This will help us track the logic flow without getting lost.

Let $Y_1$ be the output of the top NAND gate. The Boolean expression for a NAND gate is the inverse of an AND gate:

Let $Y_2$ be the output of the bottom OR gate. Its expression is simply:

Now, the middle AND gate takes $Y_1$ and $Y_2$ as inputs. Let's call its output $Y_3$:

Finally, the NOR gate takes $Y_1$ and $Y_3$ as inputs to produce $Z$. A NOR gate is an OR gate followed by an inversion:

With our master equations ready, we can now evaluate the output $Z$ for each given input combination $(A,B)$.

Case 1: $A=1,B=0$ First, we find $Y_1=\overline{1\cdot 0}=\overline{0}=1$. Next, $Y_2=1+0=1$. Then, $Y_3=1\cdot 1=1$. Finally, $Z=\overline{1+1}=\overline{1}=0$.

Case 2: $A=0,B=0$ Here, $Y_1=\overline{0\cdot 0}=1$. $Y_2=0+0=0$. $Y_3=1\cdot 0=0$. And $Z=\overline{1+0}=0$.

Case 3: $A=1,B=1$ For this case, $Y_1=\overline{1\cdot 1}=0$. $Y_2=1+1=1$. $Y_3=0\cdot 1=0$. And $Z=\overline{0+0}=1$.

Case 4: $A=0,B=1$ Lastly, $Y_1=\overline{0\cdot 1}=1$. $Y_2=0+1=1$. $Y_3=1\cdot 1=1$. And $Z=\overline{1+1}=0$.

Putting it all together, the sequence of outputs is $0,0,1,0$.

While evaluating each case step-by-step is a foolproof method, there is a much more elegant way to solve this problem using Boolean Algebra.

Let's look closely at our final expression for $Z$:

We know that $Y_3=Y_1\cdot Y_2$. Let's substitute this into the equation:

Now, we can factor out $Y_1$:

In Boolean algebra, anything ORed with $1$ is simply $1$. Therefore, $(1+Y_2)=1$. This is known as the Absorption Law. Our equation simplifies dramatically:

Since $Y_1=\overline{A\cdot B}$, we can substitute it back:

The double inversion cancels out, leaving us with:

This is a mind-blowing revelation! The entire complex network of four gates is logically equivalent to a single AND gate. If we had realized this from the start, we could have instantly determined the outputs just by looking at the inputs: only the $(1,1)$ case produces a $1$.

This beautifully illustrates the power of Boolean algebra in simplifying and optimizing digital circuits.