Animated Solution for Physics - Semiconductors: Boolean relation at the output stage $Y$ for the following circuit is

When you first look at a circuit diagram filled with diodes, resistors, and transistors, it can feel a bit overwhelming. But the beauty of digital electronics is that complex systems are just combinations of very simple, atomic building blocks.

In this problem, we are looking at a classic example of Diode-Transistor Logic (DTL). To understand what this circuit does, we need to mentally slice it into two distinct stages: the left side containing the diodes, and the right side containing the transistor.

Let's focus on the left side first. We have two inputs, $A$ and $B$, connected to the p-sides (anodes) of two diodes. The n-sides (cathodes) of these diodes are tied together and connected to ground through a resistor.

Imagine what happens if we apply a high voltage (let's say $5\text{V}$, representing logic $1$) to input $A$. The diode connected to $A$ becomes forward-biased and acts like a closed switch. Current flows through the diode and the resistor, creating a voltage drop across the resistor. This means the common point between the diodes and the resistor is now at a high voltage (logic $1$).

The exact same thing happens if input $B$ is high, or if both $A$ and $B$ are high. The only way the common point stays at $0\text{V}$ (logic $0$) is if both $A$ and $B$ are low, keeping both diodes reverse-biased (off).

This behavior perfectly describes an OR gate. The intermediate output at the common point is $A+B$.

Now, let's trace that intermediate signal into the right side of the circuit. The common point of the diodes is connected to the base of an NPN transistor.

This transistor is configured as a simple electronic switch.

1. When the base is HIGH: The transistor turns ON (enters saturation). It acts like a short circuit between its collector and emitter. Since the emitter is tied to ground, the collector (which is our output $Y$) is pulled down to $0\text{V}$. So, a HIGH input gives a LOW output. 2. When the base is LOW: The transistor turns OFF (enters cut-off). It acts like an open circuit. No current flows through the collector resistor, so the output $Y$ is pulled up to $+5\text{V}$ by the power supply. So, a LOW input gives a HIGH output.

This is the exact definition of a NOT gate (or inverter).

We have successfully decoded the two stages. The circuit takes inputs $A$ and $B$, passes them through an OR gate, and then passes that result through a NOT gate.

An OR gate followed by a NOT gate is a NOR gate.

Mathematically, we can write the Boolean expression for the output $Y$ as:

We have our expression, but if we look at the multiple-choice options, $\overline{A+B}$ isn't explicitly listed. This is a classic trap! We need to manipulate our expression into an equivalent form using Boolean algebra.

This is where De Morgan's Laws come into play. The specific law we need states that the complement of a sum is equal to the product of the complements. A handy mnemonic for this is: "Break the line, change the sign."

Applying this to our expression:

Looking at our options, we can clearly see that option (d) matches our derived expression perfectly. The circuit implements the logic $Y=\overline{A}\cdot \overline{B}$.