Animated Solution for Physics - Semiconductors: In the adjacent circuit, $A$ and $B$ represent two inputs and $C$ represents the output. The circuit represents

Imagine you are looking at a simple electronic circuit. We have two inputs, $A$ and $B$, which are connected to the p-sides of two separate diodes.

The n-sides of these two diodes are tied together. From this common junction, the circuit branches out: one path goes to our output terminal $C$, and the other path goes through a resistor down to the ground.

To figure out which logic gate this circuit represents, we need to analyze its behavior for all possible input combinations. This means we are going to build its truth table from scratch.

Diode Biasing

Before we dive into the cases, let's establish the ground rules. In digital logic, a logic $1$ corresponds to a high positive voltage (like $5\text{V}$), and a logic $0$ corresponds to zero volts, or ground.

A diode is like a one-way valve for electricity. It will only conduct current if its p-side is at a higher electrical potential than its n-side. When this happens, we say the diode is forward-biased, and it acts like a closed switch.

If the p-side is at a lower or equal potential compared to the n-side, the diode is reverse-biased. It acts like an open switch, blocking any current from flowing.

Case 1: Both Inputs Low Let's test the first scenario where both $A=0$ and $B=0$. Since both inputs are at $0\text{V}$, neither diode has a positive voltage on its p-side.

Both diodes are reverse-biased and act as open switches. Because no current can flow through the circuit, there is no voltage drop across the resistor. Consequently, the output $C$ remains firmly at $0\text{V}$, which means $C=0$.

Case 2: One Input High Now, what if $A=0$ and $B=1$? Input $B$ is now at a high voltage. This makes the second diode forward-biased, turning it into a closed switch.

Current flows from input $B$, passes through the second diode, and travels down through the resistor to ground. According to Ohm's Law, this current creates a voltage drop across the resistor. This pulls the output $C$ up to a high voltage, giving us $C=1$.

By symmetry, if we flip the inputs so that $A=1$ and $B=0$, the exact same logic applies. The first diode becomes forward-biased, current flows from $A$, and the output $C$ again becomes $1$.

Case 3: Both Inputs High Finally, let's consider when both $A=1$ and $B=1$. In this state, both diodes are receiving a high voltage on their p-sides.

Both diodes become forward-biased simultaneously. Current flows from both inputs, merging at the junction and passing through the resistor. The voltage at output $C$ is maintained at a high level, so $C=1$.

Let's step back and look at the complete truth table we have constructed.

When the inputs are $(0,0)$, the output is $0$. For any other combination—$(0,1)$, $(1,0)$, or $(1,1)$—the output is $1$.

This behavior perfectly matches the definition of an OR gate. An OR gate outputs a high signal if at least one of its inputs is high. Therefore, our diode-resistor circuit is a physical realization of an OR gate!