Animated Solution for Physics - Semiconductors: If $V_A$ and $V_B$ are the input voltages (either 5V or 0V) and $V_0$ is the output voltage then the two gates represented in the following circuit (A) and (B) are

In the fascinating world of digital electronics, the abstract logic gates we study are physically built using fundamental analog components like diodes and transistors. This problem challenges us to reverse-engineer two such circuits and identify the logic gates they represent.

The Diode Logic Gate

Let's first look at Circuit A. It consists of two diodes, $D_1$ and $D_2$, connected in parallel. The inputs $V_A$ and $V_B$ are applied to the anodes of these diodes, and their cathodes are tied together to a common output node $V_0$, which is pulled to ground via a resistor $R$.

To determine the logic function, we must construct a truth table by testing all possible combinations of inputs (0V for Logic 0, and 5V for Logic 1).

If both inputs are at 0V ($V_A=0,V_B=0$), neither diode has a positive voltage across it. Both diodes remain reverse-biased (OFF). Since no current flows through the resistor $R$, the voltage drop across it is zero, making the output $V_0=0$.

Now, what if we apply 5V to $V_A$ and keep $V_B$ at 0V? Diode $D_1$ becomes forward-biased and acts like a closed switch. Current flows from $V_A$, through $D_1$, and down through the resistor $R$. This current creates a voltage drop across $R$, pulling the output $V_0$ up to approximately 5V (Logic 1). The exact same logic applies if we reverse the inputs ($V_A=0,V_B=5\text{V}$); diode $D_2$ will conduct, and $V_0$ will again be 5V.

Finally, if both inputs are 5V ($V_A=5\text{V},V_B=5\text{V}$), both diodes are forward-biased. The output $V_0$ is still held at approximately 5V (Logic 1).

Looking at the resulting truth table, we see that the output is high if any of the inputs are high. This is the exact definition of an OR gate.

The Transistor Logic Gate

Now let's shift our focus to Circuit B. This circuit features an NPN transistor configured in the common-emitter mode. The input $V_B$ is applied to the base through a resistor $R_B$, and the output $V_0$ is taken from the collector, which is connected to a 5V supply via a load resistor $R_C$.

Let's analyze its behavior under different input conditions.

If the input $V_B$ is 0V (Logic 0), there is no voltage to drive current into the base of the transistor. The base current $I_B$ is zero, keeping the transistor in the cut-off region. In this state, the transistor acts like an open switch between the collector and emitter. Because no collector current ($I_C$) flows, there is no voltage drop across the resistor $R_C$. Therefore, the output voltage $V_0$ is equal to the supply voltage, which is 5V (Logic 1).

Conversely, if we apply 5V to the input $V_B$ (Logic 1), a significant base current flows through $R_B$. This base current is amplified by the transistor (given $\beta =150$), driving it deep into the saturation region. In saturation, the transistor acts almost like a short circuit between the collector and emitter. A large collector current flows, causing a maximum voltage drop across $R_C$. This pulls the output voltage $V_0$ down to almost 0V (Logic 0).

We observe a clear inversion: a low input yields a high output, and a high input yields a low output. This is the fundamental behavior of a NOT gate (or inverter).

By systematically analyzing the physical behavior of the diodes and the transistor under different voltage conditions, we have successfully deduced that Circuit A functions as an OR gate and Circuit B functions as a NOT gate.