Animated Solution for Physics - Semiconductors: A full wave rectifier circuit along with the output is shown in figure. The contribution (s) from the diode 1 is (are)

Imagine you have an alternating current (AC) source that constantly flips its direction, but you need a steady, one-way flow of electricity (DC) to power your devices. This is where the full-wave rectifier comes to the rescue.

The circuit consists of three main components: an AC input source, a center-tapped transformer, and two p-n junction diodes. The transformer's job is to step down the voltage and provide a center tap, which acts as a neutral reference point. The diodes act as one-way valves, ensuring that no matter which way the AC input swings, the current through the load always flows in the same direction.

Diode 1 Takes Charge

Let's trace the journey of the current. During the first positive half-cycle of the AC input, the top of the transformer's secondary coil becomes positive relative to the center tap, while the bottom becomes negative.

Because the top is positive, Diode 1 becomes forward-biased (its p-side is at a higher potential than its n-side). It opens its gates and allows current to flow through it, down through the load resistor, and back to the center tap. This creates the first positive pulse in our output waveform, labeled as pulse A.

Meanwhile, Diode 2 is reverse-biased and acts like an open switch, blocking any current from flowing through the bottom path.

Diode 2 Steps In

Now, the AC input flips. During the negative half-cycle, the bottom of the secondary coil becomes positive relative to the center tap, and the top becomes negative.

This time, Diode 2 becomes forward-biased and starts conducting. The current flows through Diode 2, and crucially, it flows through the load resistor in the exact same direction as before! This creates the second positive pulse in our output, labeled as pulse B.

This elegant dance continues. In the next positive half-cycle, Diode 1 takes over again, producing pulse C. In the subsequent negative half-cycle, Diode 2 produces pulse D.

Therefore, we can clearly see that Diode 1 is responsible for the odd pulses (A and C), while Diode 2 is responsible for the even pulses (B and D).

You might wonder why the official answer key accepts both options (b) and (c). The reason lies in a subtle detail: the diagram does not explicitly specify the winding direction of the transformer coils.

If the transformer were wound in the opposite direction, the top of the secondary coil would become negative during the first positive half-cycle of the primary input. In that scenario, Diode 2 would conduct first (producing pulse A), and Diode 1 would conduct second (producing pulse B). Thus, Diode 1 would be responsible for pulses B and D. Because both scenarios are physically possible depending on the transformer's internal construction, both (b) and (c) are considered technically correct.