Applications Of Diode

Step 1: Understanding the Concept:
Forward bias reduces the depletion layer width and lowers the potential barrier, allowing majority carriers to flow easily.

Step 2: Detailed Explanation:
In forward bias, the positive terminal of the battery is connected to the p-side and the negative terminal to the n-side. This reduces the depletion layer width and the potential barrier, making it easy for majority carriers to cross the junction. The resistance of the p-n junction becomes very low (order of ohms), and a large current (milliamperes) flows due to majority carriers.
• (A) Correct: Very low resistance.
• (B) Incorrect: Depletion width decreases.
• (C) Incorrect: Potential barrier decreases.
• (D) Incorrect: Current is in milliamperes, not microamperes.
• (E) Incorrect: Current is due to majority carriers.

Step 3: Final Answer:
In forward bias, the p-n junction provides very low resistance.

Concept: Reverse bias widens depletion region.

Step 1: Effect.
Potential barrier increases.

Concept:

Step 1: Peak voltage.


Step 2: Total resistance.


Step 3: Current.

Step 1: Understanding the Concept:
A full-wave rectifier converts both halves of AC cycle to DC. Two types: centre-tap and bridge.

Step 2: Detailed Explanation:
In a centre-tap full-wave rectifier, a centre-tapped transformer is essential. For positive cycle, one diode conducts; for negative cycle, the other conducts.
(A) Incorrect: Output appears for both cycles.
(B) & (C) Incorrect: Only one diode conducts per half-cycle.
(D) Incorrect: Diodes alternate being forward/reverse biased.
(E) Correct: A centre-tapped transformer is a key component.

Step 3: Final Answer:
A full wave rectifier uses a centre tapped transformer.

A pn junction diode without bias behaves as a resistor. This is because, when no bias is applied to the pn junction diode, it doesn't conduct current effectively, behaving as a high-resistance component. However, when forward bias is applied, the diode allows current to flow, and its behavior changes.
Thus, the correct answer is .

Concept: Electronics - Rectifiers and Ripple Frequency. Ripple frequency is the number of output pulses produced per second after rectification.
Step 1: Understand full-wave rectifier working. A full-wave rectifier converts both the positive and negative half cycles of AC input into positive output pulses.
Step 2: Count pulses in one cycle. For one complete AC cycle: * Positive half cycle gives one pulse. * Negative half cycle also gives one pulse. So, total output pulses in one cycle = 2.
Step 3: Relate to frequency. If input AC frequency is , then there are cycles per second. Since each cycle gives 2 pulses: $ $
Hence, correct option is (C).

Step 1: Understanding the Concept:
A rectifier converts alternating current (AC) to direct current (DC). The output is not perfectly smooth DC but contains fluctuations called ripples. The frequency of these ripples depends on the type of rectifier used.

Step 2: Key Formula or Approach:
Understand the operation waveform of a full-wave rectifier. It "flips" the negative half-cycles of the input AC sine wave into positive half-cycles.

Step 3: Detailed Explanation:
Let the input AC signal have a frequency . This means it completes one full cycle (one positive half and one negative half) in time .
A full-wave rectifier conducts current during both half-cycles. The output waveform consists of a series of positive pulses.
For every single full cycle of the input AC, the full-wave rectifier produces two identical positive output pulses.
Therefore, the fundamental frequency of the repeating pattern in the output (the ripple frequency) is twice the frequency of the input signal.

For example, if the standard mains supply frequency is 50 Hz, the ripple frequency from a full-wave rectifier will be 100 Hz.

Step 4: Final Answer:
The ripple frequency is .

Concept:
• Discharge of capacitor:

Step 1: Rate of fall


Step 2: Interpretation
Larger → slower discharge
Smaller → faster drop Final Conclusion:
Option (E)