NEOCODE

PN Junction Diode MCQs

1. Basics of PN Junction Diode

1. A PN junction diode is formed by joining:

Correct Answer: (C) P-type and N-type semiconductors

Explanation:
A PN junction diode is formed by joining a P-type semiconductor (with acceptor impurities creating holes) and an N-type semiconductor (with donor impurities creating free electrons). The interface where they meet is called the PN junction, which forms the basis for diode operation.

2. In a PN junction, the depletion region is formed due to:

Correct Answer: (C) Recombination of electrons and holes

Explanation:
The depletion region forms when electrons from the N-side diffuse across the junction and recombine with holes from the P-side. This creates a region depleted of mobile charge carriers near the junction, leaving behind fixed positive donor ions on the N-side and fixed negative acceptor ions on the P-side.

3. The width of the depletion region increases when:

Correct Answer: (B) Reverse biased

Explanation:
Under reverse bias, the applied voltage increases the potential barrier, causing more charge carriers to be pulled away from the junction. This widens the depletion region. In forward bias, the opposite occurs - the depletion region narrows as carriers are pushed toward the junction.

4. The potential barrier in a silicon PN junction is approximately:

Correct Answer: (B) 0.7V

Explanation:
For silicon diodes, the typical barrier potential is about 0.7V at room temperature. For germanium diodes, it's approximately 0.3V. This potential must be overcome by forward bias voltage before significant current flows. The exact value depends on doping concentrations and temperature.

2. PN Junction Diode Characteristics

5. In forward bias, a PN junction diode behaves as:

Correct Answer: (C) A conductor with low resistance

Explanation:
When forward biased (voltage > barrier potential), the diode conducts current easily with relatively low resistance. The depletion region narrows, allowing majority carriers to flow across the junction. The forward resistance is typically in the range of a few ohms to tens of ohms.

6. In reverse bias, the diode current is mainly due to:

Correct Answer: (B) Minority carriers

Explanation:
Under reverse bias, the small reverse saturation current (IS) is primarily due to thermally generated minority carriers (electrons in P-region and holes in N-region). Majority carriers are blocked by the widened depletion region. This current is typically in the nanoampere range for silicon diodes.

7. In the IV characteristics of a diode, the reverse saturation current is:

Correct Answer: (B) Increases with temperature

Explanation:
Reverse saturation current approximately doubles for every 10°C rise in temperature because it depends on intrinsic carrier concentration (ni), which is highly temperature dependent. This is why diode characteristics change significantly with temperature variations.

8. The knee voltage of a PN junction diode refers to:

Correct Answer: (B) The voltage at which the diode starts conducting appreciably in forward bias

Explanation:
The knee voltage is the forward voltage (about 0.7V for Si, 0.3V for Ge) where the diode current begins increasing rapidly. Below this voltage, current is minimal; above it, current increases exponentially with voltage according to the diode equation: I = IS(eV/nVT - 1).

3. Diode Applications

9. A Zener diode is primarily used for:

Correct Answer: (B) Voltage regulation

Explanation:
Zener diodes are specially designed to operate in the reverse breakdown region, maintaining a nearly constant voltage (VZ) across their terminals despite changes in current. This makes them ideal for voltage regulation, voltage reference, and overvoltage protection applications.

10. In a half-wave rectifier, the output frequency of the rectified signal is:

Correct Answer: (A) Same as input AC frequency

Explanation:
A half-wave rectifier conducts during only one half-cycle of the input AC waveform, producing one output pulse per input cycle. Thus, the output ripple frequency equals the input frequency (e.g., 50Hz input → 50Hz output). Full-wave rectifiers double this frequency.

11. A full-wave rectifier converts:

Correct Answer: (A) AC to DC

Explanation:
Full-wave rectifiers (center-tapped or bridge configuration) convert alternating current (AC) to pulsating direct current (DC) by conducting during both half-cycles of the input waveform. With filtering, this produces nearly constant DC output. They are more efficient than half-wave rectifiers.

12. In a clipper circuit, the function of a diode is to:

Correct Answer: (C) Remove specific portions of the waveform

Explanation:
Clipper circuits use diodes to "clip" or remove portions of a waveform above or below certain voltage levels. They are used for waveform shaping, noise removal, and signal conditioning. Series clippers place the diode in series with the load, while parallel clippers place it across the load.

13. A clamping circuit is used to:

Correct Answer: (A) Change the DC level of a waveform

Explanation:
Clamper circuits (DC restorers) shift the entire waveform up or down without changing its shape. They typically consist of a diode, capacitor, and sometimes a DC bias source. Applications include radar systems, TV receivers, and signal processing where DC restoration is needed.

14. Which type of diode is used in light-emitting applications?

Correct Answer: (C) LED

Explanation:
Light Emitting Diodes (LEDs) emit photons when electrons recombine with holes in the forward-biased PN junction. The color depends on the semiconductor material's bandgap (e.g., GaAsP for red, GaN for blue). LEDs are highly efficient, durable, and used in displays, lighting, and indicators.

4. Special Types of Diodes

15. A Zener diode operates in which region during voltage regulation?

Correct Answer: (B) Reverse breakdown region

Explanation:
Zener diodes are designed to operate in the reverse breakdown region, where they maintain a nearly constant voltage (VZ) across a wide range of reverse currents. This makes them ideal for voltage regulation applications. The breakdown can be either Zener effect (low VZ) or avalanche effect (high VZ).

16. A Schottky diode has:

Correct Answer: (C) A low forward voltage drop

Explanation:
Schottky diodes (metal-semiconductor junction) have lower forward voltage drop (~0.15-0.45V) than PN junction diodes (~0.7V for Si). They also have faster switching speeds due to majority carrier operation (no minority carrier storage). Used in high-frequency applications, power supplies, and clamping circuits.

17. Which diode is light-sensitive?

Correct Answer: (B) Photodiode

Explanation:
Photodiodes generate current when exposed to light due to the photoelectric effect. They operate in reverse bias (photoconductive mode) or zero bias (photovoltaic mode). Applications include light sensors, optical communications, and solar cells. Responsivity (A/W) measures their sensitivity to different wavelengths.

18. The main application of a varactor diode is:

Correct Answer: (C) Variable capacitance tuning

Explanation:
Varactor diodes (varicaps) act as voltage-controlled capacitors. Their junction capacitance varies with reverse bias voltage (C ∝ 1/√VR). Used in tuning circuits for radios, TV receivers, frequency multipliers, and voltage-controlled oscillators (VCOs). Quality factor (Q) and tuning ratio are key parameters.

19. A tunnel diode is used for:

Correct Answer: (A) High-frequency applications

Explanation:
Tunnel diodes (Esaki diodes) exploit quantum tunneling in heavily doped junctions. They show negative resistance in part of their operating range, enabling high-frequency (microwave) oscillators, amplifiers, and fast switching circuits (nanosecond range). Operate at very low voltages (~0.1V).

5. Breakdown Mechanisms

20. Which of the following is a mechanism of breakdown in a PN junction?

Correct Answer: (D) Both (A) and (C)

Explanation:
PN junctions exhibit two main breakdown mechanisms:
1. Zener breakdown: Dominates in heavily doped junctions (<5V), where strong electric fields cause direct electron tunneling.
2. Avalanche breakdown: Dominates in lightly doped junctions (>7V), where accelerated carriers create electron-hole pairs through impact ionization.
Between 5-7V, both mechanisms contribute.

21. Zener breakdown occurs in:

Correct Answer: (B) Heavily doped junctions

Explanation:
Zener breakdown occurs in heavily doped junctions because:
1. Narrow depletion regions create very strong electric fields at low reverse voltages
2. The small depletion width allows quantum tunneling of electrons from valence to conduction band
This mechanism dominates at breakdown voltages below about 5V.

22. Avalanche breakdown is dominant in:

Correct Answer: (A) Lightly doped junctions with high voltage

Explanation:
Avalanche breakdown occurs in lightly doped junctions because:
1. Wider depletion regions require higher voltages to achieve breakdown
2. Carriers gain enough energy to create electron-hole pairs through impact ionization
3. This multiplicative process creates an "avalanche" of carriers
Dominates at breakdown voltages above about 7V.