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Semiconductor Electronics: Materials, Devices, and Simple Circuits – CBSE NCERT Study Resources

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12th

12th - Physics

Semiconductor Electronics: Materials, Devices, and Simple Circuits

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Overview of the Chapter

This chapter introduces the fundamental concepts of semiconductor materials, their properties, and their applications in electronic devices and simple circuits. It covers topics such as energy bands in solids, types of semiconductors, p-n junction diodes, transistors, and basic logic gates.

Semiconductor: A material whose electrical conductivity lies between that of a conductor and an insulator. Its conductivity can be altered by doping or by applying external stimuli like light or heat.

Energy Bands in Solids

Solids can be classified based on their energy band structures:

  • Conductors: Have overlapping valence and conduction bands, allowing free electron movement.
  • Insulators: Have a large energy gap between valence and conduction bands, preventing electron flow.
  • Semiconductors: Have a small energy gap, enabling conductivity under certain conditions.

Types of Semiconductors

Semiconductors are categorized into two types:

  • Intrinsic Semiconductors: Pure semiconductors (e.g., Silicon, Germanium) with equal numbers of electrons and holes.
  • Extrinsic Semiconductors: Doped semiconductors with added impurities to enhance conductivity.
    • n-type: Doped with pentavalent impurities (e.g., Phosphorus), increasing free electrons.
    • p-type: Doped with trivalent impurities (e.g., Boron), increasing holes.

p-n Junction Diode

A p-n junction is formed by joining p-type and n-type semiconductors. Key concepts include:

  • Depletion Region: A region near the junction where mobile charge carriers are depleted.
  • Biasing:
    • Forward Bias: Reduces the depletion width, allowing current flow.
    • Reverse Bias: Increases the depletion width, blocking current flow.

Diode: A two-terminal device that allows current to flow primarily in one direction, used in rectifiers, LEDs, and photodiodes.

Transistors

Transistors are three-terminal devices used for amplification and switching. Types include:

  • Bipolar Junction Transistor (BJT): Consists of emitter, base, and collector. Operates in active, cutoff, or saturation modes.
  • Field-Effect Transistor (FET): Voltage-controlled device with gate, source, and drain terminals.

Logic Gates

Basic logic gates perform Boolean operations in digital circuits:

  • AND Gate: Output is high only if all inputs are high.
  • OR Gate: Output is high if at least one input is high.
  • NOT Gate: Output is the inverse of the input.
  • NAND and NOR Gates: Universal gates combining basic operations.

Applications

Semiconductor devices are used in:

  • Rectifiers (converting AC to DC).
  • Amplifiers (increasing signal strength).
  • Digital circuits (computers, microcontrollers).
  • Optoelectronic devices (LEDs, photodiodes).

All Question Types with Solutions – CBSE Exam Pattern

Explore a complete set of CBSE-style questions with detailed solutions, categorized by marks and question types. Ideal for exam preparation, revision and practice.

Very Short Answer (1 Mark) – with Solutions (CBSE Pattern)

These are 1-mark questions requiring direct, concise answers. Ideal for quick recall and concept clarity.

Question 1:
Define intrinsic semiconductor.
Answer:
Definition: Pure semiconductor with equal electrons and holes.
Question 2:
Name the majority charge carriers in n-type semiconductor.
Answer:
Electrons are majority carriers.
Question 3:
What is the role of a depletion region in a diode?
Answer:

Prevents current flow in reverse bias.

Question 4:
Identify the material used in LED for yellow light emission.
Answer:
Gallium Phosphide (GaP).
Question 5:
State the biasing condition for a Zener diode to act as a voltage regulator.
Answer:
Reverse bias beyond breakdown voltage.
Question 6:
What is the energy gap of silicon at 300K?
Answer:
1.1 eV.
Question 7:
Name the process by which holes move in a semiconductor.
Answer:
Drift under electric field.
Question 8:
Define doping in semiconductors.
Answer:
Definition: Adding impurities to alter conductivity.
Question 9:
Why is germanium less preferred than silicon for devices?
Answer:

Higher leakage current due to smaller bandgap.

Question 10:
What happens to resistance of a thermistor when heated?
Answer:
Decreases (NTC type).
Question 11:
Name the logic gate represented by the symbol: [Diagram: OR gate].
Answer:
OR gate.
Question 12:
State the output of a NAND gate if both inputs are 1.
Answer:
Output is 0.
Question 13:
Identify the majority carrier in p-type Ge doped with boron.
Answer:
Holes.
Question 14:
What is the barrier potential of a silicon diode at room temperature?
Answer:
0.7 V.
Question 15:
What is the role of a depletion region in a p-n junction?
Answer:

The depletion region is a charge-neutral area near the junction where mobile charges are depleted. It creates an electric field that opposes further diffusion of charge carriers, establishing equilibrium.

Question 16:
Name the majority charge carriers in an n-type semiconductor.
Answer:

In an n-type semiconductor, the majority charge carriers are electrons due to donor impurities (e.g., Phosphorus) providing extra free electrons.

Question 17:
Why is silicon preferred over germanium for semiconductor devices?
Answer:

Silicon has a higher bandgap (1.1 eV) than germanium (0.7 eV), making it more stable at higher temperatures and less prone to leakage currents.

Question 18:
What happens to the width of the depletion region when a p-n junction is forward biased?
Answer:

The width of the depletion region decreases in forward bias as the applied voltage reduces the barrier potential, allowing majority carriers to cross the junction.

Question 19:
State the principle of a solar cell.
Answer:

A solar cell converts light energy into electrical energy using the photovoltaic effect, where photons generate electron-hole pairs in a p-n junction, creating a potential difference.

Question 20:
What is the function of a Zener diode in a circuit?
Answer:

A Zener diode acts as a voltage regulator by maintaining a constant output voltage (Zener voltage) in reverse bias, even if the input voltage varies.

Question 21:
Why does a LED emit light when forward biased?
Answer:

In a LED, forward bias causes electrons and holes to recombine at the junction, releasing energy as photons (light) due to the semiconductor's bandgap.

Question 22:
What is the significance of the knee voltage in a diode?
Answer:

The knee voltage is the minimum forward voltage (≈0.7 V for Si, ≈0.3 V for Ge) required for significant current flow, marking the onset of conduction.

Question 23:
How does a transistor amplify signals?
Answer:

A transistor amplifies signals by transferring a small current/voltage from the base-emitter circuit to control a larger current in the collector-emitter circuit, maintaining signal proportionality.

Question 24:
Distinguish between avalanche breakdown and Zener breakdown.
Answer:
  • Avalanche breakdown: Occurs at high reverse voltage due to impact ionization, common in lightly doped diodes.
  • Zener breakdown: Occurs at low reverse voltage in heavily doped diodes due to tunneling effect.

Very Short Answer (2 Marks) – with Solutions (CBSE Pattern)

These 2-mark questions test key concepts in a brief format. Answers are expected to be accurate and slightly descriptive.

Question 1:
Define intrinsic semiconductor and give an example.
Answer:

An intrinsic semiconductor is a pure semiconductor without any doping. It has equal numbers of electrons and holes.
Example: Pure silicon (Si) or germanium (Ge).

Question 2:
What is the role of a p-n junction diode in rectification?
Answer:

A p-n junction diode allows current to flow only in one direction, converting AC to DC (rectification).
It blocks reverse current, ensuring unidirectional flow.

Question 3:
State the difference between forward bias and reverse bias in a diode.
Answer:
  • Forward bias: Positive terminal connected to p-side, negative to n-side. Current flows easily.
  • Reverse bias: Positive terminal connected to n-side, negative to p-side. Minimal current flows.
Question 4:
Why is silicon preferred over germanium for semiconductor devices?
Answer:

Silicon has a higher bandgap (1.1 eV) than germanium (0.7 eV), making it more stable at higher temperatures.
It also forms a stronger oxide layer, useful for device fabrication.

Question 5:
What is the function of a Zener diode?
Answer:

A Zener diode is used for voltage regulation. It operates in reverse breakdown, maintaining a constant voltage across it.

Question 6:
Explain the term doping in semiconductors.
Answer:

Doping is the process of adding impurities to a semiconductor to increase its conductivity.
Donor (n-type) or acceptor (p-type) atoms are added.

Question 7:
How does a photodiode work?
Answer:

A photodiode converts light energy into electrical current. When light falls on the p-n junction, electron-hole pairs are generated, producing current.

Question 8:
What is the significance of the depletion region in a p-n junction?
Answer:

The depletion region is a charge-free zone formed at the junction.
It creates an electric field that opposes further diffusion of charge carriers, establishing equilibrium.

Question 9:
Name the majority charge carriers in an n-type semiconductor.
Answer:

In an n-type semiconductor, electrons are the majority charge carriers due to donor impurities.

Question 10:
What is the purpose of a LED? Give one application.
Answer:

An LED (Light Emitting Diode) emits light when forward-biased.
Application: Used in indicator lights, displays, and lighting systems.

Question 11:
Define avalanche breakdown in a diode.
Answer:

Avalanche breakdown occurs when high reverse voltage accelerates charge carriers, creating more electron-hole pairs by collision.
This leads to a sudden increase in current.

Short Answer (3 Marks) – with Solutions (CBSE Pattern)

These 3-mark questions require brief explanations and help assess understanding and application of concepts.

Question 1:
Explain the formation of a depletion region in a p-n junction diode.
Answer:

When a p-type semiconductor is joined with an n-type semiconductor, the following occurs:

  • Electrons from the n-region diffuse into the p-region, leaving behind positively charged donor ions.
  • Holes from the p-region diffuse into the n-region, leaving behind negatively charged acceptor ions.

This creates an immobile layer of positive and negative charges called the depletion region, which acts as a barrier to further diffusion. The electric field in this region opposes the flow of majority carriers.

Question 2:
Differentiate between intrinsic and extrinsic semiconductors with one example of each.
Answer:

Intrinsic semiconductors are pure semiconductors (e.g., silicon or germanium) with equal numbers of electrons and holes. Their conductivity is low and depends on temperature.

Extrinsic semiconductors are doped with impurities to increase conductivity. Examples include:

  • n-type (e.g., silicon doped with phosphorus).
  • p-type (e.g., silicon doped with boron).
Question 3:
Describe the working principle of a Zener diode as a voltage regulator.
Answer:

A Zener diode operates in reverse breakdown mode to maintain a constant voltage across a load:

  • When the input voltage exceeds the Zener voltage, the diode conducts heavily.
  • The excess voltage is dropped across the series resistor, keeping the output voltage stable.

This property makes it useful in voltage regulation circuits to protect sensitive components.

Question 4:
What is the significance of the energy band gap in semiconductors? How does it affect conductivity?
Answer:

The energy band gap is the energy difference between the valence band and the conduction band:

  • A smaller band gap (e.g., in germanium) allows more electrons to jump to the conduction band, increasing conductivity.
  • A larger band gap (e.g., in silicon) requires more energy for conduction, resulting in lower intrinsic conductivity.

Doping reduces the effective band gap, enhancing conductivity.

Question 5:
Explain the function of a transistor as an amplifier in common-emitter configuration.
Answer:

In a common-emitter amplifier:

  • A small input signal at the base-emitter junction controls a larger output current between collector and emitter.
  • The transistor's current gain (β) amplifies the signal.

Steps:
1. Input AC signal modulates the base current.
2. This causes a larger variation in collector current.
3. The load resistor converts this current into an amplified voltage.

Question 6:
Why is a full-wave rectifier more efficient than a half-wave rectifier? Draw the output waveform for both.
Answer:

Full-wave rectifier advantages:

  • Utilizes both halves of the AC input cycle, doubling efficiency.
  • Produces a smoother output with less ripple.

Waveforms:
1. Half-wave: Pulsating DC with gaps (only positive/negative half cycles).
2. Full-wave: Continuous pulsating DC (all half cycles converted).

Diagrams would show the input AC sine wave and the respective rectified outputs.

Question 7:
Explain the difference between intrinsic and extrinsic semiconductors with examples.
Answer:

Intrinsic semiconductors are pure semiconductors without any doping, where the number of electrons and holes are equal. Example: Pure Silicon (Si) or Germanium (Ge).
Extrinsic semiconductors are doped with impurities to increase conductivity. They are of two types:

  • n-type (doped with pentavalent impurities like Phosphorus)
  • p-type (doped with trivalent impurities like Boron)
Extrinsic semiconductors have higher conductivity due to added charge carriers.

Question 8:
Describe the working principle of a p-n junction diode under forward bias.
Answer:

In a p-n junction diode, forward bias reduces the depletion layer width by applying positive voltage to the p-side and negative to the n-side.
This lowers the potential barrier, allowing majority charge carriers (holes in p-side and electrons in n-side) to cross the junction.
Current flows easily due to reduced resistance, and the diode conducts electricity.

Question 9:
What is the role of a Zener diode in voltage regulation? Explain with a diagram if necessary.
Answer:

A Zener diode operates in reverse breakdown region to maintain a constant voltage across a load.
When the input voltage exceeds the Zener voltage, it conducts heavily, diverting excess current to ground.
This stabilizes the output voltage, making it ideal for voltage regulation in circuits.
(Diagram: Show input voltage, Zener diode in reverse bias, resistor, and stable output across load.)

Question 10:
Compare the characteristics of LED and a normal p-n junction diode.
Answer:

  • LED emits light when forward-biased due to recombination of electrons and holes, while a normal diode does not.
  • LEDs are made of direct bandgap materials (e.g., GaAs), whereas normal diodes use Si or Ge.
  • LEDs have a higher forward voltage drop (~2V) compared to normal diodes (~0.7V).
  • LEDs are used in displays/lighting, while normal diodes are for rectification.

Question 11:
Explain how a transistor acts as an amplifier in common-emitter configuration.
Answer:

In common-emitter configuration, a small change in base current (IB) causes a large change in collector current (IC) due to transistor action.
The input signal is applied to the base-emitter junction, and the amplified output is taken from the collector.
Voltage gain is high because β (current gain) multiplies the input signal.
Phase inversion occurs between input and output signals.

Question 12:
Define logic gates and list the truth table for a NAND gate.
Answer:

Logic gates are electronic circuits performing Boolean operations (AND, OR, NOT, etc.).
Truth table for NAND gate (output is 0 only when all inputs are 1):

  • A=0, B=0 → Output=1
  • A=0, B=1 → Output=1
  • A=1, B=0 → Output=1
  • A=1, B=1 → Output=0
NAND is a universal gate used to construct other gates.

Long Answer (5 Marks) – with Solutions (CBSE Pattern)

These 5-mark questions are descriptive and require detailed, structured answers with proper explanation and examples.

Question 1:
Explain the working principle of a p-n junction diode under forward and reverse bias conditions. How does it act as a rectifier?
Answer:
Theoretical Framework

A p-n junction diode is formed by joining p-type and n-type semiconductors. In forward bias, the external voltage reduces the depletion layer, allowing current flow. In reverse bias, the depletion layer widens, blocking current.

Evidence Analysis
  • Forward Bias: Majority carriers (holes in p-side, electrons in n-side) recombine, enabling current.
  • Reverse Bias: Minority carriers contribute to negligible reverse saturation current.
Critical Evaluation

As a rectifier, it converts AC to DC by allowing current only in forward bias. Our textbook shows full-wave rectifiers use four diodes for efficient conversion.

Future Implications

Diodes are vital in power supplies and signal demodulation, with advancements like Schottky diodes improving efficiency.

Question 2:
Describe the energy band diagrams for conductors, insulators, and semiconductors. How does doping alter these bands?
Answer:
Theoretical Framework

Energy bands consist of valence and conduction bands separated by a bandgap. Conductors have overlapping bands, insulators have a large bandgap (>5eV), and semiconductors have a small bandgap (~1eV).

Evidence Analysis
  • Doping introduces donor/acceptor levels, reducing the effective bandgap.
  • n-type doping adds free electrons near the conduction band, p-type adds holes near the valence band.
Critical Evaluation

Our textbook shows silicon’s bandgap is 1.1eV, making it ideal for devices. Doping enhances conductivity, as seen in LEDs and transistors.

Future Implications

Band engineering enables optoelectronic devices like solar cells, with research focusing on tunable bandgap materials.

Question 3:
Explain the construction and working of a BJT transistor in common-emitter configuration. Why is it called a current amplifier?
Answer:
Theoretical Framework

A BJT has three regions: emitter, base, and collector. In common-emitter mode, the base-emitter junction is forward-biased, and the base-collector junction is reverse-biased.

Evidence Analysis
  • Small base current controls larger collector current, with current gain β = Ic/Ib.
  • Our textbook shows β ranges from 20-200, enabling amplification.
Critical Evaluation

It amplifies signals in audio devices and radios. The configuration’s high power gain makes it versatile.

Future Implications

Advancements like heterojunction transistors improve frequency response for 5G and IoT applications.

Question 4:
Compare LEDs and photodiodes in terms of working principles and applications. How does bandgap influence their operation?
Answer:
Theoretical Framework

LEDs emit light when electrons recombine with holes, while photodiodes generate current when photons excite electrons across the bandgap.

Evidence Analysis
  • LEDs use direct bandgap materials (e.g., GaAs) for efficient emission.
  • Photodiodes operate in reverse bias, with sensitivity dependent on bandgap (e.g., Si for visible light).
Critical Evaluation

Our textbook shows LEDs are used in displays, and photodiodes in light sensors. Bandgap determines the wavelength of emitted/detected light.

Future Implications

Research focuses on organic LEDs (OLEDs) and quantum dot photodiodes for flexible electronics.

Question 5:
Analyze the truth table and logic symbol of a NAND gate. How can it function as a universal gate? Provide two examples.
Answer:
Theoretical Framework

A NAND gate outputs LOW only when all inputs are HIGH. Its truth table shows it’s a combination of AND followed by NOT.

Evidence Analysis
  • Universal Gate: NAND can mimic NOT, AND, and OR gates by reconfiguring inputs.
  • Examples: NOT gate (both inputs tied), AND gate (NAND followed by NOT).
Critical Evaluation

Our textbook demonstrates its use in digital circuits due to minimal transistor count. It simplifies IC design.

Future Implications

NAND-based memory (e.g., SSDs) dominates storage technology, with 3D NAND pushing capacity limits.

Question 6:
Explain the working principle of a p-n junction diode under forward and reverse bias with a labeled diagram. Discuss its I-V characteristics.
Answer:
Theoretical Framework

A p-n junction diode conducts current easily under forward bias (p-side positive) due to reduced depletion width, while it blocks current under reverse bias (n-side positive) as the depletion layer widens.


Evidence Analysis
  • Forward Bias: Majority carriers cross the junction, causing exponential rise in current (Ohmic region).
  • Reverse Bias: Minority carriers generate negligible reverse saturation current (~µA).
[Diagram: I-V curve showing knee voltage (0.7V for Si)]
Critical Evaluation

Our textbook shows breakdown occurs at high reverse voltage due to Zener or avalanche effect. Example: LEDs use forward bias, while photodiodes operate in reverse bias.

Question 7:
Compare intrinsic and extrinsic semiconductors with respect to conductivity, charge carriers, and temperature dependence. Provide two examples of each.
Answer:
Theoretical Framework

Intrinsic semiconductors (e.g., pure Si/Ge) have equal electrons and holes, while extrinsic semiconductors have doped impurities (n-type/p-type) altering conductivity.


Evidence Analysis
ParameterIntrinsicExtrinsic
ConductivityLow (~10⁻⁶ S/m)High (dose-dependent)
Carrierse⁻ = h⁺Majority: donor/acceptor ions

Critical Evaluation

We studied that extrinsic conductivity peaks at moderate temperatures (Example: n-type Si with P doping). At very high T, intrinsic behavior dominates.

Question 8:
Describe the construction and working of a BJT in common-emitter configuration. How does it function as an amplifier?
Answer:
Theoretical Framework

A BJT has emitter, base, and collector layers. In CE mode, small base current controls large collector current (β = Ic/Ib).


Evidence Analysis
  • Construction: Heavily doped emitter, thin base, large collector.
  • Working: Forward-biased EB junction injects carriers; reverse-biased CB junction collects them.
[Diagram: CE circuit with input/output waveforms]
Critical Evaluation

Our textbook shows voltage gain (Av = β × Rc/Rin). Example: Audio amplifiers use this configuration for high gain (~100).

Question 9:
Analyze the role of depletion region in a p-n junction diode. How does its width vary with biasing?
Answer:
Theoretical Framework

The depletion region is a charge-neutral zone formed by diffusion of carriers, creating an internal electric field (barrier potential ~0.7V for Si).


Evidence Analysis
  • Forward Bias: Width decreases as applied voltage opposes barrier potential.
  • Reverse Bias: Width increases due to strengthened electric field.
[Diagram: Depletion layer under different biases]
Critical Evaluation

We studied that abrupt junctions (e.g., Zener diodes) have thinner depletion layers than graded ones. Example: Varactor diodes exploit width modulation.

Question 10:
Explain the working of a full-wave rectifier with center-tapped transformer. Derive its ripple factor and compare with half-wave rectifier.
Answer:
Theoretical Framework

A full-wave rectifier uses two diodes and center-tapped transformer to rectify both half-cycles of AC input.


Evidence Analysis
  • Working: D1 conducts during positive half-cycle; D2 during negative half-cycle.
  • Ripple factor (γ) = 0.482 (vs 1.21 for half-wave). Efficiency doubles (81.2%).
[Diagram: Circuit with output waveform]
Critical Evaluation

Our textbook shows it requires bulky transformer but gives smoother DC. Example: Power supplies in TVs use this configuration.

Question 11:
Discuss the principle of photodiode and LED as optoelectronic devices. How do their energy band diagrams differ?
Answer:
Theoretical Framework

Photodiodes convert light to current (reverse bias), while LEDs emit light via recombination (forward bias). Both involve bandgap transitions.


Evidence Analysis
  • Photodiode: Incident photons create e-h pairs in depletion region.
  • LED: Direct bandgap materials (e.g., GaAs) emit photons when e⁻ fall into holes.
[Diagram: Band diagrams showing absorption vs emission]
Critical Evaluation

We studied that IR LEDs (λ ~940nm) use GaAs, while photodiodes need high sensitivity (Example: Remote controls vs solar cells).

Question 12:
Explain the working principle of a p-n junction diode under forward and reverse bias conditions with the help of a circuit diagram. Discuss its V-I characteristics.
Answer:

A p-n junction diode is a semiconductor device formed by joining p-type and n-type materials. Its working depends on the biasing condition:

Forward Bias: When the positive terminal of the battery is connected to the p-side and the negative terminal to the n-side, the potential barrier reduces. This allows majority charge carriers (holes in p-side and electrons in n-side) to cross the junction, resulting in current flow. The diode conducts easily in this mode.

Reverse Bias: When the battery connections are reversed, the potential barrier increases. Majority carriers are pulled away from the junction, and only a tiny reverse saturation current (due to minority carriers) flows. The diode acts as an open circuit.

V-I Characteristics: The graph shows:

  • Forward bias: Current increases exponentially after the knee voltage (~0.7V for Si).
  • Reverse bias: Minimal current flows until the breakdown voltage is reached.

Applications include rectification, signal demodulation, and voltage regulation.

Question 13:
Describe the construction and working of a Zener diode. Explain how it acts as a voltage regulator with a suitable circuit diagram.
Answer:

A Zener diode is a specially designed p-n junction diode operated in reverse breakdown region for voltage regulation.

Construction: It is heavily doped to achieve a sharp breakdown voltage (Zener voltage). The depletion layer is very thin, allowing a strong electric field.

Working: Under reverse bias, once the applied voltage exceeds the Zener voltage, a large current flows due to Zener breakdown (avalanche effect in higher voltages). The voltage across the diode remains constant.

Voltage Regulation: In the circuit:

  • The Zener diode is connected in parallel to the load.
  • A series resistor limits the current.
  • If input voltage or load resistance varies, the Zener diode adjusts its current to maintain a constant output voltage.

This property makes it ideal for voltage stabilization in power supplies.

Question 14:
Explain the working principle of a p-n junction diode under forward and reverse bias conditions. Also, draw the characteristic I-V curve and label its important regions.
Answer:

A p-n junction diode is formed by joining a p-type semiconductor (with majority holes) and an n-type semiconductor (with majority electrons). The working principle depends on the biasing conditions:

1. Forward Bias: When the positive terminal of the battery is connected to the p-side and the negative terminal to the n-side, the external voltage opposes the depletion layer potential. This reduces the barrier width, allowing majority carriers to cross the junction easily. Current flows due to the movement of holes (in p-region) and electrons (in n-region). The diode conducts heavily once the applied voltage exceeds the threshold voltage (~0.7V for Si).

2. Reverse Bias: Here, the battery's positive terminal is connected to the n-side and the negative to the p-side. The external voltage widens the depletion layer, increasing the barrier potential. Majority carriers are pulled away from the junction, and only a tiny reverse saturation current (due to minority carriers) flows. If the reverse voltage exceeds the breakdown voltage, avalanche breakdown occurs.

I-V Curve: The graph shows:

  • Forward Bias Region: Current rises exponentially after the threshold voltage.
  • Reverse Bias Region: Minimal current until breakdown.
  • Breakdown Region: Sudden increase in reverse current.

Question 15:
Explain the working principle of a p-n junction diode under forward and reverse bias conditions with the help of a suitable diagram. Also, discuss its V-I characteristics.
Answer:

A p-n junction diode is a semiconductor device formed by joining p-type and n-type materials. Its working principle depends on the biasing condition:

1. Forward Bias:
When the p-side is connected to the positive terminal and the n-side to the negative terminal of a battery, the diode is in forward bias.
The applied voltage reduces the depletion layer width, allowing majority charge carriers (holes in p-side and electrons in n-side) to cross the junction.
This results in a forward current due to the movement of charge carriers.

2. Reverse Bias:
When the p-side is connected to the negative terminal and the n-side to the positive terminal, the diode is in reverse bias.
The depletion layer widens, preventing majority carriers from crossing the junction.
A very small reverse saturation current flows due to minority carriers (electrons in p-side and holes in n-side).

V-I Characteristics:
The graph of voltage (V) vs. current (I) for a diode shows:

  • In forward bias, current increases exponentially after the threshold voltage (~0.7V for Si).
  • In reverse bias, a negligible current flows until the breakdown voltage is reached, where a sudden increase in current occurs.

Diagram: (Draw a labeled diagram showing the p-n junction diode under forward and reverse bias, along with the V-I curve.)

Application: Diodes are used in rectifiers, LEDS, and voltage regulators due to their unidirectional current flow property.

Question 16:
Explain the working principle of a p-n junction diode under forward and reverse bias conditions with the help of a suitable diagram. Also, discuss its I-V characteristics.
Answer:

A p-n junction diode is a semiconductor device formed by joining a p-type and an n-type semiconductor. Its working principle depends on the biasing condition:

1. Forward Bias: When the positive terminal of the battery is connected to the p-side and the negative terminal to the n-side, the diode is forward-biased. The applied voltage reduces the depletion layer width, allowing majority charge carriers (holes in p-side and electrons in n-side) to cross the junction. This results in a forward current.

2. Reverse Bias: When the positive terminal is connected to the n-side and the negative terminal to the p-side, the diode is reverse-biased. The depletion layer widens, preventing majority carriers from crossing. However, a small reverse saturation current flows due to minority carriers.

I-V Characteristics: The graph of current (I) vs. voltage (V) shows:

  • In forward bias, current increases exponentially after the threshold voltage (~0.7V for Si).
  • In reverse bias, a negligible current flows until the breakdown voltage is reached, where a sudden increase occurs.

Diagram: (Draw a labeled diagram showing the p-n junction diode under both bias conditions and the corresponding I-V curve.)

Application: Diodes are used in rectifiers, LEDS, and voltage regulators due to their unidirectional current flow property.

Question 17:
Explain the working principle of a p-n junction diode under forward and reverse bias conditions with the help of a circuit diagram. Also, discuss the I-V characteristics of the diode.
Answer:

A p-n junction diode is a semiconductor device formed by joining a p-type and an n-type semiconductor. Its working principle depends on the biasing condition applied across it.

Forward Bias: When the positive terminal of the battery is connected to the p-side and the negative terminal to the n-side, the diode is in forward bias. The applied voltage reduces the depletion layer width, allowing majority charge carriers (holes in p-side and electrons in n-side) to cross the junction. This results in a current flow.

Reverse Bias: When the positive terminal is connected to the n-side and the negative terminal to the p-side, the diode is in reverse bias. The depletion layer widens, preventing majority carriers from crossing. Only a small reverse saturation current (due to minority carriers) flows.

I-V Characteristics:
The graph of current (I) vs. voltage (V) shows:
1. Forward Bias Region: Current increases exponentially after the threshold voltage (≈0.7V for Si).
2. Reverse Bias Region: A very small reverse current flows until the breakdown voltage is reached.
3. Breakdown Region: Beyond the breakdown voltage, the current increases sharply due to avalanche or Zener effect.

Circuit Diagram: (Draw a simple circuit with a diode, battery, resistor, and ammeter in forward and reverse bias configurations.)

Key Points:

  • The diode acts as a one-way valve for current.
  • Forward bias allows current flow, while reverse bias blocks it (ideally).
  • The I-V curve is nonlinear, showing the diode's unidirectional behavior.

Question 18:
Explain the working principle of a p-n junction diode under forward and reverse bias conditions. Also, draw the I-V characteristics of the diode and label the important regions.
Answer:

A p-n junction diode is a semiconductor device formed by joining a p-type and an n-type semiconductor. Its working principle depends on the biasing conditions:

1. Forward Bias: When the positive terminal of the battery is connected to the p-side and the negative terminal to the n-side, the diode is forward-biased. The applied voltage reduces the depletion layer width, allowing majority charge carriers (holes in p-side and electrons in n-side) to cross the junction. This results in a current flow, which increases exponentially with voltage after overcoming the barrier potential (~0.7V for Si).

2. Reverse Bias: When the positive terminal is connected to the n-side and the negative terminal to the p-side, the diode is reverse-biased. The depletion layer widens, preventing majority carriers from crossing. However, a small reverse saturation current flows due to minority carriers. If the reverse voltage exceeds the breakdown voltage, the diode conducts heavily due to avalanche or Zener breakdown.

I-V Characteristics: The graph shows:

  • Forward Bias Region: Current rises sharply after the knee voltage.
  • Reverse Bias Region: Small saturation current until breakdown.
  • Breakdown Region: Sudden increase in reverse current.

Diagram: (Draw a labeled I-V curve with axes showing Forward/Reverse Bias regions, Knee Voltage, and Breakdown Voltage.)

Application: Diodes are used in rectifiers, voltage regulators, and signal demodulation due to their unidirectional conductivity.

Question 19:
Explain the working principle of a p-n junction diode under forward and reverse bias conditions with the help of a circuit diagram. Also, describe how the I-V characteristics are obtained experimentally.
Answer:

A p-n junction diode is a semiconductor device formed by joining p-type and n-type materials. Its working principle depends on the biasing condition:

Forward Bias: When the positive terminal of the battery is connected to the p-side and the negative terminal to the n-side, the potential barrier at the junction decreases. This allows majority charge carriers (holes in p-side and electrons in n-side) to cross the junction, resulting in a current flow. The diode conducts electricity easily in this condition.

Reverse Bias: When the battery connections are reversed, the potential barrier increases, preventing majority carriers from crossing. Only a negligible reverse saturation current flows due to minority carriers. The diode acts as an open circuit in this condition.

I-V Characteristics: To obtain the I-V characteristics experimentally:
1. Connect the diode in forward bias with a variable power supply, ammeter, and voltmeter.
2. Gradually increase the voltage and record corresponding current values.
3. Repeat the process for reverse bias.
4. Plot the graph of current (I) vs. voltage (V) to observe the non-linear behavior, showing low resistance in forward bias and high resistance in reverse bias.

Circuit Diagram: (Draw a simple circuit with diode, battery, ammeter, and voltmeter for both forward and reverse bias conditions.)

Question 20:
Describe the construction and working of a Zener diode as a voltage regulator. Explain how it maintains a constant output voltage despite variations in input voltage or load resistance, with a suitable circuit diagram.
Answer:

A Zener diode is a specially designed p-n junction diode that operates in the reverse breakdown region to regulate voltage. Its construction involves heavy doping to create a thin depletion layer, enabling a sharp breakdown at a specific voltage called the Zener voltage (VZ).

Working as a Voltage Regulator:
1. The Zener diode is connected in reverse bias parallel to the load.
2. When the input voltage (Vin) exceeds VZ, the diode enters the breakdown region, allowing current to flow through it.
3. The excess voltage is dropped across the series resistor (RS), keeping the output voltage (Vout) constant at VZ.
4. If the load resistance (RL) changes, the Zener diode adjusts its current to maintain Vout = VZ.

Circuit Diagram: (Draw a circuit with input voltage, series resistor, Zener diode in reverse bias, and load resistor in parallel.)

Key Points:

  • The Zener diode acts as a voltage clamp by conducting excess current.
  • The series resistor limits the current to prevent damage.
  • This regulation is crucial in power supplies to protect sensitive electronic components.

Question 21:
Describe the construction and working of a Zener diode as a voltage regulator. Derive the expression for the series resistance required to maintain a constant output voltage.
Answer:

A Zener diode is a specially doped p-n junction diode designed to operate in the reverse breakdown region with a sharp breakdown voltage (Zener voltage, VZ).

Construction: It has a heavily doped depletion layer, causing a thin junction and a strong electric field. This allows Zener breakdown (due to tunneling effect) at low voltages.

Working as Voltage Regulator: When connected in reverse bias parallel to the load:

  • If input voltage (Vin) increases, the diode maintains VZ across the load by conducting excess current.
  • If Vin decreases, the diode reduces its current to keep VZ stable.

Series Resistance (RS) Calculation: To limit the current through the diode:
Let Vin = Input voltage, VZ = Zener voltage, IL = Load current, IZ = Zener current.
The total current through RS is I = IZ + IL.
Voltage drop across RS = Vin - VZ.
Using Ohm's Law: RS = (Vin - VZ) / (IZ + IL).
This ensures the diode operates in the breakdown region without overheating.

Case-based Questions (4 Marks) – with Solutions (CBSE Pattern)

These 4-mark case-based questions assess analytical skills through real-life scenarios. Answers must be based on the case study provided.

Question 1:
In a lab experiment, students observed that a p-n junction diode conducts only when forward-biased. However, a small current flows even in reverse bias. Analyze this behavior and explain the role of minority carriers and breakdown voltage.
Answer:
Case Deconstruction

We studied that a diode conducts primarily in forward bias due to majority carrier movement. However, a tiny reverse current exists due to minority carriers (electrons in p-side and holes in n-side).

Theoretical Application
  • Minority carriers are thermally generated and contribute to reverse saturation current.
  • Breakdown voltage occurs when high reverse bias causes avalanche multiplication of carriers.
Critical Evaluation

Our textbook shows that silicon diodes have lower reverse current than germanium due to higher energy gap. Example: Zener diodes exploit breakdown for voltage regulation.

Question 2:
A student constructs a half-wave rectifier but notices output ripple. Compare its efficiency with a bridge rectifier and suggest methods to reduce ripple using filters.
Answer:
Case Deconstruction

Half-wave rectifiers use only one half of AC cycle, resulting in 40.6% efficiency and high ripple. Bridge rectifiers utilize full cycle with 81.2% efficiency.

Theoretical Application
  • Capacitor filters store charge during peaks and discharge during gaps, smoothing output.
  • LC filters combine inductors and capacitors for better ripple reduction.
Critical Evaluation

Our textbook shows bridge rectifiers are preferred for high-current applications. Example: Mobile chargers use capacitor filters after bridge rectification.

Question 3:
Explain how doping concentration affects the conductivity of semiconductors. Compare the I-V characteristics of lightly and heavily doped diodes with reference to depletion region width.
Answer:
Case Deconstruction

Doping increases charge carriers, enhancing conductivity. Heavily doped semiconductors have more free electrons/holes than lightly doped ones.

Theoretical Application
  • Heavy doping creates narrow depletion regions, lowering breakdown voltage.
  • Light doping results in wider depletion zones and higher reverse bias tolerance.
Critical Evaluation

Our textbook shows Zener diodes are heavily doped for sharp breakdown. Example: Solar cells use light doping for wider depletion regions to absorb more photons.

Question 4:
Analyze why LEDs emit light when forward-biased while normal diodes don't. Discuss the significance of band gap in determining the emitted light's color.
Answer:
Case Deconstruction

LEDs are made of direct bandgap materials where electron-hole recombination releases photons. Normal diodes use indirect bandgap materials where energy is lost as heat.

Theoretical Application
  • Bandgap energy (Eg) determines photon wavelength: E = hc/λ.
  • GaAs (Eg=1.43eV) emits infrared, while GaP (Eg=2.26eV) emits green light.
Critical Evaluation

Our textbook shows blue LEDs use GaN with 3.4eV bandgap. Example: Traffic signals use different semiconductor materials for distinct colors.

Question 5:
A student constructs a half-wave rectifier using a silicon diode. The input AC voltage has a peak value of 12V.

(i) Sketch the output waveform across the load resistor.
(ii) Calculate the DC output voltage.
Answer:
Case Deconstruction

We studied that a half-wave rectifier allows only one half-cycle of AC to pass, blocking the reverse cycle due to the diode's forward bias property.


Theoretical Application
  • Output waveform: [Diagram: Pulsating DC with 12V peaks during positive half-cycles, 0V during negative half-cycles]
  • DC output voltage = Vpeak/π = 12/3.14 ≈ 3.82V (neglecting diode drop).

Critical Evaluation

Our textbook shows silicon diodes have ~0.7V forward voltage drop, reducing practical output to ~3.12V. Example: Mobile chargers use such circuits for low-power conversion.

Question 6:
An LED emits red light (λ = 620nm) when forward biased.

(i) Determine its band gap energy.
(ii) Why does GaAsP (Gallium Arsenide Phosphide) suit this application better than silicon?
Answer:
Case Deconstruction

We know LEDs emit light when electrons recombine with holes, releasing energy equal to the band gap.


Theoretical Application
  • Eg = hc/λ = (6.63×10-34 × 3×108)/(620×10-9) ≈ 2.0eV.
  • GaAsP has a direct band gap (~1.8-2.2eV), enabling efficient light emission, unlike silicon's indirect gap.

Critical Evaluation

Example: Traffic signals use GaAsP LEDs for high brightness. Our textbook confirms indirect gap materials (e.g., Si) emit mostly heat, not light.

Question 7:
A zener diode (VZ = 6V) regulates voltage in a circuit with input varying from 8-12V.

(i) Draw the circuit diagram.
(ii) Explain how it maintains constant output despite input fluctuations.
Answer:
Case Deconstruction

We studied zener diodes operate in reverse breakdown, maintaining fixed voltage across the load.


Theoretical Application
  • [Diagram: Input → Series resistor → Zener (parallel to load)]
  • Excess voltage drops across the series resistor. At Vin > 6V, the zener conducts, clamping output to 6V.

Critical Evaluation

Example: Voltage regulators in power supplies use this principle. Our textbook shows the series resistor limits current to prevent damage.

Question 8:
A transistor amplifier has β = 100 and operates in common-emitter mode.

(i) Derive its current gain expression.
(ii) Why is the output signal 180° phase-shifted?
Answer:
Case Deconstruction

We know common-emitter amplifiers provide both voltage and current gain.


Theoretical Application
  • Current gain (Ai) = ΔIC/ΔIB = β = 100.
  • Phase inversion occurs because increased IB reduces VCE (output), creating 180° shift.

Critical Evaluation

Example: Audio amplifiers use this configuration. Our textbook confirms the phase shift via input-output waveform diagrams.

Question 9:
A student constructs a half-wave rectifier using a silicon diode and a load resistor. The input AC voltage is 12V (peak). Explain why the output DC voltage across the load is less than 12V and calculate its approximate value.
Answer:
Case Deconstruction

In a half-wave rectifier, only one half of the AC cycle is converted to DC due to the diode's forward bias property. The output is pulsating DC.

Theoretical Application

For silicon diodes, a barrier potential of 0.7V exists. Thus, peak output voltage = Input peak voltage - Barrier potential = 12V - 0.7V = 11.3V.

Critical Evaluation

Our textbook shows that practical diodes cause voltage drops, reducing efficiency. Example: Germanium diodes (0.3V drop) yield higher output but are less temperature-stable.

Question 10:
An LED emits light when forward-biased but not when reverse-biased. Analyze the energy band diagram and explain this behavior with reference to electron-hole recombination.
Answer:
Case Deconstruction

LEDs are p-n junction devices where light emission occurs due to radiative recombination of electrons and holes.

Theoretical Application

In forward bias, electrons cross the junction, recombine with holes, and release energy as photons. Reverse bias widens the depletion layer, preventing recombination.

Critical Evaluation

We studied that material bandgap determines emitted light color. Example: GaAs (infrared) vs. GaP (green). [Diagram: Forward-biased LED band diagram]

Question 11:
A zener diode with breakdown voltage 6V is used in a voltage regulator circuit. Compare its operation in normal diode mode vs. breakdown mode, citing two applications of each mode.
Answer:
Case Deconstruction

Zener diodes operate in reverse breakdown for voltage regulation but function as normal diodes in forward bias.

Theoretical Application
  • Normal mode: Rectification (AC-DC conversion), signal clamping.
  • Breakdown mode: Voltage regulators, surge protectors.
Critical Evaluation

Our textbook shows zeners maintain stable voltage despite current variations. Example: 5V zeners in USB power circuits.

Question 12:
Contrast the majority and minority charge carriers in n-type and p-type semiconductors. How does doping concentration affect conductivity?
Answer:
Case Deconstruction

In n-type, electrons are majority carriers; in p-type, holes dominate. Minority carriers arise due to thermal generation.

Theoretical Application

Higher doping increases majority carriers, enhancing conductivity. Example: Phosphorus-doped Si (n-type) has more free electrons than intrinsic Si.

Critical Evaluation

We studied that excessive doping reduces mobility due to impurity scattering. [Diagram: Carrier concentration vs. doping density]

Question 13:
A student constructs a half-wave rectifier using a silicon diode and a load resistor. The input AC voltage is 12V (rms). Explain the working principle and calculate the peak output voltage. Discuss why a filter capacitor is often added.
Answer:
Case Deconstruction

A half-wave rectifier converts AC to pulsating DC by allowing current only during the positive half-cycle. The silicon diode has a forward bias voltage drop of 0.7V.

Theoretical Application
  • Peak input voltage = √2 × Vrms = √2 × 12V ≈ 17V
  • Peak output voltage = Vpeak - 0.7V = 16.3V
Critical Evaluation

A filter capacitor smoothens the output by storing charge during peaks and discharging during gaps. Without it, the output remains pulsating, unsuitable for most electronic devices.

Question 14:
Compare intrinsic and extrinsic semiconductors with respect to conductivity and charge carriers. How does doping alter the behavior of pure silicon? Provide two examples of dopants.
Answer:
Case Deconstruction

Intrinsic semiconductors like pure silicon have equal electrons and holes, while extrinsic ones have added impurities to dominate conductivity.

Theoretical Application
  • Doping increases conductivity by adding majority carriers (e.g., pentavalent arsenic for n-type, trivalent boron for p-type).
  • Extrinsic semiconductors show higher conductivity at room temperature.
Critical Evaluation

Our textbook shows doping creates predictable charge carriers: n-type (electrons) or p-type (holes). Examples: Phosphorus (n-type), Aluminum (p-type).

Question 15:
Analyze the I-V characteristics of a Zener diode. Why is it used as a voltage regulator? Sketch the circuit diagram for regulation.
Answer:
Case Deconstruction

Zener diodes operate in reverse breakdown, maintaining nearly constant voltage despite current variations.

Theoretical Application
  • Beyond knee voltage, current increases sharply while voltage stabilizes.
  • Used in regulators to protect circuits from voltage spikes (e.g., 5V supply for microcontrollers).
Critical Evaluation

[Diagram: Zener in parallel with load resistor.] The diode shunts excess current when input exceeds breakdown voltage, ensuring stable output.

Question 16:
A common-emitter transistor amplifier has a collector current of 2mA and base current of 20µA. Calculate current gain (β) and discuss how biasing ensures proper amplification.
Answer:
Case Deconstruction

Current gain (β) is the ratio of collector current (IC) to base current (IB).

Theoretical Application
  • β = IC/IB = 2mA/20µA = 100
  • Proper biasing (e.g., fixed bias circuit) keeps the transistor in active region for linear amplification.
Critical Evaluation

Without biasing, distortion occurs. Our textbook shows examples like audio amplifiers where β must remain stable for signal fidelity.

Question 17:

In an experiment, a student constructs a p-n junction diode circuit with a battery, resistor, and the diode in forward bias. The student observes that the current increases exponentially with voltage up to a certain point, then stabilizes. Based on this:

  • Explain the behavior of the p-n junction diode in forward bias.
  • Why does the current stabilize after a certain voltage?
Answer:

Behavior of p-n junction diode in forward bias:

When a p-n junction diode is forward-biased, the external voltage reduces the depletion layer width. This allows majority charge carriers (holes in p-region and electrons in n-region) to cross the junction, resulting in current flow. Initially, the current increases exponentially with voltage because the potential barrier is progressively overcome.


Current stabilization:

At higher voltages, the current stabilizes due to ohmic resistance of the semiconductor material and external circuit components (like the resistor). The diode's internal resistance and the external resistor limit the current, preventing further exponential growth. This region is called the ohmic region.


Additional insight: The exponential relationship is governed by the diode equation: I = I0(eV/ηVT - 1), where I0 is reverse saturation current, VT is thermal voltage, and η is the ideality factor.

Question 18:

A student tests a Zener diode in reverse bias and notices that beyond a specific voltage (breakdown voltage), the current increases sharply while the voltage remains nearly constant. The student uses this diode in a voltage regulator circuit. Answer the following:

  • What causes the Zener breakdown in the diode?
  • How does the Zener diode stabilize voltage in the regulator circuit?
Answer:

Zener breakdown mechanism:

In reverse bias, when the applied voltage exceeds the Zener breakdown voltage, the high electric field across the depletion layer causes avalanche breakdown (for higher voltages) or Zener effect (for heavily doped diodes at lower voltages). This leads to a sharp increase in current due to electron-hole pair generation by impact ionization or quantum tunneling.


Voltage stabilization:

In a regulator circuit, the Zener diode is connected in reverse bias parallel to the load. When input voltage or load resistance varies, the diode maintains a constant voltage across it (equal to Zener voltage) by adjusting its current. For example:
1. If input voltage increases, the diode conducts more current, dropping excess voltage across the series resistor.
2. If load current decreases, the diode shunts the extra current to maintain voltage.


Key point: The Zener diode operates in the breakdown region, acting as a voltage clamp due to its nearly constant voltage characteristic beyond breakdown.

Question 19:
A student sets up an experiment to study the I-V characteristics of a p-n junction diode in forward bias. The circuit consists of a diode, a variable power supply, a resistor, and a voltmeter-ammeter setup. During the experiment, the student observes that the current remains negligible until a certain voltage is reached, after which it increases rapidly. Explain this behavior and discuss the significance of the knee voltage in this context.
Answer:

The observed behavior in the experiment is due to the potential barrier at the p-n junction. Initially, the applied forward bias voltage must overcome this barrier for current to flow significantly. Below the knee voltage, the current is negligible because the majority carriers lack sufficient energy to cross the barrier.


The knee voltage is the minimum voltage required to reduce the barrier significantly, allowing a rapid increase in current. For silicon diodes, it is typically around 0.7 V, while for germanium diodes, it is approximately 0.3 V.


Significance of knee voltage:

  • It helps identify the threshold for conduction in a diode.
  • It is crucial for designing circuits where precise control of diode operation is needed, such as in rectifiers or voltage regulators.
Question 20:
In a physics lab, students are given a Zener diode and asked to demonstrate its use as a voltage regulator. They connect it in reverse bias with a series resistor and a variable load. Explain the working principle of the Zener diode in this setup and how it maintains a constant voltage across the load despite variations in input voltage or load resistance.
Answer:

The Zener diode operates in reverse breakdown region when used as a voltage regulator. In this region, a small change in reverse voltage causes a large change in current, but the voltage across the diode remains nearly constant.


Working principle:

  • When the input voltage increases, the current through the Zener diode rises sharply, dropping excess voltage across the series resistor.
  • If the load resistance decreases, the Zener diode conducts more current to maintain the voltage across the load.

Key points:

  • The Zener voltage (VZ) is the regulated output voltage.
  • The series resistor limits the current to prevent damage to the diode.
  • This property is widely used in power supplies to stabilize voltage for sensitive electronic components.
Question 21:
A student sets up an experiment to study the I-V characteristics of a p-n junction diode. The circuit consists of a diode, a variable power supply, a resistor, and a voltmeter and ammeter connected appropriately. The student observes the following readings:

Forward Bias: Voltage (V) | Current (mA)
0.2 | 0.5
0.4 | 2.0
0.6 | 15.0
0.8 | 80.0

Reverse Bias: Voltage (V) | Current (µA)
-2 | -0.1
-4 | -0.1
-6 | -0.1
-8 | -50.0

Analyze the observations and answer the following:
1. Why does the current remain nearly constant in reverse bias until a certain voltage?
2. What causes the sudden increase in reverse current at -8V?
Answer:

1. In reverse bias, the current remains nearly constant (very small reverse saturation current) because the majority charge carriers (electrons in n-side and holes in p-side) are pulled away from the junction, creating a depletion region. Only minority carriers contribute to the current, which is negligible and independent of voltage until breakdown voltage is reached.


2. The sudden increase in reverse current at -8V indicates the breakdown of the diode. This occurs due to one of two mechanisms:

  • Avalanche breakdown: High electric field accelerates minority carriers, causing collisions that free more electrons.
  • Zener breakdown: Strong field directly breaks covalent bonds (common in heavily doped diodes).
In this case, the abrupt rise suggests Zener breakdown, as it occurs at a fixed voltage.

Question 22:
A manufacturer tests a Zener diode (rated 5V, 200mW) for use in a voltage regulator circuit. The test setup includes a variable input voltage, a series resistor, and the Zener connected in reverse bias. During testing, the input voltage is varied from 0V to 10V while monitoring the output.

Observations:
- Below 5V input: No regulated output.
- At 5V input: Output stabilizes at 5V.
- Above 5V (up to 10V): Output remains 5V, but at 8V input, the Zener gets excessively hot.

Explain:
1. Why does the output stabilize at 5V only when input ≥5V?
2. What causes overheating at 8V input, and how can it be prevented?
Answer:

1. The output stabilizes at 5V (Zener voltage) only when input ≥5V because the Zener diode operates in breakdown region beyond this voltage. Here, it maintains a constant voltage drop (5V) by allowing excess current to flow through it, acting as a voltage regulator.


2. Overheating at 8V occurs due to excessive power dissipation in the Zener diode. Calculations:
Power dissipated = (Input voltage - Zener voltage) × Current
= (8V - 5V) × (Current through Zener).
If current exceeds 200mW/5V = 40mA, the diode overheats.

Prevention:

  • Use a series resistor with appropriate value to limit current.
  • Choose a Zener with higher power rating or parallel heat sinks.

Question 23:
A student sets up a circuit using a p-n junction diode in forward bias, a resistor, and a battery. The diode starts conducting at 0.7V, and the battery provides 5V. The resistor is 1kΩ.

Explain the working of the circuit and calculate the current flowing through it. Also, discuss why the diode does not conduct below 0.7V.

Answer:

The given circuit consists of a p-n junction diode in forward bias, a resistor (1kΩ), and a 5V battery. When the diode is forward-biased, it starts conducting only after the applied voltage exceeds the threshold voltage (0.7V for silicon diodes). Below this voltage, the diode acts as an open circuit due to the depletion region preventing current flow.


To calculate the current:


1. The voltage across the resistor = Total voltage - Diode voltage drop = 5V - 0.7V = 4.3V.
2. Using Ohm's Law (V = IR), the current (I) = V/R = 4.3V / 1000Ω = 4.3mA.

The diode does not conduct below 0.7V because the potential barrier at the junction prevents majority charge carriers from crossing until the external voltage overcomes this barrier. This property ensures controlled conduction in electronic circuits.

Question 24:
In a half-wave rectifier circuit, an input AC voltage of 12V (peak) is applied across a p-n junction diode and a load resistor of 2kΩ.

Explain the output waveform observed across the load resistor. What is the average DC voltage across the load? How does the inclusion of a capacitor filter improve the output?

Answer:

The half-wave rectifier allows only the positive half-cycles of the input AC (12V peak) to pass through the p-n junction diode, while blocking the negative half-cycles. Thus, the output across the 2kΩ load resistor is a pulsating DC waveform with peaks at 12V (minus the 0.7V diode drop, i.e., 11.3V) and zero during negative cycles.


The average DC voltage (Vavg) for a half-wave rectifier is calculated as:
Vavg = Vpeak / π = 11.3V / 3.14 ≈ 3.6V.


Adding a capacitor filter improves the output by:

  • Smoothing the pulsations: The capacitor charges during the positive peaks and discharges during the gaps, maintaining a steadier voltage.
  • Increasing the average DC voltage closer to the peak value (11.3V) by reducing the ripple factor.
This results in a more stable and usable DC output for electronic devices.

Question 25:
A student sets up a circuit with a p-n junction diode, a resistor, and an AC power supply. The diode is connected in forward bias. The student observes that the output voltage across the resistor is not a perfect half-wave rectified signal.

Explain the possible reasons for this observation and suggest how the circuit can be modified to improve the rectification process.

Answer:

The imperfect half-wave rectified signal observed could be due to the following reasons:

  • The forward voltage drop (~0.7V for Si diode) across the diode prevents small input voltages from conducting, causing distortion near zero crossings.
  • Presence of reverse leakage current in the diode allows a tiny unwanted current during negative half-cycles.
  • If the input AC voltage is too low, the diode may not reach sufficient forward bias to conduct properly.

To improve rectification:

  • Use a step-up transformer to increase input voltage, ensuring it always exceeds the diode's forward voltage.
  • Add a filter capacitor parallel to the load resistor to smoothen the output.
  • Select a diode with lower forward voltage (e.g., Schottky diode) or higher peak inverse voltage rating if needed.

Note: For precise rectification of small signals, op-amp based precision rectifiers can be used, though this is beyond the scope of basic diode circuits.

Question 26:
In an experiment to study the characteristics of a Zener diode, a student plots the I-V graph and notices that beyond the breakdown voltage, the current increases sharply while the voltage remains nearly constant.

Explain the physical phenomenon responsible for this behavior and describe two practical applications of Zener diodes based on this property.

Answer:

The constant voltage despite current increase is due to Zener breakdown (for diodes with breakdown voltage <5V) or avalanche breakdown (for higher voltages). In both cases:

  • Strong electric fields cause electron-hole pair generation (Zener effect) or collision-induced carrier multiplication (avalanche effect).
  • This creates abundant charge carriers, allowing high current flow without further voltage increase.

Applications:

  • Voltage regulation: In Zener voltage regulators, the diode maintains constant output voltage across loads despite input variations.
  • Overvoltage protection: Connected in reverse parallel to sensitive components, it clamps transient voltages to safe levels by conducting excess current.

Additional note: Zener diodes are always operated in reverse bias for these applications, unlike normal diodes.

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