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Semiconductor Electronics • Chapter Page 03

PN Junction
Diode

Study PN junction formation, depletion region, barrier potential, forward & reverse bias, V-I characteristics, junction capacitance, Zener & avalanche breakdown and nuclear stability concepts.

CBSE NEET JEE Main JEE Advanced IB Physics IGCSE ICSE A-Level

Section 01

PN Junction Diode

Fundamentals, definition, importance and applications of the PN junction diode.

Definition

A PN Junction Diode is a two-terminal semiconductor device formed when a p-type semiconductor is joined with an n-type semiconductor, creating a junction at the interface. It allows current to flow predominantly in one direction (forward bias) and blocks it in the other (reverse bias).

🔬 What is a PN Junction?

When a p-type semiconductor (doped with acceptor impurities, majority carriers = holes) is brought in contact with an n-type semiconductor (doped with donor impurities, majority carriers = electrons), a metallurgical junction forms between the two regions.

P-type: majority carriers are holes; minority carriers are electrons
N-type: majority carriers are electrons; minority carriers are holes
At the junction, diffusion of charge carriers occurs spontaneously
An internal electric field develops opposing further diffusion

⚡ Why is a Diode Important?

Rectification: converts AC to DC (rectifier circuits)
Signal clipping and clamping circuits
Voltage regulation (Zener diode application)
Logic gates and switching circuits
Light emission (LED) and detection (photodiode)
Solar cells (photovoltaic effect)
Frequency mixing and detection in communication
Protection against reverse polarity in circuits

🧪 Semiconductor Device Basics

Silicon (Si) and Germanium (Ge) are most common semiconductor materials
Si bandgap ≈ 1.12 eV; Ge bandgap ≈ 0.67 eV
Doping introduces controlled impurity atoms
Trivalent dopants (B, Al, In) → p-type
Pentavalent dopants (P, As, Sb) → n-type
Conductivity increases enormously with doping

    Key Concept: A diode is the simplest semiconductor device. It has only two terminals: the Anode (A) connected to the p-side and the Cathode (K) connected to the n-side. Current flows from anode to cathode in forward bias.
  

Section 02

PN Junction Formation

How diffusion, recombination and drift combine to form the PN junction.

Diffusion Process

Electrons (majority carriers in n-type) diffuse toward the p-side due to concentration gradient.
Holes (majority carriers in p-type) diffuse toward the n-side for the same reason.
Electrons recombine with holes near the junction, annihilating both.
This leaves positive donor ions exposed on the n-side and negative acceptor ions on the p-side.
These immobile fixed charges create an internal electric field E pointing from n-side to p-side.
The electric field opposes further diffusion (drift current opposes diffusion current).
Equilibrium is reached when drift current = diffusion current for each carrier type.

Carrier Movement Summary

Carrier	From → To	Type of Current
Electrons	N-side → P-side	Diffusion current
Holes	P-side → N-side	Diffusion current
Electrons	P-side → N-side	Drift current (minor)
Holes	N-side → P-side	Drift current (minor)

PN Junction Formation — Diffusion and Recombination

Section 03

Depletion Region

Formation, width, fixed charges and its role in junction physics.

Definition

The Depletion Region (also called the space charge region or transition region) is the thin layer at the PN junction that is depleted of mobile charge carriers (both electrons and holes) and contains only fixed ionised donor and acceptor atoms.

Key Properties

Fixed positive charges on n-side: ionised donor atoms (e.g., As⁺)
Fixed negative charges on p-side: ionised acceptor atoms (e.g., B⁻)
No mobile charge carriers inside the depletion region
Acts as an insulating barrier at equilibrium
Width ≈ 0.5 µm for Si at typical doping levels
Width is thinner in heavily doped junctions
Width increases under reverse bias; decreases under forward bias

Built-in (Contact) Potential

V₀ = (kT/q) × ln(N_A × N_D / n_i²)

Note: The depletion width W ∝ 1/√N (doping concentration). Heavily doped junctions have a very thin depletion layer, which is critical in Zener diodes.

Depletion Region — Fixed Charges and Electric Field

Section 04

Barrier Potential

Built-in potential, energy band representation and its effect on current flow.

What is Barrier Potential?

The barrier potential (also called contact potential or built-in voltage V₀) is the potential difference that develops across the depletion region at equilibrium due to the separation of positive and negative space charges.

It opposes the further diffusion of majority carriers across the junction
It is NOT a source of EMF and cannot drive external current by itself
For Silicon (Si): V₀ ≈ 0.6 to 0.7 V
For Germanium (Ge): V₀ ≈ 0.2 to 0.3 V
Barrier potential decreases as temperature increases

Effect on Current Flow

Majority carriers (holes from p, electrons from n) face a potential barrier and cannot cross easily
Minority carriers (electrons in p, holes in n) are accelerated across by the built-in field → drift current
At equilibrium: diffusion current = drift current → net current = zero
To conduct external current, barrier must be reduced (forward bias) or increased (reverse bias)

Energy Band Diagram — PN Junction at Equilibrium

🔵

Silicon

V₀ ≈ 0.6 – 0.7 V at room temperature

🟠

Germanium

V₀ ≈ 0.2 – 0.3 V at room temperature

🌡️

Temperature Effect

V₀ decreases ≈ 2.5 mV/°C as T rises

Section 05

Forward Bias

External voltage reduces barrier potential, enabling majority carrier current flow.

Definition

When the positive terminal of an external battery is connected to the p-side and the negative terminal to the n-side, the diode is said to be in forward bias. The applied voltage opposes the built-in potential, reducing the barrier and allowing large current to flow.

Mechanism

External voltage V applied with + to p-side, − to n-side.
Applied electric field opposes the built-in field E₀.
Net barrier reduces from V₀ to (V₀ − V).
Depletion width narrows.
Majority carriers gain enough energy to cross the junction.
Large forward current I_F flows (milliamperes to amperes).

Diode Current Equation (Shockley)

I = I₀ (e^V/ηV_T − 1)

        Where: I0 = reverse saturation current, V = applied voltage, η = ideality factor (1 for Ge, 2 for Si at low current), VT = thermal voltage = kT/q ≈ 26 mV at 300 K.
      

⚡

Threshold Voltage

Si: ~0.6–0.7 V
Ge: ~0.2–0.3 V

📉

Barrier Reduction

New barrier = V₀ − V_forward

🔋

Current Direction

Conventional current: P → N (Anode → Cathode)

Forward Biased PN Junction Diode

Section 06

Reverse Bias

External voltage increases barrier, widens depletion region, allows only tiny minority carrier current.

Definition

When the negative terminal of a battery is connected to the p-side and the positive terminal to the n-side, the diode is in reverse bias. The applied voltage aids the built-in potential, increasing the barrier height and depletion width, blocking majority carrier flow.

Mechanism

External voltage V_R applied with − to p-side, + to n-side.
Applied field adds to built-in field E₀.
Barrier height increases to (V₀ + V_R).
Depletion region widens significantly.
Majority carriers are pushed away from junction.
Only minority carriers cross → tiny reverse saturation current I₀ (µA range).

Important Values

Parameter	Silicon	Germanium
Reverse saturation current I₀	nA range	µA range
Reverse breakdown voltage	Higher (~50-200 V)	Lower (~25-100 V)
Temperature sensitivity of I₀	Doubles/10°C rise	Doubles/10°C rise

Reverse Biased PN Junction — Wide Depletion Region

Important: In reverse bias, only minority carriers contribute to current. This reverse saturation current I₀ is very small (nA for Si, µA for Ge) and nearly independent of reverse voltage — until breakdown occurs.

Section 07

V-I Characteristics

Forward and reverse bias current-voltage curves for Silicon and Germanium diodes.

Forward Bias Characteristics

Cut-in / Knee voltage: Si ≈ 0.6–0.7 V; Ge ≈ 0.2–0.3 V
Below knee voltage: negligible current flows
Above knee: current rises exponentially
Dynamic forward resistance r_f = ΔV/ΔI (low, few ohms)
Static resistance R_F = V_F/I_F

Reverse Bias Characteristics

Tiny reverse saturation current I₀ flows (µA/nA)
I₀ nearly constant until breakdown
At breakdown voltage V_BR: current suddenly increases
Dynamic reverse resistance r_r = ΔV/ΔI (very high, MΩ)
For Si: V_BR typically 50–200 V

V-I Characteristics — Silicon and Germanium Diodes

Section 08

Junction Capacitance

Transition capacitance in reverse bias and diffusion capacitance in forward bias.

Transition (Depletion Layer) Capacitance — C_T

The depletion region acts like the dielectric of a parallel-plate capacitor. This capacitance is dominant in reverse bias.

Transition Capacitance

C_T = ε A / W

ε = permittivity of semiconductor
A = junction cross-sectional area
W = depletion width (increases with reverse voltage)
C_T decreases as reverse voltage increases (W increases)
C_T ∝ 1/√V_R for abrupt junctions
Used in varactor (varicap) diodes for tuning

Diffusion Capacitance — C_D

Due to minority carrier charge storage on either side of the junction. Dominant in forward bias.

Diffusion Capacitance

C_D = τ × g_m = τ × (I_F/V_T)

τ = minority carrier lifetime
g_m = transconductance of the diode
C_D increases with forward current I_F
Much larger than C_T under forward bias
Limits high-frequency switching performance

Junction Capacitance vs Reverse Voltage

Property	Transition C_T	Diffusion C_D
Bias condition	Reverse bias	Forward bias
Origin	Depletion layer charges	Minority carrier storage
Magnitude	Small (pF range)	Large (nF range)
Varies with voltage as	Decreases with V_R	Increases with I_F
Application	Varactor diode tuning	Limits switching speed

Section 09

Reverse Breakdown

When reverse voltage exceeds breakdown voltage, current surges due to Zener or avalanche effect.

    Breakdown occurs when a sufficiently large reverse voltage VBR is applied. The reverse current increases abruptly. Two mechanisms: Zener breakdown (heavily doped, thin depletion, quantum tunnelling) and Avalanche breakdown (lightly doped, wide depletion, impact ionisation).
  

Section 10

Zener Breakdown

Mechanism

Occurs in heavily doped junctions with very thin depletion regions (≈5–10 nm)
High electric field (≈10⁷ V/m) across thin depletion layer
Valence electrons tunnel directly through the thin barrier (quantum tunnelling)
Breakdown voltage V_Z < 6 V for Si at room temperature
V_Z has a negative temperature coefficient: V_Z decreases as temperature rises
Used intentionally in Zener diodes for voltage regulation

Zener Electric Field

E = V_Z / W ≈ 10⁷ V/m

Zener Diode V-I Characteristic

Section 11

Avalanche Breakdown

Mechanism

Occurs in lightly doped junctions with wide depletion regions
Minority carriers accelerated by large reverse electric field
High-energy carriers collide with lattice atoms → impact ionisation
Each collision creates new electron-hole pairs
New carriers also accelerate → avalanche multiplication
Breakdown voltage V_AV > 6 V
V_AV has a positive temperature coefficient: increases as temperature rises
Multiplication factor M = 1 / [1 − (V/V_BR)ⁿ]

Property	Zener Breakdown	Avalanche Breakdown
Doping level	Heavy (both sides)	Light (both sides)
Depletion width	Very thin	Wide
Mechanism	Quantum tunnelling	Impact ionisation
Breakdown voltage	< 6 V (for Si)	> 6 V (for Si)
Temperature coefficient	Negative (V_Z ↓ as T ↑)	Positive (V_AV ↑ as T ↑)
Electric field required	~10⁷ V/m	~10⁶ V/m
Application	Voltage regulators	Avalanche photodiodes

Comparison Tables

Important Comparison Tables

Forward Bias vs Reverse Bias

Property	Forward Bias	Reverse Bias
Battery connection	+ to p-side, − to n-side	− to p-side, + to n-side
Effect on barrier	Reduces barrier potential	Increases barrier potential
Depletion width	Decreases	Increases
Current magnitude	Large (mA to A)	Tiny (µA/nA)
Resistance	Very low (few Ω)	Very high (MΩ)
Carrier flow	Majority carriers cross junction	Only minority carriers cross
Diode condition	ON (conducting)	OFF (non-conducting)
Threshold voltage (Si)	~0.7 V needed	No threshold; immediate blocking

Silicon vs Germanium Diode

Property	Silicon Diode	Germanium Diode
Bandgap energy	1.12 eV	0.67 eV
Knee (cut-in) voltage	0.6–0.7 V	0.2–0.3 V
Reverse saturation current I₀	nA (very small)	µA (larger)
Operating temperature range	Up to ~150°C	Up to ~75°C
Reverse breakdown voltage	Higher	Lower
Leakage current	Very low	Relatively higher
Most widely used	Yes (preferred)	Less common today

Solved Numericals

Important Numericals

Step-by-step solved problems — CBSE, NEET, JEE Main, JEE Advanced, IB, IGCSE, A-Level.

CBSEQ1 — Diode Forward Current

Question:A silicon diode has a reverse saturation current I₀ = 10 nA at 300 K. Calculate the forward current when a forward voltage of 0.65 V is applied. (η = 1, V_T = 26 mV)

Given:I₀ = 10 × 10⁻⁹ A, V = 0.65 V, η = 1, V_T = 26 mV = 0.026 V

Formula

I = I₀(e^V/ηV_T − 1)

Substitution

V/ηV_T = 0.65 / (1 × 0.026) = 25
e²⁵ ≈ 7.2 × 10¹⁰

Calculation

I = 10 × 10⁻⁹ × (7.2 × 10¹⁰ − 1) ≈ 10 × 10⁻⁹ × 7.2 × 10¹⁰ = 720 mA ≈ 0.72 A

✅ Final Answer: I_F ≈ 0.72 A

💡 Exam Tip: e²⁵ ≈ 7.2 × 10¹⁰ is a standard value to remember for Si diode calculations.

NEETQ2 — Dynamic Resistance of Diode

Question:A diode is forward biased with V_F = 0.7 V and I_F = 50 mA. Find the dynamic forward resistance r_f.

Given:V_T = 26 mV, I_F = 50 mA

Formula

r_f = η V_T / I_F = 26 mV / 50 mA

Calculation

r_f = 0.026 / 0.050 = 0.52 Ω

✅ Final Answer: r_f = 0.52 Ω

💡 Exam Tip: Dynamic resistance = V_T/I_F. At room temperature, V_T = 26 mV.

JEE MainQ3 — Transition Capacitance

Question:A PN junction has area A = 1 mm² = 10⁻⁶ m², depletion width W = 0.5 µm = 5 × 10⁻⁷ m, and ε = 11.7 × 8.85 × 10⁻¹² F/m (Si). Calculate transition capacitance C_T.

Formula

C_T = ε A / W

Calculation

ε = 11.7 × 8.85 × 10⁻¹² = 1.035 × 10⁻¹⁰ F/m
C_T = (1.035 × 10⁻¹⁰ × 10⁻⁶) / (5 × 10⁻⁷) = 2.07 × 10⁻¹⁰ F ≈ 207 pF

✅ Final Answer: C_T ≈ 207 pF

💡 Exam Tip: C_T is in pF range. As reverse voltage increases, W increases and C_T decreases.

JEE AdvancedQ4 — Built-in Potential

Question:A Si PN junction has N_A = 10¹⁷ cm⁻³, N_D = 10¹⁵ cm⁻³, n_i = 1.5 × 10¹⁰ cm⁻³ at 300 K. Find the contact potential V₀. (k = 8.62 × 10⁻⁵ eV/K)

Formula

V₀ = (kT/q) × ln(N_A × N_D / n_i²) = V_T × ln(N_A N_D / n_i²)

Calculation

V_T = 0.026 V at 300 K
N_A × N_D / n_i² = (10¹⁷ × 10¹⁵) / (1.5 × 10¹⁰)² = 10³² / (2.25 × 10²⁰) = 4.44 × 10¹¹
ln(4.44 × 10¹¹) = ln(4.44) + 11 × ln(10) = 1.49 + 25.33 = 26.82
V₀ = 0.026 × 26.82 ≈ 0.697 V

✅ Final Answer: V₀ ≈ 0.70 V

💡 Exam Tip: This confirms the well-known result that Si built-in potential ≈ 0.6–0.7 V.

NEETQ5 — Half-wave Rectifier Output

Question:An AC source of peak voltage 12 V is applied to a half-wave rectifier circuit with a load resistance R_L = 1 kΩ. Neglecting diode drop, find (a) peak output voltage, (b) dc (average) output voltage, (c) peak inverse voltage.

Solution

(a) V_peak = 12 V (same as input, diode drop neglected)
(b) V_dc = V_peak / π = 12 / 3.14 ≈ 3.82 V
(c) PIV = V_peak = 12 V

✅ V_dc ≈ 3.82 V; PIV = 12 V

💡 Exam Tip: For half-wave: V_dc = V_m/π. For full-wave: V_dc = 2V_m/π.

CBSEQ6 — Full-wave Rectifier

Question:A full-wave rectifier uses two diodes. Input AC has peak voltage V_m = 20 V. Find the dc output voltage and ripple frequency if input frequency is 50 Hz.

Solution

V_dc = 2V_m/π = 2 × 20 / 3.14 = 12.74 V
Ripple frequency = 2 × input frequency = 2 × 50 = 100 Hz

✅ V_dc ≈ 12.74 V; Ripple frequency = 100 Hz

💡 Exam Tip: Full-wave ripple frequency is twice the supply frequency. This is a standard CBSE question.

JEE MainQ7 — Reverse Saturation Current vs Temperature

Question:The reverse saturation current of a Ge diode is I₀ = 2 µA at 27°C. Find I₀ at 47°C, given that I₀ doubles for every 10°C rise.

Solution

Temperature increase = 47 − 27 = 20°C
Number of 10°C intervals = 20/10 = 2
I₀(47°C) = I₀(27°C) × 2² = 2 × 4 = 8 µA

✅ I₀ at 47°C = 8 µA

💡 Exam Tip: I₀ doubles every 10°C for both Si and Ge. This is a frequently tested rule.

IB PhysicsQ8 — Diode in Series with Resistor

Question:A silicon diode (V_knee = 0.7 V) is connected in series with a 330 Ω resistor and a 5 V supply. Find the current through the circuit and the voltage across the resistor.

Solution

Voltage across resistor = Supply − V_diode = 5 − 0.7 = 4.3 V
Current I = V_R/R = 4.3 / 330 = 13.0 mA

✅ I ≈ 13.0 mA; V_R = 4.3 V

💡 Exam Tip: Always subtract the diode voltage drop (0.7 V for Si, 0.3 V for Ge) from supply before applying Ohm's law.

IGCSEQ9 — LED Series Resistor

Question:An LED requires a forward voltage of 2.0 V and a current of 20 mA. It is connected to a 9 V supply. Calculate the series resistor required.

Solution

V_R = 9 − 2.0 = 7.0 V
R = V_R/I = 7.0 / (20 × 10⁻³) = 350 Ω

✅ R = 350 Ω (use nearest standard: 360 Ω)

💡 Exam Tip: This is a standard IGCSE LED circuit question. Always check the LED's forward voltage specification.

A-LevelQ10 — Rectification Efficiency

Question:Define rectification efficiency η and show that for a half-wave rectifier η ≈ 40.6%.

Solution

η = (P_dc/P_ac) × 100
P_dc = I_dc² × R_L = (I_m/π)² × R_L
P_ac = I_rms² × R_L = (I_m/2)² × R_L
η = [(I_m/π)²]/[(I_m/2)²] = 4/π² ≈ 0.406 = 40.6%

✅ Half-wave rectifier efficiency ≈ 40.6%

💡 Exam Tip: Full-wave rectifier efficiency ≈ 81.2% (double that of half-wave).

Past Year Questions

Previous Year Questions

NEET PYQs

NEET 2024

1. In a PN junction diode, the width of the depletion region in reverse bias is:

Ans: Wider than in forward bias. In reverse bias the built-in field is augmented, pushing majority carriers away and increasing the depletion width.

NEET 2023

2. The V-I characteristics of a PN junction diode show that in forward bias, current increases exponentially. The relation is I = I₀(e^V/V_T−1). At room temperature V_T equals:

Ans: V_T = kT/q ≈ 26 mV at 300 K (thermal voltage).

NEET 2022

3. For a silicon PN junction diode, the barrier potential at room temperature is approximately:

Ans: 0.7 V. (For Germanium ≈ 0.3 V.)

NEET 2022

4. In a half-wave rectifier, if the input frequency is 50 Hz, the output ripple frequency is:

Ans: 50 Hz (same as input, since only one half-cycle is used).

NEET 2021

5. In reverse bias, the minority carrier current in a PN junction is called:

Ans: Reverse saturation current I₀. It is small (nA for Si) and nearly independent of reverse voltage until breakdown.

NEET 2021

6. Zener breakdown occurs when the reverse voltage is:

Ans: Below approximately 6 V for silicon, where heavy doping creates a very thin depletion layer and strong tunnelling field.

NEET 2020

7. The knee voltage for a germanium diode is approximately:

Ans: 0.3 V (compared to 0.7 V for silicon).

NEET 2020

8. In a full-wave rectifier using two diodes, the ripple frequency for 50 Hz AC input is:

Ans: 100 Hz (twice the input frequency).

NEET 2019

9. The diffusion capacitance of a PN junction is dominant under:

Ans: Forward bias. It arises from minority carrier charge storage near the junction and is much larger than transition capacitance in forward bias.

NEET 2019

10. If the reverse saturation current of a diode is 1 µA at 27°C, at 47°C it becomes:

Ans: 4 µA (doubles every 10°C, so 20°C rise → ×4).

NEET 2018

11. Which of the following is an application of a PN junction diode in reverse bias?

Ans: Zener diode voltage regulator. Zener diodes are operated in controlled reverse breakdown for regulation.

NEET 2018

12. The energy gap of silicon is approximately:

Ans: 1.12 eV. Germanium: 0.67 eV; GaAs: 1.42 eV.

NEET 2017

13. In a PN junction, the diffusion current is due to:

Ans: Concentration gradient of majority carriers — holes diffuse from p to n; electrons diffuse from n to p.

NEET 2017

14. A PN junction diode does NOT conduct when:

Ans: It is reverse biased OR forward biased below the cut-in voltage.

NEET 2016

15. Avalanche breakdown in a PN junction is caused by:

Ans: Impact ionisation by high-energy minority carriers in the wide depletion region of a lightly doped junction under large reverse voltage.

NEET 2016

16. The static resistance of a diode operating at forward voltage 0.7 V and current 35 mA is:

Ans: R_F = V/I = 0.7/0.035 = 20 Ω.

NEET 2015

17. Which type of doping produces p-type semiconductor?

Ans: Trivalent impurity (e.g., Boron, Aluminium, Indium) — creates holes as majority carriers.

NEET 2015

18. The depletion region of a PN junction consists of:

Ans: Fixed positive donor ions on the n-side and fixed negative acceptor ions on the p-side; no mobile charge carriers.

NEET 2014

19. The temperature coefficient of Zener breakdown voltage is:

Ans: Negative — V_Z decreases as temperature rises (opposite to avalanche).

NEET 2013

20. In forward biased PN junction, if applied voltage is V and barrier potential is V₀, the net barrier is:

Ans: V₀ − V (barrier is reduced by the forward voltage).

JEE Main PYQs

JEE Main 2024

1. In a PN junction diode connected in reverse bias, if the reverse saturation current is 2 µA at 27°C, find the current at 57°C.

Ans: I₀ doubles per 10°C; 30°C rise → ×2³ = 8 times → 16 µA.

JEE Main 2024

2. The V-I characteristic of an ideal PN junction diode in forward bias is best described by:

Ans: I = I₀(e^V/V_T − 1) — an exponential rise beyond the cut-in voltage.

JEE Main 2023

3. The dynamic resistance of a diode at a forward current of 40 mA at 300 K is:

Ans: r_f = V_T/I = 26 × 10⁻³/40 × 10⁻³ = 0.65 Ω.

JEE Main 2023

4. In a Zener diode circuit, the Zener voltage is 5.6 V. Input varies from 8 to 12 V and series resistor R = 200 Ω. Find the range of Zener current.

Ans: I_Z(min) = (8−5.6)/200 = 12 mA; I_Z(max) = (12−5.6)/200 = 32 mA.

JEE Main 2022

5. A PN junction diode has transition capacitance C_T = 40 pF at V_R = 4 V. What is C_T at V_R = 16 V for an abrupt junction?

Ans: C_T ∝ 1/√V_R; C_T2/C_T1 = √(4/16) = 1/2; C_T2 = 20 pF.

JEE Main 2022

6. Holes in p-type semiconductor are:

Ans: Majority carriers. They are the dominant current carriers in p-type material.

JEE Main 2021

7. The peak inverse voltage (PIV) across each diode in a full-wave bridge rectifier with peak input V_m is:

Ans: PIV = V_m (for bridge rectifier; for centre-tap it is 2V_m).

JEE Main 2021

8. The ratio of forward to reverse resistance of a silicon diode is approximately:

Ans: About 10⁻⁵:1 or 1:10⁵ — forward resistance is a few Ω, reverse is MΩ.

JEE Main 2020

9. An LED emits light when forward biased. The wavelength of emitted light is related to its bandgap E_g by:

Ans: λ = hc/E_g. For E_g = 1.9 eV (red LED): λ = (6.63×10⁻³⁴ × 3×10⁸)/(1.9 × 1.6×10⁻¹⁹) ≈ 654 nm.

JEE Main 2020

10. Drift current in a PN junction at equilibrium is due to:

Ans: Minority carriers moving under the influence of the built-in electric field across the depletion region.

JEE Main 2019

11. In a Zener diode used as a voltage regulator, the Zener is operated in:

Ans: Reverse breakdown region — specifically Zener breakdown for V_Z < 6V.

JEE Main 2019

12. The contact potential V₀ of a PN junction with equal doping N_A = N_D = 10¹⁶ cm⁻³ and n_i = 10¹⁰ cm⁻³ at 300 K is:

Ans: V₀ = 0.026 × ln(10³²/10²⁰) = 0.026 × 12 × ln10 = 0.026 × 27.63 ≈ 0.72 V.

JEE Main 2018

13. For a PN junction in thermal equilibrium, the Fermi level is:

Ans: Constant throughout the junction (the same in p-region, depletion region, and n-region).

JEE Main 2018

14. Form factor of a half-wave rectifier is:

Ans: Form factor = I_rms/I_dc = (I_m/2)/(I_m/π) = π/2 ≈ 1.57.

JEE Main 2017

15. In a photodiode, the depletion region is used to:

Ans: Separate photogenerated electron-hole pairs; the built-in field sweeps carriers apart, creating a photocurrent proportional to light intensity.

JEE Main 2017

16. The temperature at which the intrinsic carrier concentration n_i of Si equals the carrier concentration of a doped Si sample (N_D = 10¹⁵ cm⁻³) is called:

Ans: Intrinsic temperature — above this, n_i ≫ N_D and the doped behaviour is lost.

JEE Main 2016

17. Ripple factor of a full-wave rectifier is:

Ans: γ = √[(I_rms/I_dc)² − 1] = √[(π/2√2)/(2/π)]² − 1 ≈ 0.48.

JEE Main 2015

18. In a solar cell, the p-side is illuminated. The photocurrent flows from:

Ans: From p-side to n-side inside the cell (conventional current), or n-side to p-side through the external circuit.

JEE Main 2014

19. A Zener diode has V_Z = 6.2 V and I_Z(max) = 50 mA. The maximum power it can dissipate is:

Ans: P_max = V_Z × I_Z(max) = 6.2 × 0.050 = 0.31 W.

JEE Main 2013

20. The Shockley diode equation is valid when the diode is:

Ans: Moderately forward biased or reverse biased below breakdown. It is not valid at very high forward currents (where series resistance dominates) or at breakdown.

JEE Advanced PYQs

JEE Advanced 2023

1. In a heavily doped PN junction, the depletion layer capacitance C_T is found to vary as C_T ∝ (V₀ − V)^−m. The value of m for an abrupt junction and a graded junction respectively is:

Ans: m = 1/2 for abrupt (step) junction; m = 1/3 for linearly graded junction.

JEE Advanced 2022

2. Derive an expression for the depletion width W in terms of doping concentrations N_A, N_D, built-in voltage V₀ and permittivity ε for a one-sided abrupt junction.

Ans: W = √(2εV₀/q × (1/N_A + 1/N_D)). For one-sided (N_D ≫ N_A): W ≈ √(2εV₀/(qN_A)).

JEE Advanced 2021

3. A PN junction solar cell of area 4 cm² is illuminated by a uniform light source. The short-circuit current density J_sc = 15 mA/cm². The open-circuit voltage V_oc = 0.55 V. Find the maximum power output if fill factor FF = 0.70.

Ans: I_sc = 15 × 4 = 60 mA; P_max = FF × V_oc × I_sc = 0.70 × 0.55 × 0.060 = 23.1 mW.

JEE Advanced 2020

4. In a diode circuit with two ideal diodes and resistors, determine which diode conducts and find the current in each branch for a given biasing configuration.

Ans: Apply Kirchhoff's laws; an ideal diode conducts (zero drop) when forward biased and blocks (infinite resistance) when reverse biased. Determine bias direction from circuit polarity, then solve the resulting resistor network.

JEE Advanced 2019

5. Explain why the reverse saturation current I₀ is virtually independent of reverse voltage but strongly temperature-dependent.

Ans: I₀ is limited by minority carrier generation rate near the depletion edge — fixed by temperature (thermal excitation across bandgap) and independent of how large the reverse voltage is. As T increases, more minority carriers are generated thermally, increasing I₀ exponentially.

JEE Advanced 2018

6. Prove that at thermal equilibrium in a PN junction, the Fermi level is constant throughout the device.

Ans: At equilibrium, net current = 0. Since J = σ(E + (1/q)∇E_F), J = 0 requires ∇E_F = 0, meaning E_F is spatially constant.

JEE Advanced 2017

7. The multiplication factor M in avalanche breakdown is given by M = 1/[1 − (V/V_BR)ⁿ]. As V → V_BR, M →?

Ans: M → ∞ as V → V_BR, representing an unlimited carrier avalanche.

JEE Advanced 2016

8. Two Zener diodes Z₁ (V_Z = 5 V) and Z₂ (V_Z = 8 V) are connected back-to-back in series with a 100 Ω resistor across a 20 V supply. Find the voltage across the combination.

Ans: One Zener is forward biased (≈0.7 V drop), one is in breakdown (its V_Z). Total across pair = V_Z1 + 0.7 = 5.7 V or V_Z2 + 0.7 = 8.7 V depending on polarity. Common configuration: 5 + 0.7 = 5.7 V or 8 + 0.7 = 8.7 V.

JEE Advanced 2015

9. Explain the physical significance of the ideality factor η in the diode equation. When is η = 1 and when is η = 2?

Ans: η = 1 when diffusion current dominates (moderate forward bias in Ge, or high forward current in Si). η = 2 when recombination-generation current in the depletion region dominates (low forward bias in Si).

JEE Advanced 2014

10. A photodiode is operated in reverse bias. Explain why reverse bias is preferred over forward bias for photodetection.

Ans: Reverse bias widens the depletion region, increasing the active absorption volume. It also reduces junction capacitance C_T, giving faster response. The large electric field efficiently separates photogenerated electron-hole pairs, minimising recombination and improving quantum efficiency.

CBSE PYQs

CBSE 2024

1. Draw the V-I characteristic of a PN junction diode and label: forward bias region, reverse bias region, knee voltage, and reverse saturation current.

Ans: See V-I characteristic graph above. Key labels: knee voltage ~0.7 V (Si), saturation current I₀ in reverse region, steep exponential rise in forward region.

CBSE 2024

2. With a labelled circuit diagram, explain the working of a half-wave rectifier.

Ans: One diode in series with load R_L. During positive half-cycle, diode conducts; during negative half-cycle, diode blocks. Output is pulsating DC with same frequency as input. Average output = V_m/π.

CBSE 2023

3. Explain why silicon is preferred over germanium for making semiconductor devices.

Ans: Si has wider bandgap (1.12 eV vs 0.67 eV), can operate at higher temperatures (~150°C vs ~75°C), has lower reverse leakage current, and has a natural oxide (SiO₂) useful for fabrication.

CBSE 2023

4. Define (a) depletion region (b) barrier potential in a PN junction.

Ans: (a) Region around junction depleted of mobile carriers, containing only fixed ionised charges. (b) Potential difference across depletion region opposing majority carrier diffusion; ~0.7 V for Si.

CBSE 2022

5. What is the effect of forward bias on the depletion region and barrier potential of a PN junction?

Ans: Forward bias reduces the depletion width and lowers the barrier potential from V₀ to (V₀ − V), allowing majority carriers to flow.

CBSE 2022

6. In a PN junction, what is drift current and how does it arise?

Ans: Drift current arises due to minority carriers (electrons in p-side, holes in n-side) being swept across the junction by the built-in electric field. It is small and flows opposite to diffusion current.

CBSE 2021

7. Draw the energy band diagram of a PN junction at thermal equilibrium and indicate the Fermi level.

Ans: Conduction and valence bands bend at the junction. The Fermi level E_F is horizontal (constant) throughout the device at equilibrium. Band bending equals eV₀.

CBSE 2021

8. What is a Zener diode? How does it differ from a regular PN junction diode?

Ans: A Zener diode is a heavily doped PN junction designed to operate in reverse breakdown at a precise voltage V_Z. Unlike regular diodes, it maintains a nearly constant voltage across itself in the Zener region, used for voltage regulation.

CBSE 2020

9. Calculate the current through a circuit with a Si diode (V_f = 0.7 V) and a 470 Ω resistor connected to a 5 V supply in forward bias.

Ans: I = (5 − 0.7)/470 = 4.3/470 ≈ 9.15 mA.

CBSE 2020

10. Distinguish between avalanche and Zener breakdown.

Ans: Zener — heavily doped, thin depletion, tunnelling mechanism, V_Z < 6 V, negative temperature coefficient. Avalanche — lightly doped, wide depletion, impact ionisation, V > 6 V, positive temperature coefficient.

CBSE 2019

11. Explain with a circuit diagram how a PN junction diode is used as a full-wave rectifier.

Ans: Centre-tap transformer with two diodes. During positive half-cycle: D₁ conducts; during negative half-cycle: D₂ conducts. Load always sees same polarity. Output frequency = 2 × input frequency.

CBSE 2019

12. What is the function of a filter capacitor in a rectifier circuit?

Ans: The capacitor smooths the pulsating DC output by charging to peak voltage when diode conducts and discharging slowly through R_L between pulses, reducing ripple.

CBSE 2018

13. Why does reverse saturation current increase with temperature?

Ans: I₀ depends on minority carrier generation. As temperature rises, more covalent bonds break (thermal energy > bandgap), generating more minority carriers, increasing I₀ exponentially.

CBSE 2018

14. A PN junction diode has I₀ = 5 nA. Find the forward current at V = 0.5 V, T = 300 K, η = 1.

Ans: V/V_T = 0.5/0.026 = 19.23; e^19.23 ≈ 2.24 × 10⁸; I = 5 × 10⁻⁹ × 2.24 × 10⁸ ≈ 1.12 A.

CBSE 2017

15. What is the peak inverse voltage (PIV) for a half-wave rectifier?

Ans: PIV = V_m (the peak AC voltage). The diode must withstand this reverse voltage during the non-conducting half-cycle without breaking down.

IB Physics Questions

IB Physics HL

1. Explain the formation of the depletion region in a PN junction from first principles, referring to diffusion, drift and equilibrium.

Ans: Concentration gradients drive holes from p→n and electrons from n→p (diffusion). Recombination at junction exposes fixed ions, creating a field E₀ that drives drift in the opposite direction. Equilibrium: J_diffusion = J_drift, net current = 0.

IB Physics SL

2. A diode and 1 kΩ resistor are connected in series to a 6 V supply. The diode has V_f = 0.6 V. Calculate current and power dissipated in resistor.

Ans: I = (6 − 0.6)/1000 = 5.4 mA; P_R = I²R = (5.4 × 10⁻³)² × 1000 = 29.2 mW.

IB Physics HL

3. The V-I characteristic of a real diode deviates from the ideal Shockley equation at high forward currents. Explain why.

Ans: At high currents, the ohmic (series) resistance of the bulk semiconductor (r_s) causes significant voltage drop V = I×r_s, making the actual curve less steep than e^V/V_T.

IB Physics SL

4. State two differences between a silicon diode and a germanium diode.

Ans: (i) Si knee voltage ≈ 0.7 V; Ge ≈ 0.3 V. (ii) Si has lower reverse leakage current (nA) compared to Ge (µA). (iii) Si operates at higher temperatures.

IB Physics HL

5. Explain why a photodiode is more sensitive when reverse biased.

Ans: Reverse bias widens depletion region (larger absorption region), reduces C_T (faster response), and the large built-in + applied field efficiently separates photogenerated pairs before recombination.

IB Physics SL

6. A Zener diode (V_Z = 4.7 V) is used to regulate a 9 V supply. Series resistance R = 100 Ω, load R_L = 470 Ω. Find (a) Zener current (b) power in Zener.

Ans: V_out = 4.7 V. I_L = 4.7/470 = 10 mA. I_series = (9 − 4.7)/100 = 43 mA. I_Z = 43 − 10 = 33 mA. P_Z = 4.7 × 0.033 = 0.155 W.

IB Physics HL

7. Define transition capacitance C_T and explain its dependence on reverse voltage.

Ans: C_T = ε A/W. As V_R increases, depletion width W increases, so C_T ∝ 1/√V_R decreases. This is exploited in varactor diodes for voltage-controlled tuning.

IB Physics SL

8. A half-wave rectifier has peak AC input of 15 V and load resistance 500 Ω. Find peak current, average current, and average power dissipated.

Ans: I_peak = 15/500 = 30 mA. I_dc = I_peak/π = 9.55 mA. P_dc = I_dc² × R = (9.55 × 10⁻³)² × 500 = 45.6 mW.

IB Physics HL

9. Compare the energy band diagrams of a PN junction in (a) equilibrium (b) forward bias (c) reverse bias.

Ans: (a) Flat Fermi level, bands bent by eV₀. (b) Bands flatten — barrier reduces to e(V₀−V). (c) Bands tilt more steeply — barrier increases to e(V₀+V_R).

IB Physics SL

10. State the purpose of the depletion region in the normal operation of a PN junction diode.

Ans: It provides the potential barrier that prevents majority carrier flow at equilibrium, and controls conduction by being narrowed (forward bias) or widened (reverse bias) by an external voltage.

IGCSE Questions

IGCSE

1. A diode is connected in a circuit. The anode is at +3 V and the cathode at +2 V. Is the diode forward or reverse biased?

Ans: Forward biased (anode more positive than cathode by 1 V > 0.7 V), so the diode conducts.

IGCSE

2. Explain what happens in a half-wave rectifier during the positive and negative half-cycles of AC.

Ans: Positive half: diode forward biased → conducts → current flows through R_L. Negative half: diode reverse biased → no conduction → zero output.

IGCSE

3. A green LED has a forward voltage of 2.1 V and requires 25 mA. Connected to 5 V supply. Find series resistor value.

Ans: R = (5 − 2.1)/0.025 = 2.9/0.025 = 116 Ω (use 120 Ω standard value).

IGCSE

4. What is a diode used for in a power supply circuit?

Ans: Rectification — converting AC to DC by allowing current in only one direction.

IGCSE

5. Draw the symbol for a diode and mark the anode and cathode.

Ans: Triangle pointing in direction of conventional current, flat base = cathode (marked with vertical bar). Anode = pointed tip side.

IGCSE

6. An ideal diode has zero voltage drop. A 12 V source with 600 Ω in series drives an ideal forward-biased diode. Find current.

Ans: I = 12/600 = 20 mA.

IGCSE

7. Explain why a bridge rectifier produces a smoother output than a half-wave rectifier.

Ans: A bridge rectifier utilises both half-cycles of the AC, producing pulses at twice the frequency. More frequent pulses are easier to smooth with a capacitor filter.

IGCSE

8. In which direction does conventional current flow through a forward-biased diode?

Ans: From anode (p-side) to cathode (n-side) — same direction as hole flow.

IGCSE

9. What is the main difference in function between an LED and a photodiode?

Ans: LED converts electrical energy to light (forward biased). Photodiode converts light to electrical current (reverse biased as detector).

IGCSE

10. A student connects a diode in reverse bias with a 9 V battery and 1 kΩ resistor. What current flows? (Assume no breakdown)

Ans: Approximately zero — only tiny reverse saturation current (nA range) flows, negligible for practical purposes.

A-Level Questions

A-Level

1. Sketch and explain the V-I characteristic of a PN junction diode for both forward and reverse bias. Comment on the asymmetry.

Ans: Forward: exponential rise after cut-in voltage (0.7 V Si). Reverse: tiny near-constant saturation current until breakdown. Asymmetry arises because majority and minority carrier densities differ by many orders of magnitude.

A-Level

2. Derive that the dc output voltage of a full-wave rectifier is V_dc = 2V_m/π.

Ans: V_dc = (1/π)∫₀^π V_msinθ dθ = (V_m/π)[−cosθ]₀^π = (V_m/π)(1+1) = 2V_m/π.

A-Level

3. Calculate the ripple factor of a full-wave rectifier and compare it to that of a half-wave rectifier.

Ans: Full-wave: γ = √[(π²/8)−1] ≈ 0.483. Half-wave: γ = √[(π²/4)−1] ≈ 1.21. Full-wave has much lower ripple.

A-Level

4. A Zener diode (V_Z = 6.8 V, P_max = 0.5 W) is used with a 12 V supply. Find maximum series resistance R to just protect the Zener.

Ans: I_Z(max) = P_max/V_Z = 0.5/6.8 = 73.5 mA. R_min = (12−6.8)/0.0735 = 70.7 Ω.

A-Level

5. Explain the operation of a bridge rectifier, naming which diodes conduct during each half-cycle.

Ans: Positive half-cycle: D₁ and D₃ conduct (current top→right→load→bottom→left). Negative half-cycle: D₂ and D₄ conduct. Current always flows in same direction through load R_L.

A-Level

6. Explain how a capacitor smooths the output of a rectifier and define ripple voltage.

Ans: Capacitor charges to peak and discharges through R_L between pulses. Ripple voltage = peak-to-trough variation in output. V_ripple ≈ I_dc/(f C) where f is ripple frequency.

A-Level

7. Distinguish between the temperature coefficients of Zener and avalanche breakdown and explain the physical reason.

Ans: Zener: negative TC (tunnelling probability increases with temperature). Avalanche: positive TC (mean free path decreases with temperature, carriers gain less energy per mean free path, need higher voltage to start avalanche).

A-Level

8. A silicon diode has I₀ = 0.1 nA at 20°C. Estimate I₀ at 80°C.

Ans: Rise = 60°C = 6 × 10°C intervals; I₀ doubles each interval. I₀(80°C) = 0.1 × 2⁶ = 0.1 × 64 = 6.4 nA.

A-Level

9. Explain why diffusion capacitance C_D limits the high-frequency performance of a diode.

Ans: C_D is large in forward bias. The RC time constant τ = C_D × r_f limits switching speed. At frequencies f >> 1/(2πτ), the capacitor shorts out the signal, so the diode cannot switch fast enough.

A-Level

10. The ideality factor η of a silicon diode is measured at η = 1.8 at low forward current. What does this suggest about the dominant current mechanism?

Ans: η close to 2 indicates that recombination-generation current in the depletion region dominates over pure diffusion current (η = 1). This is typical for Si at low forward bias where the recombination current proportional to e^V/2V_T dominates.

Case Studies

Case Study Questions

Case Study 1

PN Junction in a Power Supply Unit

A power supply unit (PSU) converts 230 V AC mains to a stable 5 V DC output for powering digital circuits. The PSU uses a step-down transformer (230 V to 12 V rms), followed by a bridge rectifier consisting of four silicon diodes, a 1000 µF smoothing capacitor, and a 7805 Zener-based voltage regulator IC. Each silicon diode has a forward voltage drop of 0.7 V. The bridge rectifier produces a full-wave rectified output with ripple frequency of 100 Hz. A student notices that when the load current is 0.5 A, the output voltage is steady at 5 V, but when the load is removed, the output voltage rises slightly.

Q1. What is the peak AC voltage after the transformer? (V_rms = 12 V)

Ans: V_peak = V_rms × √2 = 12 × 1.414 = 16.97 V ≈ 17 V

Q2. What is the effective DC voltage at the output of the bridge rectifier before the regulator? (Account for two diode drops)

Ans: In a bridge rectifier, two diodes conduct at once: V_dc = 2V_m/π − 2 × 0.7 = (2 × 16.97)/π − 1.4 = 10.81 − 1.4 = 9.41 V

Q3. Why does the output voltage rise slightly when the load is removed?

Ans: With no load, the capacitor charges to the full peak voltage and cannot discharge. With load, C discharges between pulses, lowering average voltage. No load → V_out approaches V_peak − 2×0.7 ≈ 15.57 V (before regulator). The regulator clamps to 5 V in both cases, but without load there is less voltage drop across the regulator's internal pass transistor.

Q4. What would happen to the ripple voltage if the smoothing capacitor were replaced by a 100 µF capacitor?

Ans: V_ripple ≈ I_dc/(f C). Reducing C from 1000 µF to 100 µF (×10 smaller) increases ripple by ×10. New V_ripple = 0.5/(100 × 100 × 10⁻⁶) = 50 V — which is unrealistic and means the capacitor cannot smooth the output at all. The regulator would lose regulation.

Case Study 2

Zener Diode Voltage Regulator

A Zener diode voltage regulator circuit uses a Zener diode with V_Z = 5.1 V and maximum power dissipation P_max = 400 mW. The unregulated input voltage varies between 8 V and 14 V. A series resistance R_S = 150 Ω is used. A variable load R_L can take values from 500 Ω to infinity (open circuit). The Zener must maintain 5.1 V across the load for all input and load variations.

Q1. Find the maximum current through the Zener diode.

Ans: I_Z(max) = P_max/V_Z = 400 × 10⁻³/5.1 = 78.4 mA

Q2. Find the series current when input = 14 V and R_L is open (worst case for Zener).

Ans: I_S = (V_in − V_Z)/R_S = (14 − 5.1)/150 = 8.9/150 = 59.3 mA. All of this flows through Zener (no load). P_Z = 5.1 × 0.0593 = 302 mW < 400 mW → Safe.

Q3. Find the minimum series current when input = 8 V and R_L = 500 Ω.

Ans: I_L = 5.1/500 = 10.2 mA. I_S = (8 − 5.1)/150 = 19.3 mA. I_Z = 19.3 − 10.2 = 9.1 mA. Zener still conducts → regulation maintained.

Q4. Explain why the Zener diode maintains a constant voltage despite changing input and load.

Ans: In the Zener breakdown region, the diode's V-I curve is nearly vertical — large changes in current produce negligible change in voltage. When V_in or I_L changes, the Zener current adjusts automatically to absorb the excess, keeping V_out = V_Z = 5.1 V.

Case Study 3

Solar Cell — Illuminated PN Junction

A silicon PN junction solar cell of area 100 cm² is illuminated by sunlight (1000 W/m² irradiance). The cell parameters are: short-circuit current density J_sc = 30 mA/cm², open-circuit voltage V_oc = 0.6 V, fill factor FF = 0.75, and the cell temperature is 25°C. The cell is connected to a load of 20 Ω.

Q1. Calculate the short-circuit current I_sc and maximum power P_max.

Ans: I_sc = J_sc × A = 30 × 100 = 3000 mA = 3 A. P_max = FF × V_oc × I_sc = 0.75 × 0.6 × 3 = 1.35 W

Q2. Calculate the efficiency η of the solar cell.

Ans: P_incident = 1000 W/m² × 100 × 10⁻⁴ m² = 10 W. η = P_max/P_incident = 1.35/10 = 13.5%

Q3. How does the operating voltage of the cell when connected to the 20 Ω load compare to V_oc?

Ans: At the operating point, V = I × R_L. Using Shockley: I ≈ I_sc − I₀e^V/V_T. For ideal cell with R_L = 20 Ω: V ≈ 0.55 V < V_oc = 0.6 V. The operating voltage is between 0 (short circuit) and V_oc (open circuit).

Q4. If the temperature rises to 55°C, what happens to V_oc and I_sc?

Ans: V_oc decreases (~2 mV/°C for Si), so drops ≈ 60 mV → 0.54 V. I_sc increases very slightly (~0.05%/°C). Net effect: efficiency decreases with temperature. Solar panels operate less efficiently on hot days.

Case Study 4

Varactor Diode in FM Radio Tuner

An FM radio tuner uses a varactor diode (varicap) as a voltage-controlled capacitor. The varactor has a zero-bias capacitance C₀ = 30 pF and a doping exponent m = 0.5 (abrupt junction). The resonant LC circuit has L = 100 nH. A tuning voltage V_tune can be varied from 1 V to 10 V reverse bias. Contact potential V₀ = 0.7 V.

Q1. Write the expression for varactor capacitance as a function of V_R.

Ans: C_T(V_R) = C₀ / (1 + V_R/V₀)^m = 30 / (1 + V_R/0.7)^0.5 pF

Q2. Calculate C_T at V_R = 1 V and V_R = 9 V.

Ans: At 1 V: C = 30/√(1 + 1/0.7) = 30/√2.43 = 30/1.56 = 19.2 pF. At 9 V: C = 30/√(1 + 9/0.7) = 30/√13.86 = 30/3.72 = 8.06 pF

Q3. Find the resonant frequencies of the LC circuit at these two capacitance values. (L = 100 nH)

Ans: f = 1/(2π√(LC)). At C = 19.2 pF: f = 1/(2π√(10⁻⁷ × 19.2 × 10⁻¹²)) = 1/(2π × 1.386 × 10⁻⁹) = 114.8 MHz. At C = 8.06 pF: f = 1/(2π√(10⁻⁷ × 8.06 × 10⁻¹²)) = 177 MHz.

Q4. Why is a varactor preferred over a mechanically variable capacitor in modern radio tuners?

Ans: Varactors are solid-state (no moving parts), can be voltage-controlled by a digital signal (electronic tuning), are smaller, more reliable, faster in response, and easier to integrate into ICs. Mechanical capacitors are bulky and wear over time.

Case Study 5

Breakdown and Diode Protection

A circuit designer is testing two PN junction diodes: Diode A (heavily doped Si, V_BR = 4.7 V) and Diode B (lightly doped Si, V_BR = 75 V). When the temperature is increased from 25°C to 85°C, the designer observes that the breakdown voltage of Diode A decreases by about 0.6 V, while that of Diode B increases by about 6 V. A third Zener diode with V_Z = 6.2 V is used for circuit protection across a sensitive IC. A transient voltage spike of 20 V lasting 10 µs appears on the supply line.

Q1. Identify the breakdown mechanism in Diode A and Diode B and explain the opposite temperature dependence.

Ans: Diode A (V_BR = 4.7 V < 6 V): Zener breakdown (tunnelling). V_Z has negative TC — higher T increases tunnelling probability → lower V_Z. Diode B (V_BR = 75 V > 6 V): Avalanche breakdown. V_AV has positive TC — higher T reduces carrier mean free path → less energy per collision → higher voltage needed to start avalanche.

Q2. What is the purpose of the 6.2 V Zener diode across the IC supply?

Ans: It acts as a voltage clamp. When supply is within normal range (<6.2 V), Zener is reverse biased and non-conducting. When a spike exceeds 6.2 V, the Zener conducts and clamps the voltage at 6.2 V, protecting the IC from over-voltage damage.

Q3. How much energy is dissipated in the Zener during the 20 V spike? (Zener current during spike ≈ 50 mA)

Ans: During spike, V_Z = 6.2 V (clamped), I_Z ≈ 50 mA, duration t = 10 µs. Energy E = V_Z × I_Z × t = 6.2 × 0.05 × 10 × 10⁻⁶ = 3.1 × 10⁻⁶ J = 3.1 µJ.

Q4. The designer wants a protection diode at exactly 5.0 V that is temperature-stable. How can this be achieved using a combination of Zener and junction diodes?

Ans: Combine a Zener diode (negative TC, e.g., 4.3 V Zener) in series with forward-biased PN junction diodes (positive TC, ~2 mV/°C each). A 4.3 V Zener + 2 forward diodes (2 × 0.7 V = 1.4 V) ≈ 5.7 V total; or use a 4.7 V Zener + 1 diode = 5.4 V. Fine-tune the combination until the TCs cancel, giving a temperature-compensated reference.

Assertion-Reason

Assertion-Reason Questions

Select the correct option: (A) Both A and R are true; R is correct explanation of A. (B) Both true; R is NOT correct explanation. (C) A is true; R is false. (D) A is false; R is true. (E) Both false.

Assertion (A): In a PN junction at thermal equilibrium, no net current flows.

Reason (R): The diffusion current and drift current are equal and opposite, so they exactly cancel.

Answer: (A) — Both are true and R correctly explains A. At equilibrium, J_diffusion = J_drift for both electrons and holes.

Assertion (A): Forward biasing a PN junction increases the depletion width.

Reason (R): The applied electric field in forward bias adds to the built-in field.

Answer: (D) — A is false (forward bias REDUCES depletion width); R is also false (forward bias opposes the built-in field).

Assertion (A): Silicon diodes are preferred over Germanium diodes for high-temperature applications.

Reason (R): Silicon has a larger bandgap (1.12 eV) than Germanium (0.67 eV), resulting in lower leakage current at elevated temperatures.

Answer: (A) — Both true; R is the correct explanation. Larger bandgap → fewer thermally generated minority carriers → lower I₀ at high T.

Assertion (A): The transition capacitance C_T of a PN junction decreases when reverse voltage increases.

Reason (R): Increasing reverse voltage widens the depletion region, which reduces the effective capacitor plate separation.

Answer: (C) — A is true; R is false (widening the separation INCREASES plate separation, which REDUCES capacitance C = εA/W, so C decreases — but the stated reason incorrectly says "reduces plate separation").

Assertion (A): The reverse saturation current I₀ is nearly independent of reverse voltage.

Reason (R): I₀ depends only on the rate of minority carrier generation near the junction, which is fixed by temperature.

Answer: (A) — Both true and R correctly explains A. Minority carrier generation is a thermal process independent of applied field.

Assertion (A): Zener breakdown has a negative temperature coefficient.

Reason (R): As temperature increases, the tunnelling probability through the thin depletion layer increases, requiring a lower voltage for breakdown.

Answer: (A) — Both true; R correctly explains the negative TC of Zener breakdown.

Assertion (A): Avalanche breakdown occurs in lightly doped junctions at high reverse voltages (>6 V for Si).

Reason (R): Lightly doped junctions have wide depletion regions; carriers gain enough kinetic energy over the long depletion width to cause impact ionisation.

Answer: (A) — Both true and R is the correct explanation.

Assertion (A): A full-wave rectifier produces a smoother DC output than a half-wave rectifier.

Reason (R): In a full-wave rectifier, both half-cycles are utilised, giving output pulses at twice the input frequency, which are easier to filter.

Answer: (A) — Both true and R correctly explains A. Higher ripple frequency → ripple voltage ∝ 1/f → lower ripple.

Assertion (A): The dynamic resistance of a diode decreases as forward current increases.

Reason (R): Dynamic resistance r_f = V_T/I_F, which is inversely proportional to I_F.

Answer: (A) — Both true and R is the correct explanation. At 26 mV thermal voltage, doubling I_F halves r_f.

Q10

Assertion (A): The Fermi level is constant throughout a PN junction in thermal equilibrium.

Reason (R): At thermal equilibrium, there is no current flow, which requires zero gradient of the quasi-Fermi levels.

Answer: (A) — Both true; at equilibrium, quasi-Fermi levels merge into a single flat Fermi level with zero spatial gradient.

Q11

Assertion (A): The contact potential V₀ of a PN junction cannot drive an external current on its own.

Reason (R): V₀ is a built-in potential that arises to balance diffusion current with drift current internally; connecting external wires introduces equal and opposite contact potentials at the metal-semiconductor interfaces.

Answer: (A) — Both true and R is the correct (though advanced) explanation. V₀ is a thermodynamic equilibrium quantity, not an EMF source.

Q12

Assertion (A): An LED emits light only when forward biased.

Reason (R): In forward bias, electrons and holes are injected across the junction and recombine, releasing energy as photons.

Answer: (A) — Both true and R correctly explains the electroluminescence mechanism in LEDs.

Q13

Assertion (A): Diffusion capacitance C_D is much larger than transition capacitance C_T in forward bias.

Reason (R): In forward bias, significant minority carrier charge is stored on both sides of the junction, creating a large equivalent capacitance.

Answer: (A) — Both true and R explains why C_D ≫ C_T under forward bias.

Q14

Assertion (A): A photodiode is operated in reverse bias for light detection.

Reason (R): Reverse bias widens the depletion region, increasing the volume for photon absorption and improving sensitivity and response speed.

Answer: (A) — Both true; R correctly explains why reverse bias is preferred for photodetection.

Q15

Assertion (A): The depletion region in a PN junction contains no mobile charge carriers.

Reason (R): Mobile carriers (electrons and holes) have recombined in this region, leaving behind only fixed ionised dopant atoms.

Answer: (A) — Both true and R is the correct explanation of why the depletion region is depleted of mobile carriers.

Q16

Assertion (A): In a bridge rectifier, the peak inverse voltage (PIV) across each diode is V_m.

Reason (R): In a centre-tap full-wave rectifier, the PIV across each diode is 2V_m.

Answer: (B) — Both statements are true, but R describes the centre-tap configuration, not explaining the bridge PIV. The PIV for bridge is V_m and for centre-tap is 2V_m; these are separate facts.

Q17

Assertion (A): The reverse saturation current of germanium is much higher than that of silicon at the same temperature.

Reason (R): Germanium has a smaller bandgap, leading to more thermally generated minority carriers.

Answer: (A) — Both true; smaller bandgap (0.67 eV vs 1.12 eV) means far more minority carriers generated thermally in Ge, giving µA vs nA leakage.

Q18

Assertion (A): The V-I characteristic of an ideal diode is asymmetric with respect to voltage polarity.

Reason (R): The diode has different resistances for forward and reverse bias — near zero in forward and infinite in reverse.

Answer: (A) — Both true and R explains the asymmetric (rectifying) behaviour of the ideal diode.

Q19

Assertion (A): The capacitor in a rectifier circuit reduces ripple by storing charge.

Reason (R): A larger capacitance results in a smaller ripple voltage for the same load current and ripple frequency.

Answer: (A) — Both true; V_ripple = I_dc/(f × C), so larger C → smaller V_ripple. R correctly explains A.

Q20

Assertion (A): In a Zener diode voltage regulator, the Zener current increases when the load resistance increases.

Reason (R): The series resistor current I_S = (V_in − V_Z)/R_S remains constant; if load current decreases (higher R_L), the difference I_S − I_L flows through the Zener, increasing I_Z.

Answer: (A) — Both true; R correctly explains the current redistribution mechanism in the Zener regulator.

Quick Revision Box

All essential formulas, definitions and constants for PN Junction Diode — exam ready

Diode Current Equation

I = I₀(e^V/ηV_T − 1)

η = 1 (Ge/high I); η = 2 (Si/low I). V_T = kT/q ≈ 26 mV at 300 K

Thermal Voltage

V_T = kT/q

= 26 mV at T = 300 K; k = 1.38×10⁻²³ J/K; q = 1.6×10⁻¹⁹ C

Built-in Potential

V₀ = V_T ln(N_AN_D/n_i²)

Si: ~0.7 V; Ge: ~0.3 V at 300 K

Dynamic Resistance

r_f = ηV_T/I_F

At I_F = 26 mA and η = 1: r_f = 1 Ω

Transition Capacitance

C_T = εA/W

W increases with V_R; C_T ∝ 1/√V_R for abrupt junction

Diffusion Capacitance

C_D = τI_F/V_T

τ = minority carrier lifetime; dominant in forward bias

Half-wave DC Output

V_dc = V_m/π

Ripple frequency = f (input). Ripple factor ≈ 1.21

Full-wave DC Output

V_dc = 2V_m/π

Ripple frequency = 2f. Ripple factor ≈ 0.48

Rectification Efficiency

η = P_dc/P_ac × 100%

Half-wave ≈ 40.6%; Full-wave ≈ 81.2%

Ripple Voltage

V_r ≈ I_dc/(f × C)

f = ripple frequency; C = filter capacitance

Zener Power

P_Z = V_Z × I_Z

Must not exceed P_max rating of Zener diode

I₀ Temperature Rule

I₀ doubles / 10°C rise

Valid for both Si and Ge diodes

Key Values at a Glance

Si cut-in voltage: 0.7 V

Ge cut-in voltage: 0.3 V

Si bandgap: 1.12 eV

Ge bandgap: 0.67 eV

V_T at 300 K: 26 mV

Zener TC: Negative

Avalanche TC: Positive

Zener V_BR: < 6 V

Avalanche V_BR: > 6 V

Half-wave ripple f: = f_in

Full-wave ripple f: = 2f_in

Si max temp: ~150°C

⚠ Students Often Under-prepare This Chapter

Many students focus extensively on Optics and Electromagnetism while treating PN Junction Diode as a "small topic." This is a serious mistake.

NEET, JEE Main, JEE Advanced, CBSE, IB, IGCSE and A-Level examinations consistently include questions on the V-I characteristic, depletion region, barrier potential, rectifier circuits and breakdown mechanisms. Numerical questions on dynamic resistance, ripple voltage, Zener regulation and junction capacitance appear regularly at JEE Advanced and IB levels.

This chapter forms the foundation of all semiconductor electronics — transistors, op-amps, logic circuits and communication systems all build upon the PN junction. A weak foundation here creates cascading difficulties throughout the course.

Still Confused About PN Junction Diode?

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PN JunctionDiode

PN Junction Diode

PN Junction Formation

Depletion Region

Barrier Potential

Silicon

Germanium

Temperature Effect

Forward Bias

Threshold Voltage

Barrier Reduction

Current Direction

Reverse Bias

V-I Characteristics

Junction Capacitance

Reverse Breakdown

Zener Breakdown

Avalanche Breakdown

Important Comparison Tables

Forward Bias vs Reverse Bias

Silicon vs Germanium Diode

Important Numericals

Previous Year Questions

NEET PYQs

JEE Main PYQs

JEE Advanced PYQs

CBSE PYQs

IB Physics Questions

IGCSE Questions

A-Level Questions

Case Study Questions

Assertion-Reason Questions

Quick Revision Box

⚠ Students Often Under-prepare This Chapter

Still Confused About PN Junction Diode?

PN Junction
Diode