JEE Advanced 2026 Paper 2 — Full Mock Test (54 Questions, 180 Marks)
Mathematics · Physics · Chemistry
This MCQ module is based on: Special Purpose Diodes
TOPIC 26 OF 27
Special Purpose Diodes
🎓 Class 12
Physics
CBSE
Theory
Ch 14 – Semiconductor Electronics
⏱ ~14 min
🌐 Language:
🧠 AI-Powered MCQ Assessment
▲
📝 Worksheet / Assessment
▲
This assessment will be based on: Special Purpose Diodes
Upload images, PDFs, or Word documents to include their content in assessment generation.
Special Purpose Diodes
14.8 Special-Purpose p-n Junction Diodes
Beyond rectification, the p-n junction can be engineered to perform a variety of specialised functions. By tuning the doping levels, the band gap of the semiconductor and the device geometry, we get four important "special diodes":
| Device | Operating bias | Function |
|---|---|---|
| Zener diode | Reverse (at breakdown) | Voltage regulation / reference |
| Photodiode | Reverse | Light → current (sensor) |
| LED (Light-Emitting Diode) | Forward | Current → light (emitter) |
| Solar cell | No bias (illumination only) | Light → DC voltage (energy harvester) |
14.8.1 Zener Diode — Voltage Regulator
A Zener diode is a heavily doped p-n junction designed to operate in the breakdown region under reverse bias. Two distinct breakdown mechanisms can occur:
- Zener breakdown (heavy doping, V_br < 6 V): the very thin depletion region produces a huge electric field that rips electrons out of covalent bonds. Tunnelling current rises sharply.
- Avalanche breakdown (lighter doping, V_br > 6 V): high-energy minority carriers ionise lattice atoms by impact, multiplying the carrier population.
Both produce the same effect: once V exceeds V_z, the diode current can change by orders of magnitude while V across the diode remains nearly constant at V_z. This is the property exploited for voltage regulation.
How regulation works
If V_in increases, the extra voltage drops across R_s while V across Zener stays at V_z. If V_in decreases below V_z, the Zener stops conducting and acts like an open circuit — so V_in must always exceed V_z. The Zener current adjusts itself to keep V_out fixed.
Worked Example — Designing a regulator
A 6.2 V Zener regulates a 9 V input. Load draws 20 mA at 6.2 V. Choose R_s so the Zener carries at least 5 mA.
Total current through R_s = I_load + I_Zener = 20 + 5 = 25 mA.
Voltage drop across R_s = V_in − V_z = 9.0 − 6.2 = 2.8 V.
\[ R_s = \frac{V_{in} - V_z}{I_{R_s}} = \frac{2.8}{0.025} = \mathbf{112\ \Omega} \]Choose the nearest standard value: 100 Ω or 120 Ω.
14.8.2 Optoelectronic Devices — Photodiode and LED
Both rely on the relationship: energy of a photon = E_g of the semiconductor. If hν ≥ E_g, a photon striking the depletion region creates an electron-hole pair (light → current). Conversely, an electron-hole pair recombining releases a photon of energy ≈ E_g (current → light).
Photodiode
A photodiode is a junction diode operated under reverse bias and exposed to light. The transparent window above the junction lets photons reach the depletion region. Each absorbed photon creates an e-h pair which the field separates, producing additional reverse current proportional to the light intensity.
Why operate under reverse bias and not no-bias? Because reverse bias gives a wider depletion region, and the change in fractional current is much larger and more easily measured. Photodiodes are used in TV remote receivers, light meters, optical fibre detectors and barcode scanners.
Light-Emitting Diode (LED)
An LED is a heavily doped p-n junction operated under forward bias. Injected minority carriers recombine with majority carriers near the junction; the released energy emerges as a photon of wavelength λ ≈ hc/E_g.
The colour of an LED is set by the band gap of the semiconductor:
| Material | E_g (eV) | Wavelength (nm) | Colour |
|---|---|---|---|
| GaAs | 1.43 | ~870 | IR (invisible) |
| GaAs0.6P0.4 | ~1.9 | ~660 | Red |
| GaP | 2.26 | ~550 | Green |
| GaN / InGaN | ~3.4 | ~470 | Blue |
| White LED | — | — | Blue LED + yellow phosphor |
LEDs have several advantages over incandescent bulbs: low operating voltage, low power, long life (~10⁵ h), fast switching, no warm-up time, ruggedness. They have replaced filament bulbs in nearly every general-lighting application.
14.8.3 Solar Cell
A solar cell is a large-area p-n junction that converts sunlight directly into a DC voltage without any external bias. Sunlight (with photon energies above E_g) creates electron-hole pairs in the depletion region; the built-in field separates them — electrons drift to n-side, holes to p-side. This separation makes the n-side the (−) terminal and the p-side the (+) terminal, generating an EMF.
Design considerations:
- The semiconductor must have a band gap matched to the solar spectrum (~1.0-1.8 eV is optimal). Si (1.12 eV) and GaAs (1.43 eV) are common.
- The top junction must be very near the illuminated surface so photons reach the depletion region.
- Anti-reflection coatings reduce surface losses; metal contact grids let light in while collecting current.
Why is the n-region kept thin? Because most photons are absorbed within a few μm of the surface. The depletion region must be close to where carriers are generated; otherwise the e-h pairs recombine before being separated.
Comparison Table — All Four Special Diodes
| Device | Bias | Input | Output | Key role of E_g |
|---|---|---|---|---|
| Zener | Reverse, >V_z | Voltage | Constant V_z | Sets V_z (high-doping ⇒ low V_z) |
| Photodiode | Reverse | Light (hν ≥ E_g) | Reverse current ∝ intensity | Cutoff wavelength λ_c = hc/E_g |
| LED | Forward | Current | Light (λ ≈ hc/E_g) | Sets emission colour |
| Solar cell | Unbiased (illuminated) | Sunlight | EMF + current | Optimal E_g ≈ 1-1.8 eV |
Worked Example — LED wavelength
A red LED uses a semiconductor with E_g = 1.9 eV. Find the wavelength of emitted photons.
\[ \lambda = \frac{hc}{E_g} = \frac{1240\ \text{eV·nm}}{1.9\ \text{eV}} \approx \mathbf{653\ nm} \]
Indeed the red part of the visible spectrum.
Activity 14.4 — TV-remote photon detector
Point a smartphone camera at the front of any TV remote and press a button. Most phone cameras detect near-IR light and you'll see the LED at the front of the remote glow brightly on screen — even though it looks dark to your eye.
What does this demonstrate? What is the band gap of the LED material if its emission peaks at 950 nm?
It demonstrates that LEDs can emit photons outside the visible range (here, near-IR), and that camera CMOS sensors are sensitive into the IR. Required band gap: E_g = hc/λ = 1240/950 ≈ 1.31 eV. (Close to GaAs.)
Interactive — Zener Voltage Regulator
Watch a regulator clamp the output
Adjust V_in and the load resistance. The simulator shows V_out, current through Zener, and current through load.
9.0 V
120 Ω
500 Ω
V_out: 6.2 V
I_load: 12.4 mA
I_Zener: 10.9 mA
State: regulating
Competency-Based Questions
Q1 (MCQ). A Zener diode is used as a voltage regulator. It is operated in:
(d) The Zener regulator works only when reverse-biased into the breakdown region — that's where V is independent of I.
Q2 (MCQ). For a semiconductor with E_g = 1.55 eV, the longest photon wavelength that creates an e-h pair is approximately:
(d) λ_c = 1240/E_g = 1240/1.55 = 800 nm.
Q3 (Short Answer). Why is Si (E_g = 1.12 eV) NOT used to make LEDs that emit visible light?
Visible light requires photon energies of 1.8-3.1 eV (corresponding to 400-700 nm). Si's band gap of 1.12 eV gives photons of ~1100 nm, which lies in the IR — invisible. Also Si is an indirect-band-gap semiconductor, making radiative recombination very inefficient compared to direct-band-gap materials like GaAs and GaN.
Q4 (Numerical). A photodiode with E_g = 1.43 eV (GaAs) — what is the longest wavelength it can detect?
λ_max = hc/E_g = 1240 nm·eV / 1.43 eV ≈ 867 nm (near-IR).
Q5 (HOTS). Why must the n-region of a solar cell be kept very thin compared to the p-region?
Sunlight is incident on the n-side. Most photons are absorbed within a few μm of the surface. If the n-region were thick, e-h pairs would be generated far from the depletion region, and would recombine before the built-in field could separate them. Keeping n thin ensures the depletion region (with its strong field) is close to where most carriers are generated, maximising collection efficiency.
Assertion–Reason Questions
Options: (A) Both true, R correct explanation. (B) Both true, R not the correct explanation. (C) A true, R false. (D) A false, R true.
Assertion: A photodiode is operated in reverse bias.
Reason: Reverse bias widens the depletion region and makes the fractional change in current due to incoming photons more pronounced.
(A) Both correct, and the reason explains the assertion.
Assertion: White LEDs are usually made by combining a blue LED with a yellow phosphor.
Reason: No single semiconductor band gap can produce all visible wavelengths simultaneously.
(A) Both correct, and the reason explains the assertion. The blue LED excites the phosphor, which down-converts to a broad yellow emission; mixing blue + yellow gives white perception.
Assertion: A solar cell does NOT need an external battery to produce a current through a load.
Reason: The built-in junction field separates photo-generated electron-hole pairs and develops an EMF on its own.
(A) Both correct, and the reason explains the assertion. The solar cell is the EMF source.
Frequently Asked Questions - Special Purpose Diodes
What is the main concept covered in Special Purpose Diodes?
In NCERT Class 12 Physics Chapter 14 (Semiconductor Electronics), "Special Purpose Diodes" covers the core principles and equations students need for board exam success. The MyAiSchool lesson explains the topic with definitions, derivations, worked examples, and interactive simulations. Key formulas and dimensional analysis are included to build conceptual depth and problem-solving skills aligned with the CBSE 2025-26 syllabus.
How is Special Purpose Diodes useful in real-life applications?
Real-life applications of "Special Purpose Diodes" from NCERT Class 12 Physics Chapter 14 include electronics, communication systems, medical imaging, solar energy, semiconductor devices, and modern technology. The MyAiSchool lesson links every concept to a tangible example so students see physics as a problem-solving framework for the physical world, not as abstract formulas.
What are the key formulas in Special Purpose Diodes?
Key formulas in "Special Purpose Diodes" (NCERT Class 12 Physics Chapter 14 Semiconductor Electronics) are derived step-by-step in the MyAiSchool lesson. Students should memorize the final formula AND understand its derivation for full board marks. Each formula is listed with its dimensional formula, SI unit, applicability range, and common pitfalls. The Summary section at the end of each part includes a quick-reference formula card.
How does this part connect to other parts of Chapter 14?
NCERT Class 12 Physics Chapter 14 (Semiconductor Electronics) is structured so each part builds on the previous one. "Special Purpose Diodes" connects directly to neighbouring parts via shared definitions, units, and methodology. The MyAiSchool lesson cross-references related concepts with internal links so students can navigate the whole chapter as one connected story rather than disconnected fragments.
What types of CBSE board questions come from Special Purpose Diodes?
CBSE board questions from "Special Purpose Diodes" typically include: (1) 1-mark MCQs on definitions and formulas, (2) 2-mark short-answer derivations or applications, (3) 3-mark numerical problems with units, (4) 5-mark long-answer derivations followed by application. The MyAiSchool lesson tags each Competency-Based Question (CBQ) with Bloom level (L1-L6) so students know how to study for each weight.
How can students use the interactive simulation effectively?
The interactive simulation in the "Special Purpose Diodes" lesson allows students to adjust input parameters (sliders or selectors) and see physical quantities update in real time. To use it effectively: (1) try extreme values to understand limiting cases, (2) compare with the analytical formula, (3) check unit consistency, (4) test special configurations from worked examples. The simulation reinforces conceptual intuition that pure formula manipulation cannot.
AI Tutor
Physics Class 12 Part II – NCERT (2025-26)
Ready
🎯 Practise Physics
Sit a full paper on what you have been studying, marked question by question.
Board exam sample papers
Physics
CBSE
Physics — CBSE Class XII Sample Paper 1 (2025-26)
Section A · Section B · Section C · Section D · Section E
Mock exams
JEE Advanced 2026 Paper 2 — Full Mock Test (54 Questions, 180 Marks)
Mathematics · Physics · Chemistry
JEE Main Mock Test — Full Paper (75 Questions, 300 Marks)
Part A - Physics (Section A) · Part A - Physics (Section B) · Part B - Chemistry (Section A) · Part B - Chemistry (Section B) · Part C - Mathematics (Section A) · Part C - Mathematics (Section B)
🎁 Join our community and get free AI credits!