
Explain:
(A) High reverse voltage does not appear across the LED.
(B) For the junction of the zener diode, the electric field is high even for a small reverse bias voltage of about 5V.
Answer
133.5k+ views
Hint (A) A light-emitting diode is a two-lead semiconductor light source. It is a p–n junction diode that emits light when activated. When a suitable voltage is applied to the leads, electrons are able to react
(B) A Zener diode is a silicon semiconductor device that permits current to flow in either a forward or reverse direction. ... Additionally, the voltage drop across the diode remains constant over a wide range of voltages, a feature that makes Zener diodes suitable for use in voltage regulation combined with electron holes within the device, releasing energy in the form of photons.
Complete Step by Step Solution
(A) If you apply +1 volts to the anode and 0 volts to the cathode, then current will flow. However, reversing the voltages to apply 0 volts to the anode and +1V to the cathode, prevents current from flowing! A “Light Emitting Diode” (LED) is a variant of the standard diode with the same characteristics.
(B) LED is a heavily doped p-n junction, when a reverse voltage is applied across the LED, the minority carriers increase near the junction and current starts flowing. Due to it, the high reverse voltage does not appear across the junction.
(C) The heavy doping of p and n sides of p-n junction, makes the depletion region very thin, hence for a small reverse bias voltage, electric field is very high
Since depletion width of the junction of zener diode is very small ($ < {10^{ - 7}}m$), even a small voltage say 5V can set up a very high electric field as
$Electricfield\left( E \right) = \dfrac{{Voltage\left( V \right)}}{{width(d)}}$
Here, and$d = {10^{ - 7}}$$V = 5$
$E = \dfrac{5}{{{{10}^{ - 7}}}} = 5 \times {10^7}$
$E = 5 \times {10^7}V/m$ (which is really high)
Note
(A) The simplest solution is to put an ordinary diode in parallel with the LED, pointing the other way. This allows the reverse current to bypass the LED, limiting the reverse voltage across it to about 0.65 V, which is within its rating.
(B) The major difference between PN junction and the Zener diode is that the PN junction diode allows current to pass only in the forward direction, whereas the Zener diode allows the current to flow both in the forward and the reversed direction.
(B) A Zener diode is a silicon semiconductor device that permits current to flow in either a forward or reverse direction. ... Additionally, the voltage drop across the diode remains constant over a wide range of voltages, a feature that makes Zener diodes suitable for use in voltage regulation combined with electron holes within the device, releasing energy in the form of photons.
Complete Step by Step Solution
(A) If you apply +1 volts to the anode and 0 volts to the cathode, then current will flow. However, reversing the voltages to apply 0 volts to the anode and +1V to the cathode, prevents current from flowing! A “Light Emitting Diode” (LED) is a variant of the standard diode with the same characteristics.
(B) LED is a heavily doped p-n junction, when a reverse voltage is applied across the LED, the minority carriers increase near the junction and current starts flowing. Due to it, the high reverse voltage does not appear across the junction.
(C) The heavy doping of p and n sides of p-n junction, makes the depletion region very thin, hence for a small reverse bias voltage, electric field is very high
Since depletion width of the junction of zener diode is very small ($ < {10^{ - 7}}m$), even a small voltage say 5V can set up a very high electric field as
$Electricfield\left( E \right) = \dfrac{{Voltage\left( V \right)}}{{width(d)}}$
Here, and$d = {10^{ - 7}}$$V = 5$
$E = \dfrac{5}{{{{10}^{ - 7}}}} = 5 \times {10^7}$
$E = 5 \times {10^7}V/m$ (which is really high)
Note
(A) The simplest solution is to put an ordinary diode in parallel with the LED, pointing the other way. This allows the reverse current to bypass the LED, limiting the reverse voltage across it to about 0.65 V, which is within its rating.
(B) The major difference between PN junction and the Zener diode is that the PN junction diode allows current to pass only in the forward direction, whereas the Zener diode allows the current to flow both in the forward and the reversed direction.
Recently Updated Pages
JEE Main 2025 Session 2 Form Correction (Closed) – What Can Be Edited

Sign up for JEE Main 2025 Live Classes - Vedantu

JEE Main Books 2023-24: Best JEE Main Books for Physics, Chemistry and Maths

JEE Main 2023 April 13 Shift 1 Question Paper with Answer Key

JEE Main 2023 April 11 Shift 2 Question Paper with Answer Key

JEE Main 2023 April 10 Shift 2 Question Paper with Answer Key

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

JEE Main 2025: Conversion of Galvanometer Into Ammeter And Voltmeter in Physics

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

Class 11 JEE Main Physics Mock Test 2025

Current Loop as Magnetic Dipole and Its Derivation for JEE

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

Units and Measurements Class 11 Notes: CBSE Physics Chapter 1

Important Questions for CBSE Class 11 Physics Chapter 1 - Units and Measurement

NCERT Solutions for Class 11 Physics Chapter 2 Motion In A Straight Line

Motion In A Plane: Line Class 11 Notes: CBSE Physics Chapter 3

Waves Class 11 Notes: CBSE Physics Chapter 14
