
Barrier potential of P – n junction diode does not depend on –
$\left( A \right)$ Diode design
$\left( B \right)$ Temperature
$\left( C \right)$ Forward bias
$\left( D \right)$ Doping density
Answer
134.1k+ views
- Hint: The concept that barrier potential used to render P – n junction diode whether it is silicon (Si) or germanium (Ge) was used in this issue, and the concept that forward bias makes current through a PN junction was also used. Bias voltage provides the free electrons with sufficient energy to overcome the barrier potential and move into the P region, so use this concept to solve the problem.
Complete step-by-step solution -
The electric field formed in the region of depletion serves as a barrier.
To get the electrons to pass through the boundary of the electric field, external energy must be applied.
The potential difference needed for moving the electrons through the electric field is called the potential barrier.
A PN junction's barrier potential depends on the form of semiconductor material, the amount of doping, temperature and forward bias, and the reverse bias.
That is about 0.7V for silicon, and about 0.3V for germanium.
And forward bias allows for current through a junction with a PN. Bias tension provides the free electrons with sufficient energy to overcome the barrier potential and move into the P region.
Hence the potential barrier depends on the quantity of doping, temperature and forward bias, as well as reverse bias.
But this is not based on the nature of the diode.
So this is the required answer.
Hence option (A) is the correct answer.
Note – The potential barrier in the PN-junction diode is the barrier where additional force is needed for the load to cross the area. In other words, the area through which the obstructive force has stopped the charge carrier is known as the potential barrier.
Complete step-by-step solution -
The electric field formed in the region of depletion serves as a barrier.
To get the electrons to pass through the boundary of the electric field, external energy must be applied.
The potential difference needed for moving the electrons through the electric field is called the potential barrier.
A PN junction's barrier potential depends on the form of semiconductor material, the amount of doping, temperature and forward bias, and the reverse bias.
That is about 0.7V for silicon, and about 0.3V for germanium.
And forward bias allows for current through a junction with a PN. Bias tension provides the free electrons with sufficient energy to overcome the barrier potential and move into the P region.
Hence the potential barrier depends on the quantity of doping, temperature and forward bias, as well as reverse bias.
But this is not based on the nature of the diode.
So this is the required answer.
Hence option (A) is the correct answer.
Note – The potential barrier in the PN-junction diode is the barrier where additional force is needed for the load to cross the area. In other words, the area through which the obstructive force has stopped the charge carrier is known as the potential barrier.
Recently Updated Pages
JEE Main 2025 Session 2 Form Correction (Closed) – What Can Be Edited

What are examples of Chemical Properties class 10 chemistry JEE_Main

JEE Main 2025 Session 2 Schedule Released – Check Important Details Here!

JEE Main 2025 Session 2 Admit Card – Release Date & Direct Download Link

JEE Main 2025 Session 2 Registration (Closed) - Link, Last Date & Fees

JEE Mains Result 2025 NTA NIC – Check Your Score Now!

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

Wheatstone Bridge for JEE Main Physics 2025

Electric field due to uniformly charged sphere class 12 physics JEE_Main

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

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

Diffraction of Light - Young’s Single Slit Experiment

Dual Nature of Radiation and Matter Class 12 Notes: CBSE Physics Chapter 11

Formula for number of images formed by two plane mirrors class 12 physics JEE_Main

JEE Advanced 2024 Syllabus Weightage

JEE Main Chemistry Question Paper with Answer Keys and Solutions
