| ⇦ | 
| ⇨ |
A Zenor diode has a contact potential of 1 V in the absence of biasing. It undergoes Zenor breakdown for an electric field of 10⁶ V/m at the depletion region of p-n junction. If the width of the depletion region is 2.5 μm, what should be the reverse biased potential for the Zenor breakdown to occur?
Options
(a) 3.5 V
(b) 2.5 V
(c) 1.5 V
(d) 0.5 V
Correct Answer:
2.5 V
Explanation:
No explanation available. Be the first to write the explanation for this question by commenting below.
Related Questions: - Angle of minimum deviation for a prism of reractive index 1.5 equal to the angle
- The transverse nature of electromagnetic waves is proved by which of the following?
- The heat dissipated in a resistance can be obtained by measuring of resistance,
- The dimensions of pressure are equal to those of
- In radioactive element, β-rays are emitted from
Topics: Electronic Devices
(124)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- Angle of minimum deviation for a prism of reractive index 1.5 equal to the angle
- The transverse nature of electromagnetic waves is proved by which of the following?
- The heat dissipated in a resistance can be obtained by measuring of resistance,
- The dimensions of pressure are equal to those of
- In radioactive element, β-rays are emitted from
Topics: Electronic Devices (124)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
18000+ students are using NEETLab to improve their score. What about you?
Solve Previous Year MCQs, Mock Tests, Topicwise Practice Tests, Identify Weak Topics, Formula Flash cards and much more is available in NEETLab Android App to improve your NEET score.
Share this page with your friends
V=E*d
E=10^6 v/m
d=2.5 μm
Therefor V=2.5