| ⇦ | 
| ⇨ |
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: - A particle is executing a simple harmonic motion. Its maximum acceleration
- Two slits in Young’s experiment have widths in the ratio 1:25. The ratio of intensity
- Pure Si at 500 K has equal number of electron (nₑ) and hole (nₕ) concentrations
- A beam of light of wavelength 590 nm is focussed by a converging lens of diameter
- A missile is fired for maximum range with an initial velocity of 20 m/s
Topics: Electronic Devices
(124)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- A particle is executing a simple harmonic motion. Its maximum acceleration
- Two slits in Young’s experiment have widths in the ratio 1:25. The ratio of intensity
- Pure Si at 500 K has equal number of electron (nₑ) and hole (nₕ) concentrations
- A beam of light of wavelength 590 nm is focussed by a converging lens of diameter
- A missile is fired for maximum range with an initial velocity of 20 m/s
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