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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.
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Topics: Electronic Devices
(124)
Subject: Physics
(2479)
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- The logic behind NOR gate is that it gives
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- The magnetic field at the centre of a circular coil carrying current I ampere is B.
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V=E*d
E=10^6 v/m
d=2.5 μm
Therefor V=2.5