Question:
A Zener diode has a contact potential of $1\, V$ in the absence of biasing. It undergoes Zener breakdown for an electric field of $10^{6} \,V / m$ at the depletion region of $p-n$ junction. If the width of the depletion region is $2.5 \,\mu\, m$, what should be the reverse biased potential for the Zener breakdown to occur?
A Zener diode has a contact potential of $1\, V$ in the absence of biasing. It undergoes Zener breakdown for an electric field of $10^{6} \,V / m$ at the depletion region of $p-n$ junction. If the width of the depletion region is $2.5 \,\mu\, m$, what should be the reverse biased potential for the Zener breakdown to occur?
Updated On: Apr 15, 2024
- 3.5 V
- 2.5 V
- 1.5 V
- 0.5 V
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The Correct Option is B
Solution and Explanation
Reverse biased potential for the zener breakdown
$V_{r} =E d $
$=10^{6} \times 2.5 \times 10^{-6}$
$=2.5$ volt
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Concepts Used:
Electric Field
Electric Field is the electric force experienced by a unit charge.
The electric force is calculated using the coulomb's law, whose formula is:
\(F=k\dfrac{|q_{1}q_{2}|}{r^{2}}\)
While substituting q2 as 1, electric field becomes:
\(E=k\dfrac{|q_{1}|}{r^{2}}\)
SI unit of Electric Field is V/m (Volt per meter).