Question

A potential barrier of 0.5V we exist across a P-N junction, If the depletion region is $5.0 \times {10^{ - 7}}\,m$ wide, the intensity of the electric field in the region is
A. $1.0 \times {10^6}\,V/m$
B. $1.0 \times {10^5}\,V/m$
C. $2.0 \times {10^5}\,V/m$
D. $2.0 \times {10^6}\,V/m$

Answer 1

Hint: The electric field is the force experienced per unit charge. It can be calculated by dividing the potential with the length. Since the potential of the barrier and width of the depletion region is given, we can directly substitute the values to get the value of the electric field at the junction.

Complete step by step answer:
It is given that the potential barrier in a P-N junction is $0.5\,V$
$ \Rightarrow V = 0.5\,\,V$
The width of the depletion region is given as $5.0 \times {10^{ - 7}}\,m$
$ \Rightarrow l = 5.0 \times {10^{ - 7}}\,m$
We need to find the intensity of the electric field in this region.
We know that a PN junction diode is formed by combining a p-type semiconductor with an n-type semiconductor. At the junction the electrons from the n-type flow into the p-type and holes from the p type flow into the n type. They cancel each other and a layer having no mobile chargers is made. This region is called the depletion layer. The potential barrier is the potential drop that is created due to the diffusion of these ions to form the depletion layer.
The electric field is the force experienced by a unit test charge.
Electric fields can be calculated by finding the voltage per unit length.
The relationship between electric field and the voltage can be written as
$E = \dfrac{V}{l}$
Where, V is the potential and l is the length.
Let us substitute the given potential of the potential barrier and the width of the depletion region in this equation. Then we get
$ \Rightarrow E = \dfrac{V}{l}$
$ \Rightarrow E = \dfrac{{0.5\,V}}{{5.0 \times {{10}^{ - 7}}\,m}}$
$ \Rightarrow E = 1.6 \times {10^6}\,V/m$

Therefore, the value of the electric field at the junction is $E = 1.6 \times {10^6}\,V/m$. The correct answer is option A.

Note:
Electric field is a vector quantity. That is, it has both magnitude and direction. The direction of the electric field will always be from positive charge to negative charge. It is the electric field in the depletion region that prevents further diffusion of charge across the junction.