Question

In a p.n junction, the depletion region is 400 nm wide and an electric field of 5 × 10⁵ V m⁻¹ exists in it. (a) Find the height of the potential barrier. (b) What should be the minimum kinetic energy of a conduction electron which can diffuse from the n-side to the p-side?

Answer 1

Let:
Depletion region width, d = 400 nm = 4 × 10⁻⁷ m
Electric field, E = 5 × 10⁵ Vm⁻¹

(a) Let the potential barrier be V.
The relation between the potential and the electric field is given by V = Ed 

\[\Rightarrow\]V = E × d = 5 × d

\[\Rightarrow\] V = 5 × 10⁵ × 4 × 10⁻⁷ = 0.2 V

(b) To find: Kinetic energy required
Energy of any electron accelerated through a potential of V = eV
Also, the minimum energy of the electron should be equal to the band gap of the material.
∴ Potential barrier × e = 0.2 eV   (e = Charge of the electron)