In a p-n junction, a potential barrier of 250 meV exists across the junction. A hole with a kinetic energy of 300 meV approaches the junction. Find the kinetic energy of the hole when it crosses the junction if the hole approached the junction (a) from the p-side and (b) from the n-side.
Solution
Given:
Potential barrier, d = 250 meV
Initially,
Kinetic energy of a hole = 330 meV
We know that the kinetic energy of a hole decreases when the junction is forward biassed (because of the energy loss in crossing the junction).
Also, the kinetic energy of a hole increases when the junction is reverse biassed (because reverse bias voltage pushes the hole on the n-side towards the junction) in the given the p-n junction diode.
(a) The kinetic energy of a hole decreases under forward bias.
∴ Final kinetic energy = (300 − 250) meV
= 50 meV
(b) The kinetic energy of a hole increases under reverse bias.
∴ Final kinetic energy = (300 + 250) meV
= 550 meV
