Question

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.

Answer 1

Solution:

It is given:

Potential barrier, d = 250 meV

At first, the hole’s kinetic energy = 330 meV

It is to be knows that the hole’s kinetic energy declines when we have the forward biased junction. This is due to the loss of energy while junction crossing.  

Additionally, the hole’s kinetic energy raises when the junction is biased in the reverse direction. This is due to the push given by the reverse bias voltage on the n-side towards the junction. This is shown in the p-n junction diode.

(a) The hole’s kinetic energy declines under forward bias.

Therefore, the final kinetic energy = (300 − 250) meV = 50 meV

(b) The hole’s kinetic energy under the reverse bias, increases.

Therefore, the final kinetic energy = (250 + 300) meV

= 550 meV