The leakage current in a diode is due to

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This question was previously asked in

ISRO (VSSC) Technician Electrician Feb 2019 Official Paper

Answer 1

When a diode is reversed bias, the width of the depletion region increases. This restricts the current carrier accumulation near the junction.
Majority current carriers are primarily negated in the depletion region and hence the depletion region acts as an insulator. Normally, current carriers do not pass through an insulator.
It is seen that in a reverse-biased diode, some current flows through the depletion region. This current is called leakage current.
Leakage current is dependent on minority current carriers. the minority carriers are electrons in the P type material and holes in the N type material.
The following figure shows how current carriers react when a diode is reverse biased.

F2 J.P Madhu 22.05.20 D1

Minority carriers of each material are pushed through the depletion zone to the junction. This action causes a very small leakage current to occur. Generally, leakage current is so small that it can be considered as negligible.
Here, in case of leakage current, temperature plays an important role. The minority current carriers are mostly temperature dependent.
At room temperatures of 25°C or 78°F, there is negligible amount of minority carriers present in a reverse bias diode.
When the surrounding temperature rises, it causes significant increase in minority carrier creation and as a result it causes a corresponding increase in leakage current.
In all reverse-biased diodes, occurrence of leakage current is normal to some extent.
In Germanium and Silicon diodes, leakage current is only of few micro-amperes and nano-amperes, respectively.
Germanium is much more susceptible to temperature than silicon. For this reason, mostly Silicon is used in modern semiconductor devices.