91Ó°ÊÓ

In a \(P-N\) junction, the P-side has an abundance of holes, while the N-side has an abundance of electrons. When the junction is in equilibrium, that is, when there is no external bias applied, there will be a balance of electron-hole recombination and generation. The region near the junction where recombination and generation occur is called the depletion region.

When no bias is applied, there is no external electric field acting on the carriers. However, there is a built-in electric field that is created between the P-side and the N-side of the junction due to the difference in the doping levels. This built-in electric field opposes the diffusion of charge carriers across the junction. Consequently, the net flow of charge carriers across the junction becomes zero.

Based on our explanation of the behavior of charge carriers in a \(P-N\) junction when no bias is applied, we can determine the correct answer as follows: (A) Is zero because the number of charge carriers flowing on both sides is same. This statement is TRUE because the net flow of charge carriers across the junction is zero due to the built-in electric field. (B) Is zero because the charge carriers do not move. This statement is FALSE because charge carriers do move, but their net flow across the junction becomes zero due to the built-in electric field opposing their motion. (C) Is non-zero. This statement is FALSE because the net current in the junction is zero when there is no applied bias. (D) None of these. This statement is FALSE because option (A) is the correct answer. So, when no bias is applied to a \(P-N\) junction, the current is zero because the number of charge carriers flowing on both sides is the same. Therefore, the correct answer is (A).