The current through a p-n junction diode under reverse bias is mainly…

The correct answer is Option (3) → drift of the minority carriers

Under reverse bias, the applied voltage adds to the barrier potential. So the barrier height rises to \( V_0+V \) and holds back the flow of majority carriers across the junction.

Option (1) drift of the majority carriers — Incorrect. The junction field sweeps carriers only from their minority side to their majority side. It pushes electrons on the n-side and holes on the p-side away from the junction. So they give no drift current across it.

Option (2) diffusion of the majority carriers — Incorrect. This is the main current in forward bias. In reverse bias, the higher barrier holds back the flow of electrons from n to p and holes from p to n. So the diffusion current becomes very small.

Option (3) drift of the minority carriers — Correct. Electrons on the p-side and holes on the n-side that come close to the junction are swept across by the junction field. This drift of minority carriers gives the small reverse current, of the order of a few μA.

Option (4) diffusion of the minority carriers — Incorrect. The minority carriers cross the junction because the electric field sweeps them, and this is drift. Diffusion is motion caused by a difference in concentration.

NCERT (2026-27):
p-n Junction Diode under Reverse Bias – “The electric field direction of the junction is such that if electrons on p-side or holes on n-side in their random motion come close to the junction, they will be swept to its majority zone. This drift of carriers gives rise to current. The drift current is of the order of a few μA. This is quite low because it is due to the motion of carriers from their minority side to their majority side across the junction.”