When a p-n junction diode is reverse biased,
(A) the height of the potential barrier decreases.
(B) the height of the potential barrier increases.
(C) the junction width increases.
(D) the current is essentially voltage independent up to the breakdown voltage.
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → (B), (C) and (D) only
In reverse bias, the applied voltage \( V \) adds to the built-in barrier. This widens the depletion region and stops majority carriers from crossing. Only the small current of minority carriers flows.
\( \text{barrier height} = V_{0} + V \)
(A) the height of the potential barrier decreases. — Incorrect. The barrier decreases in forward bias; in reverse bias it increases.
(B) the height of the potential barrier increases. — Correct. It becomes \( V_{0} + V \).
(C) the junction width increases. — Correct. The field pulls electrons and holes away from the junction, so the depletion region widens.
(D) the current is essentially voltage independent up to the breakdown voltage. — Correct. The reverse current comes from minority carriers, which depend on temperature, not on the voltage. It stays at a small saturation value (about μA) until breakdown.
NCERT (2026-27):
p-n Junction Diode Under Reverse Bias – “When an external voltage (V) is applied across the diode such that n-side is positive and p-side is negative, it is said to be reverse biased [Fig.14.15(a)]. … The direction of applied voltage is same as the direction of barrier potential. As a result, the barrier height increases and the depletion region widens due to the change in the electric field.”
p-n Junction Diode Under Reverse Bias – “… The current under reverse bias is essentially voltage independent upto a critical reverse bias voltage, known as breakdown voltage (Vbr).”