Q.When a forward bias is applied to a p-n junction, it

The key idea is P-N junction biasing: forward bias reduces the built-in potential barrier, allowing majority carriers to flow across the junction.

1. In a p-n junction, the built-in potential barrier opposes the diffusion of majority carriers (holes from p-side, electrons from n-side). …

Understanding P-N Junction Biasing

A p-n junction is formed when p-type and n-type semiconductors are joined. At the junction, electrons from the n-side diffuse into the p-side, and holes from the p-side diffuse into the n-side. This diffusion leaves behind immobile charged ions, creating a depletion region with an internal electric field. This field opposes further diffusion and gives rise to a potential barrier (typically about 0.7 V for silicon).

Now, biasing means applying an external voltage across the junction. The effect depends on the polarity:

• Forward bias: p-side connected to positive terminal, n-side to negative terminal.
• Reverse bias: p-side connected to negative terminal, n-side to positive terminal.

The key question is: what happens to the potential barrier in each case?

Step-by-Step Reasoning

1. What the potential barrier represents

   The potential barrier is the voltage difference across the depletion region that prevents majority carriers from crossing freely. For a p-n junction, the built-in potential $V_0$ is determined by the doping concentrations and temperature. In equilibrium (no external bias), this barrier is fixed.

2. Effect of forward bias on the barrier

   When forward bias is applied, the external voltage $V_f$ opposes the internal electric field. The positive terminal repels holes in the p-side toward the junction, and the negative terminal repels electrons in the n-side toward the junction. This reduces the width of the depletion region and lowers the effective potential barrier to $V_0 - V_f$.

   $V_{\text{barrier (forward)}} = V_0 - V_f$

   The barrier decreases as forward voltage increases.

3. Consequence of lowering the barrier …

Method: Energy-Band / Barrier-Height Analysis

This is the most direct way to think about biasing effects on a p-n junction. The key idea is that the potential barrier at the junction is what prevents majority carriers from crossing freely.

Step 1 – Recall the unbiased state.

In an unbiased p-n junction, diffusion of majority carriers (holes from p-side, electrons from n-side) creates a depletion region. The built-in potential $V_0$ (typically 0.6–0.7 V for silicon) acts as a barrier that stops further net diffusion.

Step 2 – Apply forward bias.

Forward bias means connecting the p-side to the positive terminal of a battery and the n-side to the negative terminal. This external voltage $V_F$ opposes the built-in field.

Step 3 – Determine the net barrier.

The effective barrier height becomes $V_0 - V_F$. Since $V_F$ is positive, the barrier decreases. For example, if $V_0 = 0.7\ \text{V}$ and $V_F = 0.5\ \text{V}$, the net barrier is only $0.2\ \text{V}$.

Step 4 – Consequence.

A lower barrier allows more majority carriers to diffuse across the junction, producing a large forward current. The barrier is not raised — it is lowered. …

The most common mistake here is picking (a) — "raises the potential barrier." That error comes from mixing up forward and reverse bias. In forward bias, the external voltage opposes the built-in field, so the barrier drops, not rises. Students often memorise "bias increases barrier" without checking direction.

Another frequent error is choosing (b) — "reduces the majority carrier current to zero." That would describe a reverse bias condition where current is nearly zero. In forward bias, majority carriers are pushed across the junction, so current actually increases sharply.

The correct answer is (c) — forward bias lowers the potential barrier.

To avoid these mistakes: …