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The p–n Junction

Explain how a p–n junction forms a depletion region and why it conducts under forward bias but not under reverse bias (A Level Physics).

  • A-Level Semiconductor Physics extension
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Learning objectives

  • Explain p-n junction formation and bias, and analyse diode, LED, Zener, rectifier-smoothing, sensor and transistor-switching circuits.

1. Definitions (Must Know)

A. p–n junction

A p–n junction is formed when p-type and n-type semiconductors are joined together.

It conducts easily in one direction (forward bias) but not the other (reverse bias).

B. Depletion region

The depletion region is the region near the junction where mobile charge carriers are depleted (few free electrons/holes).

It contains fixed ions:

  • positive donor ions on the n-side (where electrons have diffused away)
  • negative acceptor ions on the p-side (where holes have been filled)

C. Forward bias and reverse bias

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

2. Key Ideas (What Earns Marks)

  • When a p–n junction forms, electrons and holes diffuse across the junction and recombine, leaving fixed ions behind → depletion region.
  • The depletion region creates an electric field (potential barrier) that opposes further diffusion.
  • Forward bias reduces the barrier and depletion width → significant current can flow.
  • Reverse bias increases the barrier and depletion width → current is (approximately) zero (ignoring small leakage).
Fast check

Forward bias: p to +, n to −. Reverse bias: p to −, n to +.

3. Detailed Explanations

A. How the depletion region forms

Before contact:

  • n-type has many electrons (majority carriers)
  • p-type has many holes (majority carriers)

After contact:

  • electrons diffuse from n to p and recombine with holes
  • holes effectively diffuse from p to n (equivalently, electrons move the other way and fill holes)

Recombination near the junction removes mobile carriers and leaves behind fixed dopant ions, creating the depletion region.

Forward and reverse bias of a p–n junctionTwo junction diagrams compare a narrow depletion region and substantial current under forward bias with a wide depletion region and negligible current under reverse bias.Forward bias: p to +, n to −+++++−−−−−barrier reduced • substantial currentReverse bias: p to −, n to +++++−−−−barrier increased • negligible currentsmall leakage and breakdown omittedconventional current
Scroll diagram horizontally to read all labels.
Forward bias opposes the junction field and narrows the depletion region, whereas reverse bias reinforces the field and widens the region. The drawing is schematic, not a literal carrier-density scale.

B. Why the junction blocks current without forward bias

The fixed ions create an electric field that pushes:

  • electrons back towards the n-side
  • holes back towards the p-side

This opposes further diffusion. At equilibrium, diffusion and drift balance, so there is no net current.

C. Reverse bias (why current is small)

Reverse bias pulls:

  • electrons towards the positive terminal on the n-side
  • holes towards the negative terminal on the p-side

This widens the depletion region and strengthens the barrier, so majority carriers cannot cross the junction.

D. Forward bias (why current is large)

Forward bias pushes:

  • electrons towards the junction from the n-side
  • holes towards the junction from the p-side

This reduces the depletion width and barrier, allowing carriers to cross and a significant current to flow.

Diode I–V characteristic (qualitative, silicon-like)

A qualitative I–V curve: tiny reverse leakage current and a steep rise in forward current beyond about 0.7 V.

Scroll across the graph to read all labels.

A qualitative I–V curve: tiny reverse leakage current and a steep rise in forward current beyond about 0.7 V.A qualitative I–V curve: tiny reverse leakage current and a steep rise in forward current beyond about 0.7 V.
Reverse bias gives only a small leakage current (approximately zero in many exam questions). Forward bias reduces the barrier so current rises rapidly beyond a “turn-on” region.
Open full-size graph
View figure data
Values for Diode I–V characteristic (qualitative, silicon-like)
Potential difference across diode, V (V)Diode I–V (schematic)
-5-0.02
-3-0.02
-1-0.015
-0.2-0.01
00
0.40.02
0.60.12
0.70.35
0.80.7
0.90.92
11

4. Common Mistakes

  • Swapping the forward and reverse bias connections.
  • Saying “no current” in reverse bias without noting “approximately” (there can be small leakage).
  • Thinking the depletion region contains lots of free charges (it contains mostly fixed ions).

5. Exam Tips

  • Use a labelled diagram: show p-side, n-side, and battery polarity.
  • State what happens to depletion width (increases for reverse bias, decreases for forward bias).
  • If the question moves to circuit behaviour, link to the diode I–V characteristic (covered in the circuits topic).
Connect carriers to circuit behaviour

Use the Semiconductor Devices Lab to compare depletion width under forward and reverse bias, then connect that model to diode conduction and rectification.

6. Worked Examples

Modelled example 1

Identify the bias condition

Core

Problem

A p–n junction has its p-side connected to the negative terminal and its n-side to the positive terminal. Identify the bias and explain what happens to the depletion region and current.
Study the worked solution

  1. Match the polarity

    Method

    The junction is reverse biased.

    Reason

    Reverse bias connects p to negative and n to positive.

    Working

    p → -; n → +

  2. Track majority carriers

    Method

    Electrons and holes are pulled away from the junction.

    Reason

    The applied field drives n-side electrons towards the positive terminal and p-side holes towards the negative terminal.

    Working

    majority carriers move away from junction

  3. Infer the junction state

    Method

    The depletion region widens and current is approximately zero, apart from small leakage.

    Reason

    The potential barrier increases and blocks majority-carrier crossing.

    Working

    w_depletion↑ ⇒ I ≈ 0

Common misconception 2

Current direction in forward bias

Find and correct the mistake

Learner claim

A forward-biased diode conducts. A learner says conventional current must flow from n to p because electrons move from n towards p. Diagnose the direction claim.

Try this before viewing the solution

Conventional current through diode

View solution step by step

  1. Separate electron drift

    Method

    Electrons crossing from n towards p move opposite to conventional current.

    Reason

    Electrons carry negative charge.

    Working

    vector I opposite vector vₑ

  2. Track holes

    Method

    Holes move from p towards n and behave as positive carriers.

    Reason

    Their effective drift direction matches conventional current.

    Working

    vector vₕ parallel to vector I

  3. Correct the direction

    Method

    Conventional current flows from the p-side to the n-side through the forward-biased diode.

    Reason

    Both carrier descriptions correspond to that same conventional-current direction.

    Working

    p → n

7. Mind Stretchers

Mind stretcher 1: Question 1Extension

Explain why forward bias can lead to a large current even though the depletion region exists.

Show Answer

Forward bias reduces the potential barrier and narrows the depletion region. Majority carriers are pushed towards the junction and can cross it more easily, so the current increases greatly.

Mind stretcher 2: Question 2Extension

Why does reverse bias not “suck” electrons across the junction and create current?

Show Answer

Reverse bias increases the barrier and widens the depletion region, so majority carriers cannot cross. Only a small leakage current (due to minority carriers) may flow, which is usually negligible in basic circuit questions.

8. Optional (Enrichment)

A. Diode equation (beyond typical A Level needs)

Real diode I–V behaviour can be modelled with an exponential relationship (Shockley diode equation). A Level questions typically focus on direction, approximate conduction/blocking, and I–V graph interpretation rather than deriving the equation.