The p-n Junction: Formation, Depletion Region & Barrier

The Key to All Devices

The p-n junction is the basic building block of diodes, transistors and beyond — NCERT calls it 'the key to all semiconductor devices'. Understanding its formation and its response to applied voltage (bias) unlocks everything that follows.

How a junction is made (and how it isn't)

Take a thin p-type Si wafer and add a precisely controlled small quantity of pentavalent impurity to part of it — that part converts to n-type. One continuous crystal now contains a p-region, an n-region, and a metallurgical junction between them.

Why not just press a p-slab against an n-slab? (NCERT Example 14.3): any real surface has roughness far larger than the interatomic spacing (~2-3 angstrom), so continuous contact at the atomic level is impossible — the interface would act as a discontinuity for carriers. The junction must be grown within a single crystal.

Diffusion vs Drift: The Two Competing Currents

Two processes govern junction formation:

1. Diffusion (concentration-driven)

Electrons are abundant on the n-side, holes on the p-side. Concentration gradients drive holes p → n and electrons n → p: the diffusion current (p → n direction conventionally).

2. The space-charge region builds

• Each electron leaving the n-side abandons an ionised donor (+) — immobile, locked in the lattice → a layer of positive space-charge on the n-side.
• Each hole leaving the p-side abandons an ionised acceptor (-) → a layer of negative space-charge on the p-side.

This double layer, emptied of free carriers, is the depletion region — thickness of order one-tenth of a micrometre (~0.1 $\mu$m).

3. Drift (field-driven)

The space charge creates an electric field from n to p (positive → negative). This field sweeps any electron on the p-side to n, any hole on the n-side to p: the drift current, opposite in direction to diffusion.

Equilibrium

Initially diffusion is large and drift small. As diffusion proceeds, the space-charge widens, the field strengthens, drift grows — until diffusion current = drift current. The junction is formed; under equilibrium there is no net current.

Key Point: The loss of electrons from n and gain by p leaves n positive relative to p — a potential difference across the junction whose polarity opposes further carrier flow. This is the barrier potential $V_0$ (NCERT Fig. 14.11).

[NEET Important] Directions to lock in: diffusion = majority carriers crossing (holes p→n, electrons n→p); drift = minority carriers swept back by the field; field points n → p; n-side is the positive side of the barrier. [JEE Tip] Typical barrier values (for later circuit work): ~0.7 V for Si, ~0.2-0.3 V for Ge — the same numbers reappear as cut-in voltages.

Solved Examples

Example 1: Why pressing slabs together fails (NCERT Example 14.3)

Can a p-type slab be physically joined to an n-type slab to make a p-n junction?

Solution:

1. No. Interatomic spacing is ~2-3 angstrom.
2. Any slab surface, however polished, has roughness much larger than this.
3. Continuous atomic-level contact is impossible; the interface behaves as a discontinuity for flowing carriers.
4. Takeaway: junctions are grown inside one crystal (by converting part of a wafer through doping) — never assembled mechanically.

Example 2: Who moves in diffusion? [Board Conceptual]

During junction formation, which carriers diffuse, which way, and why?

Solution:

1. Holes: p → n; electrons: n → p.
2. Driver: concentration gradients — each species flows from where it is abundant to where it is scarce.
3. These are majority carriers crossing over; their motion constitutes the diffusion current.
4. Takeaway: diffusion is statistics, not fields — it would happen even with no charge on the carriers.

Example 3: The immobile charges [Board Conceptual]

Why does the depletion region carry net charge although it has almost no carriers?

Solution:

1. Departing electrons leave ionised donors (+) on the n-side; departing holes leave ionised acceptors (-) on the p-side.
2. These ions are bonded into the lattice — immobile.
3. The region is depleted of FREE carriers but charged by FIXED ions: positive layer (n-side) facing negative layer (p-side).
4. Takeaway: 'depletion' refers to mobile carriers only; the fixed space charge is the whole point — it generates the field and barrier.

Example 4: Direction of the built-in field [NEET Rapid]

State the direction of the depletion-region electric field and of the drift current it drives.

Solution:

1. Field: from the positive space charge (n-side) toward the negative (p-side): n → p.
2. Drift: the field pushes p-side electrons → n and n-side holes → p — i.e. minority carriers swept 'home'.
3. Drift current direction (conventional): n → p… more precisely, opposite to the diffusion current.
4. Takeaway: diffusion sends majority carriers OUT; drift sends minority carriers BACK. Balance = equilibrium.

Example 5: Why equilibrium is inevitable [Board Reasoning]

Explain why the diffusion process cannot continue indefinitely.

Solution:

1. Every diffusing carrier adds to the space charge, widening the depletion region.
2. Wider space charge → stronger field → larger drift current opposing diffusion.
3. The process self-limits: growth continues until drift = diffusion, after which no net charge crosses.
4. Takeaway: negative feedback builds the junction and then freezes it — equilibrium with zero net current but two equal, opposing currents still flowing.

Example 6: The barrier's polarity [NEET Conceptual]

Which side of an unbiased junction is at higher potential, and what does the barrier oppose?

Solution:

1. The n-side lost electrons (and gained none back); the p-side gained electrons.
2. So n is positive relative to p — the barrier potential $V_0$.
3. Its polarity opposes further diffusion of majority carriers (it would push holes back to p, electrons back to n).
4. Takeaway: the junction erects exactly the wall needed to stop its own construction — 'barrier potential' is literal.

Example 7: Depletion width order of magnitude [Numerical]

The depletion region is ~0.1 $\mu$m thick and the barrier ~0.7 V (Si). Estimate the electric field inside.

Solution:

1. Field: $E \approx \dfrac{V_0}{d} = \dfrac{0.7}{0.1 \times 10^{-6}}$.
2. Compute: $E \approx 7 \times 10^6$ V/m.
3. Takeaway: megavolt-per-metre fields live inside every diode — strong enough to sweep any minority carrier instantly, which is why drift needs no encouragement.

Example 8: What 'no net current' hides [Conceptual]

At equilibrium the junction carries zero net current. Does that mean nothing is moving?

Solution:

1. No — two currents flow continuously: diffusion (majority carriers over the barrier) and drift (minority carriers swept down it).
2. They are equal and opposite, cancelling exactly.
3. Bias will later upset this balance: forward bias boosts diffusion; reverse bias exposes bare drift.
4. Takeaway: equilibrium is a tug-of-war at stalemate, not an empty field — the whole diode story is about tilting this balance.

Example 9: Sequence the formation story [Board Ordering]

Arrange: (a) drift equals diffusion, (b) majority carriers diffuse across, (c) space-charge layers build, (d) field and drift current grow.

Solution:

1. (b) concentration gradients drive majority-carrier diffusion.
2. (c) departing carriers leave immobile ions — space-charge (depletion) layers form.
3. (d) the growing space charge strengthens the field, increasing drift.
4. (a) drift catches up to diffusion — equilibrium; junction formed.
5. Takeaway: b → c → d → a; the four-step chain is a favourite Board sequencing question.

Example 10: Barrier height and doping [JEE Insight]

Qualitatively, how would heavier doping on both sides affect the barrier potential and depletion width?

Solution:

1. Heavier doping → larger concentration gradients → more diffusion before balance → larger barrier potential $V_0$.
2. But denser dopant ions supply the needed space charge in a thinner layer → narrower depletion region.
3. Takeaway: doping turns the barrier up and the width down — the extreme case (very heavy doping, very thin depletion) is exactly what makes a Zener diode's sharp breakdown possible (Section 7).