p-n junction

Take a single crystal of silicon. Dope one half with trivalent atoms so it teems with holes, dope the other half with pentavalent atoms so it teems with free electrons, and grow them as one unbroken lattice. At the seam where p-type meets n-type, something happens that no isolated piece of doped silicon can do alone: a thin, self-built wall of electric field appears out of nowhere, with no battery in sight. That wall is the p-n junction, and it is arguably the single most consequential few hundred nanometres in modern technology — the seed structure inside every diode, every rectifier, every solar-charged calculator, every logic chip on the planet. The story is one of patient equilibrium, not violent reaction. Left alone, carriers diffuse across the boundary purely because there are more of them on one side than the other — the same reason a drop of ink spreads through water. But unlike ink, these carriers leave a fingerprint behind: fixed, immovable ions that were electrically neutral before a mobile partner walked away. Those exposed ions build an internal electric field, and the field pushes back against the very diffusion that created it, until the two processes reach a silent stalemate. No current flows, yet a barrier potential of a few tenths of a volt sits there permanently, waiting. This article rebuilds that stalemate step by step: why a junction must be grown as one continuous crystal rather than glued together, how diffusion and recombination carve out a carrier-free depletion region, why the internal field points from n to p, and how a diffusion current and a drift current cancel exactly at equilibrium. Everything here stays qualitative, exactly as the CBSE 2025-26 syllabus intends — no depletion-width formula, no Poisson-equation derivation — but the reasoning is rigorous enough that you could rebuild the whole picture from scratch in an exam answer.

Key takeaways

A p-n junction is one continuous semiconductor crystal doped p-type on one side and n-type on the other — never two separate crystals pressed together.
Right after formation, a large concentration gradient of holes and electrons across the junction drives diffusion: holes flow p to n, electrons flow n to p.
Diffusing carriers recombine near the boundary, leaving behind a carrier-free strip called the depletion region, typically well under a micrometre wide.
What's left behind are fixed dopant ions — negative acceptor ions on the p-side, positive donor ions on the n-side — that were not able to diffuse away.
These exposed ions create an internal electric field pointing from n to p, which in turn produces a small drift current of minority carriers flowing opposite to the diffusion current.
At equilibrium the diffusion current and drift current are exactly equal and opposite, giving zero net current with no external battery, and a built-in barrier potential of about $0.7\ \text{V}$ for silicon ( $0.3\ \text{V}$ for germanium).
Key idea: diffusion of majority carriers across a single continuous p-n crystal leaves exposed dopant ions that build an internal field from n to p, and that field drives a drift current which exactly cancels diffusion at equilibrium, fixing a barrier potential of about 0.7 V for silicon.

The core insight

A p-n junction is one continuous crystal where majority-carrier diffusion across the p-n boundary leaves behind fixed dopant ions whose internal field, pointing from n to p, drives an opposing drift current that exactly balances diffusion at equilibrium, leaving zero net current and a built-in barrier potential of roughly $0.7\ \text{V}$ for silicon.

Where it fits in the syllabus

The p-n junction sits at the heart of Unit IX, "Semiconductor Electronics: Materials, Devices and Simple Circuits," coming right after energy bands and intrinsic/extrinsic semiconductors, and right before the semiconductor diode and its I-V characteristics. CBSE 2025-26 keeps this entirely qualitative: you must be able to explain why a depletion region and barrier potential form, but you are never asked to derive the depletion width or barrier-potential formula from Poisson's equation — that mathematics belongs to a later course. Note too that the older devices historically bundled into this chapter — Zener diodes, LEDs, photodiodes, solar cells, and transistors — have been removed from the current syllabus, so this section stays strictly on the junction itself and the physics that feeds directly into the plain diode.

The physical picture

Before any junction forms, picture the two halves in isolation. The p-type half is electrically neutral overall: it has a large population of mobile holes plus an equal population of fixed negative acceptor ions buried in the lattice, so the positive and negative charges balance out. The n-type half is neutral in the same way, with mobile electrons balanced by fixed positive donor ions. Neutral does not mean uniform, though — and the moment the two neutral halves are joined into one crystal, that non-uniformity in carrier concentration is what sets everything in motion.

p_p \gg p_n \qquad \text{and} \qquad n_n \gg n_p \tag{1}

where: $p_p$ is the hole concentration on the p-side (large, since holes are majority carriers there), $p_n$ is the hole concentration on the n-side (small, since holes are minority carriers there), $n_n$ is the electron concentration on the n-side (large), and $n_p$ is the electron concentration on the p-side (small). The physical reading of equation (1) is simply that each carrier type is common on its "home" side and rare on the other — and any time a concentration gradient like this exists inside a single connected material, thermal motion will spread that species from the crowded side toward the sparse side, exactly as a perfume slowly fills an empty room. That spreading is diffusion, and it needs no battery, no external field — only the random thermal jostling every carrier already has at room temperature. As holes diffuse from p to n and electrons diffuse from n to p, the two species cross paths right at the junction and, being oppositely charged and highly mobile there, annihilate each other through recombination. Both carriers vanish as mobile charge in a thin layer straddling the original boundary. What is left is not empty space — the silicon lattice atoms are still there, and so are the dopant ions — but the layer no longer has enough loose carriers to conduct like the bulk material does on either side of it. This carrier-starved strip is the depletion region, and because it has given up its mobile charges, it behaves almost like an insulator sandwiched between two conducting slabs.

Figure

A p-n junction cross section showing p-type silicon on the left full of circled holes and n-type silicon on the right full of circled electrons, a narrow depletion region in the middle containing only fixed negative acceptor ions on the p side and fixed positive donor ions on the n side, with field lines and an arrow labelled internal field pointing from the n side toward the p side

The two rows of exposed ions do not sit passively. A negative acceptor ion on the p-side and a positive donor ion on the n-side, facing each other across the thin depletion region, are exactly the geometry of a parallel-plate arrangement — and it produces an internal electric field pointing from the positive (n) side to the negative (p) side. Any charge sitting inside the depletion region feels a force from this field: a hole is pushed back toward p, an electron is pushed back toward n. In other words, the field opposes the very diffusion current that created it. As more carriers diffuse and recombine, more ions get exposed, the depletion region widens, and the internal field grows stronger — until the field is strong enough to stop essentially all further net diffusion. At that point the junction has reached equilibrium, and the potential difference built up across the depletion region is called the barrier potential (or junction potential), roughly $0.7\ \text{V}$ for silicon and $0.3\ \text{V}$ for germanium — a fixed, no-battery-needed voltage sitting invisibly inside every diode you will meet in the next node.

Derivation AWhy the net current across an unbiased p-n junction is exactly zero

The goal of this derivation is to explain, using only the qualitative ideas of diffusion and drift that the CBSE syllabus prescribes, why an isolated p-n junction with no external battery attached settles into a state of true equilibrium in which absolutely no net current flows across it in either direction, even though two distinct physical currents — one due to diffusion of majority carriers and one due to drift of minority carriers under the internal field — are continuously present and continuously active inside the depletion region; understanding this cancellation is the conceptual key to everything that follows about forward and reverse bias in the diode.

Step 1. Identify the diffusion current. Because $p_p \gg p_n$ and $n_n \gg n_p$ (equation 1), majority holes diffuse p to n and majority electrons diffuse n to p. Conventional current follows the direction of positive charge motion, so hole diffusion contributes a current flowing p to n; electron diffusion, though the electrons move n to p, is a flow of negative charge n to p, which by convention also counts as current flowing p to n. Both contributions add in the same direction: the diffusion current flows from the p-side to the n-side.

Step 2. Identify the drift current. The internal field inside the depletion region, pointing from n to p, acts on whatever free carriers happen to wander into that region. It cannot pull majority carriers across, since the field opposes exactly that motion — but it readily sweeps minority carriers downhill: a stray electron on the p-side (a minority carrier there) is swept by the field toward the n-side, and a stray hole on the n-side (a minority carrier there) is swept toward the p-side. Electron motion p to n is negative charge moving p to n, which is conventional current n to p; hole motion n to p is positive charge moving n to p directly. Both contributions give a drift current flowing from the n-side to the p-side — opposite in direction to the diffusion current found in Step 1.

Step 3. Track what happens as the depletion region grows. Early on, right after the junction forms, diffusion dominates because the concentration gradients are steep and few ions have yet been exposed, so the internal field and hence the drift current both start out weak. As diffusion proceeds, more ions are exposed, the depletion region widens, and the internal field strengthens — which does two things at once: it increases the drift current (more field to sweep minority carriers) and it increasingly opposes further diffusion (fewer majority carriers have enough thermal energy to climb the growing potential barrier).

Step 4. Reach the equilibrium condition. Because increasing the internal field simultaneously raises the drift current and lowers the diffusion current, there is exactly one depletion width — and one corresponding barrier potential — at which the two currents become equal in magnitude. Beyond that point neither current can grow further relative to the other, so the junction settles there permanently, with the diffusion current (p to n) and the drift current (n to p) equal and opposite at every instant.

Step 5. Conclude. Since the two currents are equal in magnitude and opposite in direction, they cancel exactly, so the net current measured across an unbiased p-n junction is zero — not because nothing is happening, but because two opposing processes are happening at precisely matched rates. This dynamic, self-regulating balance, not a static absence of activity, is what "equilibrium" means for a p-n junction, and it is the baseline state that forward and reverse bias will each disturb in opposite ways in the next node.

Worked examples

Board level - explaining depletion-region formation

Question: A p-type silicon block and an n-type silicon block are grown as a single continuous crystal to form a p-n junction. Explain, in sequence, why a depletion region forms even though no external voltage has been applied, and state which ions are exposed on which side.

Step 1. Start from the carrier imbalance. Immediately after the p-type and n-type regions are joined into one crystal, the p-side has abundant holes and very few electrons, while the n-side has abundant electrons and very few holes — a sharp concentration gradient exists right at the junction, matching equation (1).

Step 2. Diffusion begins. Driven by this gradient, majority carriers diffuse across the junction — holes move from p to n, and electrons move from n to p — purely because of random thermal motion spreading each species from where it is crowded to where it is scarce, with no field or battery involved yet.

Step 3. Recombination removes mobile carriers near the boundary. As diffusing electrons and holes meet close to the junction, they recombine and cease to exist as free carriers, so a narrow strip on either side of the original boundary loses essentially all of its mobile charge.

Step 4. Fixed ions are exposed. What remains in that narrow strip are the immobile dopant ions that were left without a mobile partner: negative acceptor ions on the p-side (their compensating holes have diffused away and recombined) and positive donor ions on the n-side (their compensating electrons have diffused away and recombined). This carrier-free strip is the depletion region.

Step 5. Equilibrium is reached. The exposed ions set up an internal field that opposes further diffusion, and diffusion continues only until this field is strong enough to balance it, at which point the depletion region stops growing. Answer: negative acceptor ions are exposed on the p-side, positive donor ions on the n-side, and the depletion region reaches a fixed width once diffusion and the opposing internal field balance — all without any external voltage.

Competitive level - comparing diffusion and drift currents at equilibrium

Question: For an isolated silicon p-n junction at equilibrium (no external voltage), the depletion region carries an internal electric field pointing from the n-side to the p-side. Identify the two currents present, state their directions in terms of conventional current, explain why the net current is zero, and state the approximate barrier potential.

Step 1. Diffusion current. Carried by majority carriers moving down their own concentration gradient — holes diffusing p to n and electrons diffusing n to p — both contribute to a conventional diffusion current flowing from p to n, as established in Step 1 of the derivation above.

Step 2. Drift current. The internal field, pointing n to p, acts on the small population of minority carriers near the junction: it sweeps minority holes on the n-side across to the p-side, and minority electrons on the p-side across to the n-side. This field-driven motion produces a conventional drift current flowing from n to p — opposite to the diffusion current.

Step 3. Balance condition. As the depletion region widens during formation, the internal field strengthens, which simultaneously increases the drift current and suppresses the diffusion current, so the junction settles at the unique width where the two currents become equal in magnitude, exactly as shown in Steps 3-4 of the derivation.

Step 4. Net result and barrier potential. Since the diffusion current (p to n) and the drift current (n to p) are equal and opposite, they cancel exactly, giving zero net current across the unbiased junction. Answer: diffusion current flows p to n, drift current flows n to p, the two are equal and opposite so the net current is zero, and the built-in barrier potential is about $0.7\ \text{V}$ for silicon (about $0.3\ \text{V}$ for germanium). Timing note: budget 6-7 minutes for this five-mark conceptual answer, spending the first minute sketching a labelled depletion-region diagram before writing — a quick sketch showing both current directions earns marks fast and keeps you from mixing them up under time pressure.

Common traps

Thinking a p-n junction is made by pressing two separately grown crystals together — a mechanical join would leave a defective, discontinuous lattice; a real junction must be one continuous crystal, doped differently on each side during growth or diffusion.
Believing the depletion region is empty space with no atoms in it — it still contains the full silicon lattice and the fixed dopant ions; only the mobile charge carriers (free electrons and holes) are absent.
Mixing up which ion is exposed on which side — remember it is the majority carriers that diffuse away and recombine, so the ions left behind are of the opposite sign to what you might first guess: negative acceptor ions remain on the p-side, positive donor ions on the n-side.
Getting the internal field direction backwards — it points from the positive donor ions on the n-side to the negative acceptor ions on the p-side, i.e. from n to p, which is why it opposes further hole and electron diffusion.
Assuming equilibrium means nothing is happening — in fact both diffusion current and drift current are continuously present; equilibrium means they are exactly equal and opposite, not absent.
Quoting the barrier potential without matching it to the material — about $0.7\ \text{V}$ is for silicon; germanium junctions have a noticeably smaller barrier potential of about $0.3\ \text{V}$ .

You should be able to...

Explain why a p-n junction must be one continuous crystal, not two crystals joined together.
Describe, in the correct sequence, how carrier diffusion, recombination, and exposed dopant ions produce a depletion region.
State which ions are exposed on which side of the junction, and identify the direction of the resulting internal electric field.
Distinguish diffusion current from drift current at a p-n junction, including their directions and which carriers (majority or minority) drive each.
Explain why the net current across an unbiased p-n junction is zero, using the equal-and-opposite balance of diffusion and drift currents.
Quote the approximate barrier potential for silicon ( $0.7\ \text{V}$ ) and germanium ( $0.3\ \text{V}$ ) junctions.

Ask the tutor: If the equal-and-opposite balance between diffusion current and drift current still feels hand-wavy, ask your AI tutor to walk through what happens to each current the instant the depletion region is briefly disturbed and has to re-settle back to equilibrium.

Why does the depletion region stop growing instead of eating the whole crystal?

Because every carrier that diffuses across and recombines exposes one more ion, and every exposed ion adds a little more to the internal field. That field pushes back against diffusion, so the process is self-limiting: the wider the depletion region gets, the harder it becomes for more carriers to diffuse across it. Growth stops exactly when the field is strong enough to balance diffusion completely — a natural equilibrium, not something that runs away to consume the whole crystal.

If diffusion current and drift current cancel exactly, how can a diode ever conduct current when we use it in a circuit?

Good instinct — that cancellation is specifically for the unbiased junction with no external voltage. Once you connect a battery (forward or reverse bias, covered in the next node), you tip that balance deliberately: forward bias narrows the depletion region and lets diffusion current win, so real current flows; reverse bias widens it and lets only the tiny drift current survive. The unbiased equilibrium in this section is the starting point both directions get compared against.

Are the acceptor and donor ions actually charged particles floating in the depletion region?

Not floating — they are fixed in place inside the rigid silicon lattice, unable to move the way free electrons and holes can. An acceptor ion is a boron-type atom that has captured an extra electron from a neighbour, making it negatively charged; a donor ion is a phosphorus-type atom that has given away its spare electron, leaving it positively charged. They contribute to the internal field precisely because they stay put while the mobile carriers diffuse away.

Why is silicon's barrier potential about 0.7 V but germanium's is only about 0.3 V?

It comes down to each material's band structure and how tightly its valence electrons are held — germanium has a smaller energy gap between its valence and conduction bands than silicon, so less potential difference is needed before diffusion and drift can settle into balance. For CBSE purposes you just need to know and quote the two approximate values ( $0.7\ \text{V}$ for silicon, $0.3\ \text{V}$ for germanium) rather than derive why they differ, since that reasoning goes beyond the qualitative treatment prescribed for this chapter.