Formation of the p-n Junction and Its Energy Band Diagram

By the end you should be able to: Explain how the depletion region and built-in potential form at a p-n junction, draw and interpret the energy band diagram, and describe the balance of drift and diffusion currents at equilibrium.

One boundary between p-type and n-type silicon produces a device that conducts one way and not the other. Nothing was added — no mechanism, no moving part. The asymmetry comes entirely from what happens at the join.

What happens when the two meet

A p-n junction is not two pieces pressed together — the crystal is continuous, with doping changed from acceptors to donors across a plane. That matters: a mechanical contact would have surface states and no junction behaviour.

At the instant of formation there is a violent concentration gradient. Electrons are $10^{12}$ times more numerous on the n side, holes equally lopsided the other way.

1. Diffusion. Electrons diffuse from n to p; holes diffuse from p to n. Pure statistics — particles spread from where they are crowded.

2. Recombination near the boundary. An electron arriving on the p side is a minority carrier surrounded by holes, and recombines quickly. Same for holes crossing the other way. A thin region either side of the junction is stripped of mobile carriers.

3. Fixed ions are exposed. The dopant atoms cannot move. Where the n side has lost electrons, positively charged donor ions remain; where the p side has lost holes, negatively charged acceptor ions remain.

This carrier-free layer of exposed fixed charge is the depletion region (or space charge region), typically well under a micrometre wide.

4. A built-in field appears. Positive ions on the n side, negative on the p side — so an electric field points from n to p, across the junction.

5. Equilibrium. The field opposes the very diffusion that created it: it pushes electrons back towards n and holes back towards p. The process is self-limiting. Diffusion stops growing the region exactly when the field is strong enough to balance it.

Two currents in balance

At equilibrium there are two opposing currents of each carrier type:

Diffusion current — majority carriers crossing against the field, driven by the concentration gradient;
Drift current — minority carriers swept across by the field.

Note which carriers do which. Drift current is carried by minority carriers, and any minority carrier that wanders into the depletion region is swept across immediately. It is small because minority carriers are scarce, not because the field is weak.

At equilibrium these cancel exactly, and the net current is zero — as it must be, since a junction sitting on a bench with no connections cannot pass current. That would be a perpetual motion machine.

The built-in potential

The field across the depletion region corresponds to a potential difference, the built-in or contact potential $V_{bi}$ :

V_{bi} = \frac{k_BT}{e}\,\ln\!\left(\frac{N_A N_D}{n_i^{2}}\right)

For silicon it comes out at 0.6–0.8 V.

You cannot measure it with a voltmeter, and this is worth understanding rather than memorising. Connecting probes creates two new metal-semiconductor contacts, each with its own contact potential, and around the complete loop they cancel exactly. If they did not, you could build a battery from a diode and a wire, and it would deliver power for ever.

The energy band diagram

The key principle: at equilibrium the Fermi level is flat across the entire structure. A gradient in $E_F$ means a net flow of carriers, which is not equilibrium.

But the last topic established that $E_F$ sits high in the gap on the n side and low on the p side. Both cannot be true with a flat $E_F$ unless something else moves — so the bands bend.

Reading the diagram:

Deep in the p region, $E_C$ and $E_V$ sit high relative to the n side.
Deep in the n region they sit low.
Across the depletion region they bend smoothly down from p to n.
$E_F$ runs flat and level throughout.
The total bend equals $eV_{bi}$ .

The bend is a potential energy barrier for majority carriers. An electron on the n side wanting to reach the p side must climb $eV_{bi}$ ; only those in the high-energy tail of the distribution can. That is the diffusion current.

For minority carriers the same bend is a downhill slope. An electron generated on the p side rolls down to the n side unaided. That is the drift current.

A single barrier explains both currents, one exponentially sensitive to its height and one indifferent to it. Hold that thought: the next topic changes the barrier height with an applied voltage, and the whole diode equation follows.