What is a P-N Junction?

Inside virtually every piece of modern electronics — your phone, your laptop, the LED in your lamp, the solar panel on a roof — lies the same fundamental structure: the p-n junction. It is where a piece of semiconductor material doped to have excess electrons meets a piece doped to have electron holes. This interface, a few nanometers thick, is the transistor's heart, the diode's core, and the photovoltaic cell's engine. Understanding it means understanding the electronic revolution.

Semiconductor Basics

A pure silicon crystal has four valence electrons per atom, all tied up in covalent bonds. At absolute zero, silicon is a perfect insulator — no free electrons to carry current. At room temperature, thermal energy kicks a few electrons out of their bonds, creating free electron-hole pairs: a freed electron that can move through the crystal, and the vacancy it left behind (the hole), which other electrons can flow into, effectively moving as a positive charge carrier.

Doping is the deliberate introduction of impurity atoms to dramatically increase the number of free carriers:

N-type: Dope silicon with phosphorus (5 valence electrons). The extra electron is not needed for bonding — it's loosely held and easily freed. Result: many more free electrons (negative carriers) than holes. The phosphorus atoms become positive donor ions fixed in the crystal lattice.

P-type: Dope silicon with boron (3 valence electrons). Boron can only form 3 bonds with silicon, leaving one bond site empty — a hole. Result: many more holes (positive carriers) than electrons. The boron atoms become negative acceptor ions fixed in the lattice.

Formation of the Junction

When p-type and n-type silicon are brought into contact, a remarkable sequence of events occurs at the interface:

Step 1 — Diffusion: Free electrons in the n-type region near the boundary see a high concentration of holes on the p-side. They diffuse across (as do holes in the opposite direction) — not because of any electric field, but simply because of the concentration gradient. This is the same random walk that causes any diffusing species to spread from regions of high concentration to low.

Step 2 — Charge buildup: As electrons leave the n-side, they leave behind positively charged donor ions (which can't move — they're fixed in the lattice). As holes leave the p-side, they leave behind negatively charged acceptor ions. A region of net charge builds up on both sides of the junction — positive on the n-side, negative on the p-side.

Step 3 — Built-in electric field: These fixed charges create an electric field pointing from the positive n-side to the negative p-side (i.e., from n to p). This field opposes further diffusion: it pushes electrons back to the n-side and holes back to the p-side.

Step 4 — Equilibrium: Diffusion continues until the electric field exactly balances the diffusion tendency. No net current flows. This creates the depletion region — a zone depleted of free carriers, with net fixed charge on both sides.

The electric potential difference across the depletion region is called the built-in potential $V_{bi}$:

$V_{bi} = \frac{kT}{q} \ln\left(\frac{N_A N_D}{n_i^2}\right)$

where $N_A$ and $N_D$ are acceptor and donor concentrations, $n_i$ is the intrinsic carrier concentration, $k$ is Boltzmann's constant, $T$ is temperature, and $q$ is the electron charge. For silicon at room temperature with typical doping, $V_{bi}$ is about 0.6–0.7 V.

Worked Examples

Use the calculator below to explore how doping concentrations and temperature affect the built-in potential and depletion width.

Example 1 — Shockley Diode Equation

Problem: A silicon diode has $I_0 = 1\,\text{pA}$. What current flows at $V = 0.55\,\text{V}$, $T = 300\,\text{K}$?

Step 1 — Thermal voltage:

$V_T = \frac{kT}{q} = \frac{(1.381 \times 10^{-23})(300)}{1.602 \times 10^{-19}} \approx 25.85\,\text{mV}$

Step 2 — Apply Shockley equation:

$I = I_0 \left(e^{V/V_T} - 1\right) = 1\,\text{pA} \times \left(e^{0.55/0.02585} - 1\right) \approx 1.73\,\text{mA}$

Insight: The current multiplies by roughly 10 for every 60 mV increase in forward voltage — exponential growth that accounts for the diode's sharp turn-on threshold.

Example 2 — Breakdown Voltage Estimate

Problem: The critical electric field for silicon is $E_\text{crit} \approx 3 \times 10^7\,\text{V/m}$. At what reverse voltage does a $0.5\,\mu\text{m}$ depletion region break down?

$V_\text{breakdown} \approx E_\text{crit} \times W = 3 \times 10^7\,\text{V/m} \times 0.5 \times 10^{-6}\,\text{m} = 15\,\text{V}$

Insight: Zener diodes use very heavy doping to produce a narrow depletion width, which drives $V_\text{breakdown}$ down to a precise, low value (often 3–6 V) — making them ideal as voltage references.

Bias: Controlling the Junction

Applying an external voltage — biasing the junction — tips the equilibrium and allows us to control whether current flows.

Forward Bias

Connect the positive terminal to the p-side and negative terminal to the n-side. The external voltage opposes the built-in field, narrowing the depletion region. Once the applied voltage exceeds $V_{bi}$ (about 0.6 V for silicon), the depletion zone nearly collapses and carriers flood across the junction. Current rises exponentially with voltage — described by the Shockley diode equation:

$I = I_0 \left(e^{qV/kT} - 1\right)$

where $I_0$ is the tiny reverse saturation current and $kT/q \approx 0.026$ V at room temperature. This exponential behavior is why a diode has a very sharp "turn-on" voltage.

Reverse Bias

Connect positive to n-side and negative to p-side. The external voltage reinforces the built-in field, widening the depletion region. No free carriers can cross it — only the tiny thermally-generated current $I_0$ (typically nanoamps) flows. The junction blocks current.

At high enough reverse voltages, the junction breaks down: either through avalanche multiplication (carriers gain enough energy to knock out other carriers in a chain reaction) or quantum mechanical Zener tunneling (carriers tunnel through the depletion zone). These aren't always destructive — Zener diodes are specifically designed to operate in reverse breakdown for voltage regulation.

How It's Made

P-n junctions are created during semiconductor fabrication through doping processes:

Ion implantation: A beam of impurity ions (boron, phosphorus, arsenic) is accelerated to high energy and fired into the silicon wafer. The implantation depth is precisely controlled by the ion energy, and the dose by the beam current. After implantation, the wafer is annealed (heated) to repair crystal damage and activate the dopants.

Diffusion: The wafer is exposed to a vapor of the dopant species at high temperature (900–1200 °C). Dopant atoms diffuse into the surface. The depth and concentration profile depends on temperature and time — controlled to produce exactly the desired doping profile.

Modern CMOS fabrication uses ion implantation almost exclusively for its precision and repeatability.

Carrier Behavior and Band Diagrams

The p-n junction is most clearly understood in terms of the band diagram — a plot of energy vs. position across the device, showing the conduction band (where free electrons live) and the valence band (where holes live) relative to a reference energy level.

At equilibrium, the Fermi level — the energy at which the probability of electron occupation is 50% — must be constant throughout the device (no net current). This requires the bands to bend across the depletion region, creating the energy barrier that blocks carrier flow.

Forward bias lowers this barrier (making it easier for carriers to cross). Reverse bias raises it (impossible for thermally-excited carriers to cross).

Applications

The p-n junction is the foundation of:

Device	Principle
Rectifier diode	Passes current in one direction only
Zener diode	Voltage regulation via controlled breakdown
LED	Electrons and holes recombine, emitting photons
Laser diode	Stimulated emission from a p-n junction
Solar cell	Photons create electron-hole pairs; junction separates them
Bipolar transistor	Two p-n junctions in series, with a thin base
MOSFET	Gate voltage controls a p-n junction channel
Photodiode	Detects light via photogenerated current

The transistor as two junctions: A bipolar junction transistor (BJT) is simply two p-n junctions placed back-to-back (either p-n-p or n-p-n). The thin middle region — the base — is lightly doped, allowing carriers injected from one junction to diffuse to the other before recombining. A small current into the base controls a much larger current between emitter and collector: transistor action. This amplification principle, discovered at Bell Labs in 1947, launched the digital age.

FAQ

Why does silicon have a turn-on voltage of ~0.6 V?▼

The 0.6 V threshold is the built-in potential $V_{bi}$ for typical silicon doping levels. It arises from the $\ln(N_A \cdot N_D / n_i^2)$ term in the V_bi formula — silicon's intrinsic carrier concentration $n_i$ is very low (~$10^{10}$ cm$^{-3}$), so the log ratio is large. Germanium has a higher $n_i$, which is why Ge diodes turn on at only ~0.3 V.

What's the difference between diffusion current and drift current?▼

Diffusion current is driven by concentration gradients — carriers move from regions of high concentration to low. Drift current is driven by the electric field — carriers are accelerated by the field. At equilibrium in a p-n junction, these two exactly cancel across the depletion region, so no net current flows even though both are non-zero.

Why does a tiny current flow in reverse bias?▼

Thermal energy continuously generates a small number of electron-hole pairs even inside the depletion region. The strong reverse-bias electric field immediately sweeps these apart before they can recombine, producing the reverse saturation current $I_0$. It's typically in the nanoamp–picoamp range and roughly doubles for every 10 °C increase in temperature.

Why must the Fermi level be flat at equilibrium?▼

A sloping Fermi level would mean electrons at one point have a different electrochemical potential than electrons elsewhere — a thermodynamic imbalance that would drive a net current. Since no external voltage is applied at equilibrium, no net current flows, so the Fermi level must be spatially constant throughout the entire device.

What's the difference between Zener and avalanche breakdown?▼

Zener breakdown (below ~5 V) occurs via quantum tunneling — electrons tunnel directly through a very narrow depletion region under a strong field. Avalanche breakdown (above ~7 V) occurs when field-accelerated carriers gain enough energy to knock out new electron-hole pairs in a chain reaction. Both mechanisms occur in Zener diodes: tunneling dominates at low breakdown voltages, avalanche at high voltages.

How does a p-n junction emit light (LED)?▼

In forward bias, electrons from the n-side and holes from the p-side are injected across the junction. When an electron recombines with a hole, it releases energy equal to the semiconductor's band gap $E_g$. In direct band-gap materials (GaAs, GaN), this energy is released as a photon with wavelength $\lambda = hc/E_g$. Silicon has an indirect band gap, so it releases the energy as heat instead — which is why LEDs use compound semiconductors, not silicon.

Can the depletion region conduct electricity?▼

No — that is its defining characteristic. The depletion region is stripped of free carriers, leaving only fixed ions. It behaves like an insulator. The junction's junction capacitance $C_j$ arises from this structure: two conducting regions (p-bulk and n-bulk) separated by a charged insulating layer — exactly like a parallel-plate capacitor. Since $C_j \propto 1/\sqrt{W}$ and $W$ increases with reverse bias, the capacitance decreases as reverse voltage increases — a property exploited in varactor diodes for voltage-controlled tuning.

How does temperature affect a p-n junction?▼

Higher temperature increases $n_i$ (more thermally generated carriers), which reduces $V_{bi}$ and increases $I_0$. The forward voltage drop decreases by approximately 2 mV per °C — a property used to build silicon diode thermometers. At high enough temperatures, thermally generated carriers overwhelm the intentional doping, and the junction loses its rectifying behavior entirely.