The p-n junction diode

Fuse a p-type semiconductor to an n-type one, and something remarkable happens right at the join: a microscopically thin region forms that lets current flow easily in one direction and almost completely blocks it in the other. That region is the p-n junction, and the two-terminal device built around it — the junction diode — is the single most important building block in electronics, the ancestor of every rectifier, every logic gate, every solar cell. The physics is all about what happens when mobile charges meet a concentration difference. Electrons pour from the electron-rich n-side toward the hole-rich p-side, and holes pour the other way, until the traffic stops itself by leaving behind a wall of fixed charge. That wall — the depletion region — is the entire story of how a diode decides which way current is allowed to go. This article builds that story from the ground up: how the depletion region and potential barrier form, what forward and reverse bias do to them, how to read the V-I characteristic, and how to turn a biased diode into a solvable circuit.

The core insight

A p-n junction diode conducts almost freely once forward bias exceeds a fixed cut-in voltage (about 0.7 V for silicon) but blocks current in reverse bias except for a tiny, near-constant leakage current, because biasing controls the width of the internal depletion region and its potential barrier.

The core idea

Diffusion builds the barrier. When p-type and n-type crystal are joined into one continuous lattice, the huge concentration difference at the junction drives diffusion: free electrons wander from the n-side into the p-side and recombine with holes there, while holes wander from the p-side into the n-side and recombine with electrons. Each recombination event does not just cancel one carrier — it uncovers a fixed charge that was electrically neutralised by that carrier. Donor atoms on the n-side, having lost the electron they contributed, become fixed positive ions; acceptor atoms on the p-side, having lost the hole they contributed, become fixed negative ions. These ions are locked in the crystal lattice and cannot move, so a narrow strip on either side of the junction — typically only about 10−6 m10^{-6}\ \text{m} wide — ends up almost completely stripped of mobile carriers. This is the depletion region (also called the depletion layer or space-charge region).

The barrier stops diffusion. The exposed positive ions on the n-side and negative ions on the p-side set up an internal electric field pointing from n to p across the depletion region. This field opposes further diffusion of majority carriers, and it corresponds to an internal potential barrier V0V_0 — roughly 0.3 V0.3\ \text{V} for germanium and roughly 0.7 V0.7\ \text{V} for silicon. Diffusion and the opposing field reach a dynamic equilibrium: as many carriers still drift across per second as diffuse across, so the net current is zero with no external voltage applied.

Bias tips the balance. An external voltage across the junction can add to or fight against this internal field, which is what makes the diode a one-way valve — the exact mechanism explored below.

Reading the curve. The device's entire personality is captured in one plot: the V-I characteristic, current II through the diode against voltage VV across it. In forward bias the curve stays nearly flat and near zero until a threshold, then rises steeply, almost exponentially; in reverse bias it stays flat and near zero (the small reverse saturation current I0I_0) until a large breakdown voltage, beyond which it turns sharply vertical.

I≈I0(eV/VT−1)I \approx I_0\left(e^{V/V_T} - 1\right)

where: II is the current through the diode, I0I_0 is the reverse saturation current (a small constant set by the material and temperature), VV is the voltage applied across the junction (positive in forward bias), and VTV_T is a temperature-dependent voltage constant. This equation is the source of the exponential forward rise and the near-flat reverse branch; the curriculum-level treatment below does not require using it directly — instead it replaces the curve with a single fixed cut-in voltage for circuit calculations.

Figure
V-I characteristic curve of a silicon p-n junction diode showing a flat near-zero segment below the cut-in voltage of about 0-7 volts then a steep almost vertical rise in forward current, and a flat near-zero reverse current until breakdown
V-I characteristic of a silicon p-n junction diode V (forward right, reverse left) I (forward current, up) cut-in ~0.7 V breakdown reverse current ~0 (I0) steep forward rise

Forward and reverse bias

Forward bias: the battery's positive terminal connects to the p-side, negative to the n-side. The applied field opposes the internal field, so the depletion region narrows and the barrier drops. Once the applied voltage exceeds the cut-in voltage (≈0.7 V\approx 0.7\ \text{V} for silicon, ≈0.3 V\approx 0.3\ \text{V} for germanium), majority carriers cross the junction freely and a large forward current flows, rising roughly exponentially with voltage beyond that point.

Reverse bias: the battery's positive terminal connects to the n-side, negative to the p-side. The applied field reinforces the internal field, so the depletion region widens and the barrier increases. Majority carriers are pulled away from the junction, so almost no current flows — only the tiny, essentially voltage-independent reverse saturation current I0I_0 due to minority carriers survives. Push the reverse voltage far enough and the diode breaks down, current rising sharply; this effect is deliberately exploited later in the Zener diode.

Derivation AForward-biased diode as a circuit element

Given a silicon diode of cut-in voltage V0V_0 connected in series with a resistor RR across a battery of EMF EE, and treating the diode as an ideal switch that drops exactly V0V_0 once conducting, derive an expression for the steady current II flowing in the loop so that any forward-biased single-diode series circuit can be solved directly from EE, RR, and V0V_0 without redrawing the exponential characteristic curve each time.

Step 1. Once E>V0E > V_0, the diode conducts, and by the ideal-switch approximation it maintains a fixed voltage drop V0V_0 across its terminals regardless of the current through it — the entire steep part of the real exponential curve is collapsed into this one constant.

Step 2. Apply Kirchhoff's voltage law around the single loop: the EMF supplied by the battery must equal the sum of the voltage drops around the loop, one across the diode and one across the resistor.

E=V0+IRE = V_0 + IR

Step 3. Solve this linear equation for the current II by isolating it on one side.

IR=E−V0⇒I=E−V0RIR = E - V_0 \quad\Rightarrow\quad I = \frac{E - V_0}{R}

where: II is the steady forward current (A), EE is the battery EMF (V), V0V_0 is the diode's cut-in voltage (V), and RR is the series resistance (ohm). The result says the resistor absorbs whatever EMF is left over once the diode has taken its fixed share V0V_0, and it is the working formula behind every forward-biased series numerical in this chapter.

Worked example

Question: A silicon diode with cut-in voltage 0.7 V0.7\ \text{V} is connected in forward bias in series with a 220 Ω220\ \Omega resistor to a 6 V6\ \text{V} battery. Find the current in the circuit, then find what the current becomes if a second identical diode is added in series and the battery is changed to 9 V9\ \text{V} with a 470 Ω470\ \Omega resistor.

Step 1. For the first circuit, check that the diode conducts: the supply 6 V6\ \text{V} exceeds the cut-in voltage 0.7 V0.7\ \text{V}, so the diode is forward biased and holds a constant 0.7 V0.7\ \text{V} drop.

Step 2. Apply the derived formula I=(E−V0)/RI = (E - V_0)/R with E=6 VE = 6\ \text{V}, V0=0.7 VV_0 = 0.7\ \text{V}, R=220 ΩR = 220\ \Omega:

I=6−0.7220=5.3220≈0.0241 A=24.1 mAI = \frac{6 - 0.7}{220} = \frac{5.3}{220} \approx 0.0241\ \text{A} = 24.1\ \text{mA}

Step 3. For the second circuit, two identical series diodes each hold their own cut-in drop, so the total diode drop is V0=0.7+0.7=1.4 VV_0 = 0.7 + 0.7 = 1.4\ \text{V}. Apply the same formula with E=9 VE = 9\ \text{V} and R=470 ΩR = 470\ \Omega:

I=9−1.4470=7.6470≈0.0162 A=16.2 mAI = \frac{9 - 1.4}{470} = \frac{7.6}{470} \approx 0.0162\ \text{A} = 16.2\ \text{mA}

Step 4. Check the second circuit with Kirchhoff's law: the resistor drop is VR=IR≈0.0162×470≈7.6 VV_R = IR \approx 0.0162 \times 470 \approx 7.6\ \text{V}, and 1.4+7.6=9 V1.4 + 7.6 = 9\ \text{V}, which matches the supply exactly.

The current is approximately 24.1 mA in the single-diode circuit and approximately 16.2 mA in the two-diode circuit, with each diode always taking a fixed 0.7 V and the resistor absorbing the rest.

Common traps

  • Thinking the depletion region contains free charge carriers — it does not. It contains only fixed, immobile ions left behind after diffusion, and that is exactly why it acts as a barrier rather than a conducting channel.
  • Mixing up which terminal biasing does what: forward bias means p-side to positive, n-side to negative (barrier shrinks, current flows easily); reverse bias is the opposite (barrier grows, current is blocked). Swapping these flips every prediction about current and barrier width.
  • Assuming reverse current is exactly zero — a small reverse saturation current always flows due to minority carriers; it is negligible for most circuit calculations but never truly zero, and it is not the same thing as breakdown current.
  • Treating the cut-in voltage as a hard on/off switch at every point on the curve — it is really an approximation of a smooth, roughly exponential rise, valid once the diode is clearly conducting well above the knee, not near it.

Ask the tutor: Want to work through a reverse-bias breakdown numerical, or see how the diode equation's exponential term connects to the cut-in approximation used here — ask the AI tutor to walk through it with you.

Why does the depletion region form only near the junction and not spread through the whole crystal?

Diffusion is driven by the concentration difference right at the junction, and every electron or hole that crosses over recombines almost immediately with a carrier of the opposite type nearby, uncovering a fixed ion exactly where it crossed. As ions accumulate, the internal field they create grows strong enough, within a very short distance, to stop further diffusion — so the exposed-ion region self-limits to a thin strip only about 10−6 m10^{-6}\ \text{m} wide rather than spreading indefinitely.

Is the cut-in voltage a real physical threshold, or just something we assume for exam calculations?

It is a simplification of a real, smooth curve. The actual forward current rises roughly exponentially with voltage from the very start, with no sudden switch-on. The cut-in voltage marks the point on that curve where the current has become large enough that the diode behaves, for practical circuit purposes, like a nearly constant voltage drop. Treating it as a fixed 0.7 V (silicon) or 0.3 V (germanium) switch is accurate enough for series-circuit numericals, which is why the syllabus uses it, but the underlying physics is continuous.

Why does reverse saturation current not increase much even if we increase the reverse voltage further, until breakdown?

The reverse current is carried by minority carriers — electrons in the p-side and holes in the n-side — that happen to wander close enough to the junction to be swept across by the field. Their supply rate is fixed by thermal generation, not by how strong the field is, so once the field is strong enough to sweep away essentially all of them, adding more reverse voltage cannot recruit more current. That supply-limited current is I0I_0, and it stays flat until the field grows so large that a different mechanism, breakdown, takes over.

Before this

Where this leads