In short

A p-n junction is formed where a p-type semiconductor region (holes majority) meets an n-type region (electrons majority) inside the same silicon crystal. At the moment of formation, the two sides are out of equilibrium: the n side is rich in electrons, the p side is rich in holes. Diffusion takes over. Electrons flow from n into p, holes flow from p into n, and each pair that meets recombines and annihilates as a current-carrier.

In about a picosecond this diffusion creates a depletion region — a thin layer (typical width W \sim 1\ \mu\text{m}) straddling the junction, empty of mobile carriers and containing only the fixed, ionised donor atoms (positive on the n side) and acceptor atoms (negative on the p side). These fixed charges build up a built-in potential V_0 that points from n to p and stops further diffusion. For a silicon junction at room temperature, V_0 \approx 0.7\ \text{V}; for germanium, V_0 \approx 0.3\ \text{V}; for a gallium-nitride LED, V_0 \approx 3\ \text{V}.

An external forward bias (p more positive than n) shrinks the depletion region and lowers the barrier. Current rises exponentially once the applied voltage exceeds a "turn-on" voltage near V_0. A reverse bias (p more negative than n) widens the depletion region and raises the barrier. Almost no current flows except a tiny temperature-dependent reverse saturation current I_s of nanoamperes or less.

The Shockley diode equation captures both regimes in one formula:

I \;=\; I_s\left(e^{eV/(\eta k_B T)} - 1\right),

where V is the bias voltage (positive for forward, negative for reverse), \eta \approx 1–2 is the ideality factor, and k_B T/e = 25.85\ \text{mV} at T = 300\ \text{K} is the thermal voltage.

The I-V curve of this equation is the defining property of a diode — a two-terminal device that conducts heavily in one direction and hardly at all in the other. Every rectifier in every power adapter plugged into an Indian 230 V AC socket tonight is built on this asymmetry; so is every LED bulb in every home; so is every solar cell on every rooftop in Telangana; so is every MOSFET on every radiation-hardened controller aboard an ISRO spacecraft.

A raw slice of silicon, doped on one side with phosphorus and on the other with boron, is a physics experiment in a silicon-foundry format. When the two doped halves are fabricated into one continuous crystal at an ISRO or Tata Electronics fab — which happens in practice by diffusing dopants into a clean wafer at about 1000 °C — the free electrons that had been swimming around the phosphorus side and the free holes that had been swimming around the boron side suddenly find themselves next to each other, separated by nothing but crystal lattice. Diffusion does what diffusion always does: it moves things from where there are a lot of them to where there are fewer. Electrons, excess on the n side, drift leftward into the p side. Holes, excess on the p side, drift rightward into the n side. When an electron meets a hole, they annihilate — the electron falls into the hole's empty valence state, releasing the binding energy as heat or (in a properly-designed LED) as a photon.

In the handful of picoseconds it takes for this diffusion to run, a thin layer has been swept of mobile carriers on both sides of the interface. On the n side of this depletion region there are now phosphorus atoms that have donated their extra electron and been left with a net positive charge — frozen in place in the crystal lattice, because the phosphorus atom itself cannot move. On the p side there are boron atoms that have captured an extra electron into their missing-bond state, now carrying a net negative charge and equally frozen in place. The result: a thin slab of positive charge next to a thin slab of negative charge, with an electric field pointing from n to p. That field is exactly the built-in electric field that stops further diffusion — new electrons trying to wander from n to p are pushed back by the field, and similarly for holes. Equilibrium.

This article derives what the depletion region looks like, why the built-in voltage is about 0.7 V for silicon and nothing else, what happens when you apply an external voltage, and the Shockley diode equation that ties forward bias, reverse bias, and the ideal I-V curve into one expression.

Formation of the depletion region

Start with two chunks of silicon — one n-type (with donor density N_d \sim 10^{22}\ \text{m}^{-3}, meaning about one phosphorus atom for every 5 \times 10^6 silicon atoms) and one p-type (acceptor density N_a \sim 10^{22}\ \text{m}^{-3}). Imagine bringing them into atomic-scale contact so a single crystal lattice runs through the junction.

Before equilibrium. On the n side, free electrons have density n_n \approx N_d. On the p side, free holes have density p_p \approx N_a. The n side has essentially no free holes and the p side essentially no free electrons (the minority-carrier densities, set by the law of mass action np = n_i^2, are tiny: n_p = n_i^2/N_a \approx 10^{10}\ \text{m}^{-3} for typical doping).

The imbalance sets up two diffusion currents:

As carriers cross over and recombine, the charged dopant ions they leave behind form two parallel slabs of opposite charge — the depletion layer.

Cross-section of a p-n junction showing depletion region and built-in fieldA horizontal rectangle divided into three zones. The left zone is labelled p type neutral with free holes shown as small plus circles. The middle zone, spanning from x equals minus x p to x equals plus x n, is the depletion region. On the p side of the depletion region, negative minus signs mark the ionised acceptors. On the n side, plus signs mark the ionised donors. The right zone is labelled n type neutral with free electrons shown as small minus circles. An arrow inside the depletion region points from right to left, labelled built-in electric field. p type (neutral) n type (neutral) depletion region −x_p +x_n 0 ++++++ ++++++ ++++++ −−−−−− −−−−−− −−−−−− −−−−−− ++++++ built-in E field (points n → p, opposes further diffusion)
The depletion region. On the p side ($-x_p < x < 0$), the mobile holes have diffused away and only the ionised acceptor atoms (negative fixed charges) remain. On the n side ($0 < x < x_n$), the mobile electrons have diffused away and only the ionised donor atoms (positive fixed charges) remain. The charge separation sets up an electric field pointing from n to p — the built-in field, which stops further diffusion and holds the junction in equilibrium.

Built-in potential — the barrier height

The two slabs of ionised dopants set up a potential difference V_0 across the depletion region, with the n side at higher potential than the p side. This is the built-in potential or barrier potential. It can be computed from the requirement that in equilibrium the Fermi level must be flat across the junction — a standard thermodynamic argument, the algebra of which is left to the going-deeper section. The result:

V_0 \;=\; \frac{k_B T}{e}\,\ln\!\left(\frac{N_a N_d}{n_i^2}\right). \tag{1}

Why: the built-in potential depends on the product of doping densities (heavier doping on both sides → larger built-in potential) divided by the squared intrinsic density (which depends on the band gap through n_i^2 \propto e^{-E_g/k_B T}). Larger band gap → smaller n_i → larger V_0. This is why a silicon diode has V_0 \approx 0.7\ \text{V} while a germanium diode (smaller E_g) has only \approx 0.3\ \text{V}, and a GaN LED (larger E_g) has about 3\ \text{V}.

Numerical check for silicon at 300 K: k_B T/e = 0.02585\ \text{V} (the thermal voltage), n_i = 1.5 \times 10^{16}\ \text{m}^{-3}. For symmetric doping N_a = N_d = 10^{22}\ \text{m}^{-3}:

V_0 \;=\; 0.02585\,\ln\!\left(\frac{10^{22} \times 10^{22}}{(1.5\times 10^{16})^2}\right) \;=\; 0.02585 \times \ln(4.4\times 10^{11}) \;=\; 0.02585 \times 26.8 \;\approx\; 0.69\ \text{V}.

Close to the widely-quoted 0.7 V for silicon. The exact value depends on doping, but the order of magnitude is stiff: you cannot easily build a silicon diode with V_0 very different from 0.7 V without changing the materials.

Width of the depletion region

The depletion region has non-zero width because the charged slabs have to add up to a total voltage drop of V_0. The width W can be found by solving Poisson's equation with the space charge \rho(x) = -eN_a on the p side and \rho(x) = +eN_d on the n side. For a step junction,

W \;=\; \sqrt{\frac{2\varepsilon_s V_0}{e}\left(\frac{1}{N_a} + \frac{1}{N_d}\right)}, \tag{2}

where \varepsilon_s = \varepsilon_r \varepsilon_0 is the permittivity of silicon (\varepsilon_r = 11.7, so \varepsilon_s \approx 1.04 \times 10^{-10}\ \text{F/m}).

Why: Poisson's equation relates charge density to the curvature of the potential. Integrate it twice across the junction; enforce that the potential matches V_0 across the full depletion region and that the total charge on the two sides is equal and opposite (global neutrality). The square root is the signature of a region whose size grows only slowly with voltage — doubling V_0 increases W by only \sqrt{2}.

For symmetric N_a = N_d = 10^{22}\ \text{m}^{-3}, V_0 = 0.7\ \text{V}:

W \;=\; \sqrt{\frac{2 \times 1.04 \times 10^{-10} \times 0.7}{1.6 \times 10^{-19}}\times \frac{2}{10^{22}}} \;=\; \sqrt{\frac{1.456 \times 10^{-10}}{1.6 \times 10^{-19}} \times 2 \times 10^{-22}} \;\approx\; \sqrt{1.82 \times 10^{-13}}\ \text{m}.
W \;\approx\; 4.3 \times 10^{-7}\ \text{m} \;=\; 0.43\ \mu\text{m}.

Why: a depletion region that is less than half a micrometre wide. Roughly five hundred atoms across — astonishingly thin, thinner than the wavelength of visible light. All the interesting physics of a diode is happening in this tiny slab.

Forward and reverse bias

Connect an external voltage source across the diode. The p side of the diode has a terminal called the anode, the n side has one called the cathode. Define the bias voltage V as the voltage at the anode minus the voltage at the cathode. Two cases:

The diode under forward and reverse biasTwo cross-section diagrams stacked. The top shows the diode under forward bias, with a battery plus terminal connected to the p side and minus to the n side. A narrow depletion region is shown, and arrows indicate electron and hole flow across the junction giving a forward current. The bottom shows the diode under reverse bias, with the battery reversed. A wider depletion region is shown, and the current is negligible. Forward bias (V > 0): current flows p n thin depletion + − holes electrons Reverse bias (V < 0): current ≈ 0 p n wide depletion − +
Top: under forward bias, the external battery (plus to p, minus to n) pushes holes and electrons toward the junction. The depletion region is thin, the barrier is small, and majority carriers cross in large numbers. Bottom: under reverse bias, the external battery (minus to p, plus to n) pulls carriers away from the junction. The depletion region widens, the barrier rises, and the current drops to a tiny reverse-saturation value.

A cartoon calculation of the forward current

Here is an intuitive derivation of the current–voltage relation, good enough to land on the correct functional form without the full semiconductor-transport machinery.

Step 1. In equilibrium, majority carriers on each side have a thermal population with a small fraction above the barrier V_0. The fraction of electrons in the n side that have enough kinetic energy to climb the barrier is given by the Boltzmann factor:

f_\text{n→p}(0) \;=\; \exp\!\left(-\frac{eV_0}{k_B T}\right).

Why: classical statistics — the fraction of particles with energy greater than eV_0 in a thermal bath at temperature T. This assumes non-degenerate statistics, valid whenever the semiconductor is not too heavily doped.

Step 2. These "lucky" electrons cross the depletion region and constitute a tiny diffusion current. Similarly, a tiny number of minority electrons on the p side diffuse "down" the barrier in the opposite direction. In equilibrium, these two currents exactly cancel — no net current, by definition.

Step 3. Apply forward bias V > 0. The barrier for electrons going from n to p is reduced from eV_0 to e(V_0 - V). The fraction that can now climb it is

f_\text{n→p}(V) \;=\; \exp\!\left(-\frac{e(V_0 - V)}{k_B T}\right) \;=\; \exp\!\left(-\frac{eV_0}{k_B T}\right)\exp\!\left(+\frac{eV}{k_B T}\right).

Why: the barrier is now smaller, so more electrons have enough energy to cross. The Boltzmann exponential factors cleanly into an equilibrium piece (the \exp(-eV_0/k_B T) factor) and a bias-dependent piece (\exp(eV/k_B T)).

Step 4. The reverse current — minority electrons from p going to n — is unaffected by the bias, because those electrons are falling down the barrier regardless of its height. They simply arrive at the depletion edge by diffusion and get swept across. Call their contribution I_s/2, an electron-only piece of the total reverse-saturation current.

Step 5. The net electron current is the forward flux minus the reverse flux:

I_n \;=\; \tfrac{1}{2}I_s\left(\exp\!\left(\frac{eV}{k_B T}\right) - 1\right).

Why: subtract the constant reverse-saturation flow from the bias-enhanced forward flow. The -1 in the parenthesis comes from the fact that at V = 0 the total current must vanish — forward and reverse are equal.

Holes do the mirror-image calculation and contribute an identical term. Adding them:

\boxed{\; I \;=\; I_s\left(\exp\!\left(\frac{eV}{k_B T}\right) - 1\right). \;} \tag{3}

This is the ideal Shockley diode equation. In real diodes the exponent often has an empirical ideality factor \eta in the denominator (so V/(\eta k_B T/e) with \eta between 1 and 2), accounting for recombination inside the depletion region. For a clean silicon junction, \eta \approx 1 at moderate currents.

The reverse saturation current I_s is tiny — for a silicon diode of area 1\ \text{mm}^2 at room temperature, I_s \sim 10^{-12}\ \text{A} (one picoampere). But it scales as \exp(-E_g/k_B T) and thus doubles every 10 K of temperature rise, which is why silicon diodes become noticeably leaky when the junction heats up.

The thermal voltage and the turn-on shortcut

The combination V_T = k_B T/e appears everywhere in diode equations. At T = 300\ \text{K}:

V_T \;=\; \frac{k_B T}{e} \;=\; \frac{1.38 \times 10^{-23} \times 300}{1.6 \times 10^{-19}} \;=\; 0.02585\ \text{V} \;\approx\; 25.85\ \text{mV}. \tag{4}

Because V_T is so small (\ll V_0), the Shockley equation predicts an extremely sharp turn-on. The current goes up by a factor of e \approx 2.718 for every 25.85\ \text{mV} of forward bias — or equivalently by a factor of 10 for every 60\ \text{mV}. In practice this means that below about V = 0.5\ \text{V}, a silicon diode passes nanoampere-level currents (essentially zero for practical purposes); between V = 0.6\ \text{V} and V = 0.7\ \text{V} the current rises through its useful milliampere range; and above V = 0.8\ \text{V} the diode is strongly conducting. The colloquial "silicon diode drops 0.7 V" is a rough statement of where this turn-on happens.

Explore the I-V curve

Interactive: I–V curve of a silicon diode A curve plotting diode current in milliamperes versus bias voltage in volts. The curve is nearly zero for reverse bias and for small forward bias, then rises steeply starting near zero point six volts. Readouts display the current voltage and the corresponding current, computed from the Shockley diode equation with saturation current one picoampere and thermal voltage 25.85 millivolts at 300 kelvin. V (V) I (mA) −1.0 −0.5 0 0.5 0.9 0 25 50 75 100 I_s = 1 pA, V_T = 25.85 mV (ideal, η = 1, T = 300 K) drag red dot to change bias voltage
Ideal silicon-diode I–V curve, $I = I_s(e^{V/V_T} - 1)$ with $I_s = 1$ pA and $V_T = 25.85$ mV. Drag the red point to change the bias voltage. In reverse bias the current sits pinned at essentially zero (on this scale — the actual value is a microscopic $-I_s \approx -1$ pA). In forward bias, nothing happens until about $V \approx 0.55$–$0.6$ V, then the current rises precipitously — an order of magnitude for every 60 mV. Above 0.8 V the diode is strongly conducting and the ideal equation starts to fail (series resistance of the neutral regions becomes important).

Two things are worth noticing as you drag. First, the reverse bias region looks like I = 0 because the scale is in milliamperes — the actual reverse current is -I_s = -1 pA, nine orders of magnitude smaller. Second, the forward-bias curve is essentially vertical above V \approx 0.7\ \text{V}. You cannot easily push a silicon diode much past this — the current would go astronomical before the voltage rises by another 50\ \text{mV}. This is why you always put a current-limiting resistor in series with a diode in a real circuit.

Worked examples

Example 1: Current through a silicon diode at 0.65 V

A silicon diode at room temperature (300 K) has a reverse saturation current I_s = 1.0\ \text{nA}. Compute the forward current when V = 0.65\ \text{V}. Assume the ideal Shockley equation with ideality factor \eta = 1.

Reading the current off the diode curve at V equals 0.65 voltsA plot of current versus voltage. A vertical dashed line at V equals 0.65 volts intersects the red I-V curve at a point I equals about 0.08 amperes. A horizontal dashed line connects that point to the current axis. V (V) I (A) 0 0.05 0.10 0 0.2 0.4 0.6 0.8 0.65 V, ≈ 0.077 A V = 0.65
The I–V curve of a silicon diode with $I_s = 1\ \text{nA}$. At $V = 0.65\ \text{V}$, the current is about 77 mA — well into the diode's strongly-conducting regime.

Step 1. Compute the thermal voltage at 300 K.

V_T \;=\; \frac{k_B T}{e} \;=\; \frac{1.38 \times 10^{-23} \times 300}{1.6 \times 10^{-19}} \;=\; 0.02585\ \text{V} \;=\; 25.85\ \text{mV}.

Why: V_T is the natural voltage unit of the diode equation. Every factor of e increase in current corresponds to a V_T increase in voltage.

Step 2. Plug into the Shockley equation.

I \;=\; I_s\left(\exp\!\left(\frac{V}{V_T}\right) - 1\right) \;=\; 1.0 \times 10^{-9}\left(\exp\!\left(\frac{0.65}{0.02585}\right) - 1\right).
I \;=\; 10^{-9}\left(\exp(25.14) - 1\right) \;=\; 10^{-9}\left(8.25 \times 10^{10} - 1\right) \;\approx\; 10^{-9} \times 8.25 \times 10^{10}.

Why: the exponential dominates completely — the -1 is utterly negligible. A bias of 25 thermal voltages (25 \times 25.85\ \text{mV} = 646\ \text{mV}) gives a current enhancement factor of e^{25} \approx 7.2 \times 10^{10}. A little more voltage (0.65 V instead of 0.646 V) pushes the factor a bit higher.

Step 3. Compute.

I \;=\; 8.25 \times 10^{10} \times 10^{-9}\ \text{A} \;=\; 8.25 \times 10^{1}\ \text{A} \;\times\ 10^{-3} \;=\; 82.5\ \text{mA}.

(More carefully: I = 0.0825\ \text{A} = 82.5\ \text{mA}.)

Step 4. Note the sensitivity.

If the voltage rose by a further V_T \approx 26 mV to V = 0.676 V, the current would multiply by e \approx 2.72 — jumping to 224 mA. If it fell by 26 mV to V = 0.624 V, the current would drop to 82.5/2.72 \approx 30 mA.

Result: At V = 0.65 V, the diode conducts 82.5 mA (of order 0.1 A — a useful current for driving a small LED or a transistor base). The current is exponentially sensitive to voltage — any circuit that feeds a diode a fixed voltage will see wildly varying currents with small changes in temperature or voltage.

What this shows: Real diode circuits therefore feed a fixed current (via a series resistor), not a fixed voltage. The diode's own voltage drop self-adjusts to sit around 0.6–0.8 V depending on how much current you push through. This is why you calculate LED series resistors using "assume the LED drops V_f, use Ohm's law for the resistor on the remaining voltage."

Example 2: LED series resistor for a torchlight

A red LED has a turn-on voltage of V_f = 2.0\ \text{V} when conducting I_f = 20\ \text{mA} (its rated operating current). You want to run it from a 3-cell AA battery pack delivering V_s = 4.5\ \text{V}. What value of series resistor R should you use? And what power does R dissipate?

LED driver circuit with series resistorA simple circuit loop. A battery labelled 4.5 volts on the left, a resistor labelled R in the middle, and an LED on the right, all in series. Arrows showing the direction of conventional current flow. Labels indicate the voltage across each element: 2.5 volts across the resistor and 2.0 volts across the LED. + − 4.5 V R 125 Ω, 2.5 V, 50 mW LED 2.0 V I = 20 mA
An LED driver circuit: 4.5 V battery, 125 Ω series resistor, and an LED with a 2 V forward drop. The series resistor drops the remaining $4.5 - 2.0 = 2.5$ V, fixing the current at the LED's design point of 20 mA.

Step 1. Apply Kirchhoff's voltage law around the loop.

V_s \;=\; I_f R + V_f \quad\Longrightarrow\quad R \;=\; \frac{V_s - V_f}{I_f}.

Why: every volt from the battery ends up either across the resistor or across the LED (no other element in the loop). Solve for R.

Step 2. Plug in numbers.

R \;=\; \frac{4.5 - 2.0}{0.020} \;=\; \frac{2.5}{0.020} \;=\; 125\ \Omega.

Why: the resistor drops the excess voltage the battery provides beyond the LED's 2 V, and limits the current to the LED's design value. Using a 120 Ω or 150 Ω resistor (standard E12 values) in practice gives 20.8 mA or 16.7 mA respectively — either is fine for an LED rated for 20 mA.

Step 3. Compute the power dissipated in R.

P_R \;=\; I_f^2 R \;=\; (0.020)^2 \times 125 \;=\; 0.05\ \text{W} \;=\; 50\ \text{mW}.

Why: any resistor carrying current dissipates heat. 50 mW is small — a quarter-watt resistor handles this comfortably. But if you forgot the resistor and connected the LED directly to the battery, it would try to conduct a current limited only by its internal series resistance (typically \sim 5\ \Omega), drawing roughly (4.5 - 2.0)/5 = 500\ \text{mA} — 25 times its rated current — and burn out within seconds.

Step 4. Efficiency check. The battery delivers V_s \cdot I_f = 4.5 \times 0.020 = 90\ \text{mW} total. The LED receives V_f \cdot I_f = 2.0 \times 0.020 = 40\ \text{mW} (some as light, some as heat). The resistor wastes 50\ \text{mW} as pure heat.

Result: A 125 Ω series resistor limits the LED current to 20 mA and wastes 50 mW as heat. The efficiency of this circuit is 44% (LED power over total power).

What this shows: This "drop voltage through a resistor" approach works but wastes more than half the battery energy as heat in the resistor. For battery-powered applications (torches, indicator lights), modern circuits use a switching regulator or a constant-current driver chip that achieves 80–90% efficiency instead. But for the simplest jugaad LED circuit — a button battery, a resistor, and a red LED — the Ohm's law method is what you use.

Common confusions

If you came here to design a diode circuit, equations (3), (4), and the turn-on rules-of-thumb are enough. What follows is the formal derivation of the built-in potential, the Shockley equation from minority-carrier diffusion, and a sketch of junction breakdown.

Built-in potential from the Fermi-level alignment

In equilibrium, the Fermi level E_F is flat across the junction — otherwise carriers would still be flowing. On the n side, E_F is close to the conduction band edge; on the p side, close to the valence band edge. When the two sides equilibrate, the bands must bend to line up.

For non-degenerate semiconductors, the carrier densities are given by Boltzmann expressions:

n \;=\; N_c\,\exp\!\left(-\frac{E_c - E_F}{k_B T}\right), \qquad p \;=\; N_v\,\exp\!\left(-\frac{E_F - E_v}{k_B T}\right),

where N_c, N_v are the effective densities of states in the conduction and valence bands, and E_c, E_v are the band edges (see semiconductors-intrinsic-and-extrinsic).

Deep in the n-type region, n = N_d (majority carrier density), so the distance of the Fermi level from the conduction band is

E_c^n - E_F \;=\; k_B T \ln(N_c/N_d).

Deep in the p-type region, p = N_a:

E_F - E_v^p \;=\; k_B T \ln(N_v/N_a).

The total band bending across the junction is eV_0 = E_c^p - E_c^n. Adding the two equations and using E_c - E_v = E_g (the band gap):

eV_0 \;=\; E_g - k_B T\ln(N_c N_v/N_a N_d).

Using n_i^2 = N_c N_v \exp(-E_g/k_B T) (from the law of mass action), this simplifies to

V_0 \;=\; \frac{k_B T}{e}\ln\!\left(\frac{N_a N_d}{n_i^2}\right),

which recovers equation (1). The derivation shows that V_0 is a thermodynamic equilibrium property — it has nothing to do with carriers "crossing the barrier." It is set by the Fermi-level alignment condition, period.

The Shockley equation from minority-carrier diffusion

The full derivation (Shockley, 1949) goes like this. Under forward bias V, the majority carriers on each side enter the depletion region and diffuse across to the other side. Because the depletion region is assumed narrow compared to the diffusion length, we can treat it as if carriers are "injected" at the depletion-region edges on each side.

The minority carrier density at the edge of the depletion region on the p side, call it n_p(-x_p), is enhanced above its equilibrium value n_{p,0} = n_i^2/N_a by a factor \exp(eV/k_B T):

n_p(-x_p) \;=\; n_{p,0}\exp(eV/k_B T).

Why: the Boltzmann factor across the now-reduced barrier height e(V_0 - V) gives this enhancement. Under forward bias, the depletion-edge minority carrier density is pushed above equilibrium — "injected" into the p-type region.

Deep in the p-type region the minority carrier density relaxes back to n_{p,0}. In between, the excess minority density obeys the steady-state diffusion equation with a recombination term:

D_n \frac{d^2\delta n}{dx^2} \;=\; \frac{\delta n}{\tau_n},

where D_n is the electron diffusion coefficient and \tau_n the electron recombination lifetime. Solve with the boundary conditions \delta n(-x_p) = n_{p,0}(e^{eV/k_B T} - 1) and \delta n(-\infty) = 0:

\delta n(x) \;=\; n_{p,0}(e^{eV/k_B T} - 1)\exp\!\left(\frac{x + x_p}{L_n}\right), \quad L_n \;=\; \sqrt{D_n \tau_n}.

The corresponding diffusion current at the depletion edge is

J_n \;=\; -eD_n \left.\frac{d\delta n}{dx}\right|_{x=-x_p} \;=\; -\frac{eD_n n_{p,0}}{L_n}(e^{eV/k_B T} - 1).

The hole side gives an identical contribution from p-injected holes. Adding the two and reversing the sign convention (current from p to n is positive):

J \;=\; \left(\frac{eD_n n_{p,0}}{L_n} + \frac{eD_p p_{n,0}}{L_p}\right)(e^{eV/k_B T} - 1) \;\equiv\; J_s(e^{eV/k_B T} - 1).

Multiply by the junction area to get the current:

I \;=\; I_s(e^{eV/k_B T} - 1), \qquad I_s \;=\; Ae\left(\frac{D_n n_i^2}{L_n N_a} + \frac{D_p n_i^2}{L_p N_d}\right).

The reverse saturation current depends on the intrinsic density squared — it inherits the \exp(-E_g/k_B T) dependence of n_i^2, explaining its sharp temperature sensitivity (doubles every 10 K for silicon).

Junction breakdown

At sufficiently large reverse bias (tens to hundreds of volts, depending on doping), the p-n junction breaks down and conducts heavily. Two mechanisms:

Zener breakdown (at high doping, low breakdown voltage \lesssim 5 V). The large electric field in the narrow depletion region (up to 10^8 V/m) directly tunnels valence electrons from the p side into the conduction band on the n side. This is quantum tunnelling, not an avalanche.

Avalanche breakdown (at moderate doping, breakdown voltage \gtrsim 8 V). The field accelerates stray carriers to energies large enough that their next collision with a lattice atom ionises it — freeing an electron-hole pair. The new carriers are accelerated and ionise more lattice atoms. The current multiplies by a factor that diverges at breakdown — hence "avalanche."

A Zener diode is a diode engineered to break down at a specific, sharply-defined reverse voltage, used as a voltage reference. A transient voltage suppressor is an avalanche diode designed to protect sensitive electronics from voltage spikes.

The quasi-Fermi levels under bias

Under bias, the single Fermi level splits into two quasi-Fermi levels E_{F,n} for electrons and E_{F,p} for holes. Their separation equals eV. This is a more general way of saying "forward bias raises the minority-carrier density by e^{eV/k_B T}." It is the framework used to treat non-equilibrium semiconductor devices (LEDs, lasers, solar cells), all of which depend on the quasi-Fermi-level separation to describe radiative recombination and photocurrent.

Real-world non-idealities

Real diodes deviate from the ideal Shockley equation in several ways:

  • Series resistance R_s. At high forward currents, the neutral (non-depleted) regions on each side have finite resistance. The I-V curve bends to the right — the voltage across the device is V + IR_s, not just V, so the effective slope flattens.
  • Recombination in the depletion region. At low currents, recombination inside the depletion region (not modelled by the diffusion equation above) adds an extra contribution with ideality factor \eta = 2. The full empirical form is I = I_s(e^{eV/\eta k_B T} - 1) with \eta varying between 1 and 2 depending on current.
  • Leakage. Surface states and impurities add a small ohmic leakage path. Real diodes have a small dI/dV slope in reverse bias rather than a truly constant I_s.
  • Temperature. I_s doubles every 10 K; V_T rises linearly with T. The net effect on the forward I–V is that V_f (the voltage at a fixed current) drops by about 2\ \text{mV/K} — a useful thermometer and a trap for circuits that rely on a fixed 0.7 V drop.

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