P-n junction

A p-n junction is formed where a region of semiconductor doped with acceptors (p-type, with holes as majority carriers) meets a region doped with donors (n-type, with electrons as majority carriers). Electrons diffuse from the n side into the p side and holes the other way, leaving behind a thin layer of fixed, ionized dopants: positive donors on the n side and negative acceptors on the p side. This depletion region, typically a fraction of a micrometer to a few micrometers wide, holds an electric field that opposes further diffusion, and the potential across it is the built-in voltage, about 0.7 V for a silicon junction doped at $10^{16}$ cm⁻³ on both sides. Most semiconductor light sources and detectors are built around one: LEDs and laser diodes operate it in forward bias, photodiodes in reverse or zero bias.

Built-in voltage

In equilibrium the Fermi level is flat through the device, so the conduction and valence bands bend across the depletion region by $qV_\text{bi}$. For non-degenerate doping $N_A$ and $N_D$,

$$V_\text{bi} = \frac{k_BT}{q}\,\ln\frac{N_A N_D}{n_i^2},$$

where $n_i$ is the intrinsic carrier density. For silicon at 300 K, with $k_BT/q = 25.85$ mV, $n_i = 10^{10}$ cm⁻³ and $N_A = N_D = 10^{16}$ cm⁻³, $V_\text{bi} = 0.02585 \times \ln(10^{12}) = 0.714$ V. Raising one side to $10^{18}$ cm⁻³ gives 0.833 V; the dependence on doping is only logarithmic.

Depletion width and capacitance

For an abrupt junction,

$$W = \sqrt{\frac{2\varepsilon\,(V_\text{bi} - V)}{q}\left(\frac{1}{N_A} + \frac{1}{N_D}\right)},$$

with $V$ the applied voltage, positive in forward bias. For the symmetric silicon example ($\varepsilon_r = 11.7$), $W = 0.43$ µm at zero bias, shared equally between the two sides, and the peak field is 33 kV/cm. The depletion region acts as a capacitor of $\varepsilon/W$ per unit area, 24 nF/cm² or 241 pF per mm² here. At 5 V of reverse bias the width grows to 1.2 µm and the capacitance falls in proportion. When one side is much more heavily doped, the depletion region lies almost entirely in the lighter side, which then sets $W$.

The diode equation

An applied forward voltage lowers the barrier, and minority carriers diffuse across it in numbers that grow exponentially with the voltage. The current is

$$I = I_0\left[\exp\!\left(\frac{qV}{n k_BT}\right) - 1\right],$$

where $I_0$ is the saturation current, proportional to junction area and to $n_i^2$, and $n$ is the ideality factor, 1 for diffusion current and near 2 for recombination in the depletion region. At 300 K with $n = 1$ the current rises tenfold for about every 60 mV of forward voltage. With $I_0 = 1$ pA, 0.60 V gives 12 mA and 0.66 V gives 122 mA. Under reverse bias the exponential vanishes and the current saturates at $-I_0$ plus generation current in the depletion region, which are the main sources of photodiode dark current, along with surface leakage and, at high fields, tunneling.

Forward bias: LEDs and lasers

Forward bias injects carriers: electrons into the p side and holes into the n side, where they recombine. In a direct-gap material a large fraction of that recombination is radiative, which is how a light-emitting diode works, and the forward voltage at operating current is close to the photon energy divided by $q$, plus drops across series resistance: the photon energy is 1.46 eV at 850 nm and 0.946 eV at 1310 nm. In a homojunction the injected carriers spread over a diffusion length of micrometers, which keeps their density low. Laser diodes and efficient LEDs therefore use heterojunctions: a narrow-gap active layer between wider-gap p and n claddings confines both carriers and light to a layer tens to hundreds of nanometers thick, often a set of quantum wells.

Reverse bias: photodiodes

A photon absorbed in or near the depletion region creates an electron-hole pair that the junction field separates, giving a photocurrent in the reverse direction. A photodiode can run at zero bias, as a solar cell does; reverse bias lowers the capacitance and speeds collection. The PIN photodiode inserts a lightly doped intrinsic layer between p and n so that the absorbing region is wide and fully depleted at a small bias; an avalanche photodiode adds a high-field region where carriers multiply.

Pitfalls

The abrupt-junction formulas assume uniform doping and a step at the interface; diffused and implanted junctions are graded, and $W$ then grows as roughly the cube root of the voltage instead of the square root. Series resistance flattens the measured I-V curve at high current and gives a spurious ideality factor if it is not removed.

Common questions

What is the depletion region in a p-n junction?

The layer on both sides of the junction from which mobile carriers have diffused away, leaving fixed ionized dopants and an electric field. In a silicon junction doped at $10^{16}$ cm⁻³ it is 0.43 µm wide at zero bias.

Why is the built-in voltage about 0.7 V in silicon?

It is the difference in Fermi level between the two sides before contact, which for moderate doping is a little over half the 1.12 eV bandgap. Because it depends on doping only through a logarithm, it stays between about 0.6 V and 0.95 V for silicon doped from $10^{15}$ to $10^{18}$ cm⁻³ on both sides.

What is the difference between a p-n and a PIN junction?

A PIN junction has an undoped or lightly doped layer between the p and n regions. The depletion region then extends across that whole layer, which makes it wider and better defined, giving lower capacitance and more efficient collection for photodiodes.

References: S. M. Sze and K. K. Ng, Physics of Semiconductor Devices, 3rd ed. (Wiley, 2007); B. E. A. Saleh and M. C. Teich, Fundamentals of Photonics, 3rd ed. (Wiley, 2019); W. Shockley, Bell Syst. Tech. J. 28, 435 (1949).