Semiconductors, Diodes and LEDs

Beginner

A copper wire carries charge easily because some of its electrons belong to no particular atom: it is a conductor. In glass every electron stays with its atom: an insulator. Silicon, the material of computer chips, sits in between: a semiconductor.

Step 1. In pure silicon almost every electron is held in a bond between neighbouring atoms. A little heat or light shakes a few loose, and each leaves a gap called a hole. When a neighbour jumps into the gap, the gap moves the other way, so a hole travels like a positive charge.

Step 2. Engineers mix in a pinch of another element; this is called doping. One kind of impurity brings spare electrons (n-type, n for negative); another kind leaves holes (p-type, p for positive).

Step 3. Put the two types side by side and you have a p–n junction. In the animation the blue dots on the left are spare electrons and the rings on the right are holes. A thin empty layer between them acts like a wall. While the voltage is low, nothing crosses; at about 2 volts both kinds of charge stream towards the middle.

Step 4. An electron that drops into a hole loses energy, and that energy leaves as a tiny flash of light, just as an atom gives out a photon when one of its electrons falls to a lower level. This is a light-emitting diode (LED). The size of the drop, and so the colour, is set by the material.

Check yourself: connect the battery the wrong way round. Does the LED light up? No: the wall gets thicker and almost no current flows. A junction is a one-way valve for current, called a diode. Shine light on it instead and it drives a current: a solar cell.

Intermediate

Bands, gaps and holes

A copper wire carries an electric current easily and glass hardly at all, while silicon, the material of computer chips, sits in between. An electron bound to a single atom can have only certain energies. In a crystal every atomic level spreads into a band of closely spaced levels. The bonding electrons fill the valence band; above it lies the conduction band, where electrons move freely, and between the two the band gap $E_g$, energies that no electron in a perfect crystal can have.

A full band carries no current, since each electron moving one way is matched by one moving the other way. A metal conducts because its highest occupied band is only partly filled. An insulator has a full valence band and a wide gap, 5.5 eV in diamond; a semiconductor has a narrow one: 1.12 eV in silicon, 1.42 eV in gallium arsenide (GaAs), 1.9 eV in the red LED of the animation.

At room temperature the thermal energy $k_{B}T\approx 0.026\mathrm{eV}$ lifts only about $10^{10}$ electrons per cubic centimetre across the gap of silicon, out of $5\times 10^{22}$ atoms. Each leaves an empty place in a bond, a hole; a neighbouring electron slips into it and the hole moves the other way, like a positive charge $+e$. The number of pairs doubles about every 10 K, so a pure semiconductor conducts better when hot, unlike a metal.

Doping: n-type and p-type

Doping controls the carriers. Phosphorus has five outer electrons, one more than the four bonds of silicon need; the spare one, held by only 0.045 eV, is freed by thermal motion at once. Phosphorus is a donor and makes n-type silicon, with electrons as majority carriers. Boron, with three, takes an electron from a neighbouring bond and leaves a hole: p-type silicon.

One phosphorus atom per five million silicon atoms gives $10^{16}$ free electrons per cubic centimetre, a million times the pure value. The holes become rarer, because so many electrons fill them, and the product of the two densities stays fixed:
$np=n_i^2,n_i\propto \exp \left(-\frac{E_g}{2k_{B}T}\right)$
Here $n_i\approx 10^{10}\mathrm{c}{\mathrm{m}}^{-3}$ is the density in pure silicon, so the holes drop to $p=10^{20}/10^{16}=10^4\mathrm{c}{\mathrm{m}}^{-3}$: they are the minority carriers.

The p–n junction and its band diagram

Join n-type and p-type regions in one crystal. Electrons diffuse into the p side and holes into the n side, where they recombine, leaving near the boundary only fixed ions: positive donors on the n side, negative acceptors on the p side. This depletion layer, a fraction of a micrometre thick, has no free carriers, and the field of its ions pushes carriers back until diffusion stops at a built-in voltage $V_{\mathrm{bi}}$: about 0.7 V in silicon, 1.8 V in the animation's LED.

The band diagram in the animation draws this as a step, with electron energy upward: the upper line is the conduction band edge, the lower one the valence band edge, 1.9 eV apart (gold arrow). Both edges are higher on the p side, so an electron (blue dot) from the n side must climb the step, while a hole (ring), which floats up like a bubble, would have to sink down it. Unbiased, the step is 1.8 eV high and almost nothing crosses.

Bias, current and light

Forward bias (p side to +) works against the built-in voltage: the step shrinks, the depletion layer narrows and the current grows exponentially. Reverse bias raises the step, and only a tiny saturation current $I_s$ leaks through. The junction is a diode, a one-way valve. Four formulas do most of the work:

• $hf\approx E_g$
• $\lambda =\frac{hc}{E_g}\approx \frac{1240\mathrm{eV}\cdot \mathrm{nm}}{E_g}$
• $I=I_s\left[\exp \left(\frac{eV}{\eta k_{B}T}\right)-1\right]$
• $R=\frac{V_s-V_F}{I}$

A conduction electron that falls into a hole drops by about $E_g$ and emits one photon; since $hc\approx 1240\mathrm{eV}\cdot \mathrm{nm}$, dividing by the gap in electronvolts gives the wavelength in nanometres. In the diode equation the thermal voltage $V_T=k_{B}T/e$ is 25.9 mV at 300 K, and the ideality factor $\eta$ lies between 1 and 2: each extra $2.3\eta V_T$ (60 mV for $\eta =1$, 120 mV for $\eta =2$) multiplies the current by ten.

Worked example: the LED in the animation

The chip in the animation has $E_g=1.9\mathrm{eV}$. Each photon carries about that energy, $1.9\cdot 1.602\times 10^{-19}=3.04\times 10^{-19}\mathrm{J}$, and its wavelength is
$\lambda \approx \frac{1240\mathrm{eV}\cdot \mathrm{nm}}{1.9\mathrm{eV}}\approx 653\mathrm{nm}$
which the animation rounds to 650 nm: red light.

As the supply voltage $V$ rises, the readout $I$ stays below half a milliampere up to about 1.5 V, while the caption says the barrier is still high; it passes 1 mA at about 1.56 V and reaches 20 mA at 2.0 V. Half a volt multiplies the current more than fiftyfold. Meanwhile the band step shrinks from 1.8 eV to about 0.1 eV (the chip's resistance takes the other 0.3 V), the depletion layer narrows to a quarter of its width, and the $e^{-}$ and $h^{+}$ arrows show carriers streaming in from both sides.

Each crossing electron gives at most one photon, so the current counts the photons: at 20 mA, $I/e=0.020/(1.602\times 10^{-19})\approx 1.25\times 10^{17}$ per second, carrying at most $1.25\times 10^{17}\cdot 3.04\times 10^{-19}\approx 0.038\mathrm{W}$ of light from $2.0\cdot 0.020=0.040\mathrm{W}$ of electrical power. Whatever does not escape as light warms the chip.

Worked example: a resistor for a blue LED

A blue LED has a wider gap, $E_g\approx 2.7\mathrm{eV}$ ($\lambda \approx 460\mathrm{nm}$), so it needs about 3.0 V rather than 2 V to carry 20 mA. You have a 5.0 V USB supply. A series resistor carries the same current and takes the rest of the voltage:
$R=\frac{V_s-V_F}{I}=\frac{5.0-3.0}{0.020}=100\Omega$
It dissipates $2.0\cdot 0.020=0.040\mathrm{W}$, and the LED $3.0\cdot 0.020=0.060\mathrm{W}$.

Without it, the LED's own curve, far too steep, would set the current: in the animation half a volt changes the current fiftyfold, so 5 V across a bare LED would destroy it. With the resistor, an LED whose forward voltage is 3.1 V instead of 3.0 V still gets $1.9/100=0.019\mathrm{A}$: the resistor, not the LED, fixes the current.

Common mistakes

• n-type silicon is not negatively charged: each phosphorus atom enters neutral, and when its fifth electron wanders off, a fixed positive ion stays behind.
• A brighter LED does not give bluer light. The current sets the number of photons per second, about $1.25\times 10^{17}$ at 20 mA; the gap sets the energy of each.
• A diode has no fixed resistance: the LED of the animation has $V/I=1.5/0.00037\approx 4\mathrm{k\Omega }$ at 1.5 V but $2.0/0.020=100\Omega$ at 2.0 V.
• Reverse bias does not block at any voltage: beyond breakdown the reverse current shoots up, and LED data sheets allow only about 5 V in reverse.

Where you meet it

LED lamps, screens and traffic lights use the light of this lesson. A solar cell is the same junction doing the opposite job: light creates electron–hole pairs, and the field of the depletion layer drives them apart as a current, as in camera sensors and fibre-optic receivers. Transistors, built from junctions or a field-controlled channel, switch and amplify the signals in every chip. Compare Photoelectric Effect and Atomic Models: the Bohr Atom for the photon side, and DC Circuits and Ohm's Law for the circuit side.

Advanced

Bands from hopping between atoms

Everything a semiconductor does comes from one fact: its electrons are not tied to single atoms. To see how that produces bands of allowed energies separated by a gap, follow one electron in the simplest crystal there is, a straight chain of identical atoms.

Legend

$a$	Spacing of the atoms in the chain.	$\mathrm{m}$
$A$	Hopping energy: the coupling between two neighbouring atoms.	$\mathrm{J}$
$E$	Energy of one wave state of the chain.	$\mathrm{J}$
$k$	Wave number of that state: its amplitude on the atom at $x_n$ is proportional to $e^{ik{x}_n}$.	${\mathrm{m}}^{-1}$
$E_0$	Energy level of an isolated atom; the band is centred on it.	$\mathrm{J}$
$4A$	Width of the band, from $E_0-2A$ at $k=0$ to $E_0+2A$ at the edge.	$\mathrm{J}$
$\pi /a$	Edge of the range of $k$ needed: a $k$ beyond it puts the same amplitudes on the atoms as one inside.	${\mathrm{m}}^{-1}$
$m^{\ast }$	Effective mass at the bottom of the band, set by the curvature of the green parabola there; at the top the curvature is reversed, giving $-m^{\ast }$.	$\mathrm{kg}$

The top of Figure 1 shows the chain. The atoms are a distance $a$ apart, so atom number $n$ sits at $x_n=na$. On its own, each atom would hold the electron at the energy $E_0$. Because the atoms are close, the electron can also hop to either neighbour; the curved double arrow marks this, and the hopping energy $A$ measures how strongly two neighbouring atoms are coupled.

Since the electron can always leak to a neighbour, a state with the electron sitting on one atom cannot keep a fixed energy. A state of definite energy must give the electron an amplitude $C_n$ on every atom. Its Schrödinger equation says: the energy times the amplitude on atom $n$ equals $E_0$ times that amplitude, minus $A$ times the amplitudes on the two neighbours,

$E{C}_n=E_0{C}_n-A(C_{n+1}+C_{n-1})$

Every atom has the same surroundings, so try a wave, $C_n=e^{ik{x}_n}$, with wave number $k$. The two neighbours then carry $e^{ik{x}_n}{e}^{ika}$ and $e^{ik{x}_n}{e}^{-ika}$. The common factor $e^{ik{x}_n}$ cancels from both sides, and since $e^{ika}+e^{-ika}=2\cos ka$, what is left is

$E=E_0-2A\cos ka$

This is the curve in the lower part of Figure 1: the energy $E$ of each wave against its wave number $k$. Only $k$ from $-\pi /a$ to $\pi /a$ is needed. Adding $2\pi /a$ to $k$ multiplies the amplitude on atom $n$ by $e^{i2\pi n}=1$, so it gives back the same wave. At $k=0$ every atom has the same amplitude, and the energy is lowest, $E_0-2A$: the bottom of the band. At $k=\pm \pi /a$ the sign of the amplitude flips from one atom to the next, and the energy is highest, $E_0+2A$. The single atomic level $E_0$, the dashed line halfway up, has spread into a band of width $4A$, the bracket on the right.

Each atomic level spreads into its own band in this way, and between two bands lie energies that no wave in the chain can have: a band gap. In a semiconductor at absolute zero the electrons exactly fill one band, the valence band, and leave the next one, the conduction band, empty. The distance from the top of the valence band to the bottom of the conduction band is the gap $E_g$, 1.12 eV in silicon.

Near the bottom of a band only small $k$ matter, and there $\cos ka\approx 1-\frac{1}{2}(ka{)}^2$. Putting this into the band gives

$E\approx E_0-2A+A{a}^2{k}^2$

Apart from the constant, this has the form of the kinetic energy of a free particle, ${\hslash }^2{k}^2/2m$, where $\hslash$ is the reduced Planck constant. So an electron near the bottom of the band moves through the crystal like a free particle, but with an effective mass

$m^{\ast }=\frac{{\hslash }^2}{2A{a}^2}$

The stronger the hopping, the lighter the electron. A coupling of 1 eV between atoms 0.3 nm apart gives $m^{\ast }\approx 0.42m_e$, where $m_e$ is the mass of a free electron; measured values run from $0.07m_e$ (electrons in GaAs) to about $m_e$. In Figure 1 the green parabola at the bottom, labelled $m^{\ast }$, is this approximation. It hugs the band near $k=0$ and parts from it farther out, where the cosine flattens.

At the top of the band the cosine curves the other way. The second green parabola, labelled $-m^{\ast }$, fits the band there: an electron near the top has a negative effective mass, so a force accelerates it backwards. Now take a band that is full except for one electron missing near its top. A full band carries no current, because every electron moving one way is matched by one moving the other way. The band with a gap in it therefore behaves like the missing electron with both its charge and its mass reversed: charge $+e$ instead of $-e$, mass $+m^{\ast }$ instead of $-m^{\ast }$. This missing electron is a hole.

The hole is a genuine quasiparticle, a particle-like excitation of the crystal: the Hall voltage of p-type material has the sign expected for positive carriers. Lifting an electron from the top of the valence band to the bottom of the conduction band makes one free electron and one hole, so $E_g$ is the least energy that creates an electron–hole pair.

Counting carriers: the law of mass action

A pure semiconductor at room temperature has few free carriers, all made by thermal vibrations. Adding impurities, called doping, changes their numbers by many orders of magnitude. One simple rule links the densities of electrons and holes in every case, and the rest of the lesson relies on it.

Write $n$ for the number of conduction electrons per unit volume and $p$ for the number of holes. Pairs are created when a thermal vibration lifts an electron across the gap. That costs the energy $E_g$, so the rate of creation is proportional to the Boltzmann factor $e^{-E_g/k_{B}T}$, with $k_B$ the Boltzmann constant and $T$ the absolute temperature; at 300 K, $k_{B}T=0.0259\mathrm{eV}$. Pairs disappear when an electron meets a hole and falls into it, which is called recombination. The rate of meetings is proportional to $np$, since doubling either density doubles the number of meetings. In equilibrium creation and recombination balance, so the product $np$ depends only on the gap and the temperature:

$np=n_i^2$

Here $n_i$, the intrinsic density, is the value of both $n$ and $p$ in a pure crystal, where every electron in the conduction band has left a hole behind it. Since $n_i^2\propto e^{-E_g/k_{B}T}$, its square root is

$n_i\propto \exp \left(-\frac{E_g}{2k_{B}T}\right)$

where a slower $T^{3/2}$ factor is left out. Doping does not change this argument. A donor atom, such as phosphorus in silicon, gives an electron to the conduction band; an acceptor, such as boron, takes one from the valence band and leaves a hole. Neither alters how fast pairs are created across the gap, nor the chance that a given electron meets a given hole, so $np=n_i^2$ still holds.

For silicon at 300 K, $n_i\approx 10^{10}\mathrm{c}{\mathrm{m}}^{-3}$. Add donors at a density $N_D=10^{16}\mathrm{c}{\mathrm{m}}^{-3}$, a million times $n_i$. Nearly every donor gives up its electron, so $n\approx N_D$, and the law of mass action fixes the holes at $p=n_i^2/N_D=10^4\mathrm{c}{\mathrm{m}}^{-3}$. Doping raises one density and lowers the other: in this n-type silicon the electrons are the majority carriers and the holes the minority carriers.

The exponent makes $n_i$ very sensitive. Warming silicon from 300 K to 310 K doubles it. A wider gap makes it far smaller: for the 1.9 eV gap of the animation's LED the exponent grows by $(1.9-1.12)/(2\cdot 0.0259)\approx 15$, so $n_i$ is multiplied by about $e^{-15}\approx 3\times 10^{-7}$, down to a few thousand carriers per cubic centimetre.

The p–n junction: built-in voltage and depletion layer

A p–n junction is one crystal doped n-type on one side and p-type on the other. Diodes, LEDs, solar cells and transistors are all built from such junctions. When the two sides meet, electrons diffuse from the n side into the p side, where they are rare, and holes diffuse the other way. Each carrier that crosses and recombines leaves behind a fixed ion whose charge is no longer cancelled: a positive donor on the n side, a negative acceptor on the p side. The charge of these ions sets up a field that pushes the carriers back, and diffusion stops when the field balances it. The strip around the junction that has lost its free carriers is the depletion layer.

Legend

$x$	Position across the junction, from the n side to the p side.	$\mathrm{m}$
$W$	Width of the depletion layer.	$\mathrm{m}$
$\rho$	Charge density of the fixed ions: positive donors on the n half of the layer, negative acceptors on the p half.	$C/{\mathrm{m}}^3$
$e{N}_D$	Charge density of the ionised donors: $e$ times their number density $N_D$.	$C/{\mathrm{m}}^3$
$e{N}_A$	Size of the charge density of the ionised acceptors: $e$ times their number density $N_A$, equal to $e{N}_D$ here.	$C/{\mathrm{m}}^3$
$E$	Electric field, pointing from the n side to the p side.	$V/m$
$E_{\max }$	Peak of the field, at the junction.	$V/m$
$V$	Electric potential along the junction.	$\mathrm{V}$
$V_{bi}$	Built-in voltage: the drop of $V$ across the layer, equal to the shaded area under $E$.	$\mathrm{V}$
$n$	The n-type region: free electrons, with positive donor ions.
$p$	The p-type region: holes, with negative acceptor ions.

Figure 2 follows the junction along the position $x$, from the n side on the left to the p side on the right, in three graphs that share the same $x$. The depletion layer, of width $W$, lies between the dashed lines.

The top graph is the charge density $\rho$: two blocks. The n half of the layer holds the donor ions, of charge density $e{N}_D$, where $N_D$ is the number of donors per unit volume. The p half holds the acceptor ions, of charge density $-e{N}_A$. Here the doping is the same on both sides, $N_A=N_D$, as in the silicon example below, so the two halves are equal, each $W/2$ wide. Outside the layer the free carriers cancel the ions, and $\rho =0$.

The middle graph is the field $E$. Gauss's law in one dimension, $dE/dx=\rho /\epsilon$ with $\epsilon$ the permittivity of the crystal, says that the slope of the field is the charge density divided by $\epsilon$. A constant $\rho$ therefore gives a straight line. $E$ is zero outside the layer, rises steadily across the positive half, peaks at $E_{\max }$ at the junction and falls back to zero across the negative half. The result is a triangle, and the field points from n to p all the way across, as it must to push electrons back to the n side and holes back to the p side.

The bottom graph is the potential $V$. The potential is minus the integral of the field, $dV/dx=-E$, so $V$ is flat wherever $E=0$ and falls across the layer by the area under the field triangle, $\frac{1}{2}{E}_{\max }W$. That area is shaded in the middle graph, and the drop it produces is bracketed at the right of the bottom graph: it is the built-in voltage $V_{\mathrm{bi}}$.

The size of $V_{\mathrm{bi}}$ follows from equilibrium. To reach the p side, an electron on the n side must climb a potential-energy step $e{V}_{\mathrm{bi}}$. In equilibrium the electrons obey the Boltzmann factor across that step, so their density on the p side equals their density on the n side times $e^{-e{V}_{\mathrm{bi}}/k_{B}T}$. On the n side the density is $N_D$; on the p side, by the law of mass action, it is $n_i^2/N_A$. Setting $n_i^2/N_A=N_D{e}^{-e{V}_{\mathrm{bi}}/k_{B}T}$ and taking logarithms,

$V_{\mathrm{bi}}=\frac{k_{B}T}{e}\ln \left(\frac{N_A{N}_D}{n_i^2}\right)$

For silicon with $N_A=N_D=10^{16}\mathrm{c}{\mathrm{m}}^{-3}$ and $k_{B}T/e=0.02585\mathrm{V}$, this is $0.02585\cdot \ln (10^{12})\approx 0.71\mathrm{V}$. Since $n_i^2$ carries the factor $e^{-E_g/k_{B}T}$, $e{V}_{\mathrm{bi}}$ equals $E_g$ minus a term that shrinks only slowly as the doping rises, so the step approaches $E_g/e$ but cannot much exceed it. The heavily doped LED of the animation comes close, with $V_{\mathrm{bi}}=1.8\mathrm{V}$ against $E_g/e=1.9\mathrm{V}$.

The width of the layer follows from the same three graphs. With equal doping $N$ on both sides, each half has width $W/2$. Across that distance the field rises with slope $eN/\epsilon$, so its peak is $E_{\max }=eN(W/2)/\epsilon$. The voltage across the layer is the area of the triangle:

$\frac{1}{2}{E}_{\max }W=\frac{eN{W}^2}{4\epsilon }$

At zero bias this voltage is $V_{\mathrm{bi}}$. A forward bias $V$, which makes the p side more positive, opposes the built-in field and leaves $V_{\mathrm{bi}}-V$ across the layer; a reverse bias is a negative $V$. The voltage across the layer is proportional to $W^2$, so

$W\propto \sqrt{V_{\mathrm{bi}}-V}$

For the silicon junction above, with $\epsilon \approx 1.0\times 10^{-10}F/m$, the layer is $W\approx 0.43\mu \mathrm{m}$ wide at zero bias, and the peak field $E_{\max }$ is near $3\times 10^{6}V/m$. A reverse bias of 5 V widens it to $1.2\mu \mathrm{m}$. In the animation the forward bias lowers the barrier from 1.8 eV to 0.1 eV, so the layer shrinks to $\sqrt{0.1/1.8}\approx 0.24$ of its width.

A junction with a built-in voltage looks like a battery, yet a wire joining its two ends carries no current. The wire makes two new contacts, metal to n and metal to p, each with its own potential step, and at a single temperature the steps around the loop add up to zero. If they did not, the loop would turn heat from surroundings at one temperature into work for ever, which the second law of thermodynamics forbids. Like a ratchet and pawl in a gas at one temperature, a junction cannot rectify its own thermal noise. Heat one contact and the balance breaks: that is the thermoelectric effect used in a thermocouple.

Rectification: two streams out of balance

A diode passes current easily in one direction and hardly at all in the other. To see why, follow the holes that cross the junction in each direction; the electrons behave in the same way and add to the same current.

Legend

$U$	Potential energy of a hole at each position across the junction.	$\mathrm{J}$
$n$	The n side, where holes are rare (minority carriers).
$p$	The p side, where holes are plentiful (majority carriers).
$V$	Forward bias: the applied voltage, positive on the p side.	$\mathrm{V}$
$e{V}_{bi}$	Height of the step for holes at zero bias: $e$ times the built-in voltage.	$\mathrm{J}$
$eV$	How far the forward bias raises the level of the p side.	$\mathrm{J}$
$e(V_{bi}-V)$	Height of the step under forward bias.	$\mathrm{J}$
$I_s$	Saturation current: the stream of minority holes sliding down the step; at zero bias the climbing stream equals it.	$\mathrm{A}$
$I_s{e}^{eV/k_{B}T}$	The climbing stream under forward bias: $e^{eV/k_{B}T}$ times larger than $I_s$.	$\mathrm{A}$

Figure 3 draws the potential energy $U$ of a hole at each point across the junction, n side on the left and p side on the right; the rings are holes. The field of the depletion layer pushes holes towards the p side, so a hole has more potential energy on the n side: higher by $e{V}_{\mathrm{bi}}$ when no voltage is applied (top panel). A hole going from p to n must climb this step; one going from n to p slides down it.

Two streams of holes therefore cross the junction. The first climbs. Holes are plentiful on the p side, where they are the majority carriers, but only the few with thermal energy above the step get over it. The second slides down. Holes are rare on the n side, where they are minority carriers made thermally, but any that wander into the layer are swept down the step by its field. This downhill stream depends only on how many minority holes reach the layer, not on how high the step is; it is the saturation current $I_s$. At $V=0$ there is no net current, so the climbing stream must equal it, and the top panel shows one arrow labelled $I_s$ each way.

In the bottom panel a forward bias $V$ raises the p side by $eV$; the dashed line beneath it marks its old level. The step falls to $e(V_{\mathrm{bi}}-V)$, and the layer narrows to match. The number of holes energetic enough to climb grows by the Boltzmann factor $e^{eV/k_{B}T}$, so the climbing stream becomes $I_s{e}^{eV/k_{B}T}$, while the downhill stream is still $I_s$. The figure draws the larger stream as three arrows instead of one; the true factor is vastly larger, about $6\times 10^{11}$ at 0.7 V in silicon, which is why the figure is not to scale.

In symbols, the climbing stream is proportional to $p_p{e}^{-e(V_{\mathrm{bi}}-V)/k_{B}T}$, where $p_p$ is the density of holes on the p side. A forward bias multiplies it by $e^{eV/k_{B}T}$, while the downhill stream keeps its value $I_s$. The net hole current is the difference, $I_s{e}^{eV/k_{B}T}-I_s$. The electrons give two streams of the same form, so the total current obeys the diode equation of William Shockley:

$I=I_s\left[\exp \left(\frac{eV}{\eta k_{B}T}\right)-1\right]$

Here the ideality factor is $\eta =1$, for this picture in which carriers diffuse across the layer and recombine beyond it. A reverse bias, $V<0$, raises the step instead: the exponential dies away and $I\to -I_s$, only the small downhill stream. That is rectification. A forward bias of a few tenths of a volt multiplies the current many times over, while a reverse bias of any size, up to breakdown, lets through only $I_s$.

When most carriers instead recombine inside the depletion layer, the current grows more slowly. Inside the layer $np=n_i^2{e}^{eV/k_{B}T}$, and recombination is fastest where the two densities are equal, both about $n_i{e}^{eV/2k_{B}T}$. The current then grows only as $e^{eV/2k_{B}T}$, which means $\eta \approx 2$, the value used in the animation. For the animation's LED, $I_s$ is about $10^{-16}\mathrm{A}$.

The gap, more than anything else, sets the voltage at which a diode turns on. For diffusion $I_s\propto n_i^2\propto e^{-E_g/k_{B}T}$, and for recombination $I_s\propto n_i\propto e^{-E_g/2k_{B}T}$. Either way, once the current is well above $I_s$,

$I\approx C{e}^{-(E_g-eV)/\eta k_{B}T}$

with a prefactor $C$ that changes only slowly with doping and temperature. The current is set by how far $eV$ lies below $E_g$, so the gap, far more than the doping, decides where a diode turns on.

Reading the live current–voltage curve

The expert layer of the animation plots the LED's current $I$ against the supply voltage $V$ as the voltage rises. The results above explain the shape of that curve, and its numbers are a good check of them.

Write $V_T=k_{B}T/e=25.85\mathrm{mV}$ at 300 K. With $\eta =2$, $\eta V_T=2\cdot 25.85=51.7\mathrm{mV}$. The current grows tenfold each time the exponent grows by $\ln 10\approx 2.3$, that is, every $2.3\cdot 51.7\approx 119\mathrm{mV}$. So the curve hugs the axis up to about 1.5 V (0.37 mA), bends sharply between 1.6 and 1.8 V, and reaches 20 mA at 2.0 V. The dashed gold line marks $E_g/e=1.9\mathrm{V}$.

Why does the junction work about 0.2 V below $E_g/e$? For this LED $C\approx 0.96\mathrm{A}$. A current of 20 mA lies ${\log }_{10}(0.96/0.020)\approx 1.7$ decades below $C$, so by the formula above the junction voltage $V_j$ sits $1.7\cdot 119\approx 200\mathrm{mV}$ below $E_g/e$, at 1.70 V. A silicon diode, with $\eta \approx 1$, has $C$ about eight decades above its working current, so its knee falls near 0.6–0.7 V, about half a volt under its 1.12 V gap.

The supply must also push the current through the chip's series resistance of 15 Ω. At 20 mA that takes $0.020\cdot 15=0.30\mathrm{V}$, the rest of the 2.0 V. This resistance shapes the top of the curve. The junction's own dynamic resistance, $d{V}_j/dI=\eta V_T/I$, is only about $2.6\Omega$ at 20 mA, much less than 15 Ω, so the curve straightens into a line of slope about $1/(17.6\Omega )$. On the band diagram the barrier label follows $e(V_{\mathrm{bi}}-V_j)$, which falls from 1.8 eV to 0.10 eV as the voltage rises.

Worked problem: the LED as a solar cell

Run the junction backwards: instead of putting a current in to get light out, shine light in and take power out. The same equations then describe a solar cell, and the animation's LED shows how much power one junction can deliver and why its voltage stays below $E_g/e$.

Legend

$V$	Voltage across the junction, positive in the forward direction.	$\mathrm{V}$
$I$	Current through the lit junction, $I=I_d-I_L$.	$\mathrm{A}$
$I_d$	Current of the junction in the dark: the diode equation.	$\mathrm{A}$
$I_L$	Photocurrent: the current the light drives backwards through the junction, which shifts the curve down.	$\mathrm{A}$
$V_{oc}$	Open-circuit voltage: where the lit curve crosses $I=0$.	$\mathrm{V}$
$V_{mp}$	Voltage at the maximum-power point, the corner of the largest rectangle.	$\mathrm{V}$
$I_{mp}$	Current delivered at the maximum-power point; the lit curve is at $-I_{\mathrm{mp}}$ there.	$\mathrm{A}$
$P_{\max }$	Greatest power the cell delivers: the area of the shaded rectangle.	$\mathrm{W}$
$E_g/e$	The band gap divided by $e$: the voltage that $V_{\mathrm{oc}}$ approaches but stays below.	$\mathrm{V}$

Figure 4 plots the current $I$ through the junction against the voltage $V$ across it, both positive in the forward direction. The dashed curve $I_d$ is the junction in the dark, given by the diode equation, with its knee just below $E_g/e$, the dashed vertical line.

Now light falls on the junction and creates electron–hole pairs at a rate equivalent to $I_L=20\mathrm{mA}$, that is, $I_L/e$ pairs per second. The field of the depletion layer sweeps each pair apart, the electron to the n side and the hole to the p side. That is a current through the junction in the reverse direction, and its size does not depend on $V$. The lit curve $I$ is therefore the dark curve moved down by $I_L$, as the arrow labelled $I_L$ shows.

The lit curve cuts the $I$ axis at $-I_L$ and the $V$ axis at the open-circuit voltage $V_{\mathrm{oc}}$. Between these two points $V$ is positive and $I$ negative: the current leaves by the positive terminal, as from a battery, so the junction delivers power. At any point of that stretch the power is the area of the rectangle between the axes and the point on the curve. The shaded rectangle, with its corner at $V_{\mathrm{mp}}$ and $-I_{\mathrm{mp}}$, is the largest, and its area is $P_{\max }$.

In equations, the lit current is

$I=I_s\left[\exp \left(\frac{eV}{\eta k_{B}T}\right)-1\right]-I_L$

With the terminals shorted, $V=0$ and $I=-I_L=-20\mathrm{mA}$: the whole photocurrent flows round the wire, backwards through the diode. With the terminals open, $I=0$, so the diode current must equal $I_L$, which gives

$V_{\mathrm{oc}}=\eta V_T\ln \left(1+\frac{I_L}{I_s}\right)=1.70\mathrm{V}$

This is exactly the junction voltage at 20 mA in the animation, because in both cases the diode current is 20 mA.

The power delivered is $P=V(I_L-I_d)$, where $I_d$ is the dark current at the voltage $V$. It is zero at both ends of the stretch and largest in between, where $dP/dV=0$: the corner of the largest rectangle in Figure 4. Solving numerically gives $V_{\mathrm{mp}}\approx 1.52\mathrm{V}$ and $I_{\mathrm{mp}}=I_L-I_d\approx 19.3\mathrm{mA}$, so $P_{\max }\approx 29.5\mathrm{mW}$. The fill factor compares this rectangle with the box of area $V_{\mathrm{oc}}{I}_L$ around the curve: $29.5/(1.70\cdot 20)\approx 0.87$.

Each pair absorbed at least 1.9 eV but delivers only 1.52 eV: the voltage stays below $E_g/e$ by the same logarithm that put the knee below it. With the 15 Ω series resistance included, the best output falls to about 24 mW. Real solar cells use gaps near 1.1–1.4 eV, a compromise: a smaller gap absorbs more of sunlight, a larger one delivers more energy per electron.

Limits of validity

The diode equation assumes low injection: far fewer injected carriers than majority ones. So the junction voltage must stay well below $V_{\mathrm{bi}}$; the animation's 1.70 V against 1.8 V is at the edge, where the series resistance already dominates. It assumes Boltzmann statistics, which fail above roughly $10^{19}\mathrm{c}{\mathrm{m}}^{-3}$ of doping in silicon. It treats the densities as smooth averages, yet a region 10 nm across doped at $10^{18}\mathrm{c}{\mathrm{m}}^{-3}$ holds a single dopant atom on average.

In reverse, $I=-I_s$ holds only up to breakdown. In heavily doped junctions the barrier is so thin that electrons tunnel through it at a few volts (Zener breakdown); in lightly doped ones, carriers accelerated by the field knock out new pairs in an avalanche at tens or hundreds of volts.

Light emission has its own limit. In silicon the bottom of the conduction band and the top of the valence band lie at different crystal momenta $\hslash k$, which differ by about $1.0\times 10^{-24}\mathrm{kg}\cdot m/s$: some 1700 times the momentum of a 1.12 eV photon. The photon cannot carry the difference, so each recombination must also create or absorb a phonon, a quantum of lattice vibration. That takes up to milliseconds, other processes win, and the energy ends as heat. In direct-gap materials, such as GaAs, AlGaInP for red and InGaN for blue, both band edges lie at the same $k$, and electrons and holes recombine by emitting light within nanoseconds.

Where this leads

Two junctions back to back, n–p–n, make a bipolar transistor: a small current into the thin middle layer controls a much larger one through it. In a field-effect transistor a gate voltage opens or closes a conducting channel instead. With light in and power out, a junction is the solar cell of the worked problem; with mirrors and strong injection, stimulated emission turns it into a laser. Zener and tunnel diodes rely on tunnelling through a thin depletion layer, the subject of Schrödinger's Equation and Tunnelling, and the bands come from electron waves interfering on a lattice, as in Wave–Particle Duality. The Boltzmann factors are those of Kinetic Theory of Gases, and the junction that drives no current is The Second Law of Thermodynamics and Entropy at work.