The glow that carries a voltage

The gas that nobody counted is a box of thermal radiation, and its defining property is that the number of photons in it is not fixed. The walls absorb photons and emit new ones without restriction, so the number settles wherever the free energy is least. The same argument that keeps a reaction from going all the way then says something about the photons’ chemical potential: a quantity that nothing holds fixed must have a chemical potential of zero at equilibrium. That is why Planck’s law, $1/(e^{h\nu/kT} - 1)$ photons per mode, has no chemical potential in it.

A light-emitting diode is a body in which something does push on the number of photons. Under a forward voltage, electrons and holes are driven into the same region of a crystal in numbers far above those the temperature alone would sustain, and they recombine to make light. The light that results is in equilibrium not with the lattice alone but with those excess electrons and holes, and its occupation of each mode is

$$\frac{1}{e^{(h\nu - qV)/kT} - 1},$$

with $T$ the temperature of the crystal and $V$ the voltage across the junction. It is Planck’s law with the voltage appearing as the photons’ chemical potential. Peter Würfel wrote it down in this form in 1982, and every property of an LED that looks like a paradox follows from taking it literally.

A voltage multiplies the glow without changing its slope #

An ideal semiconductor absorbs every photon above its band gap and none below, and a body that absorbs must emit. With no voltage across it, it glows at its own temperature, and above the gap that glow is the tail of a blackbody spectrum — for a gallium arsenide gap of 1.42 electronvolts at 300 K, a few thousand photons per square metre per second per electronvolt. Nothing a camera could detect.

Put a voltage across it and every point of that spectrum rises by the same factor, $e^{qV/kT}$. At one volt that is 16.8 powers of ten; at 1.3 volts, 21.8. The thermal glow at the bottom of the figure and the working LED at the top are the same curve, shifted.

What the voltage does not change is the slope. Above the gap, every spectrum in the figure falls by a factor of $e$ for each 25.9 millielectronvolts of photon energy, which is $kT$ for the lattice at room temperature. The light of a working LED has the spectral shape of a room-temperature body and the brightness of something vastly hotter. That combination is impossible for any single temperature, and it is exactly what a chemical potential produces: $\mu$ shifts the occupation of every mode by the same factor, while $T$ sets how fast the occupation falls from one mode to the next.

As bright as a surface thousands of kelvin hot #

The occupation of a mode is the most direct measure of how bright light is at that frequency — it is the quantity no lens can increase — and it gives a way to say how hot an LED’s light looks.

With 1.3 volts across it, the diode’s light just above the gap has the occupation of a blackbody at 2,571 K — about the temperature of a domestic lamp filament. At every other photon energy the match fails, in the direction the slopes predict: the diode’s occupation falls with the lattice’s $kT$ of 25.9 millielectronvolts and the blackbody’s with 222. A thermometer at that single photon energy would read 2,571 K; one at a slightly higher energy, less; and the general answer is the brightness temperature $T\,h\nu/(h\nu - qV)$, which is different at every frequency.

That formula also shows where the description ends. As the voltage approaches the photon energy, the brightness temperature grows without limit. Past it, the occupation formula turns negative, which is not a meaningless result but the statement that stimulated emission now outruns absorption — the condition for gain, and for a laser. The brightness temperature has passed through infinity to negative values, which is the same passage a temperature makes when a population is pushed into its upper states. A laser diode is an LED driven past the point where its light’s chemical potential exceeds its photons’ energy.

Where the voltage comes from #

That the voltage should appear in the light as a chemical potential is not a coincidence of notation. It is an equilibrium condition, and it is the same condition that fixes where a chemical reaction stops.

In a biased semiconductor the electrons in the conduction band come to equilibrium among themselves in picoseconds, far faster than they recombine, and the same is true of the holes in the valence band. Each population therefore has a temperature, the lattice’s, and a chemical potential of its own — its quasi-Fermi level. The two are separated by $qV$: that separation is what the voltage across a junction is, measured from inside.

Recombination is then a reaction, electron plus hole goes to photon, and a reaction at equilibrium has its chemical potentials balanced: $\mu_e + \mu_h = \mu_\gamma$. With the electrons’ and holes’ potentials separated by $qV$, the photons’ chemical potential is $qV$. When no voltage is applied the quasi-Fermi levels merge, $\mu_e + \mu_h = 0$, and the light returns to ordinary thermal radiation with no chemical potential, as it must.

A solar cell runs the same relation backwards. Sunlight separates the quasi-Fermi levels, the cell’s own emission rises until its chemical potential matches that separation, and the voltage across an open-circuited cell is the chemical potential of the light it is emitting back. That is why the work a diluted beam of sunlight can do is limited by the cell’s own glow, and why concentrating the light raises the voltage by the logarithm of the concentration.

Concentrating light past the limit a lens obeys #

The chemical potential of light also explains a device that seems to break a theorem. No arrangement of lenses and mirrors can make light brighter than its source: the occupation per mode is conserved along every ray, so the concentration of sunlight is bounded by how small the Sun looks. Yet a flat sheet of plastic doped with a fluorescent dye, lit from above by diffuse daylight, delivers light out of its edges at a brightness no passive optic could produce from the same light.

The resolution is that the dye is not passive. Each molecule absorbs a photon and, after its excited state has shed a little energy into vibrations, emits another at a slightly lower frequency. The emitted light is in equilibrium with the dye’s excited population, not with the incoming sunlight, and it carries a chemical potential set by how strongly that population is pumped. A chemical potential raises the occupation of every mode by the same factor, and a small loss of photon energy — a few $kT$ per photon, given up as heat — buys a large increase in the number of modes that can be crowded, because each $kT$ of downshift pays for a factor of $e$ in the ratio.

Eli Yablonovitch set out the thermodynamic limit of such luminescent concentrators in 1980, and it has exactly this form: the concentration can exceed the passive limit by a factor exponential in the frequency shift divided by $kT$. The second law is satisfied the same way it is in the diode, with the entropy of the lost energy paying for the order of the concentrated light. What looked like a violation of a brightness theorem is the brightness theorem applied to light whose chemical potential was never zero.

More light out than power in #

The electrical power a diode draws is its current times its voltage. The light power it emits is the number of photons times the mean photon energy. If every electron becomes an escaping photon, the ratio of the two is simply the mean photon energy divided by $qV$ — and the mean photon energy is fixed by the gap and the temperature, not by the voltage.

For a 0.58 electronvolt gap at 408 K, each photon carries 0.619 electronvolts on average. At a voltage of 0.15 volts each electron is given 0.15 electronvolts and becomes a photon of 0.619, and the diode emits 4.1 times the electrical power it consumes. The ratio exceeds one at every voltage below the photon energy. It is largest where the light is faintest, because the output rises exponentially with voltage while the gain falls only as its reciprocal.

The measurement exists. In 2012 a mid-infrared diode with a gap of about this size, heated to 135 °C to raise the thermal contribution, was reported to emit 69 picowatts of light while consuming about 30 picowatts of electrical power — an efficiency of 230 per cent. The light is faint by any practical standard. The principle is not in doubt, and it violates nothing, because the energy has to come from somewhere and the figure below shows where.

Real diodes lose current that does not become light: electrons and holes that recombine without emitting, and photons that are emitted but trapped inside the crystal by total internal reflection at its surface. A device that turns half its current into escaping light crosses an efficiency of one at 0.31 volts, and one that manages 90 per cent stays above one across the whole range drawn. Commercial LEDs lose far more than that at low currents, which is why the effect had to be sought with care.

A heat pump whose working fluid is light #

If a photon leaves with more energy than its electron was given, the difference is taken from the crystal. The diode is cooling itself, and the ledger per photon says by how much.

The photon’s energy is the same at every voltage — 0.619 electronvolts — because the spectrum’s shape belongs to the lattice. What changes is how much electrical energy is spent per photon, which is $qV$ divided by the fraction of the current that becomes light. At 0.15 volts the crystal supplies 405 millielectronvolts of heat per photon. At 0.42 volts it supplies 19. At 0.46 volts the electrical energy has overtaken the photon’s, and the crystal absorbs 38 millielectronvolts per photon as heat, which is the ordinary behaviour of every lamp.

Below the crossover the diode is a heat pump. It draws heat from its lattice and dumps it into the room as light, and the second law is satisfied because the light carries entropy away with it. A photon emitted with chemical potential $qV$ from a body at temperature $T$ takes away an entropy of at least $(h\nu - qV)/T$ — exactly the entropy of the heat it drew from the lattice — so the total never falls, and the voltage marks how much of the photon’s energy arrived as ordered electrical work rather than as heat. The same accounting governs a refrigerator driven by heat rather than work, and it governs laser cooling of solids by anti-Stokes fluorescence, in which a crystal doped with suitable ions absorbs light just below a transition and re-emits it at a higher mean energy. That route has cooled crystals to below a hundred kelvin. The diode’s version has not yet produced net cooling of an emitting device at a useful power.

A voltage that makes a surface look colder #

Nothing in the generalised Planck law requires the voltage to be positive. Reverse the bias, and the quasi-Fermi levels separate the other way: electrons and holes are drawn out of the junction, fewer than the temperature alone would provide, and the light’s chemical potential is negative.

For a gap of 0.25 electronvolts, in the mid-infrared near five micrometres where thermal cameras work, the effect is large at ordinary voltages. With 0.1 volts of reverse bias the device reads 220 K to a camera at the band edge; at 0.3 volts, 144 K. Its lattice stays at room temperature throughout. It is emitting less than its own thermal glow — negative luminescence — and a camera, which measures brightness and infers temperature, reports a cold object.

This is the most direct demonstration that a glow says nothing about the surface’s temperature unless the light’s chemical potential is known to be zero. The effect has been developed for switchable cold references in infrared instruments, because a surface that can be made to look forty kelvin colder with a voltage, and back again in microseconds, is easier to build than a surface that is forty kelvin colder. In 2019 the same physics was used to cool something else: a reverse-biased photodiode held tens of nanometres from a small calorimeter, close enough for light to tunnel across the gap, drew heat out of the calorimeter by radiating less at it than it received.

What the ideal diode assumes #

Every photon above the gap is absorbed and none below. Real absorption rises gradually through the gap, with a tail of weak absorption below it, and the emission follows the absorption exactly — so the real spectrum is the ideal one multiplied by the absorptivity, and its low-energy edge is soft rather than a cliff. The multiplication by $e^{qV/kT}$ survives unchanged.

Electrons and holes are in equilibrium within their bands. That needs their scattering to be much faster than their recombination, which is true by orders of magnitude in ordinary semiconductors at ordinary injection, and fails in devices driven so hard that carriers are hotter than the lattice.

The voltage in the formula is the separation of the quasi-Fermi levels in the emitting region. The voltage at the terminals is larger, by the drop across the contacts and the bulk material, and every loss of that kind is electrical energy that becomes heat without becoming light. At the low currents where the gain exceeds one those drops are small; at the high currents where LEDs are used they dominate.

The chemical potential stays below the photon energy. The figures stop within a few $kT$ of the gap, because beyond it the device amplifies light and the thermal description with a smooth occupation per mode no longer applies.

The light an emitter keeps inside itself #

Every figure describes light per unit area of an idealised emitting surface, with all the photons that are made escaping. A real LED is a small, bright region inside a slab of high-index material, and most of the light it makes is trapped, reabsorbed, re-emitted and eventually lost as heat. How much of the generalised Planck law’s light reaches the outside is a matter of geometry and not of thermodynamics, and it decides most of the efficiency of a commercial device.

Nor do the figures show the power involved. The efficiency plot runs over voltages at which the light emitted ranges over many powers of ten, and the region where the efficiency is largest is the region where there is almost no light at all. A diode cooling itself at 230 per cent efficiency does so at picowatts, and no drawing of a ratio can make that visible.

Still open: whether light can cool a chip #

Electroluminescent cooling would be a solid-state refrigerator with no moving parts and no fluid, pumping heat out of a device as infrared light that leaves the system entirely. The physics above shows it is allowed; the question is whether it can be made to work at a useful power. The difficulty is a race between two exponentials. The light, and so the cooling, grows as $e^{qV/kT}$; every non-radiative loss and every photon trapped inside the crystal takes a fraction of the current and returns it as heat. Net cooling requires the fraction of current that becomes escaping light to be very close to one — well above 99 per cent in most analyses of power densities that would matter — and to stay there at currents high enough to pump appreciable heat.

Several routes are pursued: coupling the emitter to a photodiode across a gap narrower than a wavelength, so that light that would be trapped tunnels out through the evanescent field and is recycled as electrical power; operating at elevated temperature, where the thermal contribution is larger; and materials with extraordinarily low non-radiative recombination. Measured net cooling of a macroscopic device by its own electroluminescence has not been reported at a level that settles whether the approach can become practical.

The habit worth carrying away is to ask of any light what fixes the number of photons in it. Light whose photon number is free has no chemical potential and is described by a temperature; light whose photon number is pushed has one, and needs both. A thermal glow, a laser, a solar cell’s emission and the faint dark of a reverse-biased diode are the same law with different values of that second number.