One level, and the field that bends the bands

Doping a semiconductor does not move its band edges. It moves the Fermi level.

Add donors and there are electrons in the conduction band, so the level that marks where states stop being filled sits high, near the conduction edge. Add acceptors and there are holes in the valence band, so it sits low. The gap between the two edges is $1.12$ eV in both pieces — that is a property of silicon and not of what has been added to it — and the Fermi level is at $+0.357$ V in one and $-0.417$ V in the other, measured from the middle.

The difference is $0.774$ volts, and it is what the rest of this essay is about, because two things joined together cannot keep two different Fermi levels.

Why one level #

The Fermi level is not a level in the sense the band edges are. It is a chemical potential — the energy at which adding one more electron costs nothing — and a difference in chemical potential between two connected systems is exactly the condition under which particles flow from one to the other.

So the moment the two pieces touch, electrons flow from the $n$ side, where their electrochemical potential is high, to the $p$ side, where it is low. They do not flow for long. Leaving the $n$ side exposes the positively charged donors they came from; arriving on the $p$ side and filling acceptors exposes negative charges there. A field builds up across the junction opposing the flow, with the exposed dopants as its source, and it grows until the flow stops.

The condition for stopping is that the electrochemical potential is the same everywhere, which is to say that there is now one Fermi level, flat across the whole structure. That is not a result to be derived; it is what equilibrium means, and everything below is a consequence of imposing it.

If the Fermi level is flat and the band edges sat at different heights relative to it on the two sides, then the band edges must now bend. They bend by exactly the amount the Fermi levels differed by:

$$V_{\text{bi}} = \frac{kT}{q}\ln\frac{N_a N_d}{n_i^2},$$

which for $10^{17}$ acceptors and $10^{16}$ donors per cubic centimetre in silicon is $0.774$ V. That expression is not a new physical input. It is the difference of two Fermi levels, each computed from a doping density, and the generator checks that the two routes agree.

What has to be there for the bending to exist #

A potential that varies in space requires charge, by Poisson’s equation, and the charge is the dopants left behind.

In the region either side of the metallurgical junction, the mobile carriers have gone. What remains is the ionised dopants — negative acceptors on the $p$ side, positive donors on the $n$ side — fixed in the lattice and uncompensated. That is the depletion layer, and it is the space charge that holds the potential up.

Three checks make the picture a calculation rather than a sketch.

The charge balances. The junction as a whole is neutral, so $N_a x_p = N_d x_n$ — the two rectangles of exposed charge have equal areas. That is why the layer reaches ten times further into the lightly doped side: the same charge is reached sooner where there is more of it per unit volume.

The field integrates to the barrier. Integrating the charge gives a triangular field peaking at $4.7\times10^5$ V/m at the junction, and the area under that triangle must be the built-in potential. It is, to nine figures, and it is that requirement rather than any separate formula that fixes the width at $332$ nanometres.

And the two parabolas meet. The potential is quadratic on each side and the two branches must agree at the metallurgical junction, which they do exactly.

The generator finds the width twice — once from the closed-form square root and once by bisecting on the potential integral — and requires them to agree, because a depletion width taken from a formula and a field drawn to look plausible would pass every visual inspection while failing to be a solution of anything.

Two currents that cancel, not one that stops #

The description above — charge flows until a field stops it — is right and slightly misleading, because nothing stops.

At equilibrium there are two currents across the junction in each direction and they cancel. Electrons on the $n$ side diffuse toward the $p$ side because there are more of them here than there; that is a diffusion current, driven by a concentration gradient. Electrons that wander into the depletion region from the $p$ side are swept across by the field; that is a drift current, driven by the field. Equilibrium is the condition that the two are equal and opposite, for each carrier type separately.

This matters for two reasons. It is why the built-in potential has the form it does — setting drift equal to diffusion and integrating gives exactly $kT\ln(N_aN_d/n_i^2)/q$, with the $kT$ arriving from the Einstein relation between mobility and diffusivity. And it is why applying a voltage does what it does: the bias changes the diffusion current exponentially, by changing the barrier the diffusing carriers must climb, and leaves the drift current essentially alone, because the carriers it sweeps are supplied by generation rather than by the field.

A picture of a barrier that opens and closes gets the rectification right and this structure wrong, and the structure is what predicts the temperature dependence, the ideality factor and everything else that a designer actually uses.

The layer answers a bias #

Apply a voltage and it appears almost entirely across the depletion layer, because that is the only region with any appreciable resistance — everywhere else is full of carriers.

Reverse bias adds to the built-in barrier. Holding a larger potential requires exposing more charge, so the layer widens, as the square root of the total barrier. Forward bias subtracts, and the layer narrows.

Two consequences follow that are used constantly.

The peak field grows as the square root too, so a junction driven hard enough in reverse eventually reaches the field at which the material breaks down — by avalanche, where a carrier gains enough energy between collisions to knock out another, or at very high doping by direct tunnelling across the gap. Either way the reverse current rises abruptly at a voltage set by the doping, and a diode built to do this deliberately is a voltage reference.

A depletion layer is a capacitor whose plate separation is set by the bias, which makes a reverse-biased junction a capacitor with a knob on it. That is not a curiosity: it is how a varactor tunes a radio, and it is why the capacitance of a junction has to be counted against the speed of any circuit built from one.

And a region without carriers, between two regions with them, is a capacitor. Its capacitance is the permittivity over the width, so the bias controls it — that is the varactor, used to tune every radio receiver made since the 1960s. Run backwards, the same relation is a measurement: plotting one over the capacitance squared against bias gives a straight line whose slope is the doping density, which is how a doping profile is measured without cutting anything open.

The current, which is one exponential #

An electron on the $n$ side that wants to reach the $p$ side must climb the barrier. The fraction with enough energy is the Boltzmann factor of the barrier height, so the current in that direction is proportional to $e^{-qV_{\text{bi}}/kT}$ — which at equilibrium is exactly cancelled by the opposite flow.

Apply a forward bias $V$ and the barrier becomes $V_{\text{bi}} - V$. The uphill current is multiplied by $e^{qV/kT}$; the downhill current is unchanged, because carriers rolling down the barrier were never held back by its height. The difference is

$$I = I_0\left(e^{qV/kT} - 1\right),$$

and the whole of rectification is in the asymmetry between an exponential and a constant.

Two numbers make it concrete. The exponential rises a decade every $kT\ln 10/q = 59.5$ millivolts at room temperature, which is the steepest current–voltage relation any two-terminal device can have and is a hard limit on how abruptly a transistor of this kind can be switched. And at $\pm 0.5$ V the ratio of forward to reverse current is $2.5\times10^8$.

The famous “$0.7$ volt drop” is that steepness read on a linear axis. Between a microamp and an amp — six decades — the voltage moves by $0.36$ V, so any current a circuit is likely to pass lands in the same narrow window and the device looks like a fixed voltage. It is not one, and the temperature dependence gives it away: the drop falls by about $2$ mV per kelvin, which is a nuisance in a power supply and a useful thermometer in a chip.

The current is one exponential, and it is the ordinary one. How many carriers have enough energy to climb a barrier falls exponentially with the barrier’s height over $kT$, and everything about the forward characteristic is that curve with the height being adjusted by the applied voltage. The famous 60 mV per decade at room temperature is nothing but $kT/e$ times $\ln 10$, and it is the same number for every diode ever made because nothing about the material enters it.

Why no voltage appears across the terminals #

A junction with three-quarters of a volt built into it, connected to nothing, measures zero. That is worth taking seriously rather than waving away, because the reason is a general one.

A voltmeter measures a difference in electrochemical potential — a difference in Fermi level — between its two terminals. At equilibrium the Fermi level is flat through the whole loop by construction, so any voltmeter reads zero however many junctions the loop contains. The built-in potential of the $p$–$n$ junction is exactly cancelled by the contact potentials where the metal leads meet the two ends, and the cancellation is not approximate: it is the same statement as the flatness of the Fermi level.

The rule generalises. Any closed loop of materials at one temperature, in equilibrium, has no net electromotive force, whatever the materials are. Getting a voltage out requires taking the system out of equilibrium — illuminating it, so that the electron and hole populations are described by two separate quasi-Fermi levels; or holding the junctions at different temperatures, which is a thermocouple. Both are ways of arranging that “the Fermi level” is no longer one thing.

The alternative would be a loop of dissimilar metals driving a current round itself for ever, which is a perpetual-motion machine of the second kind. That it is forbidden here by the flatness of a line on a band diagram is a pleasingly concrete form of the general prohibition.

Where it stops #

The reverse current is not really constant. The ideal expression says it saturates, and a real junction’s reverse current rises slowly with bias, because the widening depletion layer contains more volume in which carriers are being generated thermally. That generation current dominates in silicon at room temperature and is what actually sets the leakage of a real diode.

The forward current is not really one exponential. Recombination inside the depletion region contributes a term going as $e^{qV/2kT}$, so at low currents the slope is $119$ mV per decade rather than $59.5$; at high currents the injected carrier density becomes comparable with the doping and the analysis above, which assumed it did not, fails. A measured characteristic is usually fitted with an ideality factor between one and two precisely to paper over this range.

The depletion approximation is a caricature. It assumes the carrier density falls abruptly from its full value to zero at the layer’s edges, where in fact it falls over a few Debye lengths. For the doping used here that is tens of nanometres against a layer of hundreds, so the error is modest; for a heavily doped junction it is not, and the width has to be computed by solving Poisson’s equation with the carriers in it.

Equilibrium is assumed and then abandoned. The band diagram with one flat Fermi level describes a junction with nothing connected to it. The moment a current flows the system is not in equilibrium and there is no single Fermi level to draw — the standard repair is two quasi-Fermi levels, one for electrons and one for holes, which are separately flat over the region where each carrier is plentiful and split apart across the junction by the applied voltage. Every diagram of a working diode or solar cell has two lines where this essay’s has one, and the splitting between them is the voltage the device delivers.

And none of it is a quantum calculation. Everything here uses the band structure as a given — the gap, the effective masses, the density of states — and then does classical electrostatics and Boltzmann statistics on top. Where the bands themselves come from is the quantum part, and it has been assumed rather than derived.

Where it stops is heavy doping. The Boltzmann approximation used throughout this essay assumes the occupied states are a thin tail far below the Fermi level; dope the semiconductor hard enough and the Fermi level enters the band itself, the electron gas becomes degenerate, and the exponential is replaced by the statistics of a filled sea. The junction still works. The arithmetic on this page does not.

The numbers, and how small the region is #

It is worth collecting the scales, because the structure is much smaller than its consequences suggest.

The depletion layer here is $332$ nanometres — about six hundred lattice spacings — and it holds three-quarters of a volt, giving a field of nearly half a megavolt per metre at its centre. The exposed charge is $4.8\times10^{-4}$ coulombs per square metre, which is $3\times10^{15}$ elementary charges per square metre or about one per three thousand surface atoms.

The doping itself is the striking figure. At $10^{17}$ per cubic centimetre there is one acceptor per half a million silicon atoms; at $10^{16}$, one donor per five million. A material whose electrical behaviour is decided by an impurity at a level of parts per million is a material whose purity has to be controlled at parts per billion before the deliberate doping can mean anything, and that requirement is the reason semiconductor electronics had to wait for zone refining rather than arriving with quantum mechanics.

And the built-in potential contains the intrinsic carrier density, $10^{10}$ per cubic centimetre in silicon at room temperature — one thermally generated pair per $10^{13}$ atoms. Everything above is the difference between two logarithms of very small numbers, which is why the answer comes out to less than a volt from ratios of $10^{13}$.

What was found by accident #

Russell Ohl was investigating why some silicon rectifiers worked better than others at Bell Labs in 1940, and had a rod of silicon with a crack in it — a boundary between two accidentally differently-doped regions. He found it produced a voltage when illuminated: half a volt, from a piece of silicon and a lamp.

That was the first $p$–$n$ junction, and nobody had been looking for one. The rectifying action of contacts between metals and semiconductors had been known and used since the 1900s in crystal radio detectors and was thoroughly not understood; what Ohl’s sample showed was that the interesting boundary could be inside one crystal, where no two electrons may share a state and the filling decides everything, between two regions differing only in a trace impurity.

Shockley’s theory of 1949 supplied the account above and did it properly, with drift and diffusion currents balancing at equilibrium and the exponential emerging from the barrier. The interval is worth noticing: nine years between an accidental observation and an explanation, and both were needed before anything could be designed rather than found.

What the diagram is and is not #

A band diagram is a plot of energy against position, and both axes deserve a word because neither means quite what it appears to.

The energy axis is the energy of an electron, so the curves bend upward where the electrostatic potential is lower — the electron’s charge is negative, and forgetting the sign inverts every conclusion. That is the single commonest error in reading these pictures, and the check is that the electrons must end up on the side where their energy is lowest, which is the $n$ side, which is where the bands are drawn low.

The position axis carries no information about the third dimension or about time. A junction is a plane and the diagram is a section through it; nothing in the picture says how wide the device is, and the currents computed from it are current densities. Nor is the diagram a snapshot of anything moving: it is the potential landscape, and the carriers traversing it appear nowhere on it.

The ladder from here #

Later rungs on this anchor: the bipolar transistor, which is two of these junctions sharing a thin base and whose gain is a ratio of two diffusion lengths; the photovoltaic effect in the same structure, where light shifts the two quasi-Fermi levels apart and the built-in field separates the carriers; the metal–semiconductor contact and the Schottky barrier, which rectifies for a related reason and switches faster because no minority carriers are stored; and the heterojunction, where the two sides are different materials and the band edges are discontinuous at the interface as well as bent near it.

The neighbouring ladders are what happens when the wells get close, which is where the bands come from; the mass a curve decides, which is how a carrier in a band responds to the field drawn here; and the exponential that decides everything, which is the factor the whole characteristic is made of.