The current a vacuum will not carry

The imbalance of charge that would outweigh gravity found what a small net charge would do on the scale of a planet, by Gauss’s law. The same law at the scale of a millimetre puts a ceiling on the current that can cross an empty gap. In one sense, a vacuum is the best conductor there is: an electron released in it meets nothing and accelerates without obstruction. Nothing about it should limit a current, apart from the number of electrons available. The earliest engineers of radio valves, around 1910, expected exactly that, and found otherwise. A hot filament could emit far more electrons than the valve would carry, and the current followed a clean power law in the voltage with no sign of the filament’s temperature in it.

That law was written down twice, by Clement Child in 1911 for ions and by Irving Langmuir in 1913 for electrons. It takes only a few lines, and those lines are a striking application of the law of counting what comes out. The charge inside a gap counts in the same way that the charge on its plates does. Followed a little further, the same calculation gives the upper limit to an ion engine, and produces a surprising result: the mass of the ion does not appear in the thrust.

What the first electrons do to the next #

Picture two parallel plates a distance $d$ apart, with a voltage $V$ across them and a hot emitter as the negative plate. If there is no charge in the gap, the field is the uniform $V/d$ of an empty capacitor, as the inside of a conductor would have it, and any electron arriving at the emitter’s surface feels it at once.

Let electrons leave. They move slowly just after leaving, so they spend a long time near the emitter and accumulate there. Their negative charge pushes the next electrons back toward the plate. The field at the emitter, which draws electrons off, is therefore reduced by the electrons already out. As long as the emitter offers more electrons than the field removes, the current rises until the field at the surface has fallen to zero. A further electron would then be sent back, so that is the limit. It is self-regulating, and it applies however copious the emitter.

Writing this down needs three equations. Poisson’s equation links the potential to the charge density, $d^2\phi/dx^2 = \rho/\varepsilon_0$, in magnitude. Energy conservation gives a particle’s speed at potential $\phi$, $v = \sqrt{2q\phi/m}$. Continuity says that the current density $J = \rho v$ is the same at every $x$ in the steady state, since charge does not accumulate. Eliminating $\rho$ and $v$ leaves one equation for $\phi$:

$$\frac{d^2\phi}{dx^2} = \frac{J}{\varepsilon_0}\sqrt{\frac{m}{2q\phi}},$$

with the requirement that $\phi$ and its slope both vanish at $x = 0$. Its solution is the four-thirds power, $\phi = V(x/d)^{4/3}$, and matching the constant gives

$$J = \frac{4\varepsilon_0}{9}\sqrt{\frac{2q}{m}}\,\frac{V^{3/2}}{d^2}.$$

There is the law, with its strange three-halves power. For electrons the numerical constant is $2.33 \times 10^{-6}\ \text{A V}^{-3/2}$. A 1 kV diode with a 1 cm gap carries about 74 mA per square centimetre, with no resistor anywhere. This is a vacuum behaving as a non-ohmic resistance, and its origin is purely electrostatic.

Gauss’s law, read as a running total #

The second figure below shows the same result in terms of the law that underlies it. Because a charged particle begins at rest and speeds up, the density is highest near the emitter, where it goes as $x^{-2/3}$, a true singularity. Further out, the same current needs far less charge.

Draw a box from the emitter out to any point $x$ in the gap. The field on the emitter side is zero by the limiting condition. So the flux through the box is $E(x)$ per unit area, and Gauss’s law says that equals the enclosed charge per unit area, divided by $\varepsilon_0$. The field at any point is, in this sense, a running total of the charge that has been passed. It takes nothing else. Integrating the density numerically reproduces the field of the first figure, within the precision of the sum, with total charge per unit area $\tfrac43\varepsilon_0 V/d$. An empty capacitor at the same voltage has $\varepsilon_0 V/d$ on its plates, and the charge in flight is therefore a third greater. All of those electrons are in transit: each is replaced by another as it reaches the far plate, in a transit time $3d/v_{\text{final}}$, which is half as long again as it would be with no space charge.

This is why the name space charge is apt. In an ordinary capacitor, charge rests on the surfaces, and the region between holds only field. In a working vacuum tube the charge is spread out through the volume and is moving.

The two limits a vacuum tube runs between #

The temperature of the emitter enters through a separate law. Owen Richardson showed that a metal at temperature $T$ emits a current density proportional to $T^2 e^{-W/kT}$, where $W$ is the work function, the energy needed to free an electron. For tungsten, with $W$ about 4.5 eV, that rises steeply with temperature, by about 260 times between 2200 K and 2800 K. So the current cannot exceed either of two values: what the gap will carry and what the emitter can offer.

On the left of the figure the emitter is not the limit, and temperature has vanished from the result. A valve running here has a current fixed by its voltage and geometry, and the cathode can age, cool or drift with no effect on the current. That is what allowed valve circuits to be designed reliably. Valves were commonly operated with the filament hot enough to reach well above the $V^{3/2}$ line at working voltage. The small electron cloud sitting just outside the cathode acts as a reservoir, with the excess electrons returning to the surface. It is an instructive contrast with the current a superconductor stops with its own field, a different system in which a current also limits itself through its own field.

In a real valve, a third electrode slips into the gap to manipulate the field at the emitter. A small voltage on a grid close to the cathode has a strong effect on the current, since the current depends on the field at the cathode, and this was what made amplification possible. Lee de Forest had built the triode by 1906 without any clear idea of the physics. Langmuir’s law provided the explanation for what it was doing, and the design of a century of amplifiers and of every electron gun in a CRT or a klystron has had to work with this limit. The beam that stops pushing itself apart takes up what happens to such a beam after it leaves the gap, and the wall a plasma builds against itself shows the same three-halves law reappearing in a sheath, where ions rather than electrons carry the current.

An ion engine has the same limit #

The same result holds for any charge, with the mass entering only through the square root. A positive ion such as xenon, 131 atomic mass units, is 240,000 times heavier than an electron, so its limiting current density at the same voltage and gap is 490 times less. Ion engines work with this limit, and it is the single constraint that sets their size.

Such an engine ionises a gas in a chamber and extracts the ions between two perforated grids. Ions pass through the holes of the second grid and leave at high speed, with electrons injected downstream to keep the spacecraft from charging. The rocket that leaves its fuel at home gave the attraction of very high exhaust speeds; a kilovolt on xenon gives 38 km/s, an order of magnitude more than any chemical rocket’s. The trouble with this kind of rocket is that its thrust is very small. Now the calculation becomes interesting.

Thrust per unit area is the rate at which momentum leaves. A current density $J$ carries $J/q$ ions per second per unit area, each with momentum $mv$. So

$$\frac{F}{A} = \frac{J}{q}\,m\sqrt{\frac{2qV}{m}}.$$

Substitute the Child–Langmuir $J$. One factor of $\sqrt{m}$ is in the denominator of $J$, and one in the numerator of the momentum, so the mass cancels.

$$\frac{F}{A} = \frac{8}{9}\varepsilon_0\left(\frac{V}{d}\right)^2.$$

The charge has gone too. Neither the mass nor the charge of the ion appears.

The two factors that would normally distinguish fuels have disappeared. A heavier ion moves more slowly, so it takes longer to cross and more of it can sit in the gap for a given field, and this is compensated exactly by the smaller current each charge can carry. At 1 kV over a millimetre the limit is about 7.9 N/m². A real thruster has grids with a large fraction of hole, and the ions must not strike the second grid, so the practical figure is rather lower. NASA’s NSTAR engine on Deep Space 1 had a 30 cm beam and made 92 mN at full power, about 1.3 N/m². Scaling that up requires larger area, not a different propellant, and this is why an ion engine is large for what it does.

The force is on the grid #

A second reading of this result is more surprising than the first. Where exactly does the force act on the spacecraft? A chemical rocket pushes on its combustion chamber and nozzle by gas pressure. In an ion engine, nothing touches the ions; they are in a vacuum, and they leave without colliding with anything. What pushes back?

The electric field. The ions are pushed by a field that ends on charges sitting on the metal grids, and a charged surface in a field feels a force per unit area of $\tfrac12\varepsilon_0 E^2$, as the pressure a charge puts on its own metal calculated in the setting of a charged sphere. At the emitter side the field is zero, so the emitter feels nothing. At the far grid the field is the $\tfrac43 V/d$ of the Child–Langmuir solution. So the force per unit area on the grid is

$$\frac12\varepsilon_0\left(\frac{4V}{3d}\right)^2 = \frac{8}{9}\varepsilon_0\left(\frac{V}{d}\right)^2.$$

That is identical with the momentum flux in the exhaust, and the identity is exact. It is no coincidence, since momentum balance in a steady state must make them equal, and the force read off a surface that touches nothing showed why this always happens: the Maxwell stress on a surface surrounding a charge distribution gives the force on those charges. The surprising fact is how simple it is here. An ion engine is an electrostatic pull on a perforated plate, of known size. All the talk of exhaust is one description, and the stress on the grid is another, of one physical quantity.

This also yields a sharp ceiling. The stress is set by the field, and the field is limited by sparking: above about a few million volts per metre between real grids in a real engine’s environment, discharges start. Raising the voltage across a fixed gap would not help past that. So the upper bound on thrust density at a given gap is set by $\varepsilon_0 E^2$ at the breakdown field. It is roughly the same thing that limits how much force any electrostatic device can produce, and the reason why electric motors are almost all magnetic: at practical fields, the magnetic stress $B^2/2\mu_0$ is thousands of times larger than the electric one. The push at the end of a pair of rails makes the comparison from the magnetic side.

Where the mass does matter #

The mass has dropped out of the thrust, but it has not dropped out of the design.

For a given voltage, a lighter ion leaves faster, so each newton of thrust carries more kinetic energy away with it. The power per unit thrust is $v/2$. A hydrogen ion at 1 kV has $v = 438$ km/s, and each newton of it needs 219 kW. Xenon at the same voltage needs 19 kW. A spacecraft’s solar panels are limited, so the mass matters where the energy budget is decided. A second reason for heavy ions is that a heavier ion at a fixed exhaust speed would need a higher voltage, which raises the thrust density as $V^2$, so that for a chosen exhaust speed the thrust per area actually goes as $m^2$. Xenon’s large ion mass is thus what makes an engine compact at a given specific impulse, and this is the proper sense in which mass matters. Xenon also ionises easily, is not chemically reactive and does not condense on the spacecraft’s surfaces. Its main defect is cost, and krypton is sometimes used instead on the largest constellations for this reason. Choosing the voltage and thus the exhaust speed is then a separate trade between propellant mass and power-plant mass, one that mission designers work through, and it does not concern this law at all.

A limit that is also a measure of a machine #

Because the law has only geometry in it apart from the voltage, a device’s capacity can be summarised by one number. Electron-gun designers divide the current by $V^{3/2}$ and call the result the perveance. A planar diode’s perveance is $\tfrac{4\varepsilon_0}{9}\sqrt{2e/m}\,A/d^2$, and it says nothing about the voltage at which the gun runs. A gun scaled up in size at fixed proportions keeps its perveance, since $A/d^2$ does not change. That scaling explains something surprising about high-power tubes. A klystron with a million-volt beam can be no larger than a smaller tube and carry enormously more current, since the current scales as the voltage to the three-halves power and a device can simply be run at higher voltage. A million-volt gun of one microperveance carries a thousand amperes.

It also explains why very large electron beams at low voltage are so hard to make. A welding gun or a microwave oven’s magnetron wants a large current at a modest voltage, and it must therefore have a small gap or a large area. Cathode-ray tubes went the other way, with a small current at high voltage, and they could use very fine beams. Reducing $d$ raises the limit as $1/d^2$, and that is how vacuum microelectronics has tried to rescue the vacuum tube: at a gap of 100 nanometres a few volts can give a very high current density, up to the point where the electrons a field pulls from cold metal take over as the source.

What the law assumes #

The derivation is one-dimensional, with infinite plates, particles emitted at rest, a single species, and no collisions. Each of these fails somewhere. Electrons or ions leaving with some thermal energy can push the point of zero field slightly off the surface, forming what is called a virtual cathode, and this modifies the law by a small amount at ordinary voltages. A narrow emitting strip can carry more than the planar limit, since the field can reach round its edges. Near relativistic energies, the speed saturates at $c$ and the law flattens from $V^{3/2}$ toward $V^{1}$, with a result due to Jory and Trivelpiece in 1969. In a very small gap, approaching the electron’s de Broglie wavelength, quantum effects take over, and the current can exceed the classical limit by a modest factor.

An ion engine also has beam divergence, a second grid not drawn and holes that behave differently than a uniform plate. The picture’s 7.9 N/m² is an ideal. Those limits matter for design, but they do not alter the conclusion: the law is one that bounds, and a real device runs somewhat below it.

Still open: how high the limit can be pushed #

The space-charge limit assumes a gap with only one sign of charge. Add a neutralising species, and the limit disappears, which is the idea behind Hall thrusters, in which electrons trapped by a magnetic field neutralise the ions’ space charge as they are accelerated. Such devices carry current densities far above the Child–Langmuir value for their voltage, and thrust densities several times greater. The trapped electrons, however, drift across the field far faster than classical theory predicts, by a mechanism that remains under dispute: it limits the thruster’s efficiency and its lifetime, and a quantitative theory that would let such thrusters be designed from first principles does not exist.

Experiments have also found that pulsed short gaps can carry more than the static limit for a short time, by not giving the charge time to accumulate. Whether time-structured beams can exceed the limit in a sustained way, by exploiting the time it takes for space charge to arrive, has been studied, with limits derived on how far this can go. The steady law itself is very well understood, and it stands as an absolute ceiling only so long as the space charge stands unneutralised.

A vacuum carries only as much current as leaves the field at its emitter just above zero, which by Gauss’s law gives $J \propto V^{3/2}/d^2$. An ion engine at that limit makes a thrust per area of $\tfrac89\varepsilon_0(V/d)^2$, with the ion’s mass and charge both cancelled, and that thrust is exactly the electric stress on its grid. What the ion’s mass decides is the power, not the push.