gas discharge and vacuum current

In metals and electrolytes free charge carriers always exist, so a current appears as soon as a voltage is applied. In gases under ordinary conditions molecules are neutral, there are almost no free carriers, and the gas behaves as a dielectric: $I \approx 0$.

An external ioniser — a flame, ultraviolet light, radiation, cosmic rays — creates ion–electron pairs. Under the action of the field the ions and electrons move to the electrodes, the gas becomes a conductor, and the discharge is called non-self-sustaining.

The current–voltage characteristic of a gas discharge is not described by Ohm's law as a whole. At first $I \approx U/R$ holds approximately (the linear region), then the growth slows down because of recombination and the finite ionisation rate, and the current reaches saturation:

$$I_s = N e,$$

where $N$ is the number of ion pairs produced by the ioniser per second and $e$ is the elementary charge.

As the voltage rises further, an electron acquires enough energy between two collisions to ionise a molecule by impact; an electron avalanche starts, the current grows sharply, and the discharge becomes self-sustaining (corona, glow, arc discharge).

In vacuum there are no charge carriers at all, so a current is possible only if carriers are injected from outside: by thermionic emission, by the photoelectric effect, or by field emission. In a vacuum diode the current is limited by the space charge of the electron cloud, $j \propto U^{3/2}$, and at high voltages it reaches the saturation current $j_s = A T^{2} e^{-\Phi/(kT)}$.

So Ohm's law is only a special case: in gases and in vacuum the shape of the I–V curve is set by how carriers are created and how they are transported.

In a gas-discharge tube an external ioniser creates $N = 5{,}0\cdot10^{15}$ ion pairs per second. Find the saturation current. What will the current be if the ioniser output is halved while the voltage stays in the saturation region?

A vacuum diode with a heated cathode has a saturation current $I_s = 5{,}0$ mA. In the space-charge region its I–V curve is $I = k U^{3/2}$ with $k = 1{,}0\cdot10^{-4}\ \text{A/V}^{3/2}$. Find the current at $U = 20$ V and the voltage at which the three-halves law stops working.

In a gas the electron mean free path is $\lambda = 0{,}5$ µm, and ionising a molecule requires an energy $W = 15$ eV. Estimate the field strength at which an electron gains this energy between two collisions, and compare it with the field in a vacuum diode at $100$ V across a $1$ cm gap. Explain why a vacuum diode needs no breakdown but does need a hot cathode.

Safe self-study plan (simulation only).

1. Open an interactive gas-discharge-tube simulation or a virtual lab with an adjustable voltage source, a tube, and an ammeter (for example, an online circuit simulator). Do not build a real high-voltage or discharge-tube setup.
2. Increase the voltage smoothly from zero, recording the pairs $(U, I)$ in a table.
3. Plot $I(U)$. Mark the linear region, the saturation region, and the region of sharp current growth (breakdown). Check that in saturation $I_s \approx N e$.
4. Reduce the intensity of the external ioniser (in the model) and confirm that the saturation level drops while the breakdown voltage does not change.
5. Replace the tube with a model of a vacuum diode with a heated cathode: observe the saturation plateau and check the dependence $I \propto U^{3/2}$ in the space-charge region.
6. Write a 3–4 sentence conclusion: where Ohm's law works and where it does not.

Show how the current–voltage characteristic of a gas discharge differs from the straight Ohm's-law line and from the saturation plateau of a vacuum diode.

A labelled graph of $I$ versus $U$: I — linear region ($I \approx U/R$), II — saturation with the horizontal line $I_s = N e$, III — sharp rise at impact ionisation (breakdown). Dashed for comparison: the Ohm's-law straight line and the vacuum-diode plateau. Beside it, a sketch of a tube with two electrodes, arrows for ions and electrons, and the label «external ioniser».

During a thunderstorm the base of a cloud carries about $-15\ \mathrm{C}$ at a potential roughly $10^8\ \mathrm{V}$ above the ground $2\ \mathrm{km}$ below. When the field at a protruding point reaches about $3\times10^6\ \mathrm{V/m}$, air molecules are ionised by electron impact, a stepped leader grows towards the ground, and the return stroke delivers a peak current of about $3\times10^4\ \mathrm{A}$ lasting roughly $100\ \mathrm{\mu s}$.

The stepped leader is a self-sustained discharge: every accelerated electron frees further electrons, an avalanche builds up, and the resistance of the ionised channel drops from about $10^9\ \Omega$ to a few ohms. So $I=U/R$ with a constant $R$ fails here: the volt-ampere characteristic of a gas discharge is strongly nonlinear, and the current grows faster than the voltage.

A $1.2\ \mathrm{m}$ tube runs from the $230\ \mathrm{V}$ mains through a ballast. Once struck, the voltage across the tube is only about $110\ \mathrm{V}$ at a current of $0.35\ \mathrm{A}$; electrons accelerated over a mean free path of a few millimetres at a mercury-vapour pressure near $0.5\ \mathrm{Pa}$ ionise mercury atoms, and the phosphor coating converts the ultraviolet light into visible light.

This is a glow discharge: the gas sustains the current itself through impact ionisation, and its differential resistance is negative — more current means more ionisation and a lower tube voltage. Ohm's law applies only to the metal ballast: $R=\dfrac{230-110}{0.35}\approx340\ \Omega$, and it is the ballast that limits the current.

In an old television picture tube the pressure is pumped down to $10^{-5}\ \mathrm{Pa}$. A tungsten cathode heated to about $1000\ \mathrm{K}$ emits electrons by thermionic emission; an anode voltage of $25\ \mathrm{kV}$ accelerates them to $9.4\times10^7\ \mathrm{m/s}$ over $0.4\ \mathrm{m}$, and a beam current of $1\ \mathrm{mA}$ draws a bright spot on the screen. At this pressure the mean free path is hundreds of metres, far longer than the tube, so the electrons fly without collisions.

In vacuum there is nothing to ionise, so a current is possible only if a source of free electrons exists — thermionic emission. Such a current does not obey Ohm's law: at low voltages it is limited by space charge and then saturates, so $I=U/R$ with a constant $R$ does not describe its volt-ampere characteristic.