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The following paragraph is from Chapter 3 of the book Principles of Electronics.

When cathode is heated, it emits a large number of electrons. These electrons form a cloud of electrons near the cathode, called space charge. If anode is made positive w.r.t. cathode, the electrons (magenta dots) from the space charge speed towards the anode and collide with gas molecules (cyan circles) in the tube. Fig. 3.5 If the anode-cathode voltage is low, the electrons do not possess the necessary energy to cause ionisation of the gas. Therefore, the plate current flow in the tube is only due to the electrons emitted by the cathode. As the anode-cathode voltage is increased, the electrons acquire more speed and energy and a point–called ionisation voltage is reached, where ionisation of the gas starts. The ionisation of gas produces free electrons and positive gas ions (cyan circles with +ve signs). The additional free electrons flow to the anode together with the original electrons, thus increasing plate current. However, the increase in plate current due to these added electrons is practically negligible. But the major effect is that the positive gas Gas-Filled Tubes ions slowly drift towards the cathode and neutralise the space charge. Consequently, the resistance of the tube decreases, resulting in large plate current. Hence, it is due to the neutralisation of space charge by the positive gas ions that plate current in a gas tube is too much increased.

Why does Resistance decrease resulting in large plate current? My thinking is that if you have a cloud of electrons acting as charge carriers - that would be good for conduction so neutralizing them should increase resistance

(iii) Once ionisation has started, it is maintained at anode-cathode voltage much lower than ionisation voltage. However, minimum anode-cathode voltage, called deionising voltage, exists be-low which ionisation cannot be maintained. Under such conditions, the positive gas ions combine with electrons to form neutral gas molecules and conduction stops. Because of this switching action, a gas-filled tube can be used as an electronic switch.

Could someone explain why 'conduction stops' with the formation of gas molecules - shouldn't plate current continue to flow?

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  • \$\begingroup\$ On the second question: as soon as the ions re-combine into molecules, the gas stops being ionized (as there aren't ions anymore) and as such, doesn't conduct anymore. Note that an ion is nothing more than a molecule with one (or more) electrons stripped from it making it positively charged. \$\endgroup\$
    – RJR
    Jun 18 '14 at 4:29
  • \$\begingroup\$ Note btw that afaik, most common vacuum tubes used nowadays are not gas filled. \$\endgroup\$
    – RJR
    Jun 18 '14 at 4:38
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Regarding the first question, the reason is that the negatively charged space charge itself creates an electric field, which is in opposition to the electric field generated by the positive anode voltage. Increasing the number of negative charge carriers available for conduction is offset by the increase in the opposition electric field these charge carriers generate.

This puts a theoretical limit on the maximum anode current density for a given anode voltage when a space charge is present, and for a planar plate geometry is given by the Langmuir-Child equation:

$$1.67×10^{-3}\left(\frac{q}{mc^2}\right)\frac{V_a^{\frac{3}{2}}}{d^2} \frac{A}{m^2} $$

Where \$d\$ is the distance between the plates, \$V\$ is the anode voltage, \$q\$ is the particle charge, \$m\$ is the mass, and \$c\$ is the speed of light (all SI units).

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