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Tubes: neutralization of space-charge causes drop in tube resistance? Deionizing voltage re-combination?

formatting link
ronics/CH-03.pdf

"If the anode-cathode voltage is low, the electrons do not possess the nece ssary 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 ionisat ion of the gas starts. The ionisation of gas produces free electrons and p ositive gas ions (cyan circles with +ve signs). The additional free electr ons 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 pra ctically negligible. But the major effect is that the positive gas"

My understanding is that, when the metal gets really hot, it causes 'Thermi onic Emission' and this is related to the 'work function'. Why then does ne utralization of the electron-cloud cause a DROP in resistance?

Aren't electrons the charge-carriers here - so to speak? You'd suppose that if you had a nice little cloud of electrons that were just neutralized, Re sistance would increase!!

Furthermore "(iii) Once ionisation has started, it is maintained at anode-cathode volta ge much lower than ionisation voltage. However, minimum anode-cathode volt age, called deionising voltage, exists be-low which ionisation cannot be ma intained. 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 s witch."

Why do +ve gas ions combine with electrons below the de-ionizing Voltage?

Reply to
vek.m1234
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Also, once you apply the E field, why doesn't the whole electron cloud move to the anode?

What exactly is going on here? (I do have an intutive idea of what's going on but I'd rather have someone independent confirm my views). If I said what I thought was happening - I might get fobbed off with a yes yes - that's what's happening.

Reply to
vek.m1234

It does, if the grid lets it through.

The cathode emits electrons with energies way below 1 eV. They're attracted back to the cathode by the image charge effect, so in the absence of an applied E field, they soon come into equilibrium. (There are other things that can happen, e.g. Barkhausen oscillations, but that's a fine point.)

If there's a plate, and you make it positive with respect to the cathode, the electrons nearest to it will accelerate towards it. Electrons nearer the cathode are still nearly in equilibrium, because the mobile charges shield out the E field after a short distance.

If you keep cranking up the voltage, so that the E field region reaches the cathode, you reach the cathode-limited condition. The high E field at the cathode surface leads to fairly rapid damage, so you want to avoid this with an oxide cathode. (I don't think it hurts tungsten or thoriated-tungsten cathodes, but their emission is much lower.)

The grid controls how much of the electron cloud feels the plate voltage, by repelling most of the electrons before they get through it.

Cheers

Phil Hobbs

--
Dr Philip C D Hobbs 
Principal Consultant 
ElectroOptical Innovations LLC 
Optics, Electro-optics, Photonics, Analog Electronics 

160 North State Road #203 
Briarcliff Manor NY 10510 

hobbs at electrooptical dot net 
http://electrooptical.net
Reply to
Phil Hobbs

Thanks. Just to be clear - for a gas tube:

  1. Cathode gets hot. Thermal-energy > Work-function imparted to electron. E lectron leaves metal and becomes space-charge cloud.
  2. Cloud has near-side facing-metal and far-side facing-anode.

  1. Near-side electrons fall back into the cathode because of image charge e ffect (a +ve charge is created within the metal equal to the -ve charge of the electron and that draws the near-side electrons in to the metal, like a mirror-image +z/-z distance apart)

  2. Far-side electrons are drawn to anode (plate current). On the way they i onize gas molecules (collision - if ionization-voltage is applied) which ar e attracted to the cloud.

  1. Extra electrons reach anode - contribute to increased plate current.

  2. Ions move towards e-cloud and neutralize cloud.

******************************** Resistance of Tube drops! Why!???

********************************
  1. Because, for electron to leave the metal/cathode surface it has to have energy greater than the work-function AND it also has to overcome electron cloud repulsion (like charges repel) - is this point correct (the cloud-rep ulsion bit)???

  1. Assuming 7's correct: neutralizing the cloud with ions will cause a drop in Resistance since you only need work-function energy to push the electro n out of the metal. - Correct?

***********************************
  1. If the Ionization voltage drops below de-ionization voltage: "Under such conditions, the positive gas ions combine with electrons to for m neutral gas molecules and conduction stops."

If the ionization voltage is reduced below de-ionization voltage, electrons can no longer ionize molecules - therefore they combine instead to form ne utral molecules and conduction stops? Huh? But why doesn't plate-current fl ow???

***********************************

Section 3.2, Pages: 40/41 of Chapter 3

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ronics/CH-03.pdf

or this one if you want all chapters:

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Reply to
vek.m1234

Gas tubes are different. You have avalanche ionization to worry about.

Cheers

Phil Hobbs

--
Dr Philip C D Hobbs 
Principal Consultant 
ElectroOptical Innovations LLC 
Optics, Electro-optics, Photonics, Analog Electronics 

160 North State Road #203 
Briarcliff Manor NY 10510 

hobbs at electrooptical dot net 
http://electrooptical.net
Reply to
Phil Hobbs

It's possible for electrons to diffuse within the cloud, and go back into the cathode; that's important when calculating the available charge. This is part of the pseudo-equilibrium state, not really an event in itself...

The heated cathode has a limited rate-of-emission of electrons; when you add ionization of gas, in bulk, to the current flow, you get much higher current capability (it's not precisely a resistor effect, that's just a figure of speech in this case). I wouldn't assume ions 'neutralize cloud' electrons, they could directly strike the cathode to dump their charge!

Reply to
whit3rd

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