Just a thought...I mean, I know the plates are the hot parts in the tube and we have the plate dissipation. I may have a brain fart at the moment, but thinking about the workings in there...
The metal plate inside the bulb surely has no voltage drop. (ok a few uV perhaps). (I am thinking if you could measure the ends of the plate inside the vacuum tube). The voltage drop is between the plate and the cathode, that means the voltage is in the vacuum, the space between the plate and cathode. But surely vacuum cannot be heated from current running thru it?
I understand the plate gets heated by the bombardment of electrons and the current running thru it. But since there is no voltage drop on the plate, is all the heating due to bombardment of electrons?
Probably a confusing question. And I am only asking out of curiosity, I am not worried about tube design, but for some reason I want to understand this a bit better (obviously I dont understand at all).
The metal plate inside the bulb surely has no voltage drop. (ok a few uV perhaps). (I am thinking if you could measure the ends of the plate inside the vacuum tube). The voltage drop is between the plate and the cathode, that means the voltage is in the vacuum, the space between the plate and cathode. But surely vacuum cannot be heated from current running thru it?
I understand the plate gets heated by the bombardment of electrons and the current running thru it. But since there is no voltage drop on the plate, is all the heating due to bombardment of electrons?
Probably a confusing question. And I am only asking out of curiosity, I am not worried about tube design, but for some reason I want to understand this a bit better (obviously I dont understand at all).
The heating of the anode (and screen grid) is indeed caused by the electron bombardment on the plate (and screen grid). Ofcourse the filament also contributes to the total dissipation of the tube.
The higher the voltage, the higher the kinetic energy of the electrons when arriving at the anode. The dissipation at the anode is therefore a function of both voltage and current.
Thnx.
Well yes indeed P=UxI. I knew this before I knew how to get a buzz in a controllable manner. And have always known the tube plate diss is plate voltage times current.
I know this aint a dumb question but yeah I guess confusing if you dont want to think past the data on your piece of paper.
Well yes indeed P=UxI. I knew this before I knew how to get a buzz in a controllable manner. And have always known the tube plate diss is plate voltage times current.
I know this aint a dumb question but yeah I guess confusing if you dont want to think past the data on your piece of paper.
In a solid material electrons are also hitting the atoms inside as they pass through and this cause the heat .
If you want to see a resistor inside the plate maybe you can imagine that the surface that is actually hit is very very thin , the first atom layers , like the current has to flow through a very thin wire first and then in the plate itself .
If you want to see a resistor inside the plate maybe you can imagine that the surface that is actually hit is very very thin , the first atom layers , like the current has to flow through a very thin wire first and then in the plate itself .
Last edited:
Effects of the electrons impacting on the plate:
1. Heat
2. Secondary electron emission (Suppressor grid, that is why they are there, to collect secondary electrons).
Beam Formers of Beam Power tubes can do the same thing.
3. X-rays
The impulse energy of the electron impact determines which one, or two, or all happen.
Of course, it takes very high energy impact to get Xrays.
I seem to remember a formula about impact momentum, 1/2 x (MV)squared.
When your car impacts a brick wall at 60 MPH, heat is generated (you are bending steel, and that heats it up).
Try it again with another car at 30 MPH, and there will only be 1/4 of the bending.
Sometimes a higher plate voltage can affect the major paths of the electrons, and if they concentrate in one area, it can cause a red-plating hot spot,
even though the Plate voltage x Plate current is Less than the tube's rated plate dissipation.
Leave the tube burnouts and meltdowns to someone else.
Thanks Tubelab_com!
Just my opinions
1. Heat
2. Secondary electron emission (Suppressor grid, that is why they are there, to collect secondary electrons).
Beam Formers of Beam Power tubes can do the same thing.
3. X-rays
The impulse energy of the electron impact determines which one, or two, or all happen.
Of course, it takes very high energy impact to get Xrays.
I seem to remember a formula about impact momentum, 1/2 x (MV)squared.
When your car impacts a brick wall at 60 MPH, heat is generated (you are bending steel, and that heats it up).
Try it again with another car at 30 MPH, and there will only be 1/4 of the bending.
Sometimes a higher plate voltage can affect the major paths of the electrons, and if they concentrate in one area, it can cause a red-plating hot spot,
even though the Plate voltage x Plate current is Less than the tube's rated plate dissipation.
Leave the tube burnouts and meltdowns to someone else.
Thanks Tubelab_com!
Just my opinions
Last edited:
The suppressor grid in a pentode doesn't collect secondary electrons but repels them back to the plate, and by doing so prevents them from going to the screen grid.
PFL200,
I made a mistake.
Thanks for the correction.
You have successfully 'suppressed' my error.
And . . . that re-collection of the secondary electrons will generate a little heat too.
I made a mistake.
Thanks for the correction.
You have successfully 'suppressed' my error.
And . . . that re-collection of the secondary electrons will generate a little heat too.
The re-collection itself will create a little heat too. But if I understand it correctly, that heat doesn't 'add up' to the total anode dissipation. It takes energy to create secondary emission. That energy is delivered by the primary electron bombardment. Secondary emitted electrons just temporarily cary part of the energy in kinetic form, but they 'give it back' (well, atleast most of it) to the anode in the form of heat once they return to the anode.
Got nothing to do with it.
The heat comes from the fact the internal structure of the electrodes from anode to cathode causes an internal resistance. (can be anything from 800ohms to typically 22K ohms+.)
This resistance varies according to the control grid voltage, ie. at 650V with 0.01A it will have a resistance of 65K ohms.
Vsquared over R then gives you 6.5W anode dissipation.
It's a semi conductor then, normal enough much like a transistor...
Typically a large output valve will be idling at 0.035A at 650V (18.5K) or around 23W. (807 typically prefer about this figure).
Being as the current is being conducted by the ANODE into this internal resistance, it's the anode that has the highest potential difference.
The cathode has nothing because it's at zero volts, so the anode gets hot because of the internal resistance of the valve.
Where this relates to heat, is simply because an electron flow can be allowed or prevented via the grid.
The more the electron flow the more current flows and the lower the anode voltage goes, so electron flow has zilch to do with generating heat, it's all to do with the internal resistance of the device, which has everything to do with total current conducted and voltage across the device.
If the valve suddenly is invited to flow 100m/a instead of 10, but the anode voltage drops to 225V....it then has a resistance of 2250ohms, at which point it's dissipating 22.5W.
A typical big sweep tube would flow 0.35A at 100V, so we are then approaching 285ohms at which point it will hit a Pdiss of 35W.
As an example, my AB2 807 amp according to graphs typically can manage about 350-400m/a (250-400ohms) at that voltage, much like the sweep tube they eventually became, (ie.150-200 in ab1) (700-800ohms) , so around the same 25-30W as at idle.
There are times when the anode is conducting way past it's stated maximum value but in a push pull amp it's got a 50% duty cycle as has the screen grid which means the total is derated.
...to which you have to add another 10-20m/a for the control grid at +20V and the aligned screen grid at roughly the same, giving a cathode current closer to 450m/a.
So electron flow has nothing to do with heat.
Member
Joined 2009
Paid Member
Boy, it doesn’t take long for these simple questions to fall off the rails with statements that are just sufficiently wrong in too many ways to be worth clarifying.
The simple answer: no matter what heats up in the (almost) vacuum, the heat radiates mostly by anodes, some by fins welded on top to screen grid rods. Yes, some IR rays are being captured by glass and the glass gets hot. Sometimes to 250 degrees C, like it is common for EL84 tubes. But the main way of dissipation is IR rays that go through the glass heating up transformers and capacitors that are mounted near tubes, decreasing their life expectancy...
Also note that the plate structure ends up re-radiating a fair % of the heat radiated by the cathode/filament structure, as the cathode is much hotter than the plate, and so the temperature of the plate is not just related to the operating V.I dissipation.
There are many valves that are quite happy running at 250C, mostly mil spec & transmitting valves with much higher quality glass, and much more rugged construction.
Some of them were Russian, some US made and quite a few British..
Consumer quality components like those horrible Mullard EL34 and EL84, were designed to fail quite quickly so they could sell you lots of new ones.
The EL37 which is pretty much identical to EL34 but with a 2x larger bulb does not fail.but then the EL34 (and EL84) was sold on the back of making smaller and smaller more compact amplifiers.
EL84 are well replaced by the Russian 6P14P-EV, 6П14П-ЕВ because it's supposed to be a mil spec glass etc.
Fact is, the valve dissipates all the power at the large anode, because that is where the greatest PD is.
Potential difference + internal resistance = thermal watts generated.
Large PD, with inadequately sized envelope = short life.
No, YOU are wrong.
Valves don't work in quite the way you suggest, and that article was from the 40s from way before major advances in thermionic valve design were made.
In a typical beam tetrode or output pentode the electron velocity doesn't fundamentally change, for the simple reason, the REAL anode is the screen grid.
The large heavy gauge anode is a "virtual anode" which is constantly varying in potential so it only collects electrons from the already accelerated stream.
This proves what I said beyond doubt, the the electron stream is NOT responsible for heating the anode.
It's only a vector for allowing current to pass, exactly as it does in a transistor.
A transistor heats up for exactly the same reasons, except the minimum junction loss on a transistor/FET etc full on is less than 1V, so with a rail value of 50V you get high efficiency and high power.
...Whereas on a valve you can rarely (with some very rare exceptions) get a linear voltage drop of less than 100V full on from a rail value of 400-700V, before you hit the knee, with all the major distortion that comes with it.
That means from a thermal point of view, a valve compared with a modern semi conductor (never mind class D) is a thermal and efficiency disaster. 😱
Sarcastic, are you saying that electron bombardment is not the cause of plate power dissipation (along with cathode radiation reaching the plate) ?