Hi Rich,
> >You say a sharp pulse of VHF current heats the resistor from the
> >inside out,
You replied:
> This is my guess as to what took place.
Nice guess. But I've never seen reverse skin effect and the thermal
conductivity can not be ignored.
> >while a longer HF current heats the resistor and
> >severely damages the outside.
> >
> The outside is not severely damaged because it is still intact and doing
> its job of supporting the wire leads and carbon element.
I don't follow that logic. What causes it to not be damaged?
> >That seems to disagree with what you say about other conductors.
> >
> >For example you claim gold coatings on grids can only be
> >damaged by VHF parasitics, and not by dc or lower frequency
> >currents.
>
> Correct. Boiling gold requires a higher temperature than is needed to
> melt iron. For example, to increase the temperature of the entire >50g.
> grid in an 8877 to the boiling point of gold would require an amount of
> energy that is far beyond the capability of the typical anode supply.
Let me see if I understand you.
1.) The anode supply in an amplifier has enough energy to heat a
large anode with airflow to the point of damage.
2.) The anode supply does not have enough available energy to
heat a grid.
Following that logic, people are wasting time watching screen
current in a tetrode amplifier, because the screen supply could
never overheat the screen grid.
> However, thanks to skin-effect, the energy available from the anode supply
> is able to remove thin layers of gold.
The gold is bonded to the grid. There is thermal conductivity at play.
Have you ever calculated the heating of the grid by I^2 R losses in
the conductor?
You claim above heating is primarily due to I^2 R losses, yet that
is absolutely incorrect. That is not the dominant mechanism that
heats anodes and grids at all, it isn't even a factor worth
considering.
ing to Eimac's Mr. Willis B.
> Foote, during the development of the 8877, the design team discovered that
> grids could be gold-sputtered in thin layers by an "oscillation
> condition".
Please tell me where I can see that claim first hand.
> RF travels on the surface of conductors. Gold is a conductor. The
> phenolic case of a carbon-comp resistor is not a conductor.
You and I were talking about the inside and outside of the element,
not the case.
> >How can the resistor get that hot
> >inside when the saturated anode current is only about ten
> >amperes,
>
> P = I^2 x R, and I = 9 or 10 amperes, P could be 81 or so x 50 to 100
> (ohms) watts. Approx. 4kW is seeming quite a bit for a 2w-rated
> resistor.
You ignored duty cycle above.
> > the duty cycle is in nanoseconds,
>
> This is not known.
You say it is. You say the parasitic is so "intermittent" it doesn't
even show up on the meters.
Which is it? Is the parasitic a long sustained oscillation that can
heat the resistor, or is it a sudden momentary burst?
> >This almost sounds like your theory that photons arriving from outer
> >space can make amplifiers on standby explode because the photons hit the
> >amplifier so hard they make the standby relay arc, and the arcing relay
> >starts a parasitic in what is an otherwise stable amplifier that is just
> >sitting there on standby!
>
> Borrow a geiger counter, Mr. Rauch, and tune in on what's happening on the
> upper frequencies. Be not surprised if you occasionally encounter some
> humungous signals.
The detector in a geiger counter is specifically designed to respond
to photons (it uses pressurized gas) when exposed to radiation.
Even when exposed to high levels of background radiation current
is in picoamperes.
You are proposing that a high vacuum tubes respond with many
amperes of current.
If that is the case geiger counters would work better with 3-500Z
detectors than with gas-filled geiger-muller tubes.
Please explain the apparent discrepancy between your theory and
how things really work.
73, Tom W8JI
w8ji@contesting.com
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