>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
? say what? Does this have anything to do with your revelation that the
resistance of nichrome actually increases as freq. decreases?
>and the thermal
>conductivity can not be ignored.
? Phenolic is not a good thermal conductor. .
>
>> >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?
? The resistor is found to be intact and the resistance is still in the
ballpark. Bubbles in the phenolic indicate that the case has been hot.
>
>> >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.
? How much temperature does it take to boil gold?. Are you aware of
Stephen's Law?
>
>Following that logic, people are wasting time watching screen
>current in a tetrode amplifier, because the screen supply could
>never overheat the screen grid.
? In my tetrode amplifiers, the screen supply could never overheat the
screen grid because the primary of the screen supply is fused
accordingly. Watching screen current is hardly a waste of time because
this is how one tunes/loads a tetrode amplifier.
>
>> 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.
>
? I know it seems hard to believe, but I have seen many examples of gold
sputtering through my microscope. I agree with the 8877 design team. ,
>Have you ever calculated the heating of the grid by I^2 R losses in
>the conductor?
? How would you measure R ?
>
>You claim above heating is primarily due to I^2 R losses, yet that
>is absolutely incorrect.
? I agree completely.
>That is not the dominant mechanism that
>heats anodes and grids at all, it isn't even a factor worth
>considering.
[chortle]
>
>> 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.
? Foote's letter is on my Web site. Some of the details came from my
conversation with Mr. Foote.
>
>> 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.
? "inside and outside"? The carbon comp element is a homogenous
arrangement with wire leads at the ends.
>
>
>> >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.
>
? Rich does not know it. Tom does not know it.
>> > 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.
[chortle]
>
>Which is it? Is the parasitic a long sustained oscillation that can
>heat the resistor, or is it a sudden momentary burst?
>
? Push-pull parasitics are pretty much steady-state events that make
much heat but disappear as soon as the amplifier is unkeyed. Push-pull
parasites tend to be stentorian.
>>> 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.
? grasping at straws
>
>Please explain the apparent discrepancy between your theory and
>how things really work.
>
[guffaw]
>
later, Tom
--
FAQ on WWW: http://www.contesting.com/ampsfaq.html
Submissions: amps@contesting.com
Administrative requests: amps-REQUEST@contesting.com
Problems: owner-amps@contesting.com
Search: http://www.contesting.com/km9p/search.htm
|