[AMPS] Network analysis of suppressors

[measures]

>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