----- Original Message -----
From: "R.Measures" <r@somis.org>
>>
>>> Au has c. 60% of the conductance of Cu or Ag. Just because the grid
>>> has C does that mean there is no P dissipation in the grid?
>>
>> No, of course not, Rich, but showing that a thin piece of gold plated
>> wire
>
> It's not thin wire, it's rectangular bar stock and there are 108
> segments - each of which appear to have the surface area of approx.
> #16-AWG wire. .
>
Well....yes..... but the argument is still the same. And if grid dissipation
were a function simply of conductor ESR, why would even 350mA let
alone 100mA be a big deal for a #16 AWG equivalent conductor? A
#16 AWG conductor could carry 350mA all day long even in a vacuum.
>> can carry 40 amps of RF current doesn't say anything about how much
>> power it can dissipate. The anode structure of a tube has ESR too, but
>> we don't calculate plate dissipation based on
>> I^2*ESR. We calculate plate dissipation based on Ep x Ip - Pout.
>> Changing the ESR of either structure won't necessarily change their
>> dissipation significantly (e.g. if I could make the ESR of the grid
>> or anode conductor zero, they would still dissipate power and get
>> hot due to the kinetic energy of the electrons hitting them).
>
> For the anode, the velocity of electrons is great, but for the grid it
> seems not.
In either case its a function of the accelerating voltage. For the anode
its a function of Ep x Ip, for the grid it is a function of Eg x Ig.
73 Mike, W4EF........................
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