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[AMPS] Grid dissipation

To: <amps@contesting.com>
Subject: [AMPS] Grid dissipation
From: w8jitom@postoffice.worldnet.att.net (w8jitom@postoffice.worldnet.att.net)
Date: Fri, 8 Aug 1997 20:56:14 +0000
> From:          Rich Measures <measures@vc.net>
To: <amps@contesting.com>
> Date:          Fri, 8 Aug 97 16:41:58 +0000

> >You really should not give false advice like this to people trying to 
> >learn.  Twisted statements like you made above ...
> ...snip...
> Taking a closer look at the allegations of false and twisted:  

> - According Eimac, the 8877 has about 10pF of C between the anode and the 
> grid.  At 250MHz, 10 pF has 64 ohms of capacitive reactance.  The AC 
> current through the grid = 2600V rms/64 ohms = 40.6A rms.  ....... Sure, 
> 40A is a fair amount of current.  However, the 8877 has over 100 
> rectangular grid wires, so the current is only about 400mA per 
> gold-plated grid wire.  Since gold is a fairly good conductor and gold 
> does not begin to evaporate in a vacuum until its temp. exceeds 1800 
> degrees F, why should there be a problem?

You are mixing conduction currents with cathode-grid emission 
currents, in the attempt to mislead people into thinking a few 
hundred mA of cathode-grid emission current is "nothing to worry 
about". 

The heating effects are VERY different.

The RF current relates to I^2 R losses and since R is extremely low 
the heating is low. Besides the low resistance, much of the 
current never even flows in the grid wires. Grid to anode capacitance 
also includes capacitance outside the grid structure (such as through 
the ceramic from the anode to the grid ring, and from the anode to 
the grid support). 

The emission related heating is cathode to grid voltage times grid 
current integrated over time. That is the effect heating the grid.

It matters very little if the element is graphite or gold when 
discussing kinetic heating caused by emission currents. The heat 
produced is almost exactly the same, since the vast majority of heat 
is caused by the kinetic energy of electrons striking a "target". 

Using your I^2 R analysis on an anode hewating problem, people might 
conclude they could run 40 amperes of dc anode current at 3 Kv. 
Nearly everyone knows that would produce 120 kW of anode heating.

Grid dissipation, just like anode dissipation, is almost 
exclusively the time integrated value of current times cathode to 
element voltage over the entire RF cycle.

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