At 05:23 AM 8/11/97 -0800, you wrote:
>>At 05:55 PM 8/10/97 -0800, you wrote:
>>
>>>How much grid current equates to 8.2W of grid dissipation?
>>
>> 55 ma per tube.
>...
>55mA divided into 8.2 watts equals 149V rms. Since the grid/cathode
>driving potential for an 8877 is around 50V rms, where does the extra 99V
>rms come from, Phil?
>Rich---
The 55 ma is only the DC component of the total grid current. This allows
the use of a simple DC amp meter to sample the TOTAL complex AC, DC, and RF
currents flowing in the grid. By following the Ip and Ig curves in Care
and Feeding of Power grid Tubes, one can get the instantaneous (snap shot)
values of total currents as the grid and plate currents swing positive and
negative. Also from these curves, we can get the corresponding DC current
values that will make up the DC component that we need to watch closely
so as not to exceed maximum grid dissipation. The 55 ma DC is not the
culprit that ruins grids. It is exceeding the TOTAL AC/DC limit of 20/25
watts (your choice) grid dissipation that does the grid in. Think of the
grid current meter on an 8877 as a "grid dissipation meter" if you wish.
I guess one could rescale a 0-300 ma meter to "grid watts" if so desired,
by using the Eimac curves given a certain power input.
(((73)))
Phil, K5PC
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