>> From: Peter Chadwick <Peter.Chadwick@gpsemi.com>
>>
>> OK, I see what you mean. I made the unwarrantable assumption that one
>> automatically drives the grid up to zero volts to get the most swing -
>> but you don't have to, although that seems to be the recommended design
>> procedure for maximum output from any particular tube in AB1.
>
>That's right Peter. I'm sorry I didn't explain that better.
...snip.........
>...just like
>there is no restriction against grid current in any letter class of
>PA. And contrary to claims, there is NO requirement peak anode
>current be reached at zero grid volts. It might be reached, but it
>more than likely it is not.
There is no requirement that the emission capability of the tube be fully
utilized by the designer.
>A 3CX3000F1 will work in AB2.....
According to Eimac Curve #3452, in Class AB2 g-g, an F1 (Mu = 5),, the
RF grid-cathode potential needs to swing from neg. 990v to +140v =1130v
p-p. The cathode current is about 2.8a avg. OTOH, according to Eimac
Curve #3373, the F7 (Mu = 200), the RF grid-cathode potential needs to
swing from roughly 0v to +140v = 140v p-p It seems to me that 140v p-p
volts of cathode drive would be a bit easier to come by than 1130 p-p
volts of cathode drive. Eimac lists the g-g cathode driving power at
410w for the F7, but provides no such data for the F1. Based on the fact
that the driving V needed is about 8-times higher for an F1, and that the
F1's and F7's cathode currents are quite similar, the driving power
requirement for the F1 should be a close match for a 3000w output
transceiver. However, since a goodly portion of the drive power in g-g
appears in the output, the F1 would undoubtedly make more suds than an
F7. OTOH, coming up with the 990v of bias for an F1 would seemingly be
harder than coming up with the 0v of bias for the F7. .
> just as a 3CX3000F7 will work in
>AB1, or vice versa.
For an F7 in Class AB1, using a 5.5kV anode supply, and neg. 15v of grid
bias, the peak anode current under maximum signal conditions would be
roughly 30mA---which equates to an average anode current of about 11mA,
yielding an input power of about 60w. Assuming 60% efficiency, power
output would be roughly 30-something watts. However, the power needed to
drive the grid would be minimal since no grid current flows and the
driving potential is 30v p-p. .
- - The bottom line is that Mr. Rauch is indeed "EXACTLY correct".
Rich---
R. L. Measures, 805-386-3734, AG6K
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