>> From: Rich Measures <measures@vc.net>
To: <amps@contesting.com>
>> Date: Tue, 12 Aug 97 04:30:43 +0000
>
>> >range.......for a the ETO Q of 5.3 into 50 ohms..... and pi- tank
>> >inductor fixed at 21.57 uH and L inductor fixed at 7.255 uH.....
>> >
>> >760 ohm load, Cp 600 pF, ...
>> ...snip...
>> The output R (RL) of the pair of 8877s is approx. 3750V/(2A*2)=938 ohms.
>
>Earlier you gave 1000 ohms, I used that figure.
1000 ohms is the approx. RL for an anode V of 4k. However, this has
nothing to do with the issue at hand.
>
>> Since Q is defined as RL/XC1, and Q is given as 5.3, the reactance of C1
>> (which Mr. Rauch refers to as 'Cp') would be 938 ohms/5.3 = 177 ohms. At
>> 1.9MHz, C1 is 477pF.
>>
>> - 600pF does not check out.
>
>It certainly does. This operational example is for a "Q" of approx
>5.3 into a 50 ohm load (using your Rp value of 1000 ohms), ...
Does this wash?:
- For a Q of 5.3 and RL=1000 ohms, the reactance of the tune capacitor,
XC1 = 1000 ohms/5.3 = 188.7 ohms. ...... However, @ 1.9MHz, 600pF has a
reactance of 139 ohms.
*Sorry, Charlie*.
Operating Q has nothing to do with the external load presented to the
tank. Q is DEFINED as RL/XC1, where RL is the calculated, optimal anode
load R for the tube(s) and XC1 is the reactance of the tune C.
Rich---
R. L. Measures, 805-386-3734, AG6K
--
FAQ on WWW: http://www.contesting.com/ampfaq.html
Submissions: amps@contesting.com
Administrative requests: amps-REQUEST@contesting.com
Problems: owner-amps@contesting.com
Search: http://www.contesting.com/km9p/search.htm
|