> From: Rich Measures <measures@vc.net>
To: <amps@contesting.com>
> Date: Tue, 12 Aug 97 13:28:57 +0000
> >> - 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---
Rich,
Let me walk you through this step by step.
1.) I calculated the Q based on the ETO values you bantered about, of
a 500 pF cap and 1000 ohm Rp.
2.) The anode C for 1.9 MHz and a Q of 5.3 was 444.5 pF.
I set the intermediate Z as 200 ohms in the pi-L, and calculated the
inductor values and loading cap value to match 50 ohms at that Q.
3.) I then calculated impedance matching range with fixed inductor
values, with an upper limit of 500 pF and 600 pF in the plate
capacitor and NO inductor changes (since they normally are fixed
values for each band).
4.) The results were a tank Q of 5.3 at a design center of 50 ohms
allows matching 22.8 to 325 ohms with 500 pF maximum at the anode,
and if that maximum value were extended to 600 pF---with no change in
the 50 ohm Q of 5.3--- loads from 9 to 750 ohms could be
accommodated.
Pretty basic stuff, but then so is grid dissipation and IMD. I hope
you follow this now.
73 Tom
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