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Re: [Amps] 807 transmitter

To: "The Bowers" <>, <>
Subject: Re: [Amps] 807 transmitter
From: "Tom W8JI" <>
Date: Mon, 5 Feb 2007 18:54:18 -0500
List-post: <>
> Has anyone built "K5DH's 807 CW transmitter" (Google) ? 
> I've built just the
> amp section but can't get it to load. I'm driving it with 
> a sig gen at 3.9
> KC.
> According to the on line calculators, L3 is at 10.3 uh, 
> but should be 18.5
> uh., for 75 meters.


There is a VERY wide range of tank Q's that will work just 
fine. People go way overboard fussing around to get a 
certain Q that is really just a wild approximation anyway 
and often wrong.

The minimum Q you can use and have the network behave like a 
Pi is the square root of the impedance ratio plus a tiny 
bit. This would be a phase shift of about 90 degrees.
The highest Q is really limited only by the components and 
how narrow you want the tuning.

If the tank components are reasonable quality you won't 
notice much difference in performance over a VERY wide range 
of Q's, so don't get caught up in the "Q-steria".

Now I'm assuming you are running deep class C and you meant 
3.9 MHz and not kHz  :-)  and your generator has enough 
output to drive the tube (a few watts anyway if it is 
running directly into the grid circuit) and anode voltage is 
about 750 volts.

If so, the plate voltage would swing about 500 volts RMS. 
That would be about 5500 ohms plate loadline with 50 watts. 
Let's say you want a pi that would match 20 ohms as a 
minimum impedance.  The tank Q would have to be a minimum of 
sqrt 5500/20 = 16.5 plus a little. Let's say 17 would be a 
good number.

You can see from this how the idea all tanks need a Q of 12 
falls apart.

The plate C would  be 125pF, peak voltage just over 700.
The inductor  would be 13.3uH and current would be 1.6 
amperes RMS.
The loading cap 500pF

Then when you have a 50 ohm load (or higher) all you need to 
do is close up the loading cap and re-dip the plate. 
Loading C would be  1155pF and plate C 135pF and Q would be 

With 75 ohms the settings would be loading 1050pF, tuning 
138pf, and L would still be 13.3 uH. Now the loading moves 
in a normal direction because Q is significantly higher than 
the minimum needed (phase shift 150 degrees).

If you use the values I just gave with a tank inductance of 
13.3 uH  and  capacitors of 150pF and 1300pF you could match 
any pure resistance between 20 and a hundred ohms or more. 
The reason the load cap works "backwards" at low Z is we are 
running right on the edge of minimum Q. Phase shift is only 
around 100 degrees.

So you see the fellow who wrote the article is probably 
closer than the on-line calculators. The tank inductor would 
have to be 13.3 uH or LESS to have enough Q to match 20 ohms 
to a few hundred ohms. he probably had a little more Q than 
necessary, assuming his cap values were correct (I didn't 
read it).

Remember the rule. You need a tank Q of at least the square 
root of the largest impedance ratio you expect to match plus 
a small safety factor. That would set the maximum L value at 
the highest frequency for any band. using less L than that 
would be safe, and more L means it would not tune under the 
impedance extremes.

73 Tom 

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