A subtle point I think all have missed is that 50 ohms only exists in
the lab!! However, both techniques will work equally well, I think if
I didn't have a $100k network analyser at my disposal I would opt for
the VSWR meter cos' I already have one.
John ZS5JF

> From: Jon Ogden <jono@enteract.com>
> To: 'amps' <amps@contesting.com>
> Subject: RE: [AMPS] MFJ259 and matching circuits
To: <amps@contesting.com>
> Date: 03 February 1999 05:39
>
>
> >? The cathode is like a diode that is driven with pos. and neg.
> >potentials. Heavy current flows on neg. No current flows on pos.
A
> >resistor this is not.
>
> Agreed. However, even a diode has an equivalent "impedance." Yes,
it is one that varies with drive, but it is an impedance none the less
and still has an "average" value over the drive cycle.
> >
> >Well, I used what the data sheets say is the
> >>driving impedance which equals 110 Ohms. So on paper, you use 110
Ohms in
> >>parallel with 27 pF (which on any data sheet you find is spec'd as
the
> >>input capacitance). You absorb the 27 pF into the value of C2 in
your
> >>Pinet and then solve the equation for a pi matching network.
> >
> >? 110 ohms could be close. In this case, the swr meter in the
radio
> >will work as well as the MFJ259.
>
> That's correct. You want the radio to see a good match into the
amplifier. So I think you are validating my point, but you might not
know it.
> >>
> >>Now if what they taught me about matching networks in college is
all
> >>wrong, I would like to know.
> >
> >? In college we learned how to match resistive loads. A cathode is
not.
> >
>
> I know that it is not resistive. I even stated that. However, one
can simulate an average impedance for that device as an approximation.
I have spoken to others who have done this in addition to myself and
they have confirmed that it works.
> >
> >>I KNOW that there are some additional real
> >>world effects inside the tube that I can't completely account for
by
> >>simulating it, but it should get one relatively close. Otherwise
what
> >>good is working out networks by any form of calculation. Should
all of
> >>our engineering be hit or miss?
> >>
> >? Not at all. Set the input pi network Q for 2. (XC1= 25ohms).
With
> >full drive, adj. L and C2 for a match to the actual cathode.
>
> Well, we are effectively solving the exact same problem. You are
matching 50 Ohms to the input impedance of the tube. And since the
cathode impedance varies over time and is not a "resistive" load as
we've agreed upon, you still end up tuning to an "average" value. The
only difference is we start with different known and unknown variables.
In your case, the value of XC1 is known and Zcathode is unkown. In my
case, I assume a value of Zcathode (based on mfg's data sheet) and XC1
is unkown. Both cases work out the same problem. Yours might work a
little better since you are actually driving the tube under real
conditions. I won't argue that a simulation is better than the real
thing. However, your way is convenient of C2 is adjustable. In my
case, my caps are all fixed and inconvenient to get to, so I gotta
approximate.
>
> I'll let everyone know my "real" results regardless of how they turn
out.
>
> 73,
>
> Jon
> KE9NA
>
>
>
>


> Jon Ogden
>
> jono@enteract.com
> www.qsl.net/ke9na
>
> "A life lived in fear is a life half lived."
>
>
> 
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