[AMPS] Pi-Net math

[Ian White, G3SEK]

bob alexander wrote:
>Looking at the equations for this "exact" method, I have no problem with
>the equation for Q1, or how Q2 is determined or with the equations for
>Xc1 and Xc2.  
>There is, however, a problem with the equation for XL.  It yields a
>value of XL that results in insufficient inductance in the network.
>That is why you end up with a higher resonant frequency than you
>designed for when you do the math using the values derived from this
>"exact" method.  Your circuit for 7 MHz, Q=10, Rl=2000 and Ra=50 is a 
>good example.  The values calculate to a resonant freq of 7.36 MHz not
>7.00 MHz.  
>The solution is  rather simple.   Delete the equation for XL...
>XL = (RL*Qo)/(Q1^2 +1)... and replace it with
>XL = Xc1 + Xc2 (absolute values).
>
>Again, using your example... recalculate XL using the above equation and
>you get an inductance of 6.1729 uH.  Now, do the calcs for resonant
>frequency and you get 7.00 MHz.   The input L network transforms 
>RL = 2000 ohms to 24.55 ohms ( Q=8.97).  The output L network transforms
>24.55 ohms up to 50 ohms (Q=1.03).   Total Q = 10.
>
>Now it really is "exact".

Unfortunately not... Your method gives the correct resonance when both
input and output are unloaded, but when the output is loaded with 50
ohms, the resonance moves down to about 6.68MHz. 

I think we've gotten ourselves overly preoccupied with "resonance". The
bottom line is: 
* At the design frequency and when terminated with 50 ohms, what load
impedance does the network present to the tube? 
* What is the loaded network Q in this condition?

At 7.0MHz, your network presents the tube with a load of 2504 ohms in
parallel with 10pF, and not the 2000 ohms resistive we were aiming for.
The "New Handbook" method gives exactly 2000 ohms resistive and a Q of
exactly 10. In the sense that it presents a purely resistive load
impedance, that network IS resonant. 

However, the resonance that you'd find with an idealized GDO (when all
the reactances in the loop cancel out) is at 7.04MHz, and when you take
the 50 ohm load off it moves up to 7.36MHz. If you put a 2000 ohm load
on the input, then the GDO resonance does move down to exactly 7.0MHz...
but the dip wouldn't be so good with the network loaded at both ends,
because the Q then comes down to 5. 

By the way, I'm doing these calculations with an extended version of the
"New Handbook" design spreadsheet that's on my web site. Unfortunately I
can't post the extended version yet, because it seems to change with
every new batch of AMPS messages!

73 from Ian G3SEK          Editor, 'The VHF/UHF DX Book'
                          'In Practice' columnist for RadCom (RSGB)
                           http://www.ifwtech.demon.co.uk/g3sek

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