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[AMPS] Pi-Net math

To: <amps@contesting.com>
Subject: [AMPS] Pi-Net math
From: G3SEK@ifwtech.demon.co.uk (Ian White, G3SEK)
Date: Thu, 5 Aug 1999 09:49:19 +0100
Rich Measures wrote:
>>>RL = 2000 ohms.  R out equals 50-ohms.  Freq = 7.00MHz.  Q = 10
>>>Therefore, C1 = 2000 ohms/10 = 200 ohms.  From calculation, C1 = 114pF,
>>>C2 = 562pF, L = 5.06uH.
>>
>>Rich evidently used an approximate formula - it fails the analysis test,
>>which indicates a Q of 11.4 at the design frequency, with the load
>>present.
>>
>?  There is more than one way to figure Q.

Evidently - like to tell us your definition?

The definition I'm using is Q = XL/Rs, where Rs is the transformed
effect of all resistive components, when made to appear in a series loop
with L.

[ For kibitzers: 
The terminating resistors are originally in parallel with the capacitors
C1 and C2. To add these R and C components together correctly, you have
to transform each parallel RC into the series equivalent. This changes
the *effective* values of all R's and C's. 

Once you've got all the components into one big series loop (two
modified R's, two modified C's and the original L), then you can add the
two R's together, and also add the reactances of the two C's together. 

Note that the whole calculation is frequency-dependent, so finding the
resonant frequency of the loop is a tedious cut-and-try process.
Fortunately that's what computers are good for. ]

>>
>>It makes a big difference to the resonant frequency whether the load is
>>connected or not. The tank is designed to operate with a load of 50 ohms
>>in parallel with C2. If the load is not present, it changes the
>>effective value of C2 in series with C1, so the resonance will indeed
>>shift.
>
>?  To do it right, we really need to connect 2000-ohms across C1 and
>50-ohms across C2.  

I don't think necessarily agree, but do not wish to open that much
larger debate here and now (it's currently playing on rec.radio.
amateur.antenna).

>Want to make a wild guess as to whether this will
>increase or decrease resonance?

If you do wish to include that resistance, I will tell you - no
guesswork - that it moves the resonant frequency *down*. 

With an accurately designed pi-network, loads at both ends will put the
resonant frequency *exactly* on the design frequency. The loaded Q is
then exactly *half* the design Q that you had with only the 50 ohm load.

With an inaccurately designed network, the Q and resonant frequency are
both slightly wrong, so those exact relationships do not apply. 

>>
>>With either the accurate formula or the inaccurate one, the resonant
>>frequency with the load present is between 7.00 and 7.05MHz.
>>
>?  hardly, Ian.   The calculated resonance is now 7.371MHz.
>
Care to share your working? 
With no loads, you correctly calculated 7.266MHz. Adding one load at the
C2 end, the resonance moves down. Adding a second load at the C1 end, it
moves... up? Try 6.990MHz.

I'm checking all these calculations using both a spreadsheet and an
older BASIC program. Phil, VA3UX, has an independently derived
spreadsheet of his own, and is also checking as we go.


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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