>
>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?
>
? I don't have one. Eimac defines Q as the ratio between RL and the
reactance of C1. None of the Eimac formulae can be used without choosing
a value of Q - as defined by their 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.
>
? A definition which does not work with Eimac's formulae.
>[ 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. ]
>
? So is it your position that the resonance of the 7.00MHz Pi-network
that you came up with ( C1=102.1pF, L=5.57uH, C2=463.9pF) has a resonant
frequency that will be changed by removing a 2000-ohm resistor in
parallel with C1 and removing a 50-ohm resistor in parallel with C2? .
>>>
>>>It makes a big difference to the resonant frequency whether the load is
>>>connected or not.
? Shouldn't an RL equiv.R be connected to the input?.
>> > 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).
>
? thanks, Ian.
>>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*.
>
? Assuredly, however there are two resistances, and that is a wholly
different ballgame.
>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.
>
? I think not. At this point, we need a totally unbiased volunteer to
construct the Pi-network according to Ian's calculations ( C1=102.1pF,
L=5.57uH, C2=463.9pF, 7.00MHz, 2000 ohms in/50 ohms out) and measure
resonance with and without the appropriate resistances
>With an inaccurately designed network, the Q and resonant frequency are
>both slightly wrong, so those exact relationships do not apply.
>
? Agreed. We will use your calculated values.
>>>
>>>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.
? doubtful. My guess is that having both resistances of the correct
values will not cause a frequency shift.
>
>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.
>
? Do we agree that an L-network is never resonant? If the answer is
yes, why should two L-networks in series (a Pi-network) be resonant?
Rich...
R. L. Measures, 805-386-3734, AG6K, www.vcnet.com/measures
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