>
>
>-----Original Message-----
>From: Rich Measures <measures@vcnet.com>
>To: amps@contesting.com <amps@contesting.com>
To: <amps@contesting.com>
>Date: 11 August 1999 01:25
>Subject: Re: [AMPS] Pi-Net math
>
>
>
>
>Dick
>
>>Rich...
>>
>> ........
>>........... Time to do your homework.
>>
>? I did the homework, Dick. I built the Pi-network test
>model. I used
>Ian's calculated values for: 2000-ohms RL / Ro=50-ohms /
>7.00MHz / Q=10
>or so.
>- Ian's results:
> "The accurate formula - in ARRL Handbooks since 1995 -
>gives
>C1=102.1pF, L=5.57uH, C2=463.9pF. When analysed with the
>load present,
>this checks out with Q=9.98. "
>
>- My results . Without RL and Ro, the resonant freq. was
>7.37MHz, which
>checks out mathematically. When I connected the 2000 ohm
>and 50 ohm
>resistors, the resonant frequency did not appear to drop to
>7.00MHz as
>had been predicted.
>Is there perhaps something that I am doing wrong, Dick? .
>Ian? . Tom?
>. Peter? Any suggestions would be greatly appreciated.
>
>
>
>In normal use, you don't have resistors loading the circuit
>at both ends - you put a load on one end and use the circuit
>to transform the impedance so that looking into the network
>you see the desired impedance.
However, the tube is a time-varying resistance.
> If you remove one of the
>resistors, I think the resonant frequency will decrease.
Sure, but the tube is there.
>Without doing any maths, I think I agree that the resonant
>frequency doesn't change if you load the circuit with the
>designed resistor values at both ends.
>
ok
- later, Steve
>
>
>--
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>
-
- Rich...
R. Measures, 805-386-3734, AG6K, www.vcnet.com/measures
--
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