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[AMPS] Re: Parasitics

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Subject: [AMPS] Re: Parasitics
From: measures@vc.net (Rich Measures)
Date: Fri, 15 May 98 17:37:01 -0800
>Rich Measures wrote:
>
>>G3SEK wrote:
>
>>>1. 100nH at 100MHz is +j63.83; 
>>
>>I got 62.83 
>>
>Sorry for my bad t6ping when copying results off the screen: 62.83 is
>correct.
>
>>>200nH at 100MHz is +j125.66
>>>
>>ok
>>
>>>2. Now transform 100 ohms paralleled by +j63.82 transforms into its
>>>series-equivalent. 
>Sri, more bad trping - read "62.83" again.
>
>>The equations (as posted last weekend) are:
>>>
>>>Rs = Rp*Xp^2 / (Rp^2 + Xp^2)       Xs = Rp^2*Xp / (Rp^2 + Xp^2)
>>>
>>>Answer: 28.30 in series with +j48.05
>>I got 28.32 ohm R, however I came up with  +j45.06 ohms of X {cos 57.85 
>>deg. x 53.22 ohms of Z}  
>>
>My typing still hasn't improved, but your method agrees with the correct
>answer to at least 3 sig figs, Rich. Let's assume it's rounding errors
>between our two calculations, and press onwards.
>
>
>>>3. Add the +j125.66 in series, to get 28.30 +j170.71 (still series-
>>>connected)
>>>
>>hello
>>
>>>4. This is now  NEW NETWORK that includes the external 200nH. Transform
>>>this new network back into its parallel-equivalent:
>>>
>>It is my opinion that this is not a legal move.
>
>Why not? It's a series R-L network, same as any other.
>
>>   In other words, adding 
>>external X to a parallel L-R circuit does not change the Admittance (Y) 
>>of the parallel circuit.    In other words, it appears that you are 
>>trying to add oranges and apples, Mr. White.  
>>
>The original parallel circuit disappeared in step 2 when we transformed
>it into its series-equivalent. There is absolutely no bar to treating
>the series-equivalent circuit just like any other series circuit.

I do not believe it.  
>
>By the way, if the move I made isn't "legal", all your pi-tanks just
>stopped working!
>
A pi-network tank is not the same thing
>--------------------------------------------
>
>>Your calculation of how the 100ohm suppressor R became 49 ohms at 10MHz 
>>in Wes' measurements would interest me. 
>
>That's presumably your pure-nichrome/100-ohms suppressor, which had a
>measured Rp of 49.26 ohms at 10MHz. (I'm copy-and-pasting these figures
>straight off Wes's table in your web page, 

Wes' table is not in my Web page.

>to avoid any more typing
>errors.)
>
>On its own, the pure-nichrome inductor looked like 95.6nH (Lp) in
>parallel with 93.46 ohms (Rp).

Rp is 49.26 ohms.  The 100 ohm suppressor R is in parallel with the 
95.6nH suppressor inductor, which Wes calls Ls.  93.46 ohms is not the 
resistance of the suppressor R.  
>...........


Rich...

R. L. Measures, 805-386-3734, AG6K, www.vcnet.com/measures  


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