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.
By the way, if the move I made isn't "legal", all your pi-tanks just
stopped working!
--------------------------------------------
>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, 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).
When shunted with a 100-ohm resistor, the Rp of the network is simply
calculated like resistors in parallel - after all, it *is* resistors in
parallel. That would be 1/(1/93.46 + 1/100) which is 48.31 ohms. In
principle, Lp would be unchanged from its previous value of 95.6nH.
Given that we've assumed ideal behavior, that prediction agrees very
well with the measured values for the shunted suppressor:
Rp = 49.26 ohms, and Lp = 93.9nH. Actually that is better agreement than
we're entitled to hope for, given the importance of stray series
inductance... just lucky, I guess.
BTW, I put NETCALC up on my web pages this evening (Friday) evening, but
the service provider may not release the new material until as late as
Monday. I'd be grateful if someone can tell me when it's accessible in
the USA - the updated home page will have a NEW flash and a link to the
new NETCALC page.
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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