>Rich Measures wrote:
>>>>Sample problem: Part I: Ls = 200nH, Rs = 200 ohms, f = 100MHz. What
>>>>is the Admittance, Y, Mr. White? A running description of your solution
>>>>would be helpful.
>>>
>>>No, I won't bite. There is absolutely no need to drag admittance into
>>>this discussion.
When L is in parallel with R, admittance Is the name of the game. When L
is in series with R, impedance Is the name of the game.
>>Then on to Part II: what is the parallel-equivalent resistance of 200nH
>>in parallel with 200 ohms at 100MHz? (presumably using the method
>>described below)
>>
>200 ohms, at ANY frequency, by definition - no calculation is required.
>
.....not the definition of Rp in Wes' measurements.
>Part III - it's my turn: what is the series-equivalent form, ie what are
>Rs and Xs at 100MHz?
>
>XL = 2*pi*100E6*200E-9 = 125.66 ohms
>
>XL is equal to Xp by definition, because L and R are connected in
>parallel.
>
However, when an L, or a C, and an R are connected in parallel, we have
susceptance and conductance in Siemens instead of reactance (X) and
resistance in Ohms, which, at a discrete frequency result in an
admittance (Y) instead of an impedance (Z).
However Y can be converted to Z, and vice-versa.
>R = 200 ohms = Rp by definition as we hopefully agreed above.
>
We do not because we are discussing Wes' measurements, not the *ARRL
Handbook* (which used to be called the *Radio Amateur's Handbook*).
>Then, using the equations I gave last time (out of the ARRL Handbook
>since time immemorial):
>
>
>Rs = Rp * Xp * Xp / (Rp * Rp + Xp * Xp)
>
> = 56.61 ohms
>
>Xs = Rp * Rp * Xp / (Rp * Rp + Xp * Xp)
>
> = 90.10 ohms - at 100MHz that's 143.40nH
>
>Q = Xs/Rs = 1.59
>
>(Check arithmetic by calculating Q directly from the input values:
>1/Q = Xp/Rp = 0.63, so Q = 1.59 which is correct.)
>
It looks to me like Q is pretty close to 0.63, which is lower than Wes'
measured Q of 1.3 for the W8JI suppressor, which used 100nH and 109
ohms.. I seem to recall that you and Wes stated that decreasing R would
decrease Q, however, increasing R to 200 ohms seems to have decreased Q.
.
>What do you make Xs, Rs and Q, Rich?
>
Q is not 1.59. "Xs & Rs" look ok, however, Wes and I use "Rp" where you
use "Rs" to designate the parallel-equivalent R.
>
>What does this mean in real-world terms?
>It means that at exactly 100MHz, a network analyser or R-X bridge cannot
>tell the difference between {200 ohms in parallel with 200nH}
agreed
>and its series equivalent which is {56.61 ohms in series with 143.40nH}.
. In real world terms, it means that, at 100MHz, this VHF parasitic
suppressor presents a load to the anode of 56.6 ohms in series with 143nH
to ground, which beats the resistance-wire (AG6K) suppressor's 101 ohms,
as well as beating the copper-wire (W8JI) suppressor's 166 ohms, which,
in my opinion, will result in minimal voltage amplification at 100MHz -
__Provided__ one can find a 45w, 200 ohm,12nH resistor. . . However,
the sticky-wicket is that if C-Tune has a resonance near 100MHz, all bets
are off, which is very probably why the lowest-Q vhf suppressor sometimes
can't do its job properly.
- (note: power dissipation calculations are for a 3-500Z at 28MHz.
However, for an 8877, the dissipation in the suppressor resistor would be
approx. 4x more)
>
>The two networks have exactly the same properties at that frequency. That's
what
>"equivalent" means in this context.
well put, Mr. White.
cheers
Rich...
R. L. Measures, 805-386-3734, AG6K
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