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On 9/25/2013 4:27 AM, John S wrote:
On 9/24/2013 9:04 PM, David Ryeburn wrote: In article , (David Platt) wrote: When you lengthened the hairpin, you added inductance... probably too much, so you've not only cancelled out the capacitive reactance from the DE, but have left some excess inductance shunted across the DE. Backwards. Too much inductance to resonate with the effective (parallel) capacitance would have resonated with a somewhat smaller capacitance than you actually have. (The product of inductance and capacitance has to be the same, for a given resonant frequency.) So you can think of the actual capacitance present as consisting of however much would be needed to resonate with the (too large) inductance, in parallel with more capacitance which does NOT get cancelled out by the inductance. Result: the actual hairpin, plus the effective (parallel) capacitance the too short driven element presents, is capacitive, not inductive (in parallel with the desired 50 ohms). This is just the opposite from a series resonant circuit where too much inductance gives an overall inductive result. Otherwise, I agree with everything David Platt wrote. David, VE7EZM and AF7BZ No, David Platt is correct. You are assuming the effective capacitance is in parallel. The antenna equivalent circuit is a *series* RC. Put in too much *shunt* inductance and the combination looks inductive at the feed point. John KD5YI I'm wrong. Please disregard. Sorry. John |
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