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
Actually, thinking more about this, I believe a hairpin is a shorted
transmission line. So, using a Smith Chart, I investigated a 25-j25 load
and played with the chart to see what happened.
It turns out that the feed impedance is indeed inductive if the stub is
too long. So, I have now reverted to agreeing with Mr. Platt.
I think that the stub (hairpin) will have no effect if its length is a
quarter wave. Shorter than that, it becomes inductive. Very short and it
is highly inductive. Longer is less inductive. See where this is going?
Cheers es 73,
John