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Old March 7th 13, 02:50 PM posted to sci.electronics.design,rec.radio.amateur.antenna,rec.radio.amateur.homebrew
rickman rickman is offline
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Default Antenna Simulation in LTspice

On 3/6/2013 8:13 PM, Tim Williams wrote:
wrote in message
...
Yes, I plan to use a shielded loop. I have found some contradictory
info on the effectiveness of the "shield". One reference seems to have
measurements that show it is primarily E-field coupled in the longer
distance portion of the near-field.


I trust this resource:
http://vk1od.net/antenna/shieldedloop/
He's got gobs of analytical articles.


Yes, I've seen this page. Thanks.


Yes, that is loop antenna 101 I think. It was when I added a coupling
transformer with 100:1 turns ratio that I was told I needed to consider
the parasitics. I have found it is not useful to go much above 25 or
33:1 on the turns ratio. I am receiving a single frequency, 60 kHz.
There is no need for a wide bandwidth. Ultimately, I prefer a Q of
100 for the higher gain. If it gets too high, the off tuning by
variations (drift) in the parasitic capacitance affects the antenna gain
appreciably.


High Q isn't the goal, high radiation resistance is -- the bigger the
loop, the better it couples with free space, until it's a wave length
around.


I'm not clear on why you keep referring to radiation resistance for a
receiving antenna. Does this result in a larger received signal? I am
concerned with maximizing the voltage at the input to the receiver.


You can go ahead and make a teeny coil out of polished silver litz wire,
and push the Q up into the hundreds, but all you'll see is internal
resistance, hardly anything attributable to actual radiation. Since the
losses dominate over radiation, it makes a crappy antenna. But you know
that from looking at it -- it's a tiny lump, of course it's not going to
see the outside world.


I have no idea why you are talking about Litz wire and tiny coils. I
never said I was looking to maximize the Q. I said I wanted a Q of over
100. I should have said, slightly over 100. A higher Q clearly does
increase the voltage on the input in my simulations. Is there something
wrong with my simulations?


It is true, however, that a small coil, with low losses, will have low
noise. AM radios rely on this, which is how they get away with tiny hunks
of ferrite for picking up radio.

Of course, it doesn't hurt that AM stations are 50kW or so, to push over
atmospheric noise.

Transmission line? What transmission line? The antenna is directly
connected to the receiver which has a very high input impedance. Why do
I need to consider radiation resistance? I have not read that
anywhere.


Ok, then you can merge the matching transformer, transmission line and
receiver input transformer into one -- an even larger stepup into whatever
impedance it's looking at (what's "very high", kohms? Mohms?) will get you
that much more SNR.


Yes, a higher stepup ratio gets larger signal up to a point. That point
is determined by the parasitic capacitance of the receiver input. That
capacitance is reflected back through the transformer and affects the
antenna tuning. In my simulations it creates a filter with two resonances.


I'm not familiar with the concept of voltage transformer vs. current
transformer. How do you mean that?


Current transformer measures current (its winding is in series), potential
transformer measures voltage (in parallel).


Series and parallel with what? I'm not following this. I have trouble
with series and parallel resonance, but I'm starting to get the concept.
Sometimes it is hard to tell how a circuit is being stimulated.


How did you get the 1:64 impedance ratio and the 1:8 turns ratio? I
don't follow that. Are you saying the line impedance should match the
ESR? Why exactly would it need to match the ESR?


ESR (and Q) measured on the coil corresponds to radiation resistance
(series equivalent) *plus* internal losses (also series equivalent). You
can't separate the two components, so you can only get the best power
match by the good old impedance theorem.


Internal losses of what? How do you determine the internal losses?


~1:64 is 50 ohm / 0.78 ohm, and N2/N1 = sqrt(Z2/Z1), or 8:1 turns ratio.


Ok, so you were matching the hypothetical ESR to the hypothetical line
impedance.

--

Rick