You'll be much better off simply using the conventional radio approach
than trying to simulate everything, especially when circuit equivalents
are nebulous like this.

After all, if you can't quite tell what it *should* look like, how would
you know if you could implement your model once you've found a
satisfactory result?

What kind of antenna are you looking at, loop? The first thing to know
about a loop is, if it's a very small loop (I'm guessing, at this
frequency, it is), its radiation resistance is very low, meaning, you can
treat it as a nearly pure inductance (Q 10 I think is typical), and its
bandwidth (even with a matched load) will be correspondingly narrow.

The nature of the incoming signal could be modeled as a voltage or current
source; how doesn't really matter, because it isn't really either, it's a
power source that couples in. Again, you don't have voltage without
current and vice versa, it's all about power flow, and the matching that
allows the power to flow.

Since the loop is inductive, your first priority is to resonate it with a
capacitor at the desired frequency. This will require a very precise
value, and even for a single frequency, may require a variable capacitor
to account for manufacturing tolerances. In the AM BCB, a Q of 10 gets
you 50-160kHz bandwidth, so you only get a few channels for any given
tuning position. And if the Q is higher, you get even fewer.

Now that you've got a high Q resonant tank, you can do two things: couple
into the voltage across the capacitor, or the current through the
inductor. You need only a small fraction of either, because the Q is
still going to be large. This can be arranged with a voltage divider
(usually the capacitor is split into a huge hunk and a small variable
part, e.g., 300pF variable + 10nF, output from across the 10nF), a
transformer (a potential transformer across the cap, or a current
transformer in series with the inductor), an inductive pickup (the big
loop carries lots of volts, but you only need a few, so a much smaller
loop can be placed inside the big loop), an impractically large inductor
(like in my example circuit, which models radiation resistance as a
parallel equivalent), etc. Whatever the case, you need to match
transmission line impedance (e.g., 50 ohms) to radiation resistance
(whichever series or parallel equivalent you have).

Once you get the signal into a transmission line, with a reasonable match
(Z ~= Z_line, or alternately, SWR ~= 1), you can do whatever you want with
it. Put it into an amplifier (don't forget to match it, too), etc. Yes,
you're going to have funny behavior at other frequencies, and if you're
concerned about those frequencies, you'll have to choose the coupling
circuit and adjustable (or selectable) components accordingly. But for
the most part, you completely ignore any frequency that you aren't tuning
for, usually enforcing that concept by inserting filters to reject any
stragglers.

Example: suppose you have a loop of 5uH and need to tune it to 500kHz. It
has a reactance of 15.7 ohms. Suppose further it has Q = 20. The ESR
(not counting DCR and skin effect) is X_L / Q, or 0.78 ohms; alternately,
the EPR is X_L * Q, or 314 ohms. The capacitor required is 20.3nF. If we
use a current transformer to match to a 50 ohm line, it needs an impedance
ratio of 1:64, or a turns ratio of 1:8. If we use a voltage transformer,
it's of course 8:1. (A capacitor divider is unsuitable for resonant
impedances less than line impedance, since it can only divide the
impedance down. If the inductance were a lot larger, it could be used.)
To a rough approximation, a smaller inductive loop, of 1/8 diameter of the
larger, I think, would also work.

Tim

--
Deep Friar: a very philosophical monk.
Website: http://seventransistorlabs.com

"rickman" wrote in message
...
On 2/28/2013 6:40 PM, Tim Williams wrote:
wrote in message
...
A higher frequency would imply a smaller L and/or C. How do you
combine
them to produce that? Consider the two caps to be in series???

Sure. If you bring the 10p over to the primary, it looks like 10p *
(30m
/ 5u), or whatever the ratio was (I don't have it in front of me now),
in
parallel with the primary. (I misspoke earlier, you can safely ignore
Ls,
because k = 1. There's no flux which is not common to both windings.)

Reflecting the capacitance through the transformer changes it by the
square of the turns ratio assuming the coupling coefficient is
sufficiently high. I am simulating K at 1.

This is also true for the inductance, but in the opposite manner. So
going from the 25 turn side to the 1 turn side, the effective
capacitance is multiplied by 625 and the effective inductance (or
resistance) is divided by 625. In fact, in LTspice you indicate the
turns ratio by setting the inductance of the two coils by this ratio.

I see now that the reflected secondary capacitance is in parallel with
the primary, rather than in parallel with the primary capacitor. That
explains a lot... I'll have to hit the books to see how to calculate
this new arrangement. I found a very similar circuit in the Radiotron
Designer's Handbook. In section 4.6(iv)E on page 152 they show a
series-parallel combination that only differs in the placement of the
resistance in the parallel circuit. It need to be placed inline with
the inductor... or is placing it parallel correct since this is the
reflected resistance of the secondary? I'll have to cogitate on that a
bit. I'm thinking it would be properly placed inline with the capacitor
in the reflection since it is essentially inline in the secondary.
Either way I expect it will have little impact on the resonant frequency
and I can just toss all the resistances simplifying the math.

I do see one thing immediately. The null in Vcap I see is explained by
the parallel resonance of the secondary cap with the secondary inductor.
If you reflect that cap back to the primary in parallel with the primary
inductor (resonating at the same frequency) it explains the null in the
capacitor C1 voltage I see. C2' (reflected) and L1 make a parallel
resonance with a high impedance dropping the primary cap current and
voltage to a null. This null is calculated accurately.

What I need to do is change the impedance equation from Radiotron to one
indicating the voltage at Vout relative to the input signal. I think I
can do that by treating the circuit as a voltage divider taking the
ratio of the impedance at the input versus the impedance at the primary
coil. No?


Inductors effectively in parallel also increase the expected resonant
frequency. If you have this,

. L1
. +-----UUU--+------+------+
. | + | | |
. ( Vsrc ) === C R 3 L2
. | - | 3
. | | | |
. +----------+------+------+
. _|_ GND

You might expect the resonant frequency is L2 + C, but it's actually
(L1
|| L2) = Leq. If L1 is not substantially larger than L2, the resonant
frequency will be pulled higher.

I see, L1 and L2 are in parallel because the impedance of Vsrc is very
low. That is not the circuit I am simulating however. The loop of the
antenna and the loop of the inductor are in series along with the
primary capacitor. I'm not sure what the resistor is intended to
represent, perhaps transformer losses? The resistance of L1 was added
to the simulation model along with the resistance of the secondary coil
which you have not shown... I think. It seems to me you have left out
the tuning capacitor on the primary.


Incidentally, don't forget to include loss components. I didn't see
any
explict R on the schematic. I didn't check if you set the LTSpice
default
parasitic ESR (cap), or DCR or EPR (coil) on the components. Besides
parasitic losses, your signal is going *somewhere*, and that "where"
consumes power!

The actual transmitter is most certainly not a perfect current source
inductor, nor is the receiver lossless. This simulation has no
expression
for radiation in any direction that's not directly between the two
antennas: if all the power transmitted by the current source is
reflected
back, even though it's through a 0.1% coupling coefficient, it has to
go
somewhere. If it's coming back out the antenna, and it's not being
burned
in the "transformer", it's coming back into the transmitter. This is
at
odds with reality, where a 100% reflective antenna doesn't magically
smoke
a distant transmitter, it simply reflects 99.9% back into space. The
transmitter hardly knows.

Interesting point. My primary goal with this is to simulate the
resonance of the tuning so I can understand how to best tune the
circuit. In many of the simulations I run the Q ends up being high
enough that a very small drift in the parasitic capacitance on the
secondary detunes the antenna and drops the signal level. It sounds
like there are other losses that will bring the Q much lower.

I would also like to have some idea of the signal strength to expect. My
understanding is that the radiation resistance of loop antennas is
pretty low. So not much energy will be radiated out. No?

You make it sound as if in the simulation, even with a small coupling
coefficient all the energy from antenna inductor will still couple back
into the transmitter inductor regardless of the K value. Do I
misunderstand you? It seems to result in the opposite, minimizing this
back coupling. Or are you saying that the simulation needs to simulate
the radiation resistance to show radiated losses?


In this example, if you set R very large, you'll see ever more voltage
on
the output, and ever more current draw from Vsrc. You can mitigate
this
by increasing L1 still further, but the point is, if the source and
load
(R) aren't matched in some fashion, the power will reflect back to the
transmitter and cause problems (in this case, power reflected back
in-phase causes excessive current draw; in the CCS case, reflected
power
in-phase causes minimal voltage generation and little power
transmission).

Power is always coming and going somewhere, and if you happen to forget
this fact, it'll reflect back and zap you in the butt sooner or later!

Tim

Actually, my goal was to build the receiver and I realized that my
design would require the largest signal I could get from the antenna. I
never realized I would end up having to learn quite so much about
antenna design.

I've been planning to create a PCB with lots of options so I can test a
number of configurations. Nothing about the simulation makes me doubt
the utility of this idea.

One thing that continues to bug me is that nothing I have seen gives me
a hint on how to factor in the distributed capacitance of the antenna
shield. I am using RG6 with 16 pF/Ft and likely will end up with 100
foot of coax total. At some point I'll just have to make some
measurements and see what the real world does.

--

Rick