Antenna Simulation in LTspice

I am working on a simulation for a loop antenna in LTspice and I can't 
figure out why the signal strength features are what they are.  The 
model uses a pair of loosely coupled inductors to model the transmitter 
and antenna loop with a separate pair of tightly coupled inductors to 
model the coupling transformer.  A cap on the primary circuit is the 
tuning cap and a cap on the secondary is parasitic effects of the 
circuit board leading to the inputs on the IC.

There is a resonance near the frequency I would expect, but it is not so 
close actually.  I can't figure why it is about 5% off.  There is a 
second resonance fairly high up that I can't figure at all.  None of the 
component values seem to combine appropriately to produce this peak. 
When looking at the tuning capacitor voltage there is an anti-resonance 
that is exactly at the frequency corresponding to the secondary 
resonance with the transformer and the parasitic capacitance.  That 
makes sense to me, but it is pretty much the only part that jibes with 
what I can figure out.

I have uploaded a zip file with the schematic and a measurement file.

http://arius.com/temp/Antenna_trans_LTspice.zip

-- 

Rick

I expect if you reflect the CT secondary stuff (don't forget Lsec) back to 
the primary, your answer will appear.  Offhand I can't reason out which 
sum of L and C makes the resonance, but it's a four pole series-parallel 
resonant circuit, analysis should lay it bare.

Tim

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

"rickman" <gnuarm@gmail.com> wrote in message 
news:kgmhkd$rq3$1@dont-email.me...
>I am working on a simulation for a loop antenna in LTspice and I can't 
>figure out why the signal strength features are what they are.  The model 
>uses a pair of loosely coupled inductors to model the transmitter and 
>antenna loop with a separate pair of tightly coupled inductors to model 
>the coupling transformer.  A cap on the primary circuit is the tuning cap 
>and a cap on the secondary is parasitic effects of the circuit board 
>leading to the inputs on the IC.
>
> There is a resonance near the frequency I would expect, but it is not so 
> close actually.  I can't figure why it is about 5% off.  There is a 
> second resonance fairly high up that I can't figure at all.  None of the 
> component values seem to combine appropriately to produce this peak. 
> When looking at the tuning capacitor voltage there is an anti-resonance 
> that is exactly at the frequency corresponding to the secondary 
> resonance with the transformer and the parasitic capacitance.  That 
> makes sense to me, but it is pretty much the only part that jibes with 
> what I can figure out.
>
> I have uploaded a zip file with the schematic and a measurement file.
>
> http://arius.com/temp/Antenna_trans_LTspice.zip
>
> -- 
>
> Rick 

On 2/27/2013 11:40 PM, Tim Williams wrote:
> I expect if you reflect the CT secondary stuff (don't forget Lsec) back to
> the primary, your answer will appear.  Offhand I can't reason out which
> sum of L and C makes the resonance, but it's a four pole series-parallel
> resonant circuit, analysis should lay it bare.

Are you sure about this?  When you say to reflect the secondary back to 
the primary that means the primary inductance would be doubled?  I 
believe the coupling of the two coils means they are one and the same 
for the purposes of the circuit analysis, no?

You can't reason which sum of L and C makes the resonance and I can't 
either.  The calculation is off by about 5% and I can't explain that.  I 
can explain a null at about 290 kHz.  That is the resonance of the 
secondary with the secondary capacitance.  I can't explain the other 
peak at 363 kHz at all.  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???

-- 

Rick

"rickman" <gnuarm@gmail.com> wrote in message 
news:kgoo4e$ts5$1@dont-email.me...
> 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.)

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.

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.

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

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

On Wed, 27 Feb 2013 22:07:51 -0500, rickman <gnuarm@gmail.com> wrote:

<snip>
>There is a resonance near the frequency I would expect, but it is not so 
>close actually.  I can't figure why it is about 5% off.  There is a 
>second resonance fairly high up that I can't figure at all.  None of the 
>component values seem to combine appropriately to produce this peak. 
<snip>

Pulling out the old reactance paper, there are a couple of expected
interactions using the values present:

Around 50KHz (89.42uH+48uH) with 50.42nF  (L3+L1) with C1
Around 290KHz (89.42uH+48uH) with 6.25nF  (L3+L1) with C2*N^2 
Around 360KHz 48uH with 6.25nF  L1 with C2*N^2
nL1/nL2=N=25

The mid-resonance is a dip or rejection.

What's the issue?

RL

On Wed, 27 Feb 2013 22:07:51 -0500, rickman wrote:

> I am working on a simulation for a loop antenna in LTspice and I can't
> figure out why the signal strength features are what they are.  The
> model uses a pair of loosely coupled inductors to model the transmitter
> and antenna loop with a separate pair of tightly coupled inductors to
> model the coupling transformer.  A cap on the primary circuit is the
> tuning cap and a cap on the secondary is parasitic effects of the
> circuit board leading to the inputs on the IC.
> 
> There is a resonance near the frequency I would expect, but it is not so
> close actually.  I can't figure why it is about 5% off.  There is a
> second resonance fairly high up that I can't figure at all.  None of the
> component values seem to combine appropriately to produce this peak.
> When looking at the tuning capacitor voltage there is an anti-resonance
> that is exactly at the frequency corresponding to the secondary
> resonance with the transformer and the parasitic capacitance.  That
> makes sense to me, but it is pretty much the only part that jibes with
> what I can figure out.
> 
> I have uploaded a zip file with the schematic and a measurement file.
> 
> http://arius.com/temp/Antenna_trans_LTspice.zip

Re-read what Tim Williams said.

A way to translate what he's saying into your simulation is to include 
the radiation resistance of the antenna into your simulation, and reduce 
the coupling -- 1e-6 is probably good enough.  I am, frankly, not sure 
where this is best put in your radiation resistance, but just increasing 
the series resistance on L3 is probably sufficient; putting it in as a 
parallel resistance in L3 is probably more accurate, but would be more 
useful as a way of separating the radiation resistance effect from the 
winding resistance of L3 (which is, I assume, where your figure comes 
from).

Do you have the ability to measure the Q of your antenna as built, and 
compare it to the Q calculated from the known L, C, and winding 
resistance?  That should give you a good estimate of the radiation 
resistance, or at least radiation resistance + other losses.

-- 
My liberal friends think I'm a conservative kook.
My conservative friends think I'm a liberal kook.
Why am I not happy that they have found common ground?

Tim Wescott, Communications, Control, Circuits & Software
http://www.wescottdesign.com

On 2/28/2013 6:40 PM, Tim Williams wrote:
> "rickman"<gnuarm@gmail.com>  wrote in message
> news:kgoo4e$ts5$1@dont-email.me...
>> 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

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" <gnuarm@gmail.com> wrote in message 
news:kgr44f$5lr$7@dont-email.me...
> On 2/28/2013 6:40 PM, Tim Williams wrote:
>> "rickman"<gnuarm@gmail.com>  wrote in message
>> news:kgoo4e$ts5$1@dont-email.me...
>>> 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 

On 3/1/2013 11:43 AM, legg wrote:
> On Wed, 27 Feb 2013 22:07:51 -0500, rickman<gnuarm@gmail.com>  wrote:
>
> <snip>
>> There is a resonance near the frequency I would expect, but it is not so
>> close actually.  I can't figure why it is about 5% off.  There is a
>> second resonance fairly high up that I can't figure at all.  None of the
>> component values seem to combine appropriately to produce this peak.
> <snip>
>
> Pulling out the old reactance paper, there are a couple of expected
> interactions using the values present:
>
> Around 50KHz (89.42uH+48uH) with 50.42nF  (L3+L1) with C1
> Around 290KHz (89.42uH+48uH) with 6.25nF  (L3+L1) with C2*N^2
> Around 360KHz 48uH with 6.25nF  L1 with C2*N^2
> nL1/nL2=N=25
>
> The mid-resonance is a dip or rejection.
>
> What's the issue?

290 kHz matches the calculations you just gave. But 290 kHz is the null 
(or dip as you call it) from C2 and L2 (or L1 and C2 reflected with N^2).

I thought I wasn't getting the 60 kHz resonance, but I was mistakenly 
adding the two capacitances together.  So that is closer.  Using L3+L1 
with C1 I get 60.46 kHz while it is measured at 60 dead on in 
simulation.  That's nearly a 1% error.

I solved the equations finally.  I found some info on the impedance of 
series and parallel circuits.  With that info I wrote the equation for 
the ratio of Vout/Vin and found the roots.  Turns out it is not so bad. 
  The equation is a fourth order, but it has no x^3 or x^1 terms and so 
is actually a quadratic of x^2.  Solving the quadratic gives the exact 
figures for 60 kHz and 393 kHz peaks.  Since this is from taking the 
square root of x^2, there are also solutions at the negative values... duh!

Reflecting C2 through the transformer to create C2', the two nulls I 
found can be calculated by the resonance of L1 and C2' (290 kHz null on 
C1) or L1 with C1 and C2' (96500 Hz null on L3).

-- 

Rick

On 3/1/2013 8:53 PM, Tim Williams wrote:
> 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.

I don't know what you mean by the "conventional radio approach".


> 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?

I was simulating a specific circuit for a specific purpose.  I got the 
answer I was looking for.


> 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.

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 am aware of the low radiation resistance and have not included that 
factor in my simulation.  The Q of just the antenna loop is around 100 
as calculated from the ratio of reactance to resistance.


> 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.

A friend in a loop antenna Yahoo group suggested the use of the 
transformer coupling with a low k to model the signal reception.


> 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.

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.


> 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).

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.


> 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.

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

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?

-- 

Rick