Antenna Simulation in LTspice

"rickman" <gnuarm@gmail.com> wrote in message 
news:kh83nl$uqo$1@dont-email.me...
> 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, 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.

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.

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.

> 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).

> 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.

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

Tim

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

On 6.3.13 9:00 , rickman wrote:

>> 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.

Please note that high Q will destroy the modulation sidebands on
the signal you're listening to.

In aviation, there are non-directional beacons which are transmitting
in a frequency around 300 kHz (1 km wavelength). The antennas cannot
obviously be of efficient length (250 m / 800 ft), so they are short
(20 m / 70 ft) force-tuned to the transmitting frequency. This creates
so high Q that the identification modulation sidebands for the customary
1050 Hz audio do not fit in, and the ID is modulated using 400 Hz audio.

-- 

Tauno Voipio, avionics engineer (also OH2UG)

On 3/7/2013 2:15 AM, Tauno Voipio wrote:
> On 6.3.13 9:00 , rickman wrote:
>
>>> 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.
>
> Please note that high Q will destroy the modulation sidebands on
> the signal you're listening to.

I appreciate the advice from everyone, but much of it is not in the 
proper context and way off target.  "High" Q is how high?  Where are the 
modulation sidebands?  My point is that I have already considered this. 
  The modulation sidebands of this signal are on the order of low 10's 
of Hz.  This signal is modulated at a 1 bit per second rate.  I will be 
demodulating a 30 Hz sample rate.  So a bandwidth of 100 Hz is plenty 
which corresponds to a Q of around 500.

I said I was looking for a Q over 100, maybe I should have said a Q of a 
bit over 100.  By the time it gets to 300 it is to peaky to hold a tune 
setting.  That is the problem I am concerned with.


> In aviation, there are non-directional beacons which are transmitting
> in a frequency around 300 kHz (1 km wavelength). The antennas cannot
> obviously be of efficient length (250 m / 800 ft), so they are short
> (20 m / 70 ft) force-tuned to the transmitting frequency. This creates
> so high Q that the identification modulation sidebands for the customary
> 1050 Hz audio do not fit in, and the ID is modulated using 400 Hz audio.

Ok, but that is nothing like my application, receiving WWVB.

-- 

Rick

On 3/6/2013 8:13 PM, Tim Williams wrote:
> "rickman"<gnuarm@gmail.com>  wrote in message
> news:kh83nl$uqo$1@dont-email.me...
>> 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

On 7.3.13 4:30 , rickman wrote:
> On 3/7/2013 2:15 AM, Tauno Voipio wrote:
>> On 6.3.13 9:00 , rickman wrote:
>>
>>>> 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.
>>
>> Please note that high Q will destroy the modulation sidebands on
>> the signal you're listening to.
>
> I appreciate the advice from everyone, but much of it is not in the
> proper context and way off target.  "High" Q is how high?  Where are the
> modulation sidebands?  My point is that I have already considered this.
>   The modulation sidebands of this signal are on the order of low 10's
> of Hz.  This signal is modulated at a 1 bit per second rate.  I will be
> demodulating a 30 Hz sample rate.  So a bandwidth of 100 Hz is plenty
> which corresponds to a Q of around 500.
>
> I said I was looking for a Q over 100, maybe I should have said a Q of a
> bit over 100.  By the time it gets to 300 it is to peaky to hold a tune
> setting.  That is the problem I am concerned with.
>
>
>> In aviation, there are non-directional beacons which are transmitting
>> in a frequency around 300 kHz (1 km wavelength). The antennas cannot
>> obviously be of efficient length (250 m / 800 ft), so they are short
>> (20 m / 70 ft) force-tuned to the transmitting frequency. This creates
>> so high Q that the identification modulation sidebands for the customary
>> 1050 Hz audio do not fit in, and the ID is modulated using 400 Hz audio.
>
> Ok, but that is nothing like my application, receiving WWVB.
>

I'd still be wary of high Q. The antenna is, by definition, in close
interaction with its surroundings, and a high-Q thing is quickly
detuned.

At those low frequencies, the atmospheric and other outside noise is
far larger than the internal noise of an amplifier, so in my opinion,
the way to go is a loop tuned to 60 kHz with as low Q as easily comes
without extra attenuation and a good pre-amplifier. The preamp can
then contain a tuned interstage tank for interference suppression.

-- 

Tauno Voipio

On 3/7/2013 12:55 PM, Tauno Voipio wrote:
> On 7.3.13 4:30 , rickman wrote:
>>
>> Ok, but that is nothing like my application, receiving WWVB.
>>
>
> I'd still be wary of high Q. The antenna is, by definition, in close
> interaction with its surroundings, and a high-Q thing is quickly
> detuned.
>
> At those low frequencies, the atmospheric and other outside noise is
> far larger than the internal noise of an amplifier, so in my opinion,
> the way to go is a loop tuned to 60 kHz with as low Q as easily comes
> without extra attenuation and a good pre-amplifier. The preamp can
> then contain a tuned interstage tank for interference suppression.

I understand.  But this is intended to be *very* low power and I haven't 
found an amp I can use that is in the low double digits uW power 
consumption range.  I plan to use no amp and go straight to digital.

-- 

Rick

On Thu, 07 Mar 2013 13:11:03 -0500, rickman <gnuarm@gmail.com> wrote:

>I understand.  But this is intended to be *very* low power and I haven't 
>found an amp I can use that is in the low double digits uW power 
>consumption range.  I plan to use no amp and go straight to digital.

I don't think that's possible.  Unless your input A/D converter can
operate in the microvolt region, it's going to have a difficult time
dealing with the low signal levels.  Fortunately, WWVB is on-off
keying with no amplitude component, so there's no incentive to add an
AGC controlled input amplifier in order to maximize the A/D converters
dynamic range.  Still, you need to work with something more than a few
bits above the noise level.  Incidentally, after midnight, you WWVB
delivers about 100 uV/meter or more to continental US.  
<http://tf.nist.gov/tf-cgi/wwvbmonitor_e.cgi>  (Java required)
I've seen it strong enough that I can see the waveform on an
oscilloscope after a 60Khz passive filter.

As for bandwidth, the code is sent at 1 baud (1 bit/sec) which
produces about a 2Hz occupied bandwidth.  Therefore, the maximum Q of
the antenna would need to be:
  60Khz/ 2Hz = 30,000
before the antenna bandwidth becomes a problem.  

Incidentally, while Googling away merrily, I found this on SPICE
models for a loop antenna.  It's not quite in your xformer format, but
it might be useful:
<http://sidstation.loudet.org/antenna-theory-en.xhtml>
I won't pretend to understand what the author is doing until I read it
more carefully.

Incidentally, I used a WWVB code simulator driving a signal generator
to test my receiver:
<http://www.leapsecond.com/notes/wwvb2.htm>

If you're seriously into this, I suggest asking questions on the
time-nuts mailing list:
<https://www.febo.com/mailman/listinfo/time-nuts>

-- 
Jeff Liebermann     jeffl@cruzio.com
150 Felker St #D    http://www.LearnByDestroying.com
Santa Cruz CA 95060 http://802.11junk.com
Skype: JeffLiebermann     AE6KS    831-336-2558

"rickman" <gnuarm@gmail.com> wrote in message 
news:kha9e1$9qt$1@dont-email.me...
>> 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're also not concerned about that -- you're concerned about maximizing 
SNR at the receiver.

A Q of a million will get you gobs of "gain", but if it doesn't couple 
into free space, it's only the thermal noise of the loss generating that 
signal.

An antenna with high (expressed as ESR) radiation resistance might have a 
modest Q, but gives far better SNR because it couples to free space.

Raw volts don't matter, you can always throw more amplifiers at it (as 
long as they don't corrupt the SNR also!).

> 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.

Oooh, capacitance!  I like capacitance.  Capacitance is easy to 
cancel...inductors are good at that. :)

What's a nearby inductor working against that capacitance?  The current 
transformer in your simulation, if its inductance can be controlled, would 
be an excellent candidate.  The circuit effectively becomes a double tuned 
interstage transformer, like,

http://www.jrmagnetics.com/rf/doubtune/doubccl_c.php
This is two resonators coupled with a cap, but any coupling method will 
do.  Capacitive, magnetic (putting the coils end-to-end) or 
electromagnetic (coils side-by-side) coupling does equally well; normal 
arrangements have them all in phase, so in practice, unshielded coils will 
need smaller coupling capacitance than designed, etc.

If you line up that 10p resonance with the operating frequency, you should 
get gobs more gain.  In fact, because the reactances cancel, the driven 
impedance will be much higher than you were expecting, and so will the 
gain.  The CT might go from, say, 1:8 up to, who knows, 1:20?  1:100?

The bandwidth of that coupling (not necessarily of the antenna itself, so 
they should be similar bandwidths) is determined by the coupling 
coefficient (in the coupled-inductors case, simply k) and Q of the 
components.

If your receiver datasheet specifies an equivalent input circuit, you 
might be able to estimate the equivalent loss and optimize gain.

Tim

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

On 3/7/2013 5:14 PM, Tim Williams wrote:
> "rickman"<gnuarm@gmail.com>  wrote in message
> news:kha9e1$9qt$1@dont-email.me...
>>> 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're also not concerned about that -- you're concerned about maximizing
> SNR at the receiver.

SNR would be good, but I am concerned with maximizing the signal actually.


> A Q of a million will get you gobs of "gain", but if it doesn't couple
> into free space, it's only the thermal noise of the loss generating that
> signal.

I think you aren't reading what I am writing.  I said I wanted a Q over 
100, not 1 million.  I don't get why you keep talking in hyperbole. 
What you are describing is not even a tradeoff between signal strength 
and SNR.  If there is no coupling, there is no signal.


> An antenna with high (expressed as ESR) radiation resistance might have a
> modest Q, but gives far better SNR because it couples to free space.

I have not found anything to indicate this produces a better receive 
antenna.  I have a formula for the effective height of a loop antenna 
which is what determines the received signal strength at the antenna. It 
does not calculate the radiation resistance, it uses the coil parameters 
and the wire resistance.  Is that a wrong formula?


> Raw volts don't matter, you can always throw more amplifiers at it (as
> long as they don't corrupt the SNR also!).

Maybe you didn't read my other posts.  I am not using an amplifier.  I 
am running the antenna and coupler output directly into a digital input.


>> 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.
>
> Oooh, capacitance!  I like capacitance.  Capacitance is easy to
> cancel...inductors are good at that. :)
>
> What's a nearby inductor working against that capacitance?  The current
> transformer in your simulation, if its inductance can be controlled, would
> be an excellent candidate.  The circuit effectively becomes a double tuned
> interstage transformer, like,
>
> http://www.jrmagnetics.com/rf/doubtune/doubccl_c.php
> This is two resonators coupled with a cap, but any coupling method will
> do.  Capacitive, magnetic (putting the coils end-to-end) or
> electromagnetic (coils side-by-side) coupling does equally well; normal
> arrangements have them all in phase, so in practice, unshielded coils will
> need smaller coupling capacitance than designed, etc.
>
> If you line up that 10p resonance with the operating frequency, you should
> get gobs more gain.  In fact, because the reactances cancel, the driven
> impedance will be much higher than you were expecting, and so will the
> gain.  The CT might go from, say, 1:8 up to, who knows, 1:20?  1:100?
>
> The bandwidth of that coupling (not necessarily of the antenna itself, so
> they should be similar bandwidths) is determined by the coupling
> coefficient (in the coupled-inductors case, simply k) and Q of the
> components.
>
> If your receiver datasheet specifies an equivalent input circuit, you
> might be able to estimate the equivalent loss and optimize gain.

The receiver input is high impedance, approximately 10 MOhms with a low 
capacitance between the differential inputs of not more than 10 pF.

Your description of what is happening is very terse and full of 
shortened terms that I don't understand.  What do you mean "line up that 
10p resonance with the operating frequency"?  I assume you are referring 
to the 10 pF input capacitance.  How does this get "lined up" with 
anything?

When you talk about reactances canceling, that sounds a lot like a tuned 
circuit at resonance.  That is what I *am* doing and where this thread 
started.  One problem with that is the lack of precision or stability of 
the parasitic capacitance.  Any idea how to deal with that?

Have you looked at the simulation data I had posted?  I think you are 
describing exactly the circuit we are simulating which I believe is an 
accurate representation of the circuit I plan to build.  Is that not 
correct?

-- 

Rick

On Thu, 07 Mar 2013 09:50:11 -0500, rickman wrote:

> On 3/6/2013 8:13 PM, Tim Williams wrote:
>> "rickman"<gnuarm@gmail.com>  wrote in message

<snip>
> 
> 
>>> 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.
>
<snip>

An electric circuit consists of a source of power, a load, and something 
(like wires) connecting them.  Transformers can be used if the source is 
providing alternating current.  A voltage transformer is connected in 
parallel with the load so that the source, the transformer, and the load 
all see the same voltage.  It can also be used to match a load to a 
source.  A common example of a voltage transformer is the power 
transformer in a piece of equipment that changes the AC line voltage to 
whatever other voltages are required by the equipment.

A current transformer, on the other hand, is connected in series with the 
load so that the source, load, and transformer all have the same current 
flowing through them.  The most common use of a current transformer is to 
measure the current flowing into a load.  A clamp-on ammeter is a common 
example.

Historical examples of voltage and current transformers are the "picture 
tube brighteners" that were commonly used in TV sets to prolong the 
useful life of the CRT.  There were two types, parallel and series.  The 
parallel types were used in transformer operated TVs and consisted of a 
step-up transformer to raise the heater voltage of the CRT above normal 
to increase emission.  The series type was used in sets with the tube 
heaters in series and consisted of a step-down transformer that raised 
the heater current above normal.  Of course, raising either the voltage 
or the current also raised the other.  These were, respectively, voltage 
and current transformers.

A loop antenna is a distributed source with the voltage being generated 
along the length of the wire and also having a magnetic field so that it 
can be used as part of a transformer.  This blurs the distinction between 
a current and voltage transformer.


-- 
Jim Mueller wrongname@nospam.com

To get my real email address, replace wrongname with dadoheadman.
Then replace nospam with fastmail.  Lastly, replace com with us.