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.

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Rick
Reply to
rickman
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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

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Reply to
Tim Williams

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???

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Rick
Reply to
rickman

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

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Reply to
Tim Williams

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

Reply to
legg

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.

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Reply to
Tim Wescott

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?

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.

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?

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.

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Rick
Reply to
rickman

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

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Reply to
Tim Williams

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).

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Rick
Reply to
rickman

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

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

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.

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

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.

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.

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?

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Rick
Reply to
rickman

I trust this resource:

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He's got gobs of analytical articles.

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.

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.

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

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

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Reply to
Tim Williams

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.
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Tauno Voipio, avionics engineer (also OH2UG)
Reply to
Tauno Voipio

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.

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

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Rick
Reply to
rickman

Yes, I've seen this page. Thanks.

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.

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?

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.

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.

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

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

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Rick
Reply to
rickman

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.

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Tauno Voipio
Reply to
Tauno Voipio

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.

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Rick
Reply to
rickman

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. (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: 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:

If you're seriously into this, I suggest asking questions on the time-nuts mailing list:

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Jeff Liebermann     jeffl@cruzio.com 
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Reply to
Jeff Liebermann

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!).

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,

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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

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Deep Friar: a very philosophical monk. 
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Reply to
Tim Williams

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.

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Reply to
Jim Mueller

Oops. WWVB does have an amplitude component and is not quite on-off keying (OOK). There's a -17dB drop in RF signal level at the beginning of each 1 second marker pulse. It was -10dB prior to 2005.

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Reply to
Jeff Liebermann

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