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Ferrite loopstick antenna

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vaka85

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Hi,

I'm on this problem from few weeks, it seems to be very easy, as there are a lot of DIY project on this stuff, but it's not so easy for me :)

I'd like to transmit a sine wave. It's obtained by a square wave generated by a quartz oscillator, filtered, amplified a little (but it's not so important) and then sent to a ferrite loopstick antenna:




with an inductance/capacitance meter I've measured impedance (22 uH), and the capacitance of a variable capacitor taken from an old am radio (from 10 to 130 pF), and I've set them to resonate at 4 MHz.

Then I connected the L C in parallel, and with a spectrum analyzer (in parallel with LC) I checked if the LC filter works. It doesn't!
The spectrum is exactly the same if I put the LC or if I remove it!
Also if I vary the value of the capacitor, nothing changes.

So, some questions:
1. What I'm doing wrong?
2. What's the best configuration of LC? in series or in parallel? Because I've tried also in series, it works a little better, but not so much..
3. I'm doing it right connecting the spectrum analyzer in parallel with LC?
4. Is possibile that the parasitic components in L and C are so high that the formula
F = 1/2*pi*sqrt(LC)
is wrong? Because I'm using low cost instrument to measure L and C...

Any other suggestion is also welcome ;)

Thank you
 

betwixt

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The output impedance of IC1A is very low so you are effectively connecting R1 across the tuned circuit and making it's Q factor very low. Also ferrite is not a good material to use in a transmitting circuit, an air-cored inductor will work better.

Instead of connecting R1 to the top of L1, try connecting it to a point maybe 10% of the turns from the bottom or alternatively, connect a capacitor in series with R1 (Try 47pF) and see if that works better.

You would still find a conventional RF crystal oscillator works better.

Brian.
 
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vaka85

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thank for your reply.

The output impedance of IC1A is very low so you are effectively connecting R1 across the tuned circuit and making it's Q factor very low. Also ferrite is not a good material to use in a transmitting circuit, an air-cored inductor will work better.
I haven't understood this issue. :oops: Why R1 makes the Q factor very low?
I've also tried without R1, but no changes.
Anyway I'll try with an air core inductor...

Instead of connecting R1 to the top of L1, try connecting it to a point maybe 10% of the turns from the bottom
Why?
In that why I'd decrease the L value, right?

You would still find a conventional RF crystal oscillator works better.
Yes but I've chosen this one because I need also a line at 2 MHz, with a zero phase difference between the two lines, and a square wave it's the only way (in my opinion and knowledge) to do so..

thanks ;)
 

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The Q of a resistive loaded parallel resonant circuit is R/X, X being the characteristic impedance (impedance at resonance frequency) X = |XL| = |XC|. You get Q << 1 with R = 10 ohm, but would want Q >> 1.

Brian made two reasonable suggestions to achieve a higher Q, you should review it. Particularly the tapping idea is good to increase the output. But a ferrite antenna isn't bad, basically.
 

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Thanks for the confirmation FvM.

I'll try to explain the 'Q' problem. The IC output is very low impedance, especially with such a large amount of feedback. It means the ICs output will attempt to follow the input signal as close as it can, this is NOT what you want to happen. The LC circuit should resonate, this will produce a higher voltage and clean up the square wave edges so they look more like a sine wave. With only 10 Ohms between the IC output and the LC circuit you do not have sufficient isolation for the circuit to resonate properly, the IC is trying to compensate for the LC signal and any extra voltage is being absorbed into R1. What you need to do is isolate the signal driving the LC more so it can resonate freely without losing signal back into the IC. There are several ways to do this, you could increase the value of R1 but that would also limit the power you can put into the LC circuit. You can use a capacitor as I suggested which will provide isolation at the expense of some phase shift (if you do it on the 2MHz signal as well they will still be in phase). Or, you can match the low impedance of the IC into a low impedanace part of the LC circuit by feeding the power into a tap near the bottom of the inductor. You don't increase L, you split the same inductance into two parts, a 10% tap will leave 90% of the circuit unloaded.

Remember that at resonance, a parallel LC circuit has high impedance. The higher the better, it will give more voltage and a purer sine output. Connecting 10 Ohms across it is almost like shorting it out.

Brian.
 
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vaka85

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thank you very much for the explanation.
Now it's clear...(even if I'm still wondering how the LC can resonate at 4MHz if I connect the output at the bottom of L, so the value of L "seen" by the output should change...).

I've tried with the series capacitor, but no changes, I'll try also with connecting the signal to the bottom of the L.

However, it seems to work better with LC series. But my question is: if I attach the spectrum analyzer (or the scope) across the series, what should I see? Because the signal is high until it reaches 4 MHz and after that frequency, but at 4 MHz it goes very low.
So: is it correct? On the analyzer should I see the "remaining" power on the series, as the "correct" power goes away? :oops:

thanks
 

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A series LC circuit is in fact a way to get highest field strength without using an inductor tap. To monitor the field strength , you should use a sense coil.

Your post sugests, that you aren't understanding yet the behaviour of resonant circuits. The series circuit has a low impedance in resonance and nearly shorts the amplifier output, which results in maximum energy transfer.
 
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betwixt

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And I didn't say connect R1 to the bottom of L, I said connect it to a point about 10% from the bottom of L. That means LC stays as it is but has a third connection from the 10% tap to R1.

Brian.
 

RCinFLA

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On a wild guess, the parallel Rp of the ferrite tank will be in the range of 5k to 15k ohms. If you want to drive it with a ten ohm source then configure a matching circuit to take 10 ohms to 5k ohms.

If your ferrite coil is 22 uH @ 4 MHz then you have +j553 ohms in parallel with about 5k ohms. The series equivalent is 60.5 + j546 ohms. If you add -j546 capacitor (72.8 pF cap) you would have 60.5 ohms real impedance at 4 MHz. The loaded Q will be 2500/553 = 4.5 or BW3db of 885 kHz at 4 MHz. The 2500 number assume 5k due to coil Rp and 5k in parallel due to matching loading.

I suggest you reserve some of the loading capacitance for tuning. Reserve about 10 pF for a parallel tuning cap, leaving 62 pF for matching. Change the 10 ohm to 56 ohms and connect a 62 pF in parallel with 15 pF trimmer then in series to coil. Adjust trimer to get 4 Mhz resonance.

I think a 10 ohm load on op amp may be too heavy.
 
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vaka85

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Your post sugests, that you aren't understanding yet the behaviour of resonant circuits. The series circuit has a low impedance in resonance and nearly shorts the amplifier output, which results in maximum energy transfer.
sorry, maybe I wasn't clear. I understand that behavior, but I didn't understand if my measurement was correct. Now it's clear.

And I didn't say connect R1 to the bottom of L, I said connect it to a point about 10% from the bottom of L. That means LC stays as it is but has a third connection from the 10% tap to R1.
yes, I meant 10% from the bottom ;)

On a wild guess, the parallel Rp of the ferrite tank will be in the range of 5k to 15k ohms. If you want to drive it with a ten ohm source then configure a matching circuit to take 10 ohms to 5k ohms. .................................................................
very interesting calculations!!! only one thing: how you calculate the value of the parallel Rp of the L?


However I think I have solved the transmitter part, now comes the receiver... I have some problems with a 3 MHz undesidered component that arrives very strong to the receiver.. maybe I need a good isolation...

Thank you all guys, step by step I'm doing good improvements ;)
 

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Hi,

I'm still on the trasmitter antenna issue...
Some questions...

I use the LC series configuration, tuned correctly and connected to the opamp output with a 100 ohm resistance. At the receiver the signal at 1 meter is very low, BUT there's a strange behavior!
if I connect a simple wire to one side of the transmitter inductance, boom! The received signal is huge!

I think there is some kind of impedance matching problem...

Here come the questions:
- If the LC series is at low impedance (at resonance), so the only load seen from the opamp output is the resistance, 100 ohm! So, since the opamp out is at low impedance, why there should be some impedance matching problem?

- The other possibility is that I've forgotten to consider some components in the calculations for the tuning... Should I consider the parasitical components of L and C? There should be any resistive component that I've not considered?
For example, the "the parallel Rp of the ferrite tank" said by RCinCLA in a previous post, how can I calculate it?

Thank you very much...
 

RCinFLA

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You can get an ideal of the ferrite tank Rp (or Rs if you prefer) by measuring the unloaded 3 db bandwidth. Resonate the coil and very lightly couple into the tank with small coupling caps that you applied an RF source and RF detector (could be spectrum analyzer and tracking gen or network analyzer).

An RF ferrite tank is very susceptable to any magnetizing fields. You can measure the unload Q, wave a magnet near ferrite and Q will be destroyed. Only way to restore ferrite is to take it back up above its Curie temperature.

A ferrite loop is an H field dominate antenna. You can couple into a near by metal rod or other conductive body. This will increase the capture area of antenna. Putting a wire on the coil will increase E field emission but will detune the tank circuit.

For transmitting you have to be sure you don't drive core so hard you drive it into saturation region which will permainently de-Q the ferrite just as waving a magnet across it.

FYI, Ferrite is not used for antenna loop. It is actually powered iron and permability needs to be selected based on operating frequency. Only bars are used. Torroid are not used as they would confine the field to within the torroid.

We used to use powered iron bar cores for paging receiver antennas through VHF range. They coupled into the human body to give much better performance for a small antenna then an E field wip antenna.
 
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FvM

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Ferrite is not used for antenna loop. It is actually powered iron and permability needs to be selected based on operating frequency.
Ferrite and iron powder are different materials. Ferrite is ferrimagnetic oxide. Antenna rods are mostly using RF ferrites, but may be also iron powder in some cases.
An RF ferrite tank is very susceptable to any magnetizing fields. You can measure the unload Q, wave a magnet near ferrite and Q will be destroyed. Only way to restore ferrite is to take it back up above its Curie temperature.
I wasn't aware of a strong effect (magnetic remanence affecting Q or small signal permeability) with RF ferrites. Considering the fact, that most classical AM radios are using the ferrite rod inductor in it's tuned input tank, it sounds unlikely. I would need to check, if I can achieve a similar effect with ferrite rods from my parts stock.
 

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You can get an ideal of the ferrite tank Rp (or Rs if you prefer) by measuring the unloaded 3 db bandwidth. Resonate the coil and very lightly couple into the tank with small coupling caps that you applied an RF source and RF detector (could be spectrum analyzer and tracking gen or network analyzer).
do you mean in this way?



However, why if I try to transmit a 4 MHz wave, instead of 2 MHz, it works fine? with the exact way of tuning that I used with the 2 MHz... :-(
The same circuit, the same way of tuning the antennas, but 2 MHz works very bad..
 

biff44

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also, i do not think ic1 is going to be all too happy with a dc short circuit at its output.
 

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Do parallel resonance. Coupling caps should be as small as possible to get a measurable response (5 to 10 pF). Make sure you resonating cap is good quality.

With such a high inductance value you will probably only have a Q unloaded in the order 50.
 
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vaka85

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ok, thank you very much.
I'll try the next week and then I'll let you know...

Have a good weekend ;)
 

vaka85

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Hi all,
I'm back on this project.

After some measurements and a lot of papers reading, I've concluded that maybe ferrite antenna for transmitting a sine wave with an amplitude of 2 V (peak to peak) isn't so good. The ferrite permeability seems to become very low with "strong" signals, and so it's impossibile to reach distances higher than 1 meter or so..
Am I right?
If it isn't good for this amplitude, I could use a lower one. But what about the maximum link distance?

Thanks
 

FvM

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I've concluded that maybe ferrite antenna for transmitting a sine wave with an amplitude of 2 V (peak to peak) isn't so good. The ferrite permeability seems to become very low with "strong" signals, and so it's impossibile to reach distances higher than 1 meter or so..
Sounds very unlikely related to your application. You'll hardly achieve considerable ferrite magnetization in an open magnetic path, particularly with mW power. The problems are exclusively caused by inappropriate matching, as discussed before.
 

vaka85

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Ok, so.. could you help me in the matching process?
I've done all the measurement you guys have suggested. Please tell me if I've done them in the right way:

Parallel LC, bandwidth measured with the method suggest by RCinFLA.
- µ = 102 : measured the inductance value with an inductance-meter. From L I've found µ.
- Q = 87: measured the -3dB bandwidth (i.e. 23 KHz).
- R+RR+Rf = 2*π*f0*L/Q = 16 ohm.

Is it correct?
Now I should have this:


What's the next step to do the correct matching? :oops:
Thank you very much

---------- Post added at 12:25 ---------- Previous post was at 11:21 ----------

Meantime I've done some calculation.
If I am correct, the equivalent schematic should be the one above. So:
ZL = RTOT + jωL = 16 + j 1320
ZC = -j 1/ωC = -j 1540
ZTOT = ZL // ZC = 723 + j9086

Now... to remove the complex part I add a cap in series with ZTOT, its value is -j9086 = 8.1 pF. Hence remains only the real part, i.e. 723 ohm.

So I have to connect, at the output of the opamp, and in series with the LC parallel, a resistor of about 723 ohm.

Finish! I've done it right?
Thanks
 

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