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how to design an antenna ?

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schneiderj

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Hello,

I would like to create an application to record my sport activity (bicycling). It will be based on a PIC, but I have a problem with the cardiofrequencemeter :
the thoracic belt emit at a frequency of 5 khz and I don't know hos to design an antenna to receive that signal.

Can you help me ?
Jean-Marie
 
5 kHz = 5,000 cycles per second is going to require a very long antenna. The wavelength is computed as follows
Code:
c = λ*f

3*10^8 = λ * 5000
3*10^8 / 5000 = λ
λ = 60,000 meters
So a half-wave dipole will be about 30 km long. I think there is an error of omission somewhere.
 
5Khz is in the audible range. Recheck that.

Regards,
Wayne
As a longitudinal pressure wave that is true. As a TEM wave I believe you would have some trouble haering it. LOL
 
Hello,

I would like to create an application to record my sport activity (bicycling). It will be based on a PIC, but I have a problem with the cardiofrequencemeter :
the thoracic belt emit at a frequency of 5 khz and I don't know hos to design an antenna to receive that signal.

Can you help me ?
Jean-Marie

Jean-Marie, can you please provide a manufacturer's advertisement or data sheet for this thoracic belt that you have? We need to understand it a bit better and so require the manufacturers description or specification for it.
 
Thanks all for your reply.

It is not so easy to find this informaiton but what I have is :
Polar Power Output Sensor
A sensor measuring speed, distance and chain speed. The data is wirelessly transferred to the cycling computer with 5 kHz transmission. Provides valuable cycling data like power output, cycling efficiency, pedaling index and L/R balance.
from : **broken link removed**

and
The following information can be given to the manufacturer of pacemaker,
defibrillator or another implanted electronic device when they are
evaluating the risks involved in the simultaneous use of the Polar HRM and
their devices.

The Polar Transmitters send short signals for each heartbeat to the
receiver. These signals are low-frequency 5 kHz (kilohertz) electromagnetic
fields at very low power, less than 2nW (nanowatts). The magnetic field
strength at the transmitter is 5 uT (microtesla) and 1 nT (nanotesla) at a
1 m distance.


Polar Non-Coded and T31 Transmitters :

Transmission frequency: 5kHz
Means of transmission: On/Off -modulated electromagnetic field with a
5-7ms burst for each heart beat

Polar Coded, T31 C, WearLink 31 C and T61 Transmitters:

Transmission frequency: 5kHz
Means of transmission: On/Off -modulated electromagnetic field with a
train of three 5ms bursts for each heart beat. Time interval coding.

The strength of the electromagnetic field at the transmitter is 5 uT
(microtesla) and 1 nT (nanotesla) at 1 m distance.

from this link : **broken link removed**

Jean-Marie
 
That was exactly the information I was looking for. The frequency is so low that to call the transducer an "antenna" is a bit misleading. An antenna is usually used to launch a propagating, or far field, electromagnetic wave. In this case we don't care about launching such a wave, we are only interested in near field magnetic coupling from the transmitter to your receiver.

The most practical way to couple magnetic energy into your receiver is to use a coil, or inductor, made up of many turns of wire. It may also be a good idea to use a ferrite core on this inductor to increase the magnetic flux through the coil. So, you are interested in something similar to the type of antennas used in simple AM receivers. These are made by winding many turns of very fine wire onto a ferrite bar. Typically the larger the coil (and the larger the ferrite bar) the better it works to pick up weak signals.

The other problem that you face is how to connect this coil to your receiver. I think the best kind of interface is to use a JFET as an impedance buffer, that is, a simple JFET amplifier with very high input impedance. I suggest a JFET instead of an opamp because you want this amplifier to be a low noise type. An op amp would be ok if you were to choose one that is a low noise type.
 
That was exactly the information I was looking for. The frequency is so low that to call the transducer an "antenna" is a bit misleading. An antenna is usually used to launch a propagating, or far field, electromagnetic wave. In this case we don't care about launching such a wave, we are only interested in near field magnetic coupling from the transmitter to your receiver.

The most practical way to couple magnetic energy into your receiver is to use a coil, or inductor, made up of many turns of wire. It may also be a good idea to use a ferrite core on this inductor to increase the magnetic flux through the coil. So, you are interested in something similar to the type of antennas used in simple AM receivers. These are made by winding many turns of very fine wire onto a ferrite bar. Typically the larger the coil (and the larger the ferrite bar) the better it works to pick up weak signals.

The other problem that you face is how to connect this coil to your receiver. I think the best kind of interface is to use a JFET as an impedance buffer, that is, a simple JFET amplifier with very high input impedance. I suggest a JFET instead of an opamp because you want this amplifier to be a low noise type. An op amp would be ok if you were to choose one that is a low noise type.

Thanks a lot for this explanation : I was unable to see how this receptor was able to "see" this radiofrequence.
To realise this coil, did you have some interresting site where I can find explanation on how to calculted them ?

Initialy I would like to use a INA129 from TI to interface that transducteur (https://focus.ti.com/lit/ds/symlink/ina129.pdf). Did you think that enought ?

Jean-Marie
 
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Your choice of amplifier seems OK, but I wonder why you have chosen an instrumentation amplifier. An instrumentation amplifier is a bit more complex than a basic op amp and its virtues are mainly in much better balance and offset precision. For interfacing your coil, such precise balance is not necessary. If I were searching for an op amp, I would focus my attention on the noise performance specification for each one and choose the lowest noise model I can find. Here is an application note about op amp noise that you might find useful:
http://www.linear.com/pc/downloadDocument.do?navId=H0,C1,C1154,C1009,C1026,D6539

As for the coil itself, I looked around on the net for some instruction but only found these, which are not complete and they tend to discuss the issue backwards from your application. I mean that they look at magnetic coupling as a source of noise and wish to minimize, whereas in your case you want to maximize it. But the theory is still useful.

Can I suggest that experimentation is a practical way to learn how to realize the coil? You are probably limited to a certain maximum size, which puts an upper bound on the coil diameter. Find a ferrite bar or unsheilded ferrite core that might fit this size and then find some fine magnet wire (copper wire with only a varnish insulator to keep it very thin) and go ahead and wind as many turns as you can around a single axis (not in a toroid shape). Then, to see how effective it is, attach an oscilloscope to the terminals of the coil and measure the output voltage with your coil at a chosen distance from the transmitter.

Here's the links:
https://www.electro-tech-online.com/custompdfs/2008/10/MagFieldCoupling.pdf
https://www.electro-tech-online.com/custompdfs/2008/10/emc99-w.pdf
As you can see, the theory is very simple. The output voltage from your receiver coil is directly related to coil area, number of coil turns, the permeability of the core, the mutual inductance between transmitter and receiver (which is affected by distance of course and don't forget that the coil orientation is also important) and the rate of change of the current in the transmitter.

Another source of information and perhaps coils too might be those who make those time-clock receivers. These receivers operate at 60 KHz and are put inside those so-called "atomic watches". They usually use a very small ferrite bar coil. Here is a very good application note:
Technology - Antenna Design
You can search for more using keywords "radio controlled time" to find more examples and information about these low frequency magnetic receivers.

Another source of info on the coil can be found by googling using keywords like "magnetic field probe" Such probes are used to measure magnetic fields, working similarly to what you are doing, and so the theory of how they work and how to calculate output voltage would be useful.

I also wonder if perhaps a "hall effect" device might not be able to act as a receiving sensor for your magnetic field. I have not worked with these except with DC fields, so I'm not sure they are useful here.
 
Hello, it is me again.

I perform the first test with an old electromagnet and record the following :
see attachement base time = 200 ms

and attachement base time = 1 ms

Now I have a question regarding the information you give me : in the last link they determine the value of the capacitance with this equation :

**broken link removed**

I understand that the AOP should have an hight impedance, but why they take an impedance for the coil of 100 kΩ (arround). Sorry if this question is basic, but a learn electronic by myself and I have a lot of think to learn...

Thanks you for your help,
Jean-Marie
 

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If I understand well the resistance is the resitance of the coil. But I read eslwhere that this resitance should be as low as possible. I did the calculation with various Q factor and resistance for 5500 Hz. That give :
**broken link removed**

That means that I have to had a resitance to the coil to obtain for exemple 50 kΩ ?

Thanks for your help !
Jean-Marie
 
If I understand well the resistance is the resitance of the coil. But I read eslwhere that this resitance should be as low as possible. I did the calculation with various Q factor and resistance for 5500 Hz. That give :
**broken link removed**

That means that I have to had a resitance to the coil to obtain for exemple 50 kΩ ?

Thanks for your help !
Jean-Marie

If this transmitter runs at 5KHz, your best off to tie your transmitter output to a pluming pipe in your house or use the 3rd prong on the AC power outlet.

Really at the frequency, the transducer would be a speaker.
 
If I understand well the resistance is the resitance of the coil. But I read eslwhere that this resitance should be as low as possible. I did the calculation with various Q factor and resistance for 5500 Hz. That give :
**broken link removed**

That means that I have to had a resitance to the coil to obtain for exemple 50 kΩ ?

Thanks for your help !
Jean-Marie

No, this is not their meaning. You are not supposed to add resistance to the coil to achieve the 50Kohms.

The equivalent circuit model of a real inductor is an ideal inductance in series with a resistance. When you measure the resistance with a DC ohmmeter you get this series resistance at least for low frequency applications. If we were talking about RF frequencies, than the DC ohmmeter cannot measure the resistance because resistance of a conductor changes with frequency. Anyways, we are talking about 5 KHz so the measurement with your DC ohmmeter is ok. We expect to measure a small value, perhaps a few ohms. This is normal.

In the quoted link, they are talking about a parallel equivalent circuit, not a series equivalent circuit (but they are not being very clear about it). Electrical network theory tells us that every series circuit can be changed to an equivalent parallel circuit through a mathematical transformation. When they discuss the Rres of the antenna to be 180Kohms, they are talking about its parallel equivalent resistance.

The parallel equivalent resistance is much higher than the series equivalent resistance of a resonant circuit.

I notice that they make the common assumption that the capacitor is lossless and that all of the circuit resistance is in the coil. This is a reasonably accurate assumption. I also notice that their terms are confusing. In equation 2, the term R(res) is the parallel equivalent resonant resistance. In equation 5 the term R(RES) is the series equivalent resonant resistance. The two are related in that R(res) is = Q^2 x R(RES). So, when they choose that the R(res) is 180Kohms, and you have a Q of 100, then R(RES) is 18 ohms.

When you measure your coil resistance you are measuring R(RES). For best Q factor, the series resistance of the coil should be as low as possible, and in this case the parallel equivalent resistance will be as high as possible.
 
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If this transmitter runs at 5KHz, your best off to tie your transmitter output to a pluming pipe in your house or use the 3rd prong on the AC power outlet.

Really at the frequency, the transducer would be a speaker.

To go outside that will be not very easy...
I try a speacker, but it sensibility was very low (a 8 Ω and 0.5 W).

Jean-Marie
 
No, this is not their meaning. You are not supposed to add resistance to the coil to achieve the 50Kohms.
Thanks for these very clear explanation.

I notice that they make the common assumption that the capacitor is lossless and that all of the circuit resistance is in the coil. This is a reasonably accurate assumption. I also notice that their terms are confusing. In equation 2, the term R(res) is the parallel equivalent resonant resistance. In equation 5 the term R(RES) is the series equivalent resonant resistance. The two are related in that R(res) is = Q^2 x R(RES). So, when they choose that the R(res) is 180Kohms, and you have a Q of 100, then R(RES) is 18 ohms.

When you measure your coil resistance you are measuring R(RES). For best Q factor, the series resistance of the coil should be as low as possible, and in this case the parallel equivalent resistance will be as high as possible.

OK, that make sens to what I can have with a coil : few ohms.

One more question : can you explain me why the coil ae made on a rod and not a tore ?

Thanks again,
Jean-Marie
 
To go outside that will be not very easy...
I try a speacker, but it sensibility was very low (a 8 Ω and 0.5 W).

Jean-Marie

What you need is a "trans-match". You use a random length piece of wire and feed it through the transmatch. Look them up. They usually consist of a large air coil with several selectable taps ans a couple of variable capacitors (air dielectric).
 
Thanks for these very clear explanation.



OK, that make sens to what I can have with a coil : few ohms.

One more question : can you explain me why the coil ae made on a rod and not a tore ?

Thanks again,
Jean-Marie


We want the magnetic field to couple into the antenna from surrounding space. The rod does this much better than the toroid. In the toroid, most of the magnetic field lines stay within the core because there is a complete magnetic path for the flux and so very little magnetic field gets out of the core. Conversely, very little magnetic field can get into the core from the outside then.

The bar does not provide a complete magnetic circuit and so it couples to the outside much better.
 
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