LoRa - Long-Range Radio for IoT | Arduino, ESP32, RPI Pico

@inq Uh oh, now you have done it.

Posted by: @zander

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@inq Uh oh, now you have done it.

🤣 🤣 🤣 That is too funny!

I don't want to sound negative, but Ron, you're giving me rules-of-thumb that I can get in a thousand Internet links.  Although there is a practical side of this for this LoRa project and your posts are helpful, my bent is to understand why.  It's a curse that I have to live with.  @davee, seems to have a pretty good handle on what I can absorb (and knows when to avoid the subject when he feels I can't).  😉 

I also know there will be a thousand links that if I searched would be way too thin or way too much of a leap.  I'm hoping he (or someone) may remember one that might be in the Goldilocks zone.   

@inq Dave and I are a team. I give you the pragmatic get it done, Dave gives you the college level 101. The best of both worlds. 😀 

The real answer is I have forgotten all the nitty gritty details since I never worked the field once I discovered computers.

The ESP8266 started out (in our hobbyist level) as merely a communications module to be added to an Arduino.  Then some gifted hacker realized... we're attaching a 80MHz, 32bit transceiver to a 16MHz, 8bit CPU.  Now, I'm disabling the WiFi on an ESP8266 and using this LoRa chip as a communications module.  I see there are some 5 pins labeled "Digital I/O, software configured".  I wonder how long it will take for another gifted hacker to hack an Arduino boards library for this RFM95W.

Hi @inq,

 RE: Maybe you don't want to get this deep into teaching.

  Not sure  I can .. although I have encountered some aspects of this, it can become a very specialised area. In addition, whilst much of 'school level' physics isn't too hard to visualise in an everyday fashion, some of this I find a struggle as I am not a specialist in this area.

Still, not one to willingly ignore a challenge ... with the usual caveats about all of my answers.. and sticking to antenna/feeder type issues

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Coax is short for co-axial ... the core and the screen have the same axis. I guess you understood that, but just in case..

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First, let's try to introduce a few terms in the ladder of confusion:

Resistance:

You probably don't have any problem with 'resistance' ... Ohm's law does a very credible job of relating voltage, current and resistance, providing the 'resistance' element in a circuit has a constant value, unaffected by the voltage, current, temperature, etc.

I = V / R .... where I is current in Amps, V is Voltage in Volts, and R is resistance in Ohms.

Reactance:

Reactance is analogous to resistance, and is also measured in Ohms, but is frequency dependent.

Reactance can be Inductive or Capacitive

Inductive Reactance:

Inductive reactance is usually discussed in terms of a coil of wire. A coil of copper wire will have a resistance, measured by passing a dc current through the wire, measuring the voltage across the ends of the wire coil, and applying Ohm's Law R = V/I, to calculate the resistance.

With copper, the resistance of a coil will usually be modest, albeit a coil with many turns of very thin wire, it can be significant. e.g. the coils in relays can be 100s of Ohms.

The coil will also have Inductance. For a steady DC current, this is 'invisible', but if the applied voltage changes, then the inductance effect will become 'visible'. The case of steady current DC will have resulted in a steady magnetic field. If the current through the coil is changed (e.g. by increasing the applied voltage), the strength of the magnetic field will also change, and a changing magnetic field, passing through a coil, induces a voltage. And this applies even when the coil is itself producing the changing magnetic field.

This induced voltage will always be of the sign (in a positive vs negative sense) to oppose the voltage creating the change, so that it reduces the "effective" voltage across the coil, and hence the current flow.

For a given coil, and assuming the applied voltage is a sine wave, the higher the frequency, the higher the reactance (in Ohms).

The formula for a coil reactance (XL) in Ohms (conveniently assuming zero DC resistance) is

XL = 2 * Pi * frequency * inductance

where frequency is in Hz, and inductance in Henrys.

NB if the coil has a non zero DC resistance, this is effectively in series with the reactance value. However as  the current and voltage are out of phase for the reactive part, and in phase for the resitive component, complex arithmetic is used to keep the reactive and resistive voltages and currents 'apart'. Most text books would probably delight you with the details, but for now, I will just invite you to do your own research if you really want to know more.)

Capacitance Reactance:

The 'classic' capacitor consists of two parallel metal plates separated by an insulator.

Since the two plates are separated by an insulator, the DC resistance between the plates will ideally be infinite.

However if a voltage is applied across the plates, it will set up an electric field as the positive plate attracts electrons on the negative plate. At the moment the voltage is applied, a current (of electrons) will flow along the wire from the positive plate towards the voltage source, and similarly a current (of electrons) will flow into the negative plate.

As soon as the applied voltage stabilises to a constant value, the current flow stops, but if the applied voltage is an AC sine wave, then the voltage is continually changing, and hence the current will flow in and out of the plates.

Thus, although NO current flows through the insulator separating the plates, it appears as if a current flows through the capacitor.

Like the inductance case, the capacitor has a reactance measured in Ohms, only as the frequency increases, the reactance reduces.

The formula for capacitive reactance in Ohms, when a sine wave is applied, is:

XC = 1 / (2 * Pi * F * C)

   where F is frequency in Hz and C is capacitance in Farads.

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Impedance:

Impedance is the combined effect of Reactance and (DC) Resistance. In 'real' circuits, every coil exhibits some resistance as the wire is not a superconductor and every capacitor will pass some current due to leakage and also exhibit dielectric loss.  But, there are many circumstances in which these effects are small compared to the reactance, and for '1st order' estimates, it is reasonable to consider such reactance.

However the term impedance is often used when describing an AC circuit, to cover all cases and combinations, involving resistance, capacitive reactance and inductive reactance, regardless of whether the user is carefully including all terms or just concentrating on the most significant one.

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Also, in any circuit involving AC signals, every component, and even tiny lengths of wire or PCB track, exhibit capacitance, inductance and resistance. At low frequencies, say up to a few 10s of kHz, components often appear to behave in the expected way, with a capacitor being essentially 'just' capacitive, etc. However, as frequencies increase, the other characteristics become increasingly important. e.g. at 20kHz, a 3cm piece of wire is just that, but at 2.5GHz it is (roughly) a quarter wave aerial.

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I don't know if the above helps or just confuses. I am hoping it gives some insight into what the terms are hinting at when they appear. I have tried to put them in a simplistic way. I am sure a quick Google could quickly give a more rigourous description. I hope I haven't oversimplified or distorted a more rigourous approach, but I have tried to give some insight.

I think that is enough for now ... I know you have asked more questions .. with luck I'll find a moment to revisit them, albeit my list of things to revisit is getting ever longer.

Best wishes, Dave

@davee, Excellent explainer, I think that is one of your best. I think the level of detail was right for this audience, not too much and not too little. Bookmark this one folks!

@davee,

I agree completely with @zander's observation.  I wish the forum had a categorized type Wiki location that was on a strict filtering by @dronebot-workshop, @codecage or someone willing and able.  This is definitely worthy of being in it and not hidden down in the bowls of a thread.  Yes, this needs bookmarking at the very least. 

I greatly appreciate you taking your time to do this.  For my sake alone, this is going way above and beyond.  I hope you recognize, the wider audience - those reading now, or years from now or even Internet searches that might hit this post - how much your words are/will be helping far more than my request.

I was solid through Resistance and V = IR.  I've heard and probably was introduced to those next terms back in EE 101 for non-EE engineers.  The years flushed them long ago.  I've never seemed to need them in my simplistic, monkey-see, monkey-do digital hardware designs.  I surmise these terms are the building blocks I need to understand for Antenna Impedance Matching. 

Posted by: @davee

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complex arithmetic is used to keep the reactive and resistive voltages and currents 'apart'. Most text books would probably delight you with the details, but for now, I will just invite you to do your own research if you really want to know more.)

Complex Math - I was afraid that four-letter word was on my horizon!  Math always came easy for me up until my Waterloo with Complex Math in graduate school.  Maybe, I'll take another stab at it... since I actually have something to apply it to.  I wonder if I burned that book??? 🤔 

Posted by: @davee

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I don't know if the above helps

Certainly does!  At the very least... it greases the skids.

Thank you.

Inq

@davee, Indeed, you hit the nail on the head..

Yes @inq, impedance matching in antenna theory is based on the fundamentals of reactance plus circuit resistance. This can get quite complicated, but as with everything else, is learnable if one desires the knowledge.

Good thing your a math wiz.. Most of us struggle with that part. lol

Regards,

LouisR

Posted by: @inst-tech

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reactance plus circuit resistance

Are you saying that the antenna design is based on not only the frequency, but what it's attached to???

If so, ... that sure opens a can of worms!

In the case of this LoRa, would it only be dependent on the RM95 LoRa module or would even the MPU and sensors attached to the MPU influence the antenna design?

Posted by: @inst-tech

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Good thing your a math wiz..

Someone asked me the other day what is the cube-root of 64.  Uh... I finally said... all the math I know has letters and squiggles.  If it has numbers, I pull out a calculator. 😜 🤣  

VBR,

Inq

Hi @inq (Ron @zander et al),

  Thanks for the comments above. I'll try my luck at extending the above theme a little .. fasten your safety belts ..

Yet again, messages have crossed in the ether:

Thanks once again for your contributions, Louis @inst-tech. Many of your, Inq's and others' contributions would benefit from Wiki approach. I would like to think the 'likely contributors' could be trusted to make and maintain their own contributions, but as Bill pays for the forum, as well as the rest of the site, I would not presume to make him poorer or have more responsibilities.

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And @Inq, in detail, yes antenna design is affected by source (if it is a transmitter) or load (if it is a receiver load impedances). In practice, a 'standard' value like 50 Ohms is usually adopted for a group/class of equipment, so it becomes more a case of reading spec sheets, until you want to design your own... Part of the black magic is matching an antenna to the chosen impedance. Some aerials include some kind of transformer, but don't imagine the kind of transformer you might use in a power supply ... it can be just some shaping of the metal work near to the cable connection points, but I have never been on that payscale!

In the case of modules with the Ipex or SMA connector (or expectation that one of these can be added), I would assume 50 Ohm, which happens to correspond to the natural quarter wave impedance ... hence a piece of wire of the appropriate length, either directly onto the PCB or via a coax lead  (also 50 Ohm) should be fine. Note the screen of the Coax should be connected to ground plane of the PCB at a point as physically close as possible to the ouput point used for the core connection.

And the cube root of 64 is .... um.....ummm ...  4... should I drag up a spreadsheet to check?

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As I mentioned above, Impedance is a combination of resistance and reactance, and if you actually know the know the values of resistance, inductance and capacitance, then in principle you can dust off the complex maths book, and do some 'proper sums'. However, for now, lets just pretend that resistance, reactance and impedance are near synonyms, so that if I talk of (say) impedance, think of how resistance would play out in similar circumstances.

Of course they are not synonyms, but maybe thinking of impedance as a magic kind of resistance is a starting point for getting a gut feel of what is happening. You may have already remembered complex numbers have two orthogonal components, real and imaginery, with resistance related to the real part, reactance the imaginery part, and impedance the combination of the two. And of course, real and imaginery parts are orthogonal and not interconvertible. So please don't blame me if you get the wrong answer by pushing the 'gut feel' approach too far!

Impedance Matching:

    Maximum power transfer occurs when the load impedance matches the source impedance.

Consider a simple 'two box' circuit consisting of a power source and a load, connected together. (e.g. a battery and a resistor)

If we have a power source, like a battery or power supply output, we usually assume it is a 'constant voltage' source. That is, providing we avoid heading for magic smoke territory, such as applying a short circuit somewhere, then we assume the source voltage will be the same, regardless of the load demand.

Now, if the source voltage is constant, then the load current will be inversely proportional to the impedance of the load.

And in our fictional 1-dimensional world, where resistance is interchangeable with impedance, we can use Ohm's Law to calculate the current, knowing V, the source voltage and R, the load resistance/impedance.

I = V / R

This can be extended to calculate the power P (in Watts) delivered to the load .. P = (V * V) / R

It follows, for constant V, the power dissipated in the load is inversely proportional the load impedance R, theoretically without limit, heading towards an infinite power as R approaches zero.

...........

But now consider the source has an 'internal' impedance Rs ... this is often modelled as constant voltage source Vs and a resistance Rs (impedance) in series, within the source 'box'.

In cases like aerials, receivers and transmitters, we often wish to transfer as much power from one stage to the next, starting from our source that has an impedance of Rs, what value of load R impedance will be optimal?

In the previous case, with zero source impedance, the power could head towards infinite power.

But if the source has a non-zero (positive) impedance, then as current passes through source impedance, it will have a voltage drop, and the voltage actually reaching the load will be less than Vs.

Intuitively:

If the load impedance is high (compared to the source impedance), most of the voltage will reach the load resistor, but the current, and hence the power dissipated, in the load will be small.

Similarly, if the load impedance is low (compared to the source impedance), most of the voltage will drop across the source resistor, so only a small voltage will reach the load, and once again, the power dissipated, in the load will be small.

The optimal case of maximum power transfer is when the source and load impedances are equal.

(Showing this to be true for some real values, say Vs = 2V,  5 Ohm source impedance, and load impedance values 1..10 Ohms, is left as an exercise for the reader. 😎 😎 )

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Whilst it is easy to imagine this scenario, with just a constant (DC) voltage source, and two resistors, the optimal case of matched impedances is equally valid for AC systems with reactances as well as resistances, albeit the sums are now complex ... arguably in both senses of the word 'complex'.

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But in practice for our purposes, this tends to mean looking for components with inputs and outputs matched to a common standard, often 50 Ohms, but as mentioned previously, other values are used in appropriate circumstances.

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I'll suggest we both take a short break at this point..

Best wishes, Dave

Posted by: @davee

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I'll suggest we both take a short break at this point.

Sure... I don't want to impose on you too much and wear you out.  I am not in any hurry as the project is progressing well with just the wire antennas.  And your post has given me several things to look up.  I do want to load up the hopper for when/if you come back... (follows)

Posted by: @davee

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Thanks for the comments above. I'll try my luck at extending the above theme a little .. fasten your safety belts ..

Strapped in!

Posted by: @davee

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And @Inq, in detail, yes antenna design is affected by source (if it is a transmitter) or load (if it is a receiver load impedances).

😯 Eeeh gads... you mean I might add a sensor to the MPU that is attached to the LoRa unit that is attached to the antenna and totally screw up what was tuned before the addition?  That seems kind of brutal!

Posted by: @davee

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And the cube root of 64 is .... um.....ummm ...  4... should I drag up a spreadsheet to check?

It was kind of embarrassing coming from my pottery instructor. 🙄 

Posted by: @davee

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You may have already remembered complex numbers have two orthogonal components, real and imaginery, with resistance related to the real part, reactance the imaginery part, and impedance the combination of the two.

That seems to be bouncing on a few brain cells that haven't been used in... 40 years. 🤔 

Posted by: @davee

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It follows, for constant V, the power dissipated in the load is inversely proportional the load impedance R, theoretically without limit, heading towards an infinite power as R approaches zero.

...........

I think I'm following the analogy down to here... and replacing the magic resistance portion where the "i" part of the complex number will be varying based on the 915MHz 

XL = 2 * Pi * frequency * inductance & XC = 1 / (2 * Pi * F * C).

I'm starting to get fuzzy after this.  I feel like there is some key concept I'm missing (besides the complex math and frequency).  The transition point I think you are talking about (on the transmit side) the solder joint for the 8.19cm wire antenna.  The source is the power generated by the circuitry on the RFM95W board.  The load is the antenna. 

1. The voltage is cycling at ~915MHz at the pin
2. The wire goes nowhere.  No continuity.
3. So no current flows through the wire
4. Yet, I know it emits power
5. I can more easily comprehend the coax acting as a capacitor and using power by the XC component, but there's no coax in this example and its the part doesn't transmit in reality.
6. I know you said that even a strait piece of wire has Resistance, Henrys and Farads, but I'm failing to see the mechanism that explains that.

I'm hitting a brick wall between 3 and 4... or what about the piece of wire that magickly converts cyclic circuit electricity into radio waves and/or how the power is "burning off"?

I will take a break also and ponder.  Maybe it'll come to me in a vision. 😜  

Thank you for your supreme patience.

VBR,

Inq

@inq . I found this explanation of how power in an antenna is radiated into radio waves.

the resource link can be found at: https://electronics.stackexchange.com/questions/73998/how-does-an-antenna-radiate-how-do-currents-flow-through-the-wire

regards,

LouisR

Hi @inq,

A pile of random comments .. sorry some are don't knows and best guesses, but hopefully it is helpful overall.

  Hitting a natural frequency can be a very powerful thing - Tacoma Narrows Bridge.  I don't know if that model in my head is relevant to tuning of antenna's.

There are similarities, but it is more of an analogy than a direct equivalence. An antenna deals with electrical current flow, whilst the bridge would have been the mechanical resonance.

I now understand from the previous posts about the coax shielding the actual core ( .. etc.

The co-ax cable (or equivalent feeder) is a transmission line, roughly analogous to hose pipe for carrying water. The outer layer is indeed a shield, but more importantly it is half of the transmission line. It plays a major part of the characteristics of the cable, including the impedance.

The shield being connected to a ground pin on the radio unit.

In principle, the transmission line, e.g. the core and shield of the coax cable, should be 'intact' until it reaches the input stage of a receiver or output stage of a transmitter. The criticality of keeping the feeder 'intact' for a receiver tends to increase as the wavelength reduces. For (say) a Medium wave (the old 'AM' amplitude modulation band), with a wavelenth range of about 200 to 500 metres, a few millimetres (or inches) is unlikely to significantly change its sensitivity. Furthermore, a typical 'long wire' aerial would only be small fraction of a wavelength, whilst a 'ground' (signal) wire, if connected, would probably take a long path (several metres) before reaching a buried conductor.

By contrast, at say 2.4GHz, with a wavelength of around 12.5 cm, a few mm can change significantly change things.

50 ohm Low Loss

The 50 Ohm characteristic is an AC impedance characteristic. The actual value for a coax cable depends upon the relative diameter sizes of the core and the shield, plus the material the insulator is made of. As mentioned previously. Ideally, the source and load impedances will match that of the cable. Any mismatch will cause a fraction of the electrical energy to be reflected back up the line, where it will 'suffer' constructive and destructive interference, in the same way that a sea wave hitting a vertical wall will bounce back to interact with the incoming waves. The worst case (possibly) being when waves constructively add to extent that the voltage experienced can exceed the voltage rating ... and shorts, magic smoke, etc. appear soon after.

The insulator between the core and the shield is often described as the 'dielectric'. Any material that could be used, e.g. polythene, will be affected by the electric field between the core and the shield, which will tend to cause the molecules to align with the field. As the field is oscillating, it will tend to cause the molecules to move in some form of oscillatory manner. The molecular motion will tend to absorb a portion of the oscillatory energy of the radio wave, which will be a loss. The description 'low loss' implies the insulation material has been chosen to minimise such 'dielectric loss'. In practice, there are several loss mechanisms, which also include the DC resistance of the core and the shield, and emissions that 'escape' through the shield. A commercial co-ax cable specification should include the dB loss per unit length (say per 100 metres), so you can estimate what fraction of the signal will emerge at the far end.

From an earlier post, you mentioned that it didn't have to be coax. It could simply be two wires next to each other. So a half-wave dipole could be as simple as:

The 'model' feeder and dipole yu showed is correct in general form, but note that the feeder wire (even 'bell wire', if that is what you used) will have an impedance depending on the spacing between the wires, etc., which will probably be higher than the impedance at the centre of the dipole, so there will be some mismatch. Whether it is close enough I don't have enough experience to say ... it will cause some energy to be fed back into the transmitter which might result in a component failure. Chances are, it will survive, though it might be some way from optimal. You did buy a few spares?????? 😏 😏 😏 

Which brings up why do some antenna's come in pairs and they are 90° to each other instead of 180°?

Sorry, I am not clear what you are referring to. Maybe you are thinking of something like the 'home' WiFi routers that come with two or more rotatable quarter-wave aerials.

In the case of a vertical aerial, the electrical field is in the vertical plane. For a receiver to receive the maximum signal, it too should be aligned vertically, as theoretically, there is no signal in the horizontal plane, although local reflections, and some 'stray' emissions from the aerial, etc.  may reorientate some of the signal towards the horizontal plane.

If the transmitting aerial is horizontal, then the maximum signal, at the same height, will be received by a horizontal aerial, in a line perpendicular to the transmitting aerial.

As aerials in laptops, moobiles, etc. are often not aligned to a particular aerial orientation, or directions, it can sometimes help to have transmissions form two or more aerials at different angles. Longer distance transmissions, with fixed installations, usually adopt a convention that both receiver and transmitter follow.

Re: And this 75/50 ohm helps Impedance Matching in some way for TV's?

Sorry, I suspect the 'convention that TV uses 75 Ohm probably has some historical base, but I don't know what it is. If my grey cell remembers correctly, the same 75 Ohm connector was in use in the mid 1960s and probably further back, when in the UK, some transmissions were at frequencies down to about 50 MHz, which implies rather large dipole aerials. Perhaps it was thought easier to match to?

I think some domestic VHF 'FM' receivers have used 75 Ohm as well, whilst others have used a 300 Ohm balanced input.

Most 'professional' equipment, seems to use 50 Ohm, albeit with several different connector styles.

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Sorry, I don't have any convenient reference books on this subject. Beyond radio amateurs, until recently, most radio transmitter and receiver designs were probably largely confined to specialist companies and others buying modules, etc., so I guess the market for any specialist books is going to be small, and hence expensive. I too am hoping someone else will make some suggestions.

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But since the core (antenna on the end) isn't going TO anything, how is current flowing to cause all this impedance / inductance aspects?

Sorry, but with the quarter-wave especially,where the shield stops abruptly, I am not totally clear how, but somehow, it seems that the physical earth takes the role. Wikipedia has several articles with useful information, including:

https://en.wikipedia.org/wiki/Antenna_(radio)#Effect_of_ground

and https://en.wikipedia.org/wiki/Monopole_antenna

which provide:

The monopole is often used as a resonant antenna. The rod functions as an open resonator for radio waves and oscillates with standing waves of voltage and current along its length. The length of the antenna, therefore, is determined based on the wavelength of the desired radio waves. The most common form is the quarter-wave monopole, in which the antenna length is approximately one quarter of the wavelength of the radio waves.

Also, in the particular case of a monopole antenna, the ground (or an artificial ground plane) serves as the return connection for the antenna current thus having an additional effect, particularly on the impedance seen by the feed line.

A further piece of the magic, is that even 'free space' has an impedance, Again using Wikipedia..

https://en.wikipedia.org/wiki/Impedance_of_free_space

which includes the following gem of information to ponder, if you are having trouble sleeping:

In electromagnetism, the impedance of free space, Z₀, is a physical constant relating the magnitudes of the electric and magnetic fields of electromagnetic radiation travelling through free space. That is,

Z0=|E||H|,

where |E| is the electric field strength and |H| is the magnetic field strength. Its presently accepted value is^[1]

 

Z₀ = 376.730313668(57) Ω.

Where Ω is the ohm, the SI unit of electrical resistance. The impedance of free space (that is the wave impedance of a plane wave in free space) is equal to the product of the vacuum permeability μ₀ and the speed of light in vacuum c₀. Before 2019, the values of both these constants were taken to be exact (they were given in the definitions of the ampere and the metre respectively), and the value of the impedance of free space was therefore likewise taken to be exact. However, with the redefinition of the SI base units that came into force on 20 May 2019, the impedance of free space is subject to experimental measurement because only the speed of light in vacuum c₀ retains an exactly defined value.

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So yes, the power eminating from a piece of wire that apparently isn't connected to anything, is 'genuine'. Remember, visible light is also an electromagnetic wave. The only thing that differentiates it, is that the frequency is a lot higher ... covering about 400 - 790 TerraHz, and the wavelength correspondingly shorter .. say 750 - 380 nanometres. We have no trouble believing it is being launched from a surface, and travelling through clear solids, liquids, gases and vacuum.

Plus if we look at conducting frequencies in the GHz and above range, waveguides are commonly selected, which (somewhat facetiously), are more reminiscent of water plumbing than electronic wiring. e.g.

https://en.wikipedia.org/wiki/Waveguide_(radio_frequency)

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In the case of this LoRa, would it only be dependent on the RM95 LoRa module or would even the MPU and sensors attached to the MPU influence the antenna design?

In general, I would hope the radio performance would be essentially depdendent on connecting to the antenna together with the nearby 'antenna ground', using an appropriate feeder and antenna. However, bear in mind, this is a physically small transmitter (implying some level of power) and receiver (implying a highly sensitive input, etc.), plus other digital circuits, all of which 'accidentally' radiate and intercept radio signals in the vicinity. This is a pretty ambitious set of potential problems, all squeezed into a tiny space.

This implies, it might be found necessary to adopt screening, filtering of power leads, 'sensible' layout with accompanying electronic units and so on, for an optimal design. I would try to avoid having PCBs stacked closely above each other, to minimise the chance of accidental cross-coupling, and take care with routing cables together, as there will be some (small) level of cross-coupling. Sorry, I wouldn't know where to start in predicting all the possible gotchas.

That does not mean, you are doomed to failure, but be prepared for some unexpected events and problems. Of course, if you are not looking for ultimate performance etc., which I don't think you are, then the chances of it working are generally much better. I suggest you try to proceed in small steps, starting with the simplest, etc. and build up the system, continually ensuring nothing has 'broken' on the way.

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If Bill's video was anything to go by, it seemed like a robust radio system, so it will probably be fine.

Best wishes and fingers crossed it all works perfectly, Dave

Hi @inq,

 A quick-ish spot from your comments.

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XL = 2 * Pi * frequency * inductance & XC = 1 / (2 * Pi * F * C).

If you have a reliable, accurate, precise,  resistor with 10 Ohm written on it (in some way), then if you check it with your multimeter, it will show a number close to 10 Ohm.

Furthermore, if you rig up a little circuit, with the same adequately power rated, resistor and current meter in series, connected to an equally good power supply set to 20V, then the current will be given by I = V / R = 20/10 = 2 amps.

(Of course, Don't try it with a 0.25 W resistor.!!)

Now switch your power source to AC (sine wave), again 20V, and providing your meter can accurately read AC current at any frequency you choose to generate, the current measurement will still be 2 amps.

In other words, the resistance is the same at any frequency, including DC.

==============

You may spot a few practical problems with your kit at home, like your power supply doesn't have a variable frequency AC output, your multimeter probably starts to struggle as the frequency increases, and if you push the frequency high enough, the resistor also starts to exhibit some inductive and capacitive traits, but just assume you have been given this lab with a billion dollar equipment budget ... in which case, hopefully you can see life is simple so far, even at high-ish frequencies.

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Now we try the same kind of trick with an inductor. It says it is 3 H (Henry) inductance, and being made of a coil of copper wire, has a DC resistance of 2 Ohms.

Initially, we put in the same 20Vdc supply circuit, replacing the resistor with the inductor.

I = V / R    = 20 / 2 = 10 Amps

Nothing much different so far.

Now, lets imagine the 2 Ohms DC resistance is magically reduced to zero, but the inductor is still 3 H ... and switch the power supply to AC ... start at 1000 Hz.

There is no DC resistance, but what about the inductive reactance?

Here is where we need XL = 2 * Pi * frequency * inductance

XL = 2 * Pi * 1000 (Hz) * 3 (H)

If we approximate Pi to 3.1 ...

XL = 6.2 * 1000 * 3 = 18.6 * 1000 (Ohms) = 18.6 kOhms

Then using the current becomes I = V /XL = 20 / 18.6 * 1000 = (approx) 1.1 milliAmps.

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And if we switched the supply to 1 MegaHz, the same calculation would yield approximately 18.6 MegOhms and 1.1 microAmps

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Notes:

• With a pure inductance (ie no DC resistance or capacitance), for the AC case
• • the current waveform will be 90 degrees behind the voltage waveform
  •  
  • although there is a current flowing through the coil and voltage across the coil, the average power dissipation is zero ... that is the coil takes in power and then 'gives' it back, so that the total power over a whole cycle becomes zero. (That doesn't happen with resistance, because the voltage and current are in phase.)
  •  
  • If we reintroduce the 2 Ohm DC coil resistance, as well as the 3 H inductance, we will need to dust off the complex arithmetic book, etc., but you can intuitively guess that 2 Ohms is going to be largely 'lost' in comparison with XL values of kOhms and MegOhms. This is the basis of my 'magic' thinking ... there will be occasions, when one characteristic will largely swamp the others. Of course, sometimes they are comparable, and the magic falls apart without doing the proper sums.
  •  
  • Finding a 3 H inductor, with zero DC resistance and/or zero capacitance means you will need to start mining unobtainium from another universe... 😎 😎 ... i.e. they don't exist, but hopefully their contributions can be kept small enough to only be a minor inconvnience.

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It is too late tonight for me to go through the equivalent calculation for capacitance, but hopefully you can see the principle. Note in this case, XC goes down, as the frequency increases. Let me know if you have any queries.

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Best wishes, Dave

Posted by: @inst-tech

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@inq . I found this explanation of how power in an antenna is radiated into radio waves.

-- attachment is not available --

the resource link can be found at: https://electronics.stackexchange.com/questions/73998/how-does-an-antenna-radiate-how-do-currents-flow-through-the-wire

regards,

LouisR

Outstanding!  That totally clicked!  The charge being cyclically repelled, then drawn towards the antenna base resulting in a current flow back and forth.

From that explanation... the leap for why the speed-of-light constant is used to find the natural frequency of an antenna's length, now seems totally obvious.  It is almost exactly analogous to a mechanical system of a weight on the end of a spring and its natural frequency.  If the antenna is longer than the frequency, the charge hasn't gotten to the end and built-up its full rebounding strength because its already started its attraction portion of the cycle.  Likewise, if its too short, it's reached its full rebounding strength, but is still being pushed (without success).  Only in the sweet-spot does it complement and use the least energy, yet deliver the most.  GOT IT! 

Although, with that analogy, it seems like a 1/2 wave would be optimum, and not a 1/4  wave and certainly not a full wave length antenna.  Got to digest on this more.  Might need to dig back into the Maxwell Equations your paper suggested.  

Well... at least it isn't total black magic anymore. 

Thanks Louis.

VBR,

Inq