Does this look like a scam?

zero_coke

Joined Apr 22, 2009
294
Yes Ghar, that sums it up well enough. I was thinking about this the entire time in regards to separate the magnetic field from the RF radiation. I don't know if much people on this forum understand that I'm not trying to do this through RF, but rather through magnetic field, so losing power from my transmitter is really not that much of a big deal. Even if I did lose power my input current will be very very low compared to the induced current in the secondary coil. So as long as I am within an acceptable vicinity of the magnetic field the efficiency should not be a major issue . I was hoping to create this massive magnetic field around a house by making the house itself a transformer, so any device inside the house would be able to be inside the magnetic field, and hence any device can be inducted to receive current and etc.

Like I said, I am not trying to do this through radio, but rather magnetic fields only. My objective is to have a small input current, and then create this massive strong magnetic field that can induce a current into any device within an acceptable vicinity dictated by the cube law. I know that the field is even worse when it comes to losing its strength: inverse cube law > inverse square law, but at least you don't lose hundreds of watts of power just to send 60 watts to a lightbulb right? You'll only be sending milliwatts of power and let the magnetic field do the transfer efficiently, rather than send 300 watts and get 60 watts at 30 feet using RF, you might as well use 70-80 watts and get 60 watts at, say, 10 feet, through magnetic field coupling, no?

???? confirm physicists! :) hehe
 
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Ghar

Joined Mar 8, 2010
655
Your not propagating electromagnetic energy?

Any wave that disapates by the inverse-cube law will be more inefficient.

Shipley's Thesis was for 30 feet.
It doesn't dissipate by inverse cube and it's not a wave.
It's a stationary field whose amplitude decreases with inverse cube. It is not energy loss, it's more like a reduction in potential.
 

zero_coke

Joined Apr 22, 2009
294
The only energy loss would be to CREATE the magnetic field right? So this is where evanescent magnetic coupling comes into play if I'm not wrong. Here's how I think it works like:

Evanescent (rapidly diminishing) fields are the best method to induce a current into a secondary coil because it "rapidly" diminishes, meaning it is created with "minimal" current and that "minimal" current is current transferred at some %, and lost some %. Because it is so minimal, it is better than creating a non-evanescent field that requires more current to set up the field and maintain it, and then have it diminish with some % lost and some % transferred. Does it make sense? E.g. 1 mA power used to on/off rapidly a magnetic field vs. 100mA used to on/off field slower than previous setup. You lose a lot more using the 100mA approach I think. It's like sort of comparing AC vs DC current. If you let the electrons go further, more energy is lost. Likewise, if you let the current flow longer and field not collapsing fast enough, you lose more % of current that can otherwise be used to power your secondary coil.

So I'm saying that higher frequency, lower current input, lower the cumulative resistance of the circuit (capacitor and inductor together), and you should be able to transfer power at good efficiency for a couple of feet or any desired distance so long as your magnetic field is big enough to encompass that target distance. I'm going to try this as soon as I get home and my function generator arrives. Gee, won't be home for another 2-3 months, but oh well, will give me a lot more time to continue on in my research on this!
 
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Ghar

Joined Mar 8, 2010
655
I think you have the basic idea but still sounds like there's some confusion (not that I know exactly how it works either...)

Induced voltage is of course:
\(V = \frac{d\phi}{dt}\)

You can either increase the magnitude of the field or increase the rate of change to increase your received potential.
As you go up in frequency you're more likely to radiate and incur a loss there so you can't do that indefinitely.
As you go up in field strength you need to increase your current and incur a loss in the generation of the field (like \(I^2R\)) so you don't want that either.

Also, I'm not sure if anyone told you yet but a function generator is very low power output. You'll need to have a high frequency power amplifier.
 

zero_coke

Joined Apr 22, 2009
294
I think you have the basic idea but still sounds like there's some confusion (not that I know exactly how it works either...)

Induced voltage is of course:
\(V = \frac{d\phi}{dt}\)

You can either increase the magnitude of the field or increase the rate of change to increase your received potential.
As you go up in frequency you're more likely to radiate and incur a loss there so you can't do that indefinitely.
As you go up in field strength you need to increase your current and incur a loss in the generation of the field (like \(I^2R\)) so you don't want that either.

Also, I'm not sure if anyone told you yet but a function generator is very low power output. You'll need to have a high frequency power amplifier.
I think induced voltage for a coil of wire is:

\(V = N*\frac{d\phi}{dt}\)

So I don't think induced voltage is really what I'm looking for, but induced current. That would be

\(I = N*\frac{d\phi}{dt}* \frac{1} {R}\)

R would be my transmitter's resistance or my receiver's resistance?

You can either increase the magnitude of the field
How do you "increase the magnitude of the field" without touching the current or frequency? Isn't the flux

\( B = \frac{\mu*N*I}{l}\)

\( \phi = B* A_loop\)

\( \phi = \frac{\mu*N*I}{l}* A_loop\) --> magnetic flux of solenoid

? There is a current term in the magnetic field, so they are dependent on each other, no?

And why am I more likely to radiate if I increase the frequency? Isn't increasing the frequency good so I can build and collapse the field faster, thus increasing the rate at which the current is being induced to increase?
 
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Ghar

Joined Mar 8, 2010
655
To increase field strength you do need to increase current, you're right. I did say that but poor wording on my part, should have said "To go up in field strength"

The N doesn't need to be there, it's simply how you define your flux. If you consider phi as 'total flux' then N is implicitly in there. When you include N you're saying phi is your flux per winding.

I did say that higher frequency is good, because like you said you get a higher induced voltage for the same field magnitude.
Your equation for current assumes a purely resistive loop, which it won't be, but it will still be related to induced voltage. Increasing one will increase the other.
In your equation R would be the resistance of the receiver because that's where your induced voltage is. Think spatially - where is the induced voltage, where is the resistance, where is the current? You're relating induced current to induced voltage so you must be talking about the receiving loop. You always need to keep in mind what's where when you get into coupling because you will have several loops and several currents etc.

You're more likely to radiate because your wavelength will become smaller. Radiating efficiency and pretty much everything to do with this stuff is related to wavelength. Near field vs far field are defined in terms of wavelength. By definition of what you're trying to do you need to stay in the near field so you need it to extend at least a few meters.
At the same time you need your structure to be a poor radiator which means having it small compared to the wavelength along with some more complicated features.
Take a look at this:
http://en.wikipedia.org/wiki/Radiation_resistance
A high radiation resistance means you're radiating effectively so you need to keep that quantity small. That means keeping the fraction of physical size to wavelength small.

Remember the MIT numbers they used. 10 MHz (30m wavelength) with 60cm diameter loop. One simple calculation of the range of the near field is just this:
\(\frac{\lambda}{2\pi} = 4.8m\)
There's a transition area and it's fairly complicated but that's a simple definition.
(Also note that by going from a distance of 1m to a distance of 5m your field magnitude decreased by a factor of 5^3 = 125)

Pretty much the main point of my previous post was how it's a compromise between efficiency and working at all.
 

zero_coke

Joined Apr 22, 2009
294
Beenthere: thanks for the link for EMF and electromagnetic induction!

Ghar: Very well put. I understand what you are saying. Thank you for the physics knowledge behind this all. My experiment will consist of a function generator that will oscillate at 171.2 KHz. I used 171 KHz because like you said, if I increase frequency I decrease wavelength then I would be radiating or losing energy out at a rate of \(P = I^2 R\) , which is not good I'm thinking.

Using this frequency, I calculated that my wavelength is 1752 m, which is much larger than my transmitter's size, so it will have low radiation resistance. Using this wavelength, I calculated the field-range to be ~280m, and I'm looking to transfer power to 2-3 meters at most, so I'm good to go right? However, something that caught my attention was that no matter what the wavelength is the strength of it decreases by 1/R^3 anyway, so although my near-field is ~280m, it still decreases by 125 fold if I go from, say, 1m to 5m = 1/5^3 = 125, so I guess frequency/wavelength do not play that much of a role in the case of range of the field, right?

I mean, what is the flaw in this? No matter how "big" the field's range is, it still decreases with 1/R^3. I think the only way I can make the field "stronger" is with higher current, reduce the length of solenoid, or increase # of turns of the solenoid.

I can't quite grasp the importance of frequency in this sort of application. So I re-read your post and I know you said that the higher the frequency the better the radiator, and I'm thinking: if radiation was NOT a problem, so just for curiosity then you can have any frequency and it will have no effect on the coupling and induction range right? I'm using low frequency to make my transmitter radiate poorly so all the energy is contained within its magnetic field, and I understand that. I just wanted to make sure that the only role of frequency in this case is only for managing good/bad radiation of the transmitter. Is it? Is frequency only important for radiation or does it play any other role in the transmitter?

The question that comes into my mind is why MIT would use 10 MHz. It's in the VHF (Very High Frequency) spectrum and I don't know why they used it. I'm sure they know why and have a good reason for it, but so far I'm thinking it was a really bad idea. I bet there's something else to this that I do not know. In other words, something that I don't understand actually because it's written in their paper on witricity and their demo hehe.
 

beenthere

Joined Apr 20, 2004
15,819
One thing is for certain - they had a means to insure that the signal did not propagate. 10 MHz is one of the freq's reserved for WWV, the NBS' accurate time broadcast. Never radiate on a freq used by los Federales.
 

nsaspook

Joined Aug 27, 2009
13,086
One thing is for certain - they had a means to insure that the signal did not propagate. 10 MHz is one of the freq's reserved for WWV, the NBS' accurate time broadcast. Never radiate on a freq used by los Federales.
I would have used the 13.56Mhz ISM band, but maybe 10.0Mhz has some special property?
 

Ghar

Joined Mar 8, 2010
655
I mostly agree with what you said.

There is more to it yet, notice we haven't gotten into the topic of resonance at all yet.
Like I said your equation for current isn't accurate unless the loop is purely resistive and since you're picking up magnetic field it clearly isn't!
It will be some more complicated equation based on capacitance and inductance of the receiver, not only for the magnitude of your current but also the phase relationship that determines how much power you actually receive (and send)

The range does increase with frequency, because it raises the induced voltage you get at your receiver.
If you need to induce 1V then the derivative of total flux must be equal to 1.
With a sinusoidal field the derivative is of course:
\(V = \frac{d\phi}{dt} = \frac{d}{dt}Asin(wt) = Aw cos(wt)\)
So at the same distance with the same field magnitude, your induced voltage will increase if your frequency is higher, effectively raising your range.
So you're right, the frequency doesn't technically increase the field's range but it does let you receive something further away. If you increase your frequency by a factor of 8 you will be able to get the same result twice as far away:
Example;
At 1m:\(|V| = Aw\)
At 2m:\(|V| = \frac{Aw}{2^3}\)
Raise your frequency, still 2m: \(|V| = \frac{8Aw}{2^3} = Aw\)

Also notice that if you wanted to increase A by a factor of 8 you would've increased your transmitter's I by 8.
The I^2R loss would've went up by 64.

I'm pretty much nearing the extent of my physics by the way, I don't know enough about specific antenna equations to actually get into power efficiency.
 
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zero_coke

Joined Apr 22, 2009
294
I mostly agree with what you said.

There is more to it yet, notice we haven't gotten into the topic of resonance at all yet.
Like I said your equation for current isn't accurate unless the loop is purely resistive and since you're picking up magnetic field it clearly isn't!
It will be some more complicated equation based on capacitance and inductance of the receiver, not only for the magnitude of your current but also the phase relationship that determines how much power you actually receive (and send)

The range does increase with frequency, because it raises the induced voltage you get at your receiver.
If you need to induce 1V then the derivative of total flux must be equal to 1.
With a sinusoidal field the derivative is of course:
\(V = \frac{d\phi}{dt} = \frac{d}{dt}Asin(wt) = Aw cos(wt)\)
So at the same distance with the same field magnitude, your induced voltage will increase if your frequency is higher, effectively raising your range.
So you're right, the frequency doesn't technically increase the field's range but it does let you receive something further away. If you increase your frequency by a factor of 8 you will be able to get the same result twice as far away:
Example;
At 1m:\(|V| = Aw\)
At 2m:\(|V| = \frac{Aw}{2^3}\)
Raise your frequency, still 2m: \(|V| = \frac{8Aw}{2^3} = Aw\)

Also notice that if you wanted to increase A by a factor of 8 you would've increased your transmitter's I by 8.
The I^2R loss would've went up by 64.

I'm pretty much nearing the extent of my physics by the way, I don't know enough about specific antenna equations to actually get into power efficiency.

I think I know what you mean by the current not being what I wrote in my earlier equation. I should've put:

\( I = \frac{d\phi}{dt}*\frac{1}{Z}\) for AC circuits that have phasor relationships in an inductor-capacitor circuit.

Furthermore,

\( I = \frac{\epsilon_0}{\sqrt{R^2 + (X_L - X_C)^2}}\)

At resonance, my secondary loop will be purely resistive because X_L will equal X_C, and so my induced current will be:

\( I = \frac{\epsilon_0}{R}\)

The R will be some bogus resistance I will throw in there just to not get 1/0 = ∞ current and short circuit my function generator source. Is it correct so far?
 
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Ghar

Joined Mar 8, 2010
655
Well it's correct for a series RLC circuit but I don't know what the most accurate model for this would be.
(Though I don't know where you got \(\epsilon_0\) from)
 

JoeJester

Joined Apr 26, 2005
4,390
but maybe 10.0Mhz has some special property
Easier math.

Even closed systems ... like the current power distribution system's Power Line Carrier ... has been known to cause interference with other services.

Since you are using RF frequencies, you might want to consider the current rules on transmitting frequencies, which are interim, what power you may use. Of course if you wish to exceed those limits, you might want to consider building a screen room to trap all the emitted waves from reaching the outside world.

The RF spectrum is pretty busy. Interfering with other services carries a hefty fine.

Every country has the equvalent of the U.S.' FCC. FCC rules and regulations are covered by 47 CFR.
 

zero_coke

Joined Apr 22, 2009
294
Well it's correct for a series RLC circuit but I don't know what the most accurate model for this would be.
(Though I don't know where you got \(\epsilon_0\) from)
The \(\epsilon_0\) just means the peak voltage of the source. It's supposed to be the peak arrow in the phasor diagram.

And JoeJester: I am aware of the consequences of such transmission of RF frequencies. However, I don't know which frequencies are safe to transmit something in? It seems like its not allowed in any frequency spectrum except one or two which are reserved for ham radio amateurs, and I don't have a licence.
 

Wendy

Joined Mar 24, 2008
23,415
It's been said many times, but 13.57Mhz has been set aside for industrial power use for jobs other than communications.
 

JoeJester

Joined Apr 26, 2005
4,390
However, I don't know which frequencies are safe to transmit something in? It seems like its not allowed in any frequency spectrum except one or two which are reserved for ham radio amateurs, and I don't have a licence.
Ignorance of the law does not prevent prosecution.
 

zero_coke

Joined Apr 22, 2009
294
Ignorance of the law does not prevent prosecution.
I don't know if you read questions backward or something, but you didn't answer any part of the question in case you were trying to. I asked you, since you know so much about FCC and the law, what is/are the safe bands to radiate in without getting into trouble with the 'law'?
 
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