RF Energy Harvesting

Question

Why can energy invested in RF radiation not be harvested fully? What are the main factors that affect this? Is there any significant research nowadays that fully describes this process? I stand on the conclusion: "Energy moves from one form to another.", but in this case why can it not be harvested in that quantity to use it for powering low-power devices or store it in significant quantities in some batteries for later use? Why is it not profitable nowadays and why is it not used today like energy from water, wind and other energy sources?

diagram of RF energy harvesting system

I don't think it is a duplicate. The other is about a particular scheme (which really is just a scam) whereas this seems to be about why you can't capture all of the available RF. That is, why can't we reclaim all of the RF we are broadcasting and use it for power. — JRE, Jun 12 '15 at 08:26
Do you mean all frequency spectrum by saying 'all of the available RF'? — Aenid, Jun 12 '15 at 08:28
@Lazar you've added a new bounty to this so is there something about the information you have received so far that is missing? — Andy aka, Apr 14 '16 at 11:56

Andy aka · Accepted Answer · 2016-04-16T10:29:33.160

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Consider an antenna that converts all the electrical energy from a power oscillator to an electromagnetic radio transmission. Let's say 1 watt is fed to the antenna and 1 watt is the emitted radio wave power.

That power is flung out in most directions; just like a lightbulb emits light power in (virtually) all directions. The lightbulb isn't just a similar example it is exactly the same example - the light emitted is an electromagnetic wave.

Now imagine you are 1 metre from that lightbulb with a 100% efficient solar panel. Then imagine that you build several solar panels that exactly and totally shroud the lightbulb. You wire the solar panels in series and see how much power you can extract. Well, for this thought experiment, you can actually extract 100% of that power.

Then you stand 10m away and do the same. Obviously you need plenty more solar panels but, assuming the light emitted is totally hitting the panels and there is no leakage of light beyond those panels, you will receive 100% power.

So it can be done, but at great expense and inconvenience. But not too much of a problem with a pencil sharp laser beam and a 100% efficient solar panel. If you can focus the light sufficiently you can collect ALL the energy, Remember this is just a thought exercise and solar panels and lamps/lasers are far from 100% efficient.

Going back to a transmitted radio wave, you could build a parabolic dish and transmit at a sufficiently high frequency to be able to focus most of the power onto a receive antenna (another dish). This is done across the country by military and telecom people to get line-of-sight data and voice comms that are "largely" private in that it's hard to "steal" a listen to what is being transmitted unless you are up-close to the antennas and picking up a small side-lobe of energy.

For the rest of all the transmitters broadcasting, they are generally dipoles and spread their energy emitted in all directions in at least one plane. This allows music stations to be picked-up quite easily and that, of course, is the aim.

Can you easily harvest that energy? Not very well because the power is largely flying off into space and only a fraction is received by "listeners".

Each listener's antenna is effectively a net (i.e. it has a real "area" despite its long thin shape) that captures power sent from that transmitter and the amount it captures is femto watts usually. Normally, an RF receiver (at the risk of being too general) can work with about 10 micro volts received from the antenna and, averaging across a broad sample of antenna types, this voltage will have a source impedance of about 50 ohms.

The power liberated is therefore 5 uV squared / 50 = 5 femto watts.

Obviously the closer you are to the transmitter the greater the power that can be liberated but, if all you are interested in is getting usable power then use wires.

If you did the math on the sum total of energy that can be harvested at any one point on the earth you wouldn't find many places where more than a couple of milli watts can be gleaned.

Solar power and RF energy harvesting are basically both forms of the same thing - electromagnetic wave harvesting.

Even when we stand by router? I read that some routers gives up to 80mW

Routers typically use quarter wave antennas and to collect energy you could also use a quarter wave antenna. The realistic distance you can get up-close is about lambda (1 wavelength) before you start causing strange loading effects on the antenna. This is a distance of about 0.1 metres so, using the Friis transmission equation converted to dB: -

Loss (dB) = 32.45 + 20\$log_{10}\$(f) + 20\$log_{10}\$(d)

Where f is in MHz and d is in kilometres. This equation tells you how many dB of power loss you can expect at a given distance with a given carrier frequency.

For 2.45 GHz and 0.1 metres (0.0001 km) the loss is 32.45 + 67.6 - 80 = 20 dB.

But the tx and rx antennas have about 4dB gain between them (because they are semi-directional) and so the loss is only about 16 dB.

80 mW in becomes (theoretically) 12.7 mW out.

edited Apr 16 '16 at 10:29

answered Jun 12 '15 at 09:06

Andy aka

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Another thing to consider is what happens if we actually tried to harvest all of it passing through a point - that's just a fancy, expensive shield, through which nothing could pass. Might become problematic if they were on all the buildings. +1 for Andy. – Jun 12 '15 at 09:56
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@SeanBoddy What about harvesting energy from our home wifi routers? – Jun 12 '15 at 10:18
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@MasterYoda Regarding the above comment, to what point? Even if you did harvest the milliwatts coming from your router, you wouldn't be able to use it to transceive information. Also, since it is the real world, you will lose more power overall than you can recover. (Power supply wastes heat, transmitter wastes heat, so would the reciever). – R Drast Jun 12 '15 at 10:23
@RDrast Yes you're right, but I thought about device which can scan some frequency range and find where the most power is transmitted (strong signal) and then store energy to battery for some future use, not immediately for charging devices. No matter how many miliwatts, for some time, battery will be charged and can be used. – Jun 12 '15 at 10:28
@MasterYoda - a very strong signal might be 1 milli watt. – Andy aka Jun 12 '15 at 10:35
Even when we stand by router? I read that some routers gives up to 80mW - http://www.eham.net/ehamforum/smf/index.php?topic=41599.0;wap2 – Jun 12 '15 at 10:37
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@MasterYoda You might just harvest enough to stop your phone battery going down in sleep mode! Of course, then you wouldn't be able to use Wi-Fi at the same time. And you'd have to be up against the Wi-Fi access point, so why not just plug your phone into whatever the AP's plugged into? – user253751 Jun 12 '15 at 10:39
Yes, it doesn't have sense, but its a little sad, really great topic to research and maybe find some exploitable solution. Thanks for the answers. – Jun 12 '15 at 10:42
@MasterYoda - see the calc I did for the wifi router at 0.1 m between send and receive antennas. – Andy aka Jun 12 '15 at 11:07
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@masteryoda, the point is that if you harvest the power from an RF transmission, the transmission is no longer there. In a very real way, you would gain more net energy by unplugging the transmitter. – Jun 12 '15 at 18:06
@SeanBoddy Harvesting the RF energy is more like placing an obstacle in the path of the waves, producing a "shadow". All antennas have an equivalent capture area. But they cannot pull energy from regions outside their near field, or perhaps their "light cone", as an analogy. The energy that was turned in to useful power by harvesting would have either 1) radiated away to infinity or 2) been turned in to heat by hitting something that absorbs it. This is identical to the behavior of light, only the size scale is different. If you don't "harvest" light from a light bulb, does anything "care"? – Apr 03 '16 at 14:05
@nocomprende, you forgot 3) have its equivalent electrical power consumed by one of the devices it was actually trying to communicate to. If the wall between you and the router ate half the power, your device can't get a usable signal, and worse, you got virtually nothing in return as power. I can do this math. Unplug the router. Seriously. – Apr 03 '16 at 14:11
@Andyaka I want to point that solar power IS NOT RF energy harvesting. RF and solar light are electromagnetic waves, but the frequencies of RF are completely different from the frequencies of visible light – Claudio Avi Chami Apr 16 '16 at 10:06
@ClaudioAviChami you are absolutely correct and I stretched the truth a little too much. Correction on its way. – Andy aka Apr 16 '16 at 10:27
I'd like to point out that the various losses and parasitics of the components themselves are largely being ignored here. Every electronic component is lossy and therefore will subtract from these theoretical maximums. Even rectification diodes are significantly lossy - meaning that a 12.7mW theoretical maximum is probably more like half that in reality. – rdtsc Apr 18 '16 at 16:39
@rdtsc all excellent points but "more like half that in reality" is only a drop of 3dB - I reckon more than that but, these are also approximately the same losses for other "reclamation" schemes. – Andy aka Apr 18 '16 at 22:15

score 8 · Answer 2 · answered Jan 29 '16 at 17:44

If we generalize your question a bit to cover not just RF energy harvesting, but electromagnetic energy harvesting in general, we do that all the time.

The sun is essentially an electromagnetic transmitter with a power output of approximately 4 × 10²⁶ watts. A very large solar farm has a (peak) output of 6 × 10⁸ watts.

To put this in perspective, if you had 8 kilograms of gold, and you removed a single atom, that's about the same fraction as the power received from the sun.

That's $287,000 of gold, and you are able to harvest one-half of one trillionth of a cent.

If you took all the people in the world, you'd harvest about a fingernail trimming's worth of human.

The reason of course is because most of the sun's energy output doesn't even hit the Earth.

The problem is much the same with RF energy harvesting. Of all the RF energy transmitted by all the world's transmitters, only a minuscule fraction of it will hit your antenna.

If I have my figures correct (from memory), the sun turns 4,000 tons of its mass into light every second, and the earth intercepts 2 ounces of photons per second. Er, 60 grams. — , Jan 29 '16 at 22:41

score 2 · Answer 3 · answered Jun 12 '15 at 14:01

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Related: I heard that farmers with long wire fences running parallel to power lines could "steal" coupled 50/60 Hz AC. Not sure if this is true or how much power would be available, but I am guessing a small amount, but enough that the power companies noticed the loss and went looking for it. After you have spent the money to buy all that wire for your scheme, you might have done better buying a solar panel or wind or water mill. No one will accuse you of stealing from them in that case.

Maybe you could try harvesting gravity energy from falling rain, etc. instead?

answered Jun 12 '15 at 14:01

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Not sure if this answer is complete enough to be considered a proper answer. But aside from that, I believe the electricity stealing is more so from the capacitance created by the high voltage and ground, not necessarily the EM interference generated by them. – Jarrod Christman Jun 12 '15 at 14:09
Ofc they can steal, its two conductor wires in parallel, one inducing electron flow in second - electro-magnetism. – Jun 12 '15 at 14:11
@JarrodChristman: a long wire parallel to another long wire is effectively a transformer, so it fits the scenario that the OP asks about, just in the ELF range of "radio". If it was a capacitance effect, then a large metal plate or box - such as a barn - would do the job. I have never heard of that being done. There would have to be a complete circuit in that case...? – Jun 12 '15 at 14:14
You're correct, after researching further, induction makes the most sense. – Jarrod Christman Jun 12 '15 at 14:30
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"Maybe you could try harvesting gravity energy from falling rain, etc. instead?" I think we call those hydroelectric plants. Seem to work pretty well. Maybe one could construct a miniature one placed at the bottom of a downspout. Probably generate more power than one of these RF schemes. – tcrosley Jan 28 '16 at 17:41
@tcrosley Yes, micropower can work, if you are willing to wait on nature to give you rain, wind, cloudless days, etc. It is an interesting notion, but rf is not energy-dense enough, and there are not reliable natural sources. Ultimately, of course, ALL the energy in the universe comes from gravity, and the big bang. Too bad we can't just use gravity as is. – Jan 29 '16 at 22:39

score 2 · Answer 4 · answered Apr 16 '16 at 20:03

The problem has to do with physics and the efficiency of energy conversion. A summary of work over frequency is in the following plot from [hemour2014towards] below:

The short answer is that you have to drive a device to capture energy, and you have losses across the board due to the realities of manufacturing, such as device mismatch. You can find everything you need for a crash course regarding the issues of RF energy harvesting by reading:

@article{hemour2014towards,
title={Towards low-power high-efficiency RF and microwave energy harvesting},
author={Hemour, Simon and Zhao, Yangping and Lorenz, Carlos Henrique Petzl and Houssameddine, Dimitri and Gui, Yongsheng and Hu, Can-Ming and Wu, Ke},
journal={Microwave theory and techniques, IEEE transactions on},
volume={62},
number={4},
pages={965--976},
year={2014},
publisher={IEEE}
}

@article{hemour2014radio,
title={Radio-frequency rectifier for electromagnetic energy harvesting:    Development path and future outlook},
author={Hemour, Simon and Wu, Ke},
journal={Proceedings of the IEEE},
volume={102},
number={11},
pages={1667--1691},
year={2014},
publisher={IEEE}
}

I could do all of the derivations for you, but just read those two documents as they are available.

Wow, we can actually harvest power measurable in milliwatts! — Tim Spriggs, Apr 18 '16 at 20:55

score 1 · Answer 5 · edited Jul 21 '22 at 16:50

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There are research groups doing this: IEEE Paper There are several more but I'm tired of googling. I remember one article\group that was capturing wireless energy to power a microcontroller. Some of the MSP430's from TI can run with very low powers (uAmps) TI MSP430 App Note and use either a very small solar cell or antenna to run.

edited Jul 21 '22 at 16:50

Glorfindel

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answered Jan 28 '16 at 17:14

Voltage Spike

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score 1 · Answer 6 · answered Apr 14 '16 at 13:07

I was asked several times similar questions. It looks like absolutely no value idea.

All above comments are true. I would add that most of RF radiation our technology produces is either absorbed by earth, walls, human bodies etc or leaves to Space. In the first case, it causes marginal heating of some objects. In the second case, it is lost forever.

The only exception is RF systems designed specifically for power transfer or keeping this functionality in mind. However, they can not be used for broadcasting or communications.

You can see a clear contradiction: broadcasting assumes that RF energy flows all directions freely. "Harvesting RF energy" assumes that some system collects all this energy, so the latest is lost for broadcasting (or communications, WiFi connections etc).

RF Energy Harvesting

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