Generating tiny bits of electricity from RF

Discussion in 'The Projects Forum' started by MMH, Mar 2, 2013.

Feb 8, 2013
143
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Hi guys,

I have made a Circuit that should generate electricity from Radio frequencies. However, it is not working. Is it because of the earthing? But it is not mentioned in the circuit!! The circuit is is from the one in this video:-http://hight3ch.com/free-electricity-from-thin-air/

Can you please tell me where the bug is? Or is the video fake?
My setup is this:

The thing in the bottom right corner is the antennae 50 feet long.thanks in advance for your assistance.

2. Meixner Member

Sep 26, 2011
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It looks like BS to me, maybe he lives next to a high power transmitter. But it for sure isnt going to charge a cell phone.

3. Papabravo Expert

Feb 24, 2006
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2,520
Two things might be worth mentioning
The power in an RF signal falls off as the inverse square of the distance.
Any recovery process you can imagine is going to be less than 100% efficient.

The power in an RF signal is often measured in dBm. That is dB with respect to 1 milliwatt. Now you must understand that a milliwatt is a very small amount of power. The noise floor for RF signals is between -135 dBm and -155 dBm. Signals just a few dB above the noise floor can be used to convey useful information, but the amount of power you can recover from let us say a loud signal at -115 dBm is too small to be useful for very much. Do the math and don't forget to account for less than 100% efficiency in the conversion process.

Feb 8, 2013
143
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Thanks for answering!! I read in grade 7 and i understood nothing you said!! So can you please explain it in a easier language? And shouldnt it at least be able to charge a 1mF cap? My circuit generates 0.000mV. Where is the bug?

5. Papabravo Expert

Feb 24, 2006
11,859
2,520
If your level of mathematical accomplishment is also at grade 7 we're going to have a problem. In any case a milliwatt is a thousandth of a watt. When we express something in dB we use a logarithmic scale instead of a linear scale. Logarithms are taught in Algebra II which is 11th grade material. I'm going to do the Math and we'll convert dBm to a linear scale.

I said the noise floor is at -135 dBm, so
Code ( (Unknown Language)):
1.
2. -135 = 10 * log (P/1 mW)
3. -13.5 = log(P / .001)
4. 31.6 x 10^-15 = P / .001
5. 31.6 x 10 ^-18 = P
6.
So the noise floor represents a power of 31.6 attowatts. An attowatt is a billionth of billionth of a watt. Just for the sake of argument ask yourself what voltage and current might produce that power level. One answer is to take the square root which suggests that a current of 5.6 nanoamperes at a voltage of 5.6 nanovolts would produce a power of -- 31.6 attowatts.

Alright. So a loud radio signal will be 20 dB above that at -115 dBm , so how much power do we have to work with in a loud radio signal?
Code ( (Unknown Language)):
1.
2. -115 = 10 * log (P /.001)
3. -11.5 = log (P / .001)
4. 3.16 x 10^-12 = P / .001
5. 3.16 x 10^-15 = P
6.
This power level of 3.15 femtowatts corresponds to a voltage of 56 nanovolts at 56 nanoamps. As you can see these are still very small quantities which would be difficult to measure without sophisticated laboratory equipment. You certainly cannot measure it with a \$9.99 voltmeter from Harbor Freight.

Let us now assume that you could somehow get 1 volt out of that loud radio signal at a current of 3.16 femtoamps. How long will it take to charge a 1 microfarad capacitor to 0.98V at a current of 3.16 femtoamps. It is after 3AM and I'm sleepy so I'll leave it as an exercise for the reader. HINT: I think it is just shy of 10 years.

Last edited: Mar 3, 2013
6. DickCappels Moderator

Aug 21, 2008
5,126
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I could not see any obvious errors in your breadboard, but you might want to go over it again.

It is a hoax based on a workable idea.

Yes, you will need a pretty good antenna and a pretty good ground. The exact nature of the antenna depends on the kind of signal you are going to use. The 50' length of wire sounds like a good start.

Your ground should connect to the place where C3 and C4 connect to each other (and two diodes). The circuit will not do much without this ground, which was not shown in the video.

The idea that this would be able to supply enough power to charge a cell phone...well, it makes me smile

7. ErnieM AAC Fanatic!

Apr 24, 2011
7,969
1,827
If you can get enough energy from an antenna to charge a cell phone...

then why are their batteries inside cell phones? Why not just run them off the antenna?

MMH likes this.
8. MrChips Moderator

Oct 2, 2009
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My crystal radio set runs off the energy from the antenna.

9. gootee Senior Member

Apr 24, 2007
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Papa Bravo: Assuming your math is correct, say there are 100 such signals. That brings it down to about 0.1 year. If we could hope that you were off by a factor of ten in the right direction, somewhere , then it would almost look practical, in a very-dense-signal area.

If there is some strong source that is very nearby (power lines?), or if the energy could be collected and somehow focused to concentrate it, and an efficient antenna were then used at the focal point, then maybe the story would end differently, although I can't see how it would ever be viable, ECONOMICALLY, in most cases (except for the case where energy is stolen from power lines, which obviously should not be done, for moral, ethical, and legal reasons). But it could still be a fun project.

MMH: Considering that your reading level is only grade 7 (is that your actual grade level?), I commend you on your interest, and your development so far. Math will be a hindrance, until you learn it and your math skills catch up with your interests. I remember having exactly that problem, when I was very young. Algebra, geometry, calculus, and differential equations aren't really "difficult" to learn, if learned incrementally, and are usually much easier to learn when you are working on something for which you need them. But it does take TIME. (But after that, you can figure out "anything"!)

The field known as "energy harvesting" is alive and well. But its uses, and the ways it is implemented (built), are still rather specialized. You might want to also look into harvesting "free" energy from OTHER types of sources, besides the RF in the air.

Nov 30, 2010
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File size:
231.7 KB
Views:
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11. gootee Senior Member

Apr 24, 2007
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Thanks for posting that PDF file. That's a pretty nice device and I hadn't noticed it before.

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12. Papabravo Expert

Feb 24, 2006
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So how many orders of magnitude are there between 20 millivolts and 56 nanovolts?

13. gootee Senior Member

Apr 24, 2007
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I'd say about 5.5, if we consider an order of magnitude to be a factor of ten.

But I was only looking at your "10 years" figure and dividing it by 100.

14. takao21203 AAC Fanatic!

Apr 28, 2012
3,682
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You can actually charge small capacitors if you are close enough to a transmitting station. Obviously the circuit must be tuned.

Stellar background radiation is not going to power your heating and microwave.

15. thatoneguy AAC Fanatic!

Feb 19, 2009
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20mV to 56nV is similar to 26 meters to 56 micrometers, if thinking in distance helps.

16. gootee Senior Member

Apr 24, 2007
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What, you didn't like my answer? 10^6 is 6 orders of magnitude, isn't it?

17. thatoneguy AAC Fanatic!

Feb 19, 2009
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Doesn't have quite the same impact for some people. Pick something people can visualize, then reduce/expand it so they get an idea of relative size.

Some people think an order of magnitude is double, others think it's the first mantissa digit in exponential notation, etc.

I was visualizing your accurate statement.

Feb 8, 2013
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I now understnd that it will take a LONG time to charge a 1mF cap; I have to ground it and charge a 47-100uF cap; and many other things. I will like to repeat once more that I read in grade 7 and I understood everything partially. Anyways, thanks for answering.

Feb 8, 2013
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20. charliewilde New Member

Mar 13, 2013
13
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It seems that your PDF file has a light relevance to the problem. But it merely gives a good point and ideas.

#12 likes this.