I found a website recently about harvesting energy from the air, and I am wondering if someone could tell me why the following wouldn't work?

Their "generator" is supposed to make electricity from the ionosphere (not UV, X-Ray, etc). Some claim it can knock out your electric bill totally (a bigger version, not this example). From what I understand, this is possible (it was discovered by Nikola Tesla), however this diagram doesn't look like it would work. Any help would be appreciated.

An example from the website:

You need:

  • (4) 1N34 germanium diodes
  • (2) 100 µF 50 V electrolytic capacitors
  • 0.2 µF 50 V ceramic capacitors

Here is the electrical diagram they provide:

Diagram of generator

And they claimed to power a cell phone with it. I am not sure what kind of antenna to use.

While this seems like a sham, I just found this website recently and am looking for more info on the physics behind it.

Perhaps Teslo's patent explains things better, so here it is: Patent 685958.pdf

Found something: Here is a page that explains it. Nikola Tesla free energy: unraveling Greatest Secret

  • 26
    \$\begingroup\$ "PDF for sale" should tell you all you need to know about this - its a scam. "Threats from energy companies" is also a classic telltale sign of prople who are either fraudsters or self-delusional. \$\endgroup\$ – mikeselectricstuff Dec 27 '10 at 16:40
  • 3
    \$\begingroup\$ Virtually identical post as here: electronics.stackexchange.com/questions/4639/… This question may still be useful as a method to refute the bogus claims on the linked website, however. \$\endgroup\$ – Nick T Dec 27 '10 at 16:42
  • 5
    \$\begingroup\$ @Arlen: In most countries around the world, all works are copyrighted by the author by default unless mentioned otherwise. In the US, the relevant law would be the Berne Convention Implementation Act of 1988, I believe. \$\endgroup\$ – Nick T Dec 27 '10 at 17:55
  • 10
    \$\begingroup\$ I watched the ad video and I'm appalled. This is the biggest load of crap and lies I've seen since "The Secret". Watch it at your own risk. You could write pages about what's wrong with it, which is more or less everything. It should be an incentive for more scientific education in high school, so that more people realize why it's a load of crap, and won't be gullible enough to buy this kind of rubbish. \$\endgroup\$ – stevenvh Dec 27 '10 at 18:31
  • 8
    \$\begingroup\$ Any company that's developed a device for generating useful amounts of energy from thin air would not be selling plans for it. They'd keep the design a secret, build a bunch of power plants, and sell the energy. Free energy = free money! \$\endgroup\$ – endolith Dec 29 '10 at 20:35

To address the original question of "Is Nikola Tesla's free energy discovery...", Tesla never created a "free energy device". One of his noted ideas, however, was a system to intentionally transmit power wirelessly. Power companies don't intentionally radiate energy (as it's a pure loss for them).

As an aside, Nikola Tesla was one of the first true electrical engineers, taking arcane, hard-to-understand forces and turning them into marketable solutions. While there is no doubt he was brilliant, this revolutionary engineer would quickly tell you that if you wanted to harvest naturally occurring electrical fields (not those he intentionally radiated) it would take an antenna (or an array of them) on a truly grand scale.

Regarding the document you linked:

Chapter 4 - Tesla's Radiant Energy Device

This chapter discusses a patent by Tesla which discusses using either the photoelectric effect via "ultra-violet light [...] and Roentgen rays [X-rays]" to generate a positive charge by ejecting electrons, or cathodic rays to capture electrons and generate a negative charge.

While you might be able to use the photoelectric effect from solar UV on metals, with great care, you are going to derive an extraordinary small current, certainly far less than you would get with a photovoltaic (solar) cell. PV cells use the photoelectric effect, but within a semiconductor.

Chapter 5 - The Tesla Coil

Tesla coils are essentially antennas that can radiate and receive a great deal of power. In order to actually capture an appreciable amount, much, much more must be broadcast on the particular wavelength that the coil is tuned to. Because they are tuned, they cannot capture broadband noise

  • 3
    \$\begingroup\$ Tesla did some really cool stuff, but don't get to the point of hero worship. Some of his ideas about physics were wrong. \$\endgroup\$ – endolith Jun 23 '11 at 15:56
  • 4
    \$\begingroup\$ @endolith, and he slowly went mad over his life thinking he could get energy from the ether and such. He was however able to think through his designs in his head and was know for the ability to build something and have it work the first time. Pretty cool in my opinion. Of course many people worship da vinci, calling him an engineer. What good is an engineer whom did not share his secrets and help the world? It is our job to improve the quality of living of the populace, not to be able to and hide it. \$\endgroup\$ – Kortuk Jun 25 '11 at 9:04
  • \$\begingroup\$ If the coil is large enough, would it absorb energy from the ionosphere? Tesla was into that- he was the first to measure the voltage of ionosphere... I was thinking that the Wardenclyffe Tower was potentially a power generator (the coil was for absorbing "environmental" energy). \$\endgroup\$ – Ben Welborn Jul 14 '16 at 16:05

alt text

  • 7
    \$\begingroup\$ Clever answer. :) \$\endgroup\$ – nitro2k01 Apr 18 '11 at 21:42
  • \$\begingroup\$ Cooler version: en.wikipedia.org/wiki/Vaneless_ion_wind_generator But I don't know if it would really work. \$\endgroup\$ – endolith Jun 23 '11 at 15:56
  • \$\begingroup\$ haha a coal plant in the back \$\endgroup\$ – skyler Jun 15 '13 at 1:26
  • \$\begingroup\$ @endolith: I saw pictures of such a generator built on the campus of a Dutch technical university. A few square meters in size, I don't recall how much power it generates. (BTW, the name on WP is ill-chosen; it suggests the apparatus generates wind :)) \$\endgroup\$ – radagast Nov 22 '13 at 9:05

SPICE simulation:

Bogus Tesla Generator
Vin 1 0 0 SIN(0 1 34k 1ns 1e10)
C1  1 3 0.2u
C2  1 2 0.2u
C3  4 6 100u
C4  5 4 100u
D1  3 4 Dgermanium
D2  4 2 Dgermanium
D3  6 3 Dgermanium
D4  2 5 Dgermanium
RL  5 6 10k
.model Dgermanium D IS=200p RS=84m N=2.19 TT=144n CJO=4.82p M=0.333 VJ=0.75 EG=0.67 BV=60 IBV=15u
delete all
tran 600u 60
plot V(5,6)

34kHz was chosen for transient analysis almost arbitrarily, but it is the AC analysis that tells the story.

Vin: 1Vpp 34kHz signal, no load:

transient analysis, 1Vpp 34kHz, no load
(source: tyblu.ca)

Vin: 1Vpp 34kHz signal, 10kΩ load:

transient analysis, 1Vpp 34kHz
(source: tyblu.ca)

Let's check the AC response, from 0.1Hz to 1GHz, again with a 10kΩ load:

AC analysis, 1Vpp, 10kΩ load
(source: tyblu.ca)

It's all floating, you say? Well, here's the results with node 4 grounded:

transient analysis, 1Vpp 34kHz, no load, node 4 grounded
(source: tyblu.ca)

transient analysis, 1Vpp 34kHz, node 4 grounded
(source: tyblu.ca)

AC analysis, 1Vpp, node 4 grounded
(source: tyblu.ca)

Although 1Vpp is massive to be floating around, what happens when it is large, such as standing in front of a microwave transmitter? (Other than the blocking caps blowing up.)

Vin 1 0 SIN(0 10k 34k 1ns 1e10)

transient analysis, 10kVpp 34kHz
(source: tyblu.ca)

AC analysis, 10kVpp
(source: tyblu.ca)

And with node 4 grounded:

transient analysis, 10kVpp 34kHz
(source: tyblu.ca)

AC analysis, 10kVpp 34kHz
(source: tyblu.ca)

Nothin'. Guess it doesn't work. Any corrections or suggestions are appreciated. Code is Berkeley Spice3 compatible'ish, but really isn't fit for anything, including merchantability.

  • \$\begingroup\$ Which spice program are you using on OSX? \$\endgroup\$ – akohlsmith Dec 27 '10 at 21:03
  • \$\begingroup\$ @AndrewKohlsmith, MacSpice; netlist- and command line-based. I'm not much of a SPICE guru, but it seems to work well. \$\endgroup\$ – tyblu Dec 27 '10 at 21:17
  • \$\begingroup\$ This answer is probably the one of the most helpful ones, along with the one I marked as the answer. \$\endgroup\$ – Arlen Beiler Dec 29 '10 at 14:53
  • 1
    \$\begingroup\$ @DarenW, About as much sense as any other analysis on a bogus circuit, but what do you mean? Rectifiers and doublers have frequency dependancies when modeled with non-idealities as I did. \$\endgroup\$ – tyblu Dec 30 '10 at 7:13
  • 3
    \$\begingroup\$ Look at the multiplier on the Y-axis, @davidcary ;) \$\endgroup\$ – tyblu Dec 31 '10 at 0:40

The "air" is full of electric fields. When I touch the probe of my scope with my finger I can see a 50Hz sine wave of 6Vpp. The problem is that your antenna (your finger or anything else) doesn't capture any serious amounts of energy from the fields: once you load the detected voltage it drops to almost nothing. The same with the circuit in your question. So, electricity yes, electrical energy no.

The energy stored in a capacitor is equal to 1/2 * C * V^2. This means that for the same amount of energy the voltage over a capacitor decreases as the capacity increases. So, if you want to use a 10uF capacitor instead of a 1nF the voltage will only be 1% of your starting value, and most likely too low to overcome the diode's voltage drop, even for a germanium diode. BTW, germanium is not going to help, as its voltage drop is almost half of a silicium diode. The solution is a tin diode, which has a neglectible voltage drop. You'll have to cool it to tens of degrees belows zero, however...

  • \$\begingroup\$ I'd probably say useable energy, no. More to the point, what antenna you use will largely determine what frequency you will capture, "bulk capacitive" perhaps being the most efficient for 50-60 Hz line power. \$\endgroup\$ – Nick T Dec 27 '10 at 17:06
  • \$\begingroup\$ You are a serious coupler! I only get ~1Vpp ;) \$\endgroup\$ – tyblu Dec 27 '10 at 17:07
  • \$\begingroup\$ How much energy would 50 feet of copper tubing capture? \$\endgroup\$ – Arlen Beiler Dec 27 '10 at 17:14
  • 2
    \$\begingroup\$ @Arlen: I bet you could reach up to a power-co. kV line with 50' of copper, so potentially, lots. You need to specify more details if you're talking wireless. \$\endgroup\$ – Nick T Dec 27 '10 at 18:04
  • 1
    \$\begingroup\$ Lots of metal oxides can have diode-behavior under the right conditions - often undesired ones, such as rectification in corroded antenna connections leading harmonic distortion and thus interference on other frequencies. My great uncle claims to have made a crystal radio with a lump of coal instead of the usual galena crystal. \$\endgroup\$ – Chris Stratton Dec 28 '10 at 5:00

Is it possible? Yes. But unless you are talking about 'receiving' a man-made source such as a nearby broadcast station or a power line, and even then in conjunction with a huge receiving antenna, it's impractical as a way to obtain more than a tiny amount of power. But a tiny amount can be enough for a crystal radio, or to power a microcircuit which transmits an extremely weak signal.

For most power-generation purposes, you'd get orders of magnitude more return on investment from photovoltaics or a wind turbine.

  • 1
    \$\begingroup\$ It's worth noting that radio receivers which "pulled power out of the air" used to be widely commercially available; their performance was feeble, but getting anything better would generally require using vacuum tubes that gobbled expensive batteries. I would think it might have been possible to improve performance by adding a small battery and capacitor to bias the diode, but I don't know if that was done. I also saw in a museum a radio that used a rotating disk, some ferric beads on a string, and a diaphragm to mechanically amplify sound (the friction of the beads... \$\endgroup\$ – supercat May 3 '13 at 15:31
  • \$\begingroup\$ ...on the disk would supposedly be affected by perpendicular force induced by the incoming signal; since any friction would be added to the tension of the string on the diaphragm, this could, if the coefficient of friction was greater than one, cause small changes in magnetic force to produce larger changes in force on the diaphragm. I have no idea how well such a technique could work, since it would seem hard to control unwanted resonant behavior, but it seemed an interesting concept. \$\endgroup\$ – supercat May 3 '13 at 15:34

To answer the question in your title of "Can you harvest electrical energy from the air?", yes you can.

Can you do it efficiently and actually make it a marketable product? If someone has already done it, wouldn't you think it would be all over the news and be very popular?

  • 3
    \$\begingroup\$ It's a big-corp. conspiracy! \$\endgroup\$ – tyblu Dec 27 '10 at 17:19
  • 7
    \$\begingroup\$ If it seems too good to be true, it probably is. \$\endgroup\$ – stevenvh Dec 27 '10 at 17:25
  • \$\begingroup\$ @mark: ​​​​​​​​win. \$\endgroup\$ – Nick T Dec 27 '10 at 18:02
  • 1
    \$\begingroup\$ @markrages: sorry mark, I removed your picture because it completely alters Kellenjb's answer. Please post it as a separate answer. \$\endgroup\$ – stevenvh Dec 27 '10 at 18:37
  • \$\begingroup\$ @stevenvh, I do not think it completely changes it either. Let @kellenjb decide what he thinks. He very well might go either way. Next time flag @kellenjb and ask him. With a tiny bit more edit this could go from prose to a substantial and thought provoking answer. \$\endgroup\$ – Kortuk Jun 25 '11 at 8:51

Let's suppose the circuit you show works. I don't know if it would, but it seems at least possible that you could harvest small amounts of energy from the electrical noise in the air. How small is small?

The circuit you showed has 4 capacitors-- two 100 uF caps and two that are 500x smaller. Let's ignore the small ones, as they're not used for storage, as David Cary points out in the comments. How much energy can they hold at their peak voltage of 50 V?

The energy capacity of a capacitor is 0.5 * C * V2 joules. We have 2 of them, so the total energy is just C * V2 joules. Substituting in the actual numbers, that's 100 * 10-6 * 50 * 50 = 0.250 joules. We're talking about electricity, so let's convert that to units of kWh, which is how the electric utilities measure energy. 0.250 joules is 7 * 10-8 kWh, i.e. 0.00000007 kWh. In the USA, one kWh costs around $0.10, so this is worth around $0.000000007. If I have my zeros right, that circuit (assuming it works perfectly) can store a maximum of about 7 billionths of a dollar worth of energy.

Of course, by hooking the circuit to a cellphone battery, you'd be limiting the capacitor voltage at 3 V, or whatever the battery voltage is. In this case, the capacitors don't actually serve any purpose as their storage capacity is dwarfed by that of the battery, and they also allow some reverse current leakage as well.

The bad news is that if you remove the capacitors, all you have left is some diodes. It's actually common practice to put diodes in this configuration when driving inductive loads like motors to reduce arcing when the motor stops; they're called "flyback" or "freewheeling" diodes.

Unfortunately, I can say with authority that if you leave a lead acid battery sitting in your garage with just a flyback diode connected, it will not charge itself. With lead acid batteries, they eventually undergo a process known as sulfation, which means that they stop accepting charge. In the long run, you have to load them into your trunk and take them to a household hazardous waste dropoff on a Saturday morning.

I'll stick with the electric bill, thanks.

  • \$\begingroup\$ The circuit has an input; the antenna (or coil, plate of a cap, whatever). You're interpreting it as a closed system (with some initial conditions). \$\endgroup\$ – Nick T Dec 27 '10 at 18:53
  • \$\begingroup\$ Yeah, that's true. The storage capacity is still limited as described. If the caps aren't storage, what are they? \$\endgroup\$ – pingswept Dec 27 '10 at 18:58
  • 1
    \$\begingroup\$ -1 irrelevant. The capacitors in the PSU in my computer and in a Cockcroft-Walton voltage multiplier are connected with diodes in more or less the same circuit shown by the original poster -- a voltage doubler. Those small capacitors connect the antenna -- the source of the power -- to the rest of the circuit. "Let's ignore the small [capacitors]" is like saying "let's ignore the power cable from my computer to that mysterious socket in the wall". The small capacitors in those 3 circuits aren't used for storage; they are used to pass AC current and block DC current (aka "voltage shifting"). \$\endgroup\$ – davidcary Jan 1 '11 at 23:08
  • \$\begingroup\$ I agree with you that the small caps are used to pass AC and block DC, and I've added a note about that to my answer. My assumption, which is seems you disagree with, is that the circuit is intended for energy storage, as it is described as a replacement for a battery. Do you agree or disagree with my main point, which is that the circuit can store very little energy? \$\endgroup\$ – pingswept Jan 3 '11 at 3:07
  • 1
    \$\begingroup\$ Sure, but in the WiTricity case, they're irradiating the antenna from a short distance at a tuned frequency. The crystal radio example is perfect-- that's a very similar case to what we've been discussing. I think that a crystal radio antenna captures much less than 1 W and thus would be useless for powering a cellphone. \$\endgroup\$ – pingswept Feb 25 '11 at 23:05

The Radiant Energy device is actually different from what you describe. The 'antenna' is actually a (very) large isolated metal plate raised to high altitudes (closest you can get to the ionosphere) while isolated from the ground, and tethered through a copper wire all the way down to the ground. Assuming that the ionosphere is charged positively thanks to "cosmic rays", and the ground is negatively charged, a potential difference exists of a few hundred thousand volts. But this energy is static and won't do any work as it is. For that, Tesla proposes connecting the wire to a very large capacitor that charges then discharges into some form of oscillating or switching device that could produce usable work, before sending it down to ground.

It's like a device that can harvest lightning bolts, even when the resistance of the air is not low enough to naturally allow them (such as during a storm).

Is it possible? I don't see any reason why not. Is it free energy? Yes. But then so is oil if you happen to own an oil well. The term 'Free' energy is very misleading. The problem with oil is that it is not renewable. So a better name for Tesla's Radiant Energy Harvester would be an unlimited source of 'renewable energy'.

But then again, Solar, wind, tidal and geothermal are all renewable energy sources and are very well established and reasonably cheap to build and operate. The point is we don't really need "Free Energy sources that the big oil corps are keeping from the rest of us". Those sources are already there and no one is keeping anything from anyone. What we need is more legislation and incentives to invest more into research to optimize and further develop them harvesting (and especially storage techniques).


Not the answer you're looking for? Browse other questions tagged or ask your own question.