Can you harvest electrical energy from the air?

Question

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:

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

Answer 1

Answer 2

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

Answer 3

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
.control
delete all
tran 600u 60
plot V(5,6)
.endc
.END

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

Vin: 1Vpp 34kHz signal, no load:

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Vin: 1Vpp 34kHz signal, 10kΩ load:

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

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

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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)

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And with node 4 grounded:

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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.

Answer 4

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...

Answer 5

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?

Answer 6

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.

Answer 7

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.