Earth Electrical Energy Datalogging Experiments

[gyulasun]

Hi Bill and All,

Normally a long piece of wire erected relatively high in the air and held insulated in its position cannot really produce a DC voltage wrt the ground because I believe any voltage that is able to develop in it is by induction, hence it must be AC. 
I suggest Freepow to rectify the AC, by using a diode bridge, preferable made from 4 germanium diodes to minimize the forward voltage drop. The AC input of the bridge connects to the antenna end wire and the ground wire and DC output of the bridge should go to a puffer (electrolytic) capacitor and to this capacitor a load like a JT may be connected.
If you do not have germanium diodes like 1N34A 1N270 etc but happen to have germanium transistors in your junkbox, you may also use them as a diode.  If you have only silicon diodes at hand (1N914, 1N4148 etc) it is also good of course but the forward voltage drop (2 times 0.7V in the bridge) will reduce the useful DC output.  Schottky diodes are also a better choice than the Si ones in this respect.

rgds,  Gyula

[freepow]

  Hi all, Jeanna what do you mean when you say...My multimeter shows 0.6vdc 185mA and my scope shows 85mv ac

How do you get 185mA
Can you explain what you do to get this or a drawing !

Oh, yes my multimeter reads DC-volts with no problem, I get DC-volts out of my antenna system when it goes through the joule thief!

[gyulasun]

one more thing: such a setup behaves like a current source or current generator,  well more or less. This means that its instantenous output current and hence voltage always depends on the real value of the load you use.
The biggest current it is capable of can be measured by short-circuiting the ground and antenna end wires by an AC ampermeter if you could have a very wideband AC ampermeter from some ten Hz frequency to several ten or hundred MHz frequency,  such an instrument is very rare and expensive.  A solution is the wideband diode bridge and shorting its DC output by a DC ampermeter. The current indicated this way is the maximum that this antenna wire will be able to supply and its output voltage under this measurement is the 2 forward voltage drop out of the 4 diodes in the bridge.
Suppose you have germanium diode bridge, and the voltage drop is two times 0.12V on two of its diodes. and you measure 10mA DC current on your current meter via the bridge output, then the useful power available from this antenna is 2*0.12V*10mA= 2.4mW (milliWatt).
This way you can estimate the max output power you can use from this antenna setup, used as an energy source. 

You may speculate on what is the real source that feeds this long piece of wire?  My personal opinion is that it comes by induction from the man-made energy sources like mains power lines, radio and televison broadcast stations, military and police transmitters etc etc.

rgds, Gyula

[freepow]

Does anyone think its AC power lines?
                             cosmic rays    ?
                             radio waves    ?

[jeanna]

Hi all, Jeanna what do you mean when you say...My multimeter shows 0.6vdc 185mA and my scope shows 85mv ac

How do you get 185mA
Can you explain what you do to get this or a drawing !

I have an ordinary EEB but added to that on each end I have a stubblefield generator. This brought my voltage reading on the dmm down about .200-.300Vdc, but it also rose the amps to well above 150mA.
I still cannot light a led, however... still hoping though. 

I posted a drawing a few times.
It was around august 1 or so, maybe looking by date will help a little?

Oh, yes my multimeter reads DC-volts with no problem, I get DC-volts out of my antenna system when it goes through the joule thief!

Are you reading this across the C-E junction with a led in there? Even without a led in there, that is good news.


Didn't you get a scope?
I really believe that is the best way to show this.
I think gyula's suggestions are assuming you do not have one?

jeanna