Hall Effect Energy Generation

[Jetijs]

Hi Elias
Can you elaborate more on your test setup? Was it just a copper foil where current was flowing from right to left and the output came form the up and down side of the foil? And the magnetic field was applied perpendicular to the larger surface area of that foil?

I tested this before posting and yes I was able to produce around 0.1 volts @ 50mA, by using low impedance coupling of the Neo magnet and a copper foil. But this would not produce good amount of current as the impedance prevents it. (the current draw was about 7 Amps, and the voltage of the foil was about 0.2 mV), this system was an over-unity system already:
Input energy: 0.2 mV * 7A = 1.4 mW
Output energy: 0.1 V * 50mA = 5 mW

To get real results, semiconductors are needed. This would make us more voltage, thus increasing the COP, and making us produce usable power.

I am looking for buying an n-type semiconductor bar to test on it. Anyone can offer any suggestions for obtaining one?

This is really important people!, hall effect means creating a dipole by using the "free magnetic force", thus creating a free battery, by paying the price of some current at a very low voltage. (negligible power input, milliwatts)

[broli]

I have to say I'm quite astonished by your numbers:

I tested this before posting and yes I was able to produce around 0.1 volts @ 50mA, by using low impedance coupling of the Neo magnet and a copper foil. But this would not produce good amount of current as the impedance prevents it. (the current draw was about 7 Amps, and the voltage of the foil was about 0.2 mV), this system was an over-unity system already:
Input energy: 0.2 mV * 7A = 1.4 mW
Output energy: 0.1 V * 50mA = 5 mW

I today created the fold setup using aluminum foil which has a thickness of 0.015mm. I used around 26 folds. The folds were squares of 1cmx1cm. Input was 1V and 5A giving a resistance of 0.2ohm. The total stack is about 1cm high, so the field in the middle isn't very strong. One 1cmx1cm cylindrical neo mag on both sides.

The output voltage was around 50mv so never mind the generated current. I have also tried with a single alu foil strip and I couldn't measure any voltage.

I don't understand these results compared to yours, as aluminum is 60% as conductive as copper. Are you sure all your meters are functioning correctly? Containing non empty batteries...

And what are the dimensions of your magnets and copper strip?

[sebosfato]

I tested this before posting and yes I was able to produce around 0.1 volts @ 50mA, by using low impedance coupling of the Neo magnet and a copper foil. But this would not produce good amount of current as the impedance prevents it. (the current draw was about 7 Amps, and the voltage of the foil was about 0.2 mV), this system was an over-unity system already:
Input energy: 0.2 mV * 7A = 1.4 mW
Output energy: 0.1 V * 50mA = 5 mW


This is really important people!, hall effect means creating a dipole by using the "free magnetic force", thus creating a free battery, by paying the price of some current at a very low voltage. (negligible power input, milliwatts)

Hello elias,

I didn't read that before.

Man would you make a photo of your setup, and or give us the exact schematic of your test? Like copper thickness, length, width, the magnets size.. .

I'm interested in trying this! Within a couple of weeks i will have some more time.

I thought of making the thing in the middle of a very high current resonant tank, maybe 1000 amps so the voltage and current are out phased but than i want to combine the generated electricity into the tank as a feedback as to be able to transform the electricity generated into standard voltages for home equipments while making the resonance self sustaining...

Thanks again.

[broli]

The quality came out bad, but the below animation should clarefiy the stack concept:



Just standard metallic tape. You must have an insulated side, that's why copper or aluminum adhesive tape saves some gluing work.

You'll also notice blue sides on the copper side. This is an additional strip of regular tape you apply (around 2mm-5mm wide). The reason for this is when you compress the stack those sides will stick out and even touch, thus short out the stack. This is what happened to me, so before folding don't forget to apply a thin layer of tape on one side. Mine were sticking out 1mm, so the above numbers keep you safe.

[electr0n]

I tried with a 50mm x 30mm x 0.1mm (L,W,D) copper foil, magnets insulated via .6mm teflon
power supply was a kepco lab supply, the maximum current i could draw was around 2v 5A, various neodium magnets used
leads to the foil were 5mm solid copper wire
best reading attained was around 6mV off the sides of the foil.
maybe a few copper foil/magnet arrangments in parallel would yield a higher current at lower voltage...
I tried routing the negative to the positive in the power supply via 5mm wire and increased the current till 5A, the voltage was 2.1v.

[broli]

I discovered that in sputtering, semiconductors are used as targets. So these targets can be bought in relatively bulky dimensions. You can find them on Alibaba.com by searching for "sillicon target". Here's an example:

Single Crystal Silicon Target products, buy Single Crystal Silicon Target products from alibaba.com

I requested an inquiry and the price was not that high. The sample was 50mm diameter 3mm thick and an N type semiconductor. The price breakdown was:

sample: 30$
shipping: 60$
TT bank cost: 30$
---------------------
Total:120$

The problem is shipping cost, the same cost goes for 1 to 10 pieces. So it's cheaper to buy by the ten, but that's a little too much. I'll keep looking.

[elias]

I discovered that in sputtering, semiconductors are used as targets. So these targets can be bought in relatively bulky dimensions. You can find them on Alibaba.com by searching for "sillicon target". Here's an example:

Single Crystal Silicon Target products, buy Single Crystal Silicon Target products from alibaba.com

I requested an inquiry and the price was not that high. The sample was 50mm diameter 3mm thick and an N type semiconductor. The price breakdown was:

sample: 30$
shipping: 60$
TT bank cost: 30$
---------------------
Total:120$

The problem is shipping cost, the same cost goes for 1 to 10 pieces. So it's cheaper to buy by the ten, but that's a little too much. I'll keep looking.

Copper seems to be "difficult", Note that lower the impedance, it will be a better source of energy. It seems that the impedance of the mentioned semiconductor is 0.2 ohms/cm. it means that with 5 amps @ 1V for each cm^2. We might be able to generate about 10 volts, and it would provide us with 50 Amps at most, but to avoid heating up we need to limit the current draw.

I tried chopping up my old transistors, to use the silicon, but it was too small for experimentation.

[broli]

I agree. I have done a lot of research in the couple of days on this. It all boils down to electron mobility and resistivity. In fact it's too naive to even just buy a random doped semiconductor. A very crude formula is this one:



That equation shows that we want mu and rho to be high, but h to be low. But we can't go too high with rho either as that would mean we would need to apply a crazy big voltage for say 1A of current. So we must look for a good balance.

The reason why copper is bad lies in its resistivity. The electron mobility in copper is slightly lower than one order of magnitude (46x lower to be exact) than that of n-doped silicon. However the resistivity of copper is 1000000000000000000 or 10^18 lower than that of a reasonable doped n-type silicon. This is especially bad if the current is kept constant. Even if it didn't there's no way we could provide that much current to compensate for this low value.

I found the best balance to be a Silicon n doped crystal which has a resistivity of 1ohm cm (=0.01ohm m). This way voltage you apply doesn't have to be too large to have a significant current, and it's electron mobility is still in the same order of magnitude as pure undoped silicon (which has the highest electron mobility).


Here's a source for doping amount and the resulting parameters of the silicon:
http://www.virginiasemi.com/pdf/Resi...tinSilicon.pdf

That's also where I got that 1ohm cm from.

Edit: Oh and I forgot to ask you again about your numbers with copper. If you used a 10micron thick foil,1 Tesla field you would still need a current of about 1984.12A to generate a voltage of 0.1V. With your 7A the calculated estimate is 0.35mV.

The strange thing is that you said that the value would be a fractional mV but your data says otherwise.

[elias]

Good research on this stuff ...
The experiment I did, was using some soldering on a copper foil, and contacting it slightly with its soldered side (some ohmic contact) with a Neo Magnet, and passing current through the magnet. When I did this I noticed a voltage build up of upto 100mV (with abtout 7 amps of current) and when I measured the current, I found out that about up to 50mA could be drawn in that config.

Let me see if I can take some photos to demonstrate this ...

[broli]

Good research on this stuff ...
The experiment I did, was using some soldering on a copper foil, and contacting it slightly with its soldered side (some ohmic contact) with a Neo Magnet, and passing current through the magnet. When I did this I noticed a voltage build up of upto 100mV (with abtout 7 amps of current) and when I measured the current, I found out that about up to 50mA could be drawn in that config.

Let me see if I can take some photos to demonstrate this ...

Is that 50mA a short circuit current? Or did it go through a load? If so can you share its resistance.

[elias]

Is that 50mA a short circuit current? Or did it go through a load? If so can you share its resistance.

It was the short circuit current, but I am starting to doubt that it was from the hall effect. it might have been cross current from the source, let me recreate the experiment and I'll post the new results ...

[elias]

It was a measurement error, due to the impedance of the contact, the measured voltage was the voltage drop of the contact between the metal and the magnet, sorry ... . The real Hall voltage was about 0.6 mVolts.

[elias]

I developed an idea to build a negative resistor using the hall effect.
suppose we have a thin foil of conductor, longer the better, we need to turn it into a sin-wave like structure, by folding the copper in Brolli's diagram above then cutting it off and turning it into a sinwave structure, almost like the DNA, but the width of the strip must be maintained. Then we may put this folded sinwave structure in a magnetic field. using this method as current passes through the strip provided that the polarity of the magnetic field is right, a voltage will develop on the folded sin-wave like strip which will be of the same polarity as the applied voltage thus making a negative resistor.
In order to get a good amount of voltage, semi-conductors are a must. (Low impedance Semiconductors)

[gyula]

Hi Elias an All,

I have done some search on manufactured Hall sensors that are also called as Hall generators. For instance this firm, F.W. Bell produces some type, see this link: fw bell products

At newark.com there is some of the Bell products but with a min 26 days leading time (they do not stock it):
More Test & Laboratory Equipment | Newark.com

Let's examine the most sensitive type, the SH-410 which is US$20.39 at Newark (with 26 days Lead Time).

Here is the data sheet:
http://www.fwbell.com/PDF%20Documents/SHseries.pdf

Suppose we get such device with the following parameters:

Rin=400 Ohm (ranges between 240-550 Ohms)
Rout=400 Ohm (ranges between also 240-550 Ohms)
Sensitivity ranges between 292 - 1120 mV/kGauss, let's pick a 300mV/kG
Nominal input current In=5mA

Now let us use a magnetic field of 1Tesla (10kG), this is an assumption from me that this device is able to operate at 1Tesla, data sheet does not give this data for this device), this gives 300mV*10=3V unterminated output voltage.

Using 5mA DC input current from a current source, input power is

Pin=0.005*0.005*400=0.01W=10mW

Because we want the highest output power, the output must be terminated with the same value of resistor as the output resistance itself, so I assume a linear relationship and the output voltage will be halved: Vout=3/2=1.5V DC
So the output power is

Pout=(1.5*1.5)/400=0.005625W=5.62mW This gives a COP=5.62/10=0.562

If you receive an SH-410 which happens to have the specified 1120mV/kG sensitivity (from data sheet), then the unterminated output voltage at the 5mA input current and at B=1T comes out as 11.2V, half of which is 5.6V.
So assuming the same input and output resistance, the output power comes as:
Pout=(5.6*5.6)/400=0.0784W

This gives a COP=0.0784/0.01=7.84 sounds quite a nice COP number!

Notice that half of the output power is dissipated in the device itself i.e. in the latter case 0.0784W heats the Hall device and 0.0784W is dissipated in the 400 Ohm terminating resistor.

I have not considered pulsed input current operation that gives pulsed AC output voltage.

I hope I did not make errors in the understanding of the data sheet and in calculations. IF you notice errors, please correct it. This Hall sensor so far seems an off the shelf device that has a COP possibility higher than 1.

Remark: In case the input resistance changes to a lower value when you connect a load resistor across the output, then input power should be revised again (though using a true current generator operation, this may be minimised?).

rgds, Gyula

[elias]

This is a great find Gyula.

I read the datasheet. Because the input and output impedance is the same, and the voltage output of the hall device is dependent on the input current, then the only thing that matters is the magnetic field strength, after a certain magnetic field strength the system becomes COP>1.

For example considering SH-410, in a scenario we would provide the device with 20mA @ 10V (assuming 500ohms of input and output resistance) Pin = 200 mW.
Note that at maximum input current we have maximum sensitivity. So by considering a sensitivity of 1760 mV/kG, for 1 Tesla we have got 17.6 Volts at the output, so Pout = 17.6*17.6 / 500 = 620 mW, so COP = Pout / Pin = 3.1

But, we need to make usable power, so if we use a 500 ohm load at the output, we can produce about 154mW of usable power. The beauty is that we can increase the COP, by simply using a more powerful magnetic field.
if we double the magnetic field strength the usable power at the output would become 620 mW, thus yielding to a real COP of 3.1.

So by using a more powerful magnetic field and a more sensitive device, we have can increase the COP of the system.

This is really remarkable, somebody really needs to experiment with these devices.

Elias