Overunity.com Archives

Hi.

My latest test results :

INPUT VOLTAGE : 18.03 VDC

INPUT CURRENT : 0.15 A

INPUT WATTS  = V*I = 18.03*0.15 = 2.7045 W


LOAD : 220 Ohms


OUTPUT VOLTAGE : 29 VDC

OUTPUT CURRENT : V/R = 29/220 = 0.1318181818181818 A

OUTPUT WATTS = V*I = 29*0.1318181818181818 = 3.822727272727273 W


APPARENT C.O.P. = 3.822727272727273/2.7045 = 1.413469133935024

I have a call-out to attend to, when i get back this evening i will try to self-run.


Thanks,

Gary.

What device? Have photos?

Sounds Good DeepCut.

Can you please post some photos and a diagram?

Thanks.

.

Back from work now.

Got to eat then stick the DPDT on and go for self-running.

There's only a 27 ohm difference between the load that's on now and the load that equals the input so i'm hopeful.

It's freezing here in London and once again we are seeing all the old-age pensioners sitting in their homes wearing loads of clothes and gloves because they can't afford extra heating on the government pension.

Makes me sick.


Later,

Gary.

OK, self-running blew the transistor :(

I'm using a DPDT switch for the self-run test and i added a cap which increased output voltage to around 330 VDC.

Does anyone know if there is a 'standard' and safe way to connect for a self-run test ?


Thanks,

Gary.

http://www.radio-electronics.com/info/circuits/diode_current_limiter/power_supply_current_limiter.php

Thanks broli i'll try that tomorrow.


Gary.

Hi Gary,

I fully agree with broli to use a voltage regulator against the run-away situation.
Would like to suggest an off-the-shelf regulator that includes all the neccessary circuits shown in the link, it is produced by most big analog IC manufacturers. It is known as the three terminal 7818 linear voltage regulator, like LM7818, MC7818, uA7818 etc. Gives out exactly 18V at any current up to maximum 1A, seems fully cover your input requirement of DC 18V and .15A current.
Here is a link for a data sheet at random, out of many:
http://www.datasheetarchive.com/MC7818-datasheet.html 

One thing to consider:  linear voltage regulators waste power because of the voltage difference between their input and output. Just think about it: they need a minimum of 2.5V higher voltage than the output 18V to be able to work correctly, they are designed like that. (there are special low drop out regulators too that need only .2 or .3V voltage higher than the output but now you have 29V-19V=10V difference (when you load the output with 220 Ohm resistor) and that is what counts now. So the regulator will have a 10V times .15A=1.5Watts loss in heat form inside it, use a small meatal heat sink to defend it from overheating. You will have to consider this loss of course when estimating total power, later if looping is a success, we can discuss how to reduce this loss.
The 7818 regulator IC (any make) can work with a maximum of 35V so your 29V is perfect input voltage for it.
I suggest using still at least a 360 or 420 Ohm loading resistor across your 29V output and then connect the input of the 7818 regulator IC and only then close the loop  i.e. to connect the 7818 output to the input of your circuit, to replace the power supply.  Use a diode 1N4001 or similar in series with the positive output of your power supply voltage to separate  the 7818 output from the power supply voltage. This way the two outputs cannot get connected in parallel, the diode blocks the supply to load the output of the 7818. Hope this all is understandable.

Gyula

@DeepCut,

I would also like to know how you achieved this. Obviously it is some part of a research you've discussed elsewhere which I've missed. Could you please elaborate?

Gyula, thankyou.

I will grab this one tomorrow :

http://www.datasheetdir.com/TS7818CZ+download

The specs seem appropriate but please correct me if i'm wrong.

I'll set that all up tomorrow and then post results.


Thanks, Gary.

Quote from: DeepCut on November 30, 2010, 08:04:49 AM
Hi.

My latest test results :

INPUT VOLTAGE : 18.03 VDC

INPUT CURRENT : 0.15 A

INPUT WATTS  = V*I = 18.03*0.15 = 2.7045 W


LOAD : 220 Ohms


OUTPUT VOLTAGE : 29 VDC

OUTPUT CURRENT : V/R = 29/220 = 0.1318181818181818 A

OUTPUT WATTS = V*I = 29*0.1318181818181818 = 3.822727272727273 W


APPARENT C.O.P. = 3.822727272727273/2.7045 = 1.413469133935024

I have a call-out to attend to, when i get back this evening i will try to self-run.


Thanks,

Gary.

Before we go much further, let me please point out to you that when you say:

QuoteAPPARENT C.O.P. = 3.822727272727273/2.7045 = 1.413469133935024

You are saying it is NOT 1.413469133935025 and NOT 1.413469133935023...that is, you are giving ridiculous precision that you in no way can support with actual measurements.

Please let me remind everyone that your final answer CANNOT POSSIBLY be more precise than your LEAST PRECISE measurement.

That is, in the above experiment, DeepCut, you have measured your input voltage and current to 2 decimal places. This means your intermediate results and your final answers cannot be more than 2 decimal places in precision --- even assuming that your input data is correct in the first place. (Have you calibrated your meters?)

Please, quoting precision to all the digits that your calculator spits out only assures you of 2 things: 1) that you are wrong (that is, the true value is certainly NOT 1.413469133935024), and 2) people like me, who do this stuff for a living, pay less attention to you because we know your numbers aren't credible.

Now, if you like, we can address the issue of your apparent COP of ABOUT 1.41. First, though, perhaps you would like to check up on the difficulties associated with power measurements in oscillating circuits, particularly using consumer-grade DMMs.

For an example, I have a device here that draws 5 to 8 amps at 120VAC from the wall, and produces over 250 kV with a continuous plasma arc that can carry well over that 9 amps. Using your method of simple calculation of COP, what do I get? Does it mean I've made free energy?
Maybe...
http://www.youtube.com/watch?v=PdOhIoA1Z-k&feature=related

@Omnibus

Hi Omni, you haven't missed it, just forgotten a previous discussion we had :)

Here's  a process-list :

1. 18 VDC PSU powering a Bedini-circuit.
2. Bedini circuit linked to the usual bifilar-wound 'drive' coil.
3. A small iron rod inside the drive coil.
4. A diametrically-magnetised cylinder magnet mounted horizontally above the drive coil on a carbon-rod axle.
5. An output coil consisting of 1.15 kilometres of 0.25mm copper magnet wire, this coil is on a separate former that is larger than the drive coil's former and so can be placed around the drive coil.

I suppose it's a pulse-driven induction generator, here are some pictures, please excuse the mess :

http://qvision.pwp.blueyonder.co.uk/pic1.JPG

http://qvision.pwp.blueyonder.co.uk/pic2.JPG

http://qvision.pwp.blueyonder.co.uk/pic3.JPG

My new analogue meter should arrive tomorrow or Thursday then i can measure the current directly also.


Thanks,

Gary.

Hi Tinsel.

Yes, please excuse my ignorance and shoddy methods, i am learning all the time.

Where i have used only a couple of decimal places is because there was constant variance (!) and i took an average.

*EDIT* and my PSU only goes up to 2 decimal places. *EDIT**

I hope to get a scope next year and to grow into a more mature experimenter.


Thanks,

Gary.

Keep up the work, Gary, and don't worry about it. And especially don't let it get you down.
There's a lot to learn, electronics wise, and I'm just a chapter or two ahead of you. I suffer from a traditional education, though, and I even see the merits of a lot of it, so if I get a bit pedantic about things like significant digits and power measurements...well, just consider the source, take what you need and leave the rest.

I'd like to see the circuit you are working with, and I know others would too.

cheers--
--TK

(Often, believe it or not, a good ANALOG meter like the legendary Simpson can give you a better reading (that is, a more honest and stable average) than a typical DMM on spiky oscillating signals. If you can find one surplus, it's a great piece of kit for the FE experimenter. Of course, a fast oscilloscope is even better, but more costly and has a steeper learning curve...)

EDIT to add: OK, I see the description. Thanks ! Good luck.

Thanks TK.

I think you're probably a few kilos of volumes, rather than chapters, ahead of me !

I suffer from a lack of education, sometimes that has it's merits ;+}

The circuit is the standard SSG circuit, in the black box in the pictures i posted a couple of posts above this one.

I ordered this today :

http://www.rmcybernetics.com/files/pdf/PWM-OCXI.pdf

So i'll probably be offending your electronic sensibilities with some ridiculous coil-pulsing-related posts in the very near future.


Thanks,

Gary.

Hi Gary,

Keep up the good work.  My favorite part of your setup... "An output coil consisting of 1.15 kilometres of 0.25mm copper magnet wire, this coil is on a separate former that is larger than the drive coil's former and so can be placed around the drive coil."  Hehehe  Good to see someone is listening and actually grasps some of the concepts I have learned and spewed out from SM.  He said one would be suprised how much energy can be extracted from a magnet, based on the very principle you have used. 

Cheers,

Bruce

Hi Bruce, long time no see :)

I'm following Gyulasun's advice and will get the voltage regulator etc tomorrow.

Will post results.

*EDIT* Who's SM ? *EDIT*


Thanks,

Gary.

@Gyula

Hi. I can't get hold of a 7818 tomorrow, i think this one will do please tell me if i am correct :

http://www.maplin.co.uk/media/pdfs/Module%208067-2.pdf


Thanks,

Gary.

The LM317 is a classic (adjustable) voltage regulator that should be a good replacement for the 7818 (18v fixed voltage output). There are a few things to keep in mind. First, the maximum output voltage you can expect, will always be about 1.5 volts less than your input voltage, so if you want 18 volts out, you will need to feed the regulator with at least 19.5 volts, and ideally, a bit more. Secondly, the 317 is an adjustable regulator, so you will need to add a couple of resistors to the circuit to set the voltage. I usually use a 240 ohm resistor for R1 and an adjustable 5K (or 4.7k) trimpot for R2. If you don't have a trimpot, use a 3.3k resistor for R2, which should get you reasonably close, at 18.4 volt output.

http://www.national.com/ds/LM/LM117.pdf

Hi Gary,

Yes, this IC (http://www.maplin.co.uk/media/pdfs/Module%208067-2.pdf )  is also fine to use, another  make of what derricka just suggested.

NOW that I have seen your setup, please do a favour for yourself: check the current consumption of the 0.15A from your 18V power supply WHEN your 220 Ohm load resistor is connected to the output (unless you have done so already).

There is one more problem: as I tried to estimate the power loss in the 7818 voltage regulator (it will be the same for the TS317 too) it comes out too much loss and it eats up your COP to under the value of 1 so looping would not work....   Think it over:
your input power is 18V at 0.15A=2.7W (assuming you measured this when the 220 Ohm is actually across the output)
power loss in the 7818: 10V at 0.15A=1.5W
output power in the 220 Ohm as you calculated: 3.82W

Now 3.82-1.5=2.32W and already this is under the needed 2.7W input power.   SO one solution is NOT to use the 220 Ohm when you are looping, not to use even the 350 or 400 Ohm what I suggested last night, then the dissipation might just be enough for a COP of just over 1....

Gyula


EDIT: possibly your unloaded output voltage is higher than 40V which is the limit for the TS317 or LM317 max input alloweable so make sure not to exceed this, otherwise you burn the regulator IC....  use preloading just under 40V output first , then connect the regulator IC and make the looping for a moment and check always the voltage level across the regulator input, assuming you first built already it to give out 18V.

Thanks Derricka and Gyulasun, i will bear all your points in mind.


Gary.

Can't get what i need locally, have to mail-order :(


Gary.