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Author Topic: GENERATOR- YOU DO THE IN/OUT POWER MATH  (Read 60508 times)

Offline Rapadura

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Re: GENERATOR- YOU DO THE IN/OUT POWER MATH
« Reply #75 on: May 06, 2010, 04:42:59 AM »
Rectified "overdouble" output is good news...

Try to feedback the output to the input and take a video of it!

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Re: GENERATOR- YOU DO THE IN/OUT POWER MATH
« Reply #75 on: May 06, 2010, 04:42:59 AM »

Offline infringer

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Re: GENERATOR- YOU DO THE IN/OUT POWER MATH
« Reply #76 on: May 06, 2010, 05:22:04 AM »
indeed it does sound rather interesting...

Surely a video and full detailed information would be of the order weather it is a bad reading or not has yet to be determined but the conscious effort is there and you must continue to post your results.

Offline DeepCut

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Re: GENERATOR- YOU DO THE IN/OUT POWER MATH
« Reply #77 on: May 06, 2010, 12:12:18 PM »
Getting this today :

http://maplin.co.uk/Module.aspx?ModuleNo=2355

If it doesn't kick in then it's time to wind that second coil and perhaps double the output for free.

Bruce,  would another coil, wound as you suggested, induct ? I thought it had to be at 90 deg to the movement of the changing magnetic field ?

@Infringer

Lots more detail eralier on in thread, MagnetMan has his videos posted on the concept.

I am following in his footsteps and ahve a much cruder setup, he will be doing his load-testing this week i think and should have better results but i will post video once it's self-powering, if it ever is lol ;+}


Gary.

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Re: GENERATOR- YOU DO THE IN/OUT POWER MATH
« Reply #77 on: May 06, 2010, 12:12:18 PM »
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Offline gyulasun

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Re: GENERATOR- YOU DO THE IN/OUT POWER MATH
« Reply #78 on: May 06, 2010, 12:26:50 PM »
...
System components :
Drive circuit: Bedini, no JonnyDavro relay. Using 10k, 3-contact linear pot.
Drive coil : Bifilar, single coil.
               Trigger wire : 0.25mm.
               Induction wire : 0.28mm.
Magnet : Neo @ 6,100 gauss.
Full wave bridge rectifier, shop-bought, maximum reverse-current is 10A per diode.
Load resistor : 100k
Resistance of Bedini pot : ~1200
Peak performance result :
Input : 9.43 VDC @ 0.075 A = 0.70725 W
Output (rectified) : 41.5 VDC @ 0.035 A = 1.4525 W

Apparent COP = 2.053729232944503 ...

I ran the test 10 times and all results were similar, these are average figures.

*EDIT* What's an easy way of switching between the PSU and the rectified output ? Can i just connect the output to where the input goes to begin with or will that blow my house up lol ? *EDIT*

Hi DeepCut,

Here is what I would do:

I would not loop directly, FIRST I would load further down the DC voltage at the output of the diode bridge by using much less value resistors than your present 100 kOhm. Try using as low as anything between 100 Ohm to 1 kOhm and see that how this increased load affects the output DC voltage and the current consumption from the input (9.43 VDC @ 0.075 A).

IF and WHEN you have an output voltage across the diode bridge output higher than 10 Volt MEASURED when you use a load resistor of  say 100 Ohm, Than this means your load current is 10V/100 Ohm= 0.1 Amper, hence you have overunity PROVIDED your input conditions 9.43 VDC @ 0.075 A did not change.   

By the way I do not get when you use a 100 kOhm load resistor across the diode bridge output how can 0.035 Amper flow into it if the DC voltage is 41.5V?  Ohm's law should be valid and it calculates as 41.5/100000=0.000415 A   i.e. 0.415 mA and not 35 mA you measured.

Do you have a puffer capacitor across the diode bridge output? It is an electrolytic cap of any value between 470 uF to 2200 uF.  IF you did not have such capacitor when you did the measurements, please start it again with using one. 

Your input circuit as a load at 9.43 VDC @ 0.075 A corresponds to a R=9.43/0.075=125.7 Ohm resistor.  This may change to an even lower value when you use a 100-150 Ohm load resistor at your diode bridge output, you have to check that.  IT is good to use two meters: one ampermeter at the input in series with the 9.43V DC source and one voltage meter across the diode bridge output (where the puffer cap is also connected) , in parallel with the 100-150 Ohm load resistor.

rgds,  Gyula

Offline DeepCut

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Re: GENERATOR- YOU DO THE IN/OUT POWER MATH
« Reply #79 on: May 06, 2010, 05:34:26 PM »
Gyula thanks.

There's a lot i don't know ...

I thought a larger resistor put more load on but it's the other way around is it ? I will try it with a 100 ohm after this post.

The bridge is shop-bought, i'm not sure what it contains but max reverse-current is 10 A per diode.

About the ampere difference, in the Bedini circuit i have a 10k linear pot to 'tune' the circuit with, perhaps this is the reason ? It's set to about 1200k at peak performance.

I have a 470u cap @ 16 V and will try that in addition to the 100 o resistor.

My digi meter just died, it was cheap ... so will only be able to use analogue tonight.

Will report back.


Gary.


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Re: GENERATOR- YOU DO THE IN/OUT POWER MATH
« Reply #79 on: May 06, 2010, 05:34:26 PM »
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Offline gyulasun

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Re: GENERATOR- YOU DO THE IN/OUT POWER MATH
« Reply #80 on: May 06, 2010, 06:03:02 PM »
Hi Gary,

Yes the lower value resistor you use at the diode bridge output, the heavier loading effect it has, of course. 

The 470uF capacitor is ok but being only 16V voltage rated, please connect it across the bridge output when your 100 or 150 Ohm load resistor is ALREADY connected to the bridge,  because your 45V unloaded output voltage from the bridge will ruin it.  AND I suggest this ASSUMING that your output voltage will be reduced to 10V or so when you connect the 100-150 Ohm load resistor, ok? 

What is your output coil's DC resistance by the way? or wire gauge and its total length if you cannot measure its resistance.  Also if you do not mind saying the type of your diode bridge. The 10 Amper is probably not the reverse current data but its max forward current per diode.

Thanks,  Gyula

Offline DeepCut

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Re: GENERATOR- YOU DO THE IN/OUT POWER MATH
« Reply #81 on: May 06, 2010, 06:36:39 PM »
Hi.

Induction coil is 0.25mm thick, resistance is 61.1 ohms.

This is the bridge, it's the black, square one with four legs :

http://maplin.co.uk/Module.aspx?ModuleNo=19088

I'm not used to using analogue multimeters to test resistance.

The next lowest resistors i have are reading nearly 10 with the meter set to x10, does that mean it's a hundred ohms or one ? !


Thanks,

Gary.


*EDIT* I found a colour chart. They are 100 ohms :)*EDIT*


Free Energy | searching for free energy and discussing free energy

Re: GENERATOR- YOU DO THE IN/OUT POWER MATH
« Reply #81 on: May 06, 2010, 06:36:39 PM »
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Offline mscoffman

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Re: GENERATOR- YOU DO THE IN/OUT POWER MATH
« Reply #82 on: May 06, 2010, 07:24:35 PM »
Gyula thanks.

There's a lot i don't know ...

I thought a larger resistor put more load on but it's the other way around is it ? I will try it with a 100 ohm after this post.

About the ampere difference, in the Bedini circuit i have a 10k linear pot to 'tune' the circuit with, perhaps this is the reason ? It's set to about 1200k at peak performance.

Gary.


Gary the 100Ohm resister is lower therefore it puts *more* load on gen. or
bat. circuit. It's inverse. Looks at ohms law E=I*R THE ALGEBRAIC
manipulation says I = E/R, current therefore is proportional to
voltage and *inversely* proportional to resistance. more voltage means more
current, more resistance means less current because it is in the divisor.

0ohms = short circuit, infinity ohms= open circuit.

Also when you state resistance the K (meaning kilohms) really means
something, it is not redundant. So 1,200 Ohms = 1.2K ohms and not
1200Kohms (= 1.2Megohms).
So your second statement makes no sense if a pot is 10K then the reading
on the CT with nothing else connected should be somewhere between
0ohms and 10Kohms.  Therefore your statement should probably say
1.2KOhms, right? You can't get 1.2Megohms out of a 10K pot.

And Gyulasun is correct you really want to load the output coils
with a resistance that is close to or less than the (AC impedance)
of your drive coils. And I agree you seems to be getting the decimal
point wrong somewhere possibly when you read the current.
Try putting a 1.2K ohms resister across a 1.5Volt dry cell and
read the current. Should be slightly greater than 1.0ma You should
cross check these things on the meter scale that you think you have.

brown/black/brown = 100ohm resister (measured resistance will vary slightly)
                             one,zero, and one zero

brown/black/red =     1Kohm resister
                             one,zero and two zeroes

or use some high-current number printed resisters as a reference so
you can check that you can read your meter scales successfully.

Yes, on your mechanical meter if the resistance scale says
10 and your switch is set to resistance times 10 then it is
10 times 10 or 100ohms.

:S:MarkSCoffman

Offline DeepCut

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Re: GENERATOR- YOU DO THE IN/OUT POWER MATH
« Reply #83 on: May 06, 2010, 08:46:10 PM »
Thanks Marc.

Sorry to be so ignorant, i'm trying to get things done by learning only just enough to build things. It's time i sat down and read a little (or a lot !).

I will take care with my 'K's' now, i see what you mean when i re-read my post, doh !

I will wait until my meter arrives before testing further.

I'm really looking forward to Tom's load-testing.


Gary.



Free Energy | searching for free energy and discussing free energy

Re: GENERATOR- YOU DO THE IN/OUT POWER MATH
« Reply #83 on: May 06, 2010, 08:46:10 PM »
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Offline gyulasun

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Re: GENERATOR- YOU DO THE IN/OUT POWER MATH
« Reply #84 on: May 06, 2010, 11:52:11 PM »
Hello Gary,

Just take it easy, everybody has a learning cycle.

If you happen to connect the 100 Ohm resistance to your diode bridge output (I identified your bridge type, it seems ok) without the puffer capacitor, you could check the output voltage there with your analoge meter set into DC voltage range 100V or or whatever it has above 50V, first without the resistor, then with the resistor and write down the voltage values.  IF you measure  say  8V-10V DC voltage in the loaded case then connect the 470uF also in parallel with the bridge output and read the voltmeter again and write down.  (Do not leave the bridge output unloaded when your capacitor is there because the voltage would go up much higher than the 16V limit of your cap and get it damaged.)

If you think you are not yet comfortable with using and reading an analog meter, then just measure some battery voltages like a 9V alkali or 12V car accu etc to practice it,  ;)  change the ranges accordingly, always start with the highest voltage range that surely includes your possible voltage value to be measured.

rgds,  Gyula

Offline DeepCut

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Re: GENERATOR- YOU DO THE IN/OUT POWER MATH
« Reply #85 on: May 07, 2010, 12:19:59 AM »
Hi.

With the resistor, the voltage is 38.

Without it the voltage is 38.

Just to check i have the resistor in the right place :

I've got negative of the meter attached to neg out from bridge.
I've got resistor coming off pos out from bridge.
I've got positive of the meter connected to resistor.


Gary.


Free Energy | searching for free energy and discussing free energy

Re: GENERATOR- YOU DO THE IN/OUT POWER MATH
« Reply #85 on: May 07, 2010, 12:19:59 AM »
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Offline gyulasun

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Re: GENERATOR- YOU DO THE IN/OUT POWER MATH
« Reply #86 on: May 07, 2010, 12:29:59 AM »
One wire leg of the resistor goes directly to the positive bridge output and the other resistor leg goes directly to the negative bridge output.
Putting it otherwise: you connect the resistor directly across the bridge outputs.
Then you measure the voltage across the resistor.

Offline Bruce_TPU

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Re: GENERATOR- YOU DO THE IN/OUT POWER MATH
« Reply #87 on: May 07, 2010, 04:15:24 AM »
Getting this today :

http://maplin.co.uk/Module.aspx?ModuleNo=2355

If it doesn't kick in then it's time to wind that second coil and perhaps double the output for free.

Bruce,  would another coil, wound as you suggested, induct ? I thought it had to be at 90 deg to the movement of the changing magnetic field ?

@Infringer

Lots more detail eralier on in thread, MagnetMan has his videos posted on the concept.

I am following in his footsteps and ahve a much cruder setup, he will be doing his load-testing this week i think and should have better results but i will post video once it's self-powering, if it ever is lol ;+}


Gary.

It will work!!  Positioned like this...

Offline DeepCut

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Re: GENERATOR- YOU DO THE IN/OUT POWER MATH
« Reply #88 on: May 07, 2010, 04:52:29 PM »
OK.

I have the second coil wound now. Only half as many turns as i ran out of wire :(

When i connected that to the bridge, as well as the original coil, the motor won't start.

I'm assuming thatt's because it's two seperate outputs.

Would i be right to have the two coils connected at one end then the other two ends going onto the bridge ?

I've done that and am getting slightly better-looking results, but i can't measure volts and amps at the same time until my digi meter arrives.


Gary.





Offline gyulasun

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Re: GENERATOR- YOU DO THE IN/OUT POWER MATH
« Reply #89 on: May 07, 2010, 06:32:00 PM »
Hi Gary,

Would you mind telling what voltage did you measure across the 100 Ohm load late last night? (I described 2 posts above how to connect it to the bridge because as you described 3 posts above it would have been good for the loading current measurement, setting your analog meter into current measure range.)

By the way you could be a king with only the analog meter as the only instrument at the moment:  You can trust in the 100 Ohm it is 100 Ohm and measure the output voltage as described two posts above,  THEN you could simply calculate the current from Ohm's law like I=V/R  R=100  and V=what you measure across the R, ok?

Then you could check WHETHER the input current changes as soon as you connect the 100 Ohm resistor to the output. To do this you wish to connect the positive meter pin to the positive 9.5V pole and the negative meter pin goes to your positive circuit input (you insert the meter between the battery positive and the circuit positive, that is all).  Circuit negative remains connected to the 9.5V battery negative.
And you surely see the 75mA current draw when the 100 Ohm is NOT connected and then please connect it and write down the current change (if any), ok?   
Please leave the second coil outputs unconnected for these measurements, I will return to the 2nd coil's  "how to"  later today, ok?

rgds,  Gyula

 

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