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Author Topic: 5 Watts in 60 Watts out ...  (Read 10969 times)

Offline DeepCut

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5 Watts in 60 Watts out ...
« on: November 09, 2010, 06:38:56 PM »
Hi all.

OK i've checked this over and over but, knowing me, i've probably missed something so please check these figures.

This is my crude replication of Toms device in this thread :

http://www.overunity.com/index.php?topic=9076.0

It's a standard Bedini circuit, i haven't added a relay yet as i want to prove the concept to myslf before maximising efficiency.

The drive coil is the usual bifilar, 800 turns of 0.30mm and 0.28mm.

The output coil is just over a kilometre (1145 metres) of 0.25mm magnet wire.


Input/output figures :


Output coil resistance = 388 Ohms

INPUT POWER: 18 Volts DC @ .280 Amps = 5.04 Watts

OUTPUT VOLTAGE : 155 Volts DC (Rectified)

OUTPUT CURRENT : 155 / 388 = 0.3994845360824742 Amps

OUTPUT POWER : 155 x 0.3994845360824742 = 61.92010309278351 Watts

Please advise if my calculations are wrong, i'm sure they're not as i'm more confident than i was when i joined this forum !

I'm now going to try and get it to run itself ...

Video :

http://www.youtube.com/watch?v=8R7ehAVnp8c


Gary.



Free Energy | searching for free energy and discussing free energy

5 Watts in 60 Watts out ...
« on: November 09, 2010, 06:38:56 PM »

Offline void109

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Re: 5 Watts in 60 Watts out ...
« Reply #1 on: November 09, 2010, 07:04:33 PM »
I dont think you're measuring your output current properly.  I believe you'd need to measure the voltage over a load of some sort, then you can calculate the current with voltage / resistance = current.  You're just measuring the output voltage with no load.

Offline DeepCut

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Proper load-test ...
« Reply #2 on: November 09, 2010, 08:34:17 PM »
Thanks void.

OK i've done a proper load test.

My input resistance is 18 volts / 0.28 amps = 64 Ohms.

I put the nearest i had, 66 Ohms, across the rectifier output.

Results :

OUTPUT VOLTAGE : 138 Volts DC

OUTPUT CURRENT : 0.3994845360824742 (Calculated).

OUTPUT POWER : 55 Watts.

I remember mscoffman talking about the AC impedance of the output coil having an effect on this so i will now try to get the thing to self-run.

Looking good though ...


Gary.

Free Energy | searching for free energy and discussing free energy

Proper load-test ...
« Reply #2 on: November 09, 2010, 08:34:17 PM »
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Offline void109

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Re: 5 Watts in 60 Watts out ...
« Reply #3 on: November 09, 2010, 08:38:29 PM »
What kind of resistor are you using?  I assume its a heavy duty power resistor as 55 watts of power would turn a 1/4W-1W resistor into a molten pile of oops pretty quick :)

Free Energy | searching for free energy and discussing free energy


Offline void109

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Re: 5 Watts in 60 Watts out ...
« Reply #4 on: November 09, 2010, 08:44:40 PM »
I had to do a double take hehe, in your first post you had:

155 volts / 388 ohms = 399mA

in your reply you have:

138 volts / 66 ohms = 399mA, it should be 2.09A!  Giving 288.54W of power!

Are you measuring the 138 volts across the 66 ohm load?

Free Energy | searching for free energy and discussing free energy

Re: 5 Watts in 60 Watts out ...
« Reply #4 on: November 09, 2010, 08:44:40 PM »
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Offline DeepCut

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Re: 5 Watts in 60 Watts out ...
« Reply #5 on: November 09, 2010, 09:04:52 PM »
Ooops !

Ye that should have been :

OUTPUT VOLTAGE : 138 Volts DC

OUTPUT CURRENT : (138/388) 0.3556701030927835 (Calculated).

OUTPUT POWER : 48 Watts.

**EDIT** and i've not divided by 66 but only by the coil resistance which is 388 **EDIT**

My load resistor is across the rectifier output and i'm measuring from there.

I just tried to self run it, it stops immediately and the safety neon flashes very brightly for an instant.

I will try it with a puffer cap in a bit but i need a cup of tea and a smoke !

brb

Gary.




Offline void109

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Re: 5 Watts in 60 Watts out ...
« Reply #6 on: November 09, 2010, 09:08:17 PM »
I still think you're measuring the output current incorrectly - you place the load on the output of the rectifier, probably with a big smoothing cap to get an accurate reading with your digital meter, then you place the ground and hot lead of the meter on either side of the load to get the voltage.  Then you divide that voltage by the resistance of the load to yield the current.  Shoot me if I'm wrong, I'm new too, but that's my understanding of the situation.

I'll see if I can draw a picture in paint or something.

Free Energy | searching for free energy and discussing free energy

Re: 5 Watts in 60 Watts out ...
« Reply #6 on: November 09, 2010, 09:08:17 PM »
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Offline DeepCut

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Re: 5 Watts in 60 Watts out ...
« Reply #7 on: November 09, 2010, 09:11:30 PM »
Thanks void.

Ye that's exactly what i'm doing.

I'm just looking for a suitable cap right now.

Offline void109

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Re: 5 Watts in 60 Watts out ...
« Reply #8 on: November 09, 2010, 09:12:38 PM »
haha excuse my paint skills!

Free Energy | searching for free energy and discussing free energy

Re: 5 Watts in 60 Watts out ...
« Reply #8 on: November 09, 2010, 09:12:38 PM »
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Offline penno64

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Re: 5 Watts in 60 Watts out ...
« Reply #9 on: November 09, 2010, 09:15:33 PM »
Hi Gaz,

From one Gaz to another,

Power at the output has nothing to do with the resistance of the output coil.

What voidy is trying to tell you is -
Measure the voltage across your load resistor (66ohm)
Then use E = I x R like this
I = E/R

So if you read for example 24v -

I = 24v / 66 ohms
so I = .36 amps
and

Power = I x V
so
P = .36 x 24
8.73 Watts

Hope that helps.

Regards, Penno

Offline DeepCut

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Re: 5 Watts in 60 Watts out ...
« Reply #10 on: November 09, 2010, 09:17:13 PM »
Cheers penn.

Yup, that's how i did it.

I'm just looking for a smoothing cap now, will 450 Volts @ 150 microfarads be ok ?


Gary.

Free Energy | searching for free energy and discussing free energy

Re: 5 Watts in 60 Watts out ...
« Reply #10 on: November 09, 2010, 09:17:13 PM »
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Offline void109

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Re: 5 Watts in 60 Watts out ...
« Reply #11 on: November 09, 2010, 09:21:26 PM »
I'm too new to comment on that, but I think you'd need the frequency to nail the cap size down.  That aside, if there isn't a measurement error there, AND, if your meter isn't going crazy from either the frequency or the proximity of the magnet, that implies something like 5000% OU, and this looks like a typical circuit I've seen pounded to death around here.  I'll pore over the original thread you linked in the original post.

After you throw in the output cap, would you mind making another video? :)

Offline DeepCut

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Re: 5 Watts in 60 Watts out ...
« Reply #12 on: November 09, 2010, 09:21:56 PM »
Oh I See !

So my power is calculated using the load resistance NOT the coil resistance ?

But when i calculate it without the load the proper way is to use the coil resistance isn't it ??


Gary.

Offline penno64

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Re: 5 Watts in 60 Watts out ...
« Reply #13 on: November 09, 2010, 09:25:22 PM »
Hi Gaz,

No, you cannot make a power calculation without having a load.

Unloaded, you may only state your output voltage.

Also, I agree with Void, in that, frequency plays an important role in measurements.

Regards Penno

Offline DeepCut

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Re: 5 Watts in 60 Watts out ...
« Reply #14 on: November 09, 2010, 09:35:16 PM »
Well if the correct calculation is with 66 ohms then that would be a very large power output.

I'm just gonna solder this cap on (when i find my glasses !) and see if it self-runs.


Gary.

 

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