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Author Topic: Partnered Output Coils - Free Energy  (Read 3501560 times)

madddann

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Re: Partnered Output Coils - Free Energy
« Reply #3090 on: May 27, 2015, 05:25:03 PM »
NO-i did not take the average voltage at all-please watch video again where i show the average voltage on the scope,but use the supply voltage when doing the math,because we use average current. The average voltage was around 2.8 on the scope when we were running at 3 volts at the power supply. When i did that math,i used the supply voltage(3 volts)X the average current.
Then when we did the test on 20 volt's,the scope showed an average voltage across that globe of around 1.3 volt's,but when i did the math,i used the supply voltage of 20 volt's x the average current.

So please get it right ;)

OK, I admitt, my mistake... you are using power supply voltage, but you are still using average current - I don't know what is the result of that, but it is sure not average power.

If I remember right average power is: Pavg = Vrms x Irms
So If your scope can read Vrms and Irms values you should be close to the result - someone correct me if I'm mistaken...
Actually search on your scope for Crms (cycle rms) or something similar.

Hope it helps, good luck!

tinman

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Re: Partnered Output Coils - Free Energy
« Reply #3091 on: May 27, 2015, 05:29:41 PM »
Hi Tinman.
Here's what I'd expect .... beware, I have been known to have weird expectations at times....oh, and I'm regularly wrong!

Using a normal transformer in the circuit, D2 provides a path for kickback current to flow through L1 in a unidirectional loop, when the L1 field wants to collapse during the off time of the pulsing. Since L1 and L2 share a common core, they will share the kickback induction current capability. However the current path with least resistance will be the preferred path that the induced kickback current from the field collapse will take.

Removing D2 should give a slight increase in the total output of L2 into the load, because all the inductive kickback potential will be available to L2 and it's load. Without a diode D2, there is no path for the induced kickback potential to cause a unidirectional current to flow through L1. All the kickback energy must therefore dissipate through L2 and its load.

When D2 is in place, it should also extend the 'apparent' on time of L1, even when L1 is actually in pulse off time, because D2 allows kickback current to flow through L1 in the same (uni)-direction as the preceding supply current flowing through L1 during the pulse on time.
D2 should cause L1 current to ring down slowly during the off time.

Cheers. .... and  KneeDeep. Keep up the great work.

Yes-that would be standard transformer outcome.
But here is what i want to know-a little more detail.
Below is the schematic with test points and CSR value. The yellow trace is across the load(globe),and the blue trace is across the CSR. The scope shot shows the trace results with D2 in place. What i want to know is-
1- will the yellow trace(positive voltage value) increase with D2 removed during the on cycle,or will it decrease?.
2-Will the yellow trace(negative voltage value) increase with D2 removed during the off cycle,or will it decrease.
3- Will the blue trace(forward current value) increase with D2 removed during the on cycle,or will it decrease?.
4- Will the blue trace(reverse current value) increase with D2 removed during the off cycle,or will it decrease?.
5- Will the P/in increase with D2 removed,or will it decrease?.

P.S
I have been trying different voltage,duty cycles and frequencies,and believe i have now hit the optimum settings. The frequency seems to be best at the 7KHz mark,but i found that dropping the voltage,and increasing the duty cycle gave me better results. I think you would have to agree that the off time current and voltage values are now very very close to even until the next cycle starts.
I have not been able to achieve this with any other transformer i have tried so far.

Magluvin

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Re: Partnered Output Coils - Free Energy
« Reply #3092 on: May 27, 2015, 06:20:25 PM »
Probably a good thing to check inductance of the primary with the sec loaded then unloaded, to see what effects the sec has on the primary, loaded and unloaded. Someone asked about that before. Im sure you can set it up with a freq sweep using the scope and some resistors to measure the inductance.

Like when you have 2 toroid cores, the primary only wound around 1 side of one core, but the secondary wound around the other side of the primary core and also wound around the other core. shorting the sec increases the inductance of the primary.  But this may be different.


Mags

tinman

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Re: Partnered Output Coils - Free Energy
« Reply #3093 on: May 27, 2015, 06:31:49 PM »
OK, I admitt, my mistake... you are using power supply voltage, but you are still using average current - I don't know what is the result of that, but it is sure not average power.

If I remember right average power is: Pavg = Vrms x Irms
So If your scope can read Vrms and Irms values you should be close to the result - someone correct me if I'm mistaken...
Actually search on your scope for Crms (cycle rms) or something similar.

Hope it helps, good luck!
We are using a DC pulsed current,not AC. You either use an average voltage X instantaneous current per cycle,or you use an average current X supply voltage per cycle.

Now with your VRMS X IRMS-->we could just put a lower value CSR in there,and lower the IRMS the scope shows to any value we want.

So here is how it go's for a pulsed square wave DC.
We use a known value CSR,and obtain the instantaneous voltage value across that CSR. Using ohms law,we now can work out what the instantaneous current is flowing through that CSR.
Now we have that,and we also know our input voltage,we can do 1 of two things. We can get the average voltage of each cycle(which our scope will show us),and X's that by our instantaneous current value to get our average power -or we can average our current over a full cycle,and X's that by our supply voltage.

Lets say we use a 1 ohm CSR,and across that 1 ohm CSR we get an instantaneous voltage of 1 volt. This means we have a current flow of 1 amp. Now if we have a 5% duty cycle,we simply X's that 1 amp by 5%-->which is 50mA. So now we have an average current flow of 50mA over the whole cycle. Lets say our input voltage is 10 volts,so we then have 10v x 50mA and have an average power of 500mW
Now we do it the other way around-we use our instantaneous current X's our average voltage.
Our voltage was 10 volts,so once again we X's this by 5%,which gives us an average voltage over one full cycle of 50mV- 50mV X's our 1 amp gives us an average power of 500mW

OR-we take our instantaneous voltage and current-->1 amp X's 10 volts X's 5%,and we still get an average power of 500mW

padova

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Re: Partnered Output Coils - Free Energy
« Reply #3094 on: May 27, 2015, 06:37:40 PM »
Hi,
Try two diode in series, see if there is the neg. Voltage  difference.

regards

MarkE

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Re: Partnered Output Coils - Free Energy
« Reply #3095 on: May 27, 2015, 07:31:44 PM »
We are using a DC pulsed current,not AC. You either use an average voltage X instantaneous current per cycle,or you use an average current X supply voltage per cycle.

Now with your VRMS X IRMS-->we could just put a lower value CSR in there,and lower the IRMS the scope shows to any value we want.

So here is how it go's for a pulsed square wave DC.
We use a known value CSR,and obtain the instantaneous voltage value across that CSR. Using ohms law,we now can work out what the instantaneous current is flowing through that CSR.
Now we have that,and we also know our input voltage,we can do 1 of two things. We can get the average voltage of each cycle(which our scope will show us),and X's that by our instantaneous current value to get our average power -or we can average our current over a full cycle,and X's that by our supply voltage.

Lets say we use a 1 ohm CSR,and across that 1 ohm CSR we get an instantaneous voltage of 1 volt. This means we have a current flow of 1 amp. Now if we have a 5% duty cycle,we simply X's that 1 amp by 5%-->which is 50mA. So now we have an average current flow of 50mA over the whole cycle. Lets say our input voltage is 10 volts,so we then have 10v x 50mA and have an average power of 500mW
Now we do it the other way around-we use our instantaneous current X's our average voltage.
Our voltage was 10 volts,so once again we X's this by 5%,which gives us an average voltage over one full cycle of 50mV- 50mV X's our 1 amp gives us an average power of 500mW

OR-we take our instantaneous voltage and current-->1 amp X's 10 volts X's 5%,and we still get an average power of 500mW
If you make the pulse voltage very stiff with good decoupling capacitors and the MOSFET on voltage doesn't change much, then the average voltage will be plenty accurate.

madddann

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Re: Partnered Output Coils - Free Energy
« Reply #3096 on: May 27, 2015, 07:55:23 PM »
We are using a DC pulsed current,not AC. You either use an average voltage X instantaneous current per cycle,or you use an average current X supply voltage per cycle.

Now with your VRMS X IRMS-->we could just put a lower value CSR in there,and lower the IRMS the scope shows to any value we want.

So here is how it go's for a pulsed square wave DC.
We use a known value CSR,and obtain the instantaneous voltage value across that CSR. Using ohms law,we now can work out what the instantaneous current is flowing through that CSR.
Now we have that,and we also know our input voltage,we can do 1 of two things. We can get the average voltage of each cycle(which our scope will show us),and X's that by our instantaneous current value to get our average power -or we can average our current over a full cycle,and X's that by our supply voltage.

Lets say we use a 1 ohm CSR,and across that 1 ohm CSR we get an instantaneous voltage of 1 volt. This means we have a current flow of 1 amp. Now if we have a 5% duty cycle,we simply X's that 1 amp by 5%-->which is 50mA. So now we have an average current flow of 50mA over the whole cycle. Lets say our input voltage is 10 volts,so we then have 10v x 50mA and have an average power of 500mW
Now we do it the other way around-we use our instantaneous current X's our average voltage.
Our voltage was 10 volts,so once again we X's this by 5%,which gives us an average voltage over one full cycle of 50mV- 50mV X's our 1 amp gives us an average power of 500mW

OR-we take our instantaneous voltage and current-->1 amp X's 10 volts X's 5%,and we still get an average power of 500mW

...Well... after all that school I went to didn't teach me very much...

Thanks for the lesson, but you might be right or maybe not... not sure yet...

After a quick search on the net i found out this:

"In order to calculate average power, you need to integrate the instantaneous power (instantaneous voltage times instantaneous current) over some time interval, which yields total energy for that interval, and then divide by the time interval.
The details of doing this depend on the circuit you're considering. For example, in a purely resistive circuit, the instantaneous power is proportional to the square of the voltage (or the square of the current), and so the average power can be computed directly from the RMS voltage (or current).

In nonlinear circuits (including those with diodes), some part of the power dissipation might be proportional to current only; in this case, you could use the average value of the current to compute the average power."


Go here for more: http://electronics.stackexchange.com/questions/113578/rms-vs-dcmean-value-when-calculating-power-of-pulsed-or-rectified-signals

...so If you would put in the circuit a resistor instead of the lightbulb would the circuit categorize as purely resistive?
If this is the case you can just use the rms values as I said before...

It would also be helpful if you could post the integrated waveform that your scope produces (with resistor instead of lightbulb), at least we could have a slight clue about it, and see wich calculation method would fit/match.

Keep at it!  ;)

Drak

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Re: Partnered Output Coils - Free Energy
« Reply #3097 on: May 27, 2015, 10:53:03 PM »
 @EMJunkie
I've tried a couple different setups but I'm not getting anything. I still have some other stuff to try but I'm curious, when you got the results you did, what were you using, sine or square with what duty? You probably already answered this and I should just reread the thread. I'm trying with just an audio amp right now. I thought you said at one point that it did not have to be in resonance for it to work, is it a requirement?
http://www.draksplace.com/DSCN0755.png

Also, did you do any experiments with this:
http://www.draksplace.com/floyd.png
If so, any weird results?

@all
Can anyone recommend a “good enough” current probe for my scope (up to $300, new)? I might as well learn how to measure power that way.
Thanks
 

EMJunkie

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Re: Partnered Output Coils - Free Energy
« Reply #3098 on: May 27, 2015, 11:40:11 PM »
@EMJunkie
I've tried a couple different setups but I'm not getting anything. I still have some other stuff to try but I'm curious, when you got the results you did, what were you using, sine or square with what duty? You probably already answered this and I should just reread the thread. I'm trying with just an audio amp right now. I thought you said at one point that it did not have to be in resonance for it to work, is it a requirement?
http://www.draksplace.com/DSCN0755.png

Also, did you do any experiments with this:
http://www.draksplace.com/floyd.png
If so, any weird results?

@all
Can anyone recommend a “good enough” current probe for my scope (up to $300, new)? I might as well learn how to measure power that way.
Thanks

@Drak - Nice experiments!

I will reply to you in private. We will get you up and running.

   Chris Sykes
       hyiq.org

madddann

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Re: Partnered Output Coils - Free Energy
« Reply #3099 on: May 27, 2015, 11:47:27 PM »
Hi again!

I just found this calculator: http://www.vishay.com/resistors/pulse-energy-calculator/
...so I calculated the aproximate numbers for Tinman's scope shot in post #3097
The results are like so:

 -for the positive side of the waveforms    Pavg = 1.875W
 -for the negative side of the waveforms   Pavg = 0.4833W

...the results are aproximate as I said, and the change in the filament resistence is not accounted for...

Comments welcome.

tinman

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Re: Partnered Output Coils - Free Energy
« Reply #3100 on: May 28, 2015, 02:57:22 AM »
Hi again!

I just found this calculator: http://www.vishay.com/resistors/pulse-energy-calculator/
...so I calculated the aproximate numbers for Tinman's scope shot in post #3097
The results are like so:

 -for the positive side of the waveforms    Pavg = 1.875W
 -for the negative side of the waveforms   Pavg = 0.4833W

...the results are aproximate as I said, and the change in the filament resistence is not accounted for...

Comments welcome.

Interesting,considering that both the average voltage values on the scope shot  across the globe and CSR are both negative. ;)

Dave45

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Re: Partnered Output Coils - Free Energy
« Reply #3101 on: May 28, 2015, 03:41:29 AM »
D2 recirculates magnetization energy around L1 when the transistor turns off.  If you remove D2 then the voltage on L1 will flyback with a big spike, while the voltage on L2 will flyback without much of a spike at all.  The V*T across L2 over time is zero.
D1 is interesting it stops the recirculating energy from hitting the battery, leaving only the path through the coil.

TinselKoala

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Re: Partnered Output Coils - Free Energy
« Reply #3102 on: May 28, 2015, 06:11:47 AM »
Interesting,considering that both the average voltage values on the scope shot  across the globe and CSR are both negative. ;)

What's the matter with that? A negative number (about -2.8V) times another negative number (about -0.190 mA) is a positive number (about 0.53 W). 

Perhaps the problem is that the Vishay calculator is telling you the averages across an entire period. That is, if you look at the positive portions of the peaks, they are about 1/5 of the entire period, and the negative portions are about 4/5 of the entire period. Mentally "squaring off" the pulses we can estimate that the positive portions of the waveforms account for about 13.5 V x 0.65 A x 0.2 = about 1.8 Watts average across the entire waveform, and the negative portions account for about -2.8 V x -0.190 A x 0.8 = about 0.43 W average across the entire waveform. Very roughly, since I'm just eyeballing the values and mentally squaring off the waveshapes.


There may be something else to consider as well. How does your scope compute those averages? Is it using the data displayed on the screen? Many scopes do just use the screen display to compute those averages. In that case errors can arise because of the number of periods and partial periods displayed. As you slow down the timebase to display more and more periods across the screen, do these average numbers change?

TinselKoala

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Re: Partnered Output Coils - Free Energy
« Reply #3103 on: May 28, 2015, 06:49:12 AM »
@Drak: There are "good enough" current probes, and there are cheap (under 300 dollars) current probes. Good luck finding both qualities in the same probe.

Whatever you do get, you should do a comparison test, looking at the same current signal as measured by the probe, and simultaneously looking at the voltage drop across a good non-inductive current-viewing resistor. Has EMJunkie ever reported the results of such a calibration test for his inexpensive current probe?

MarkE

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Re: Partnered Output Coils - Free Energy
« Reply #3104 on: May 28, 2015, 07:16:40 AM »
@Drak: There are "good enough" current probes, and there are cheap (under 300 dollars) current probes. Good luck finding both qualities in the same probe.

Whatever you do get, you should do a comparison test, looking at the same current signal as measured by the probe, and simultaneously looking at the voltage drop across a good non-inductive current-viewing resistor. Has EMJunkie ever reported the results of such a calibration test for his inexpensive current probe?
It is a shame that really good non contact DC/AC current probes plus amplifiers cost about $6000. for the set.