Partnered Output Coils - Free Energy

[MarkE]

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]

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]

 @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]

@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]

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.