Partnered Output Coils - Free Energy

[madddann]

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.

OK ... so ... how any other way is it supposed to be? (I really want to learn the right method).
I actually used 1/6 for the positive waveforms and 5/6 for the negative.
The energy calculated in the first 1/6 of the period was 0.02678569mJ and in the last 5/6 of the period was 0.00690478mJ.

[tinman]

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.

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).

OK,you lost me. How do you get a positive from two negative values?. If we look at AC,and consider the positive above the ) volt line,then a negative value would mean a value below the 0 volt line. In this case,we wish to know which way the current is flowing in which direction,and when the above and below the 0 volt line are averaged out over each cycle,we see a higher average below the zero volt line per cycle-both voltage and current.

As you slow down the timebase to display more and more periods across the screen, do these average numbers change?

No,they remain the same-->is this good or bad,as you know im still learning about scope's.

[MarkE]

OK,you lost me. How do you get a positive from two negative values?. If we look at AC,and consider the positive above the ) volt line,then a negative value would mean a value below the 0 volt line. In this case,we wish to know which way the current is flowing in which direction,and when the above and below the 0 volt line are averaged out over each cycle,we see a higher average below the zero volt line per cycle-both voltage and current.

No,they remain the same-->is this good or bad,as you know im still learning about scope's.

Current and voltage multiply.  In positive current convention:  work is absorbed by a circuit branch when the sign of the voltage applied across the branch matches the sign of the current through the branch.  When the signs don't match, then the branch is supplying energy.  With AC waveforms, at any moment it is possible for a circuit branch to:

Absorb energy and dissipate it (into heat and/or a load that performs work):  A pure resistance. Voltage and current signs match.
Absorb energy storing some and dissipating some of it:  A resistance with a reactance: net capacitive, or net inductive.  Voltage and current signs match.
Release some stored energy: A resistance with a reactance: net capacitive, or net inductive.  Voltage and current signs oppose.

In order to find out how much energy the branch is dissipates over the course of some repetitive cycle, we have to integrate the power transferred during that cycle.  We have to keep track of whether the branch is absorbing or releasing energy (power sign) and the rate (power magnitude) at every instant.  When we have tallied the energy in one repetitive cycle, then we can just multiply that by the frequency to get average power. 

[nelsonrochaa]

@ all here

Below is a circuit diagram of my setup.
What i would like to know is this-->who can tell me what D2 dose in the circuit,and what would change in L2 in a normal transformer setup if i remove D2 when L2 has a load(resistive)across it.

Cheers

Hi ,
I have followed the overall progress in this theme and would like to ask you if you test already use a high side mosfet in configuration to the coil L1.
It is just a tip if you've curiosity to test;)

Good tests and thanks

[MladenStijepic]

This Atten scope is actually Siglent.Very cheap stuff.Have one on my bench.

By the way ,excellent work tinman,keep going!

   
Cheers